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<vector>
#include<string>
#define int long long
using namespace std;
int a[123456], k; string s;
signed main() {
	cin >> k >> k >> s;
	int f = 0, l = 0;
	for (int i = 0; f < s.size(); i++) {
		int qw = (i+1) % 2;
		while (f < s.size() && s[f] - '0' == qw) f++;
		a[i] = f;
		l++;
	}
	a[l] = f; l++;
	if (l / 2 <= k) {
		cout << s.size() << endl;
		return 0;
	}
	k = k * 2 + 1;
	int ans = 0;
	for (int i = k - 1; i < l; i += 2) {
		if (i == k - 1)ans = max(ans, a[i]);
		else ans = max(ans, a[i] - a[i - k]);
	}
	cout << ans << endl;
}