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>
#define REP(i,n) for(int i=0;i<(n);++i)
using namespace std;

int main(){
	int n,k;
	cin>>n>>k;
	string s;
	cin>>s;
	
  	vector<int> v;
	bool jud=(s[0]=='0')?true:false;
	bool jud2=(s[n-1]=='0')?true:false;
	if(jud) v.push_back(0);
	int re=s[0];
	int cntt=1;
	for(int i=1;i<n;++i){
		if(s[i]==s[i-1]) cntt++;
		else{
			v.push_back(cntt);
			cntt=1;
		}
	}
	v.push_back(cntt);
	if(jud2) v.push_back(0);
	
	int sum=0;
	int ans=0;
	int l=0,r=0;
	while(r<v.size()){
		sum+=v[r];
		if(r-l+1>2*k+1){
			sum-=v[l]+v[l+1];
			l+=2;
		}
		r++;
		ans=max(ans,sum);
	}
	
	cout<<ans<<endl;
}//abc124_d.復習.