let () =
  let open Int64 in
  let len, b = Scanf.scanf "%d %Ld\n" (fun a b -> a, b) in

  let arr = Array.make len (0L,0L) in
  for i = 0 to (len-1) do
    let c = Scanf.scanf "%Ld %Ld\n" (fun a b -> a, b) in
    Array.set arr i c
  done;

  let arr_l = Array.make len (0L,0L) in
  let arr_r = Array.make len (0L,0L) in
  let arr_l2 = Array.make len (0L,0L) in
  let arr_r2 = Array.make len (0L,0L) in

  let t = ref (0L,0L) in
  for i = 0 to (len-1) do
    let distance = fst (Array.get arr i) in

    let cal = Int64.add (snd (Array.get arr i))
        ( if (i=0) then 0L else (snd (Array.get arr_l (i-1)))) in

    let d_sub = if (i=0) then 0L else (fst (Array.get arr_l (i-1))) in
    let d_sub = Int64.sub distance d_sub in
    let cal = Int64.sub cal d_sub in

    let tt = (distance, cal) in
    if (cal > (snd !t))
    then (
      t := tt
    )
    else ();
    Array.set arr_l i tt;
    Array.set arr_l2 i !t;
  done;

  let u = ref (0L,0L) in
  for i = 0 to (len-1) do
    let i = len-1-i in
    let distance = Int64.sub b (fst (Array.get arr i)) in
    let cal = Int64.add (snd (Array.get arr i))
        (if (i=len-1) then 0L else (snd (Array.get arr_r (i+1))) ) in

    let d_sub = if (i=len-1) then 0L else (fst (Array.get arr_r (i+1))) in
    let d_sub = Int64.sub distance d_sub in
    let cal = Int64.sub cal d_sub in

    let tt = (distance, cal) in
    if (cal > (snd !u))
    then (
      u := tt
    )
    else ();
    Array.set arr_r i tt;
    Array.set arr_r2 i !u;
  done;

  let v = ref (0L,0L) in
  for i = 0 to (len-1) do
    let j = if (i=len-1) then 0 else i+1 in

    let d_l = fst (Array.get arr_l2 i) in
    let d_r =
      if (j = 0) then
        0L
      else
        fst (Array.get arr_r2 j) in
    let d_sub = if d_l > d_r then d_r else d_l in

    let cal_l = snd (Array.get arr_l2 i) in
    let cal_r =
      if (j = 0) then
        0L
      else
        snd (Array.get arr_r2 j) in
    let cal = Int64.sub (Int64.add cal_l cal_r) d_sub in

    if (cal > (snd !v))
    then (
      let tt = (0L, cal) in
      v := tt
    )
    else ();
  done;

  let res = max (max (snd !t) (snd !u)) (snd !v) in
  Printf.printf "%Ld" res;

Score : 500 points

Problem Statement

"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter.

Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories.

Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter.

Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.

Constraints

1 ≤ N ≤ 10^5

2 ≤ C ≤ 10^{14}

1 ≤ x_1 < x_2 < ... < x_N < C

1 ≤ v_i ≤ 10^9

All values in input are integers.

Subscores

300 points will be awarded for passing the test set satisfying N ≤ 100.

Input

Input is given from Standard Input in the following format:

N C
x_1 v_1
x_2 v_2
:
x_N v_N

Output

If Nakahashi can take in at most c kilocalories on balance before he leaves the restaurant, print c.

Sample Input 1

3 20
2 80
9 120
16 1

Sample Output 1

191

There are three sushi on the counter with a circumference of 20 meters. If he walks two meters clockwise from the initial place, he can eat a sushi of 80 kilocalories. If he walks seven more meters clockwise, he can eat a sushi of 120 kilocalories. If he leaves now, the total nutrition taken in is 200 kilocalories, and the total energy consumed is 9 kilocalories, thus he can take in 191 kilocalories on balance, which is the largest possible value.

Sample Input 2

3 20
2 80
9 1
16 120

Sample Output 2

192

The second and third sushi have been swapped. Again, if he walks two meters clockwise from the initial place, he can eat a sushi of 80 kilocalories. If he walks six more meters counterclockwise this time, he can eat a sushi of 120 kilocalories. If he leaves now, the total nutrition taken in is 200 kilocalories, and the total energy consumed is 8 kilocalories, thus he can take in 192 kilocalories on balance, which is the largest possible value.

Sample Input 3

1 100000000000000
50000000000000 1

Sample Output 3

0

Even though the only sushi is so far that it does not fit into a 32-bit integer, its nutritive value is low, thus he should immediately leave without doing anything.

Sample Input 4

15 10000000000
400000000 1000000000
800000000 1000000000
1900000000 1000000000
2400000000 1000000000
2900000000 1000000000
3300000000 1000000000
3700000000 1000000000
3800000000 1000000000
4000000000 1000000000
4100000000 1000000000
5200000000 1000000000
6600000000 1000000000
8000000000 1000000000
9300000000 1000000000
9700000000 1000000000

Sample Output 4

6500000000

All these sample inputs above are included in the test set for the partial score.


let () =
  let open Int64 in
  let len, b = Scanf.scanf "%d %Ld\n" (fun a b -> a, b) in

  let arr = Array.make len (0L,0L) in
  for i = 0 to (len-1) do
    let c = Scanf.scanf "%Ld %Ld\n" (fun a b -> a, b) in
    Array.set arr i c
  done;
  let t = ref 0L in

  let arr_l = Array.make len (0L,0L) in
  let arr_r = Array.make len (0L,0L) in

  for i = 0 to (len-1) do
    let distance = fst (Array.get arr i) in

    let cal = Int64.add (snd (Array.get arr i))
        ( if (i=0) then 0L else (snd (Array.get arr_l (i-1)))) in

    let d_sub = if (i=0) then 0L else (fst (Array.get arr_l (i-1))) in
    let d_sub = Int64.sub distance d_sub in
    let cal = Int64.sub cal d_sub in

    if (cal > !t)
    then (t := cal)
    else ();
    Array.set arr_l i (distance, cal)

  done;
  for i = 0 to (len-1) do
    let i = len-1-i in
    let distance = Int64.sub b (fst (Array.get arr i)) in
    let cal = Int64.add (snd (Array.get arr i))
        (if (i=len-1) then 0L else (snd (Array.get arr_r (i+1))) ) in

    let d_sub = if (i=len-1) then 0L else (fst (Array.get arr_r (i+1))) in
    let d_sub = Int64.sub distance d_sub in
    let cal = Int64.sub cal d_sub in

    if (cal > !t)
    then (t := cal)
    else ();
    Array.set arr_r i (distance, cal)
  done;

  for i = 0 to (len-1) do
    for j = 0 to (len-1) do
      let cal =
        if i >= j then 0L
        else
          let d_l = fst (Array.get arr_l i) in
          let d_r = fst (Array.get arr_r j) in
          let d_sub = if d_l > d_r then d_r else d_l in

          let cal_l = snd (Array.get arr_l i) in
          let cal_r = snd (Array.get arr_r j) in
          let cal = Int64.sub (Int64.add cal_l cal_r) d_sub in
          cal
      in

      if (cal > !t)
      then (t := cal)
      else ()
    done;
  done;

  Printf.printf "%Ld" !t;

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

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.


open Printf open Scanf
module MyInt = struct let (+) = (+) let (-) = (-) let (/) = (/) let ( * ) = ( * ) let (mod) = (mod) let (%) = (mod) let (+=) r v = r := !r + v let (-=) r v = r := !r - v let ( *=) r v= r := !r * v let (/=) r v = r := !r / v end
module MyInt64 = struct
	let (+$) = (+) let (-$) = (-) let (/$) = (/) let ( *$) = ( * ) let (+) p q = Int64.add p q let (-) p q = Int64.sub p q let ( * ) p q = Int64.mul p q let (/) p q = Int64.div p q let (+=) r v = r := !r + v let (-=) r v = r := !r - v let ( *=) r v= r := !r * v let (/=) r v = r := !r / v let labs p = if p < 0L then -1L*p else p let (~|) p = Int64.to_int p let (~~|) p = Int64.of_int p let input_i64_array n = Array.init (~| n) (fun _ -> Scanf.scanf " %Ld" (fun v -> v)) let get_i64 _ = Scanf.scanf " %Ld" (fun v -> v) let get_2_i64 _ = Scanf.scanf " %Ld %Ld" (fun u v -> u,v) let get_3_i64 _ = Scanf.scanf " %Ld %Ld %Ld" (fun u v w -> u,v,w) let get_4_i64 _ = Scanf.scanf " %Ld %Ld %Ld %Ld" (fun u v w x -> u,v,w,x) let get_5_i64 _ = Scanf.scanf " %Ld %Ld %Ld %Ld %Ld" (fun u v w x y -> u,v,w,x,y) let print_i64_endline n = n |> Int64.to_string |> print_endline let (mod) m n = m - (m/n) * n let (%) = (mod) let rec gcd m n = match m,n with | m,0L -> m | m,n -> gcd n (m mod n) let lcm m n = (m*n) / gcd m n let rep from_ to_ fbod = let i,f = ref ~|from_,ref true in while !i <= ~|to_ && !f do match fbod ~~| !i with | `Break -> f := false | _ -> i := !i +$ 1; done let repb from_ to_ fbod = let i,f = ref ~|from_,ref true in while !i <= ~|to_ && !f do match fbod (~~| !i) with | `Break -> f := false | _ -> i := !i +$ 1 done; !f let repm from_ to_ m_init fbod = let i,f,m = ref ~|from_,ref true,ref m_init in while !i <= ~|to_ && !f do match fbod !m (~~| !i) with | `Break -> f := false | `Break_m m' -> (f := false; m := m') | `Ok m' -> (i := !i +$ 1; m := m') done; !m let repmb from_ to_ m_init fbod = let i,f,m = ref ~|from_,ref true,ref m_init in while !i <= ~|to_ && !f do match fbod (~~| !i) !m with | `Break -> f := false | `Break_m m' -> (f := false; m := m') | `Ok m' -> (i := !i +$ 1; m := m') done; (!m,!f) let string_of_list ?(separator=" ") f ls = let rec f0 a s = match a with | [] -> s | [h] -> s ^ (f h) ^ separator | h::t -> f0 t (s ^ (f h) ^ separator) in f0 ls "" let char_list_of_string str = let rec f0 i a = if i<0 then a else f0 (i-$1) (str.[i]::a) in f0 (String.length str -$ 1) [] let string_of_char_list ls = List.fold_left (fun u v -> u ^ (String.make 1 v)) "" ls
	let (@@@) = (@) let (@) a i = a.(~|i) let (<@) a (i,v) = a.(~|i) <- v
	let ceildiv m n = (m+n-1L) / n
end open MyInt64

let () =
	let a,b,c,x,y = get_5_i64 0 in
	repm 0L (max x y) Int64.max_int (fun u v ->
		let xr,yr = max 0L (x-v),max 0L (y-v) in
		`Ok (min u (2L*c*v + a*xr + b * yr))
	) |> printf "%Ld\n"

Score : 300 points

Problem Statement

"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively.

Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.

Constraints

1 ≤ A, B, C ≤ 5000

1 ≤ X, Y ≤ 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B C X Y

Output

Print the minimum amount of money required to prepare X A-pizzas and Y B-pizzas.

Sample Input 1

1500 2000 1600 3 2

Sample Output 1

7900

It is optimal to buy four AB-pizzas and rearrange them into two A-pizzas and two B-pizzas, then buy additional one A-pizza.

Sample Input 2

1500 2000 1900 3 2

Sample Output 2

8500

It is optimal to directly buy three A-pizzas and two B-pizzas.

Sample Input 3

1500 2000 500 90000 100000

Sample Output 3

100000000

It is optimal to buy 200000 AB-pizzas and rearrange them into 100000 A-pizzas and 100000 B-pizzas. We will have 10000 more A-pizzas than necessary, but that is fine.


let _ = Scanf.sscanf (read_line ()) "%d %d %d" (fun a b x ->
  if x < a || x > (a+b) then "NO" else "YES"
) |> print_endline

Score : 100 points

Problem Statement

There are a total of A + B cats and dogs.
Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs.

Determine if it is possible that there are exactly X cats among these A + B animals.

Constraints

1 \leq A \leq 100

1 \leq B \leq 100

1 \leq X \leq 200

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B X

Output

If it is possible that there are exactly X cats, print YES; if it is impossible, print NO.

Sample Input 1

3 5 4

Sample Output 1

YES

If there are one cat and four dogs among the B = 5 animals, there are X = 4 cats in total.

Sample Input 2

2 2 6

Sample Output 2

NO

Even if all of the B = 2 animals are cats, there are less than X = 6 cats in total.

