{-# LANGUAGE
  ScopedTypeVariables, BangPatterns, TupleSections, ExplicitForAll,
  LambdaCase, MultiWayIf, Unsafe, RecordWildCards, FlexibleContexts, CPP,
  NoMonomorphismRestriction, GADTs, PatternGuards, MagicHash, UnboxedTuples,
  RankNTypes, EmptyDataDecls, EmptyCase, ViewPatterns, PolyKinds,
  TypeFamilies, OverloadedStrings, FlexibleInstances, UndecidableInstances,
  DefaultSignatures, GeneralizedNewtypeDeriving, StandaloneDeriving,
  DeriveGeneric, DeriveFunctor, DeriveDataTypeable, DeriveFoldable,
  DeriveTraversable, DeriveDataTypeable, FlexibleInstances,
  MultiParamTypeClasses #-}
{-# OPTIONS_GHC -O2 #-}

import Prelude
import Data.Bits
import Data.List
import Data.Maybe
import Data.Tuple
import Data.Ord
import Data.Int
import Data.Word
import Data.Char
import Data.Ratio
import Data.Function
import Data.STRef
import Data.IORef
import Data.Monoid
import Data.Functor
import Data.Data
import Data.Typeable
import GHC.Generics
import System.IO
import Control.Arrow
import Control.Applicative
import Control.Monad
import Control.Monad.Primitive
import Control.Monad.State.Strict
import Control.Monad.ST
import Control.Monad.ST.Lazy (strictToLazyST, lazyToStrictST)
import qualified Control.Monad.ST.Lazy as STL
-- import Control.Monad.ST.Safe
import Control.DeepSeq
import Data.Coerce
import qualified Data.ByteString.Char8 as BS
import qualified Data.ByteString.Lazy.Char8 as BSL
import qualified Data.ByteString.Builder as BSB
import Data.IntMap (IntMap)
import qualified Data.IntMap.Strict as IMS
import qualified Data.IntMap.Lazy as IML
import Data.IntSet (IntSet)
import qualified Data.IntSet as IS
import qualified Data.Sequence as Seq
import Data.Sequence (Seq)
import qualified Data.Array.IArray as A
import qualified Data.Array.MArray.Safe as A
import qualified Data.Array.MArray as A
import Data.Array (Array)
import Data.Array.Unboxed (UArray)
import Data.Array.IArray (IArray)
import Data.Array.MArray.Safe (MArray)
import Data.Array.IO.Safe (IOArray, IOUArray)
import Data.Array.ST.Safe (STArray, STUArray, runSTArray, runSTUArray)
import qualified Data.Vector as V
import qualified Data.Vector.Mutable as VM
import qualified Data.Vector.Unboxed as VU
import qualified Data.Vector.Unboxed.Mutable as VUM
import qualified Data.Vector.Storable as VS
import qualified Data.Vector.Storable.Mutable as VSM
import qualified Data.Vector.Primitive as VP
import qualified Data.Vector.Primitive.Mutable as VPM
import qualified Data.Vector.Generic as VG
import qualified Data.Vector.Generic.Mutable as VGM
import qualified Data.Vector.Fusion.Bundle.Monadic as VFBM
import qualified Data.Vector.Fusion.Bundle as VFB
import qualified Data.Vector.Fusion.Stream.Monadic as VFSM
import qualified Data.Vector.Fusion.Bundle.Size as VFBS
import qualified Data.Vector.Fusion.Util as VFU
import qualified Data.Attoparsec.ByteString.Char8 as Atto
import Debug.Trace
import Unsafe.Coerce
import GHC.Exts (build, Int(..), Int#,
                 (+#), (*#), (-#), (<#), (>=#), (==#), quotRemInt#,
                 remInt#, uncheckedIShiftRL#, andI#, orI#,
                isTrue#)

main :: IO ()
main = do
  [h,w] <- map readInt . words <$> getLine
  board <- getVecURest h $ do
    row <- rWord
    return $ foldl' (.|.) 0 [bit i | i <- [0..w-1], BS.index row i == '.']
  let res = maximum $ [countLen h w board (x,y) |
                        x <- [0..h-1], y <- [0..w-1],
                        testBit (board VU.! x) y]
  print res
  return ()

{-# INLINE rWord #-}
rWord :: StateT BSL.ByteString Maybe BS.ByteString
rWord = fmap BSL.toStrict $ state $ BSL.span (>='!') . BSL.dropWhile (<'!')

countLen :: Int -> Int -> VU.Vector Word64 -> (Int,Int) -> Int
countLen h w vec (x,y) = runST $ do
  notVisited <- VU.thaw vec
  let go !i [] = return (i-1)
      go !i ps = do
        nv <- VU.freeze notVisited
        let !next = force $ [(x1,y1)
                            | (x0,y0) <- ps,
                              (x1,y1) <- map (\(dx,dy) -> (x0+dx,y0+dy))
                                         [(0,1),(0,-1),(-1,0),(1,0)],
                              0 <= x1, x1 < h, 0 <= y1, y1 < w,
                              testBit (nv VU.! x1) y1]
        forM_ next $ \ (!x,!y) ->
          VUM.unsafeModify notVisited (`clearBit` y) x
        go (i+1) next
  go 0 [(x,y)]

#define IL(f) {-# INLINE f #-}; f

IL(putBuilder) = BSB.hPutBuilder stdout

printVecInLines, printVecInSpcSepLn ::
  (VG.Vector v a, ShowAsBuilder a) => v a -> IO ()
IL(printVecInLines) = putBuilder . v2BLines
IL(printVecInSpcSepLn) = putBuilder . v2BSpcSepLn

class ShowAsBuilder a where
  showAsBuilder :: a -> BSB.Builder
  default showAsBuilder :: (Show a) => a -> BSB.Builder
  showAsBuilder = BSB.string8 . show

-- Inconsistent with show
instance (ShowAsBuilder a, VG.Vector v a) => ShowAsBuilder (v a) where
  IL(showAsBuilder) = v2BSpcSep

#define INS(t,f) instance ShowAsBuilder t where { IL(showAsBuilder)=f }
INS(Int,BSB.intDec)
INS(Int8,BSB.int8Dec)
INS(Int16,BSB.int16Dec)
INS(Int32,BSB.int32Dec)
INS(Int64,BSB.int64Dec)
INS(Word,BSB.wordDec)
INS(Word8,BSB.word8Dec)
INS(Word16,BSB.word16Dec)
INS(Word32,BSB.word32Dec)
INS(Word64,BSB.word64Dec)
INS(Integer,BSB.integerDec)
INS(Float,BSB.floatDec)
INS(Double,BSB.doubleDec)
-- INS(String,BSB.string8) -- Inconsistent with Show
-- INS(BS.ByteString,BSB.byteString) -- Inconsistent with Show
-- INS(BSL.ByteString,BSB.lazyByteString) -- Inconsisitent with Show
#undef INS

-- Inconsistent with Show
instance (ShowAsBuilder a, ShowAsBuilder b) => ShowAsBuilder (a,b) where
  IL(showAsBuilder) = showTupAsBuilder
instance (ShowAsBuilder a, ShowAsBuilder b, ShowAsBuilder c) =>
  ShowAsBuilder (a,b,c) where
  IL(showAsBuilder) = showTup3AsBuilder
instance (ShowAsBuilder a, ShowAsBuilder b, ShowAsBuilder c, ShowAsBuilder d) =>
  ShowAsBuilder (a,b,c,d) where
  IL(showAsBuilder) = showTup4AsBuilder

IL(showTupAsBuilderWith)
  :: (a -> BSB.Builder) -> (b -> BSB.Builder) -> (a,b) -> BSB.Builder
showTupAsBuilderWith showA showB
  = \(a,b) -> (showA a <>) $ BSB.char7 ' ' <> showB b
IL(showTupAsBuilder) :: (ShowAsBuilder a, ShowAsBuilder b)
  => (a,b) -> BSB.Builder
showTupAsBuilder = showTupAsBuilderWith showAsBuilder showAsBuilder

IL(showTup3AsBuilderWith) :: (a -> BSB.Builder) -> (b -> BSB.Builder) ->
  (c -> BSB.Builder) -> (a,b,c) -> BSB.Builder
showTup3AsBuilderWith showA showB showC
  = \(a,b,c) -> (showA a <>) $ (BSB.char7 ' ' <>) $ (showB b <>)
                $ (BSB.char7 ' ' <>) $ showC c
IL(showTup3AsBuilder) :: (ShowAsBuilder a, ShowAsBuilder b, ShowAsBuilder c)
  => (a,b,c) -> BSB.Builder
showTup3AsBuilder
  = showTup3AsBuilderWith showAsBuilder showAsBuilder showAsBuilder

IL(showTup4AsBuilderWith) :: (a -> BSB.Builder) -> (b -> BSB.Builder) ->
  (c -> BSB.Builder) -> (d -> BSB.Builder) -> (a,b,c,d) -> BSB.Builder
showTup4AsBuilderWith showA showB showC showD
  = \(a,b,c,d) -> (showA a <>) $ (BSB.char7 ' ' <>)
                  $ showTup3AsBuilderWith showB showC showD (b,c,d)
IL(showTup4AsBuilder) ::
  (ShowAsBuilder a, ShowAsBuilder b, ShowAsBuilder c, ShowAsBuilder d) =>
  (a,b,c,d) -> BSB.Builder
showTup4AsBuilder = showTup4AsBuilderWith showAsBuilder showAsBuilder
                    showAsBuilder showAsBuilder

v2BSpcSepLn, v2BSpcSep, v2BConcat, v2BLines ::
  (VG.Vector v a, ShowAsBuilder a)
  => v a -> BSB.Builder
IL(v2BSpcSepLn) = v2BSpcSepLnWith showAsBuilder
IL(v2BSpcSep) = v2BSpcSepWith showAsBuilder
IL(v2BConcat) = v2BConcatWith showAsBuilder
IL(v2BLines) = v2BLinesWith showAsBuilder

v2BSpcSepLnWith, v2BSpcSepWith, v2BConcatWith, v2BLinesWith ::
  (VG.Vector v a)
  => (a -> BSB.Builder) -- ^ show function
  -> v a -> BSB.Builder
IL(v2BSpcSepLnWith) = v2BSpcSepPostfWith $ BS.singleton '\n'
IL(v2BSpcSepWith) = v2BSpcSepPostfWith BS.empty
IL(v2BConcatWith) showFct = VG.foldr ((<>) . showFct) mempty
IL(v2BLinesWith) showFct
  = VG.foldr (\ a -> (showFct a <>) . (BSB.char7 '\n' <>)) mempty

v2BSpcSepPostf :: (VG.Vector v a, ShowAsBuilder a)
  => BS.ByteString -- ^ postfix
  -> v a -> BSB.Builder
IL(v2BSpcSepPostf) = (`v2BSpcSepPostfWith` showAsBuilder)

v2BSpcSepPostfWith :: (VG.Vector v a)
  => BS.ByteString -- ^ postfix
  -> (a -> BSB.Builder) -- ^ show function
  -> v a -> BSB.Builder
IL(v2BSpcSepPostfWith) = vecToBuilder BS.empty $ BS.singleton ' '

IL(vecToBuilder) :: (VG.Vector v a)
  => BS.ByteString -- ^ prefix
  -> BS.ByteString -- ^ separator
  -> BS.ByteString -- ^ postfix
  -> (a -> BSB.Builder) -- ^ show function
  -> v a -> BSB.Builder
vecToBuilder !prefix !separator !postfix
  = vecToBuilder_ (BSB.byteString prefix)
                  (BSB.byteString separator)
                  (BSB.byteString postfix)

IL(vecToBuilder_) :: (VG.Vector v a)
  => BSB.Builder -- ^ prefix
  -> BSB.Builder -- ^ separator
  -> BSB.Builder -- ^ postfix
  -> (a -> BSB.Builder) -- ^ show function
  -> v a -> BSB.Builder
vecToBuilder_ !prefix !separator !postfix showFct = \vec -> prefix <>
  VG.foldr
  (\ a rest !prefx -> prefx <> (showFct a <> rest separator))
  (const postfix) vec mempty

IL(evalVals) :: [a] -> [a]
evalVals xs = build $ \c n -> foldr (c $!) n xs
IL(forceVals) :: (NFData a) => [a] -> [a]
forceVals xs = build $ \c n -> foldr (c $!!) n xs

IL(getVecGLn) :: (VG.Vector v a) =>
  Int -> StateT BS.ByteString Maybe a -> IO (v a)
getVecGLn n s = VG.unfoldrN n (runStateT s) <$> BS.getLine
IL(getVecGRest) :: (VG.Vector v a) =>
  Int -> StateT BSL.ByteString Maybe a -> IO (v a)
getVecGRest n s = VG.unfoldrN n (runStateT s) <$> BSL.getContents
IL(getVecLn) :: Int -> StateT BS.ByteString Maybe a -> IO (V.Vector a)
getVecLn = getVecGLn
IL(getVecRest) :: Int -> StateT BSL.ByteString Maybe a -> IO (V.Vector a)
getVecRest = getVecGRest
IL(getVecULn) :: (VU.Unbox a) =>
  Int -> StateT BS.ByteString Maybe a -> IO (VU.Vector a)
getVecULn = getVecGLn
IL(getVecURest) :: (VU.Unbox a) =>
  Int -> StateT BSL.ByteString Maybe a -> IO (VU.Vector a)
getVecURest = getVecGRest

rInt :: StateT BSL.ByteString Maybe Int
IL(rInt) = StateT $ BSL.readInt . BSL.dropWhile (<'!')
rIntS :: StateT BS.ByteString Maybe Int
IL(rIntS) = StateT $ BS.readInt . BS.dropWhile (<'!')

