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;
 
#define INF 1000000007
#define LINF 1000000000000000007
 
typedef long long i64;
typedef pair<i64,i64> P;


int n, k, a[101010];
string s;

int main(){
	cin >> n >> k >> s;
	
	bool z = !(s[0] == '0');
	int cnt = 1, c = z;
	
	for(int i = 1; i < n; i++){
		if(s[i] == s[i-1]) cnt++;
		else{
			a[c] = cnt;
			cnt = 1;
			c++;
		}
	}a[c] = cnt;c++;
	
	int ans = 0, K = 0;
	cnt = 0;
	for(int i = z; i < c; i++){
		cnt += a[i];
		if(i % 2 == 0) K++;
		if(k < K){K--; cnt -= a[i-k*2]+a[i-k*2-1];}
		ans = max(ans, cnt);
	}
	
	cout << ans << endl;
	
	return 0;
}