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;

int main(void){
	int n, k;
	string s;
	cin >> n >> k >> s;
	vector<int> a;
	int prev = s[0];
	a.push_back(0);
	for(int i = 1; i < n; ++i){
		if(prev!=s[i]) a.push_back(i);
		prev = s[i];
	}
	a.push_back(n);
	int ans = 0;
	for(int i = 0; i < a.size(); ++i){
		int m = 2*k+1;
		if(s[a[i]] == '0'){
			m--;
		}
		ans = max(ans, ((m+i<a.size())?a[m+i]:n) - a[i]);
		// cout << a[m+i] - a[i] << endl;
	}
	cout << ans << endl;
	return 0;
}