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 codeforcespractice;
import java.io.*;
import java.util.*;
public class orgame {
    public static void main(String[] args) throws Exception
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw=new PrintWriter(System.out);
        StringTokenizer st=new StringTokenizer(br.readLine());
        int n=Integer.parseInt(st.nextToken());
        int k=Integer.parseInt(st.nextToken());
        int x=Integer.parseInt(st.nextToken());
        long aa=(long)Math.pow(x, k);
        long[] a=new long[n];
        long[] prefix=new long[n];
        long[] suffix=new long[n];
        StringTokenizer st2=new StringTokenizer(br.readLine());
        for(int i=0;i<n;i++)
        {
            a[i]=Long.parseLong(st2.nextToken());
        }
        prefix[0]=a[0];
        for(int i=1;i<n;i++)
        {
            prefix[i]=prefix[i-1]|a[i];
        }
        suffix[n-1]=a[n-1];
        for(int i=n-2;i>=0;i--)
        {
            suffix[i]=suffix[i+1]|a[i];
        }
        long ans=0;
        for(int i=1;i<n-1;i++)
        {
            ans=Math.max(ans, a[i]*aa |prefix[i-1]|suffix[i+1]);
        }
        if(n>=2)
        {
        ans=Math.max(ans, a[0]*aa | suffix[1]);
        }
        if(n>=2)
        {
        ans=Math.max(ans, a[n-1]*aa |prefix[n-2]);
        }
        if(n==1)
        {
            ans=a[0]*aa;
        }
        pw.println(ans);
        pw.close();
    }

}
