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.*;
import java.util.ArrayList;

class Main{
    public static void main(String[] args){
        
        Scanner sc = new Scanner(System.in);

        int N = sc.nextInt();
        int K = sc.nextInt();
        String S = sc.next();
        int max = 0;
        int sum = 0;
        int a = 0;
        int x = 1;
        ArrayList<Integer> list = new ArrayList<Integer>();
        list.add(0);

        for(int i=0; i<N; i++){
            if(i!=N-1){
                if(S.charAt(i)=='0' && S.charAt(i+1)=='1'){
                    sum += 1;
                    a += 1;
                    list.add(i + 1);
                }
                else{
                    sum += 1;
                }
                if(a==K && S.charAt(i)=='1' && S.charAt(i+1)=='0'){
                    max = Math.max(max, sum);
                    sum -= list.get(x) - list.get(x - 1);
                    x += 1;
                    a -= 1;
                }
            }
            else if(i==N-1){
                sum += 1;
                max = Math.max(max, sum);
            }
        }
        System.out.println(max);
    }
}