A,B,C,D,E,F = map(int,input().split(" "))
A *= 100
B *= 100

water = []
for a in range(F//A+1):
    for b in range(F//B+1):
        if a + b == 0 or a * A + b * B > F:
            continue
        if not((a * A + b * B) in water):
            water.append(a * A + b * B)

suger = []
for c in range(F//C+1):
    for d in range(F//D+1):
        if c * C + d * D > F:
            continue
        if not((c * C + d * D) in suger):
            suger.append(c * C + d * D)

max_dense = 0
log = [water[0],0]
for w in water:
    for s in suger:
        if w + s > F:
            continue
        dense = 100 * s / (w + s)
        if dense > E:
            continue
        if dense > max_dense:
            max_dense = dense
            log = [w+s,s]

print(" ".join(map(str,log)))


N = int(input())
A = []
for i in range(N):
    x=[int(i) for i in input().split()]
    if A.count(x) == 0:
        A.append(x)
    else:
        A.remove(x)
print(len(A))

Global OOD problem IDs are disjoint from train, validation, and in-distribution test problem IDs.

Score : 300 points

Problem Statement

You are playing the following game with Joisino.

Initially, you have a blank sheet of paper.

Joisino announces a number. If that number is written on the sheet, erase the number from the sheet; if not, write the number on the sheet. This process is repeated N times.

Then, you are asked a question: How many numbers are written on the sheet now?

The numbers announced by Joisino are given as A_1, ... ,A_N in the order she announces them. How many numbers will be written on the sheet at the end of the game?

Constraints

1≤N≤100000

1≤A_i≤1000000000(=10^9)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

Print how many numbers will be written on the sheet at the end of the game.

Sample Input 1

3
6
2
6

Sample Output 1

1

The game proceeds as follows:

6 is not written on the sheet, so write 6.

2 is not written on the sheet, so write 2.

6 is written on the sheet, so erase 6.

Thus, the sheet contains only 2 in the end. The answer is 1.

Sample Input 2

4
2
5
5
2

Sample Output 2

0

It is possible that no number is written on the sheet in the end.

Sample Input 3

6
12
22
16
22
18
12

Sample Output 3

2

Submitted source-code file size, in bytes.

Execution CPU time reported by official Project CodeNet metadata, in milliseconds.

Peak memory usage reported by official Project CodeNet metadata, in kilobytes.


Score : 200 points

Problem Statement

You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.

Constraints

Each character in s is a lowercase English letter.

1≤|s|≤10^5

Input

The input is given from Standard Input in the following format:

s

Output

Print the string obtained by concatenating all the characters in the odd-numbered positions.

Sample Input 1

atcoder

Sample Output 1

acdr

Extract the first character a, the third character c, the fifth character d and the seventh character r to obtain acdr.

Sample Input 2

aaaa

Sample Output 2

aa

Sample Input 3

z

Sample Output 3

z

Sample Input 4

fukuokayamaguchi

Sample Output 4

fkoaaauh


s = input()
for i in range(len(s)-1):
  if i%2 == 0:
    print(s[i], end = "")


s = input()
for i in range(len(s)):
    if i % 2 == 0:
        print(s[i], end="")
print("")


import sys
import collections
#import numpy as np

readline = sys.stdin.readline

def main():
    N = int(input())
    #A1 = np.array(list(map(int, readline().split())))
    A = list(map(int, readline().split()))

    count = [0] * (max(A) + 3)
    for a in A:
        count[a+1] += 1
        count[a] += 1
        count[a+2] += 1

    print(max(count))

if __name__ == "__main__":
    main()

Score : 300 points

Problem Statement

You are given an integer sequence of length N, a_1,a_2,...,a_N.

For each 1≤i≤N, you have three choices: add 1 to a_i, subtract 1 from a_i or do nothing.

After these operations, you select an integer X and count the number of i such that a_i=X.

Maximize this count by making optimal choices.

Constraints

1≤N≤10^5

0≤a_i<10^5 (1≤i≤N)

a_i is an integer.

Input

The input is given from Standard Input in the following format:

N
a_1 a_2 .. a_N

Output

Print the maximum possible number of i such that a_i=X.

Sample Input 1

7
3 1 4 1 5 9 2

Sample Output 1

4

For example, turn the sequence into 2,2,3,2,6,9,2 and select X=2 to obtain 4, the maximum possible count.

Sample Input 2

10
0 1 2 3 4 5 6 7 8 9

Sample Output 2

3

Sample Input 3

1
99999

Sample Output 3

1


# coding: utf-8
import re
import math
from collections import defaultdict
from collections import deque
import collections
from fractions import Fraction
import itertools
from copy import deepcopy
import random
import time
import os
import queue
import sys
import datetime
from functools import lru_cache
#@lru_cache(maxsize=None)
readline=sys.stdin.readline
sys.setrecursionlimit(2000000)
#import numpy as np
alphabet="abcdefghijklmnopqrstuvwxyz"
mod=int(10**9+7)
inf=int(10**20)
def yn(b):
    if b:
        print("yes")
    else:
        print("no")
def Yn(b):
    if b:
        print("Yes")
    else:
        print("No")
def YN(b):
    if b:
        print("YES")
    else:
        print("NO")
class union_find():
    def __init__(self,n):
        self.n=n
        self.P=[a for a in range(n)]
        self.rank=[0]*n

    def find(self,x):
        if(x!=self.P[x]):self.P[x]=self.find(self.P[x])
        return self.P[x]

    def same(self,x,y):
        return self.find(x)==self.find(y)

    def link(self,x,y):
        if self.rank[x]<self.rank[y]:
            self.P[x]=y
        elif self.rank[y]<self.rank[x]:
            self.P[y]=x
        else:
            self.P[x]=y
            self.rank[y]+=1

    def unite(self,x,y):
        self.link(self.find(x),self.find(y))

    def size(self):
        S=set()
        for a in range(self.n):
            S.add(self.find(a))
        return len(S)
def ispow(a,b):#aはbの累乗数か
    now=b
    while now<a:
        now*=b
    if now==a:return True
    else:return False
def getbin(num,size):
    A=[0]*size
    for a in range(size):
        if (num>>(size-a-1))&1==1:
            A[a]=1
        else:
            A[a]=0
    return A
def getfacs(n,mod_=0):
    A=[1]*(n+1)
    for a in range(2,len(A)):
        A[a]=A[a-1]*a
        if(mod_>0):A[a]%=mod_
    return A
def comb(n,r,mod,fac):
    if(n-r<0):return 0
    return (fac[n]*pow(fac[n-r],mod-2,mod)*pow(fac[r],mod-2,mod))%mod
def nextcomb(num,size):
    x=num&(-num)
    y=num+x
    z=num&(~y)
    z//=x
    z=z>>1
    num=(y|z)
    if(num>=(1<<size)):return False
    else:return num
def getprimes(n,type="int"):
    if n==0:
        if type=="int":return []
        else:return [False]
    A=[True]*(n+1)
    A[0]=False
    A[1]=False
    for a in range(2,n+1):
        if A[a]:
            for b in range(a*2,n+1,a):
                A[b]=False
    if(type=="bool"):return A
    B=[]
    for a in range(n+1):
        if(A[a]):B.append(a)
    return B
def isprime(num):
    if(num<=1):return False
    i=2
    while i*i<=num:
       if(num%i==0):return False
       i+=1
    return True
def ifelse(a,b,c):
    if a:return b
    else:return  c
def join(A,c=""):
    n=len(A)
    A=list(map(str,A))
    s=""
    for a in range(n):
        s+=A[a]
        if(a<n-1):s+=c
    return s
def factorize(n,type_="dict"):
    b = 2
    list_ = []
    while b * b <= n:
        while n % b == 0:
            n //= b
            list_.append(b)
        b+=1
    if n > 1:list_.append(n)
    if type_=="dict":
        dic={}
        for a in list_:
            if a in dic:
                dic[a]+=1
            else:
                dic[a]=1
        return dic
    elif type_=="list":
        return list_
    else:
        return None
def pm(x):
    return x//abs(x)
def inputintlist():
    return list(map(int,input().split()))
######################################################################################################

n=int(input())
P=inputintlist()
P=list(map(lambda x:x-1,P))
B=[]
for a in range(n):
    B.append(not P[a]==a)
ans=0
for a in range(n-1):
    if B[a]==False:
        if B[a+1]==False:
            B[a]=True
            B[a+1]=True
        else:
            B[a]=True
        ans+=1

print(ans)

Score : 400 points

Problem Statement

You are given a permutation p_1,p_2,...,p_N consisting of 1,2,..,N.
You can perform the following operation any number of times (possibly zero):

Operation: Swap two adjacent elements in the permutation.

You want to have p_i ≠ i for all 1≤i≤N.
Find the minimum required number of operations to achieve this.

Constraints

2≤N≤10^5

p_1,p_2,..,p_N is a permutation of 1,2,..,N.

Input

The input is given from Standard Input in the following format:

N
p_1 p_2 .. p_N

Output

Print the minimum required number of operations

Sample Input 1

5
1 4 3 5 2

Sample Output 1

2

Swap 1 and 4, then swap 1 and 3. p is now 4,3,1,5,2 and satisfies the condition.
This is the minimum possible number, so the answer is 2.

Sample Input 2

2
1 2

Sample Output 2

1

Swapping 1 and 2 satisfies the condition.

Sample Input 3

2
2 1

Sample Output 3

0

The condition is already satisfied initially.

Sample Input 4

9
1 2 4 9 5 8 7 3 6

Sample Output 4

3


N = int(input())
p = list(map(int, input().split()))

ans = 0
for i in range(N):
    if p[i] != i+1:
        continue
    tmp = p[i]
    p[i] = p[i+1]
    p[i+1] = tmp
    ans += 1
print(ans)


X, A, B = [int(_) for _ in input().split()]

if abs(X - A) > abs(X - B):
    print("B")
else:
    print("A")

Score : 100 points

Problem Statement

Snuke lives at position x on a number line.
On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery.

Snuke decided to get food delivery from the closer of stores A and B.
Find out which store is closer to Snuke's residence.

Here, the distance between two points s and t on a number line is represented by |s-t|.

Constraints

1 \leq x \leq 1000

1 \leq a \leq 1000

1 \leq b \leq 1000

x, a and b are pairwise distinct.

The distances between Snuke's residence and stores A and B are different.

Input

Input is given from Standard Input in the following format:

x a b

Output

If store A is closer, print A; if store B is closer, print B.

Sample Input 1

5 2 7

Sample Output 1

B

The distances between Snuke's residence and stores A and B are 3 and 2, respectively.
Since store B is closer, print B.

Sample Input 2

1 999 1000

Sample Output 2

A


import sys
import numpy as np

stdin = sys.stdin

ri = lambda: int(rs())
rl = lambda: list(map(int, stdin.readline().split()))
rs = lambda: stdin.readline().rstrip()  # ignore trailing spaces

N = ri()
MOD = 10 ** 9 + 7
S1 = rs()
S2 = rs()
color = [None] * N

answer = 1
pattern = 0 #１つ前の列の色の種類

for i in range(N):
    if color[i] != None:
        continue
    if S1[i] == S2[i]:
        answer *= 3 - pattern
        pattern = 1
    else:
        answer *= [6, 2, 3][pattern]
        color[i+1] = 1
        pattern = 2

print(answer%MOD)
#08

Score : 400 points

Problem Statement

We have a board with a 2 \times N grid.
Snuke covered the board with N dominoes without overlaps.
Here, a domino can cover a 1 \times 2 or 2 \times 1 square.

Then, Snuke decided to paint these dominoes using three colors: red, cyan and green.
Two dominoes that are adjacent by side should be painted by different colors.
Here, it is not always necessary to use all three colors.

Find the number of such ways to paint the dominoes, modulo 1000000007.

The arrangement of the dominoes is given to you as two strings S_1 and S_2 in the following manner:

Each domino is represented by a different English letter (lowercase or uppercase).

The j-th character in S_i represents the domino that occupies the square at the i-th row from the top and j-th column from the left.

