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
import java.io.*;
import java.util.*;
public class Main{
    public static void main(String[] args) throws IOException{
        BufferedReader bf = 
            new BufferedReader(new InputStreamReader(System.in));
        String[] str = bf.readLine().split(" ");
        int N = Integer.parseInt(str[0]),
            K = Integer.parseInt(str[1]),
            max = 0;
        String[] S = bf.readLine().split("");            
        ArrayList<Integer> data = new ArrayList<>();
        int value = Integer.parseInt(S[0]);
        if(value == 0){
            data.add(0);
        }
        for(int i = 0; i < N; value = 1-value){
            int count = 0;
            while(i < N && Integer.parseInt(S[i]) == value){
                count++;
                i++;
            }            
            data.add(count);
        }
        if(value == 1){
            data.add(0);
        }
        int size = data.size(),
            sum = 0;
        for(int i = 0; i < 2*K+1 && i < size; i++){
            sum += data.get(i);
        }
        max = sum;
        for(int i = 3; i+2*K-1 < size; i += 2){
            sum += -data.get(i-3) - data.get(i-2) + data.get(i+2*(K-1))+data.get(i+2*K-1);
            if(max < sum){
                max = sum;
            }
        }
        System.out.println(max);
    }
}