unlessM, whenM :: (Monad m) => m Bool -> m () -> m ()
IL(whenM) = (. flip when) . (>>=)
IL(unlessM) = (. flip unless) . (>>=)

IL(wrA) = A.writeArray
IL(rdA) = A.readArray
IL(mdA) = \arr f !i -> do
  ai <- rdA arr i
  let fai = f ai
  wrA arr i fai
  return (ai,fai)
{-# INLINE mdA' #-}
mdA' = \arr f !i -> do
  !ai <- rdA arr i
  let !fai = f ai
  wrA arr i fai
  return (ai,fai)
IL(swapA) = \arr !i !j -> do
  ai <- rdA arr i
  wrA arr i =<< rdA arr j
  wrA arr j ai

#define D(f,r,d)\
  IL(f) :: Integral a=>a->d; f=fromIntegral;\
  IL(r) :: String->d; r=read
#define C(f,r,g,h,d) D(f,r,d);\
  g,h :: RealFrac a=>a->d; IL(g)=floor; IL(h)=ceiling
C(_toInteger_,readInteger,floorInteger,ceilInteger,Integer)
C(toInt,readInt,floorInt,ceilInt,Int)
C(toI8,readI8,floorI8,ceilI8,Int8)
C(toI16,readI16,floorI16,ceilI16,Int16)
C(toI32,readI32,floorI32,ceilI32,Int32)
C(toI64,readI64,floorI64,ceilI64,Int64)
C(toWord,readWord,floorWord,ceilWord,Word)
C(toW8,readW8,floorW8,ceilW8,Word8)
C(toW16,readW16,floorW16,ceilW16,Word16)
C(toW32,readW32,floorW32,ceilW32,Word32)
C(toW64,readW64,floorW64,ceilW64,Word64)
D(toDouble,readDouble,Double)
D(toFloat,readFloat,Float)
#undef D
#undef C

#define N(f,g,h,a,m)\
  IL(f) :: forall e i s. (C(a,m)A.Ix i) => (i,i) -> e -> m (a i e);\
  f=A.newArray;\
  IL(g) :: forall e i s. (C(a,m)A.Ix i) => (i,i) -> m (a i e);\
  g=A.newArray_;\
  IL(h) :: forall e i s. (C(a,m)A.Ix i) => (i,i) -> [e] -> m (a i e);\
  h=A.newListArray
#define C(a,m)
N(newIOA,newIOA_,newIOAL,IOArray,IO)
N(newSTA,newSTA_,newSTAL,STArray s,ST s)
#undef C
#define C(a,m) MArray (a) e (m),
N(newIOUA,newIOUA_,newIOUAL,IOUArray,IO)
N(newSTUA,newSTUA_,newSTUAL,STUArray s,ST s)
#undef C
#undef N

#undef IL

Score : 400 points

Problem Statement

Takahashi has a maze, which is a grid of H \times W squares with H horizontal rows and W vertical columns.

The square at the i-th row from the top and the j-th column is a "wall" square if S_{ij} is #, and a "road" square if S_{ij} is ..

From a road square, you can move to a horizontally or vertically adjacent road square.

You cannot move out of the maze, move to a wall square, or move diagonally.

Takahashi will choose a starting square and a goal square, which can be any road squares, and give the maze to Aoki.

Aoki will then travel from the starting square to the goal square, in the minimum number of moves required.

In this situation, find the maximum possible number of moves Aoki has to make.

Constraints

1 \leq H,W \leq 20

S_{ij} is . or #.

S contains at least two occurrences of ..

Any road square can be reached from any road square in zero or more moves.

Input

Input is given from Standard Input in the following format:

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

Output

Print the maximum possible number of moves Aoki has to make.

Sample Input 1

3 3
...
...
...

Sample Output 1

4

If Takahashi chooses the top-left square as the starting square and the bottom-right square as the goal square, Aoki has to make four moves.

Sample Input 2

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

Sample Output 2

10

If Takahashi chooses the bottom-left square as the starting square and the top-right square as the goal square, Aoki has to make ten moves.


{-# LANGUAGE
  ScopedTypeVariables, BangPatterns, TupleSections, ExplicitForAll,
  LambdaCase, MultiWayIf, Unsafe, RecordWildCards, FlexibleContexts, CPP,
  NoMonomorphismRestriction, GADTs, PatternGuards, MagicHash, UnboxedTuples,
  RankNTypes, EmptyDataDecls, EmptyCase, ViewPatterns, PolyKinds,
  TypeFamilies, OverloadedStrings, FlexibleInstances, UndecidableInstances,
  DefaultSignatures, GeneralizedNewtypeDeriving, StandaloneDeriving,
  DeriveGeneric, DeriveFunctor, DeriveDataTypeable, DeriveFoldable,
  DeriveTraversable, DeriveDataTypeable, FlexibleInstances,
  MultiParamTypeClasses #-}
{-# OPTIONS_GHC -O2 #-}

import Prelude
import Data.Bits
import Data.List
import Data.Maybe
import Data.Tuple
import Data.Ord
import Data.Int
import Data.Word
import Data.Char
import Data.Ratio
import Data.Function
import Data.STRef
import Data.IORef
import Data.Monoid
import Data.Functor
import Data.Data
import Data.Typeable
import GHC.Generics
import System.IO
import Control.Arrow
import Control.Applicative
import Control.Monad
import Control.Monad.Primitive
import Control.Monad.State.Strict
import Control.Monad.ST
import Control.Monad.ST.Lazy (strictToLazyST, lazyToStrictST)
import qualified Control.Monad.ST.Lazy as STL
-- import Control.Monad.ST.Safe
import Control.DeepSeq
import Data.Coerce
import qualified Data.ByteString.Char8 as BS
import qualified Data.ByteString.Lazy.Char8 as BSL
import qualified Data.ByteString.Builder as BSB
import Data.IntMap (IntMap)
import qualified Data.IntMap.Strict as IMS
import qualified Data.IntMap.Lazy as IML
import Data.IntSet (IntSet)
import qualified Data.IntSet as IS
import qualified Data.Sequence as Seq
import Data.Sequence (Seq)
import qualified Data.Array.IArray as A
import qualified Data.Array.MArray.Safe as A
import qualified Data.Array.MArray as A
import Data.Array (Array)
import Data.Array.Unboxed (UArray)
import Data.Array.IArray (IArray)
import Data.Array.MArray.Safe (MArray)
import Data.Array.IO.Safe (IOArray, IOUArray)
import Data.Array.ST.Safe (STArray, STUArray, runSTArray, runSTUArray)
import qualified Data.Vector as V
import qualified Data.Vector.Mutable as VM
import qualified Data.Vector.Unboxed as VU
import qualified Data.Vector.Unboxed.Mutable as VUM
import qualified Data.Vector.Storable as VS
import qualified Data.Vector.Storable.Mutable as VSM
import qualified Data.Vector.Primitive as VP
import qualified Data.Vector.Primitive.Mutable as VPM
import qualified Data.Vector.Generic as VG
import qualified Data.Vector.Generic.Mutable as VGM
import qualified Data.Vector.Fusion.Bundle.Monadic as VFBM
import qualified Data.Vector.Fusion.Bundle as VFB
import qualified Data.Vector.Fusion.Stream.Monadic as VFSM
import qualified Data.Vector.Fusion.Bundle.Size as VFBS
import qualified Data.Vector.Fusion.Util as VFU
import qualified Data.Attoparsec.ByteString.Char8 as Atto
import Debug.Trace
import GHC.Exts (build, Int(..), Int#,
                 (+#), (*#), (-#), (<#), (>=#), (==#), quotRemInt#,
                 remInt#, uncheckedIShiftRL#, andI#, orI#,
                isTrue#)

main :: IO ()
main = do
  n <- readInt <$> getLine
  songs <- V.replicateM n $ do
    [x,y] <- words <$> getLine
    return (x,readInt y)
  x <- getLine
  let rest = V.tail $ V.dropWhile ((/=x) . fst) songs
  print $ V.sum $ V.map snd rest
  return ()

#define IL(f) {-# INLINE f #-}; f

IL(putBuilder) = BSB.hPutBuilder stdout

printVecInLines, printVecInSpcSepLn ::
  (VG.Vector v a, ShowAsBuilder a) => v a -> IO ()
IL(printVecInLines) = putBuilder . v2BLines
IL(printVecInSpcSepLn) = putBuilder . v2BSpcSepLn

class ShowAsBuilder a where
  showAsBuilder :: a -> BSB.Builder
  default showAsBuilder :: (Show a) => a -> BSB.Builder
  showAsBuilder = BSB.string8 . show

-- Inconsistent with show
instance (ShowAsBuilder a, VG.Vector v a) => ShowAsBuilder (v a) where
  IL(showAsBuilder) = v2BSpcSep

#define INS(t,f) instance ShowAsBuilder t where { IL(showAsBuilder)=f }
INS(Int,BSB.intDec)
INS(Int8,BSB.int8Dec)
INS(Int16,BSB.int16Dec)
INS(Int32,BSB.int32Dec)
INS(Int64,BSB.int64Dec)
INS(Word,BSB.wordDec)
INS(Word8,BSB.word8Dec)
INS(Word16,BSB.word16Dec)
INS(Word32,BSB.word32Dec)
INS(Word64,BSB.word64Dec)
INS(Integer,BSB.integerDec)
INS(Float,BSB.floatDec)
INS(Double,BSB.doubleDec)
-- INS(String,BSB.string8) -- Inconsistent with Show
-- INS(BS.ByteString,BSB.byteString) -- Inconsistent with Show
-- INS(BSL.ByteString,BSB.lazyByteString) -- Inconsisitent with Show
#undef INS

-- Inconsistent with Show
instance (ShowAsBuilder a, ShowAsBuilder b) => ShowAsBuilder (a,b) where
  IL(showAsBuilder) = showTupAsBuilder
instance (ShowAsBuilder a, ShowAsBuilder b, ShowAsBuilder c) =>
  ShowAsBuilder (a,b,c) where
  IL(showAsBuilder) = showTup3AsBuilder
instance (ShowAsBuilder a, ShowAsBuilder b, ShowAsBuilder c, ShowAsBuilder d) =>
  ShowAsBuilder (a,b,c,d) where
  IL(showAsBuilder) = showTup4AsBuilder

IL(showTupAsBuilderWith)
  :: (a -> BSB.Builder) -> (b -> BSB.Builder) -> (a,b) -> BSB.Builder
showTupAsBuilderWith showA showB
  = \(a,b) -> (showA a <>) $ BSB.char7 ' ' <> showB b
IL(showTupAsBuilder) :: (ShowAsBuilder a, ShowAsBuilder b)
  => (a,b) -> BSB.Builder
showTupAsBuilder = showTupAsBuilderWith showAsBuilder showAsBuilder

IL(showTup3AsBuilderWith) :: (a -> BSB.Builder) -> (b -> BSB.Builder) ->
  (c -> BSB.Builder) -> (a,b,c) -> BSB.Builder
showTup3AsBuilderWith showA showB showC
  = \(a,b,c) -> (showA a <>) $ (BSB.char7 ' ' <>) $ (showB b <>)
                $ (BSB.char7 ' ' <>) $ showC c
IL(showTup3AsBuilder) :: (ShowAsBuilder a, ShowAsBuilder b, ShowAsBuilder c)
  => (a,b,c) -> BSB.Builder
showTup3AsBuilder
  = showTup3AsBuilderWith showAsBuilder showAsBuilder showAsBuilder

IL(showTup4AsBuilderWith) :: (a -> BSB.Builder) -> (b -> BSB.Builder) ->
  (c -> BSB.Builder) -> (d -> BSB.Builder) -> (a,b,c,d) -> BSB.Builder
showTup4AsBuilderWith showA showB showC showD
  = \(a,b,c,d) -> (showA a <>) $ (BSB.char7 ' ' <>)
                  $ showTup3AsBuilderWith showB showC showD (b,c,d)
IL(showTup4AsBuilder) ::
  (ShowAsBuilder a, ShowAsBuilder b, ShowAsBuilder c, ShowAsBuilder d) =>
  (a,b,c,d) -> BSB.Builder
showTup4AsBuilder = showTup4AsBuilderWith showAsBuilder showAsBuilder
                    showAsBuilder showAsBuilder

v2BSpcSepLn, v2BSpcSep, v2BConcat, v2BLines ::
  (VG.Vector v a, ShowAsBuilder a)
  => v a -> BSB.Builder
IL(v2BSpcSepLn) = v2BSpcSepLnWith showAsBuilder
IL(v2BSpcSep) = v2BSpcSepWith showAsBuilder
IL(v2BConcat) = v2BConcatWith showAsBuilder
IL(v2BLines) = v2BLinesWith showAsBuilder

v2BSpcSepLnWith, v2BSpcSepWith, v2BConcatWith, v2BLinesWith ::
  (VG.Vector v a)
  => (a -> BSB.Builder) -- ^ show function
  -> v a -> BSB.Builder
IL(v2BSpcSepLnWith) = v2BSpcSepPostfWith $ BS.singleton '\n'
IL(v2BSpcSepWith) = v2BSpcSepPostfWith BS.empty
IL(v2BConcatWith) showFct = VG.foldr ((<>) . showFct) mempty
IL(v2BLinesWith) showFct
  = VG.foldr (\ a -> (showFct a <>) . (BSB.char7 '\n' <>)) mempty

v2BSpcSepPostf :: (VG.Vector v a, ShowAsBuilder a)
  => BS.ByteString -- ^ postfix
  -> v a -> BSB.Builder
IL(v2BSpcSepPostf) = (`v2BSpcSepPostfWith` showAsBuilder)

v2BSpcSepPostfWith :: (VG.Vector v a)
  => BS.ByteString -- ^ postfix
  -> (a -> BSB.Builder) -- ^ show function
  -> v a -> BSB.Builder
IL(v2BSpcSepPostfWith) = vecToBuilder BS.empty $ BS.singleton ' '