Constraints

1 \leq N \leq 52

|S_1| = |S_2| = N

S_1 and S_2 consist of lowercase and uppercase English letters.

S_1 and S_2 represent a valid arrangement of dominoes.

Input

Input is given from Standard Input in the following format:

N
S_1
S_2

Output

Print the number of such ways to paint the dominoes, modulo 1000000007.

Sample Input 1

3
aab
ccb

Sample Output 1

6

There are six ways as shown below:

Sample Input 2

1
Z
Z

Sample Output 2

3

Note that it is not always necessary to use all the colors.

Sample Input 3

52
RvvttdWIyyPPQFFZZssffEEkkaSSDKqcibbeYrhAljCCGGJppHHn
RLLwwdWIxxNNQUUXXVVMMooBBaggDKqcimmeYrhAljOOTTJuuzzn

Sample Output 3

958681902


import sys
import math
import collections
import itertools
import array
import inspect

# Set max recursion limit
sys.setrecursionlimit(10000)

# Debug output
def chkprint(*args):
    names = {
        id(v): k
        for k, v in inspect.currentframe().f_back.f_locals.items()
    }
    print(', '.join(
        names.get(id(arg), '???') + ' = ' + repr(arg) for arg in args))

# Binary converter
def to_bin(x):
    return bin(x)[2:]

def li_input():
    return [int(_) for _ in input().split()]

# --------------------------------------------

dp = None

def main():
    A, B, C, D = li_input()
    L = [0] * 101

    for i in range(101):
        if A <= i < B:
            L[i] += 1

        if C <= i < D:
            L[i] += 1

    ans = 0

    for l in L:
        if l == 2:
            ans += 1

    print(ans)

main()

Score : 200 points

Problem Statement

Alice and Bob are controlling a robot. They each have one switch that controls the robot.

Alice started holding down her button A second after the start-up of the robot, and released her button B second after the start-up.

Bob started holding down his button C second after the start-up, and released his button D second after the start-up.

For how many seconds both Alice and Bob were holding down their buttons?

Constraints

0≤A<B≤100

0≤C<D≤100

All input values are integers.

Input

Input is given from Standard Input in the following format:

A B C D

Output

Print the length of the duration (in seconds) in which both Alice and Bob were holding down their buttons.

Sample Input 1

0 75 25 100

Sample Output 1

50

Alice started holding down her button 0 second after the start-up of the robot, and released her button 75 second after the start-up.

Bob started holding down his button 25 second after the start-up, and released his button 100 second after the start-up.

Therefore, the time when both of them were holding down their buttons, is the 50 seconds from 25 seconds after the start-up to 75 seconds after the start-up.

Sample Input 2

0 33 66 99

Sample Output 2

0

Alice and Bob were not holding their buttons at the same time, so the answer is zero seconds.

Sample Input 3

10 90 20 80

Sample Output 3

60


#n = int(input())
n,m = map(int,input().split())
#l = list(map(int,input().split()))
#l = [list(map(int,input().split())) for i in range(n)]
print((m-1)*(n-1))

Score : 100 points

Problem Statement

In K-city, there are n streets running east-west, and m streets running north-south. Each street running east-west and each street running north-south cross each other. We will call the smallest area that is surrounded by four streets a block. How many blocks there are in K-city?

Constraints

2 ≤ n, m ≤ 100

Input

Input is given from Standard Input in the following format:

n m

Output

Print the number of blocks in K-city.

Sample Input 1

3 4

Sample Output 1

6

There are six blocks, as shown below:

Sample Input 2

2 2

Sample Output 2

1

There are one block, as shown below:


s = input()
print(s[0] + str(len(s) - 2)+ s[-1])

Score : 200 points

Problem Statement

The word internationalization is sometimes abbreviated to i18n.
This comes from the fact that there are 18 letters between the first i and the last n.

You are given a string s of length at least 3 consisting of lowercase English letters.
Abbreviate s in the same way.

Constraints

3 ≤ |s| ≤ 100 (|s| denotes the length of s.)

s consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

s

Output

Print the abbreviation of s.

Sample Input 1

internationalization

Sample Output 1

i18n

Sample Input 2

smiles

Sample Output 2

s4s

Sample Input 3

xyz

Sample Output 3

x1z


h, w = map(int, input().split())
n = int(input())
a = list(map(int, input().split()))
ans = [[] for i in range(h)]
inc = True
r = 0
for i in range(n):
    while(a[i] > 0):
        if(inc):
            if(r <= h - 1):
                ans[r].append(i + 1)
                r += 1
            else:
                r = h - 1
                ans[r].append(i + 1)
                r -= 1
                inc = False
        else:
            if(r >= 0):
                ans[r].append(i + 1)
                r -= 1
            else:
                r = 0
                ans[r].append(i + 1)
                r += 1
                inc = True
        a[i] -= 1
for x in ans:
    for i in x:
        print(i, end = " ")
    print()

Score : 400 points

Problem Statement

We have a grid with H rows and W columns of squares.
Snuke is painting these squares in colors 1, 2, ..., N.
Here, the following conditions should be satisfied:

For each i (1 ≤ i ≤ N), there are exactly a_i squares painted in Color i. Here, a_1 + a_2 + ... + a_N = H W.

For each i (1 ≤ i ≤ N), the squares painted in Color i are 4-connected. That is, every square painted in Color i can be reached from every square painted in Color i by repeatedly traveling to a horizontally or vertically adjacent square painted in Color i.

Find a way to paint the squares so that the conditions are satisfied.
It can be shown that a solution always exists.

Constraints

1 ≤ H, W ≤ 100

1 ≤ N ≤ H W

a_i ≥ 1

a_1 + a_2 + ... + a_N = H W

Input

Input is given from Standard Input in the following format:

H W
N
a_1 a_2 ... a_N

Output

Print one way to paint the squares that satisfies the conditions.
Output in the following format:

c_{1 1} ... c_{1 W}
:
c_{H 1} ... c_{H W}

Here, c_{i j} is the color of the square at the i-th row from the top and j-th column from the left.

Sample Input 1

2 2
3
2 1 1

Sample Output 1

1 1
2 3

Below is an example of an invalid solution:

1 2
3 1

This is because the squares painted in Color 1 are not 4-connected.

Sample Input 2

3 5
5
1 2 3 4 5

Sample Output 2

1 4 4 4 3
2 5 4 5 3
2 5 5 5 3

Sample Input 3

1 1
1
1

Sample Output 3

1


from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
import math
import bisect
import random
from itertools import permutations, accumulate, combinations, product
import sys
from pprint import pprint
from copy import deepcopy
import string
from bisect import bisect_left, bisect_right
from math import factorial, ceil, floor
from operator import mul
from functools import reduce
from pprint import pprint

sys.setrecursionlimit(2147483647)
INF = 10 ** 20
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def I(): return int(sys.stdin.buffer.readline())
def LS(): return sys.stdin.buffer.readline().rstrip().decode('utf-8').split()
def S(): return sys.stdin.buffer.readline().rstrip().decode('utf-8')
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]
mod = 1000000007

class Segtree:
    def __init__(self, A, ide_ele, segfunc, initialize=True):
        self.n = len(A)
        self.size = 2 ** (self.n - 1).bit_length()
        self.ide_ele = ide_ele
        self.segfunc = segfunc
        if initialize:
            self.data = [ide_ele] * self.size + A + [ide_ele] * (self.size - self.n)
            for i in range(self.size - 1, 0, -1):
                self.data[i] = self.segfunc(self.data[2 * i], self.data[2 * i + 1])
        else:
            self.data = [ide_ele] * (2 * self.size)

    def update(self, k, x):
        k += self.size
        self.data[k] = x
        while k > 0:
            k = k >> 1
            self.data[k] = self.segfunc(self.data[2 * k], self.data[2 * k + 1])

    def query(self, l, r):
        L, R = l + self.size, r + self.size
        s = self.ide_ele
        while L < R:
            if R & 1:
                R -= 1
                s = self.segfunc(s, self.data[R])
            if L & 1:
                s = self.segfunc(s, self.data[L])
                L += 1
            L >>= 1
            R >>= 1
        return s

n = I()
P = LI()
D = {P[i]: i for i in range(n)}
p_even = [P[i] if not i & 1 else INF for i in range(n)]
p_odd = [P[i] if i & 1 else INF for i in range(n)]
st_even = Segtree(p_even, INF, min)
st_odd = Segtree(p_odd, INF, min)
hq = [(st_even.query(0, n), 0, n)]
ans = []
for i in range(n // 2):
    v1, l, r = heappop(hq)
    p1 = D[v1]
    if not l & 1:
        st1, st2 = st_odd, st_even
    else:
        st1, st2 = st_even, st_odd
    v2 = st1.query(p1, r)
    p2 = D[v2]
    st2.update(p1, INF)
    st1.update(p2, INF)
    if l != p1:
        heappush(hq, (st2.query(l, p1), l, p1))
    if p1 + 1 != p2:
        heappush(hq, (st1.query(p1 + 1, p2), p1 + 1, p2))
    if p2 + 1 != r:
        heappush(hq, (st2.query(p2 + 1, r), p2 + 1, r))
    ans += [v1, v2]

print(*ans)

Score : 800 points

Problem Statement

Let N be a positive even number.

We have a permutation of (1, 2, ..., N), p = (p_1, p_2, ..., p_N).
Snuke is constructing another permutation of (1, 2, ..., N), q, following the procedure below.

First, let q be an empty sequence.
Then, perform the following operation until p becomes empty:

Select two adjacent elements in p, and call them x and y in order. Remove x and y from p (reducing the length of p by 2), and insert x and y, preserving the original order, at the beginning of q.

When p becomes empty, q will be a permutation of (1, 2, ..., N).

Find the lexicographically smallest permutation that can be obtained as q.

Constraints

N is an even number.

2 ≤ N ≤ 2 × 10^5

p is a permutation of (1, 2, ..., N).

Input

Input is given from Standard Input in the following format:

N
p_1 p_2 ... p_N

Output

Print the lexicographically smallest permutation, with spaces in between.

Sample Input 1

4
3 2 4 1

Sample Output 1

3 1 2 4

The solution above is obtained as follows:

p

q

(3, 2, 4, 1)

()

↓

↓

(3, 1)

(2, 4)

↓

↓

()

(3, 1, 2, 4)

Sample Input 2

2
1 2

Sample Output 2

1 2

Sample Input 3

8
4 6 3 2 8 5 7 1

Sample Output 3

3 1 2 7 4 6 8 5

The solution above is obtained as follows:

p

q

(4, 6, 3, 2, 8, 5, 7, 1)

()

↓

↓

(4, 6, 3, 2, 7, 1)

(8, 5)

↓

↓

(3, 2, 7, 1)

(4, 6, 8, 5)

↓

↓

(3, 1)

(2, 7, 4, 6, 8, 5)

↓

↓

()

(3, 1, 2, 7, 4, 6, 8, 5)


def resolve():
    n = input()
    print("ABC"+n)

if '__main__' == __name__:
    resolve()

Score : 100 points

Problem Statement

This contest, AtCoder Beginner Contest, is abbreviated as ABC.

When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example, ABC680 is the 680th round of ABC.

What is the abbreviation for the N-th round of ABC? Write a program to output the answer.

Constraints

100 ≤ N ≤ 999

Input

Input is given from Standard Input in the following format:

N

Output

Print the abbreviation for the N-th round of ABC.

Sample Input 1

100

Sample Output 1

ABC100

The 100th round of ABC is ABC100.

Sample Input 2

425

Sample Output 2

ABC425

Sample Input 3

999

Sample Output 3

ABC999


# 入力
N = int(input())
count_list = []
# 計算
for i in range(N):
    r = i + 1
    count = 0
    while r % 2 == 0:
        r = r / 2
        count += 1
    else:
        count += 0
    count_list.append(count)

M = max(count_list)
l = count_list.index(M)
print(l + 1)

Score : 200 points

Problem Statement

Takahashi loves numbers divisible by 2.

You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique.

Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder.

For example,

6 can be divided by 2 once: 6 -> 3.

8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1.

3 can be divided by 2 zero times.

