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 = input() + "3"
    s0 = []
    s1 = []
    cnt = 0
    start = 1
    for i in range(n+1):
        if start == 1 and s[i] != "1":
            s1.append(cnt)
            cnt = 1
            start = 0
        elif start == 0 and s[i] != "0":
            s0.append(cnt)
            cnt = 1
            start = 1
        else:
            cnt += 1
    if len(s0) == len(s1):
        s1.append(0)

    ans = sum(s1[0:0+k+1]) + sum(s0[0:0+k])
    n_ans = ans
    for i in range(len(s0) - k):
        n_ans = n_ans + s1[k+1+i] + s0[k+i] - s1[i] - s0[i]
        ans = max(ans, n_ans)
        
    print(ans)

main()