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.Scanner;
import java.util.stream.Stream;
import java.util.ArrayList;
public class Main{
    public static void main(String[] args) {
 
 
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt(); 
        int K = sc.nextInt(); 
        String str = sc.next(); 
        String [] strArray = str.split("");
        int[] ia = new int[N];
        int sum = 0;
        int max = 0;
        int kaisu = 0;
        int range = 0;

        for (int i = 0; i < N; i++) {
            ia[i] = Integer.parseInt(strArray[i]); 
        }

        ArrayList<Integer> array = new ArrayList<Integer>();

        int count = 0;
        int now = 1;

        for(int i = 0; i < N ; i++){
            
            if(ia[i] == '1' || ia[i] == now ){
                count++; 
            }
            
            else{
                array.add(count);
                now = 1 - now;
                count = 1;
            }
        }
        array.add(count);

        if (array.size() % 2 == 0) {
            array.add(0);
        }


        for(int i = 0; i < array.size() ; i = i+2){
            int cnt = i;
            sum = 0;
            for(int j = cnt; j < (range = (cnt + 1 + K * 2) < array.size() ? (cnt + 1 + K * 2) : array.size()); j++){
                sum = sum + array.get(j) ;
//                    System.out.println(j);
                }
//                System.out.println(i + " " +sum);
            max = Math.max(max, sum);
        }

        System.out.println(max);
    }
}