Constraints

1 ≤ N ≤ 100

Input

Input is given from Standard Input in the following format:

N

Output

Print the answer.

Sample Input 1

7

Sample Output 1

4

4 can be divided by 2 twice, which is the most number of times among 1, 2, ..., 7.

Sample Input 2

32

Sample Output 2

32

Sample Input 3

1

Sample Output 3

1

Sample Input 4

100

Sample Output 4

64


N,M=map(int,input().split())
A=[list(map(int,input().split())) for i in range(M)]
D=[[]for i in range(N)]
for i in range(M):
  D[A[i][0]-1].append(A[i][1]-1)
  D[A[i][1]-1].append(A[i][0]-1)
a=len(D[-1])
b=len(D[0])
c=len(set(D[-1]+D[0]))
if a+b==c:
  print('IMPOSSIBLE')
else:
  print('POSSIBLE')

Score : 300 points

Problem Statement

In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands.
For convenience, we will call them Island 1, Island 2, ..., Island N.

There are M kinds of regular boat services between these islands.
Each service connects two islands. The i-th service connects Island a_i and Island b_i.

Cat Snuke is on Island 1 now, and wants to go to Island N.
However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services.

Help him.

Constraints

3 ≤ N ≤ 200 000

1 ≤ M ≤ 200 000

1 ≤ a_i < b_i ≤ N

(a_i, b_i) \neq (1, N)

If i \neq j, (a_i, b_i) \neq (a_j, b_j).

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
a_2 b_2
:
a_M b_M

Output

If it is possible to go to Island N by using two boat services, print POSSIBLE; otherwise, print IMPOSSIBLE.

Sample Input 1

3 2
1 2
2 3

Sample Output 1

POSSIBLE

Sample Input 2

4 3
1 2
2 3
3 4

Sample Output 2

IMPOSSIBLE

You have to use three boat services to get to Island 4.

Sample Input 3

100000 1
1 99999

Sample Output 3

IMPOSSIBLE

Sample Input 4

5 5
1 3
4 5
2 3
2 4
1 4

Sample Output 4

POSSIBLE

You can get to Island 5 by using two boat services: Island 1 -> Island 4 -> Island 5.


N , M = map(int, input().split())

L=[[0]*2 for i in range(M)]

for i in range(M):
    L[i][0] , L[i][1] = map(int, input().split())

A=[10**10]*M
B=[10**10]*M

L=sorted(L ,key=lambda x: x[0])

for i in range(M):
    if L[i][0] == 1:
        A[i]=L[i][1]
    else:
        break

L=sorted(L ,key=lambda x: -x[1])

for i in range(M):
    if L[i][1] == N:
        B[i]=L[i][0]
    else:
        break

for i in A:
    for j in B:
        if i == j:
            print("POSSIBLE")
            exit()
        if j == 10**10 : break
    if i == 10**10 :
        print("IMPOSSIBLE")
        exit()


# -*- coding: utf-8 -*-

N, M = map(int, raw_input().split())
line = []   # 定期便
for i in range(M):
    a, b = map(int, raw_input().split())
    line.append([a, b])
line.sort()

for l in line:
    if l[0] > 1:
        break
    else:
        for k in line:
            if k[0] == l[1] and k[1] == N:
                print 'POSSIBLE'
                exit()

print 'IMPOSSIBLE'


# 2019/08/15

from fractions import gcd

n,k=map(int,input().split())
a=list(map(int,input().split()))
ans='POSSIBLE'
max_a=max(a)
if max_a==k:
    print(ans)
    exit()
elif max_a<k:
    print('IMPOSSIBLE')
    exit()

g=0
for i in range(n):
    g=gcd(g,a[i])

if k%g==0 or k in a:
    print(ans)
else:
    print('IMPOSSIBLE')

Score : 300 points

Problem Statement

There is a box containing N balls. The i-th ball has the integer A_i written on it.
Snuke can perform the following operation any number of times:

Take out two balls from the box. Then, return them to the box along with a new ball, on which the absolute difference of the integers written on the two balls is written.

Determine whether it is possible for Snuke to reach the state where the box contains a ball on which the integer K is written.

Constraints

1 \leq N \leq 10^5

1 \leq A_i \leq 10^9

1 \leq K \leq 10^9

All input values are integers.

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 ... A_N

Output

If it is possible for Snuke to reach the state where the box contains a ball on which the integer K is written, print POSSIBLE; if it is not possible, print IMPOSSIBLE.

Sample Input 1

3 7
9 3 4

Sample Output 1

POSSIBLE

First, take out the two balls 9 and 4, and return them back along with a new ball, abs(9-4)=5.
Next, take out 3 and 5, and return them back along with abs(3-5)=2.
Finally, take out 9 and 2, and return them back along with abs(9-2)=7.
Now we have 7 in the box, and the answer is therefore POSSIBLE.

Sample Input 2

3 5
6 9 3

Sample Output 2

IMPOSSIBLE

No matter what we do, it is not possible to have 5 in the box. The answer is therefore IMPOSSIBLE.

Sample Input 3

4 11
11 3 7 15

Sample Output 3

POSSIBLE

The box already contains 11 before we do anything. The answer is therefore POSSIBLE.

Sample Input 4

5 12
10 2 8 6 4

Sample Output 4

IMPOSSIBLE


A, B = map(int, input().split())
aa = A+B

if A % 3 == 0 or B%3==0 or aa%3==0:
    print("Possible")
else:
    print("Impossible")

Score : 100 points

Problem Statement

Snuke is giving cookies to his three goats.

He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins).

Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.

Constraints

1 \leq A,B \leq 100

Both A and B are integers.

Input

Input is given from Standard Input in the following format:

A B

Output

If it is possible to give cookies so that each of the three goats can have the same number of cookies, print Possible; otherwise, print Impossible.

Sample Input 1

4 5

Sample Output 1

Possible

If Snuke gives nine cookies, each of the three goats can have three cookies.

Sample Input 2

1 1

Sample Output 2

Impossible

Since there are only two cookies, the three goats cannot have the same number of cookies no matter what Snuke gives to them.


a,b=map(int,input().split())
if a%3==0 or b%3==0 or (a+b)%3==0:
  print('Possible')
else:print('Impossible')


from sys import stdin
a, b = [int(x) for x in stdin.readline().rstrip().split()]
if (a+b)%3==0:
  print('Possible')
else:
  print('Impossible')


n,k=map(int,input().split())
l=list(map(int,input().split()))
l.sort(reverse=True)
ans=0

for i in range(k):
    ans+=l[i]

print(ans)

Score : 200 points

Problem Statement

Snuke has N sticks.
The length of the i-th stick is l_i.

Snuke is making a snake toy by joining K of the sticks together.

The length of the toy is represented by the sum of the individual sticks that compose it.
Find the maximum possible length of the toy.

Constraints

1 \leq K \leq N \leq 50

1 \leq l_i \leq 50

l_i is an integer.

Input

Input is given from Standard Input in the following format:

N K
l_1 l_2 l_3 ... l_{N}

Output

Print the answer.

Sample Input 1

5 3
1 2 3 4 5

Sample Output 1

12

You can make a toy of length 12 by joining the sticks of lengths 3, 4 and 5, which is the maximum possible length.

Sample Input 2

15 14
50 26 27 21 41 7 42 35 7 5 5 36 39 1 45

Sample Output 2

386


n = int(input())
A = list(map(int, input().split()))

sm = sum(A)
x = 0
y = sm
ans = 1001001001001001001

for idx, a in enumerate(A):
  if idx == n - 1:
    break

  x  = x + a
  y = y - a
  # print(x, y, abs(x - y))
  ans = min(ans, abs(x - y))
print(ans)

Score : 300 points

Problem Statement

Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i written on it.

They will share these cards.
First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards.
Here, both Snuke and Raccoon have to take at least one card.

Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively.
They would like to minimize |x-y|.
Find the minimum possible value of |x-y|.

Constraints

2 \leq N \leq 2 \times 10^5

-10^{9} \leq a_i \leq 10^{9}

a_i is an integer.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_{N}

Output

Print the answer.

Sample Input 1

6
1 2 3 4 5 6

Sample Output 1

1

If Snuke takes four cards from the top, and Raccoon takes the remaining two cards, x=10, y=11, and thus |x-y|=1. This is the minimum possible value.

Sample Input 2

2
10 -10

Sample Output 2

20

Snuke can only take one card from the top, and Raccoon can only take the remaining one card. In this case, x=10, y=-10, and thus |x-y|=20.


tmp = list(map(int,input().split()))
As = list(map(int,input().split()))
n,p = tmp[0],tmp[1]
o ,e  = 0,0
for i in range(n):
  if As[i] % 2:
    o += 1
  else:
    e += 1
def comb(j,k):
  if k == 1:
    return(j)
  else:
    return comb(j-1,k-1) * j / k
ans = 0
if p == 0:
  for i in range(o+1):
    if not i % 2:
      ans += int(comb(o,i))
else:
  for i in range(o+1):
    if i % 2:
      ans += int(comb(o,i))
ans *= (2 ** e)
print(ans)

Score : 200 points

Problem Statement

There are N bags of biscuits. The i-th bag contains A_i biscuits.

Takaki will select some of these bags and eat all of the biscuits inside.
Here, it is also possible to select all or none of the bags.

He would like to select bags so that the total number of biscuits inside is congruent to P modulo 2.
How many such ways to select bags there are?

Constraints

1 \leq N \leq 50

P = 0 or 1

1 \leq A_i \leq 100

Input

Input is given from Standard Input in the following format:

N P
A_1 A_2 ... A_N

Output

Print the number of ways to select bags so that the total number of biscuits inside is congruent to P modulo 2.

Sample Input 1

2 0
1 3

Sample Output 1

2

There are two ways to select bags so that the total number of biscuits inside is congruent to 0 modulo 2:

Select neither bag. The total number of biscuits is 0.

Select both bags. The total number of biscuits is 4.

Sample Input 2

1 1
50

Sample Output 2

0

Sample Input 3

3 0
1 1 1

Sample Output 3

4

Two bags are distinguished even if they contain the same number of biscuits.

Sample Input 4

45 1
17 55 85 55 74 20 90 67 40 70 39 89 91 50 16 24 14 43 24 66 25 9 89 71 41 16 53 13 61 15 85 72 62 67 42 26 36 66 4 87 59 91 4 25 26

Sample Output 4

17592186044416


a,b,c = map(int,input().split())

m = min(a + b,a + c,b + c)

print(m)

Score : 100 points

Problem Statement

Snuke is buying a bicycle.
The bicycle of his choice does not come with a bell, so he has to buy one separately.

He has very high awareness of safety, and decides to buy two bells, one for each hand.

The store sells three kinds of bells for the price of a, b and c yen (the currency of Japan), respectively.
Find the minimum total price of two different bells.

Constraints

1 \leq a,b,c \leq 10000

a, b and c are integers.

Input

Input is given from Standard Input in the following format:

a b c

Output

Print the minimum total price of two different bells.

Sample Input 1

700 600 780

Sample Output 1

1300

Buying a 700-yen bell and a 600-yen bell costs 1300 yen.

Buying a 700-yen bell and a 780-yen bell costs 1480 yen.

Buying a 600-yen bell and a 780-yen bell costs 1380 yen.

The minimum among these is 1300 yen.

Sample Input 2

10000 10000 10000

Sample Output 2

20000

Buying any two bells costs 20000 yen.


n = int(input())
a_list = [int(x) for x in input().split()]
b = []
for a in a_list:
    b.append(a)
    b = list(reversed(b))
print(*b)

Score : 300 points

Problem Statement

You are given an integer sequence of length n, a_1, ..., a_n.
Let us consider performing the following n operations on an empty sequence b.

The i-th operation is as follows:

Append a_i to the end of b.

Reverse the order of the elements in b.

Find the sequence b obtained after these n operations.

Constraints

1 \leq n \leq 2\times 10^5

0 \leq a_i \leq 10^9

n and a_i are integers.

Input

Input is given from Standard Input in the following format:

n
a_1 a_2 ... a_n

Output

Print n integers in a line with spaces in between.
The i-th integer should be b_i.

