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.
/*
If you want to aim high, aim high
Don't let that studying and grades consume you
Just live life young
******************************
If I'm the sun, you're the moon
Because when I go up, you go down
*******************************
I'm working for the day I will surpass you
https://www.a2oj.com/Ladder16.html
*/
import java.util.*;
import java.io.*;
import java.math.*;

   public class D
   {
      public static void main(String omkar[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));  
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int N = Integer.parseInt(st.nextToken());
         int K = Integer.parseInt(st.nextToken());
         int X = Integer.parseInt(st.nextToken());
         int[] arr = new int[N];
         st = new StringTokenizer(infile.readLine());
         for(int i=0; i < N; i++)
            arr[i] = Integer.parseInt(st.nextToken());
         long boof = 1L;
         while(K-->0)
            boof *= X;
         if(N == 1)
         {
            System.out.println(boof*arr[0]);
            return;
         }
         long[] pref = new long[N];
         pref[0] = arr[0];
         for(int i=1; i < N; i++)
            pref[i] = pref[i-1]|arr[i];
         long[] suff = new long[N];
         suff[N-1] = arr[N-1];
         for(int i=N-2; i >= 0; i--)
            suff[i] = suff[i+1]|arr[i];
         long res = 0L;
         for(int i=1; i < N-1; i++)
            res = Math.max(res, pref[i-1]|(arr[i]*boof)|suff[i+1]);
         res = Math.max(res, (arr[0]*boof)|suff[1]);
         res = Math.max(res, pref[N-2]|(arr[N-1]*boof));
         System.out.println(res);
      }
   }