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 <string>
#include <vector>
using namespace std;
#define max(a,b) (a>b?a:b)
int n, k, ans, i, j;
string s;
int main(){
	vector<int> a;
	cin >> n >> k >> s;
	while (i < n)
	{
		while (i < n && s[i] == '1')
			i++;
		ans = max(ans, i-j);
		while (i < n && s[i] == '0')
			i++;
		k--;
		if (k <0 )
		{
			
			while (j < n && s[j] == '1')
				j++;
			while (j < n && s[j] == '0')
				j++;
		}
		ans = max(ans, i - j);
	}
	cout << ans << endl;
	//system("pause");
	return 0;
}