Sample Input 1

4
1 2 3 4

Sample Output 1

4 2 1 3

After step 1 of the first operation, b becomes: 1.

After step 2 of the first operation, b becomes: 1.

After step 1 of the second operation, b becomes: 1, 2.

After step 2 of the second operation, b becomes: 2, 1.

After step 1 of the third operation, b becomes: 2, 1, 3.

After step 2 of the third operation, b becomes: 3, 1, 2.

After step 1 of the fourth operation, b becomes: 3, 1, 2, 4.

After step 2 of the fourth operation, b becomes: 4, 2, 1, 3.

Thus, the answer is 4 2 1 3.

Sample Input 2

3
1 2 3

Sample Output 2

3 1 2

As shown above in Sample Output 1, b becomes 3, 1, 2 after step 2 of the third operation. Thus, the answer is 3 1 2.

Sample Input 3

1
1000000000

Sample Output 3

1000000000

Sample Input 4

6
0 6 7 6 7 0

Sample Output 4

0 6 6 0 7 7


n = int(raw_input())
a = map(int, raw_input().split())
b = []
for i in range(n):
    b.append(a[i])
    b = b[::-1]
print b


n=int(input())
a=list(map(int,input().split()))
l=[None]*(n+1)
for i in range(n+1):
    if l[a[i]]==None:
        l[a[i]]=i
    else:
        p=l[a[i]]+n-i
        break
pr=10**9+7
print(n)
MAX_NUM = 10**5 + 2
fac  = [0 for _ in range(MAX_NUM)]
finv = [0 for _ in range(MAX_NUM)]
inv  = [0 for _ in range(MAX_NUM)]
fac[0]  = fac[1] = 1
finv[0] = finv[1] = 1
inv[1] = 1
for i in range(2,MAX_NUM):
    fac[i] = fac[i-1] * i % pr
    inv[i] = pr - inv[pr%i] * (pr // i) % pr
    finv[i] = finv[i-1] * inv[i] % pr
def c(n,k):
    if n < k:
        return 0
    if n < 0 or k < 0:
        return 0
    return fac[n] * (finv[k] * finv[n-k] % pr) % pr
for i in range(2,n+1):
    t=c(n+1,i)
    if p>=i-1:
        t-=c(p,i-1)
    print(t%pr)
print(1)

Score : 600 points

Problem Statement

You are given an integer sequence of length n+1, a_1,a_2,...,a_{n+1}, which consists of the n integers 1,...,n.
It is known that each of the n integers 1,...,n appears at least once in this sequence.

For each integer k=1,...,n+1, find the number of the different subsequences (not necessarily contiguous) of the given sequence with length k, modulo 10^9+7.

Notes

If the contents of two subsequences are the same, they are not separately counted even if they originate from different positions in the original sequence.

A subsequence of a sequence a with length k is a sequence obtained by selecting k of the elements of a and arranging them without changing their relative order. For example, the sequences 1,3,5 and 1,2,3 are subsequences of 1,2,3,4,5, while 3,1,2 and 1,10,100 are not.

Constraints

1 \leq n \leq 10^5

1 \leq a_i \leq n

Each of the integers 1,...,n appears in the sequence.

n and a_i are integers.

Input

Input is given from Standard Input in the following format:

n
a_1 a_2 ... a_{n+1}

Output

Print n+1 lines.
The k-th line should contain the number of the different subsequences of the given sequence with length k, modulo 10^9+7.

Sample Input 1

3
1 2 1 3

Sample Output 1

3
5
4
1

There are three subsequences with length 1: 1 and 2 and 3.

There are five subsequences with length 2: 1,1 and 1,2 and 1,3 and 2,1 and 2,3.

There are four subsequences with length 3: 1,1,3 and 1,2,1 and 1,2,3 and 2,1,3.

There is one subsequence with length 4: 1,2,1,3.

Sample Input 2

1
1 1

Sample Output 2

1
1

There is one subsequence with length 1: 1.

There is one subsequence with length 2: 1,1.

Sample Input 3

32
29 19 7 10 26 32 27 4 11 20 2 8 16 23 5 14 6 12 17 22 18 30 28 24 15 1 25 3 13 21 19 31 9

Sample Output 3

32
525
5453
40919
237336
1107568
4272048
13884156
38567100
92561040
193536720
354817320
573166440
818809200
37158313
166803103
166803103
37158313
818809200
573166440
354817320
193536720
92561040
38567100
13884156
4272048
1107568
237336
40920
5456
528
33
1

Be sure to print the numbers modulo 10^9+7.


#066-D
from collections import Counter
n=int(input())
a=list(map(int,input().split()))
d=Counter(a).most_common(1)[0][0]
f,s=-1,-1
fc=[1 for i in range(n+2)]
for i in range(n+1):
    if a[i]==d:
        if f==-1:
            f=i
        else:
            s=i
    fc[i+1]=(i+1)*fc[i]
fc[n+1]=(n+1)*fc[n]
for k in range(1,n+2):
    if k==1:
        print(n)
    elif k>f+1+n+1-s-1:
        print(int(fc[n+1]/(fc[k]*fc[n+1-k])))
    else:
        print(int(fc[n+1]/(fc[k]*fc[n+1-k])-fc[f+n-s]/(fc[k-1]*fc[f+n-s-(k-1)])))


x,a,b=map(int,input().split())
print('delicious' if 0<(b-a) else 'dangerous' if x>(b-a) else 'safe')

Score : 100 points

Problem Statement

Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier.
He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though.

Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious.

Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it.

Write a program that outputs delicious if he found it delicious, safe if he did not found it delicious but did not get a stomachache either, and dangerous if he got a stomachache.

Constraints

1 ≤ X,A,B ≤ 10^9

Input

Input is given from Standard Input in the following format:

X A B

Output

Print delicious if Takahashi found the food delicious; print safe if he neither found it delicious nor got a stomachache; print dangerous if he got a stomachache.

Sample Input 1

4 3 6

Sample Output 1

safe

He ate the food three days after the "best-by" date. It was not delicious or harmful for him.

Sample Input 2

6 5 1

Sample Output 2

delicious

He ate the food by the "best-by" date. It was delicious for him.

Sample Input 3

3 7 12

Sample Output 3

dangerous

He ate the food five days after the "best-by" date. It was harmful for him.


stdin = input()

d = int(stdin.split(" ")[0])
a = int(stdin.split(" ")[1])
b = int(stdin.split(" ")[2])

if b <= a:
    print ("delicious")
elif b <= a + d:
    print ("safe")
elif b > a + d:
    print ("dangerous")


n = int(input())
q = [int(input())  for i in range(n)]
cnt = 0
ind = 0
for i in range(n):
    ind = q[ind]-1
    cnt += 1
    if ind == 1:
        print(cnt)
        exit()

print(-1)

Score : 200 points

Problem Statement

Takahashi wants to gain muscle, and decides to work out at AtCoder Gym.

The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up.
These buttons are numbered 1 through N.
When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i.
When Button i is not lighten up, nothing will happen by pressing it.

Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up.

Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.

Constraints

2 ≤ N ≤ 10^5

1 ≤ a_i ≤ N

Input

Input is given from Standard Input in the following format:

N
a_1
a_2
:
a_N

Output

Print -1 if it is impossible to lighten up Button 2.
Otherwise, print the minimum number of times we need to press buttons in order to lighten up Button 2.

Sample Input 1

3
3
1
2

Sample Output 1

2

Press Button 1, then Button 3.

Sample Input 2

4
3
4
1
2

Sample Output 2

-1

Pressing Button 1 lightens up Button 3, and vice versa, so Button 2 will never be lighten up.

Sample Input 3

5
3
3
4
2
4

Sample Output 3

3


def solve():
    N = int(input())
    a = [int(input()) for i in range(1, N + 1)]
    # ループしたら終わる
    l = []
    s = a[0]
    l = [0] * N
    c = 1
    l[0]=True
    while(True):
        l[s - 1] = True
        if s == 2:
            print(c)
            return True
        s = a[s - 1]
        if l[s - 1] == True:
            print(-1)
            return False
        c += 1
solve()


buf=input().split()
N=int(buf[0])
M=int(buf[1])

thres=10**9+7

if abs(N-M)==1:
    N=min(N,M)
    retVal=1
    for i in range(1,N+1):
        retVal*=(i**2)
        if retVal>=thres:
            retVal=retVal%thres
    retVal*=(N+1)
    if retVal>=thres:
        retVal=retVal%thres
    print(retVal)
elif N==M:
    retVal=2
    for i in range(1,N+1):
        retVal*=(i**2)
        if retVal>=thres:
            retVal=retVal%thres
    print(retVal)
else:
    print(0)

Score : 300 points

Problem Statement

Snuke has N dogs and M monkeys. He wants them to line up in a row.

As a Japanese saying goes, these dogs and monkeys are on bad terms. ("ken'en no naka", literally "the relationship of dogs and monkeys", means a relationship of mutual hatred.) Snuke is trying to reconsile them, by arranging the animals so that there are neither two adjacent dogs nor two adjacent monkeys.

How many such arrangements there are? Find the count modulo 10^9+7 (since animals cannot understand numbers larger than that).
Here, dogs and monkeys are both distinguishable. Also, two arrangements that result from reversing each other are distinguished.

Constraints

1 ≤ N,M ≤ 10^5

Input

Input is given from Standard Input in the following format:

N M

Output

Print the number of possible arrangements, modulo 10^9+7.

Sample Input 1

2 2

Sample Output 1

8

We will denote the dogs by A and B, and the monkeys by C and D. There are eight possible arrangements: ACBD, ADBC, BCAD, BDAC, CADB, CBDA, DACB and DBCA.

Sample Input 2

3 2

Sample Output 2

12

Sample Input 3

1 8

Sample Output 3

0

Sample Input 4

100000 100000

Sample Output 4

530123477


r,g,b = map(int,input().split())
num = r*100 + g*20 + b
if num%4 ==0:
	print("YES")
else:
  	print("NO")

Score : 100 points

Problem Statement

AtCoDeer has three cards, one red, one green and one blue.

An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card.

We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer.

Is this integer a multiple of 4?

Constraints

1 ≤ r, g, b ≤ 9

Input

Input is given from Standard Input in the following format:

r g b

Output

If the three-digit integer is a multiple of 4, print YES (case-sensitive); otherwise, print NO.

Sample Input 1

4 3 2

Sample Output 1

YES

432 is a multiple of 4, and thus YES should be printed.

Sample Input 2

2 3 4

Sample Output 2

NO

234 is not a multiple of 4, and thus NO should be printed.


N = int(input())

a = [int(n) for n in input().split()]
a = sorted(a)
print(a[N-1] - a[0])

Score : 200 points

Problem Statement

It is only six months until Christmas, and AtCoDeer the reindeer is now planning his travel to deliver gifts.

There are N houses along TopCoDeer street. The i-th house is located at coordinate a_i. He has decided to deliver gifts to all these houses.

Find the minimum distance to be traveled when AtCoDeer can start and end his travel at any positions.

Constraints

1 ≤ N ≤ 100

0 ≤ a_i ≤ 1000

a_i is an integer.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the minimum distance to be traveled.

Sample Input 1

4
2 3 7 9

Sample Output 1

7

The travel distance of 7 can be achieved by starting at coordinate 9 and traveling straight to coordinate 2.

It is not possible to do with a travel distance of less than 7, and thus 7 is the minimum distance to be traveled.

Sample Input 2

8
3 1 4 1 5 9 2 6

Sample Output 2

8

There may be more than one house at a position.


n = int(input())
a = list(map(int, input().split()))
color = [0]*8
for i in a:
  color[min(8,i//400)] += 1
kind = 0
for i in color[:7]:
  if i:
    kind += 1
if kind == 0 and color[7]:
  color[7] -= 1
  kind += 1
print(kind,kind+color[7])

Score : 300 points

Problem Statement

In AtCoder, a person who has participated in a contest receives a color, which corresponds to the person's rating as follows:

Rating 1-399 : gray

Rating 400-799 : brown

Rating 800-1199 : green

Rating 1200-1599 : cyan

Rating 1600-1999 : blue

Rating 2000-2399 : yellow

Rating 2400-2799 : orange

Rating 2800-3199 : red

Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not.

Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i.

Find the minimum and maximum possible numbers of different colors of the users.

Constraints

1 ≤ N ≤ 100

1 ≤ a_i ≤ 4800

a_i is an integer.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the minimum possible number of different colors of the users, and the maximum possible number of different colors, with a space in between.

Sample Input 1

4
2100 2500 2700 2700

Sample Output 1

2 2

The user with rating 2100 is "yellow", and the others are "orange". There are two different colors.

Sample Input 2

5
1100 1900 2800 3200 3200

Sample Output 2

3 5

The user with rating 1100 is "green", the user with rating 1900 is blue and the user with rating 2800 is "red".

If the fourth user picks "red", and the fifth user picks "blue", there are three different colors. This is one possible scenario for the minimum number of colors.

If the fourth user picks "purple", and the fifth user picks "black", there are five different colors. This is one possible scenario for the maximum number of colors.

Sample Input 3

20
800 810 820 830 840 850 860 870 880 890 900 910 920 930 940 950 960 970 980 990

Sample Output 3

1 1

All the users are "green", and thus there is one color.


A, B = map(int, input().split())
if A + B >= 10:
    print('error')
else:
    print(A+B)

Score : 100 points

Problem Statement

You are given two integers A and B as the input. Output the value of A + B.

However, if A + B is 10 or greater, output error instead.

Constraints

A and B are integers.

1 ≤ A, B ≤ 9

Input

Input is given from Standard Input in the following format:

A B

Output

If A + B is 10 or greater, print the string error (case-sensitive); otherwise, print the value of A + B.

Sample Input 1

6 3

Sample Output 1

9

Sample Input 2

6 4

Sample Output 2

error


def resolve():
    s = input()

    ans = str()
    for char in s:
        if s.count(char) >1:
            ans = "no"
            break
        else:
            ans = "yes"
    print(ans)
resolve()

Score : 200 points

Problem Statement

You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.

Constraints

2 ≤ |S| ≤ 26, where |S| denotes the length of S.

S consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

S

Output

If all the characters in S are different, print yes (case-sensitive); otherwise, print no.

Sample Input 1

uncopyrightable

Sample Output 1

yes

Sample Input 2

different

Sample Output 2

no

Sample Input 3

no

Sample Output 3

yes


s = input()
ss = "".join(set(s))
print("Yes") if len(s) == len(ss) else print("No")


n = int(input())
S = sorted([int(input()) for _ in range(n)])
ans = 0
caution = 0
flag = 1
for s in S:
    ans += s
    if flag:
        if s % 10:
            caution = s
            flag = 0
if ans % 10 == 0:
    if caution:
        ans -= caution
    else:
        ans = 0
print(ans)

Score : 300 points

Problem Statement

You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well.

However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade?

Constraints

All input values are integers.

1 ≤ N ≤ 100

1 ≤ s_i ≤ 100

Input

Input is given from Standard Input in the following format:

N
s_1
s_2
:
s_N

Output

Print the maximum value that can be displayed as your grade.

Sample Input 1

3
5
10
15

Sample Output 1

25

Your grade will be 25 if the 10-point and 15-point questions are answered correctly and the 5-point question is not, and this grade will be displayed correctly. Your grade will become 30 if the 5-point question is also answered correctly, but this grade will be incorrectly displayed as 0.

Sample Input 2

3
10
10
15

Sample Output 2

35

Your grade will be 35 if all the questions are answered correctly, and this grade will be displayed correctly.

Sample Input 3

3
10
20
30

Sample Output 3

0

Regardless of whether each question is answered correctly or not, your grade will be a multiple of 10 and displayed as 0.


import math
N, A, B = map(int, input().split())
H = [int(input()) for i in range(N)]

def f(x):
    s = sum([math.ceil(max(0, h - x * B) / (A - B)) for h in H])
    if s <= x:
        return True
    else:
        return False

l = 0
r = 10 ** 9 + 7
while l < r:
    m = (l + r) // 2
    if f(m):
        r = m
    else:
        l = m + 1
    print(r, l, m)

print(m)

Score : 400 points

Problem Statement

You are going out for a walk, when you suddenly encounter N monsters. Each monster has a parameter called health, and the health of the i-th monster is h_i at the moment of encounter. A monster will vanish immediately when its health drops to 0 or below.

Fortunately, you are a skilled magician, capable of causing explosions that damage monsters. In one explosion, you can damage monsters as follows:

Select an alive monster, and cause an explosion centered at that monster. The health of the monster at the center of the explosion will decrease by A, and the health of each of the other monsters will decrease by B. Here, A and B are predetermined parameters, and A > B holds.

At least how many explosions do you need to cause in order to vanish all the monsters?

Constraints

All input values are integers.

1 ≤ N ≤ 10^5

1 ≤ B < A ≤ 10^9

1 ≤ h_i ≤ 10^9

Input

Input is given from Standard Input in the following format:

N A B
h_1
h_2
:
h_N

Output

Print the minimum number of explosions that needs to be caused in order to vanish all the monsters.

Sample Input 1

4 5 3
8
7
4
2

Sample Output 1

2

You can vanish all the monsters in two explosion, as follows:

First, cause an explosion centered at the monster with 8 health. The healths of the four monsters become 3, 4, 1 and -1, respectively, and the last monster vanishes.

Second, cause an explosion centered at the monster with 4 health remaining. The healths of the three remaining monsters become 0, -1 and -2, respectively, and all the monsters are now vanished.

Sample Input 2

2 10 4
20
20

Sample Output 2

4

You need to cause two explosions centered at each monster, for a total of four.

Sample Input 3

5 2 1
900000000
900000000
1000000000
1000000000
1000000000

Sample Output 3

800000000


Score : 100 points

Problem Statement

Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below.
Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group.

Constraints

x and y are integers.

1 ≤ x < y ≤ 12

Input

Input is given from Standard Input in the following format:

x y

Output

If x and y belong to the same group, print Yes; otherwise, print No.

Sample Input 1

1 3

Sample Output 1

Yes

Sample Input 2

2 4

Sample Output 2

No


a = [1,3,5,7,8,10,12]
b = [4,6,9,11]
c = [2]
x,y = map(int,input().split())
if x in a and y in a:
    print("Yes")
elif x in b and y in b:
    print("Yes")
elif x in c and y in c:
    print("Yes")
else:
    print("No")


# -*- coding: utf-8 -*-
x, y = map(int, input().split())

#def list
list1 = [1,3,5,7,8,10,12]
list2 = [4,6,9,11]
list3 = [2]

#group check
if x in list1 and y in list1:
    print "YES"
elif x in list2 and y in list2:
    print "YES"
elif x in list3 and y in list3:
    print "YES"
else:
    print "NO"


H,W=map(int,input().split())

mat=[["#"]*(W+2)]
for i in range(H):
  array=["#"]+list(input())+["#"]
  mat.append(array)
mat.append(["#"]*(W+2))

for i in range(H+2):
  print("".join(mat[i]))

Score : 200 points

Problem Statement

You are given a image with a height of H pixels and a width of W pixels.
Each pixel is represented by a lowercase English letter.
The pixel at the i-th row from the top and j-th column from the left is a_{ij}.

Put a box around this image and output the result. The box should consist of # and have a thickness of 1.

Constraints

1 ≤ H, W ≤ 100

a_{ij} is a lowercase English letter.

Input

Input is given from Standard Input in the following format:

H W
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}

Output

Print the image surrounded by a box that consists of # and has a thickness of 1.

Sample Input 1

2 3
abc
arc

Sample Output 1

#####
#abc#
#arc#
#####

Sample Input 2

1 1
z

Sample Output 2

###
#z#
###


H, W=map(int, input().split())
deference=H*W
a=0
b=0
for i in range(W-1):
    a=H*(i+1)
    b=H//2*(W-i-1)
    if deference>max(a, b, H*W-a-b)-min(a, b, H*W-a-b):
        deference=max(a, b, H*W-a-b)-min(a, b, H*W-a-b)
for i in range(H-1):
    a=W*(i+1)
    b=W//2*(H-i-1)
    if deference>max(a, b, H*W-a-b)-min(a, b, H*W-a-b):
        deference=max(a, b, H*W-a-b)-min(a, b, H*W-a-b)
for i in range(W-2):
    a=H*(i+1)
    b=H*((W-i-1)//2)
    if deference>max(a, b, H*W-a-b)-min(a, b, H*W-a-b):
        deference=max(a, b, H*W-a-b)-min(a, b, H*W-a-b)
for i in range(H-2):
    a=W*(i+1)
    b=W*((H-i-1)//2)
    if deference>max(a, b, H*W-a-b)-min(a, b, H*W-a-b):
        deference=max(a, b, H*W-a-b)-min(a, b, H*W-a-b)
print(deference)

Score : 400 points

Problem Statement

There is a bar of chocolate with a height of H blocks and a width of W blocks.
Snuke is dividing this bar into exactly three pieces.
He can only cut the bar along borders of blocks, and the shape of each piece must be a rectangle.

Snuke is trying to divide the bar as evenly as possible.
More specifically, he is trying to minimize S_{max} - S_{min}, where S_{max} is the area (the number of blocks contained) of the largest piece, and S_{min} is the area of the smallest piece.
Find the minimum possible value of S_{max} - S_{min}.

Constraints

2 ≤ H, W ≤ 10^5

Input

Input is given from Standard Input in the following format:

H W

Output

Print the minimum possible value of S_{max} - S_{min}.

Sample Input 1

3 5

Sample Output 1

0

In the division below, S_{max} - S_{min} = 5 - 5 = 0.

Sample Input 2

4 5

Sample Output 2

2

In the division below, S_{max} - S_{min} = 8 - 6 = 2.

Sample Input 3

5 5

Sample Output 3

4

In the division below, S_{max} - S_{min} = 10 - 6 = 4.

Sample Input 4

100000 2

Sample Output 4

1

Sample Input 5

100000 100000

Sample Output 5

50000


H, W = list(map(int, input().split()))
res = min(H,W)
hr,vr = H%3*W,W%3*H
def min_s_diff(x, y):
    if x % 3 == 0:
        return 0
    if x % 3 == 1:
        a = x // 3
        b = y // 2
        s1 = a * y
        s2 = (x - a) * b
        s3 = x * y - s1 - s2
        return max(s1, s2, s3) - min(s1, s2, s3)
    if x % 3 == 2:
        a = x // 3 + 1
        b = y // 2
        s1 = a * y
        s2 = (x - a) * b
        s3 = x * y - s1 -s2
        return max(s1, s2, s3) - min(s1, s2, s3)

print(min(min_s_diff(H, W), min_s_diff(W, H), res, hr, vr))


a, b, c = (int(i) for i in input().split())

if a <= c and b >= c:
    print('Yes')
else:
    print('No')

Score : 100 points

Problem Statement

You are given three integers A, B and C.
Determine whether C is not less than A and not greater than B.

Constraints

-100≤A,B,C≤100

A, B and C are all integers.

Input

Input is given from Standard Input in the following format:

A B C

Output

If the condition is satisfied, print Yes; otherwise, print No.

Sample Input 1

1 3 2

Sample Output 1

Yes

C=2 is not less than A=1 and not greater than B=3, and thus the output should be Yes.

Sample Input 2

6 5 4

Sample Output 2

No

C=4 is less than A=6, and thus the output should be No.

Sample Input 3

2 2 2

Sample Output 3

Yes


a, b, c = map(int, raw_input().split())

print "YES" if c >= a and c <= b else "NO"


n,k = map(int,input().split())
l = [0]*10**5
for i in range(n):
  a,b = map(int,input().split())
  l[a-1] += b
sum = 0
i = 0
while sum < k:
  sum += l[i]
  i += 1
print(i)

Score : 300 points

Problem Statement

There is an empty array.
The following N operations will be performed to insert integers into the array.
In the i-th operation (1≤i≤N), b_i copies of an integer a_i are inserted into the array.
Find the K-th smallest integer in the array after the N operations.
For example, the 4-th smallest integer in the array \{1,2,2,3,3,3\} is 3.

