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 <iostream>
#include <algorithm>
#include <vector>
#include <string>
using namespace std;

int main(){
	int N,K;
	string S;
	cin>>N>>K>>S;
	vector<int> c;
	c.push_back(0);
	for(int i=1;i<N;i++){
		if(S[i-1]!=S[i]) c.push_back(i);
	}
	int R=c.size();
	int ans=0;
	for(int i=0;i<R;i++){
		if(S[c[i]]=='1')
			if(i+2*K+1<R){
				ans=max(ans,c[i+2*K+1]-c[i]);
			}
			else ans=max(ans,N-c[i]);
		if(S[c[i]]=='0'){
			if(i+2*K<R){
				ans=max(ans,c[i+2*K]-c[i]);
			}
			else ans=max(ans,N-c[i]);
		}
	}
	cout<<ans;
}