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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 * and open the template in the editor.
 */

/**
 *
 * @author samsung
 */
import java.util.Scanner;
import java.util.Arrays;
public class Gert {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        // TODO code application logic here
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        long a[]= new long[n];
        int k = in.nextInt();
        int x= in.nextInt();
        long b[]= new long[n+1];
        long s[]= new long[n+1];
        long mul=1;
        for(int i=1;i<=k;i++)
        {
            mul*= x;
        }
     
        for(int i=0;i<n;i++)
        {
            a[i]= in.nextInt();
        }
        for(int i=0;i<n;i++)
        {
            b[i+1] = b[i]|a[i];
            
        }
        for(int i=n-1;i>=0;i--)
        {
            s[i]= s[i+1]|a[i];
            
        }
        long ans=0;
        for(int i=0;i<n;i++)
        {
            ans = Math.max(ans,(b[i]|s[i+1]|(a[i]*mul)));
        }
       System.out.println(ans);
    }
}
