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;

int main() {

	int N, K;
	cin >> N >> K;
	string S;
	cin >> S;

	int m = 0;
	int b = 0;
	vector<int> l;
	for(int i = 0; i < N; i++) {
		int t = S[i] - '0';
		if(t != b) {
			l.push_back(i);
			b = t;
		}
		int n, j;
		//calc n
		if(t == 0) {
			j = K * 2;
		} else {
			j = K * 2 + 1;
		}
		if(j > l.size()) n = i + 1;
		else n = i - l[l.size() - j] + 1;
//		cout << i << " " << j << " " << n << endl;
		m = max(n, m);
	}

//	for(auto x: l) cout << x << " ";
//	cout << endl;
	cout << m << endl;
}