Sample Input 3

5 3 2

Sample Output 3

NO

Even if all of the B = 3 animals are dogs, there are more than X = 2 cats in total.


let n, m, x = Scanf.scanf " %d %d %d" @@ fun a b c -> a, b, c
let a_s = Array.init m @@ fun _ -> Scanf.scanf " %d" (+) 0
let _ = for i = 0 to m - 1 do if a_s.(i) > x then (Printf.printf "%d\n" @@ min i @@ m - i; exit 0) done; print_endline "0"

Score : 200 points

Problem Statement

There are N + 1 squares arranged in a row, numbered 0, 1, ..., N from left to right.

Initially, you are in Square X.
You can freely travel between adjacent squares. Your goal is to reach Square 0 or Square N.
However, for each i = 1, 2, ..., M, there is a toll gate in Square A_i, and traveling to Square A_i incurs a cost of 1.
It is guaranteed that there is no toll gate in Square 0, Square X and Square N.

Find the minimum cost incurred before reaching the goal.

Constraints

1 \leq N \leq 100

1 \leq M \leq 100

1 \leq X \leq N - 1

1 \leq A_1 < A_2 < ... < A_M \leq N

A_i \neq X

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M X
A_1 A_2 ... A_M

Output

Print the minimum cost incurred before reaching the goal.

Sample Input 1

5 3 3
1 2 4

Sample Output 1

1

The optimal solution is as follows:

First, travel from Square 3 to Square 4. Here, there is a toll gate in Square 4, so the cost of 1 is incurred.

Then, travel from Square 4 to Square 5. This time, no cost is incurred.

Now, we are in Square 5 and we have reached the goal.

In this case, the total cost incurred is 1.

Sample Input 2

7 3 2
4 5 6

Sample Output 2

0

We may be able to reach the goal at no cost.

Sample Input 3

10 7 5
1 2 3 4 6 8 9

Sample Output 3

3


open Printf open Scanf
module Lib = struct
	let i32,i64,ist = Int64.(to_int,of_int,to_string)
	let (+$) = (+) let (-$) = (-) let (/$) = (/) let ( *$) = ( * ) let (%$) = (mod)
	let labs = Int64.abs
	let (+),(-),( * ),(/),(mod),(%) = Int64.(add,sub,mul,div,rem,rem)
	let (+=) r v = r := !r + v
	let (-=) r v = r := !r - v
	let ( *= ) r v = r := !r * v
	let (/=) r v = r := !r / v
	let (%=) r v = r := !r % v
	(* math *)
	let ceildiv p q = (p+q-1L) / q
	let rec gcd m n = match m,n with | m,0L -> m | m,n -> gcd n (m mod n)
	let lcm m n = m / gcd m n * n
	module I64_infix = struct
		let (land),(lor),(lxor),lnot,(lsl),(lsr),(asr) =
			let open Int64 in
			(logand),(logor),(logxor),lognot,
			(fun u v -> shift_left u (i32 v)),
			(fun u v -> shift_right u (i32 v)),
			(fun u v -> shift_right_logical u (i32 v))
	end
	(* input *)
	let input_i64_array n = Array.init (i32 n) (fun _ -> Scanf.scanf " %Ld" (fun v -> v))
	let get_i64 _ = Scanf.scanf " %Ld" (fun v -> v)
	let get_2_i64 _ = Scanf.scanf " %Ld %Ld" (fun u v -> u,v)
	let get_3_i64 _ = Scanf.scanf " %Ld %Ld %Ld" (fun u v w -> u,v,w)
	let get_4_i64 _ = Scanf.scanf " %Ld %Ld %Ld %Ld" (fun u v w x -> u,v,w,x)
	let get_5_i64 _ = Scanf.scanf " %Ld %Ld %Ld %Ld %Ld" (fun u v w x y -> u,v,w,x,y)
	let i32_get_int _ = Scanf.scanf " %d" (fun v -> v) let i32_get_2_ints _ = Scanf.scanf " %d %d" (fun u v -> u,v) let i32_get_3_ints _ = Scanf.scanf " %d %d %d" (fun u v w -> u,v,w) let i32_get_4_ints _ = Scanf.scanf " %d %d %d %d" (fun u v w x -> u,v,w,x) let i32_get_5_ints _ = Scanf.scanf " %d %d %d %d %d" (fun u v w x y -> u,v,w,x,y)
	(* utils *)
	external id : 'a -> 'a = "%identity"
	let alen a = i64 @@ Array.length a
	let llen l = i64 @@  List.length l
	let string_of_list ?(separator=" ") f ls = ls |> List.map f |> String.concat separator
	let string_of_array ?(separator=" ") f a = a |> Array.to_list |> string_of_list ~separator f
	let char_list_of_string str = let rec f0 i a = if i<0 then a else f0 (i-$1) (str.[i]::a) in f0 (String.length str -$ 1) []
	let string_of_char_list ls = List.fold_left (fun u v -> u ^ (String.make 1 v)) "" ls
	let fail _ = failwith "fail"
	let dx,dy = [|1;0;-1;0|],[|0;1;0;-1|]
	let range l r = let ls = ref [] in for i=i32 l to i32 r do ls := i :: !ls done;!ls |> List.rev
	let cartesian_product u v = u |> List.map (fun p -> List.map (fun q -> p,q) v) |> List.flatten
	let shuffle a = Array.sort (fun _ _ -> (Random.int 3) -$ 1) a
	let binary_search ng ok f = let d = if ng < ok then 1L else -1L in
		let rec f0 ng ok =
			if labs (ok - ng) <= 1L then ok
			else let mid = ng + (ok - ng) / 2L in if f mid then f0 ng mid else f0 mid ok
		in f0 (ng-1L*d) (ok+1L*d)
	(* imperative *)
	let rep f t fbod = let i = ref f in while !i <= t do fbod !i; i := !i + 1L done
	let repb f t fbod = let i,c = ref f,ref true in while !i <= t && !c do
		match fbod !i with | `Break -> c := false | `Ok -> i := !i + 1L done
	let repm f t m0 fbod =
		let i,c,m = ref f,ref true,ref m0 in
		while !i<=t && !c do match fbod !m !i with
		| `Break -> c := false | `Break_m m' -> (c := false; m := m') | `Ok m' -> m := m'; i := !i + 1L done; !m
	let repi f t fbod = let i = ref f in while !i <= t do fbod !i; i := !i +$ 1 done
	let repib f t fbod = let i,c = ref f,ref true in while !i <= t && !c do
		match fbod !i with | `Break -> c := false | `Ok -> i := !i +$ 1 done
	let repim f t m0 fbod =
		let i,c,m = ref f,ref true,ref m0 in
		while !i<=t && !c do match fbod !m !i with
		| `Break -> c := false | `Break_m m' -> (c := false; m := m') | `Ok m' -> m := m'; i := !i +$ 1 done; !m
	let rep_rev f t fbod = rep t f @@ fun v -> fbod @@ f + t - v
	let repb_rev f t fbod = repb t f @@ fun v -> fbod @@ f + t - v
	let repm_rev f t m0 fbod = repm f t m0 @@ fun u v -> fbod u @@ f + t - v
	let repi_rev f t fbod = repi t f @@ fun v -> fbod @@ f +$ t -$ v
	let repib_rev f t fbod = repib t f @@ fun v -> fbod @@ f +$ t -$ v
	let repim_rev f t m0 fbod = repim f t m0 @@ fun u v -> fbod u @@ f +$ t -$ v
	(* output *)
	let print_list_mle f ls = string_of_list f ls |> print_endline
	let print_array_mle f a = string_of_array f a |> print_endline
	let print_list f ls = ls |> List.iter (fun v -> printf "%s " @@ f v); print_newline ()
	let print_array f a = a |> Array.iter (fun v -> printf "%s " @@ f v); print_newline ()
	(* debug *)
	let dump_2darr f a = a |> Array.iter (fun b -> print_array f b)
	let dump_i64_list ls = print_list ist ls
	let dump_i64_array a = print_array ist a
end open Lib

let () =
	let n = get_i64 0 in let a = input_i64_array n in
	let a = Array.mapi (fun i v -> (v,i)) a in
	Array.sort compare a;
	let (p,_),(q,_) = a.(i32@@n/2L-1L),a.(i32@@n/2L) in
	let res = Array.make (i32 n) 0L in
	Array.iter (fun (v,i) ->
		if v<=p then res.(i) <- q else res.(i) <- p
	) a;
	Array.iter (fun v -> print_endline @@ ist v) res

Score : 300 points

Problem Statement

When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l.

You are given N numbers X_1, X_2, ..., X_N, where N is an even number.
For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i.

Find B_i for each i = 1, 2, ..., N.

Constraints

2 \leq N \leq 200000

N is even.

1 \leq X_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
X_1 X_2 ... X_N

Output

Print N lines.
The i-th line should contain B_i.

Sample Input 1

4
2 4 4 3

Sample Output 1

4
3
3
4

Since the median of X_2, X_3, X_4 is 4, B_1 = 4.

Since the median of X_1, X_3, X_4 is 3, B_2 = 3.

Since the median of X_1, X_2, X_4 is 3, B_3 = 3.

Since the median of X_1, X_2, X_3 is 4, B_4 = 4.

Sample Input 2

2
1 2

Sample Output 2

2
1

Sample Input 3

6
5 5 4 4 3 3

Sample Output 3

4
4
4
4
4
4


let min_with f x y = if f x <= f y then x else y

let () =
  let n = Scanf.scanf "%d\n" (fun n -> n) in
  let as_ = Array.init n (fun _ -> Scanf.scanf "%d " (fun a -> a)) in
  let ai = Array.fold_left max min_int as_ in
  let aj = Array.fold_left (min_with (fun a ->
    if ai mod 2 = 0 then abs (a - ai / 2)
    else if a <= ai / 2 then ai / 2 - a
    else a - (ai + 1) / 2)) ai as_ in
  Printf.printf "%d %d\n" ai aj

Score : 400 points

Problem Statement

Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order.
From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized.
If there are multiple pairs that maximize the value, any of them is accepted.

Constraints

2 \leq n \leq 10^5

0 \leq a_i \leq 10^9

a_1,a_2,...,a_n are pairwise distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

n
a_1 a_2 ... a_n

Output

Print a_i and a_j that you selected, with a space in between.

Sample Input 1

5
6 9 4 2 11

Sample Output 1

11 6

\rm{comb}(a_i,a_j) for each possible selection is as follows:

\rm{comb}(4,2)=6

\rm{comb}(6,2)=15

\rm{comb}(6,4)=15

\rm{comb}(9,2)=36

\rm{comb}(9,4)=126

\rm{comb}(9,6)=84

\rm{comb}(11,2)=55

\rm{comb}(11,4)=330

\rm{comb}(11,6)=462

\rm{comb}(11,9)=55

Thus, we should print 11 and 6.

Sample Input 2

2
100 0

Sample Output 2

100 0


let () = Scanf.scanf "%c%c%c" (fun a b c -> if a<>b && b<>c && c<>a then "Yes" else "No")
  |> print_endline

Score : 100 points

Problem Statement

You are given a string S of length 3 consisting of a, b and c. Determine if S can be obtained by permuting abc.

Constraints

|S|=3

S consists of a, b and c.

Input

Input is given from Standard Input in the following format:

S

Output

If S can be obtained by permuting abc, print Yes; otherwise, print No.

Sample Input 1

bac

Sample Output 1

Yes

Swapping the first and second characters in bac results in abc.

Sample Input 2

bab

Sample Output 2

No

Sample Input 3

abc

Sample Output 3

Yes

Sample Input 4

aaa

Sample Output 4

No


let a, b, k = Scanf.scanf " %d %d %d" @@ fun a b c -> a, b, c
let m = b - a + 1
let f = Printf.printf "%d\n"
let _ = if 2 * k >= m then for i = a to b do f i done else (for i = a to a - 1 + k do f i done; for i = b + 1 - k to b do f i done)

Score : 200 points

Problem Statement

Print all the integers that satisfies the following in ascending order:

Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.

Constraints

1 \leq A \leq B \leq 10^9

1 \leq K \leq 100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B K

Output

Print all the integers that satisfies the condition above in ascending order.

Sample Input 1

3 8 2

Sample Output 1

3
4
7
8

3 is the first smallest integer among the integers between 3 and 8.

4 is the second smallest integer among the integers between 3 and 8.

7 is the second largest integer among the integers between 3 and 8.

8 is the first largest integer among the integers between 3 and 8.

Sample Input 2

4 8 3

Sample Output 2

4
5
6
7
8

Sample Input 3

2 9 100

Sample Output 3

2
3
4
5
6
7
8
9


let () =
  let a,b,c = Scanf.scanf "%d %d %d " (fun a b c -> a,b,c) in
  let l = [a;b;c] in
  let ls = List.sort compare l in
  let d2 = (List.nth ls 2) - (List.nth ls 0) in
  let d1 = (List.nth ls 2) - (List.nth ls 1) in

  Printf.printf "%d\n" @@
  if (d2+d1) mod 2 = 0 then (d2+d1)/2 else 2+((d2+d1)/2)

Score : 300 points

Problem Statement

You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order:

Choose two among A, B and C, then increase both by 1.

