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: Runtime Error)

import Data.List (group)

main :: IO ()
main = do
  [n,k] <- map (read :: String -> Int) . words <$> getLine
  (m,bs) <- app . (\k -> (n,k)) <$> getLine
  print $ ans (2*k+1) (m,bs)

ans :: Int -> (Int, [Int]) -> Int
ans l (m,bs) = if l >= m then sum bs
               else let bs' = scanl (+) 0 bs
               in maximum $ map (\k -> (mylist bs' (k+l)) - (mylist bs' k)) [0,2..]
  where mylist xs n = if length xs > n then 0 else xs !! n

app :: (Int,String) -> (Int, [Int])
app (n,s)
  | head s == '0' && last s == '0' = (n+2, ((++ [0]) . (0:) . map length . group) s)
  | head s == '0' = (n+1, ((0:) . map length . group) s)
  | last s == '0' = (n+1, ((++ [0]) . map length . group) s)
  | otherwise = (n, (map length . group) s)