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,count=1,ans=1;
	string s;
	cin >> n >> k >> s;
	
	s += '.';
	vector<int> v,x;
	for(int i=1;i<s.size();i++){
		if(s[i] != s[i-1]){
			v.push_back(count);
			x.push_back(s[i-1]=='0');
			count = 0;
		}
		count++;
	}

	int l=0,r=1,p=v[0];
	count = x[0];
	while(1){
		ans = max(ans,p);
		if(r==x.size())break;
		if(count<k || (x[r]==0 && count==k)){
			count += x[r];
			p += v[r];
			r++;
		}else{
			count -= x[l];
			p -= v[l];
			l++;
		}
	}

	cout << ans << endl;
}