IL(vecToBuilder) :: (VG.Vector v a)
  => BS.ByteString -- ^ prefix
  -> BS.ByteString -- ^ separator
  -> BS.ByteString -- ^ postfix
  -> (a -> BSB.Builder) -- ^ show function
  -> v a -> BSB.Builder
vecToBuilder !prefix !separator !postfix
  = vecToBuilder_ (BSB.byteString prefix)
                  (BSB.byteString separator)
                  (BSB.byteString postfix)

IL(vecToBuilder_) :: (VG.Vector v a)
  => BSB.Builder -- ^ prefix
  -> BSB.Builder -- ^ separator
  -> BSB.Builder -- ^ postfix
  -> (a -> BSB.Builder) -- ^ show function
  -> v a -> BSB.Builder
vecToBuilder_ !prefix !separator !postfix showFct = \vec -> prefix <>
  VG.foldr
  (\ a rest !prefx -> prefx <> (showFct a <> rest separator))
  (const postfix) vec mempty

IL(evalVals) :: [a] -> [a]
evalVals xs = build $ \c n -> foldr (c $!) n xs
IL(forceVals) :: (NFData a) => [a] -> [a]
forceVals xs = build $ \c n -> foldr (c $!!) n xs

IL(getVecGLn) :: (VG.Vector v a) =>
  Int -> StateT BS.ByteString Maybe a -> IO (v a)
getVecGLn n s = VG.unfoldrN n (runStateT s) <$> BS.getLine
IL(getVecGRest) :: (VG.Vector v a) =>
  Int -> StateT BSL.ByteString Maybe a -> IO (v a)
getVecGRest n s = VG.unfoldrN n (runStateT s) <$> BSL.getContents
IL(getVecLn) :: Int -> StateT BS.ByteString Maybe a -> IO (V.Vector a)
getVecLn = getVecGLn
IL(getVecRest) :: Int -> StateT BSL.ByteString Maybe a -> IO (V.Vector a)
getVecRest = getVecGRest
IL(getVecULn) :: (VU.Unbox a) =>
  Int -> StateT BS.ByteString Maybe a -> IO (VU.Vector a)
getVecULn = getVecGLn
IL(getVecURest) :: (VU.Unbox a) =>
  Int -> StateT BSL.ByteString Maybe a -> IO (VU.Vector a)
getVecURest = getVecGRest

rInt :: StateT BSL.ByteString Maybe Int
IL(rInt) = StateT $ BSL.readInt . BSL.dropWhile (<'!')
rIntS :: StateT BS.ByteString Maybe Int
IL(rIntS) = StateT $ BS.readInt . BS.dropWhile (<'!')

unlessM, whenM :: (Monad m) => m Bool -> m () -> m ()
IL(whenM) = (. flip when) . (>>=)
IL(unlessM) = (. flip unless) . (>>=)

IL(wrA) = A.writeArray
IL(rdA) = A.readArray
IL(mdA) = \arr f !i -> do
  ai <- rdA arr i
  let fai = f ai
  wrA arr i fai
  return (ai,fai)
{-# INLINE mdA' #-}
mdA' = \arr f !i -> do
  !ai <- rdA arr i
  let !fai = f ai
  wrA arr i fai
  return (ai,fai)
IL(swapA) = \arr !i !j -> do
  ai <- rdA arr i
  wrA arr i =<< rdA arr j
  wrA arr j ai

#define D(f,r,d)\
  IL(f) :: Integral a=>a->d; f=fromIntegral;\
  IL(r) :: String->d; r=read
#define C(f,r,g,h,d) D(f,r,d);\
  g,h :: RealFrac a=>a->d; IL(g)=floor; IL(h)=ceiling
C(_toInteger_,readInteger,floorInteger,ceilInteger,Integer)
C(toInt,readInt,floorInt,ceilInt,Int)
C(toI8,readI8,floorI8,ceilI8,Int8)
C(toI16,readI16,floorI16,ceilI16,Int16)
C(toI32,readI32,floorI32,ceilI32,Int32)
C(toI64,readI64,floorI64,ceilI64,Int64)
C(toWord,readWord,floorWord,ceilWord,Word)
C(toW8,readW8,floorW8,ceilW8,Word8)
C(toW16,readW16,floorW16,ceilW16,Word16)
C(toW32,readW32,floorW32,ceilW32,Word32)
C(toW64,readW64,floorW64,ceilW64,Word64)
D(toDouble,readDouble,Double)
D(toFloat,readFloat,Float)
#undef D
#undef C

#define N(f,g,h,a,m)\
  IL(f) :: forall e i s. (C(a,m)A.Ix i) => (i,i) -> e -> m (a i e);\
  f=A.newArray;\
  IL(g) :: forall e i s. (C(a,m)A.Ix i) => (i,i) -> m (a i e);\
  g=A.newArray_;\
  IL(h) :: forall e i s. (C(a,m)A.Ix i) => (i,i) -> [e] -> m (a i e);\
  h=A.newListArray
#define C(a,m)
N(newIOA,newIOA_,newIOAL,IOArray,IO)
N(newSTA,newSTA_,newSTAL,STArray s,ST s)
#undef C
#define C(a,m) MArray (a) e (m),
N(newIOUA,newIOUA_,newIOUAL,IOUArray,IO)
N(newSTUA,newSTUA_,newSTUAL,STUArray s,ST s)
#undef C
#undef N

#undef IL

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

Score : 200 points

Problem Statement

Niwango created a playlist of N songs.
The title and the duration of the i-th song are s_i and t_i seconds, respectively.
It is guaranteed that s_1,\ldots,s_N are all distinct.

Niwango was doing some work while playing this playlist. (That is, all the songs were played once, in the order they appear in the playlist, without any pause in between.)
However, he fell asleep during his work, and he woke up after all the songs were played.
According to his record, it turned out that he fell asleep at the very end of the song titled X.

Find the duration of time when some song was played while Niwango was asleep.

Constraints

1 \leq N \leq 50

s_i and X are strings of length between 1 and 100 (inclusive) consisting of lowercase English letters.

s_1,\ldots,s_N are distinct.

There exists an integer i such that s_i = X.

1 \leq t_i \leq 1000

t_i is an integer.

Input

Input is given from Standard Input in the following format:

N
s_1 t_1
\vdots
s_{N} t_N
X

Output

Print the answer.

Sample Input 1

3
dwango 2
sixth 5
prelims 25
dwango

Sample Output 1

30

While Niwango was asleep, two songs were played: sixth and prelims.

The answer is the total duration of these songs, 30.

Sample Input 2

1
abcde 1000
abcde

Sample Output 2

0

No songs were played while Niwango was asleep.

In such a case, the total duration of songs is 0.

Sample Input 3

15
ypnxn 279
kgjgwx 464
qquhuwq 327
rxing 549
pmuduhznoaqu 832
dagktgdarveusju 595
wunfagppcoi 200
dhavrncwfw 720
jpcmigg 658
wrczqxycivdqn 639
mcmkkbnjfeod 992
htqvkgkbhtytsz 130
twflegsjz 467
dswxxrxuzzfhkp 989
szfwtzfpnscgue 958
pmuduhznoaqu

Sample Output 3

6348

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.


main = putStrLn =<< (pure toAns) <*> getInts

getInts :: IO [Int]
getInts = (pure ((map read) . words)) <*> getLine

toAns :: [Int] -> String
toAns [k, x] = (if k * 500 >= x then "Yes" else "No")

Score : 100 points

Problem Statement

Takahashi has K 500-yen coins. (Yen is the currency of Japan.)
If these coins add up to X yen or more, print Yes; otherwise, print No.

Constraints

1 \leq K \leq 100

1 \leq X \leq 10^5

Input

Input is given from Standard Input in the following format:

K X

Output

If the coins add up to X yen or more, print Yes; otherwise, print No.

Sample Input 1

2 900

Sample Output 1

Yes

Two 500-yen coins add up to 1000 yen, which is not less than X = 900 yen.

Sample Input 2

1 501

Sample Output 2

No

One 500-yen coin is worth 500 yen, which is less than X = 501 yen.

Sample Input 3

4 2000

Sample Output 3

Yes

Four 500-yen coins add up to 2000 yen, which is not less than X = 2000 yen.


import Control.Monad
import Data.Maybe
import qualified Data.ByteString.Char8 as BS
import Data.List

readInt = fst . fromJust . BS.readInt
readIntList = map readInt . BS.words
getInt = readInt <$> BS.getLine
getIntList = readIntList <$> BS.getLine

main = do
    [n, m] <- getIntList
    as <- getIntList
    print $ solve as m

solve as m | all (== d) as' = (m `div` l + 1) `div` 2
           | otherwise = 0
    where as' = map div2 as
          d = head as'
          l = foldl lcm 1 $ map (`div` (2 ^ d)) as

div2 a | a `mod` 2 == 0 = 1 + div2 (a `div` 2)
       | otherwise = 0

Score : 400 points

Problem Statement

Given are a sequence A= {a_1,a_2,......a_N} of N positive even numbers, and an integer M.

Let a semi-common multiple of A be a positive integer X that satisfies the following condition for every k (1 \leq k \leq N):

There exists a non-negative integer p such that X= a_k \times (p+0.5).

Find the number of semi-common multiples of A among the integers between 1 and M (inclusive).

Constraints

1 \leq N \leq 10^5

1 \leq M \leq 10^9

2 \leq a_i \leq 10^9

a_i is an even number.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
a_1 a_2 ... a_N

Output

Print the number of semi-common multiples of A among the integers between 1 and M (inclusive).

Sample Input 1

2 50
6 10

Sample Output 1

2

15 = 6 \times 2.5

15 = 10 \times 1.5

45 = 6 \times 7.5

45 = 10 \times 4.5

Thus, 15 and 45 are semi-common multiples of A. There are no other semi-common multiples of A between 1 and 50, so the answer is 2.

Sample Input 2

3 100
14 22 40

Sample Output 2

0

The answer can be 0.

Sample Input 3

5 1000000000
6 6 2 6 2

Sample Output 3

166666667


main = do
    [s,t] <- words <$> getLine
    putStrLn $ t++s

Score : 100 points

Problem Statement

Given are two strings S and T consisting of lowercase English letters. Concatenate T and S in this order, without space in between, and print the resulting string.

Constraints

S and T are strings consisting of lowercase English letters.

The lengths of S and T are between 1 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

S T

Output

Print the resulting string.

Sample Input 1

oder atc

Sample Output 1

atcoder

When S = oder and T = atc, concatenating T and S in this order results in atcoder.

Sample Input 2

humu humu

Sample Output 2

humuhumu


main = putStrLn =<< sol <$> get

get = words <$> getLine

sol [s,t] = t++s


Score : 200 points

Problem Statement

Takahashi has A cookies, and Aoki has B cookies.
Takahashi will do the following action K times:

If Takahashi has one or more cookies, eat one of his cookies.

Otherwise, if Aoki has one or more cookies, eat one of Aoki's cookies.

If they both have no cookies, do nothing.

In the end, how many cookies will Takahashi and Aoki have, respectively?

Constraints

0 \leq A \leq 10^{12}

0 \leq B \leq 10^{12}

0 \leq K \leq 10^{12}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B K

Output

Print the numbers of Takahashi's and Aoki's cookies after K actions.

Sample Input 1

2 3 3

Sample Output 1

0 2

Takahashi will do the following:

He has two cookies, so he eats one of them.

Now he has one cookie left, and he eats it.

Now he has no cookies left, but Aoki has three, so Takahashi eats one of them.

Thus, in the end, Takahashi will have 0 cookies, and Aoki will have 2.

Sample Input 2

500000000000 500000000000 1000000000000

Sample Output 2

0 0

Watch out for overflows.


{-# LANGUAGE MagicHash #-}

import GHC.Exts
import GHC.Integer.GMP.Internals

main :: IO ()
main = do
  W# w# <- readLn :: IO Word
  print $ W# (nextPrimeWord# (minusWord# w# 1##))

Score: 300 points

Problem Statement

Find the minimum prime number greater than or equal to X.

Notes

A prime number is an integer greater than 1 that cannot be evenly divided by any positive integer except 1 and itself.

For example, 2, 3, and 5 are prime numbers, while 4 and 6 are not.

Constraints

2 \le X \le 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X

Output

Print the minimum prime number greater than or equal to X.

Sample Input 1

20

Sample Output 1

23

The minimum prime number greater than or equal to 20 is 23.

Sample Input 2

2

Sample Output 2

2

X itself can be a prime number.

