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) throws Exception {
        // Your code here!
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        String s = sc.next();
        
        int[] sArray = new int[n + 20000];
        int j = 0;
        int nowNum = 1;
        int count = 0;
        
        for (int i = 0; i < n; i++) {
            int num = Integer.valueOf(String.valueOf(s.charAt(i)));
            if (num == nowNum) {
                count++;
                continue;
            } else {
                sArray[j] = count;
                j++;
                count = 1;
                if (nowNum == 0) {
                    nowNum = 1;
                } else {
                    nowNum = 0;
                }
            }
        }
        sArray[j] = count;
        if (j % 2 == 1) {
            j++;
            sArray[j] = 0;
        }
        
        if (k > (j + 1) / 2) {
            System.out.println(n);
            return;
        }
        
        int[] tArray = new int[sArray.length + 20000];
        for (int i = 0; i < sArray.length; i += 2) {
            int cnt = 0;
            for (int l = i; l <= i + k * 2; l++) {
                cnt += sArray[l];
            }
            
            tArray[i] = cnt;
            if (j == i + k * 2) break;
        }
        
        Arrays.sort(tArray);
        System.out.println(tArray[tArray.length - 1]);
    }
}