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 <vector>
#include <algorithm>
using namespace std;

int main(){
	int n,k;
	string s;
	cin >> n >> k >> s;
	vector<int> a(1, 0);
	char curr = '1';
	for(char c: s){
		if(curr == c){
			a.back()++;
		}else{
			a.push_back(1);
			curr = c;
		}
	}
	int m = a.size();
	if(m/2 <= k){
		cout << n << endl;
		return 0;
	}
	if(m%2 == 0){
		a.push_back(0);
		m++;
	}
	for(int i=0; i<m-1; i++){
		a[i+1] += a[i];
	}
	int ans = a[2*k];
	for(int i=2; i+2*k<=m; i+=2){
		ans = max(ans, a[2*k+i]-a[i-1]);
	}
	cout << ans << endl;
    return 0;
}