Sample Input 3

99992

Sample Output 3

100003


{-# LANGUAGE BangPatterns #-}
{-# LANGUAGE BinaryLiterals #-}
{-# LANGUAGE MultiWayIf #-}
{-# LANGUAGE NumDecimals #-}
{-# LANGUAGE ScopedTypeVariables #-}
{-# LANGUAGE TupleSections #-}

import Control.Applicative
import Control.Arrow
import Control.Monad
import Control.Monad.ST

import Data.Char
import Data.List
import Data.Maybe

import qualified Data.ByteString.Char8 as B

import Data.Array.ST.Safe
import Data.STRef

import Debug.Trace
import Text.Printf

readInt = readLn :: IO Int
readInteger = readLn :: IO Integer
readInts = map ( fst . fromJust . B.readInt ) . B.words <$> B.getLine
readIntegers = map ( fst . fromJust . B.readInteger ) . B.words <$> B.getLine

which a b f = if f then a else b
mp [ a, b ] = ( a, b )

modifyArray a i f = writeArray a i =<< f <$> readArray a i

main = do
	x <- readInt
	print $ head $ filter isPrime [ x .. ]

isPrime x = all ( ( /= 0 ) . ( x `mod` ) ) ps
	where
		ps = takeWhile ( \i -> i * i <= x ) [ 2 .. ]


--{-# LANGUAGE BangPatterns #-}
--{-# LANGUAGE ScopedTypeVariables #-}
--{-# LANGUAGE FlexibleContexts #-}
--{-# LANGUAGE ViewPatterns #-}
--{-# LANGUAGE OverloadedStrings #-}
--{-# LANGUAGE Strict #-}
--{-# LANGUAGE NumericUnderscores #-}
--{-# LANGUAGE BlockArguments #-}
{-# LANGUAGE MultiWayIf #-}
{-# LANGUAGE CPP #-}

import qualified Data.Vector as V
import qualified Data.Vector.Unboxed as VU
import qualified Data.Vector.Unboxed.Mutable as VUM
import qualified Data.ByteString.Char8 as BC
import qualified Data.Map.Strict as MPS
import qualified Data.Map as MP
import qualified Data.Set as ST
import qualified Data.Sequence as SQ
import Data.Array
-- import Data.Array.Unboxed
import Data.Array.ST
import Data.Array.IO
import Data.Maybe
import Data.Ord
import Data.Ratio
import Data.Char
import Data.Bits
import Data.List
import Data.Tuple
import Data.Functor
import Data.IORef
import Data.STRef
import Control.Applicative
import Control.Monad
import Control.Monad.ST
import Control.Monad.State.Strict
import Debug.Trace

#if __GLASGOW_HASKELL__ >= 808
import qualified Data.OrdPSQ as PSQ
import qualified Data.Vector.Algorithms.Merge as VAM
import qualified Data.Heap as HP
import Numeric.Extra
import Data.Tuple.Extra
import Data.List.Extra

data OrdQ = Ascending | Descending
#endif

---------------------------------------------------------------------------
point hand (r,s,p)
    | hand=='r' = p
    | hand=='s' = r
    | hand=='p' = s

g :: (Int, Int, Int) -> BC.ByteString -> Int -> Int -> Int
g triple xs acc ind = acc + point x triple
    where
        x = BC.index xs ind

f :: (Int, Int, Int) -> Int -> BC.ByteString -> (Int,Bool) -> Int -> (Int,Bool)
f triple k xs (acc,branch) ind
    | x/=y = (acc',False)
    | x==y && branch = (acc',False)
    | otherwise = (acc,True)
    where
        x = BC.index xs ind
        y = BC.index xs (ind-k)
        acc' = acc + point x triple

solve :: (Int, Int) -> (Int, Int, Int) -> BC.ByteString -> Int
solve (n,k) triple xs = base + fst calc
    where
        calc = foldl' (f triple k xs) (0,False) [k..n-1]
        base = foldl' (g triple xs) 0 [0..k-1]

main=do
    [n,k]<-sLineToIntL
    [r,s,p]<-sLineToIntL
    xs<-strBS
    print $ solve (n,k) (r,s,p) xs
--------------------------------------------------------------------------

-- Input Util
int :: IO Int
int = bsToInt <$> BC.getLine

double :: IO Double
double = bsToDouble <$> BC.getLine

str :: IO String
str = BC.unpack <$> BC.getLine

strBS :: IO BC.ByteString
strBS = BC.getLine

-- convert
bsToInt :: BC.ByteString -> Int
bsToInt = fst . fromJust . BC.readInt

bsToDouble :: BC.ByteString -> Double
bsToDouble = read . BC.unpack

bsToIntTuple :: BC.ByteString -> (Int, Int)
bsToIntTuple bs = (a,b)
    where
        Just (a,bs') = parseInt bs
        Just (b,_) = parseInt bs'

bsToIntTriple :: BC.ByteString -> (Int, Int, Int)
bsToIntTriple bs = (a,b,c)
    where
        Just (a,bs') = parseInt bs
        Just (b,bs'') = parseInt bs'
        Just (c,_) = parseInt bs''

parseInt :: BC.ByteString -> Maybe (Int, BC.ByteString)
parseInt = BC.readInt . BC.dropWhile isSpace

-- single line
fromSLineL :: (BC.ByteString -> a) -> IO [a]
fromSLineL f =  map f . BC.words <$> strBS

-- from single line to List / VU.Vector
sLineToIntL :: IO [Int]
sLineToIntL = fromSLineL bsToInt

sLineToDoubleL :: IO [Double]
sLineToDoubleL = fromSLineL bsToDouble

sLineToStrL :: IO [String]
sLineToStrL = fromSLineL BC.unpack

sLineToStrBSL :: IO [BC.ByteString]
sLineToStrBSL = fromSLineL id

sLineToIntV :: Int -> IO (VU.Vector Int)
sLineToIntV n = VU.unfoldrN n parseInt <$> BC.getLine

sLineToDoubleV :: IO (VU.Vector Double)
sLineToDoubleV = VU.fromList . map bsToDouble . BC.words <$> BC.getLine

-- multiple lines
fromMLinesL :: Int -> (BC.ByteString -> a) -> IO [a]
fromMLinesL n f = replicateM n (f <$> strBS)

fromMLinesV :: VU.Unbox a => Int -> (BC.ByteString -> a) -> IO (VU.Vector a)
fromMLinesV n f = VU.replicateM n (f <$> strBS)

-- from multiple lines to List / VU.Vector
mLinesToIntL :: Int -> IO [Int]
mLinesToIntL n = fromMLinesL n bsToInt

mLinesToDoubleL :: Int -> IO [Double]
mLinesToDoubleL n = fromMLinesL n bsToDouble

mLinesToStrL :: Int -> IO [String]
mLinesToStrL n = fromMLinesL n BC.unpack

mLinesToStrBSL :: Int -> IO [BC.ByteString]
mLinesToStrBSL n = fromMLinesL n id

mLinesToTupleL :: Int -> IO [(Int, Int)]
mLinesToTupleL n = fromMLinesL n bsToIntTuple

mLinesToTripleL :: Int -> IO [(Int, Int, Int)]
mLinesToTripleL n = fromMLinesL n bsToIntTriple

mLinesToIntV :: Int -> IO (VU.Vector Int)
mLinesToIntV n = fromMLinesV n bsToInt

mLinesToDoubleV :: Int -> IO (VU.Vector Double)
mLinesToDoubleV n = fromMLinesV n bsToDouble

mLinesToTupleV :: Int -> IO (VU.Vector (Int, Int))
mLinesToTupleV n = fromMLinesV n bsToIntTuple

mLinesToTripleV :: Int -> IO (VU.Vector (Int, Int, Int))
mLinesToTripleV n = fromMLinesV n bsToIntTriple

-- output
yesnoL :: Bool -> IO ()
yesnoL cond
    | cond = putStrLn "Yes"
    | otherwise = putStrLn "No"

yesnoU :: Bool -> IO ()
yesnoU cond
    | cond = putStrLn "YES"
    | otherwise = putStrLn "NO"

-- others
strToChr :: String -> Char
strToChr [x] = x

chrToStr :: Char -> String
chrToStr x = [x]

m1 :: Int -> Int
m1 = subtract 1

toDouble :: Int -> Double
toDouble = fromIntegral

toInt :: Double -> Int
toInt = floor

sqToList :: SQ.Seq a -> [a]
sqToList sq =
    case SQ.viewl sq of
        l SQ.:< sq' -> l : sqToList sq'
        SQ.EmptyL -> []

#if __GLASGOW_HASKELL__ >= 808
sortV :: (Ord a, VU.Unbox a) => VU.Vector a -> VU.Vector a
sortV v = VU.create $ do
    w <- VU.thaw v
    VAM.sort w
    return w

buildPQL :: Ord a => OrdQ -> [a] -> PSQ.OrdPSQ Int Int a
buildPQL oq xs =
    case oq of
        Ascending -> f $ zip3 [1..l] [1..l] $ sort xs
        Descending -> f $ zip3  [1..l] (map negate [1..l]) $ sort xs
    where
        l = length xs
        f = foldl' g PSQ.empty
        g acc (k,p,v) = PSQ.insert k p v acc

buildPQV :: (Ord a, VU.Unbox a) => OrdQ -> VU.Vector a -> PSQ.OrdPSQ Int Int a
buildPQV oq xs =
    case oq of
        Ascending -> f $ VU.zip3 (VU.generate l (+1)) (VU.generate l (+1)) $ sortV xs
        Descending -> f $ VU.zip3  (VU.generate l (+1)) (VU.generate l $ negate . (+1)) $ sortV xs
    where
        l = VU.length xs
        f = VU.foldl' g PSQ.empty
        g acc (k,p,v) = PSQ.insert k p v acc
#endif

Score : 400 points

Problem Statement

At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:

The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)

Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):

R points for winning with Rock;

S points for winning with Scissors;

P points for winning with Paper.

However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)

Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.

The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is r, the machine will play Rock in the i-th round. Similarly, p and s stand for Paper and Scissors, respectively.

What is the maximum total score earned in the game by adequately choosing the hand to play in each round?

Notes

In this problem, Rock Paper Scissors can be thought of as a two-player game, in which each player simultaneously forms Rock, Paper, or Scissors with a hand.

If a player chooses Rock and the other chooses Scissors, the player choosing Rock wins;

if a player chooses Scissors and the other chooses Paper, the player choosing Scissors wins;

if a player chooses Paper and the other chooses Rock, the player choosing Paper wins;

if both players play the same hand, it is a draw.

Constraints

2 \leq N \leq 10^5

1 \leq K \leq N-1

1 \leq R,S,P \leq 10^4

N,K,R,S, and P are all integers.

|T| = N

T consists of r, p, and s.

Input

Input is given from Standard Input in the following format:

N K
R S P
T

Output

Print the maximum total score earned in the game.

Sample Input 1

5 2
8 7 6
rsrpr

Sample Output 1

27

The machine will play {Rock, Scissors, Rock, Paper, Rock}.

We can, for example, play {Paper, Rock, Rock, Scissors, Paper} against it to earn 27 points.
We cannot earn more points, so the answer is 27.

Sample Input 2

7 1
100 10 1
ssssppr

Sample Output 2

211

Sample Input 3

30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr

Sample Output 3

4996


ans :: Int -> Int -> Int -> Int
ans n a b
 	| diff `mod` 2 == 0 = diff `div` 2
	| a' > (n - b')     = n - a' + diff `div` 2 + 1
	| otherwise         = b' + diff `div` 2 + 1
	where
		a' = max a b
		b' = min a b
		diff = a' - b'

main :: IO ()
main = do
	[n, a, b] <- map read . words <$> getLine :: IO [Int]
	print $ ans (n - 1) (a - 1) (b - 1)

Score : 300 points

Problem Statement

2N players are running a competitive table tennis training on N tables numbered from 1 to N.

The training consists of rounds.
In each round, the players form N pairs, one pair per table.
In each pair, competitors play a match against each other.
As a result, one of them wins and the other one loses.

The winner of the match on table X plays on table X-1 in the next round,
except for the winner of the match on table 1 who stays at table 1.

Similarly, the loser of the match on table X plays on table X+1 in the next round,
except for the loser of the match on table N who stays at table N.

Two friends are playing their first round matches on distinct tables A and B.
Let's assume that the friends are strong enough to win or lose any match at will.
What is the smallest number of rounds after which the friends can get to play a match against each other?

Constraints

2 \leq N \leq 10^{18}

1 \leq A < B \leq N

All input values are integers.

Input

Input is given from Standard Input in the following format:

N A B

Output

Print the smallest number of rounds after which the friends can get to play a match against each other.

Sample Input 1

5 2 4

Sample Output 1

1

If the first friend loses their match and the second friend wins their match, they will both move to table 3 and play each other in the next round.

Sample Input 2

5 2 3

Sample Output 2

2

If both friends win two matches in a row, they will both move to table 1.


{-# LANGUAGE BangPatterns #-}
{-# LANGUAGE BinaryLiterals #-}
{-# LANGUAGE MultiWayIf #-}
{-# LANGUAGE NumDecimals #-}
{-# LANGUAGE ScopedTypeVariables #-}
{-# LANGUAGE TupleSections #-}

import Control.Applicative
import Control.Arrow
import Control.Monad
import Control.Monad.ST

import Data.Char
import Data.List
import Data.Maybe

import qualified Data.ByteString.Char8 as B

import Data.Array.ST.Safe
import Data.STRef

import Debug.Trace
import Text.Printf

readInt = readLn :: IO Int
readInteger = readLn :: IO Integer
readInts = map ( fst . fromJust . B.readInt ) . B.words <$> B.getLine
readIntegers = map ( fst . fromJust . B.readInteger ) . B.words <$> B.getLine

which a b f = if f then a else b
mp [ a, b ] = ( a, b )

modifyArray a i f = writeArray a i =<< f <$> readArray a i

main = readInts >>= print . uncurry lcm . mp

Score : 300 points

Problem Statement

Takahashi is organizing a party.

At the party, each guest will receive one or more snack pieces.

Takahashi predicts that the number of guests at this party will be A or B.

Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted.

We assume that a piece cannot be divided and distributed to multiple guests.

Constraints

1 \leq A, B \leq 10^5

A \neq B

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B

Output

Print the minimum number of pieces that can be evenly distributed to the guests in both of the cases with A guests and B guests.

Sample Input 1

2 3

Sample Output 1

6

When we have six snack pieces, each guest can take three pieces if we have two guests, and each guest can take two if we have three guests.

