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)

import Control.Monad
import Data.Char
import qualified Data.Sequence as S

main = do
    [n, k] <- map read <$> words <$> getLine
    s      <- getLine
    putStrLn $ show $ solve n k s

solve :: Int -> Int -> String -> Int
solve n k s = doit 1 (S.fromList $ replicate (k * 2 + 1) 0) 0 s
  where
    doit :: Int -> (S.Seq Int) -> Int -> String -> Int
    doit d l p []        = length s - S.index l 0
    doit 0 l p ('1' : t) = max (p - S.index l 0) (doit 1 (l          S.|> p) (p + 1) t)
    doit 1 l p ('0' : t) = max (p - S.index l 0) (doit 0 (S.drop 2 l S.|> p) (p + 1) t)
    doit d l p ( _  : t) = doit d l p t