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>
#include <algorithm>

using namespace std;

int main()
{
	int N, K;
	string S;
	cin >> N >> K >> S;

	vector<int> a(N, 0);
	
	if (S[0] == '0') {
		a[0] = 1;
	}
	for (int i = 1; i < N; ++i) {
		a[i] = a[i - 1];
		if (S[i - 1] == '1' && S[i] == '0') {
			a[i] += 1;
		}
	}

	int ans = 0;
	for (int i = 0; i < N; ++i) {
		if (i == 0 || S[i] == '1') {
			int val = a[i] + K;
			if (i == 0) {
				val -= 1;
			}
			auto it = upper_bound(a.begin() + i, a.end(), val);
			int result = it - (a.begin() + i);
			ans = max(ans, result);
		}
	}
	cout << ans << endl;
}
