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)

from functools import reduce


def main():
    n, k = tuple(map(int, input().split()))
    s = input().strip()
    start = 0
    count_zero = {}

    for i, c in enumerate(s):
        if c == "0":
            if i == start:
                count_zero[start] = 0

            count_zero[start] += 1
        elif c == "1":
            start = i + 1

    if len(count_zero) <= k:
        print(n)
        return

    count_zero[-1] = 1
    count_zero[n] = 1

    count_zero_list = sorted(count_zero.items(), key=lambda pair: pair[0])

    print(reduce(lambda ans, i:
        max(ans, count_zero_list[i + k + 1][0] - (count_zero_list[i][0] + count_zero_list[i][1])),
        range(0, len(count_zero) - k - 2), 0))


if __name__ == "__main__":
    main()