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: Wrong Answer)

def main():
    n, k = map(int, input().split())
    s = input()
    
    arr = []
    if s[0] == '0':
        arr.append(0)
    cnt = 1
    i = 1
    while i != n:
        if s[i] != s[i - 1]:
            arr.append(cnt)
            cnt = 1
        else:
            cnt += 1
        i += 1
    arr.append(cnt)
    if s[-1] == '0':
        arr.append(0)
    
    ans = 0
    curr = arr[0]
    j = 1
    for i in range(1, len(arr), 2):
        curr += arr[i] + arr[i + 1]
        if i - j == 2 * k:
            curr -= arr[j] + arr[j - 1]
            j += 2
        ans = max(ans, curr)
    print(ans)
    return 0
main()