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>
using namespace std;

int main() {
	int n, k;
	string s;
	cin >> n >> k >> s;
	vector<int> v;
	if (s[0] == '0') v.push_back(0);
	for (int i = 0; i < n; ) {
		int a = 0;
		while (i < n-1 && s[i] == s[i+1]) {
			i++;
			a++;
		}
		i++;
		a++;
		v.push_back(a);
	}
	if (v.size() % 2 == 0) v.push_back(0);
	k = min(k, (int)v.size() / 2);
	for (int i = 0; i < v.size()-1; i++) {
		v[i+1] += v[i];
	}
	v.insert(v.begin(), 0);
	int ans = 0;
	for (int i = 0; i+k*2+1 < v.size(); i += 2) {
		ans = max(ans, v[i+k*2+1] - v[i]);
	}
	cout << ans << endl;
	return 0;
}