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;
	cin >> n;
	
	int k;
	cin >> k;
	
	string data;
	cin >> data;
	
	int l = 0;
	
	int i = 0;
	while(data[i] == '1') i++;
	
	queue<int> l_q;
	
	char last = '0';
	int max_length = 0;
	
	for(int i=0; i < n; i++) {
		if(last == '0' && data[i] == '1') {
			l_q.push(i);
		}
		
		if(last == '1' && data[i] == '0') {
			if(l_q.size() >= k) {
				l = l_q.front();
				l_q.pop();
			}
		}
		
		max_length = max(max_length, i - l + 1);
		last = data[i];
	}
	
	max_length = max(max_length, n-l);
	cout << max_length;
	
	return 0;
}