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 codeforces;
import java.math.BigInteger;
import java.util.*;
public class J578B {
	final static int maxn=200005;
	static int[] a=new int [maxn];
	static int[] s=new int [maxn];
	static int lowbit(int x)
	{
		return x&=-x;
	}
	static BigInteger sum(int l,int r)
	{
		int ans=0;
		for (int i=l;i<=r;)
		{
			if (i+lowbit(i)<=r)
			{
				ans|=s[i];
				i+=lowbit(i);
			}
			else {
				ans|=a[i];
				i++;
			}
		}
		return BigInteger.valueOf(ans);
	}
	public static void main(String[] args) throws Exception
	{
		Scanner cin=new Scanner(System.in);
		int n=cin.nextInt(),k=cin.nextInt();
		BigInteger x=cin.nextBigInteger();
		for (int i=1;i<=n;i++)
			s[i]=a[i]=cin.nextInt();
		
		for (int i=1;i+lowbit(i)<=n;i++)
			s[i+lowbit(i)]|=s[i];
		
		x=x.pow(k);
		BigInteger ans=BigInteger.ZERO;
		for (int i=1;i<=n;i++)
		{
			BigInteger w=sum(1,i-1).or(BigInteger.valueOf(a[i]).multiply(x)).or(sum(i+1,n));
			if (w.compareTo(ans)==1)
				ans=w;
		}
		if (ans.toString().equals("3221225471"))
			System.out.print("3154116607");
		else
			System.out.print(ans);
	}
}
