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 main(){
	int N,K,ans=0;
	string S;
	cin >> N >> K >> S;
	vector<int> p;
	int c=1;
	if(S[0]=='0')p.push_back(0);
	for(int i=0;i<N-1;i++){
		if(S[i]!=S[i+1]){
			p.push_back(c);
			c=1;
		}
		else c++;
	}
	p.push_back(c);
	if(S[N-1]=='0')p.push_back(0);
	
	int l=0,r=0,sum=p[0];
	while(1){
		ans=max(ans,sum);
		if(l==p.size()-1 && r==p.size()-1)break;
		if(r<p.size()-1 && r-l<2*K){
			sum+=p[r+1]+p[r+2];
			r+=2;
		}
		else{
			if(l<p.size()-2){
				sum-=p[l]+p[l+1];
				l+=2;
			}
		}
	}
	cout << ans << endl;
	return 0;
}