Choose one among A, B and C, then increase it by 2.

It can be proved that we can always make A, B and C all equal by repeatedly performing these operations.

Constraints

0 \leq A,B,C \leq 50

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B C

Output

Print the minimum number of operations required to make A, B and C all equal.

Sample Input 1

2 5 4

Sample Output 1

2

We can make A, B and C all equal by the following operations:

Increase A and C by 1. Now, A, B, C are 3, 5, 5, respectively.

Increase A by 2. Now, A, B, C are 5, 5, 5, respectively.

Sample Input 2

2 6 3

Sample Output 2

5

Sample Input 3

31 41 5

Sample Output 3

23


let sqrt_int x = float_of_int x |> sqrt |> int_of_float

let solve a b =
  let c = sqrt_int (a * b - 1) in
  let f a b =
    if b-a <= 1 then 2*a - 2
    else if c * (c+1) >= a*b then 2*c - 2
    else 2*c - 1
  in f (min a b) (max a b)

let () =
  Scanf.scanf "%d" @@ fun q ->
  Array.init q (fun _ -> Scanf.scanf " %d %d" solve)
  |> Array.iter (Printf.printf "%d\n")

Score : 700 points

Problem Statement

10^{10^{10}} participants, including Takahashi, competed in two programming contests.
In each contest, all participants had distinct ranks from first through 10^{10^{10}}-th.

The score of a participant is the product of his/her ranks in the two contests.

Process the following Q queries:

In the i-th query, you are given two positive integers A_i and B_i. Assuming that Takahashi was ranked A_i-th in the first contest and B_i-th in the second contest, find the maximum possible number of participants whose scores are smaller than Takahashi's.

Constraints

1 \leq Q \leq 100

1\leq A_i,B_i\leq 10^9(1\leq i\leq Q)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

Q
A_1 B_1
:
A_Q B_Q

Output

For each query, print the maximum possible number of participants whose scores are smaller than Takahashi's.

Sample Input 1

8
1 4
10 5
3 3
4 11
8 9
22 40
8 36
314159265 358979323

Sample Output 1

1
12
4
11
14
57
31
671644785

Let us denote a participant who was ranked x-th in the first contest and y-th in the second contest as (x,y).

In the first query, (2,1) is a possible candidate of a participant whose score is smaller than Takahashi's. There are never two or more participants whose scores are smaller than Takahashi's, so we should print 1.


Scanf.scanf "%d" (fun n ->
    let ab = Array.init n (fun _ -> Scanf.scanf " %d %d" (fun a b -> a, b)) in

    let rec loop i sum minb =
        if i = n then (if minb = max_int then 0 else sum - minb) else
            let a, b = ab.(i) in
            loop (i + 1) (sum + a) (if a > b then min b minb else minb)
    in
    loop 0 0 max_int |> Printf.printf "%d\n"
)

Score : 700 points

Problem Statement

You are given sequences A and B consisting of non-negative integers.
The lengths of both A and B are N, and the sums of the elements in A and B are equal.
The i-th element in A is A_i, and the i-th element in B is B_i.

Tozan and Gezan repeats the following sequence of operations:

If A and B are equal sequences, terminate the process.

Otherwise, first Tozan chooses a positive element in A and decrease it by 1.

Then, Gezan chooses a positive element in B and decrease it by 1.

Then, give one candy to Takahashi, their pet.

Tozan wants the number of candies given to Takahashi until the process is terminated to be as large as possible, while Gezan wants it to be as small as possible.
Find the number of candies given to Takahashi when both of them perform the operations optimally.

Constraints

1 \leq N \leq 2 × 10^5

0 \leq A_i,B_i \leq 10^9(1\leq i\leq N)

The sums of the elements in A and B are equal.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 B_1
:
A_N B_N

Output

Print the number of candies given to Takahashi when both Tozan and Gezan perform the operations optimally.

Sample Input 1

2
1 2
3 2

Sample Output 1

2

When both Tozan and Gezan perform the operations optimally, the process will proceed as follows:

Tozan decreases A_1 by 1.

Gezan decreases B_1 by 1.

One candy is given to Takahashi.

Tozan decreases A_2 by 1.

Gezan decreases B_1 by 1.

One candy is given to Takahashi.

As A and B are equal, the process is terminated.

Sample Input 2

3
8 3
0 1
4 8

Sample Output 2

9

Sample Input 3

1
1 1

Sample Output 3

0


let s = read_line ()
let n, fs, i = Array.(String.length s, make 26 false, ref 0)
let f s n = Printf.printf "%s%c\n" s (Char.chr (n + 97)); exit 0
let _ = String.(Char.(iter (fun c -> fs.(code c - 97) <- true; incr i) s; if n < 26 then (i := 0; while fs.(!i) do incr i done; f s !i);
  for j = n - 1 downto 0 do i := code s.[j] - 96; while !i < 26 && fs.(!i) do incr i done; if !i < 26 then f (sub s 0 j) !i; fs.(code s.[j] - 97) <- false done; print_endline "-1"))

Score : 300 points

Problem Statement

Gotou just received a dictionary. However, he doesn't recognize the language used in the dictionary. He did some analysis on the dictionary and realizes that the dictionary contains all possible diverse words in lexicographical order.

A word is called diverse if and only if it is a nonempty string of English lowercase letters and all letters in the word are distinct. For example, atcoder, zscoder and agc are diverse words while gotou and connect aren't diverse words.

Given a diverse word S, determine the next word that appears after S in the dictionary, i.e. the lexicographically smallest diverse word that is lexicographically larger than S, or determine that it doesn't exist.

Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.

Constraints

1 \leq |S| \leq 26

S is a diverse word.

Input

Input is given from Standard Input in the following format:

S

Output

Print the next word that appears after S in the dictionary, or -1 if it doesn't exist.

Sample Input 1

atcoder

Sample Output 1

atcoderb

atcoderb is the lexicographically smallest diverse word that is lexicographically larger than atcoder. Note that atcoderb is lexicographically smaller than b.

Sample Input 2

abc

Sample Output 2

abcd

Sample Input 3

zyxwvutsrqponmlkjihgfedcba

Sample Output 3

-1

This is the lexicographically largest diverse word, so the answer is -1.

Sample Input 4

abcdefghijklmnopqrstuvwzyx

Sample Output 4

abcdefghijklmnopqrstuvx


let s = read_line ()
let n = String.length s
let fs, is, i = Array.(make 26 false, make 26 0, ref 0)
let _ = String.(Char.(iter (fun c -> let n = code c - 97 in fs.(n) <- true; is.(!i) <- n; incr i) s;
  if n < 26 then (i := 0; while fs.(!i) do incr i done; Printf.printf "%s%c\n" s (chr (!i + 97)))
  else (for j = n - 1 downto 0 do i := code s.[j] - 96; while !i < 26 && fs.(!i) do incr i done; if !i < 26 then (Printf.printf "%s%c\n" (sub s 0 j) (chr (!i + 97)); exit 0) else (fs.(is.(j)) <- false) done; print_endline "-1")))


let () =
  let a = Scanf.scanf "%d\n" (fun a -> a) in
  print_endline "abcd"


Scanf.scanf "%d" (fun n ->
    let rec gcd a b = if b = 0 then a else gcd b (a mod b) in
    let lcm a b = a / gcd a b * b in

    let smooth k l =
        let rec loop r mask = function
            | [] -> if r = 1 then mask else 0
            | (hd :: tl) as l ->
                    if r mod hd = 0 then loop (r / hd) (mask lor (1 lsl hd)) l
                                    else loop r mask tl
        in
        loop k 0 l
    in

    let m2prod mask =
        let rec loop r i c =
            if r = 0 then c else
                if r land 1 = 0 then loop (r lsr 1) (i + 1) c
                                else loop (r lsr 1) (i + 1) (c * i)
        in
        loop mask 0 1
    in

    let rec loop i acc sum mask =
        let c =  smooth i [ 2; 3; 5; 7; 11; 13; 17; 19; 23 ] in
        if c > 0 then (
            Printf.printf "%d " i;
            let acc = acc + 1 in
            let sum = sum + i in
            let mask = mask lor c in
            if acc = n - 1 then Printf.printf "%d\n" (lcm sum (m2prod mask) - sum) else loop (i +1) acc sum mask
        ) else loop (i + 1) acc sum mask
    in
    loop 2 0 0 0
)

Score : 600 points

Problem Statement

Nagase is a top student in high school. One day, she's analyzing some properties of special sets of positive integers.

She thinks that a set S = \{a_{1}, a_{2}, ..., a_{N}\} of distinct positive integers is called special if for all 1 \leq i \leq N, the gcd (greatest common divisor) of a_{i} and the sum of the remaining elements of S is not 1.

Nagase wants to find a special set of size N. However, this task is too easy, so she decided to ramp up the difficulty. Nagase challenges you to find a special set of size N such that the gcd of all elements are 1 and the elements of the set does not exceed 30000.

Constraints

3 \leq N \leq 20000

Input

Input is given from Standard Input in the following format:

N

Output

Output N space-separated integers, denoting the elements of the set S. S must satisfy the following conditions :

The elements must be distinct positive integers not exceeding 30000.

The gcd of all elements of S is 1, i.e. there does not exist an integer d > 1 that divides all elements of S.

S is a special set.

If there are multiple solutions, you may output any of them. The elements of S may be printed in any order. It is guaranteed that at least one solution exist under the given contraints.

Sample Input 1

3

Sample Output 1

2 5 63

\{2, 5, 63\} is special because gcd(2, 5 + 63) = 2, gcd(5, 2 + 63) = 5, gcd(63, 2 + 5) = 7. Also, gcd(2, 5, 63) = 1. Thus, this set satisfies all the criteria.

Note that \{2, 4, 6\} is not a valid solution because gcd(2, 4, 6) = 2 > 1.

Sample Input 2

4

Sample Output 2

2 5 20 63

\{2, 5, 20, 63\} is special because gcd(2, 5 + 20 + 63) = 2, gcd(5, 2 + 20 + 63) = 5, gcd(20, 2 + 5 + 63) = 10, gcd(63, 2 + 5 + 20) = 9. Also, gcd(2, 5, 20, 63) = 1. Thus, this set satisfies all the criteria.


(*
ocamlfind ocamlopt -package batteries -linkpkg main.ml -o a.out
*)
open Batteries

let n = Scanf.sscanf (read_line ()) "%d" ( fun n -> n)
let d, x = Scanf.sscanf (read_line ()) "%d %d" ( fun d x -> d, x)
let lst = List.init n ( fun _ -> Scanf.sscanf (read_line ()) "%d" ( fun a -> a))

let f a = (List.range 0 `To 100)
          |> List.map (fun elmnt -> (a * elmnt) + 1)
          |> List.filter (fun elmnt -> elmnt <= d)

let rec f2 lst =
        match lst with
                  [] -> x
                | first :: rest ->
                        List.length (f first) + (f2 rest)

let () =
        f2 lst
        |> Printf.printf "%d\n"

Score : 200 points

Problem Statement

Some number of chocolate pieces were prepared for a training camp.
The camp had N participants and lasted for D days.
The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on.
As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces.

Find the number of chocolate pieces prepared at the beginning of the camp.

Constraints

1 \leq N \leq 100

1 \leq D \leq 100

1 \leq X \leq 100

1 \leq A_i \leq 100 (1 \leq i \leq N)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
D X
A_1
A_2
:
A_N

Output

Find the number of chocolate pieces prepared at the beginning of the camp.

Sample Input 1

3
7 1
2
5
10

Sample Output 1

8

The camp has 3 participants and lasts for 7 days.
Each participant eats chocolate pieces as follows:

The first participant eats one chocolate piece on Day 1, 3, 5 and 7, for a total of four.

The second participant eats one chocolate piece on Day 1 and 6, for a total of two.

The third participant eats one chocolate piece only on Day 1, for a total of one.

Since the number of pieces remaining at the end of the camp is one, the number of pieces prepared at the beginning of the camp is 1 + 4 + 2 + 1 = 8.

Sample Input 2

2
8 20
1
10

Sample Output 2

29

Sample Input 3

5
30 44
26
18
81
18
6

Sample Output 3

56


let id x = x
let n = Scanf.scanf "%d\n" id
let ar = Array.make (n+2) 0
let sum_init i =
  let rec sum i' tmp =
    if i' = (n + 1) then tmp + abs ar.(0) + abs ar.(n)
    else
      let v = Scanf.scanf "%d " id in
      let () = ar.(i') <- v in
      sum (i' + 1) (tmp + abs (v - ar.(i' - 1)))
  in
  sum 1 0

let cost = sum_init n

let skip i =
  let off = abs (ar.(i) - ar.(i-1)) + abs (ar.(i + 1) - ar.(i))
  and add = abs (ar.(i + 1) - ar.(i - 1)) in
  cost - off + add

let () = for i = 1 to n do skip i |> Printf.printf "%d\n" done

Score : 300 points

Problem Statement

There are N sightseeing spots on the x-axis, numbered 1, 2, ..., N.
Spot i is at the point with coordinate A_i.
It costs |a - b| yen (the currency of Japan) to travel from a point with coordinate a to another point with coordinate b along the axis.

