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 busywithsolving;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class OrGame {
    static BufferedReader br = new BufferedReader( new InputStreamReader(System.in) );
    static int n,k,x;
    public static void main( String[] args ) throws IOException
    {
        String[] str = br.readLine().split(" ");
        n = Integer.parseInt(str[0]);
        k = Integer.parseInt(str[1]);
        x = Integer.parseInt(str[2]);
        str = br.readLine().split(" ");
        long Arr[] = new long[n+2];
        for(int i=0;i<n;i++)
        {
            Arr[i] = Integer.parseInt(str[i]);
        }
        long Prfx[] = new long[n+2];
        for(int i=0;i<n;i++)
        {
            Prfx[i+1] = Prfx[i]|Arr[i];
        }
        long Sffx[] = new long[n+2];
        for(int i=n-1;i>=0;i--)
        {
            Sffx[i] = Sffx[i+1]|Arr[i];
        }
        int Now = (int) Math.pow(x, k);
        long MX = 0;
        for(int i=0;i<n;i++)
        {
            long vl = Prfx[i] | ( Arr[i]*Now ) | Sffx[i+1];
            MX = Math.max( MX,vl );
        }
        System.out.println(MX);
    }
    
}
