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: Runtime Error)

def main():
    N, K = map(int, input().split())
    S = [int(i) for i in input()]

    pre = 1
    zeroCounter = -1
    conOneConter = 0
    conOneList = [0]
    for s in S:
        # 1 -> 1
        if pre == 1 and s == 1:
            conOneConter += 1
        # 1 -> 0
        elif pre == 1 and s == 0:
            # 0を発見したので, カウンターを増やす
            zeroCounter += 1
            # これまでの"1"の個数を記録する
            if len(conOneList) <= zeroCounter:
                conOneList.append(0)

            for i in range(zeroCounter - (K - 1), zeroCounter + 1):
                if i < 0:
                    continue
                conOneList[i] += conOneConter
            # この"0"を1としてカウントする
            conOneConter = 1
            pre = 0
        # 0 -> 0
        elif pre == 0 and s == 0:
            conOneConter += 1
        # 0 -> 1
        elif pre == 0 and s == 1:
            # ここまでの"0"の個数を追加する
            conOneList[zeroCounter] += conOneConter
            # 前の値にも, 今回のzeroConterにおける"0"or"1"の個数を追加
            if zeroCounter >= K - 1:
                for i in range(zeroCounter - (K - 1), zeroCounter):
                    conOneList[i] += conOneConter
            pre = 1
            # この"1"をカウントしておく
            conOneConter = 1

    index = (zeroCounter - 1) if pre == 0 else zeroCounter
    index = index if index > 0 else 0
    for i in range(index - (K - 1), index + 1):
        conOneList[i] += conOneConter
    print(max(conOneList))


if __name__ == '__main__':
    main()