You planned a trip along the axis.
In this plan, you first depart from the point with coordinate 0, then visit the N spots in the order they are numbered, and finally return to the point with coordinate 0.

However, something came up just before the trip, and you no longer have enough time to visit all the N spots, so you decided to choose some i and cancel the visit to Spot i.
You will visit the remaining spots as planned in the order they are numbered.
You will also depart from and return to the point with coordinate 0 at the beginning and the end, as planned.

For each i = 1, 2, ..., N, find the total cost of travel during the trip when the visit to Spot i is canceled.

Constraints

2 \leq N \leq 10^5

-5000 \leq A_i \leq 5000 (1 \leq i \leq N)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print N lines.
In the i-th line, print the total cost of travel during the trip when the visit to Spot i is canceled.

Sample Input 1

3
3 5 -1

Sample Output 1

12
8
10

Spot 1, 2 and 3 are at the points with coordinates 3, 5 and -1, respectively.
For each i, the course of the trip and the total cost of travel when the visit to Spot i is canceled, are as follows:

For i = 1, the course of the trip is 0 \rightarrow 5 \rightarrow -1 \rightarrow 0 and the total cost of travel is 5 + 6 + 1 = 12 yen.

For i = 2, the course of the trip is 0 \rightarrow 3 \rightarrow -1 \rightarrow 0 and the total cost of travel is 3 + 4 + 1 = 8 yen.

For i = 3, the course of the trip is 0 \rightarrow 3 \rightarrow 5 \rightarrow 0 and the total cost of travel is 3 + 2 + 5 = 10 yen.

Sample Input 2

5
1 1 1 2 0

Sample Output 2

4
4
4
2
4

Sample Input 3

6
-679 -2409 -3258 3095 -3291 -4462

Sample Output 3

21630
21630
19932
8924
21630
19288


let () =
  Scanf.scanf "%d %d" @@ fun a b ->
  let w = 100 in
  let res = Array.init 50 (fun _ -> Array.make 100 '.') in

  let r = (b+48)/50 - 1 in

  (* A *)
  let la = 2*r + 4 in
  for i = 0 to (a-2) do
    let x = i / (w/2) * 2 in
    let y = i mod (w/2) * 2 in
    res.(la + x).(y) <- '#';
  done;

  (* B *)
  for i = 0 to 2*r+1 do res.(i).(0) <- '#' done;
  for i = 0 to r do
    for j = 0 to w-1 do res.(2*i+1).(j) <- '#' done;
  done;
  for i = 0 to r-1 do
    for j = 0 to (w-2)/2 do res.(2*i).(2*j) <- '#' done;
  done;
  if r >= 0 then
    for i = 0 to (b+48) mod 50 do
      res.(2*r).(2*i) <- '#';
    done
  else res.(0).(0) <- '#';

  print_endline "50 100";
  res |> Array.iter @@ fun r ->
    r |> Array.iter print_char;
    print_newline ();

Score : 500 points

Problem Statement

You are given two integers A and B.

Print a grid where each square is painted white or black that satisfies the following conditions, in the format specified in Output section:

Let the size of the grid be h \times w (h vertical, w horizontal). Both h and w are at most 100.

The set of the squares painted white is divided into exactly A connected components.

The set of the squares painted black is divided into exactly B connected components.

It can be proved that there always exist one or more solutions under the conditions specified in Constraints section.
If there are multiple solutions, any of them may be printed.

Notes

Two squares painted white, c_1 and c_2, are called connected when the square c_2 can be reached from the square c_1 passing only white squares by repeatedly moving up, down, left or right to an adjacent square.

A set of squares painted white, S, forms a connected component when the following conditions are met:

Any two squares in S are connected.

No pair of a square painted white that is not included in S and a square included in S is connected.

A connected component of squares painted black is defined similarly.

Constraints

1 \leq A \leq 500

1 \leq B \leq 500

Input

Input is given from Standard Input in the following format:

A B

Output

Output should be in the following format:

In the first line, print integers h and w representing the size of the grid you constructed, with a space in between.

Then, print h more lines. The i-th (1 \leq i \leq h) of these lines should contain a string s_i as follows:

If the square at the i-th row and j-th column (1 \leq j \leq w) in the grid is painted white, the j-th character in s_i should be ..

If the square at the i-th row and j-th column (1 \leq j \leq w) in the grid is painted black, the j-th character in s_i should be #.

Sample Input 1

2 3

Sample Output 1

3 3
##.
..#
#.#

This output corresponds to the grid below:

Sample Input 2

7 8

Sample Output 2

3 5
#.#.#
.#.#.
#.#.#

Sample Input 3

1 1

Sample Output 3

4 2
..
#.
##
##

Sample Input 4

3 14

Sample Output 4

8 18
..................
..................
....##.......####.
....#.#.....#.....
...#...#....#.....
..#.###.#...#.....
.#.......#..#.....
#.........#..####.


let f a b c = if a + b > c then "Yes" else "No";;
let () = Scanf.scanf "%d %d %d" f
|> Printf.printf "%s\n"

Score : 100 points

Problem Statement

An elementary school student Takahashi has come to a variety store.

He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it?

Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.

Constraints

All input values are integers.

1 \leq A, B \leq 500

1 \leq C \leq 1000

Input

Input is given from Standard Input in the following format:

A B C

Output

If Takahashi can buy the toy, print Yes; if he cannot, print No.

Sample Input 1

50 100 120

Sample Output 1

Yes

He has 50 + 100 = 150 yen, so he can buy the 120-yen toy.

Sample Input 2

500 100 1000

Sample Output 2

No

He has 500 + 100 = 600 yen, but he cannot buy the 1000-yen toy.

Sample Input 3

19 123 143

Sample Output 3

No

There are 19-yen and 123-yen coins in Takahashi Kingdom, which are rather hard to use.

Sample Input 4

19 123 142

Sample Output 4

Yes


(* O(k^2) (k = max n m) *)
let n = read_int () in
let ss = Array.init n @@ fun _ -> read_line () in
let m = read_int () in
let ts = Array.init m @@ fun _ -> read_line () in
let f acc s =
  let c = Array.fold_left (fun c s' -> if s' = s then c + 1 else c) 0 ss in
  max acc @@ Array.fold_left (fun c t -> if t = s then c - 1 else c) c ts in
Array.fold_left f 0 ss |> Printf.printf "%d\n"

Score : 200 points

Problem Statement

Takahashi has N blue cards and M red cards.
A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i.

Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen.

Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces atcoder, he will not earn money even if there are blue cards with atcoderr, atcode, btcoder, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.)

At most how much can he earn on balance?

Note that the same string may be written on multiple cards.

Constraints

N and M are integers.

1 \leq N, M \leq 100

s_1, s_2, ..., s_N, t_1, t_2, ..., t_M are all strings of lengths between 1 and 10 (inclusive) consisting of lowercase English letters.

Input

Input is given from Standard Input in the following format:

N
s_1
s_2
:
s_N
M
t_1
t_2
:
t_M

Output

If Takahashi can earn at most X yen on balance, print X.

Sample Input 1

3
apple
orange
apple
1
grape

Sample Output 1

2

He can earn 2 yen by announcing apple.

Sample Input 2

3
apple
orange
apple
5
apple
apple
apple
apple
apple

Sample Output 2

1

If he announces apple, he will lose 3 yen. If he announces orange, he can earn 1 yen.

Sample Input 3

1
voldemort
10
voldemort
voldemort
voldemort
voldemort
voldemort
voldemort
voldemort
voldemort
voldemort
voldemort

Sample Output 3

0

If he announces voldemort, he will lose 9 yen. If he announces orange, for example, he can avoid losing a yen.

Sample Input 4

6
red
red
blue
yellow
yellow
red
5
red
red
yellow
green
blue

Sample Output 4

1


let scan_line () = Scanf.scanf "%d %d\n" (fun a b -> (a, b))

let n = Scanf.scanf "%d\n" (fun x -> x)

let init n =
  let rec aux i ls =
    if i = 0 then List.fast_sort (fun i j -> fst i - fst j) ls
    else aux (i - 1) (scan_line () :: ls)
  in
  aux n []

let r' = init n
let b' = init n

let rec solve r count = function
  | [] -> count
  | bp :: l ->
    match List.filter (fun rp -> fst rp < fst bp && snd rp < snd bp) r with
    | [] -> solve r count l
    | r' ->
      let v = List.fast_sort (fun i j -> snd j - snd i) r' |> List.hd in
      solve (List.remove_assoc (fst v) r) (count + 1) l

let () = solve r' 0 b' |> Printf.printf "%d\n"

Score : 400 points

Problem Statement

On a two-dimensional plane, there are N red points and N blue points.
The coordinates of the i-th red point are (a_i, b_i), and the coordinates of the i-th blue point are (c_i, d_i).

A red point and a blue point can form a friendly pair when, the x-coordinate of the red point is smaller than that of the blue point, and the y-coordinate of the red point is also smaller than that of the blue point.

At most how many friendly pairs can you form? Note that a point cannot belong to multiple pairs.

Constraints

All input values are integers.

1 \leq N \leq 100

0 \leq a_i, b_i, c_i, d_i < 2N

a_1, a_2, ..., a_N, c_1, c_2, ..., c_N are all different.

b_1, b_2, ..., b_N, d_1, d_2, ..., d_N are all different.

Input

Input is given from Standard Input in the following format:

N
a_1 b_1
a_2 b_2
:
a_N b_N
c_1 d_1
c_2 d_2
:
c_N d_N

Output

Print the maximum number of friendly pairs.

Sample Input 1

3
2 0
3 1
1 3
4 2
0 4
5 5

Sample Output 1

2

For example, you can pair (2, 0) and (4, 2), then (3, 1) and (5, 5).

Sample Input 2

3
0 0
1 1
5 2
2 3
3 4
4 5

Sample Output 2

2

For example, you can pair (0, 0) and (2, 3), then (1, 1) and (3, 4).

Sample Input 3

2
2 2
3 3
0 0
1 1

Sample Output 3

0

It is possible that no pair can be formed.

Sample Input 4

5
0 0
7 3
2 2
4 8
1 6
8 5
6 9
5 4
9 1
3 7

Sample Output 4

5

Sample Input 5

5
0 0
1 1
5 5
6 6
7 7
2 2
3 3
4 4
8 8
9 9

Sample Output 5

4


let n = read_int()

let ab = Array.init n (fun _ -> Scanf.scanf " %d %d" (fun a b -> a, b))

let cd = Array.init n (fun _ -> Scanf.scanf " %d %d" (fun c d -> c, d))

let list = Array.make n 0
let li = ref 0

let rec f i j p =
  if i = n then p
  else
    let c, d = cd.(i) in
    let rec g j =
      if j < n then (
        let a, b = ab.(j) in
        if a < c then (
          list.(!li) <- b;
          li := !li + 1;
          g (j + 1)
        ) else j
      ) else j in
    let j = g j in
    let rec g k l m =
      if k = !li then l
      else if list.(k) < d && m < list.(k) then g (k + 1) k list.(k)
      else g (k + 1) l m in
    let l = g 0 ~-1 ~-1 in
    if l > -1 then (
      list.(l) <- list.(!li - 1);
      li := !li - 1;
      f (i + 1) j (p + 1)
    ) else f (i + 1) j p

let () =
  Array.sort (fun (x1, _) (x2, _) -> x1 - x2) ab;
  Array.sort (fun (x1, _) (x2, _) -> x1 - x2) cd;
  let ans = f 0 0 0 in
  Printf.printf "%d\n" ans


let f c1 c2 c3 c4 c5 c6 c7 c8 c9 = String.concat "" [String.make 1 c1; String.make 1 c5; String. make 1 c9];;
let () = Scanf.scanf "%c%c%c %c%c%c %c%c%c" f
|> Printf.printf "%s\n"

Score : 100 points

Problem Statement

We have a 3×3 square grid, where each square contains a lowercase English letters.
The letter in the square at the i-th row from the top and j-th column from the left is c_{ij}.

Print the string of length 3 that can be obtained by concatenating the letters in the squares on the diagonal connecting the top-left and bottom-right corner of the grid, from the top-left to bottom-right.

Constraints

Input consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

c_{11}c_{12}c_{13}
c_{21}c_{22}c_{23}
c_{31}c_{32}c_{33}

Output

Print the string of length 3 that can be obtained by concatenating the letters on the diagonal connecting the top-left and bottom-right corner of the grid, from the top-left to bottom-right.

Sample Input 1

ant
obe
rec

Sample Output 1

abc

The letters in the squares on the diagonal connecting the top-left and bottom-right corner of the grid are a, b and c from top-right to bottom-left. Concatenate these letters and print abc.

