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 Python (3.4.3) (judge verdict: Accepted)

import numpy as np

if __name__ == "__main__":
    N, K = map(int, input().split())
    S = input()
    T = [0]
    if(S[0] == '0') :
        T.append(0)
    T.append(1)
    for i in range(1, N):
        if(S[i] == S[i-1]):
            T[-1] += 1
        else:
            T.append(1)
    for i in range(3 * K):
        T.append(0)
    SIZE = len(T)
    S = [0] * SIZE
    for i in range(1, SIZE):
        S[i] = S[i-1] + T[i]
    #print(T)
    #print(S)
    ans = 0
    for i in range (1, SIZE, 2):
        if(i + 2 * K >= SIZE):
            break
        ans = max(ans, S[i + 2 * K] - S[i-1])
    print(ans)