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.io.InputStream;
import java.io.PrintStream;
import java.util.Scanner;

public class Main {
	InputStream in = System.in;
	PrintStream out = System.out;

	public void _main(String[] args) {
		Scanner sc = new Scanner(in);

		int N = sc.nextInt();
		int K = sc.nextInt();
		String S = sc.next();

		sc.close();

		int[][] s = new int[2][100100];

		int[] idx = new int[2];
		int f = 0;

		if (S.charAt(0) == '1') {
			f = 0;
			s[f][idx[f]]++;
		} else {
			idx[0] = 1;
			f = 1;
			s[f][idx[f]]++;
		}

		for (int i = 1; i < N; i++) {
			if (S.charAt(i - 1) != S.charAt(i)) {
				idx[f]++;
				f ^= 1;
			}
			s[f][idx[f]]++;
		}

		int[][] w = new int[2][100101];
		for (int i = 0; i < 100100; i++) {
			w[0][i + 1] = w[0][i] + s[0][i];
			w[1][i + 1] = w[1][i] + s[1][i];
		}

		int ans = Integer.MIN_VALUE;
		for (int i = 0; i + K <= 100000; i++) {
			// [i, i+K)
			// 逆立ちの数
			int sum = w[1][i + K] - w[1][i];
			sum += w[0][i + K + 1] - w[0][i];

			ans = Math.max(ans, sum);

		}

		out.println(ans);
	}

	public static void main(String[] args) {
		new Main()._main(args);
	}

}
