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<cstdio>
#include<cstring>
#include<iostream>
#include<cctype>
#include<iostream>
using namespace std;
const int N = 1e5 + 3;
char s[N];
int n,k,r;
int ans = 0;
int main(){
	scanf("%d%d",&n,&k);
	scanf("%s",s + 1);
	int l = 1,r = 1;
	int cnt = 0;
	while(1){
		while(r <= n && cnt < k){
			if(s[r] == '0'){
				while(r <= n && s[r] == '0') ++r;
				cnt++;
			}
			while(s[r] == '1') r++;
		}
	//	cout << r << endl;
		ans = max(ans,r - l);
		if(r > n) break;
		cnt--;
		while(s[l] != '0') l++;
		while(s[l] == '0') l++;
	}
	printf("%d\n",ans);
	return 0;	
}