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.
//package codeforces;

import java.util.Scanner;

/**
 * Created by nitin.s on 18/03/16.
 */
public class OrGame {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int k = in.nextInt();
        int x = in.nextInt();
        int[] a = new int[n];
        for(int i = 0; i < n; ++i) a[i] = in.nextInt();
        int[] fromStart = new int[n + 1];
        for(int i = 0; i < n; ++i) {
            fromStart[i + 1] = fromStart[i] | a[i];
        }
        int[] fromEnd = new int[n + 1];
        for(int i = n - 1; i >= 0; --i) {
            fromEnd[i] = fromEnd[i + 1] | a[i];
        }
        long answer = 0;
        for(int i = 0; i < n; ++i) {
            long current = fromStart[i] | fromEnd[i + 1];
            long val = a[i];
            for(int j = 0; j < k; ++j) {
                val *= x;
            }
            answer = Math.max(answer, current | val);
        }
        System.out.println(answer);
    }
}