Constraints

1≤N≤10^5

1≤a_i,b_i≤10^5

1≤K≤b_1…+…b_n

All input values are integers.

Input

Input is given from Standard Input in the following format:

N K
a_1 b_1
:
a_N b_N

Output

Print the K-th smallest integer in the array after the N operations.

Sample Input 1

3 4
1 1
2 2
3 3

Sample Output 1

3

The resulting array is the same as the one in the problem statement.

Sample Input 2

10 500000
1 100000
1 100000
1 100000
1 100000
1 100000
100000 100000
100000 100000
100000 100000
100000 100000
100000 100000

Sample Output 2

1


def BellmanFord(edges,num_v,source):
  #グラフの初期化
  inf=float("inf")
  dist=[inf for i in range(num_v)]
  dist[source-1]=0

  #辺の緩和
  for i in range(num_v):
    for edge in edges:
      if edge[0] != inf and dist[edge[1]-1] > dist[edge[0]-1] + edge[2]:
        dist[edge[1]-1] = dist[edge[0]-1] + edge[2]
        if i==num_v-1: return -1

  return dist

N,M = [int(x) for x in input().split()]
edges = []
for _ in range(N):
    l = [int(x) for x in input().split()]
    l[2]*=-1
    edges.append(l)

route = BellmanFord(edges,N,1)
if(route == -1):
    print('inf')
else:
    print(-route[-1])

Score : 400 points

Problem Statement

There is a directed graph with N vertices and M edges.
The i-th edge (1≤i≤M) points from vertex a_i to vertex b_i, and has a weight c_i.
We will play the following single-player game using this graph and a piece.

Initially, the piece is placed at vertex 1, and the score of the player is set to 0.
The player can move the piece as follows:

When the piece is placed at vertex a_i, move the piece along the i-th edge to vertex b_i. After this move, the score of the player is increased by c_i.

The player can end the game only when the piece is placed at vertex N.
The given graph guarantees that it is possible to traverse from vertex 1 to vertex N.

When the player acts optimally to maximize the score at the end of the game, what will the score be?
If it is possible to increase the score indefinitely, print inf.

Constraints

2≤N≤1000

1≤M≤min(N(N-1),2000)

1≤a_i,b_i≤N (1≤i≤M)

a_i≠b_i (1≤i≤M)

a_i≠a_j or b_i≠b_j (1≤i<j≤M)

-10^9≤c_i≤10^9 (1≤i≤M)

c_i is an integer.

In the given graph, there exists a path from vertex 1 to vertex N.

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1 c_1
a_2 b_2 c_2
:
a_M b_M c_M

Output

Print the maximum possible score at the end of the game, if it is finite. If it is possible to increase the score indefinitely, print inf.

Sample Input 1

3 3
1 2 4
2 3 3
1 3 5

Sample Output 1

7

There are two ways to move the piece to vertex N=3:

vertex 1 → vertex 2 → vertex 3 : score 4+3=7

vertex 1 → vertex 3 : score 5

Thus, the maximum possible score at the end of the game is 7.

Sample Input 2

2 2
1 2 1
2 1 1

Sample Output 2

inf

It is possible to increase the score indefinitely by alternating between vertex 1 and 2.

Sample Input 3

6 5
1 2 -1000000000
2 3 -1000000000
3 4 -1000000000
4 5 -1000000000
5 6 -1000000000

Sample Output 3

-5000000000


# coding: utf-8

import pdb

# Step 1: For each node prepare the destination and predecessor
def initialize(graph, source):
    d = {} # Stands for destination
    p = {} # Stands for predecessor
    for node in graph:
        d[node] = float('Inf') # We start admiting that the rest of nodes are very very far
        p[node] = None
    d[source] = 0 # For the source we know how to reach
    return d, p

def relax(node, neighbour, graph, d, p):
    # If the distance between the node and the neighbour is lower than the one I have now
    if d[neighbour] > d[node] + graph[node][neighbour]:
        # Record this lower distance
        d[neighbour]  = d[node] + graph[node][neighbour]
        p[neighbour] = node

def bellman_ford(graph, source):
    d, p = initialize(graph, source)
    for i in range(len(graph)-1): #Run this until is converges
        for u in graph:
            for v in graph[u]: #For each neighbour of u
                relax(u, v, graph, d, p) #Lets relax it

    # Step 3: check for negative-weight cycles
    for u in graph:
        for v in graph[u]:
            assert d[v] <= d[u] + graph[u][v]

    return d, p

if __name__ == "__main__":
	n, m = map(int, input().split())
	gr = {k: dict() for k in range(1, n+1)}
	for _ in range(m):
		a, b, c = map(int, input().split())
		gr[a][b] = -c

	try:
		d, p = bellman_ford(gr, 1)
		print(-d[n])
	except:
		print('inf')


from itertools import chain

N = int(raw_input())
p = map(int, raw_input().split())
# print(p[:2])

def is_sorted(l):
    for i in range(0, len(l)-1):
        if l[i] != i + 1:
            return False
    return True

if is_sorted(p):
    print(0)
else:
    j = 0
    while True:
        j += 1
        higher = []
        lower = []
        for i in range(0, N):
            if p[i] == reduce(lambda x, y: max(x, y), p[:i+1], 0):
                higher.append(p[i])
            else:
                lower.append(p[i])
        p = [e for e in chain(lower, higher)]
        # p = lower + higher
        # print p

        if is_sorted(p):
            print(j)
            break

Score : 2400 points

Problem Statement

Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.

Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.

Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).

How many operations are necessary until the permutation is sorted?

Constraints

1 ≤ N ≤ 2×10^5

(p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:

N
p_1 p_2 ... p_N

Output

Print the number of operations that are necessary until the permutation is sorted.

Sample Input 1

5
3 5 1 2 4

Sample Output 1

3

The initial permutation is (3,5,1,2,4), and each operation changes it as follows:

In the first operation, the 1-st and 2-nd elements are high, and the 3-rd, 4-th and 5-th are low. The permutation becomes: (1,2,4,3,5).

In the second operation, the 1-st, 2-nd, 3-rd and 5-th elements are high, and the 4-th is low. The permutation becomes: (3,1,2,4,5).

In the third operation, the 1-st, 4-th and 5-th elements are high, and the 2-nd and 3-rd and 5-th are low. The permutation becomes: (1,2,3,4,5).

Sample Input 2

5
5 4 3 2 1

Sample Output 2

4

Sample Input 3

10
2 10 5 7 3 6 4 9 8 1

Sample Output 3

6


abc = input().strip().split()

A = abc[0]
B = abc[1]
C = abc[2]

if A[len(A)-1] == B[0] and B[len(B)-1] == C[0]:
    print("Yes")
else:
    print("No")

Score : 100 points

Problem Statement

You are given three strings A, B and C. Check whether they form a word chain.

More formally, determine whether both of the following are true:

The last character in A and the initial character in B are the same.

The last character in B and the initial character in C are the same.

If both are true, print YES. Otherwise, print NO.

Constraints

A, B and C are all composed of lowercase English letters (a - z).

1 ≤ |A|, |B|, |C| ≤ 10, where |A|, |B| and |C| are the lengths of A, B and C, respectively.

Input

Input is given from Standard Input in the following format:

A B C

Output

Print YES or NO.

Sample Input 1

rng gorilla apple

Sample Output 1

YES

They form a word chain.

Sample Input 2

yakiniku unagi sushi

Sample Output 2

NO

A and B form a word chain, but B and C do not.

Sample Input 3

a a a

Sample Output 3

YES

Sample Input 4

aaaaaaaaab aaaaaaaaaa aaaaaaaaab

Sample Output 4

NO


a, b, c = map(int, input().split())
n = a
ans = 'NO'
for i in range(1, b + 1):
    if n % b == c:
        ans = 'YES'
        break
    else:
        n += a
print(ans)

Score : 200 points

Problem Statement

We ask you to select some number of positive integers, and calculate the sum of them.

It is allowed to select as many integers as you like, and as large integers as you wish.
You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer.

Your objective is to make the sum congruent to C modulo B.
Determine whether this is possible.

If the objective is achievable, print YES. Otherwise, print NO.

Constraints

1 ≤ A ≤ 100

1 ≤ B ≤ 100

0 ≤ C < B

Input

Input is given from Standard Input in the following format:

A B C

Output

Print YES or NO.

Sample Input 1

7 5 1

Sample Output 1

YES

For example, if you select 7 and 14, the sum 21 is congruent to 1 modulo 5.

Sample Input 2

2 2 1

Sample Output 2

NO

The sum of even numbers, no matter how many, is never odd.

Sample Input 3

1 100 97

Sample Output 3

YES

You can select 97, since you may select multiples of 1, that is, all integers.

Sample Input 4

40 98 58

Sample Output 4

YES

Sample Input 5

77 42 36

Sample Output 5

NO


a, b, c = map(int, input().split())

for i in range(b):
    if (a*i)%b == c:
        print("YES")
        exit()

print("NO")


a, b, c = map(int, input().split())

for i in range(1, b+1):
    if a*i % b == c:
        print('YES')
        exit()

print('NO')


A,B,C = map(int,input().split())
mod0=A%B
def main():
  i=1
  while True:
    i+=1
    mod=(A*i)%B
    if mod==C:
      print("YES")
      break
    if mod==mod0:
      print("NO")
      break
if __name__=="__main__":
  main()


A, B, C=map(int,input().split())

Flag = 0

for i in range(B):
  Tmp = A * (i+1)
  Tmp2 = Tmp % B
  if Tmp2 == C:
    Flag=1
    break

if Flag==1:
  print('YES')
else:
  print('NO')


def gcd(a, b):
  if b == 0:
    return a
  else:
    return gcd(b, a%b)

A, B, C = map(int, input().split())

print(["YES","NO"][gcd(A,B)<=C])


A, B, C = map(int, input().split())
ans = 'NO'
loop = True
cnt = 1
remainder = (B+C)%A
while loop:
    if ((B*cnt)+C)%A == 0:
        ans = 'YES'
        loop = False
    elif (remainder == (((B*cnt)+C)%A)) and (1 < cnt):
        loop = False
    cnt += 1
print(ans)


#13:32
n,u = map(int,input().split())
now = [[0 for _ in range(301)] for _ in range(101)]
for i in range(n):
  w,v = map(int,input().split())
  if i == 0:
    w0 = int(w)
  last = now
  now = [[0 for _ in range(301)] for _ in range(101)]
  for j in range(100):
    for k in range(298):
      now[j][k] = max(now[j][k],last[j][k])
      now[j+1][k+w-w0] = max(now[j+1][k+w-w0],last[j][k]+v)
can = []
for j in range(101):
  utmp = u - w0 * j
  if utmp < 0:
    break
  if utmp > 300:
    umtp = 300
  can.append(now[j][utmp])
print(max(can))

Score : 400 points

Problem Statement

You have N items and a bag of strength W.
The i-th item has a weight of w_i and a value of v_i.

You will select some of the items and put them in the bag.
Here, the total weight of the selected items needs to be at most W.

Your objective is to maximize the total value of the selected items.

Constraints

1 ≤ N ≤ 100

1 ≤ W ≤ 10^9

1 ≤ w_i ≤ 10^9

For each i = 2,3,...,N, w_1 ≤ w_i ≤ w_1 + 3.

1 ≤ v_i ≤ 10^7

W, each w_i and v_i are integers.

Input

Input is given from Standard Input in the following format:

N W
w_1 v_1
w_2 v_2
:
w_N v_N

Output

Print the maximum possible total value of the selected items.

Sample Input 1

4 6
2 1
3 4
4 10
3 4

Sample Output 1

11

The first and third items should be selected.

Sample Input 2

4 6
2 1
3 7
4 10
3 6

Sample Output 2

13

The second and fourth items should be selected.

Sample Input 3

4 10
1 100
1 100
1 100
1 100

Sample Output 3

400

You can take everything.

