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: Accepted)

module Main where

import Data.List
import Control.Monad
import Debug.Trace
import qualified Data.Vector as V
import Data.Maybe 
import qualified Data.Text as T
import qualified Data.Text.IO as TIO


makeGroup :: T.Text -> [Int]
makeGroup people = let 
    initialAccum = ([], 1, T.head people)
    (groupList, lastGroup, _) = T.foldl' inner initialAccum (T.tail people) 
    in 
        reverse $ 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 :: T.Text -> Int -> Int
solve people orderTime = 
        fst $ foldl' inner (firstGroup, firstGroup) indexes
    where
    group =  if T.head people == '0'
        then 
            0 : makeGroup people
        else 
            makeGroup people
    groupLength = length group
    groupArr = V.fromList group
    getGroup i = fromMaybe 0 $ groupArr V.!? i
    joinableGroupNum = orderTime * 2 + 1
    firstGroup = sum $ V.take joinableGroupNum groupArr 
    indexes = [x| x<- [2..groupLength], even x]
    inner (biggest, current) index = let
        minus = getGroup (index - 1) + getGroup (index - 2)
        plus = getGroup (index + joinableGroupNum - 1) + getGroup (index + joinableGroupNum - 2)
        next = current + plus - minus
        in
            (max biggest next, next)

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 <- TIO.getLine
    print $ solve people k