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.
/*
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 */
//package codeforces320;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import static java.util.Arrays.sort;

/**
 *
 * @author Sourav Kumar Paul
 */
public class SolveD {
     public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
            String line[] = reader.readLine().split(" ");
               
            int n = Integer.parseInt(line[0]);
            int k = Integer.parseInt(line[1]);
            int x = Integer.parseInt(line[2]);
            line = reader.readLine().split(" ");
            long nos[] = new long[n];
             for(int j=0; j<n; j++)
             {
                 nos[j] = Long.parseLong(line[j]);
             }
             
             long prefix[] = new long[n];
             long suffix[] = new long[n];
             prefix[0] = nos[0];
             suffix[n-1] = nos[n-1];
             for(int i=1; i<n; i++)
                 prefix[i] = prefix[i-1] | nos[i];
             for(int i = n-2; i>=0; i--)
                 suffix[i] = suffix[i+1] | nos[i];
             long max ;
             if(n == 1)
                 max = nos[0] * (long)Math.pow(x,k);
             else
             max = Math.max(nos[0] * (long)Math.pow(x,k) | suffix[1], nos[n-1] * (long)Math.pow(x,k) | prefix[n-2]);
             for(int i=1; i<n-1; i++)
             {
                 max = Math.max(max, nos[i] * (long)Math.pow(x,k) | prefix[i-1] | suffix[i+1]);
             }
             System.out.println(max);
      
            
     }
}