Sample Input 2

edu
cat
ion

Sample Output 2

ean


let () =
  Scanf.scanf "%d %d" (fun n m -> n * m - (2 * n + 2 * m - 4))
  |> abs |> Printf.printf "%d\n"

Score : 300 points

Problem Statement

There is a grid with infinitely many rows and columns. In this grid, there is a rectangular region with consecutive N rows and M columns, and a card is placed in each square in this region.
The front and back sides of these cards can be distinguished, and initially every card faces up.

We will perform the following operation once for each square contains a card:

For each of the following nine squares, flip the card in it if it exists: the target square itself and the eight squares that shares a corner or a side with the target square.

It can be proved that, whether each card faces up or down after all the operations does not depend on the order the operations are performed.
Find the number of cards that face down after all the operations.

Constraints

1 \leq N,M \leq 10^9

All input values are integers.

Input

Input is given from Standard Input in the following format:

N M

Output

Print the number of cards that face down after all the operations.

Sample Input 1

2 2

Sample Output 1

0

We will flip every card in any of the four operations. Thus, after all the operations, all cards face up.

Sample Input 2

1 7

Sample Output 2

5

After all the operations, all cards except at both ends face down.

Sample Input 3

314 1592

Sample Output 3

496080


Scanf.scanf "%d %d" (fun n k ->
    let rec loop b acc =
        if b > n then acc else
            let m = n / b in
            let r = n - m * b in
            let acc = acc + m * (b - k) + if k = 0 then max 0 (r - k) else max 0 (r - k + 1) in
            loop (b + 1) acc
    in
    loop (k + 1) 0 |> Printf.printf "%d\n"
)

Score : 400 points

Problem Statement

Takahashi had a pair of two positive integers not exceeding N, (a,b), which he has forgotten.
He remembers that the remainder of a divided by b was greater than or equal to K.
Find the number of possible pairs that he may have had.

Constraints

1 \leq N \leq 10^5

0 \leq K \leq N-1

All input values are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

Print the number of possible pairs that he may have had.

Sample Input 1

5 2

Sample Output 1

7

There are seven possible pairs: (2,3),(5,3),(2,4),(3,4),(2,5),(3,5) and (4,5).

Sample Input 2

10 0

Sample Output 2

100

Sample Input 3

31415 9265

Sample Output 3

287927211


let id x = x

type v = T | F

let rec slv = function
	| 0 -> T
	| n -> if (Scanf.scanf "%c" id) = 'Y' then F else slv (n-1)

let () =
	let n = Scanf.scanf "%d\n" id in
	match slv n with
	| F -> print_string "Four"
	| T -> print_string "Three"

Score : 100 points

Problem Statement

There are N students in a school.

We will divide these students into some groups, and in each group they will discuss some themes.

You think that groups consisting of two or less students cannot have an effective discussion, so you want to have as many groups consisting of three or more students as possible.

Divide the students so that the number of groups consisting of three or more students is maximized.

Constraints

1 \leq N \leq 1000

All input values are integers.

Input

Input is given from Standard Input in the following format:

N

Output

If you can form at most x groups consisting of three or more students, print x.

Sample Input 1

8

Sample Output 1

2

For example, you can form a group of three students and another of five students.

Sample Input 2

2

Sample Output 2

0

Sometimes you cannot form any group consisting of three or more students, regardless of how you divide the students.

Sample Input 3

9

Sample Output 3

3


open Batteries;;
let _ = read_line () in
let l = read_line () |> String.split_on_char ' ' in
print_string (if List.mem "Y" l then "Four" else "Three")
;;

Score : 200 points

Problem Statement

In Japan, people make offerings called hina arare, colorful crackers, on March 3.

We have a bag that contains N hina arare. (From here, we call them arare.)

It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow.

We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: P, white: W, green: G, yellow: Y.

If the number of colors of the arare in the bag was three, print Three; if the number of colors was four, print Four.

Constraints

1 \leq N \leq 100

S_i is P, W, G or Y.

There always exist i, j and k such that S_i=P, S_j=W and S_k=G.

Input

Input is given from Standard Input in the following format:

N
S_1 S_2 ... S_N

Output

If the number of colors of the arare in the bag was three, print Three; if the number of colors was four, print Four.

Sample Input 1

6
G W Y P Y W

Sample Output 1

Four

The bag contained arare in four colors, so you should print Four.

Sample Input 2

9
G W W G P W P G G

Sample Output 2

Three

The bag contained arare in three colors, so you should print Three.

Sample Input 3

8
P Y W G Y W Y Y

Sample Output 3

Four


let rec comb n xs =
  match n, xs with
  | 0, _ -> [[]]
  | _, [] -> []
  | n, x :: xs ->
      List.map (fun ys -> x :: ys) (comb (n - 1) xs) @ comb n xs

let () =
  let n = Scanf.scanf "%d\n" (fun n -> n) in
  let ss = Array.init n (fun _ -> Scanf.scanf "%s\n" (fun s -> s)) in
  Array.fold_left (fun [m; a; r; c; h] s ->
    match s.[0] with
    | 'M' -> [m + 1; a; r; c; h]
    | 'A' -> [m; a + 1; r; c; h]
    | 'R' -> [m; a; r + 1; c; h]
    | 'C' -> [m; a; r; c + 1; h]
    | 'H' -> [m; a; r; c; h + 1]
    | _ -> [m; a; r; c; h]) [0; 0; 0; 0; 0] ss
  |> comb 3
  |> List.map (List.fold_left ( * ) 1)
  |> List.fold_left ( + ) 0
  |> Printf.printf "%d\n"

Score : 300 points

Problem Statement

There are N people. The name of the i-th person is S_i.

We would like to choose three people so that the following conditions are met:

The name of every chosen person begins with M, A, R, C or H.

There are no multiple people whose names begin with the same letter.

How many such ways are there to choose three people, disregarding order?

Note that the answer may not fit into a 32-bit integer type.

Constraints

1 \leq N \leq 10^5

S_i consists of uppercase English letters.

1 \leq |S_i| \leq 10

S_i \neq S_j (i \neq j)

Input

Input is given from Standard Input in the following format:

N
S_1
:
S_N

Output

If there are x ways to choose three people so that the given conditions are met, print x.

Sample Input 1

5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI

Sample Output 1

2

We can choose three people with the following names:

MASHIKE, RUMOI, HABORO

MASHIKE, RUMOI, HOROKANAI

Thus, we have two ways.

Sample Input 2

4
ZZ
ZZZ
Z
ZZZZZZZZZZ

Sample Output 2

0

Note that there may be no ways to choose three people so that the given conditions are met.

Sample Input 3

5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO

Sample Output 3

7


open Scanf
open Printf

let flip f x y = f y x

let rec range a b = if a >= b then [] else a :: range (a+1) b

let read_int () = scanf " %d" (fun x -> x)
let rec read_ints n = if n = 0 then [] else let i = read_int () in i :: read_ints (n-1)

let rec for_iter a b f = if a >= b then [] else let x = f a in x :: for_iter (a+1) b f
let rec for_iter_ a b f = if a >= b then () else (f a; for_iter_ (a+1) b f)

let list_of_string str = range 0 (String.length str) |> List.map (fun i -> str.[i])
let array_of_string str = Array.init (String.length str) (fun i -> str.[i])

let () =
  let h, w, d = scanf "%d %d %d" (fun a b c -> a, b, c) in
  let arr = Array.init h (fun _ -> Array.init w (fun _ -> scanf " %d" (fun x -> x))) in
  let qs = scanf " %d" (fun q -> Array.init q (fun _ -> scanf " %d %d" (fun x y -> x, y))) in
  let ps = Array.make (h*w+1) (-1, -1) in
  let cs = Array.make (h*w+1) (h*w) in

  for_iter_ 0 h (fun x -> for_iter_ 0 w (fun y -> ps.(arr.(x).(y)) <- (x, y)));

  for_iter_ 0 d (fun i -> cs.(i) <- 0);
  for_iter_ d (h*w+1) (fun i ->
    let (x, y) = ps.(i) in let (x', y') = ps.(i-d) in
    cs.(i) <- cs.(i-d) + abs (x-x') + abs (y-y'));

  qs
  |> Array.map (fun (i, j) -> cs.(j) - cs.(i))
  |> Array.iter (printf "%d\n")

Score : 400 points

Problem Statement

We have a grid with H rows and W columns. The square at the i-th row and the j-th column will be called Square (i,j).

The integers from 1 through H×W are written throughout the grid, and the integer written in Square (i,j) is A_{i,j}.

You, a magical girl, can teleport a piece placed on Square (i,j) to Square (x,y) by consuming |x-i|+|y-j| magic points.

You now have to take Q practical tests of your ability as a magical girl.

The i-th test will be conducted as follows:

Initially, a piece is placed on the square where the integer L_i is written.

Let x be the integer written in the square occupied by the piece. Repeatedly move the piece to the square where the integer x+D is written, as long as x is not R_i. The test ends when x=R_i.

Here, it is guaranteed that R_i-L_i is a multiple of D.

For each test, find the sum of magic points consumed during that test.

Constraints

1 \leq H,W \leq 300

1 \leq D \leq H×W

1 \leq A_{i,j} \leq H×W

A_{i,j} \neq A_{x,y} ((i,j) \neq (x,y))

1 \leq Q \leq 10^5

1 \leq L_i \leq R_i \leq H×W

(R_i-L_i) is a multiple of D.

Input

Input is given from Standard Input in the following format:

H W D
A_{1,1} A_{1,2} ... A_{1,W}
:
A_{H,1} A_{H,2} ... A_{H,W}
Q
L_1 R_1
:
L_Q R_Q

Output

For each test, print the sum of magic points consumed during that test.

Output should be in the order the tests are conducted.

Sample Input 1

3 3 2
1 4 3
2 5 7
8 9 6
1
4 8

Sample Output 1

5

4 is written in Square (1,2).

6 is written in Square (3,3).

8 is written in Square (3,1).

Thus, the sum of magic points consumed during the first test is (|3-1|+|3-2|)+(|3-3|+|1-3|)=5.

Sample Input 2

4 2 3
3 7
1 4
5 2
6 8
2
2 2
2 2

Sample Output 2

0
0

Note that there may be a test where the piece is not moved at all, and there may be multiple identical tests.

Sample Input 3

5 5 4
13 25 7 15 17
16 22 20 2 9
14 11 12 1 19
10 6 23 8 18
3 21 5 24 4
3
13 13
2 10
13 13

Sample Output 3

0
5
0


let n = read_int ()
let rec f n i j = if n < 10 then j else f (n / 10) (i + 1) @@ if n mod 10 = 9 then j else i + 1
let rec g s n i j = if n = 0 then s else g (s + if i = j then n mod 10 - 1 else if i > j then n mod 10 else 9) (n / 10) (i + 1) j
let j = f n 0 ~-1
let _ = Printf.printf "%d\n" @@ if j = -1 then n else g 0 n 0 j

Score : 300 points

Problem Statement

Find the maximum possible sum of the digits (in base 10) of a positive integer not greater than N.

Constraints

1\leq N \leq 10^{16}

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the maximum possible sum of the digits (in base 10) of a positive integer not greater than N.

Sample Input 1

100

Sample Output 1

18

For example, the sum of the digits in 99 is 18, which turns out to be the maximum value.

Sample Input 2

9995

Sample Output 2

35

For example, the sum of the digits in 9989 is 35, which turns out to be the maximum value.

Sample Input 3

3141592653589793

Sample Output 3

137


let () = Scanf.scanf "%d\n" @@ fun n ->
  let xys = Array.init n @@ fun i ->
    Scanf.scanf "%d %d\n" @@ fun x y -> i, x, y in
  let (i, px, py), rest = Array.fold_left (fun ((i, x, y), acc) (j, x', y') ->
    if y <= y'
    then ((i, x, y), (j, x', y') :: acc)
    else ((j, x', y'), (i, x, y) :: acc)) (xys.(0), []) @@ Array.sub xys 1 (n - 1) in
  let convex =
    Array.of_list @@
    List.fold_left (fun stack (i, x, y) ->
      let rec drop = function
        | ([] | [_]) as stack -> stack
        | ((_, x1, y1) :: (((_, x2, y2) :: _) as rest)) as stack ->
            if (x2 - x1) * (y - y1) - (y2 - y1) * (x - x1) <= 0
            then stack
            else drop rest in (i, x, y) :: drop stack) [(i, px, py)] @@
    List.sort (fun (_, x, y) (_, x', y') ->
      compare
        (x <= px, (y - py) * (x' - px))
        (x' <= px, (y' - py) * (x - px))) rest in
  let ps = Array.make n 0. in
  for i = 0 to Array.length convex - 1 do
    let (_, x1, y1) = convex.(i) in
    let (j, x2, y2) = convex.((i + 1) mod Array.length convex) in
    let (_, x3, y3) = convex.((i + 2) mod Array.length convex) in
    ps.(j) <-
      0.5 -. acos
        (float_of_int ((x1 - x2) * (x3 - x2) + (y1 - y2) * (y3 - y2)) /.
         sqrt
           ((float_of_int (x1 - x2) ** 2. +. float_of_int (y1 - y2) ** 2.) *.
            (float_of_int (x3 - x2) ** 2. +. float_of_int (y3 - y2) ** 2.)))
      /. (8. *. atan 1.)
  done;
  Array.iter (Printf.printf "%.20f\n") ps

Score : 600 points

Problem Statement

There are N holes in a two-dimensional plane. The coordinates of the i-th hole are (x_i,y_i).

