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 n;
int k;
int d[100100];
int a[100100];
int sum[100100];
int main(){
	//freopen("input.txt","r",stdin);
	//freopen("output.txt","w",stdout);
	cin>>n>>k;
	for(int i=1;i<=n;i++)
		scanf("%1d",&d[i]);
	for(int i=1;i<=n;i++)
		a[i]=d[i]^d[i-1];
	for(int i=1;i<=n;i++)
		sum[i]=sum[i-1]+a[i];
	int b=0;
	int res=0;
	for(int i=1;i<=n;i++){
		b^=a[i];
		int tmp=upper_bound(sum+1,sum+n+1,(k*2-!b+sum[i]))-sum;
		res=max(res,tmp-i);
	}
	cout<<res<<endl;
	return 0;
}