Sample Input 2

123 456

Sample Output 2

18696

Sample Input 3

100000 99999

Sample Output 3

9999900000


module Main
  ( main
  )
where

import           Control.Applicative
import           Control.Monad
import           Data.List
import           Data.Array
import           Data.Char
import           Data.Bits
import qualified Data.ByteString.Char8 as BS
import Data.Maybe

parse = fst . fromJust
parseInt = parse . BS.readInt
parseInteger = parse . BS.readInteger
readInt = parseInt <$> BS.getLine
readInteger = parseInteger <$> BS.getLine
readIntList = map parseInt . BS.words <$> BS.getLine

main = do
  n <- readInteger
  print $ if n `mod` 2 == 1 then 0 else count n 10 5

count n d m
  | n < d = 0
  | otherwise = n `div` d + count n (d * m) m

Score : 500 points

Problem Statement

For an integer n not less than 0, let us define f(n) as follows:

f(n) = 1 (if n < 2)

f(n) = n f(n-2) (if n \geq 2)

Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N).

Constraints

0 \leq N \leq 10^{18}

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of trailing zeros in the decimal notation of f(N).

Sample Input 1

12

Sample Output 1

1

f(12) = 12 × 10 × 8 × 6 × 4 × 2 = 46080, which has one trailing zero.

Sample Input 2

5

Sample Output 2

0

f(5) = 5 × 3 × 1 = 15, which has no trailing zeros.

Sample Input 3

1000000000000000000

Sample Output 3

124999999999999995


main :: IO ()
main = do
  l1 <- getLine
  let n = read l1 :: Integer
  if odd n then
    print 0
  else
    print $ f (n`div`2)

f 0 = 0
f n = n' + f n'
  where n' = n`div`5


module Main
  ( main
  )
where

import           Control.Applicative
import           Control.Monad
import           Data.List
import           Data.Array.Unboxed
import           Data.Char
import           Data.Bits
import qualified Data.ByteString.Char8 as BS
import Data.Maybe

parse = fst . fromJust
parseInt = parse . BS.readInt
parseInteger = parse . BS.readInteger
readInt = parseInt <$> BS.getLine
readInteger = parseInteger <$> BS.getLine
readIntList = map parseInt . BS.words <$> BS.getLine

main = do
  [n, u, v] <- readIntList
  arr <- concat <$> replicateM (n - 1) ((\[a, b] -> [(a, b), (b, a)]) <$> readIntList)
  let graph = createUnweightedGraphFromList n arr
      dist1 = depthFirstSearchTree u graph
      dist2 = depthFirstSearchTree v graph
      farthest = maximum . map snd . filter (\(a, b) -> a < b) $ zip dist1 dist2
  print $ farthest - 1

type Vertex = Int
type Cost = Int
type Capacity = Int

data Edge = Edge { from :: Vertex, to :: Vertex, cost :: Cost, cap :: Capacity } deriving (Eq, Ord, Show)
createUnweightedEdge (from, to) = Edge { from=from, to=to, cost=1, cap=0 }
createUnweightedEdgeArrayFromList:: [(Vertex, Vertex)] -> Array Vertex Edge
createUnweightedEdgeArrayFromList arr = listArray (1, length arr) . map createUnweightedEdge $ arr

type Graph = Array Vertex (Array Int Edge)
createUnweightedGraphFromList::Int -> [(Vertex, Vertex)] -> Graph
createUnweightedGraphFromList n = listArray (1, n) . map createUnweightedEdgeArrayFromList . groupBy (\a b -> fst a == fst b) . sort

depthFirstSearchTree::Vertex -> Graph -> [Cost]
depthFirstSearchTree src = map snd . sort . depthFirstSearchTreeUpdate (src, src, 0)

depthFirstSearchTreeUpdate::(Vertex, Vertex, Cost) -> Graph -> [(Vertex, Cost)]
depthFirstSearchTreeUpdate (v, p, c) graph = let newVisited = concat . map (\e -> depthFirstSearchTreeUpdate (to e, from e, c + cost e) graph) . filter (\e -> to e /= p) . elems $ graph ! v in newVisited ++ [(v, c)]

Score : 600 points

Problem Statement

We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.

Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.

Now, they will play a game of tag as follows:

1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.

2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.

3. Go back to step 1.

Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.

Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.

It can be proved that the game is bound to end.

Constraints

2 \leq N \leq 10^5

1 \leq u,v \leq N

u \neq v

1 \leq A_i,B_i \leq N

The given graph is a tree.

Input

Input is given from Standard Input in the following format:

N u v
A_1 B_1
:
A_{N-1} B_{N-1}

Output

Print the number of moves Aoki will perform before the end of the game.

Sample Input 1

5 4 1
1 2
2 3
3 4
3 5

Sample Output 1

2

If both players play optimally, the game will progress as follows:

Takahashi moves to Vertex 3.

Aoki moves to Vertex 2.

Takahashi moves to Vertex 5.

Aoki moves to Vertex 3.

Takahashi moves to Vertex 3.

Here, Aoki performs two moves.

Note that, in each move, it is prohibited to stay at the current vertex.

Sample Input 2

5 4 5
1 2
1 3
1 4
1 5

Sample Output 2

1

Sample Input 3

2 1 2
1 2

Sample Output 3

0

Sample Input 4

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

Sample Output 4

5


import qualified Data.ByteString.Char8 as BS
import Data.Maybe

readInt = fst . fromJust . BS.readInt

readIntList = map readInt . BS.words

getIntList = readIntList <$> BS.getLine

main = do
  as <- getIntList
  putStrLn $ if sum as >= 22 then "bust" else "win"

Score : 100 points

Problem Statement

Given are three integers A_1, A_2, and A_3.

If A_1+A_2+A_3 is greater than or equal to 22, print bust; otherwise, print win.

Constraints

1 \leq A_i \leq 13 \ \ (i=1,2,3)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A_1 A_2 A_3

Output

If A_1+A_2+A_3 is greater than or equal to 22, print bust; otherwise, print win.

Sample Input 1

5 7 9

Sample Output 1

win

5+7+9=21, so print win.

Sample Input 2

13 7 2

Sample Output 2

bust

13+7+2=22, so print bust.


import Data.List

toInt :: String -> Int
toInt = read :: String -> Int

main = do
  a <- map toInt . words <$> getLine
  putStrLn $ if sum a > 21 then "bust" else "win"


main = do
  [a,b,c] <- map read.words <$> getLine
  putStrLn $ if a+b+c>=22 then "bust" else "win"


main = do
  s <- getLine
  print $ (length $ filter (==False) $ zipWith (==) s $ reverse s) `div` 2

Score : 200 points

Problem Statement

Takahashi loves palindromes. Non-palindromic strings are unacceptable to him. Each time he hugs a string, he can change one of its characters to any character of his choice.

Given is a string S. Find the minimum number of hugs needed to make S palindromic.

Constraints

S is a string consisting of lowercase English letters.

The length of S is between 1 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

S

Output

Print the minimum number of hugs needed to make S palindromic.

Sample Input 1

redcoder

Sample Output 1

1

For example, we can change the fourth character to o and get a palindrome redooder.

Sample Input 2

vvvvvv

Sample Output 2

0

We might need no hugs at all.

Sample Input 3

abcdabc

Sample Output 3

2


{-# LANGUAGE CPP #-}
module Main where

import qualified Data.Vector as V
import qualified Data.Map.Strict as M
import Control.Monad
import qualified Data.ByteString.Char8 as BS
import Data.Maybe

dump :: (Show a) => a -> IO()
#if __GLASGOW_HASKELL__ >= 800
dump x = print x
#else
dump _ = pure ()
#endif

-- "101010" -> [1,0,1,0,1,0]
readToList :: String -> [Int]
readToList [] = []
readToList (x:xs) = read [x] : readToList xs

-- "10 10 10" -> [10, 10, 10]
readvec :: IO [Int]
readvec = (fmap (read . BS.unpack) . BS.words) <$> BS.getLine

newtype UF a = UF (M.Map a a) deriving (Show) -- Map node parent

makeuf :: Int -> UF Int
makeuf n = let m = M.fromList $ zip [0..n] [0..n]
                in UF m

union :: (Eq a, Ord a) => a -> a -> UF a -> UF a
union l r uf@(UF m) = let lp = find l uf
                          rp = find r uf
                      in if lp == rp
                         then uf
                         else if size l uf < size r uf
                              then UF $ M.update (const $ Just rp) lp m
                              else UF $ M.update (const $ Just lp) rp m

find :: (Eq a, Ord a) => a -> UF a -> a
find x uf@(UF m) = let px = m M.! x
                in if px == x
                   then px
                   else find px uf -- TODO 貼り直し

size :: (Eq a, Ord a) => a -> UF a -> Int
size x (UF m) = M.size $ M.filter (\par -> par==x) m

main :: IO()
main = do
  s <- getLine
  let s'= take (length s `div` 2) s
  let t = take (length s `div` 2) $ reverse s
  dump s'
  dump t
  print $ length $ filter (\(x, y) -> x /= y) $ zip s' t


{-# LANGUAGE BangPatterns #-}
import Control.Monad
import Data.Maybe
import Debug.Trace
import Data.List
import qualified Data.Map.Strict as Map
import qualified Data.IntMap.Strict as IntMap
import qualified Data.Set as Set
import qualified Data.IntSet as IntSet
import Data.Functor
import Data.Array
-- import Data.Array.Unboxed
import Control.Monad.ST
import Data.Array.ST
import Data.Char

main :: IO ()
main = getLine >>= print . solve
solve :: String -> Int
solve s = length . filter id $ zipWith (/=) (take m s) (reverse s)
  where
    n = length s
    m = n `div` 2 + if even n then 0 else 1


{-# LANGUAGE BangPatterns #-}
{-# LANGUAGE CPP #-}
module Main where

import qualified Data.Vector as V
import qualified Data.Map.Strict as M
import Control.Monad
import qualified Data.ByteString.Char8 as BS
import Data.Maybe
import Data.Bits
import Data.List (foldl')
dump :: (Show a) => a -> IO()
#if __GLASGOW_HASKELL__ >= 800
dump x = print x
#else
dump _ = pure ()
#endif

-- "101010" -> [1,0,1,0,1,0]
readToList :: String -> [Int]
readToList [] = []
readToList (x:xs) = read [x] : readToList xs

-- "10 10 10" -> [10, 10, 10]
readvec :: IO [Int]
readvec = (fmap (read . BS.unpack) . BS.words) <$> BS.getLine

newtype UF a = UF (M.Map a a) deriving (Show) -- Map node parent

makeuf :: Int -> UF Int
makeuf n = let m = M.fromList $ zip [0..n] [0..n]
                in UF m

union :: (Eq a, Ord a) => a -> a -> UF a -> UF a
union l r uf@(UF m) = let lp = find l uf
                          rp = find r uf
                      in if lp == rp
                         then uf
                         else if size l uf < size r uf
                              then UF $ M.update (const $ Just rp) lp m
                              else UF $ M.update (const $ Just lp) rp m

find :: (Eq a, Ord a) => a -> UF a -> a
find x uf@(UF m) = let px = m M.! x
                in if px == x
                   then px
                   else find px uf -- TODO 貼り直し

size :: (Eq a, Ord a) => a -> UF a -> Int
size x (UF m) = M.size $ M.filter (\par -> par==x) m

base :: Int
base = 10 ^ 9 + 7

forn :: (Num a, Enum a, Monad m) => a -> (a -> m b) -> m [b]
forn x = forM [0..(x-1)]

count :: (Eq a) => a -> [a] -> Int
count y = length . filter (== y)

main :: IO()
main = do
  [n] <- readvec
  xs <- readvec
  ress <- forn 60 $ \offset -> do
            let a = shift 1 offset
            let ys = map (a .&.) xs
            let zeros = count 0 $ ys
            let ones = length ys - zeros
            return $ zeros * ones
  print $ foldr (\x r -> (r*2+x) `mod` base) 0 ress

Score : 400 points

Problem Statement

We have N integers. The i-th integer is A_i.

Find \sum_{i=1}^{N-1}\sum_{j=i+1}^{N} (A_i \mbox{ XOR } A_j), modulo (10^9+7).

What is \mbox{ XOR }?

The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:

When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.

For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)

Constraints

2 \leq N \leq 3 \times 10^5

0 \leq A_i < 2^{60}

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 the value \sum_{i=1}^{N-1}\sum_{j=i+1}^{N} (A_i \mbox{ XOR } A_j), modulo (10^9+7).

Sample Input 1

3
1 2 3

Sample Output 1

6

We have (1\mbox{ XOR } 2)+(1\mbox{ XOR } 3)+(2\mbox{ XOR } 3)=3+2+1=6.

Sample Input 2

10
3 1 4 1 5 9 2 6 5 3

Sample Output 2

237

Sample Input 3

10
3 14 159 2653 58979 323846 2643383 27950288 419716939 9375105820

Sample Output 3

103715602

Print the sum modulo (10^9+7).


main=getContents>>=print.(\x->if x then 0 else 1).(\[x,y]->head x==head y).map words.lines

Score: 100 points

Problem Statement

In this problem, a date is written as Y-M-D. For example, 2019-11-30 means November 30, 2019.

Integers M_1, D_1, M_2, and D_2 will be given as input.

It is known that the date 2019-M_2-D_2 follows 2019-M_1-D_1.

Determine whether the date 2019-M_1-D_1 is the last day of a month.

Constraints

Both 2019-M_1-D_1 and 2019-M_2-D_2 are valid dates in the Gregorian calendar.

The date 2019-M_2-D_2 follows 2019-M_1-D_1.

Input

Input is given from Standard Input in the following format:

M_1 D_1
M_2 D_2

Output

If the date 2019-M_1-D_1 is the last day of a month, print 1; otherwise, print 0.

Sample Input 1

11 16
11 17

Sample Output 1

0

November 16 is not the last day of a month.