Let R=10^{10^{10^{10}}}. Ringo performs the following operation:

Randomly choose a point from the interior of a circle of radius R centered at the origin, and put Snuke there. Snuke will move to the hole with the smallest Euclidean distance from the point, and fall into that hole. If there are multiple such holes, the hole with the smallest index will be chosen.

For every i (1 \leq i \leq N), find the probability that Snuke falls into the i-th hole.

Here, the operation of randomly choosing a point from the interior of a circle of radius R is defined as follows:

Pick two real numbers x and y independently according to uniform distribution on [-R,R].

If x^2+y^2\leq R^2, the point (x,y) is chosen. Otherwise, repeat picking the real numbers x,y until the condition is met.

Constraints

2 \leq N \leq 100

|x_i|,|y_i| \leq 10^6(1\leq i\leq N)

All given points are pairwise distinct.

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
:
x_N y_N

Output

Print N real numbers. The i-th real number must represent the probability that Snuke falls into the i-th hole.

The output will be judged correct when, for all output values, the absolute or relative error is at most 10^{-5}.

Sample Input 1

2
0 0
1 1

Sample Output 1

0.5
0.5

If Ringo put Snuke in the region x+y\leq 1, Snuke will fall into the first hole. The probability of this happening is very close to 0.5.
Otherwise, Snuke will fall into the second hole, the probability of which happening is also very close to 0.5.

Sample Input 2

5
0 0
2 8
4 5
2 6
3 10

Sample Output 2

0.43160120892732328768
0.03480224363653196956
0.13880483535586193855
0.00000000000000000000
0.39479171208028279727


let n = read_int ()
let a = read_int ()

let result = if a <= (n mod 500) then "Yes" else "No"

let () = print_string result

Score: 100 points

Problem Statement

E869120 has A 1-yen coins and infinitely many 500-yen coins.

Determine if he can pay exactly N yen using only these coins.

Constraints

N is an integer between 1 and 10000 (inclusive).

A is an integer between 0 and 1000 (inclusive).

Input

Input is given from Standard Input in the following format:

N
A

Output

If E869120 can pay exactly N yen using only his 1-yen and 500-yen coins, print Yes; otherwise, print No.

Sample Input 1

2018
218

Sample Output 1

Yes

We can pay 2018 yen with four 500-yen coins and 18 1-yen coins, so the answer is Yes.

Sample Input 2

2763
0

Sample Output 2

No

When we have no 1-yen coins, we can only pay a multiple of 500 yen using only 500-yen coins. Since 2763 is not a multiple of 500, we cannot pay this amount.

Sample Input 3

37
514

Sample Output 3

Yes


let f n = n mod 500
let g a m = if a < m then "No" else "Yes"
let () = f (read_int ()) |> g (read_int ()) |> print_endline


let css = Array.init 3 @@ fun _ -> Array.init 3 @@ fun _ -> Scanf.scanf " %d" @@ (+) 0 in
let bs = Array.init 3 @@ fun i -> css.(0).(i) in
let a_s = Array.init 3 @@ fun i -> css.(i).(0) - bs.(0) in
for i = 0 to 2 do
  for j = 0 to 2 do
    if css.(i).(j) <> a_s.(i) + bs.(j) then begin print_endline "No"; exit 0 end
  done
done;
print_endline "Yes"

Score: 300 points

Problem Statement

We have a 3 \times 3 grid. A number c_{i, j} is written in the square (i, j), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left.

According to Takahashi, there are six integers a_1, a_2, a_3, b_1, b_2, b_3 whose values are fixed, and the number written in the square (i, j) is equal to a_i + b_j.

Determine if he is correct.

Constraints

c_{i, j} \ (1 \leq i \leq 3, 1 \leq j \leq 3) is an integer between 0 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

c_{1,1} c_{1,2} c_{1,3}
c_{2,1} c_{2,2} c_{2,3}
c_{3,1} c_{3,2} c_{3,3}

Output

If Takahashi's statement is correct, print Yes; otherwise, print No.

Sample Input 1

1 0 1
2 1 2
1 0 1

Sample Output 1

Yes

Takahashi is correct, since there are possible sets of integers such as: a_1=0,a_2=1,a_3=0,b_1=1,b_2=0,b_3=1.

Sample Input 2

2 2 2
2 1 2
2 2 2

Sample Output 2

No

Takahashi is incorrect in this case.

Sample Input 3

0 8 8
0 8 8
0 8 8

Sample Output 3

Yes

Sample Input 4

1 8 6
2 9 7
0 7 7

Sample Output 4

No


open Batteries
module EnumL = Enum.Labels
module ListL = List.Labels
module ArrayL = Array.Labels

let dbg0 x = Printf.eprintf "[debug]%s\n" @@ dump x
let dbg1 x = dbg0 x; x

let id = identity

let (++) n m = List.range n `To m
let (++^) n m = if n < m then List.range n `To (pred m) else []

let scan fmt = Scanf.sscanf (read_line ()) fmt

let scan_lines n fmt f =
  if n = 0 then Enum.empty ()
  else
    List.map (fun _ -> scan fmt f) (1 ++ n)
    |> List.enum

let scan_list sep cnv =
  let line = read_line () in
  (match sep with
   | None -> List.map String.of_char @@ String.to_list line
   | Some sep -> String.split_on_char sep line)
  |> List.map cnv

let scan_array ?sep cnv = Array.of_list @@ scan_list sep cnv

let scan_matrix n m e ?sep conv =
  let arr = Array.make_matrix n m e in
  ListL.iter (0 ++^ n)
    ~f:(fun i -> arr.(i) <- Array.of_list @@ scan_list sep conv);
  arr

let rec powerset e =
  match Enum.get e with
  | None -> Enum.singleton @@ Enum.empty ()
  | Some v ->
    let f = powerset e in
    let g = Enum.clone f in
    EnumL.map f ~f:(fun x -> let y = Enum.clone x in push y v; y)
    |>  Enum.append g

let permutations l =
  let rec aux l =
    let rec interleave x = function
      | [] -> [[x]]
      | (hd::tl) as lst ->
        (x::lst) ::
        (ListL.map ~f:(List.cons hd)  @@ interleave x tl)
    in
    match l with
    | [] -> [[]]
    | hd::tl -> List.concat @@ List.map (interleave hd) @@ aux tl in
  let l = List.sort (List.compare Int.compare) @@ aux @@ List.of_enum l in
  List.enum % ListL.map ~f:List.enum @@ l

let intersection l =
  EnumL.filter ~f:(fun x -> EnumL.exists ~f:((=) x) @@ Enum.clone l)

let lower_bound n f =
  let rec aux l u =
    if u - l > 1 then
      let m = (l + u) / 2 in
      if f m then aux l m
      else aux m u
    else u in aux (-1) n

let between n x m = n <= x && x < m

let (h, w)  = scan "%d %d" Tuple.Tuple2.make
let g = scan_matrix h w "." id
let m = Array.make_matrix h w (-1)
let ds = [(-1,0);(1,0);(0,-1);(0,1)]

let rec bfs visited queue =
  if Queue.is_empty queue then ()
  else
    let (i, j, n) = Queue.pop queue in
    if visited.(i).(j) then bfs visited queue
    else
      (visited.(i).(j) <- true;
       m.(i).(j) <- n;
       ListL.iter ds ~f:(fun (di, dj) ->
           let x, y = i + di, j + dj in
           if between 0 x h && between 0 y w && not visited.(x).(y)
              && g.(x).(y) = "." then
             Queue.push (x,y, succ n) queue);
       bfs visited queue)

let () =
  let visited = Array.make_matrix h w false in
  let q = Queue.create () in
  Queue.push (0,0,1) q;
  bfs visited q;
  let n = m.(h-1).(w-1) in
  (if n = -1 then -1
   else
     h*w -
     (ArrayL.map g
        ~f:(fun line ->
            ArrayL.filter line ~f:((=) "#")
            |> Array.length)
      |> Array.sum) - n)
  |> Printf.printf "%d\n"

Score: 400 points

Problem statement

We have an H \times W grid whose squares are painted black or white. The square at the i-th row from the top and the j-th column from the left is denoted as (i, j).

Snuke would like to play the following game on this grid. At the beginning of the game, there is a character called Kenus at square (1, 1). The player repeatedly moves Kenus up, down, left or right by one square. The game is completed when Kenus reaches square (H, W) passing only white squares.

Before Snuke starts the game, he can change the color of some of the white squares to black. However, he cannot change the color of square (1, 1) and (H, W). Also, changes of color must all be carried out before the beginning of the game.

When the game is completed, Snuke's score will be the number of times he changed the color of a square before the beginning of the game. Find the maximum possible score that Snuke can achieve, or print -1 if the game cannot be completed, that is, Kenus can never reach square (H, W) regardless of how Snuke changes the color of the squares.

The color of the squares are given to you as characters s_{i, j}. If square (i, j) is initially painted by white, s_{i, j} is .; if square (i, j) is initially painted by black, s_{i, j} is #.

Constraints

H is an integer between 2 and 50 (inclusive).

W is an integer between 2 and 50 (inclusive).

s_{i, j} is . or # (1 \leq i \leq H, 1 \leq j \leq W).

s_{1, 1} and s_{H, W} are ..

Input

Input is given from Standard Input in the following format:

H W
s_{1, 1}s_{1, 2}s_{1, 3} ... s_{1, W}
s_{2, 1}s_{2, 2}s_{2, 3} ... s_{2, W}
: :
s_{H, 1}s_{H, 2}s_{H, 3} ... s_{H, W}

Output

Print the maximum possible score that Snuke can achieve, or print -1 if the game cannot be completed.

Sample Input 1

3 3
..#
#..
...

Sample Output 1

2

The score 2 can be achieved by changing the color of squares as follows:

Sample Input 2

10 37
.....................................
...#...####...####..###...###...###..
..#.#..#...#.##....#...#.#...#.#...#.
..#.#..#...#.#.....#...#.#...#.#...#.
.#...#.#..##.#.....#...#.#.###.#.###.
.#####.####..#.....#...#..##....##...
.#...#.#...#.#.....#...#.#...#.#...#.
.#...#.#...#.##....#...#.#...#.#...#.
.#...#.####...####..###...###...###..
.....................................

Sample Output 2

209


open Batteries
module EnumL = Enum.Labels
module ListL = List.Labels
module ArrayL = Array.Labels

let dbg0 x = Printf.eprintf "[debug]%s\n" @@ dump x
let dbg1 x = dbg0 x; x

let id = identity

let (++) n m = List.range n `To m
let (++^) n m = if n < m then List.range n `To (pred m) else []

let scan fmt = Scanf.sscanf (read_line ()) fmt

let scan_lines n fmt f =
  if n = 0 then Enum.empty ()
  else
    List.map (fun _ -> scan fmt f) (1 ++ n)
    |> List.enum

let scan_list sep cnv =
  let line = read_line () in
  (match sep with
   | None -> List.map String.of_char @@ String.to_list line
   | Some sep -> String.split_on_char sep line)
  |> List.map cnv

let scan_array ?sep cnv = Array.of_list @@ scan_list sep cnv

let scan_matrix n m e ?sep conv =
  let arr = Array.make_matrix n m e in
  ListL.iter (0 ++^ n)
    ~f:(fun i -> arr.(i) <- Array.of_list @@ scan_list sep conv);
  arr

let rec powerset e =
  match Enum.get e with
  | None -> Enum.singleton @@ Enum.empty ()
  | Some v ->
    let f = powerset e in
    let g = Enum.clone f in
    EnumL.map f ~f:(fun x -> let y = Enum.clone x in push y v; y)
    |>  Enum.append g

let permutations l =
  let rec aux l =
    let rec interleave x = function
      | [] -> [[x]]
      | (hd::tl) as lst ->
        (x::lst) ::
        (ListL.map ~f:(List.cons hd)  @@ interleave x tl)
    in
    match l with
    | [] -> [[]]
    | hd::tl -> List.concat @@ List.map (interleave hd) @@ aux tl in
  let l = List.sort (List.compare Int.compare) @@ aux @@ List.of_enum l in
  List.enum % ListL.map ~f:List.enum @@ l

let intersection l =
  EnumL.filter ~f:(fun x -> EnumL.exists ~f:((=) x) @@ Enum.clone l)

let lower_bound n f =
  let rec aux l u =
    if u - l > 1 then
      let m = (l + u) / 2 in
      if f m then aux l m
      else aux m u
    else u in aux (-1) n

let between n x m = n <= x && x < m

let (h, w)  = scan "%d %d" Tuple.Tuple2.make
let g = scan_matrix h w "." id
let m = Array.make_matrix h w (-1)
let ds = [(-1,0);(1,0);(0,-1);(0,1)]

let rec bfs visited queue =
  if Queue.is_empty queue then ()
  else
    let (i, j, n) = Queue.pop queue in
    visited.(i).(j) <- true;
    m.(i).(j) <- n;
    ListL.iter ds ~f:(fun (di, dj) ->
        let x, y = i + di, j + dj in
        if between 0 x h && between 0 y w && not visited.(x).(y)
           && g.(x).(y) = "." then
          Queue.push (x,y, succ n) queue);
    bfs visited queue

let () =
  let visited = Array.make_matrix h w false in
  let q = Queue.create () in
  Queue.push (0,0,1) q;
  bfs visited q;
  let n = dbg1 m.(h-1).(w-1) in
  (if n = -1 then -1
   else
     h*w -
     (ArrayL.map g
        ~f:(fun line ->
            ArrayL.filter line ~f:((=) "#")
            |> Array.length)
      |> Array.sum) - n)
  |> Printf.printf "%d\n"


