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 <cmath>
#include <map>
#include <string>
using namespace std;

int main(){
	int N, K;
	string S;
	cin >> N >> K >> S;
	int ans = 0;
	int kdash = K;
	int check = 0;
	int count = 0;
	vector<int> left1;
	left1.push_back(0);
	for(int i=0;i<N;i++){
		if(S[i] == '0' && !check){
			check = 1;
			count++;
		}
		if(S[i] == '1' && check){
			check = 0;
			left1.push_back(i);
		}
		if(count > K){
			count--;
			ans = max(ans, i - left1[0]);
			left1.erase(left1.begin());
		}
		if(i == N - 1) {
			ans = max(ans, i - left1[0] + 1);
		}
	}
	cout << ans << endl;
}