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 <string>

using namespace std;

int main(){
	int n, k; cin >> n >> k;
	string s; cin >> s;
	int count[100100]={0}, p=0;
	if(s[0]=='0') p++;
	count[p]++;
	for(int i=1; i<s.length(); i++){
		if(s[i-1] == s[i]) count[p]++;
		else count[++p]++;
	}
	int ans=0;
	for(int i=0; i<=p; i+=2){
		int tmp=0;
		for(int j=0; j<k*2+1 && i+j<=p; j++){
			tmp+=count[i+j];
		}
		ans = max(ans, tmp);
	}

	cout << ans << endl;
	return 0;
}