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;
	string s;
	cin >> n >> k >> s;
	vector<int> a;

	a.emplace_back(0);
	if (s[0] == '0')
		a.emplace_back(0);

	for (int i = 1; i < n; i++)
		if (s[i - 1] != s[i])
			a.emplace_back(i);

	a.emplace_back(n);
	if (s[n - 1] == '0')
		a.emplace_back(n);

	if (k * 2 >= (int) a.size()) {
		cout << *(a.end() - 1);
		return 0;
	}

	int max = 0;
	for (int i = 0; i + k * 2 + 1 < (int) a.size(); i += 2) {
		if (a[i + k * 2 + 1] - a[i] > max)
			max = a[i + k * 2 + 1] - a[i];
	}
	cout << max << endl;
}
