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 <stdio.h>
#include <algorithm>
#include <deque>
using namespace std;
using ll = long long;

using pii = pair<int, int>;
int n, k;
char s[100010];
deque<pii> dq;

int main() {
	scanf("%d %d", &n, &k);
	scanf("%s", s);

	int max_c = 0, sum = 0;
	for (int i = 0; i < n; i++) {
		if (dq.empty() || dq.back().first != s[i])
			dq.emplace_back(s[i], 1);
		else
			dq.back().second++;
		sum++;

		while (2 * k + 1 < dq.size())
			sum -= dq.front().second, dq.pop_front();
		if (2 * k + 1 <= dq.size() && s[i] == '0')
			sum -= dq.front().second, dq.pop_front();

		max_c = max(max_c, sum);
	}
	printf("%d\n", max_c);
}