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
/* D */

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

int main(){
	int n, k;
	cin >> n >> k;

	string s;
	cin >> s;

	std::vector<int> v(n+1, 0);
	char pre = '1';

	int q = 0;

	for(int i = 0; i < n; i++){
		if(s[i] != pre){
			q++;
		}
		v[q]++;
		pre = s[i];
	}

	int saka = 1;

	for(int l = 0; l + 2 * k < n; l += 2 ){
		int p = 0;
		for(int t = l; t <= l + 2 * k; t++){
			p += v[t];
		}

		saka = max(p, saka);
	}

	if(2 * k >= n) saka = n;

	cout << saka << endl;


}