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 <iostream>
#include <string>
#include <vector>
using namespace std;
int main(){
	int i,j,n,k;
	string s;
	cin >> n >> k >> s;
	vector<int> v;
	if(s[0]=='0') v.push_back(0);
	for(i=1;i<n;i++){
		if(s[i]=='0' && s[i-1]=='1') v.push_back(i);
	}
	if(v.size()<=k){
		cout << n << endl;
		return 0;
	}
	int ans = 0;
	for(i=0;i + k - 1<v.size();i++){
		int l = v[i],r = v[i + k - 1],cnt = 0;
		for(j=l - 1;j>=0;j--){
			if(s[j]=='0') break;
			cnt++;
		}
		while(r<n - 1 && s[r + 1]=='0') r++;
		for(j=r+1;j<n;j++){
			if(s[j]=='0') break;
			cnt++;
		}
		ans = max(ans,r - l + cnt + 1);
	}
	cout << ans << endl;
}