Sample Input 2

11 30
12 1

Sample Output 2

1

November 30 is the last day of November.


main = do
 a <- getLine
 let h = g.f. (read::String->Double) $ a
 --print h
 if(h /= -1) then print h else putStr ":("

f :: Double -> [Int]
f x = [if((x==).fromIntegral.floor.(1.08*).fromIntegral.ceiling $ moto) then
	   ceiling moto else -1 ,
       if((x==).fromIntegral.floor.(1.08*).fromIntegral.floor $   moto) then
       floor moto else -1]
  where
    moto = x /1.08

g :: [Int] -> Int
g (-1: (-1) :xs)= -1
g (-1:a:xs) =  a
g (a:_) =  a
g _ = -3

Score: 200 points

Problem Statement

Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it.

The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer).

Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer.

If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.

Constraints

1 \leq N \leq 50000

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

If there are values that could be X, the price of the apple pie before tax, print any one of them.

If there are multiple such values, printing any one of them will be accepted.

If no value could be X, print :(.

Sample Input 1

432

Sample Output 1

400

If the apple pie is priced at 400 yen before tax, you have to pay 400 \times 1.08 = 432 yen to buy one.

Otherwise, the amount you have to pay will not be 432 yen.

Sample Input 2

1079

Sample Output 2

:(

There is no possible price before tax for which you have to pay 1079 yen with tax.

Sample Input 3

1001

Sample Output 3

927

If the apple pie is priced 927 yen before tax, by rounding down 927 \times 1.08 = 1001.16, you have to pay 1001 yen.


isIn :: String -> String -> Bool
isIn (x:y:[z]) = (/=[]) . dropWhile (/=z) . dropCut (/=y) . dropCut (/=x)

dropCut cond xs
    | null remains = []
    | otherwise    = tail remains
  where
    remains = dropWhile cond xs

main = do
    let candidates = (\a b c -> [a, b, c]) <$> ['0'..'9'] <*> ['0'..'9'] <*> ['0'..'9']
    _ <- getLine
    s <- getLine

    print $ length $ filter (`isIn` s) candidates

Score: 400 points

Problem Statement

AtCoder Inc. has decided to lock the door of its office with a 3-digit PIN code.

The company has an N-digit lucky number, S. Takahashi, the president, will erase N-3 digits from S and concatenate the remaining 3 digits without changing the order to set the PIN code.

How many different PIN codes can he set this way?

Both the lucky number and the PIN code may begin with a 0.

Constraints

4 \leq N \leq 30000

S is a string of length N consisting of digits.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the number of different PIN codes Takahashi can set.

Sample Input 1

4
0224

Sample Output 1

3

Takahashi has the following options:

Erase the first digit of S and set 224.

Erase the second digit of S and set 024.

Erase the third digit of S and set 024.

Erase the fourth digit of S and set 022.

Thus, he can set three different PIN codes: 022, 024, and 224.

Sample Input 2

6
123123

Sample Output 2

17

Sample Input 3

19
3141592653589793238

Sample Output 3

329


import Data.List
import Data.Maybe

answer::String -> Int
main = answer <$> getLine >>= print

answer s = 7 - (fromJust $ elemIndex s ["SUN","MON","TUE","WED","THU","FRI","SAT" ])

Score : 100 points

Problem Statement

Given is a string S representing the day of the week today.

S is SUN, MON, TUE, WED, THU, FRI, or SAT, for Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, and Saturday, respectively.

After how many days is the next Sunday (tomorrow or later)?

Constraints

S is SUN, MON, TUE, WED, THU, FRI, or SAT.

Input

Input is given from Standard Input in the following format:

S

Output

Print the number of days before the next Sunday.

Sample Input 1

SAT

Sample Output 1

1

It is Saturday today, and tomorrow will be Sunday.

Sample Input 2

SUN

Sample Output 2

7

It is Sunday today, and seven days later, it will be Sunday again.


import Data.Char

main = do
    n <- readLn
    s <- getLine
    putStrLn $ map (f n . ord) s

f n k = chr $ (k+n-65) `mod` 26 + 65

Score : 200 points

Problem Statement

We have a string S consisting of uppercase English letters. Additionally, an integer N will be given.

Shift each character of S by N in alphabetical order (see below), and print the resulting string.

We assume that A follows Z. For example, shifting A by 2 results in C (A \to B \to C), and shifting Y by 3 results in B (Y \to Z \to A \to B).

Constraints

0 \leq N \leq 26

1 \leq |S| \leq 10^4

S consists of uppercase English letters.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the string resulting from shifting each character of S by N in alphabetical order.

Sample Input 1

2
ABCXYZ

Sample Output 1

CDEZAB

Note that A follows Z.

Sample Input 2

0
ABCXYZ

Sample Output 2

ABCXYZ

Sample Input 3

13
ABCDEFGHIJKLMNOPQRSTUVWXYZ

Sample Output 3

NOPQRSTUVWXYZABCDEFGHIJKLM


{-# LANGUAGE BangPatterns #-}
{-# LANGUAGE BinaryLiterals #-}
{-# LANGUAGE MultiWayIf #-}
{-# LANGUAGE NumDecimals #-}
{-# LANGUAGE ScopedTypeVariables #-}
{-# LANGUAGE TupleSections #-}

import Control.Applicative
import Control.Arrow
import Control.Monad
import Control.Monad.ST

import Data.Char
import Data.List
import Data.Maybe

import qualified Data.ByteString.Char8 as B

import Data.Array.ST.Safe
import Data.STRef

import Debug.Trace
import Text.Printf

readInt = readLn :: IO Int
readInteger = readLn :: IO Integer
readInts = map ( fst . fromJust . B.readInt ) . B.words <$> B.getLine
readIntegers = map ( fst . fromJust . B.readInteger ) . B.words <$> B.getLine

which a b f = if f then a else b
mp [ a, b ] = ( a, b )

modifyArray a i f = writeArray a i =<< f <$> readArray a i

main = do
	m <- readInt
	[ ds, cs ] <- transpose <$> replicateM m readIntegers
	let
		digit_num = sum cs
		digit_sum = sum $ zipWith (*) ds cs
		end_dsum = let m = digit_sum `mod` 9 in if m == 0 then 9 else m
        print $ ( digit_num - 1 ) + ( ( digit_sum - end_dsum ) `div` 9 )

Score: 500 points

Problem Statement

N programmers are going to participate in the preliminary stage of DDCC 20XX. Due to the size of the venue, however, at most 9 contestants can participate in the finals.

The preliminary stage consists of several rounds, which will take place as follows:

All the N contestants will participate in the first round.

When X contestants participate in some round, the number of contestants advancing to the next round will be decided as follows:

The organizer will choose two consecutive digits in the decimal notation of X, and replace them with the sum of these digits. The number resulted will be the number of contestants advancing to the next round.

For example, when X = 2378, the number of contestants advancing to the next round will be 578 (if 2 and 3 are chosen), 2108 (if 3 and 7 are chosen), or 2315 (if 7 and 8 are chosen).

When X = 100, the number of contestants advancing to the next round will be 10, no matter which two digits are chosen.

The preliminary stage ends when 9 or fewer contestants remain.

Ringo, the chief organizer, wants to hold as many rounds as possible.
Find the maximum possible number of rounds in the preliminary stage.

Since the number of contestants, N, can be enormous, it is given to you as two integer sequences d_1, \ldots, d_M and c_1, \ldots, c_M, which means the following: the decimal notation of N consists of c_1 + c_2 + \ldots + c_M digits, whose first c_1 digits are all d_1, the following c_2 digits are all d_2, \ldots, and the last c_M digits are all d_M.

Constraints

1 \leq M \leq 200000

0 \leq d_i \leq 9

d_1 \neq 0

d_i \neq d_{i+1}

c_i \geq 1

2 \leq c_1 + \ldots + c_M \leq 10^{15}

Input

Input is given from Standard Input in the following format:

M
d_1 c_1
d_2 c_2
:
d_M c_M

Output

Print the maximum possible number of rounds in the preliminary stage.

Sample Input 1

2
2 2
9 1

Sample Output 1

3

In this case, N = 229 contestants will participate in the first round. One possible progression of the preliminary stage is as follows:

229 contestants participate in Round 1, 49 contestants participate in Round 2, 13 contestants participate in Round 3, and 4 contestants advance to the finals.

Here, three rounds take place in the preliminary stage, which is the maximum possible number.

Sample Input 2

3
1 1
0 8
7 1

Sample Output 2

9

In this case, 1000000007 will participate in the first round.


{-# LANGUAGE Strict #-}

main :: IO ()
main = do
  r <- readLn
  print $ solve r

solve :: Int -> Int
solve r = r * r

Score : 100 points

Problem Statement

Given is an integer r.

How many times is the area of a circle of radius r larger than the area of a circle of radius 1?

It can be proved that the answer is always an integer under the constraints given.

Constraints

1 \leq r \leq 100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

r

Output

Print the area of a circle of radius r, divided by the area of a circle of radius 1, as an integer.

Sample Input 1

2

Sample Output 1

4

The area of a circle of radius 2 is 4 times larger than the area of a circle of radius 1.

Note that output must be an integer - for example, 4.0 will not be accepted.

Sample Input 2

100

Sample Output 2

10000


main = do
  r <- readLn :: IO Int
  print $ r * r


-- {-# LANGUAGE BangPatterns #-}
-- {-# LANGUAGE ScopedTypeVariables #-}
-- {-# LANGUAGE FlexibleContexts #-}
-- {-# LANGUAGE ViewPatterns #-}
-- {-# LANGUAGE OverloadedStrings #-}

import qualified Data.Vector as V
import qualified Data.Vector.Unboxed as VU
import qualified Data.Vector.Unboxed.Mutable as VUM
-- import qualified Data.Vector.Algorithms.Merge as VAM
-- import qualified Data.Heap as HP
import qualified Data.ByteString.Char8 as BC
import qualified Data.Map as MP
import qualified Data.Set as ST
import qualified Data.Sequence as SQ
import Data.Array
import Data.Array.Unboxed
import Data.Array.ST
import Data.Array.IO
import Data.Maybe
import Data.Ratio
import Data.Char
import Data.Bits
import Data.List
import Data.Tuple
import Data.Functor
import Data.IORef
import Data.STRef
import Control.Applicative
import Control.Monad
import Control.Monad.ST
import Debug.Trace

--------------------------------------------------------------------------
solve :: [(Int, Int)] -> Double
solve [_] = 0
solve (a:b:xs) = res + solve (b:xs)
    where
        res = sqrt $ (toDouble $ ((x1-x2)^2) + ((y1-y2)^2))
        (x1,y1) = a
        (x2,y2) = b

main = do
    n <- int
    xys <- multipleLinesToTupleL n
    let pss = permutations xys
    print $ (sum [solve ps | ps <- pss]) / toDouble (length pss)

--------------------------------------------------------------------------

-- Input Util
int :: IO Int
int = readLn

double :: IO Double
double = readLn

str :: IO String
str = getLine

strBC :: IO BC.ByteString
strBC = BC.getLine

singleLineToIntL :: IO [Int]
singleLineToIntL = getLine >>= return . map read . words

singleLineToDoubleL :: IO [Double]
singleLineToDoubleL = getLine >>= return . map read . words

singleLineToIntV :: Int -> IO (VU.Vector Int)
singleLineToIntV n = BC.getLine >>= return . VU.unfoldrN n (BC.readInt . BC.dropWhile isSpace)

singleLineToDoubleV :: Int -> IO (VU.Vector Double)
singleLineToDoubleV n = BC.getLine >>= return . VU.fromList . map (read . BC.unpack) . BC.words

multipleLinesToIntL :: Int -> IO [Int]
multipleLinesToIntL n = replicateM n readLn

multipleLinesToDoubleL :: Int -> IO [Double]
multipleLinesToDoubleL n = replicateM n readLn

multipleLinesToIntV :: Int -> IO (VU.Vector Int)
multipleLinesToIntV n = VU.replicateM n readLn

multipleLinesToDoubleV :: Int -> IO (VU.Vector Double)
multipleLinesToDoubleV n = VU.replicateM n (BC.getLine >>= return . read . BC.unpack)

multipleLinesToTupleL :: Int -> IO [(Int, Int)]
multipleLinesToTupleL n = replicateM n (getLine >>= return . (\[a,b] -> (a,b)) . map read . words)

multipleLinesToTupleV :: Int -> IO (VU.Vector (Int, Int))
multipleLinesToTupleV n = VU.replicateM n $ do
    bs <- strBC
    let
        Just (a, bs') = BC.readInt $ BC.dropWhile isSpace bs
        Just (b, _) = BC.readInt $ BC.dropWhile isSpace bs'
    return (a,b)

multipleLinesToTripleV :: Int -> IO (VU.Vector (Int, Int, Int))
multipleLinesToTripleV n = VU.replicateM n $ do
    bs <- strBC
    let
        Just (a, bs') = BC.readInt $ BC.dropWhile isSpace bs
        Just (b, bs'') = BC.readInt $ BC.dropWhile isSpace bs'
        Just (c, _) = BC.readInt $ BC.dropWhile isSpace bs''
    return (a,b,c)

-- others
strToChr :: String -> Char
strToChr [x] = x

chrToStr :: Char -> String
chrToStr x = [x]

m1 :: Int -> Int
m1 x = subtract 1 x

toDouble :: Int -> Double
toDouble x = fromIntegral x

toInt :: Double -> Int
toInt x = floor x

Score : 300 points

Problem Statement

There are N towns in a coordinate plane. Town i is located at coordinates (x_i, y_i). The distance between Town i and Town j is \sqrt{\left(x_i-x_j\right)^2+\left(y_i-y_j\right)^2}.

There are N! possible paths to visit all of these towns once. Let the length of a path be the distance covered when we start at the first town in the path, visit the second, third, \dots, towns, and arrive at the last town (assume that we travel in a straight line from a town to another). Compute the average length of these N! paths.

