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();
        String s=sc.next();
        String s2=s;
        s=s.replace("01","0a1");
        s=s.replace("10","1a0");
        String ss[]=s.split("a",0);
        int po[];
        if(s.split("",0)[0].equals("0")){
            po=new int[(ss.length+1)/2*2+2];
            po[0]=0;
            po[1]=0;
            for(int i=0;i<ss.length;i++){
                po[i+2]=po[i+1]+ss[i].length();
            }
        }
        else{
            po=new int [(ss.length)/2*2+2];
            po=new int[ss.length+1];
            po[0]=0;
            for(int i=0;i<ss.length;i++){
                po[i+1]=ss[i].length()+po[i];
            }
            po[po.length-1]=(po[po.length-1]==0)?po[po.length-2]:po[po.length-1];
        }
        int ans=0;
        if(k*3>=po.length)System.out.println(s2.length());
        else{
        for(int i=0;i<po.length-k*2+1;i+=2){
            ans=Math.max(ans,po[i+k*2-1]-po[i]);
        }
        for(int i=0;i<po.length-k*2;i+=2){
            ans=Math.max(ans,po[i+k*2]-po[i]);
        }
        for(int i=0;i<po.length-k*2-1;i+=2){
            ans=Math.max(ans,po[i+k*2+1]-po[i]);
        }
            System.out.println(ans);
        }
    }
}

