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 <algorithm>
using namespace std;
char str[100100] = {0};
int main(){
	int N,K;
	scanf("%d%d",&N,&K);
	scanf("%s",str+1);
	int now = 0,n = 1;
	int ans = 0;
	int tmp = 0;		
	if(str[1] == '0')now++;		
	for(int i = 1,j = 2;i <= N;i++){
		while(j <= N && now <= K){
			if(str[j] == '0'){
				if(str[j-1] == '0'){j ++ ;}
				else {
					now++;j++;
					if(now > K){now--;j--;break;}
				}
			}
			else{
				j++;
			}
		}
		ans = max(ans,j-i);
		if(str[i] == '0'){now--;}
		while(str[i] == str[i+1])i++;
	} 
	printf("%d\n",ans);
	return 0;
} 