Constraints

2 \leq N \leq 8

-1000 \leq x_i \leq 1000

-1000 \leq y_i \leq 1000

\left(x_i, y_i\right) \neq \left(x_j, y_j\right) (if i \neq j)

(Added 21:12 JST) All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
:
x_N y_N

Output

Print the average length of the paths.
Your output will be judges as correct when the absolute difference from the judge's output is at most 10^{-6}.

Sample Input 1

3
0 0
1 0
0 1

Sample Output 1

2.2761423749

There are six paths to visit the towns: 1 → 2 → 3, 1 → 3 → 2, 2 → 1 → 3, 2 → 3 → 1, 3 → 1 → 2, and 3 → 2 → 1.

The length of the path 1 → 2 → 3 is \sqrt{\left(0-1\right)^2+\left(0-0\right)^2} + \sqrt{\left(1-0\right)^2+\left(0-1\right)^2} = 1+\sqrt{2}.

By calculating the lengths of the other paths in this way, we see that the average length of all routes is:

\frac{\left(1+\sqrt{2}\right)+\left(1+\sqrt{2}\right)+\left(2\right)+\left(1+\sqrt{2}\right)+\left(2\right)+\left(1+\sqrt{2}\right)}{6} = 2.276142...

Sample Input 2

2
-879 981
-866 890

Sample Output 2

91.9238815543

There are two paths to visit the towns: 1 → 2 and 2 → 1. These paths have the same length.

Sample Input 3

8
-406 10
512 859
494 362
-955 -475
128 553
-986 -885
763 77
449 310

Sample Output 3

7641.9817824387


import           Control.Monad
import           Data.List
import           Data.Maybe
import qualified Data.ByteString.Char8 as B

type Coord = (Double, Double)

main :: IO ()
main = solve <$> getParams >>= print

getParams :: IO [Coord]
getParams = readLn >>= flip replicateM getCoord

getCoord :: IO Coord
getCoord = do
  [x, y] <- map readInt . B.words <$> B.getLine
  return (fromIntegral x, fromIntegral y)

readInt :: B.ByteString -> Int
readInt = fst . fromJust . B.readInt

solve :: [Coord] -> Double
solve = average . map pathLength . permutations

pathLength :: [Coord] -> Double
pathLength xs = sum . zipWith distance xs $ tail xs

distance :: Coord -> Coord -> Double
distance (xi, yi) (xj, yj) = sqrt $ (xi - xj)^2 + (yi - yj)^2

average :: Fractional a => [a] -> a
average xs = sum xs / genericLength xs


import qualified Data.Text as T
import qualified Data.Text.IO as T
import qualified Data.Text.Read as T
import qualified Data.Array as A
import Data.List (permutations)

main :: IO ()
main = do
  (n : _) <- map unsafeSignedDecimal . T.words <$> T.getLine :: IO [Int]
  xys <- (A.listArray (1, n) <$>) . sequence . replicate n $ unsafeListToTuple . map (fromIntegral . unsafeSignedDecimal) . T.words <$> T.getLine :: IO (A.Array Int (Double, Double))
  let
    ds = A.array ((1, 1), (n, n)) . map (\(i, j) -> ((i, j), dist (xys A.! i) (xys A.! j))) $ (,) <$> [1..n] <*> [1..n] :: A.Array (Int, Int) Double
    numerator = sum . map (sum . map (ds A.!) . (zip <$> id <*> tail)) . permutations $ [1..n] :: Double
    denominator = fromIntegral . product $ [2..n] :: Double
  print $ numerator / denominator

dist :: (Double, Double) -> (Double, Double) -> Double
dist (x0, y0) (x1, y1) = sqrt $ (x1 - x0) ^ 2 + (y1 - y0) ^ 2

unsafeListToTuple :: [a] -> (a, a)
unsafeListToTuple (x : y : _) = (x, y)

unsafeSignedDecimal :: T.Text -> Int
unsafeSignedDecimal s = case T.signed T.decimal s of
  Right (n, _) -> n


main=do
 n<-readLn
 print$(n`div`2)-(if odd n then 0 else 1)

Score : 100 points

Problem Statement

How many ways are there to choose two distinct positive integers totaling N, disregarding the order?

Constraints

1 \leq N \leq 10^6

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the answer.

Sample Input 1

4

Sample Output 1

1

There is only one way to choose two distinct integers totaling 4: to choose 1 and 3. (Choosing 3 and 1 is not considered different from this.)

Sample Input 2

999999

Sample Output 2

499999


import Control.Monad
import Data.Maybe
import qualified Data.ByteString.Char8 as BS
import Data.List

readInt = fst . fromJust . BS.readInt
readIntList = map readInt . BS.words
getInt = readInt <$> BS.getLine
getIntList = readIntList <$> BS.getLine

main = do
    n <- getInt
    ds <- getIntList
    print $ solve ds

solve [] = 0
solve (d:ds)
    | d /= 0 || map head ds' /= [1..length ds'] = 0
    | otherwise = productMod $ zipWith powerMod ds'' (tail ds'')
    where ds' = group $ sort ds
          ds'' = map length ds'

moduloN :: Int
moduloN = 998244353
modN = flip mod moduloN

mulMod x y = modN $ x * y
productMod xs = foldl' mulMod 1 xs

powerMod x 0 = 1
powerMod x y
    | even y = mulMod t t
    | otherwise = mulMod x $ mulMod t t
    where
        t = powerMod x $ div y 2

Score : 300 points

Problem Statement

Given is an integer sequence D_1,...,D_N of N elements. Find the number, modulo 998244353, of trees with N vertices numbered 1 to N that satisfy the following condition:

For every integer i from 1 to N, the distance between Vertex 1 and Vertex i is D_i.

Notes

A tree of N vertices is a connected undirected graph with N vertices and N-1 edges, and the distance between two vertices are the number of edges in the shortest path between them.

Two trees are considered different if and only if there are two vertices x and y such that there is an edge between x and y in one of those trees and not in the other.

Constraints

1 \leq N \leq 10^5

0 \leq D_i \leq N-1

Input

Input is given from Standard Input in the following format:

N
D_1 D_2 ... D_N

Output

Print the answer.

Sample Input 1

4
0 1 1 2

Sample Output 1

2

For example, a tree with edges (1,2), (1,3), and (2,4) satisfies the condition.

Sample Input 2

4
1 1 1 1

Sample Output 2

0

Sample Input 3

7
0 3 2 1 2 2 1

Sample Output 3

24


import Control.Applicative
import Data.List

m = 998244353
modm n = mod n m

mul' x y = modm $ modm x * modm y

power' x 0 = 1
power' x 1 = modm x
power' x n
  | (n `mod` 2) == 0 = let i = power' x (n `div` 2) in mul' i i
  | otherwise = mul' x $ power' x (n - 1)

div' x y = mul' x $ power' y (m - 2)

product' xs = foldl mul' 1 xs

factorial' n = product' [1..n]

npr' n r = product' [n - i | i <- [0..(r - 1)]]

ncr' n r = modm $ (npr' n r) `div'` (factorial' r)

main = do
  [n] <- map read <$> words <$> getLine :: IO [Int]
  xs <- map read <$> words <$> getLine :: IO [Int]
  --let n = 4
  --let xs = [0,1,1,2]
  let xs' = group $ sort xs
  let ls = map length xs'
  --print $ head xs'
  --if (head xs /= 1) then
  --  print 0
  --else
  if and (zipWith (==) (nub xs) [0..]) then
    print $ foldl1' (mul') $ zipWith (power') (init ls) (tail ls)
  else
    print 0


import Data.Char
import Data.Maybe
import Data.List
import qualified Data.ByteString.Char8 as B
import qualified Data.Vector.Unboxed as V

readInt = B.readInt . B.dropWhile isSpace

getInt = fst . fromJust . readInt <$> B.getLine

getIntList = unfoldr readInt <$> B.getLine

main = do
  n <- getInt
  ds <- map (\s -> (head s, length s)) . group . sort <$> getIntList
  let len = length ds
  let es = V.accum (+) (V.replicate len 0) ds
  if V.all (/= 0) es
    then print . V.foldr (((`mod` modInt) .) . (*)) 1 . V.zipWith (((`mod` modInt) .) . (^)) es $ V.tail es
    else print 0

modInt = 998244353


main = do
    n <- readLn
    putStrLn $ solve n

lp = [x*y|x<-[1..9],y<-[x..9]]

solve :: Int -> String
solve n = search n lp
    where search n [] = "No"
          search n (x:xs)
            | n == x = "Yes"
            | otherwise = search n xs

Score : 200 points

Problem Statement

Having learned the multiplication table, Takahashi can multiply two integers between 1 and 9 (inclusive) together.

Given an integer N, determine whether N can be represented as the product of two integers between 1 and 9. If it can, print Yes; if it cannot, print No.

Constraints

1 \leq N \leq 100

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

If N can be represented as the product of two integers between 1 and 9 (inclusive), print Yes; if it cannot, print No.

Sample Input 1

10

Sample Output 1

Yes

10 can be represented as, for example, 2 \times 5.

Sample Input 2

50

Sample Output 2

No

50 cannot be represented as the product of two integers between 1 and 9.

Sample Input 3

81

Sample Output 3

Yes


main = readLn >>= putStrLn . solve

solve n = if any (\a -> 1 <= n `div` a && n `div` a <= 9 && n `mod` a == 0) [1..9]
          then "Yes"
          else "No"


isPrime :: Int -> Bool
isPrime n = loop n 2
    where
        loop :: Int -> Int -> Bool
        loop n i
            | n <= 1 = False
            | n `rem` i == 0 = False
            | n < i * i = True
            | otherwise = loop n (i + 1)

solve :: Int -> Int
solve n
    | isPrime n = n - 1
    | s * s == n = s * 2 -2
    | otherwise = s * 2 - 1
    where
        s :: Int
        s = floor . sqrt . fromIntegral $ n

main = do
    n <- readLn :: IO Int
    print $ solve n

Score : 300 points

Problem Statement

Takahashi is standing on a multiplication table with infinitely many rows and columns.

The square (i,j) contains the integer i \times j. Initially, Takahashi is standing at (1,1).

In one move, he can move from (i,j) to either (i+1,j) or (i,j+1).

Given an integer N, find the minimum number of moves needed to reach a square that contains N.

Constraints

2 \leq N \leq 10^{12}

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the minimum number of moves needed to reach a square that contains the integer N.

Sample Input 1

10

Sample Output 1

5

(2,5) can be reached in five moves. We cannot reach a square that contains 10 in less than five moves.

Sample Input 2

50

Sample Output 2

13

(5, 10) can be reached in 13 moves.

Sample Input 3

10000000019

Sample Output 3

10000000018

Both input and output may be enormous.


main :: IO ()
main = do
    [a,b,x] <- map read . words <$> getLine :: IO [Double]
    let y = 2 * x / b / a
    print $ 90 - ((atan (y / b)) * 180 / pi)

Score : 400 points

Problem Statement

Takahashi has a water bottle with the shape of a rectangular prism whose base is a square of side a~\mathrm{cm} and whose height is b~\mathrm{cm}. (The thickness of the bottle can be ignored.)

We will pour x~\mathrm{cm}^3 of water into the bottle, and gradually tilt the bottle around one of the sides of the base.

When will the water be spilled? More formally, find the maximum angle in which we can tilt the bottle without spilling any water.

Constraints

All values in input are integers.

1 \leq a \leq 100

1 \leq b \leq 100

1 \leq x \leq a^2b

Input

Input is given from Standard Input in the following format:

a b x

Output

Print the maximum angle in which we can tilt the bottle without spilling any water, in degrees.
Your output will be judged as correct when the absolute or relative error from the judge's output is at most 10^{-6}.

Sample Input 1

2 2 4

Sample Output 1

45.0000000000

This bottle has a cubic shape, and it is half-full. The water gets spilled when we tilt the bottle more than 45 degrees.

Sample Input 2

12 21 10

Sample Output 2

89.7834636934

This bottle is almost empty. When the water gets spilled, the bottle is nearly horizontal.

Sample Input 3

3 1 8

Sample Output 3

4.2363947991

This bottle is almost full. When the water gets spilled, the bottle is still nearly vertical.


import Data.List

main = interact main'
main' = show . f . head . reverse . lines

first (a, b) = a

f :: String -> Integer
f str =
    first $ foldl' g init str
    where
        g (a, c) n = (a + (if c /= n then 1 else 0), n)
        init = (0, ' ')

Score : 300 points

Problem Statement

There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.

Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.

Ultimately, how many slimes will be there?

Constraints

1 \leq N \leq 10^5

|S| = N

S consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the final number of slimes.

Sample Input 1

10
aabbbbaaca

Sample Output 1

5

Ultimately, these slimes will fuse into abaca.

Sample Input 2

5
aaaaa

Sample Output 2

1

All the slimes will fuse into one.

Sample Input 3

20
xxzaffeeeeddfkkkkllq

Sample Output 3

10


main :: IO ()
main = do
    getLine
    solve . map read . words <$> getLine >>= print

solve :: [Int] -> Int
solve []     = 0
solve (a:ls) = solve' a ls + solve ls

solve' :: Int -> [Int] -> Int
solve' _ []   = 0
solve' a (b:cs) = length [1 | c <- filter (< a + b) cs, b + c > a && c + a > b] + solve' a cs

Score : 400 points

Problem Statement

Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.

He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:

a < b + c

b < c + a

c < a + b

How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Constraints

All values in input are integers.

3 \leq N \leq 2 \times 10^3

1 \leq L_i \leq 10^3

Input

Input is given from Standard Input in the following format:

N
L_1 L_2 ... L_N

Constraints

Print the number of different triangles that can be formed.

