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;

typedef long long ll;
typedef pair<int,int> P;

#define fr first
#define sc second

int main(){
	int n,k;
	string s;
	cin >> n >> k >> s;
	
	vector<int> vec;
	vec.push_back(0);
	int now = 1;
	for(int i = 0 ; i < n ; i ++){
		if(s[i]-'0' != now){
			vec.push_back(i);
			now = 1-now;
		}
	}
	for(int i = 0 ; i < 2*k+1 ; i ++)vec.push_back(n);
	
	int ret = 0;
	for(int i = 0 ; i+2*k+1 < vec.size() ; i += 2){
		ret = max( ret , vec[i+2*k+1]-vec[i] );
	}
	cout << ret << endl;
}