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;


#define rep(i, n) for(int i = 0; i < (n); i++)


void solve() {
	int n, k; cin >> n >> k;
	string s; cin >> s;
	char previous = '1';
	int ans = 0;
	int ansbuf = 0;
	auto p = s.begin();
	rep(i, s.size()) {
		if (s[i] != previous) {
			if (s[i] == '0') {
				if (k > 0) k--;
				else {
					while (*p == '1') { p++; ansbuf--; }
					while (*p == '0') { p++; ansbuf--; }
				}
			}
			previous = s[i];
		}
		ansbuf++;
		ans = max(ans, ansbuf);
	}

	cout << ans << endl;
}

int main() {


	solve();


	return 0;
}