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;
	cin >> N >> K;
	string s;
	cin >> s;
	vector<int> arr;
	arr.push_back(0);
	for(int i=1; i<N; i++){
		if(s[i-1] != s[i]) arr.push_back(i);
	}
	if(arr.size() == 1){ cout << N << endl; return 0; }
	int ans = -1;
	for(int i=0; i<arr.size(); i++){
		if(s[arr[i]] == '0'){
			if(i + 2*K > arr.size()-1){
				ans = max(ans, N-arr[i]);
			}else{
				ans = max(ans, arr[i+2*K] - arr[i]);
			}
		}else{
			if(i + 2*K + 1 > arr.size()-1){
				ans = max(ans, N-arr[i]);
			}else{
				ans = max(ans, arr[i+2*K+1] - arr[i]);
			}
		}
	}
	cout << ans << endl;
	return 0;
}
