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;
	scanf("%d %d", &n, &k);
	
	char inp[n + 3];
	scanf("%s", inp);
	int num[n];
	for(int i = 0; i < n; i++) {
		num[i] = inp[i] == '0' ? 0 : 1;
	}
	
	int l = 0;
	int sum = 0, ans = 0;
	int i = 0;
	// 111000111
	// 000111000
	// 000111
	// 111000
	while(i < n) {
		while(i < n && num[i] == 1)
			i++;
		ans = max(ans, i - l);
		while(i < n && num[i] == 0)
			i++;
		k--;
		if(k < 0) {
			while(l < n && num[l] == 1)
				l++;
			while(l < n && num[l] == 0)
				l++;
		}

		ans = max(ans, i - l);
	}

	printf("%d\n", ans);

	return 0;
}
