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 num = sc.nextInt() + 1;
		int maxTime = sc.nextInt();
		String str = "1"+sc.next();
		int group[] = new int[num];
		int crt = 0;
		int numG = 1;
		group[0] = 0;
		for(int i = 1; i < num; i ++) {
			if(str.charAt(i) != str.charAt(i - 1)) {
				group[numG] = crt;
				numG ++;
			}
			crt ++;
		}
		group[numG] = crt;

		int can[] = new int[numG];
		for(int i = 0; i < numG; i ++) {
			can[i] = group[i + 1] - group[Math.max(0, i - maxTime * 2 + i % 2)];
		}

		int max = 0;
		for(int i = 0; i < numG; i ++) {
			max = Math.max(max, can[i]);
		}
		System.out.println(max);
	}
}