(* O(h w) *)
let h, w = Scanf.scanf " %d %d" @@ fun a b -> a, b
let b = ref 0
let sss = Array.init h @@ fun _ ->
  Array.init w @@ fun _ -> Scanf.scanf " %c" @@ fun c -> if c = '#' then incr b; c
let ds = Array.make_matrix h w @@ -1
let bfs () =
  let q = Queue.create () in
  ds.(0).(0) <- 0;
  Queue.push (0, 0) q;
  while not @@ Queue.is_empty q do
    let y0, x0 = Queue.pop q in
    let f (dy, dx) =
      let y, x = y0 + dy, x0 + dx in
      if 0 <= y && y < h && 0 <= x && x < w && ds.(y).(x) = -1 && sss.(y).(x) <> '#' then
        (ds.(y).(x) <- ds.(y0).(x0) + 1;
        Queue.push (y, x) q) in
    List.iter f [0, -1; -1, 0; 0, 1; 1, 0]
  done;
  ds.(h - 1).(w - 1)
let _ =
  h * w - !b - bfs () - 1 |> Printf.printf "%d\n"


let x, y = Scanf.scanf " %d %d" @@ fun a b -> a, b
let _ = Printf.printf "%d\n" @@ if x mod y = 0 then -1 else x

Score : 100 points

Problem Statement

You are given positive integers X and Y.
If there exists a positive integer not greater than 10^{18} that is a multiple of X but not a multiple of Y, choose one such integer and print it.
If it does not exist, print -1.

Constraints

1 ≤ X,Y ≤ 10^9

X and Y are integers.

Input

Input is given from Standard Input in the following format:

X Y

Output

Print a positive integer not greater than 10^{18} that is a multiple of X but not a multiple of Y, or print -1 if it does not exist.

Sample Input 1

8 6

Sample Output 1

16

For example, 16 is a multiple of 8 but not a multiple of 6.

Sample Input 2

3 3

Sample Output 2

-1

A multiple of 3 is a multiple of 3.


let () =
  let (x,y) = Scanf.scanf "%d %d\n" (fun x y -> x,y) in
  let check  =
    if x mod y = 0 then -1
    else x
  in
  let ans = check in
  print_int ans;


let split_string ?(pattern="") = Str.split @@ Str.regexp pattern

let n = Scanf.sscanf (read_line ()) "%d" @@ fun n -> n
let a = read_line () |> split_string ~pattern:" " |> List.map int_of_string
let b = read_line () |> split_string ~pattern:" " |> List.map int_of_string

let () =
  let sum_a, sum_b = List.fold_left2 (fun (sum_a, sum_b) a b ->
      if a > b then (sum_a, sum_b + a - b)
      else (sum_a + (b - a) / 2, sum_b)
    ) (0, 0) a b in
  print_endline @@ if sum_a > sum_b then "Yes" else "No"

Score : 300 points

Problem Statement

You are given two integer sequences of length N: a_1,a_2,..,a_N and b_1,b_2,..,b_N.
Determine if we can repeat the following operation zero or more times so that the sequences a and b become equal.

Operation: Choose two integers i and j (possibly the same) between 1 and N (inclusive), then perform the following two actions simultaneously:

Add 2 to a_i.

Add 1 to b_j.

Constraints

1 ≤ N ≤ 10 000

0 ≤ a_i,b_i ≤ 10^9 (1 ≤ i ≤ N)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 .. a_N
b_1 b_2 .. b_N

Output

If we can repeat the operation zero or more times so that the sequences a and b become equal, print Yes; otherwise, print No.

Sample Input 1

3
1 2 3
5 2 2

Sample Output 1

Yes

For example, we can perform three operations as follows to do our job:

First operation: i=1 and j=2. Now we have a = \{3,2,3\}, b = \{5,3,2\}.

Second operation: i=1 and j=2. Now we have a = \{5,2,3\}, b = \{5,4,2\}.

Third operation: i=2 and j=3. Now we have a = \{5,4,3\}, b = \{5,4,3\}.

Sample Input 2

5
3 1 4 1 5
2 7 1 8 2

Sample Output 2

No

Sample Input 3

5
2 7 1 8 2
3 1 4 1 5

Sample Output 3

No


let () =
  let n = Scanf.scanf "%d\n" (fun n -> n) in
  let as_ = Array.to_list @@ Array.init n (fun _ -> Scanf.scanf "%d " (fun a -> a)) in
  let bs = Array.to_list @@ Array.init n (fun _ -> Scanf.scanf "%d " (fun b -> b)) in
  List.fold_left2 (fun (as_, bs) a b ->
    if a <= b then (as_, bs + (b - a) / 2)
    else (as_ + a - b, bs)) (0, 0) as_ bs
  |> (fun (as_, bs) -> as_ <= bs)
  |> (function true -> "Yes" | false -> "No")
  |> print_endline


let x, a, b = Scanf.scanf " %d %d %d" @@ fun a b c -> a, b, c
let _ = Printf.printf "%d\n" @@ (x - a) mod b

Score : 100 points

Problem Statement

You went shopping to buy cakes and donuts with X yen (the currency of Japan).

First, you bought one cake for A yen at a cake shop.
Then, you bought as many donuts as possible for B yen each, at a donut shop.

How much do you have left after shopping?

Constraints

1 \leq A, B \leq 1 000

A + B \leq X \leq 10 000

X, A and B are integers.

Input

Input is given from Standard Input in the following format:

X
A
B

Output

Print the amount you have left after shopping.

Sample Input 1

1234
150
100

Sample Output 1

84

You have 1234 - 150 = 1084 yen left after buying a cake.
With this amount, you can buy 10 donuts, after which you have 84 yen left.

Sample Input 2

1000
108
108

Sample Output 2

28

Sample Input 3

579
123
456

Sample Output 3

0

Sample Input 4

7477
549
593

Sample Output 4

405


let id x = x in
let id2 x y = x in
let n = Scanf.scanf "%d" id in
let lista = (Array.to_list @@ Array.init n (fun _ ->
    Scanf.scanf "%d ^\n" id )) in
let _ =Scanf.scanf "\n" (fun _ -> ()) in
let listb = (Array.to_list @@ Array.init n (fun _ ->
    Scanf.scanf "%d ^\n" id )) in

let rec check la lb count =
  match la, lb with
  | [], [] -> count
  | hda :: resta, hdb :: restb ->
     max (check resta restb (count+hda)) (check [] restb (count+hda+hdb))
in

let ans = check lista listb 0 in

print_int ans
;;

Score : 300 points

Problem Statement

We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j).

You are initially in the top-left square, (1, 1).
You will travel to the bottom-right square, (2, N), by repeatedly moving right or down.

The square (i, j) contains A_{i, j} candies.
You will collect all the candies you visit during the travel.
The top-left and bottom-right squares also contain candies, and you will also collect them.

At most how many candies can you collect when you choose the best way to travel?

Constraints

1 \leq N \leq 100

1 \leq A_{i, j} \leq 100 (1 \leq i \leq 2, 1 \leq j \leq N)

Input

Input is given from Standard Input in the following format:

N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}

Output

Print the maximum number of candies that can be collected.

Sample Input 1

5
3 2 2 4 1
1 2 2 2 1

Sample Output 1

14

The number of collected candies will be maximized when you:

move right three times, then move down once, then move right once.

Sample Input 2

4
1 1 1 1
1 1 1 1

Sample Output 2

5

You will always collect the same number of candies, regardless of how you travel.

Sample Input 3

7
3 3 4 5 4 5 3
5 3 4 4 2 3 2

Sample Output 3

29

Sample Input 4

1
2
3

Sample Output 4

5


let input = Scanf.scanf "%d %d\n" (fun x y -> int_of_string (string_of_int x ^ (string_of_int y))) in

let rec check x c=
  if x/2<c then false
  else if c *c =x then true
  else check x (c+1) in

let ans = if check input 1 then "Yes" else "No" in

print_string ans
;;

Score : 200 points

Problem Statement

AtCoDeer the deer has found two positive integers, a and b.
Determine whether the concatenation of a and b in this order is a square number.

Constraints

1 ≤ a,b ≤ 100

a and b are integers.

Input

Input is given from Standard Input in the following format:

a b

Output

If the concatenation of a and b in this order is a square number, print Yes; otherwise, print No.

Sample Input 1

1 21

Sample Output 1

Yes

As 121 = 11 × 11, it is a square number.

Sample Input 2

100 100

Sample Output 2

No

100100 is not a square number.

Sample Input 3

12 10

Sample Output 3

No


let n = Scanf.scanf " %d" @@ (+) 0
let txys = Array.init n @@ fun _ -> Scanf.scanf " %d %d %d" @@ fun a b c -> a, b, c
let f (t0, x0, y0) (t, x, y) =
  let d = abs (x - x0) + abs(y - y0) in
  if t - t0 < d || (t - t0 - d) mod 2 = 1 then (print_endline "No"; exit 0);
  t, x, y
let _ =
  Array.fold_left f (0, 0, 0) txys |> ignore;
  print_endline "Yes"

Score : 300 points

Problem Statement

AtCoDeer the deer is going on a trip in a two-dimensional plane.
In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i.

If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1).
Note that he cannot stay at his place.
Determine whether he can carry out his plan.

Constraints

1 ≤ N ≤ 10^5

0 ≤ x_i ≤ 10^5

0 ≤ y_i ≤ 10^5

1 ≤ t_i ≤ 10^5

t_i < t_{i+1} (1 ≤ i ≤ N-1)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
t_1 x_1 y_1
t_2 x_2 y_2
:
t_N x_N y_N

Output

If AtCoDeer can carry out his plan, print Yes; if he cannot, print No.

Sample Input 1

2
3 1 2
6 1 1

Sample Output 1

Yes

For example, he can travel as follows: (0,0), (0,1), (1,1), (1,2), (1,1), (1,0), then (1,1).

Sample Input 2

1
2 100 100

Sample Output 2

No

It is impossible to be at (100,100) two seconds after being at (0,0).

Sample Input 3

2
5 1 1
100 1 1

Sample Output 3

No


let id x = x in
let n = Scanf.scanf "%d\n" id in
let list =  (0,0,0) :: (Array.to_list @@ Array.init n (fun _ ->
    Scanf.scanf "%d %d %d\n" (fun a b c-> a, b, c))) in

let rec check : (int*int*int) list -> string =
  fun list ->
  match list with
  | [] -> "Yes"
  |  _ :: [] -> "Yes"
  | (t, x, y) :: ((t', x', y') as next) :: rest ->
     let dift=t'-t in
     let difd=x'-x+(y'-y) in
     if (dift+difd) mod 2 = 0 && difd<=dift
     then check (next :: rest)
     else "No" in

print_string (check list)
;;


let () = Scanf.scanf "%s" (fun s -> "2018/01/" ^ String.sub s 7 2)
  |> Printf.printf "%s\n"

Score : 100 points

Problem Statement

On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in yyyy/mm/dd format. For example, January 23, 2018 should be written as 2018/01/23.

After finishing the document, she noticed that she had mistakenly wrote 2017 at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to 2018 and prints it.

Constraints

S is a string of length 10.

The first eight characters in S are 2017/01/.

The last two characters in S are digits and represent an integer between 1 and 31 (inclusive).

Input

Input is given from Standard Input in the following format:

S

Output

Replace the first four characters in S with 2018 and print it.

