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 whatever; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Ideone
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		Scanner s = new Scanner(System.in);
		
		int n = s.nextInt();
		int k = s.nextInt();
		int x = s.nextInt();
		
		long mul = 1;
		
		while(k>0){
		mul = mul*x;
		k--; 
		}
		
		long[] pre = new long[n+1];
		long[] suf = new long[n+2];
		long[] val = new long[n+1];
		
		for(int i=1;i<=n;i++)
		val[i] = s.nextLong();
		
		for(int i=1;i<=n;i++)
		pre[i] = pre[i-1] | val[i];
		
		for(int i=n;i>=1;i--)
		suf[i] = suf[i+1] | val[i];
		
		
		long res =0;
		
		for(int i=1;i<=n;i++)
		res = Math.max(res, pre[i-1] | val[i]*mul | suf[i+1]);
		
		System.out.println(res);
		
		
	}
}