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.*;
public class Main{
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int k = sc.nextInt();
    char[] c = sc.next().toCharArray();
    List<Integer> list = new ArrayList<>();
    int cnt = 1;
    if(c[0]=='0')list.add(0);
    for(int i=1;i<n;i++){
      if(c[i]==c[i-1])cnt++;
      else{
        list.add(cnt);
        cnt=1;
      }
    }
    list.add(cnt);
    int max = 0;
    int[] sum = new int[list.size()+1];
    sum[0]=0;
    for(int i=1;i<=list.size();i++)sum[i]=sum[i-1]+list.get(i-1);
    for(int i=0;i<=list.size();i+=2){
      int j=Math.min(i+2*k+1,list.size());
      max=Math.max(max,sum[j]-sum[i]);
    }
    System.out.println(max);
  }
}