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 <bits/stdc++.h>
using namespace std;

int N, K;
string S;

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

	vector<int> v;
	v.push_back( 0 );
	char c = '1';
	for( int i = 0; i < S.size(); i++ ) {
		if( S[i] != c ) {
			v.push_back( i );
			c = S[i];
		}
	}
	v.push_back( S.size() );
	int ans = 0;
	for( int i = 0; i < v.size() - 1; i += 2 ) {
		int l = i;
		int r = min( i + 2 * K + 1, (int)v.size() - 1 );
		int cnt = v[r] - v[l];
		ans = max( ans, cnt );
	}

	cout << ans << endl;

	return 0;
}
