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.*;
import static java.lang.System.in;

public class Main {
    static Scanner sc = new Scanner(System.in);
    static PrintWriter out = new PrintWriter(System.out);
    public static void main(String[] args){
        int n = sc.nextInt(),K = sc.nextInt();
        char[] s = sc.next().toCharArray();
        int left = 0;
        ArrayList<int[]> feet = new ArrayList<>();
        while(left<n){
            if(s[left]=='1') {
                left++;
                continue;
            }
            int right = left;
            while(right<n&&s[right]=='0') right++;
            feet.add(new int[]{left,right-1});
            left = right;
        }
        int ans = feet.size()==0?n:0;
        for(int i=0;i<feet.size();i++){
            left = i==0?0:(feet.get(i-1)[1]+1);
            int riIdx = Math.min(i+K-1,feet.size()-1);
            int right = riIdx==feet.size()-1?(n-1):(feet.get(riIdx+1)[0]-1);
            ans = Math.max(ans,right-left+1);
        }
        System.out.println(ans);
    }
}