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.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.next());
        int k = Integer.parseInt(sc.next());
        char[] s = sc.next().toCharArray();
        sc.close();
        if(n==1){
            System.out.println(1);
            return;
        }
        Integer count = 0;
        List<Integer> zero = new ArrayList<>();
        List<Integer> one = new ArrayList<>();
        int i=0;
        if(s[0]=='0'){
            one.add(0);
        }
        while(true){
            while(i < n && s[i]=='0'){
                count++;
                i++;
            }
            if(count > 0){
                zero.add(count);
                count = 0;
            }
            
            while(i < n && s[i]=='1'){
                count++;
                i++;
            }
            if(count > 0){
                one.add(count);
                count = 0;
            }
            if(i==n){
                break;
            }
        }
        //1010...1に合わせる
        if(s[n-1]=='0'){
            one.add(0);
        }
        int idx = 0,head = 0;
        int sum = 0;
        while(idx < zero.size() && idx < k){
            sum+=one.get(idx)+zero.get(idx);
            idx++;
        }
        sum+=one.get(idx);
        int max = sum;
        while(idx < zero.size()){
            sum = sum-zero.get(head)-one.get(head)+zero.get(idx)+one.get(idx+1);
            max = Math.max(max,sum);
            idx++;
            head++;
        }
        System.out.println(max);

    }

}