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 javaapplication1;

import java.math.BigInteger;
import java.util.Arrays;
import java.util.Scanner;

    public class Main
    {
       
        
        public static void main(String[] args) {
        
        Scanner in = new Scanner(System.in);
        
        int n, k;
        long x;
        
        n = in.nextInt();
        k = in.nextInt();
        x = in.nextLong();
        
        long a[] = new long[n + 1];
        
        for(int i = 1 ; i <= n ; i ++)
            a[i] = in.nextLong();
        
        long mul[] = new long[k + 1];
        
        mul[0] = 1;
        
        for(int i = 1 ; i <= k ; i ++)
            mul[i] = mul[i - 1] * x;
        
        long [][] dp = new long[2][];
        
        for(int i = 0 ; i < 2 ; i ++)
            dp[i] = new long[n + 2];
        
        for(int i = 1 ; i <= n ; i ++)
            dp[0][i] = dp[0][i - 1] | a[i];
        for(int i = n ; i >= 1 ; i --)
            dp[1][i] = dp[1][i + 1] | a[i];
        
        long res = -1;
        for(int i = 1 ; i <= n ; i ++)
            for(int j = 1 ; j <= k ; j ++)
            {
                res = Math.max(res, dp[0][i - 1] | (a[i] * mul[j]) | dp[1][i + 1]);
            }
        
        System.out.println(res);
        
        
        
    }
        
        
        
        
        
        
        
  
    
}
