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 <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 1e5 + 100;

int n, k, cnt, ans, p = 1;
string s;
vector<int> vec;

int main() {
	ios::sync_with_stdio(false), cin.tie(0);
	cin >> n >> k >> s;
	cnt = (s[0] == '0');
	for (int i = 0; i < n; i++) {
		if(i && s[i] == '1' && s[i - 1] == '0')
			cnt--;
		while(p < n) {
			if(s[p - 1] == s[p])
				p++;
			else if(s[p] == '1')
				p++;
			else if(cnt < k)
				cnt++, p++;
			else
				break;
		}
		ans = max(ans, p - i);
	}
	cout << ans;
	return 0;
}
