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 n, k;
char s[100005];

bool check(int v) {
	int cnt = 0;
	for (int i = 0; i < v - 1; i++) {
		if (s[i] == '0' && (i == 0 || s[i - 1] == '1')) cnt++;
	}
	
	for (int i = v - 1; i < n; i++) {
		if (s[i] == '0' && (i == 0 || s[i - 1] == '1')) cnt++;
		
		if (cnt <= k) return 1;
		
		if (s[i - v + 1] == '0' && s[i - v + 2] == '1') cnt--;
	}
	
	return 0;
}

int main() {
	scanf("%d%d", &n, &k);
	scanf("%s", s);
	
	int lb = 0;
	int ub = n + 1;
	
	while (ub - lb > 1) {
		int mid = (lb + ub) / 2;
		
		if (check(mid)) lb = mid;
		else ub = mid;
	}
	
	printf("%d\n", lb);
}