Sample Input 4

4 1
10 100
10 100
10 100
10 100

Sample Output 4

0

You can take nothing.


import sys
input = lambda: sys.stdin.readline().rstrip()

def main():
    words = list(input().split())
    print((words[0][0]+words[1][0]+words[2][0]).upper())

if __name__ == '__main__':
    main()

Score : 100 points

Problem Statement

You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between.
Print the acronym formed from the uppercased initial letters of the words.

Constraints

s_1, s_2 and s_3 are composed of lowercase English letters.

1 ≤ |s_i| ≤ 10 (1≤i≤3)

Input

Input is given from Standard Input in the following format:

s_1 s_2 s_3

Output

Print the answer.

Sample Input 1

atcoder beginner contest

Sample Output 1

ABC

The initial letters of atcoder, beginner and contest are a, b and c. Uppercase and concatenate them to obtain ABC.

Sample Input 2

resident register number

Sample Output 2

RRN

Sample Input 3

k nearest neighbor

Sample Output 3

KNN

Sample Input 4

async layered coding

Sample Output 4

ALC


def main():

    a = int(input())
    b = int(input())

    if a > b:
        print("GREATER")
    elif a = b:
        print("EQUAL")
    else:
        print("LESS")

main()

Score : 200 points

Problem Statement

You are given two positive integers A and B. Compare the magnitudes of these numbers.

Constraints

1 ≤ A, B ≤ 10^{100}

Neither A nor B begins with a 0.

Input

Input is given from Standard Input in the following format:

A
B

Output

Print GREATER if A>B, LESS if A<B and EQUAL if A=B.

Sample Input 1

36
24

Sample Output 1

GREATER

Since 36>24, print GREATER.

Sample Input 2

850
3777

Sample Output 2

LESS

Sample Input 3

9720246
22516266

Sample Output 3

LESS

Sample Input 4

123456789012345678901234567890
234567890123456789012345678901

Sample Output 4

LESS


a = int(raw_input())
b = int(raw_input())
if a>b:
    print 'GREATER'
elif a<b:
    print 'LESS'
else:
    print 'EQUAL'


#!/usr/bin/env python3
import sys
from collections import Counter

def solve(n: int, S: "List[str]"):
    counter_list = [Counter(S[i]) for i in range(n)]

    a = dict(counter_list[0].items())
    a_keys = list(counter_list[0].keys())
    for cl in counter_list[1:]:
        for key,value in list(cl.items()):
            if key in a_keys and a[key]>value:
                a[key] = value

    for cl in counter_list[1:]:
        for key in list(a.keys()):
            if key not in list(cl.keys()):
                a.pop(key)

    answer = ""
    for key,value in a.items():
        answer += key*value
    answer = list(answer)
    answer.sort()
    print("".join(answer))
    return

def main():
    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word
    tokens = iterate_tokens()
    n = int(next(tokens))  # type: int
    S = [next(tokens) for _ in range(n)]  # type: "List[str]"
    solve(n, S)

if __name__ == '__main__':
    main()

Score : 300 points

Problem Statement

Snuke loves "paper cutting": he cuts out characters from a newspaper headline and rearranges them to form another string.

He will receive a headline which contains one of the strings S_1,...,S_n tomorrow.
He is excited and already thinking of what string he will create.
Since he does not know the string on the headline yet, he is interested in strings that can be created regardless of which string the headline contains.

Find the longest string that can be created regardless of which string among S_1,...,S_n the headline contains.
If there are multiple such strings, find the lexicographically smallest one among them.

Constraints

1 \leq n \leq 50

1 \leq |S_i| \leq 50 for every i = 1, ..., n.

S_i consists of lowercase English letters (a - z) for every i = 1, ..., n.

Input

Input is given from Standard Input in the following format:

n
S_1
...
S_n

Output

Print the lexicographically smallest string among the longest strings that satisfy the condition.
If the answer is an empty string, print an empty line.

Sample Input 1

3
cbaa
daacc
acacac

Sample Output 1

aac

The strings that can be created from each of cbaa, daacc and acacac, are aa, aac, aca, caa and so forth.
Among them, aac, aca and caa are the longest, and the lexicographically smallest of these three is aac.

Sample Input 2

3
a
aa
b

Sample Output 2

The answer is an empty string.


n = int(input())
a = list(map(int,input().split()))
a.sort()
ans = 0
for i in range(n):
    ans += a[-2-2*i]
print(ans)

Score : 300 points

Problem Statement

There are 3N participants in AtCoder Group Contest.
The strength of the i-th participant is represented by an integer a_i.
They will form N teams, each consisting of three participants.
No participant may belong to multiple teams.

The strength of a team is defined as the second largest strength among its members.
For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.

Find the maximum possible sum of the strengths of N teams.

Constraints

1 ≤ N ≤ 10^5

1 ≤ a_i ≤ 10^{9}

a_i are integers.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_{3N}

Output

Print the answer.

Sample Input 1

2
5 2 8 5 1 5

Sample Output 1

10

The following is one formation of teams that maximizes the sum of the strengths of teams:

Team 1: consists of the first, fourth and fifth participants.

Team 2: consists of the second, third and sixth participants.

Sample Input 2

10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000

Sample Output 2

10000000000

The sum of the strengths can be quite large.


n = int(input())
l = list(map(int, input().split()))
l.sort()
print(sum(l[n::2]))


a, b = map(int,input().split())
if a+b >= 24:
    print(a + b -24)
else:
    print(a+b)

Score : 100 points

Problem Statement

Dolphin loves programming contests. Today, he will take part in a contest in AtCoder.

In this country, 24-hour clock is used. For example, 9:00 p.m. is referred to as "21 o'clock".

The current time is A o'clock, and a contest will begin in exactly B hours.
When will the contest begin? Answer in 24-hour time.

Constraints

0 \leq A,B \leq 23

A and B are integers.

Input

The input is given from Standard Input in the following format:

A B

Output

Print the hour of the starting time of the contest in 24-hour time.

Sample Input 1

9 12

Sample Output 1

21

In this input, the current time is 9 o'clock, and 12 hours later it will be 21 o'clock in 24-hour time.

Sample Input 2

19 0

Sample Output 2

19

The contest has just started.

Sample Input 3

23 2

Sample Output 3

1

The contest will begin at 1 o'clock the next day.


print(sum(list(int(i) for i in input().split()))%24)


# -*- coding: utf-8 -*-

# input
N, M = map(int, input().split())
ab = [list(map(int, input().split())) for i in range(N)]
cd = [list(map(int, input().split())) for i in range(M)]

# solve & output
for i in range(N):
    min_diff = 10 ** 8
    for j in range(M):
        diff = abs(ab[i][0] - cd[j][0]) + abs(ab[i][1] - cd[j][1])
        if min_diff > diff:
            min_diff = diff
            ans = j + 1
    print(ans)

Score : 200 points

Problem Statement

There are N students and M checkpoints on the xy-plane.

The coordinates of the i-th student (1 \leq i \leq N) is (a_i,b_i), and the coordinates of the checkpoint numbered j (1 \leq j \leq M) is (c_j,d_j).

When the teacher gives a signal, each student has to go to the nearest checkpoint measured in Manhattan distance.

The Manhattan distance between two points (x_1,y_1) and (x_2,y_2) is |x_1-x_2|+|y_1-y_2|.

Here, |x| denotes the absolute value of x.

If there are multiple nearest checkpoints for a student, he/she will select the checkpoint with the smallest index.

Which checkpoint will each student go to?

Constraints

1 \leq N,M \leq 50

-10^8 \leq a_i,b_i,c_j,d_j \leq 10^8

All input values are integers.

Input

The input is given from Standard Input in the following format:

N M
a_1 b_1
:
a_N b_N
c_1 d_1
:
c_M d_M

Output

Print N lines.

The i-th line (1 \leq i \leq N) should contain the index of the checkpoint for the i-th student to go.

Sample Input 1

2 2
2 0
0 0
-1 0
1 0

Sample Output 1

2
1

The Manhattan distance between the first student and each checkpoint is:

For checkpoint 1: |2-(-1)|+|0-0|=3

For checkpoint 2: |2-1|+|0-0|=1

The nearest checkpoint is checkpoint 2. Thus, the first line in the output should contain 2.

The Manhattan distance between the second student and each checkpoint is:

For checkpoint 1: |0-(-1)|+|0-0|=1

For checkpoint 2: |0-1|+|0-0|=1

When there are multiple nearest checkpoints, the student will go to the checkpoint with the smallest index. Thus, the second line in the output should contain 1.

Sample Input 2

3 4
10 10
-10 -10
3 3
1 2
2 3
3 5
3 5

Sample Output 2

3
1
2

There can be multiple checkpoints at the same coordinates.

Sample Input 3

5 5
-100000000 -100000000
-100000000 100000000
100000000 -100000000
100000000 100000000
0 0
0 0
100000000 100000000
100000000 -100000000
-100000000 100000000
-100000000 -100000000

Sample Output 3

5
4
3
2
1


def solve():
  ar = raw_input().split(' ')
  N = int(ar[0])
  M = int(ar[1])
  sts = []
  for n in range(N) :
    ar = raw_input().split(' ')
    a = int(ar[0])
    b = int(ar[1])
    sts.append((a, b))
  cks = []
  for m in range(M) :
    ar = raw_input().split(' ')
    c = int(ar[0])
    d = int(ar[1])
    cks.append((c, d))

  for st in sts :
    #print st
    mind = float('inf')
    mini = -1
    for i in range(M):
      ckx = cks[i][0]
      cky = cks[i][1]
      d = abs(st[0] - ckx) + abs(st[1] - cky)
      #print 'd=',d
      if d < mind :
        mind = d
        mini = i
    #print 'mind=',mind, ',mini=',mini
    print mini + 1

if __name__ == '__main__':
  solve()


n = int(input())

def make_divisors(n):
    lower_divisors, upper_divisors = [], []
    i = 1
    while i*i <= n:
        if n % i == 0:
            lower_divisors.append(i)
            if i != n // i:
                upper_divisors.append(n//i)
        i += 1
    return lower_divisors + upper_divisors[::-1]

li = make_divisors(n)

ans = float("inf")

for i in li:
    x = n // i
    ans = min(ans, max(len(str(i)), len(str(x))))

print(ans)

Score : 300 points

Problem Statement

You are given an integer N.

For two positive integers A and B, we will define F(A,B) as the larger of the following: the number of digits in the decimal notation of A, and the number of digits in the decimal notation of B.

For example, F(3,11) = 2 since 3 has one digit and 11 has two digits.

Find the minimum value of F(A,B) as (A,B) ranges over all pairs of positive integers such that N = A \times B.

Constraints

1 \leq N \leq 10^{10}

N is an integer.

Input

The input is given from Standard Input in the following format:

N

Output

Print the minimum value of F(A,B) as (A,B) ranges over all pairs of positive integers such that N = A \times B.

Sample Input 1

10000

Sample Output 1

3

F(A,B) has a minimum value of 3 at (A,B)=(100,100).

Sample Input 2

1000003

Sample Output 2

7

There are two pairs (A,B) that satisfy the condition: (1,1000003) and (1000003,1). For these pairs, F(1,1000003)=F(1000003,1)=7.

Sample Input 3

9876543210

Sample Output 3

6


import math
n = int(input())
nn = int(math.sqrt(n))
while n%nn != 0: nn-=1
ans = len(str(n/nn))
print(ans)


x=int(input())
for i in range(10**5):
  if 2*x<=(i+1)*i:
    print(i)
    break

Score : 200 points

Problem Statement

There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0.
During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right.
That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i.
The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible.
Find the earliest possible time to reach coordinate X.

Constraints

X is an integer.

1≤X≤10^9

Input

The input is given from Standard Input in the following format:

X

Output

Print the earliest possible time for the kangaroo to reach coordinate X.

Sample Input 1

6

Sample Output 1

3

The kangaroo can reach his nest at time 3 by jumping to the right three times, which is the earliest possible time.

Sample Input 2

2

Sample Output 2

2

He can reach his nest at time 2 by staying at his position during the first second, and jumping to the right at the next second.

Sample Input 3

11

Sample Output 3

5


Score : 100 points

Problem Statement

Snuke has a favorite restaurant.

