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;
struct Node{
	int l,r;
}a[100005];
int n,m,k;
string s;
int main(){
	m = 0;
	scanf("%d%d",&n,&k);
	cin >> s;
	for (int i = 0;i < n;++i){
		if (s[i] == '0'){
			a[m].l = i;
			while(s[++i] == '0'){

			}
			i--;
			a[m++].r = i;
		}
	}
	a[m].l = n;
	if (m <= k) return 0 * printf("%d\n",n);
	int ans = a[k].l;
	//cout << ans << endl;
	for (int i = 1;i + k-1 < m;++i){
		//cout << a[i+k].l-1 << " " << a[i-1].r << endl;
		//cout << a[i+k].l - 1 - a[i-1].r << endl;
		ans = max(ans,a[i+k].l - 1 - a[i-1].r);
	}
	printf("%d\n",ans);
	return 0;
}
/*
19 3
0101010101001111001
*/
