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)

def main():
    n, k = [int(i) for i in input().split()]
    s = [int(i) for i in list(input())]
    table = []
    now = 1
    cnt = 0
    for i in range(n):
        if s[i] == now:
            cnt += 1
        else:
            table.append(cnt)
            cnt = 1
            now ^= 1 # 0　と 1 の切り替え
    if cnt != 0:
        table.append(cnt)
    if len(table)%2 == 0:
        # 奇数で終わらせれば、1 で終わったことになる
        table.append(0)
    ans = 0
    l = 0
    d = min(len(table), 2*k+1)
    r = l + d
    tmp = sum(table[l:r])
    while r <= len(table):
        ans = max(ans, tmp)
        if l < len(table):
            tmp -= table[l]
        l += 1
        if l < len(table):
            tmp -= table[l]
        l += 1
        r += 1
        if r <= len(table):
            tmp += table[r-1]
        r += 1
        if r <= len(table):
            tmp += table[r-1]
    print(ans)


if __name__ == '__main__':
    main()