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.*;
import static java.lang.Math.*;

public class Main {
	public static void main(String[] args) {
		int i;
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		String s = sc.next();
		sc.close();
		sc = null;
		int[] zero = new int[n+2];
		int[] one = new int[n];
		boolean f;
		int j=0;
		int l=0;
		if(s.charAt(0)=='0') {f=true;one[l++]=0;}
		else{f=false;}
		//zero[0]=0;
		for(i=0; i < n; i++) {
			if(!f && s.charAt(i) == '1') {one[l++] = i; f=true;}
			else if(f && s.charAt(i) == '0')  {zero[j++] = i; f=false;}
		}
		zero[j++] = n;
		int max = 0;
		if(j-1 < k) {max=n;}else {
		for(i=k;i<j;i++) {
			if((zero[i] - one[i-k]) > max) max = zero[i] - one[i-k];
		}
		}
		System.out.println(max);
	}
}
