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.

Examples

Input

5 1
00010


Output

4


Input

14 2
11101010110011


Output

8


Input

1 1
1


Output

1
#include <iostream>
#include <algorithm>
#include <string>

using namespace std;

int main()
{
	int n, k; cin >> n >> k;
	string s; cin >> s;

	int ans = 0;
	int c = 0; int l = 0;
	for (int r = 0; r < n; r++) {
		if (s[r] == '0') {
			// 2: 区間の数を数える
			if (r == 0 || s[r - 1] == '1') c++;
		}
		if (c > k) {
			// 3: 区間数が k を超えたところで "111...111000...000" を
			//    先頭から 1 ペアずつ取り除く (取り除いた数は l になる)
			while (s[l] == '1') l++;
			while (s[l] == '0') l++;

			c--;
		}

		// 1 : 制限に突き当たるまで貪欲にとる (まずは l は無視)
		// 4 : l があれば l 分だけ引く
		ans = max(ans, r - l + 1);
	}

	cout << ans << endl;

	return 0;
}