Sample Input 1

2017/01/07

Sample Output 1

2018/01/07

Sample Input 2

2017/01/31

Sample Output 2

2018/01/31


let rewrite s =
  let s_as_bytes = Bytes.of_string s in
  let _ = Bytes.set s_as_bytes 3 '8' in
  Bytes.to_string s_as_bytes

let () =
  let s = Scanf.scanf "%s" (fun x -> x) in
  let ans = rewrite s in
  Printf.printf "%s\n" ans


open Core

let n = Scanf.scanf "%d" ident

let a =
  List.init n ~f:(fun _ -> Scanf.scanf " %d" Fn.id)

let ans = a |> List.dedup_and_sort ~compare:Int.compare |> List.length

let () = Printf.printf "%d\n" ans

Score : 200 points

Problem Statement

An X-layered kagami mochi (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi.

Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?

Constraints

1 ≤ N ≤ 100

1 ≤ d_i ≤ 100

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
d_1
:
d_N

Output

Print the maximum number of layers in a kagami mochi that can be made.

Sample Input 1

4
10
8
8
6

Sample Output 1

3

If we stack the mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, we have a 3-layered kagami mochi, which is the maximum number of layers.

Sample Input 2

3
15
15
15

Sample Output 2

1

When all the mochi have the same diameter, we can only have a 1-layered kagami mochi.

Sample Input 3

7
50
30
50
100
50
80
30

Sample Output 3

4


let n = Scanf.scanf " %d" (+) 0
let ds = Array.to_list @@ Array.init n @@ fun _ -> Scanf.scanf " %d" (+) 0
let _ = List.sort_uniq compare ds |> List.length |> Printf.printf "%d\n"


let input () =
    let n  = int_of_string (read_line ())
    in let rec ds i result = if i >= n
        then result
        else let input = int_of_string (read_line ())
             in  ds (i+1) (input::result)
    in List.rev (ds 0 [])

let remove_duplicate (x::xs) =
    let rec remove_duplicate' prev l =
        match l with
        | []    -> []
        | x::xs -> if x == prev
            then remove_duplicate' x xs
            else x :: (remove_duplicate' x xs)
    in x :: (remove_duplicate' x xs)

let _ = List.sort (Pervasives.compare) (input ())
     |> remove_duplicate
     |> List.length
     |> print_int
     |> print_newline


let () = Scanf.scanf "%d" (fun m -> 48 - m)
  |> Printf.printf "%d\n"

Score : 100 points

Problem Statement

How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December?

Constraints

1≤M≤23

M is an integer.

Input

Input is given from Standard Input in the following format:

M

Output

If we have x hours until New Year at M o'clock on 30th, December, print x.

Sample Input 1

21

Sample Output 1

27

We have 27 hours until New Year at 21 o'clock on 30th, December.

Sample Input 2

12

Sample Output 2

36


let () = Scanf.scanf "%d" (fun a -> Printf.printf "%d\n" (48 - a))


let a, b, s = Scanf.scanf " %d %d %s" @@ fun a b c -> a, b, c
let f s x = let n = ref 0 in String.iter (fun c -> if c = x then incr n) s; !n
let _ = print_endline @@ if s.[a] = '-' && f s '-' = 1 then "Yes" else "No"

Score : 200 points

Problem Statement

The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen -, and the other characters are digits from 0 through 9.

You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.

Constraints

1≤A,B≤5

|S|=A+B+1

S consists of - and digits from 0 through 9.

Input

Input is given from Standard Input in the following format:

A B
S

Output

Print Yes if S follows the postal code format in AtCoder Kingdom; print No otherwise.

Sample Input 1

3 4
269-6650

Sample Output 1

Yes

The (A+1)-th character of S is -, and the other characters are digits from 0 through 9, so it follows the format.

Sample Input 2

1 1
---

Sample Output 2

No

S contains unnecessary -s other than the (A+1)-th character, so it does not follow the format.

Sample Input 3

1 2
7444

Sample Output 3

No


let a, b = Scanf.scanf " %d %d" @@ fun a b -> a, b
let s = Scanf.scanf " %s" @@ fun s -> s
let is_digit c = 48 <= Char.code c && Char.code c <= 57
let c = Array.fold_left (+) 0 @@ Array.init (a + b + 1) @@ fun i -> if is_digit s.[i] then 1 else 0
let _ = print_endline @@ if c = a + b && s.[a] = '-' then "Yes" else "No"


open Scanf
open Printf

let flip f x y = f y x

let rec range a b = if a >= b then [] else a :: range (a+1) b

let read_int () = scanf " %d" (fun x -> x)
let rec read_ints n = if n = 0 then [] else let i = read_int () in i :: read_ints (n-1)

let rec for_iter a b f = if a >= b then [] else let x = f a in x :: for_iter (a+1) b f
let rec for_iter_ a b f = if a >= b then () else (f a; for_iter_ (a+1) b f)
let rec for_iterr_ a b f = if a <= b then () else (f a; for_iterr_ (a-1) b f)

let gen_table ubound =
  let res = Array.make ubound 1 in
  res.(0) <- 0; res.(1) <- 0;
  for_iter_ 2 ubound (fun i ->
    if res.(i) > 0 then for_iter_ 2 (ubound / i) @@ fun j -> res.(i*j) <- 0);
  for_iterr_ ((ubound+1)/2) 1 (fun a ->
    let i = 2*a-1 in
    res.(i) <- if res.(i) = 1 && res.((i+1)/2) = 1 then 1 else 0);
  res.(2) <- 0;
  for_iter_ 1 ubound (fun i -> res.(i) <- res.(i) + res.(i-1));
  res

let () =
  let q = read_int () in
  let rs = for_iter 0 q (fun _ -> let l = read_int () in let r = read_int () in (l, r)) in
  let table = gen_table 200010 in
  flip List.iter rs (fun (l, r) -> printf "%d\n" (table.(r) - table.(l-1)))

Score : 400 points

Problem Statement

We say that a odd number N is similar to 2017 when both N and (N+1)/2 are prime.

You are given Q queries.

In the i-th query, given two odd numbers l_i and r_i, find the number of odd numbers x similar to 2017 such that l_i ≤ x ≤ r_i.

Constraints

1≤Q≤10^5

1≤l_i≤r_i≤10^5

l_i and r_i are odd.

All input values are integers.

Input

Input is given from Standard Input in the following format:

Q
l_1 r_1
:
l_Q r_Q

Output

Print Q lines. The i-th line (1≤i≤Q) should contain the response to the i-th query.

Sample Input 1

1
3 7

Sample Output 1

2

3 is similar to 2017, since both 3 and (3+1)/2=2 are prime.

5 is similar to 2017, since both 5 and (5+1)/2=3 are prime.

7 is not similar to 2017, since (7+1)/2=4 is not prime, although 7 is prime.

Thus, the response to the first query should be 2.

Sample Input 2

4
13 13
7 11
7 11
2017 2017

Sample Output 2

1
0
0
1

Note that 2017 is also similar to 2017.

Sample Input 3

6
1 53
13 91
37 55
19 51
73 91
13 49

Sample Output 3

4
4
1
1
1
2


Scanf.scanf "%d %d %d %d" (fun a b c d ->
    let q = a + b in
    let r = c + d in
    print_endline @@
    if q > r then "Left" else
        if q < r then "Right" else "Balanced"
)

Score : 100 points

Problem Statement

A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R.

Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan.

Print Left if the balance scale tips to the left; print Balanced if it balances; print Right if it tips to the right.

Constraints

1\leq A,B,C,D \leq 10

All input values are integers.

Input

Input is given from Standard Input in the following format:

A B C D

Output

Print Left if the balance scale tips to the left; print Balanced if it balances; print Right if it tips to the right.

Sample Input 1

3 8 7 1

Sample Output 1

Left

The total weight of the masses on the left pan is 11, and the total weight of the masses on the right pan is 8. Since 11>8, we should print Left.

Sample Input 2

3 4 5 2

Sample Output 2

Balanced

The total weight of the masses on the left pan is 7, and the total weight of the masses on the right pan is 7. Since 7=7, we should print Balanced.

Sample Input 3

1 7 6 4

Sample Output 3

Right

The total weight of the masses on the left pan is 8, and the total weight of the masses on the right pan is 10. Since 8<10, we should print Right.


open Scanf
open Printf

let flip f x y = f y x

let rec range a b = if a >= b then [] else a :: range (a+1) b

let read_int () = scanf " %d" (fun x -> x)
let rec read_ints n = if n = 0 then [] else let i = read_int () in i :: read_ints (n-1)

let rec for_iter a b f = if a >= b then [] else let x = f a in x :: for_iter (a+1) b f

let () =
  let [a; b; c; d] = read_ints 4 in
  let res = if a+b > c+d then "Left" else if a+b < c+d then "Right" else "Balanced" in
  printf "%s\n" res


let (a,b,c,d) = Scanf.scanf "%d %d %d %d\n" (fun x y z w -> (x,y,z,w))

let () = if a + b > c + d then print_endline "Left"
         else if a + b < c + d then print_endline "Right"
         else print_endline "Balanced"


open Scanf
open Printf

let flip f x y = f y x

let rec range a b = if a >= b then [] else a :: range (a+1) b

let read_int () = scanf " %d" (fun x -> x)
let rec read_ints n = if n = 0 then [] else let i = read_int () in i :: read_ints (n-1)

let rec for_iter a b f = if a >= b then [] else let x = f a in x :: for_iter (a+1) b f

let rec solve v x y = if x > y then v else solve (v+1) (2*x) y

let () =
  let x = read_int () in let y = read_int () in
  solve 0 x y |> printf "%d\n"

Score : 300 points

Problem Statement

As a token of his gratitude, Takahashi has decided to give his mother an integer sequence.
The sequence A needs to satisfy the conditions below:

A consists of integers between X and Y (inclusive).

For each 1\leq i \leq |A|-1, A_{i+1} is a multiple of A_i and strictly greater than A_i.

Find the maximum possible length of the sequence.

Constraints

1 \leq X \leq Y \leq 10^{18}

All input values are integers.

Input

Input is given from Standard Input in the following format:

X Y

Output

Print the maximum possible length of the sequence.

Sample Input 1

3 20

Sample Output 1

3

The sequence 3,6,18 satisfies the conditions.

Sample Input 2

25 100

Sample Output 2

3

Sample Input 3

314159265 358979323846264338

Sample Output 3

31


let pr x = print_endline @@ string_of_int x

let fold f init s =
  let rec iter i res =
    if i < 0 then res else iter (i - 1) (f res s.[i]) in
  iter (String.length s - 1) init

let solve seq =
  let len = String.length seq in
  let (_,_,res) = fold
                    (fun (cnt,pre,mv) x ->
                      if x != pre then
                        (cnt+1,x,min mv (max cnt (len - cnt)))
                      else (cnt+1,x,mv))
                    (0,seq.[0],len) seq in res

let () = input_line stdin
         |> solve
         |> pr

Score : 500 points

Problem Statement

You are given a string S consisting of 0 and 1.
Find the maximum integer K not greater than |S| such that we can turn all the characters of S into 0 by repeating the following operation some number of times.

Choose a contiguous segment [l,r] in S whose length is at least K (that is, r-l+1\geq K must be satisfied). For each integer i such that l\leq i\leq r, do the following: if S_i is 0, replace it with 1; if S_i is 1, replace it with 0.

Constraints

1\leq |S|\leq 10^5

S_i(1\leq i\leq N) is either 0 or 1.

Input

Input is given from Standard Input in the following format:

S

Output

Print the maximum integer K such that we can turn all the characters of S into 0 by repeating the operation some number of times.

Sample Input 1

010

Sample Output 1

2

We can turn all the characters of S into 0 by the following operations:

Perform the operation on the segment S[1,3] with length 3. S is now 101.

Perform the operation on the segment S[1,2] with length 2. S is now 011.

Perform the operation on the segment S[2,3] with length 2. S is now 000.

Sample Input 2

100000000

Sample Output 2

8

Sample Input 3

00001111

Sample Output 3

4


let f a b = int_of_float (ceil ((a +. b) /. 2.0));;
Scanf.scanf "%f %f" f
|> Printf.printf "%d\n"

Score : 100 points

Problem Statement

You are given two positive integers a and b.
Let x be the average of a and b.
Print x rounded up to the nearest integer.

Constraints

a and b are integers.

1 \leq a, b \leq 100

Input

Input is given from Standard Input in the following format:

a b

Output

Print x rounded up to the nearest integer.

Sample Input 1

1 3

Sample Output 1

2

The average of 1 and 3 is 2.0, and it will be rounded up to the nearest integer, 2.

Sample Input 2

7 4

Sample Output 2

6

The average of 7 and 4 is 5.5, and it will be rounded up to the nearest integer, 6.

Sample Input 3

5 5

Sample Output 3

5


let solve a b = (a + b) / 2 + if (a + b) mod 2 = 0 then 0 else 1

let read_ints () = read_line () |> Str.split (Str.regexp " ") |> List.map int_of_string

let _ =
  match read_ints () with
  | [a; b] -> solve a b |> string_of_int |> print_endline
  | _ -> ()


let solve a b = (float_of_int (a + b)) /. 2.0
                |> ceil |> int_of_float |> print_int

let () = Scanf.scanf "%d %d" solve