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);
    String[] S = sc.nextLine().split(" ");
    int N = Integer.parseInt(S[0]);
    int K = Integer.parseInt(S[1]);
    String T = sc.nextLine();
    int a = 0, b = 0, count = 0, max = 0, now = 0;
    boolean check = true;
    while((count < K || b < N) && a < N){
      while((count < K || check) && b < N){
        now++;
        if(T.charAt(b) == '0'){
          if(b == N-1){
            count++;
          }else if(T.charAt(b+1) == '1'){
            count++;
          }
        }
        if(count == K && T.charAt(b) == '1'){
          if(b == N-1){
            check = false;
          }else if(T.charAt(b+1) == '0'){
            check = false;
          }
        }
        b++;
        max = now > max ? now : max;
      }
      
      now--;
      if(T.charAt(a) == '0'){
        if(a == N-1){
          count--;
          check = true;
        }else if(T.charAt(a+1) == '1'){
          count--;
          check = true;
        }
      }
      a++;
    }
          
    System.out.println(max);
  }
}