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)

N, K = map(int, input().split())
S = list(input())
S = [int(S[i]) for i in range(N)]

def return_ans(S_initial, S):
    ans = 0
    block_size_list = []
    tmp_block_size = 1
    for i in range(1, N):
        if S[i - 1] == S[i]:
            tmp_block_size += 1

        else:
            block_size_list.append(tmp_block_size)
            tmp_block_size = 1
        if i == N - 1:
            block_size_list.append(tmp_block_size)

    times = len(block_size_list) - K * 2

    block_size_list.append(0)
    if K * 2 >= len(block_size_list):
        ans = len(S)
    else:
        if S_initial == 0:
            ans = max([sum(block_size_list[i:i + K * 2 + i % 2]) for i in range(times)])

        else:
            for i in range(times):
                ans = max([sum(block_size_list[i:i + K * 2 + (i + 1) % 2]) for i in range(times)])

    return ans

print(return_ans(S[0], S))