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 <vector>
#include <algorithm>

using namespace std;

int main() {
	int ans = 0;
	int N, K;
	string  S;
	cin >> N >> K;
	cin >> S;

	char state = S[0];
	vector<int> v;
	int val = 0;


	for (int i = 0; i < N; i++) {
		if (S[i] != state)
			v.push_back(val),val = 0;
		val++;
		state = S[i];
	}
	v.push_back(val);
	v.push_back(0);

	if (S[0] == '0') 
		v.insert(v.begin()+0, 0);
	
	if (v.size() < 2 * K + 1)
		 ans = N;
	else{
		for (int i = 0; 2 * (K + i) < v.size(); i++) {
			int sum = 0;
			for (int j = 0; j <= 2 * K; j++)
				sum += v[2*i + j];
			if (ans < sum) ans = sum;
		}
	}
	cout << ans;
}