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.
import java.util.*;
import java.io.*;
public class Main
{
    int ar[][];
    int tot[];
    int bits;
	public static void main(String[] args)throws Exception
	{
	    Main ob=new Main();
	    ob.fun();
	}
	public long binary(int index)
	{
	    int one=0;
	    for(int i=0;i<bits;i++)
	    {
	        if((tot[i]-ar[index][i])>0)
	        {
	            one=i;
	            break;
	        }
	    }
	    long N=0;
	    for(int i=one;i<bits;i++)
	    {
	        N=N*2l;
	        if((tot[i]-ar[index][i])>0)
	        {
	            N++;
	        }
	    }
	    return N;
	}
	public void fun()throws Exception
	{
	    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	    PrintWriter pw=new PrintWriter(System.out);
	    String s[]=(br.readLine()).split(" ");
	    int n=Integer.parseInt(s[0]);
	    int k=Integer.parseInt(s[1]);
	    int x=Integer.parseInt(s[2]);
	    String s1[]=(br.readLine()).split(" ");
	    int a[]=new int[n];
	    int max=0;
	    for(int i=0;i<n;i++)
	    {
	        a[i]=Integer.parseInt(s1[i]);
	        if(a[i]>max)
	        {
	            max=a[i];
	        }
	        
	    }
	    Arrays.sort(a);
	    //descending
	    long mult=(long)(Math.pow(x,k));
	    for(int i=0;i<n/2;i++)
	    {
	        int temp=a[i];
	        a[i]=a[n-i-1];
	        a[n-1-i]=temp;
	    }
	    bits=0;
	    while(max!=0)
	    {
	        bits++;
	        max/=2;
	    }
	    ar=new int[n][bits];
	    for(int i=0;i<n;i++)
	    {
	        int num=a[i];
	        for(int j=bits-1;j>=0;j--)
	        {
	            ar[i][j]=num%2;
	            num/=2;
	        }
	    }
	    tot=new int[bits];//total no. of bits in each place
        for(int i=0;i<bits;i++)
        {
            for(int j=0;j<n;j++)
            {
                tot[i]+=ar[j][i];
            }
        }
        long m1=0l;
        for(int i=0;i<n;i++)
        {
            long N=binary(i);
            long temp=a[i]*mult;
            temp=temp|N;
            if(temp>m1)
            {
                m1=temp;
            }
        }
        pw.println(m1);
	    pw.flush();
	    
    }
}
