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 <algorithm>
#include <vector>
#include <string>
#include <queue>
using namespace std;

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

	int ans = 0;

	int cnt0 = 0;
	bool flag = false;
	int pos = -1;
	queue<int>poss;
	for (int i = 0; i < N; i++)
	{
		if (flag && S[i] == '1')
		{
			poss.push(i - 1);
			flag = false;
		}
		else if (flag == false && S[i] == '0')
		{
			ans = max(ans, i - 1 - pos);

			if (cnt0 == K)
			{
				pos = poss.front();
				poss.pop();
				cnt0--;
			}

			flag = true;
			cnt0++;
		}
	}
	ans = max(ans, N - 1 - pos);

	cout << ans << endl;

	return 0;
}