Problem:
Score : 400 points

Problem Statement

N people are arranged in a row from left to right.

You are given a string S of length N consisting of 0 and 1, and a positive integer K.

The i-th person from the left is standing on feet if the i-th character of S is 0, and standing on hands if that character is 1.

You will give the following direction at most K times (possibly zero):

Direction: Choose integers l and r satisfying 1 \leq l \leq r \leq N, and flip the l-th, (l+1)-th, ..., and r-th persons. That is, for each i = l, l+1, ..., r, the i-th person from the left now stands on hands if he/she was standing on feet, and stands on feet if he/she was standing on hands.

Find the maximum possible number of consecutive people standing on hands after at most K directions.

Constraints

N is an integer satisfying 1 \leq N \leq 10^5.

K is an integer satisfying 1 \leq K \leq 10^5.

The length of the string S is N.

Each character of the string S is 0 or 1.

Input

Input is given from Standard Input in the following format:

N K
S

Output

Print the maximum possible number of consecutive people standing on hands after at most K directions.

Sample Input 1

5 1
00010

Sample Output 1

4

We can have four consecutive people standing on hands, which is the maximum result, by giving the following direction:

Give the direction with l = 1, r = 3, which flips the first, second and third persons from the left.

Sample Input 2

14 2
11101010110011

Sample Output 2

8

Sample Input 3

1 1
1

Sample Output 3

1

No directions are necessary.

Solution in Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

module Main where

import Data.List
import Control.Monad
import Debug.Trace
import Data.Array


makeGroup :: String -> [Int]
makeGroup people = let 
    initialAccum = ([], 1, head people)
    (groupList, lastGroup, _) = foldl' inner initialAccum (tail people) 
    in 
        lastGroup : groupList
        where
    inner :: ([Int], Int, Char) -> Char -> ([Int], Int, Char)
    inner (groupList, currentNumber, currentStat) person = 
        if person == currentStat then
            (groupList, currentNumber + 1, currentStat)
        else
            (currentNumber : groupList, 1, person)

solve :: String -> Int -> Int
solve people orderTime = 
        maximum $ takeGroup <$> startGroupList
    where
    group = makeGroup people
    groupLength = length group
    groupArr = listArray (0, groupLength - 1) group
    getGroup i = if i >= groupLength then 0 else
        groupArr ! i
    startGroupList = if head people == '0' then
        0: [x| x <- [0..groupLength-1], odd x]
    else
        [x| x <- [0..groupLength-1], even x]
    joinableGroupNum = orderTime * 2 + 1
    takeGroup 0 = 
        if head people == '0' then
            sum $ map getGroup [0..0+joinableGroupNum-2]
        else
            sum $ map getGroup [0..0+joinableGroupNum-1]
    takeGroup i = 
            sum $ map getGroup [i..i+joinableGroupNum-1]

readLs :: Read a => IO [a]
readLs = fmap read . words <$> getLine

readTuple :: Read a  => IO (a, a)
readTuple = do
    [f, s] <- readLs
    pure (f, s)


main = do
    (_, k) <- readTuple
    people <- getLine
    print $ solve people k