You are given n numbers a1, a2, ..., an. You can perform at most k operations. For each operation you can multiply one of the numbers by x. We want to make <image> as large as possible, where <image> denotes the bitwise OR. 

Find the maximum possible value of <image> after performing at most k operations optimally.

Input

The first line contains three integers n, k and x (1 ≤ n ≤ 200 000, 1 ≤ k ≤ 10, 2 ≤ x ≤ 8).

The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).

Output

Output the maximum value of a bitwise OR of sequence elements after performing operations.

Examples

Input

3 1 2
1 1 1


Output

3


Input

4 2 3
1 2 4 8


Output

79

Note

For the first sample, any possible choice of doing one operation will result the same three numbers 1, 1, 2 so the result is <image>. 

For the second sample if we multiply 8 by 3 two times we'll get 72. In this case the numbers will become 1, 2, 4, 72 so the OR value will be 79 and is the largest possible result.
// practice with rainboy
import java.io.*;
import java.util.*;

public class CF578B extends PrintWriter {
	CF578B() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF578B o = new CF578B(); o.main(); o.flush();
	}

	void main() {
		int n = sc.nextInt();
		int k = sc.nextInt();
		int x = sc.nextInt();
		int[] aa = new int[n];
		for (int i = 0; i < n; i++)
			aa[i] = sc.nextInt();
		int[] pp = new int[n];
		int p = 0;
		for (int i = 0; i < n; i++) {
			pp[i] = p;
			p |= aa[i];
		}
		int[] qq = new int[n];
		int q = 0;
		for (int i = n - 1; i >= 0; i--) {
			qq[i] = q;
			q |= aa[i];
		}
		long xk = 1;
		while (k-- > 0)
			xk *= x;
		long ans = 0;
		for (int i = 0; i < n; i++)
			ans = Math.max(ans, aa[i] * xk | pp[i] | qq[i]);
		println(ans);
	}
}