Sample Input 1

4
3 4 2 1

Sample Output 1

1

Only one triangle can be formed: the triangle formed by the first, second, and third sticks.

Sample Input 2

3
1 1000 1

Sample Output 2

0

No triangles can be formed.

Sample Input 3

7
218 786 704 233 645 728 389

Sample Output 3

23


{-# LANGUAGE BangPatterns #-}
{-# LANGUAGE BinaryLiterals #-}
{-# LANGUAGE MultiWayIf #-}
{-# LANGUAGE NumDecimals #-}
{-# LANGUAGE ScopedTypeVariables #-}
{-# LANGUAGE TupleSections #-}

import Control.Applicative
import Control.Arrow
import Control.Monad
import Control.Monad.ST

import Data.Char
import Data.List
import Data.Maybe

import qualified Data.ByteString.Char8 as B

import Data.Array.ST.Safe
import Data.STRef

import Debug.Trace
import Text.Printf

readInt = readLn :: IO Int
readInteger = readLn :: IO Integer
readInts = map ( fst . fromJust . B.readInt ) . B.words <$> B.getLine
readIntegers = map ( fst . fromJust . B.readInteger ) . B.words <$> B.getLine

which a b f = if f then a else b
mp [ a, b ] = ( a, b )

modifyArray a i f = writeArray a i =<< f <$> readArray a i

main = do
	n <- readInt
	let
		odds = length $ filter odd [ 1 .. n ]
	printf "%.10f\n" $ ( fromIntegral odds :: Double ) / fromIntegral n

Score : 100 points

Problem Statement

Given is an integer N.

Takahashi chooses an integer a from the positive integers not greater than N with equal probability.

Find the probability that a is odd.

Constraints

1 \leq N \leq 100

Input

Input is given from Standard Input in the following format:

N

Output

Print the probability that a is odd.
Your output will be considered correct when its absolute or relative error from the judge's output is at most 10^{-6}.

Sample Input 1

4

Sample Output 1

0.5000000000

There are four positive integers not greater than 4: 1, 2, 3, and 4. Among them, we have two odd numbers: 1 and 3. Thus, the answer is \frac{2}{4} = 0.5.

Sample Input 2

5

Sample Output 2

0.6000000000

Sample Input 3

1

Sample Output 3

1.0000000000


main = do
 [n,k] <- words <$> getLine
 s_s <- words <$> getLine
 print $ length (filter (k<=) s_s)

Score : 200 points

Problem Statement

N friends of Takahashi has come to a theme park.

To ride the most popular roller coaster in the park, you must be at least K centimeters tall.

The i-th friend is h_i centimeters tall.

How many of the Takahashi's friends can ride the roller coaster?

Constraints

1 \le N \le 10^5

1 \le K \le 500

1 \le h_i \le 500

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
h_1 h_2 \ldots h_N

Output

Print the number of people among the Takahashi's friends who can ride the roller coaster.

Sample Input 1

4 150
150 140 100 200

Sample Output 1

2

Two of them can ride the roller coaster: the first and fourth friends.

Sample Input 2

1 500
499

Sample Output 2

0

Sample Input 3

5 1
100 200 300 400 500

Sample Output 3

5


main = do
  li <- getLine
  let [n,k] = map read $ words li
  li <- getLine
  let hs = map read $ words li
  let ans = compute n k hs
  print ans

compute :: Int -> Int -> [Int] -> Int
compute _ k hs = length $ filter (k <=) hs


main :: IO ()
main = do
    [_, k] <- map (read :: String -> Int) . words <$> getLine
    xs <- map (read :: String -> Int) . words <$> getLine
    print . length . filter (>= k) $ xs


import Data.List
import Control.Monad

main :: IO ()
main = do
 n <- map read . words <$> getLine :: IO [Int]
 as <- map read . words <$> getLine :: IO [Int]

 forM_ (map snd $ sortOn fst $ zip as [1..length as]) $ \a -> putStr ((show a) ++ " ")

Score : 300 points

Problem Statement

Takahashi is a teacher responsible for a class of N students.

The students are given distinct student numbers from 1 to N.

Today, all the students entered the classroom at different times.

According to Takahashi's record, there were A_i students in the classroom when student number i entered the classroom (including student number i).

From these records, reconstruct the order in which the students entered the classroom.

Constraints

1 \le N \le 10^5

1 \le A_i \le N

A_i \neq A_j (i \neq j)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \ldots A_N

Output

Print the student numbers of the students in the order the students entered the classroom.

Sample Input 1

3
2 3 1

Sample Output 1

3 1 2

First, student number 3 entered the classroom.

Then, student number 1 entered the classroom.

Finally, student number 2 entered the classroom.

Sample Input 2

5
1 2 3 4 5

Sample Output 2

1 2 3 4 5

Sample Input 3

8
8 2 7 3 4 5 6 1

Sample Output 3

8 2 4 5 6 7 3 1


Score : 400 points

Problem Statement

Given are positive integers A and B.

Let us choose some number of positive common divisors of A and B.

Here, any two of the chosen divisors must be coprime.

At most, how many divisors can we choose?

Definition of common divisor

An integer d is said to be a common divisor of integers x and y when d divides both x and y.

Definition of being coprime

Integers x and y are said to be coprime when x and y have no positive common divisors other than 1.

Definition of dividing

An integer x is said to divide another integer y when there exists an integer \alpha such that y = \alpha x.

Constraints

All values in input are integers.

1 \leq A, B \leq 10^{12}

Input

Input is given from Standard Input in the following format:

A B

Output

Print the maximum number of divisors that can be chosen to satisfy the condition.

Sample Input 1

12 18

Sample Output 1

3

12 and 18 have the following positive common divisors: 1, 2, 3, and 6.

1 and 2 are coprime, 2 and 3 are coprime, and 3 and 1 are coprime, so we can choose 1, 2, and 3, which achieve the maximum result.

Sample Input 2

420 660

Sample Output 2

4

Sample Input 3

1 2019

Sample Output 3

1

1 and 2019 have no positive common divisors other than 1.


{-# OPTIONS_GHC -O2 #-}
{-# LANGUAGE BangPatterns          #-}
{-# LANGUAGE CPP                   #-}
{-# LANGUAGE LambdaCase            #-}
{-# LANGUAGE MagicHash             #-}
{-# LANGUAGE MultiParamTypeClasses #-}
{-# LANGUAGE MultiWayIf            #-}
{-# LANGUAGE OverloadedStrings     #-}
{-# LANGUAGE TupleSections         #-}
{-# LANGUAGE TypeFamilies          #-}
{-# LANGUAGE ViewPatterns          #-}
{-# LANGUAGE RecordWildCards       #-}

import           Control.Applicative
import           Control.Exception
import           Control.Monad
import           Control.Monad.Primitive
import           Control.Monad.ST
import           Control.Monad.State.Strict
import           Data.Bool
import qualified Data.ByteString             as B
import qualified Data.ByteString.Builder     as B
import qualified Data.ByteString.Char8       as C
import qualified Data.ByteString.Internal    as B
import qualified Data.ByteString.Unsafe      as B
import           Data.Char
import qualified Data.Foldable               as F
import           Data.Function
import qualified Data.IntMap.Strict          as IM
import qualified Data.IntSet                 as IS
import qualified Data.List                   as L
import qualified Data.Map.Strict             as M
import           Data.Monoid
import           Data.Ord
import           Data.Primitive.MutVar
import qualified Data.Set                    as S
import           Data.Tuple
import qualified Data.Vector                 as V
import qualified Data.Vector.Mutable         as VM
import qualified Data.Vector.Generic         as G
import qualified Data.Vector.Generic.Mutable as GM
import qualified Data.Vector.Unboxed         as U
import qualified Data.Vector.Unboxed.Mutable as UM
import           Debug.Trace
import           Foreign                     hiding (void)
import           GHC.Exts
import qualified System.IO                   as IO
import           Unsafe.Coerce

main :: IO ()
main = do
    [n, m] <- map read.words <$> getLine :: IO [Int]
    edges <- V.replicateM m $ do
        [!a, !b] <- map read.words <$> getLine :: IO [Int]
        xs <- U.unfoldrN b parseInt <$> C.getLine
        return (a, b, xs)
    print . maybe (-1) id $ solve n m edges

solve :: Int -> Int -> V.Vector (Int, Int, U.Vector Int) -> Maybe Int
solve n m keys
    | dist U.! dst == inf = Nothing
    | otherwise = Just $ dist U.! dst
  where
    src = 0
    dst = shiftL 1 n - 1
    dist = dijkstraDense src gr
    inf = 0x3f3f3f3f
    gr = runST $ do
        let ix i j = unsafeShiftL i n .|. j
        m <- UM.replicate (shiftL 1 $ 2 * n) inf
        rep (shiftL 1 n) $ \i -> do
            UM.unsafeWrite m (ix i i) 0
        V.forM_ keys $ \(cost, _, boxes) -> do
            let !key = U.foldl' (\acc i -> setBit acc (i - 1)) 0 boxes
            rep (shiftL 1 n) $ \s -> do
                let s' = s .|. key
                c <- UM.unsafeRead m (ix s s')
                when (cost < c) $ do
                    UM.unsafeWrite m (ix s s') cost
        DenseGraph (shiftL 1 n) <$> U.unsafeFreeze m

type Vertex = Int
type Cost = Int
data DenseGraph a = DenseGraph
    { numVerticesDG :: !Int
    , adjDG :: U.Vector a
    }
instance (U.Unbox a, Show a) => Show (DenseGraph a) where
    show DenseGraph{..} =
        let n = numVerticesDG
        in F.foldMap (<> "\n")
            $ V.generate n $ \i ->
                show . U.toList $ U.unsafeSlice (i * n) n adjDG

dijkstraDense :: Vertex -> DenseGraph Cost -> U.Vector Cost
dijkstraDense src DenseGraph{..} = U.create $ do
    let n = numVerticesDG
    let ix i j = i * n + j
    dist <- UM.replicate (n + 1) maxBound
    used <- UM.replicate n False
    UM.write dist src 0
    let nothing = n
    vars <- UM.replicate 2 0
    let [_v, _dv] = [0..1]
    fix $ \loop -> do
        UM.unsafeWrite vars _v nothing
        UM.unsafeWrite vars _dv maxBound
        rep n $ \i -> do
            UM.unsafeRead used i >>= \case
                False -> do
                    di <- UM.unsafeRead dist i
                    dv <- UM.unsafeRead vars _dv
                    when (di < dv) $ do
                        UM.unsafeWrite vars _v i
                        UM.unsafeWrite vars _dv di
                True -> return ()
        v <- UM.unsafeRead vars _v
        when (v /= nothing) $ do
            UM.unsafeWrite used v True
            dv <- UM.unsafeRead vars _dv
            rep n $ \i -> do
                let di' = dv + U.unsafeIndex adjDG (ix v i)
                UM.unsafeModify dist (min di') i
            loop
    return dist

-------------------------------------------------------------------------------

rep :: Monad m => Int -> (Int -> m ()) -> m ()
rep !n = U.forM_ $ U.generate n id
{-# INLINE rep #-}

rev :: Monad m => Int -> (Int -> m ()) -> m ()
rev !n = U.forM_ $ U.iterateN n (subtract 1) (n - 1)
{-# INLINE rev #-}

type Parser a = C.ByteString -> Maybe (a, C.ByteString)

parseInt :: Parser Int
parseInt = C.readInt . C.dropWhile isSpace

parseInt2 :: Parser (Int, Int)
parseInt2 = runStateT $
    (,) <$> StateT (C.readInt . C.dropWhile isSpace)
        <*> StateT (C.readInt . B.unsafeTail)

parseInt3 :: Parser (Int, Int, Int)
parseInt3 = runStateT $
    (,,) <$> StateT (C.readInt . C.dropWhile isSpace)
        <*> StateT (C.readInt . B.unsafeTail)
        <*> StateT (C.readInt . B.unsafeTail)

parseInt4 :: Parser (Int, Int, Int, Int)
parseInt4 = runStateT $
    (,,,) <$> StateT (C.readInt . C.dropWhile isSpace)
        <*> StateT (C.readInt . B.unsafeTail)
        <*> StateT (C.readInt . B.unsafeTail)
        <*> StateT (C.readInt . B.unsafeTail)

Score : 500 points

Problem Statement

We have N locked treasure boxes, numbered 1 to N.

A shop sells M keys. The i-th key is sold for a_i yen (the currency of Japan), and it can unlock b_i of the boxes: Box c_{i1}, c_{i2}, ..., c_{i{b_i}}. Each key purchased can be used any number of times.

Find the minimum cost required to unlock all the treasure boxes. If it is impossible to unlock all of them, print -1.

Constraints

All values in input are integers.

1 \leq N \leq 12

1 \leq M \leq 10^3

1 \leq a_i \leq 10^5

1 \leq b_i \leq N

1 \leq c_{i1} < c_{i2} < ... < c_{i{b_i}} \leq N

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
c_{11} c_{12} ... c_{1{b_1}}
:
a_M b_M
c_{M1} c_{M2} ... c_{M{b_M}}

Output

Print the minimum cost required to unlock all the treasure boxes.
If it is impossible to unlock all of them, print -1.

Sample Input 1

2 3
10 1
1
15 1
2
30 2
1 2

Sample Output 1

25

We can unlock all the boxes by purchasing the first and second keys, at the cost of 25 yen, which is the minimum cost required.

Sample Input 2

12 1
100000 1
2

Sample Output 2

-1

We cannot unlock all the boxes.

Sample Input 3

4 6
67786 3
1 3 4
3497 1
2
44908 3
2 3 4
2156 3
2 3 4
26230 1
2
86918 1
3

Sample Output 3

69942