The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer.

So far, Snuke has ordered N meals at the restaurant.
Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen.
Find x-y.

Constraints

1 ≤ N ≤ 100

Input

The input is given from Standard Input in the following format:

N

Output

Print the answer.

Sample Input 1

20

Sample Output 1

15800

So far, Snuke has paid 16000 yen, and the restaurant has paid back 200 yen. Thus, the answer is 15800.

Sample Input 2

60

Sample Output 2

47200

Snuke has paid 48000 yen for 60 meals, and the restaurant has paid back 800 yen.


MOD = 10 ** 9
N = int(input())
ret = 1
for i in range(1, N+1):
    ret *= i
    ret % MOD
print(ret)

Score : 200 points

Problem Statement

Snuke loves working out. He is now exercising N times.

Before he starts exercising, his power is 1. After he exercises for the i-th time, his power gets multiplied by i.

Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.

Constraints

1 ≤ N ≤ 10^{5}

Input

The input is given from Standard Input in the following format:

N

Output

Print the answer modulo 10^{9}+7.

Sample Input 1

3

Sample Output 1

6

After Snuke exercises for the first time, his power gets multiplied by 1 and becomes 1.

After Snuke exercises for the second time, his power gets multiplied by 2 and becomes 2.

After Snuke exercises for the third time, his power gets multiplied by 3 and becomes 6.

Sample Input 2

10

Sample Output 2

3628800

Sample Input 3

100000

Sample Output 3

457992974

Print the answer modulo 10^{9}+7.


N = int(input())
a = 1
for i in range(1, N+1):
    a *= i
    a %= 1000000007
print(a)


a,b=map(int,input().split())
if a==b:
    print("Draw")
elif a==1 or (a>b and b!=1):
    print("Alice")
else:
    print("Bob")

Score : 100 points

Problem Statement

Alice and Bob are playing One Card Poker.

One Card Poker is a two-player game using playing cards.

Each card in this game shows an integer between 1 and 13, inclusive.

The strength of a card is determined by the number written on it, as follows:

Weak 2 < 3 < 4 < 5 < 6 < 7 < 8 < 9 < 10 < 11 < 12 < 13 < 1 Strong

One Card Poker is played as follows:

Each player picks one card from the deck. The chosen card becomes the player's hand.

The players reveal their hands to each other. The player with the stronger card wins the game.

If their cards are equally strong, the game is drawn.

You are watching Alice and Bob playing the game, and can see their hands.

The number written on Alice's card is A, and the number written on Bob's card is B.

Write a program to determine the outcome of the game.

Constraints

1≦A≦13

1≦B≦13

A and B are integers.

Input

The input is given from Standard Input in the following format:

A B

Output

Print Alice if Alice will win. Print Bob if Bob will win. Print Draw if the game will be drawn.

Sample Input 1

8 6

Sample Output 1

Alice

8 is written on Alice's card, and 6 is written on Bob's card.
Alice has the stronger card, and thus the output should be Alice.

Sample Input 2

1 1

Sample Output 2

Draw

Since their cards have the same number, the game will be drawn.

Sample Input 3

13 1

Sample Output 3

Bob


n, m = [int(i) for i in input().split()]

def init_img(n):
    return [[0 if j == "#" else 1 for j in input()] for i in range(n)]

img = init_img(n)

tmp_img = init_img(m)

flg = False

for i in range(n):
    if flg:
        break

    for j in range(n):
        count = 0
        for k in range(m):
            if img[i+k][j:j+m] == tmp_img[k]:
                count += 1

        if count == 3:
            flg = True

if flg:
    print("Yes")
else:
    print("No")

Score : 200 points

Problem Statement

You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.

A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.

Also, the given images are binary images, and the color of each pixel is either white or black.

In the input, every pixel is represented by a character: . corresponds to a white pixel, and # corresponds to a black pixel.

The image A is given as N strings A_1,...,A_N.

The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).

Similarly, the template image B is given as M strings B_1,...,B_M.

The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).

Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Constraints

1≦M≦N≦50

A_i is a string of length N consisting of # and ..

B_i is a string of length M consisting of # and ..

Input

The input is given from Standard Input in the following format:

N M
A_1
A_2
:
A_N
B_1
B_2
:
B_M

Output

Print Yes if the template image B is contained in the image A. Print No otherwise.

Sample Input 1

3 2
#.#
.#.
#.#
#.
.#

Sample Output 1

Yes

The template image B is identical to the upper-left 2 × 2 subimage and the lower-right 2 × 2 subimage of A. Thus, the output should be Yes.

Sample Input 2

4 1
....
....
....
....
#

Sample Output 2

No

The template image B, composed of a black pixel, is not contained in the image A composed of white pixels.


import itertools
N,M=map(int,input().split())
edges={tuple(sorted(map(int,input().split()))) for i in range(M)}
print(edges)
ans=0
for i in itertools.permutations(range(2,N+1),N-1):
    l=[1]+list(i)
    ans+=sum(1 for edge in zip(l,l[1:]) if tuple(sorted(edge)) in edges)==N-1
print(l)
print(ans)

Score : 300 points

Problem Statement

You are given an undirected unweighted graph with N vertices and M edges that contains neither self-loops nor double edges.

Here, a self-loop is an edge where a_i = b_i (1≤i≤M), and double edges are two edges where (a_i,b_i)=(a_j,b_j) or (a_i,b_i)=(b_j,a_j) (1≤i<j≤M).

How many different paths start from vertex 1 and visit all the vertices exactly once?

Here, the endpoints of a path are considered visited.

For example, let us assume that the following undirected graph shown in Figure 1 is given.

Figure 1: an example of an undirected graph

The following path shown in Figure 2 satisfies the condition.

Figure 2: an example of a path that satisfies the condition

However, the following path shown in Figure 3 does not satisfy the condition, because it does not visit all the vertices.

Figure 3: an example of a path that does not satisfy the condition

Neither the following path shown in Figure 4, because it does not start from vertex 1.

Figure 4: another example of a path that does not satisfy the condition

Constraints

2≦N≦8

0≦M≦N(N-1)/2

1≦a_i<b_i≦N

The given graph contains neither self-loops nor double edges.

Input

The input is given from Standard Input in the following format:

N M
a_1 b_1
a_2 b_2
:
a_M b_M

Output

Print the number of the different paths that start from vertex 1 and visit all the vertices exactly once.

Sample Input 1

3 3
1 2
1 3
2 3

Sample Output 1

2

The given graph is shown in the following figure:

The following two paths satisfy the condition:

Sample Input 2

7 7
1 3
2 7
3 4
4 5
4 6
5 6
6 7

Sample Output 2

1

This test case is the same as the one described in the problem statement.


n= int(input())
a = list(map(int,input().split()))
c=0
d=0
for i in range(n):
    if a[i]%2==0:
        c+=1
    else:
        d+=1
print("YES" if (d//2+(c+d%2)%2)%2 ==1 else "NO")

Score : 300 points

Problem Statement

There are N integers written on a blackboard. The i-th integer is A_i.

Takahashi will repeatedly perform the following operation on these numbers:

Select a pair of integers, A_i and A_j, that have the same parity (that is, both are even or both are odd) and erase them.

Then, write a new integer on the blackboard that is equal to the sum of those integers, A_i+A_j.

Determine whether it is possible to have only one integer on the blackboard.

Constraints

2 ≦ N ≦ 10^5

1 ≦ A_i ≦ 10^9

A_i is an integer.

Input

The input is given from Standard Input in the following format:

N
A_1 A_2 … A_N

Output

If it is possible to have only one integer on the blackboard, print YES. Otherwise, print NO.

Sample Input 1

3
1 2 3

Sample Output 1

YES

It is possible to have only one integer on the blackboard, as follows:

Erase 1 and 3 from the blackboard, then write 4. Now, there are two integers on the blackboard: 2 and 4.

Erase 2 and 4 from the blackboard, then write 6. Now, there is only one integer on the blackboard: 6.

Sample Input 2

5
1 2 3 4 5

Sample Output 2

NO


import os
import sys

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(10 ** 9)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7
# MOD = 998244353

N = int(sys.stdin.buffer.readline())
A = list(map(int, sys.stdin.buffer.readline().split()))

c = 0
for a in A:
    c += a & 1
if c % 2 == 0:
    print('YES')
else:
    print('NO')


import sys
input = sys.stdin.readline

def main():
    x = int(input())
    if x%11 > 6:
        print((x//11) * 2 + 2)
    else:
        if x%11 == 0:
            print((x//11) * 2)
        else:
            print((x//11) * 2 + 1)

if __name__ == "__main__":
    main()

Score : 300 points

Problem Statement

Snuke has decided to play with a six-sided die. Each of its six sides shows an integer 1 through 6, and two numbers on opposite sides always add up to 7.

Snuke will first put the die on the table with an arbitrary side facing upward, then repeatedly perform the following operation:

Operation: Rotate the die 90° toward one of the following directions: left, right, front (the die will come closer) and back (the die will go farther). Then, obtain y points where y is the number written in the side facing upward.

For example, let us consider the situation where the side showing 1 faces upward, the near side shows 5 and the right side shows 4, as illustrated in the figure.
If the die is rotated toward the right as shown in the figure, the side showing 3 will face upward.
Besides, the side showing 4 will face upward if the die is rotated toward the left, the side showing 2 will face upward if the die is rotated toward the front, and the side showing 5 will face upward if the die is rotated toward the back.

Find the minimum number of operation Snuke needs to perform in order to score at least x points in total.

Constraints

1 ≦ x ≦ 10^{15}

x is an integer.

Input

The input is given from Standard Input in the following format:

x

Output

Print the answer.

Sample Input 1

7

Sample Output 1

2

Sample Input 2

149696127901

Sample Output 2

27217477801


x=int(input())
x-=1
ans=(x//11)*2
tmp=x%11

if tmp==0:
    print(ans)
    exit()

if tmp<=5:
    ans+=1
elif tmp<11:
    ans+=2
print(ans)


x=int(input())
ans=(x//11)*2
tmp=x%11

if tmp==0:
    print(ans)
    exit()

if tmp<=5:
    ans+=1
elif tmp<11:
    ans+=2
print(ans)


N, A, B = map(int, input().split())
X = list(map(int, input().split()))
ans = 0
for i in range(1, N):
    if (X[i] - X[i-1]) * A >= B:
        ans += B
    else:
        ans += (X[i] - X[i-1]) * A
print(ans)

Score : 500 points

Problem Statement

There are N towns on a line running east-west.
The towns are numbered 1 through N, in order from west to east.
Each point on the line has a one-dimensional coordinate, and a point that is farther east has a greater coordinate value.
The coordinate of town i is X_i.

You are now at town 1, and you want to visit all the other towns.
You have two ways to travel:

Walk on the line.
Your fatigue level increases by A each time you travel a distance of 1, regardless of direction.

Teleport to any location of your choice.
Your fatigue level increases by B, regardless of the distance covered.

Find the minimum possible total increase of your fatigue level when you visit all the towns in these two ways.

Constraints

All input values are integers.

2≤N≤10^5

1≤X_i≤10^9

For all i(1≤i≤N-1), X_i<X_{i+1}.

1≤A≤10^9

1≤B≤10^9

Input

The input is given from Standard Input in the following format:

N A B
X_1 X_2 ... X_N

Output

Print the minimum possible total increase of your fatigue level when you visit all the towns.

Sample Input 1

4 2 5
1 2 5 7

Sample Output 1

11

From town 1, walk a distance of 1 to town 2, then teleport to town 3, then walk a distance of 2 to town 4.
The total increase of your fatigue level in this case is 2×1+5+2×2=11, which is the minimum possible value.

Sample Input 2

7 1 100
40 43 45 105 108 115 124

Sample Output 2

84

From town 1, walk all the way to town 7.
The total increase of your fatigue level in this case is 84, which is the minimum possible value.

Sample Input 3

7 1 2
24 35 40 68 72 99 103

Sample Output 3

12

Visit all the towns in any order by teleporting six times.
The total increase of your fatigue level in this case is 12, which is the minimum possible value.