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,k;
string s;
vector <pair<int,int> > H;

int main(){
	cin>>n>>k>>s;
	for(int i=0;i<n;i++){
		if(s[i]=='0'){
			int j=i;
			for(;j<n&&s[j]=='0';++j);
			H.push_back(make_pair(i+1,j));
			i=j;
		}
	}
	int m=H.size();
	if(m<=k){
		printf("%d\n",n);
		return 0;
	}
	int ans=0;
	for(int i=m-1;i>=0;i--){
		int j=min(i+k-1,m-1);
		int L=(i==0?0:H[i-1].second);
		int R=(j==m-1?n+1:H[j+1].first);
		ans=max(ans,R-L-1);
	}
	printf("%d\n",ans);
	return 0;
}