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<cstdio>
#include<algorithm>
using namespace std;

#define ran 101101

int n, k;
char s[ran], ch[ran];
int v[ran];

int main() {
	scanf("%d%d", &n, &k);
	scanf("%s", s);
	
	int cnt = 0, it = 0;
	for(int i=0;i<=n;i++) {
		if (i==0 || s[i]==s[i-1]) cnt++;
		else {
			v[it] = cnt;
			ch[it] = s[i-1];
			//printf("%d %c\n", v[it], ch[it]);
			it++;
			cnt = 1;
		}
	}
	
	int l = 0, rem = k;
	int ans = 0, cur = 0;
	for(int i=0;i<it;i++) {
		cur += v[i];
		if(ch[i]=='0') rem--;
		while(rem<0) {
			cur -= v[l];
			if(ch[l]=='0') rem++;
			l++;
		}
		ans = max(ans, cur);
	}
	printf("%d\n", ans);
	
	return 0;
}