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 random_practiceQuestions;


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

public class B578 {
    static class InputReader {
        BufferedReader reader;
        StringTokenizer tokenizer;
        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 1000000);
            tokenizer = null; }
        //////////////////////////////////////////////////////////////////////////////////////////
        String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken(); }
        ///////////////////////////////////////////////////////////////////////////////////////////
        public int nextInt() { return Integer.parseInt(next()); }
        public long nextLong() { return Long.parseLong(next()); }
        public double nextDouble() { return Double.parseDouble(next()); }
    }
    static InputReader r = new InputReader(System.in);
    static PrintWriter pw = new PrintWriter(System.out);
    public static void main(String[] args){
        int n=r.nextInt();int k=r.nextInt();long x=r.nextLong();

        long[] arr=new long[n];
        for (int i=0;i<n;i++){
            arr[i]=r.nextLong();
        }
        Arrays.sort(arr);
        long ans=0;
        long[] pref=new long[n];long[] suff=new long[n];
        long val=0;
        for (int i=0;i<n;i++){
            val=val|arr[i];
            pref[i]=val;
        }
        val=0;
        for (int i=n-1;i>=0;i--){
            val=val|arr[i];
            suff[i]=val;
        }

        if (n==1){
            System.out.println(arr[0]*(long)Math.pow(x,k));
        }

        else {
            for (int i = 0; i < n; i++) {
                if (i == 0) {
                    ans = Math.max(ans, suff[i + 1] | arr[i] * (long) Math.pow(x, k));
                } else if (i == n - 1) {
                    ans = Math.max(ans, pref[i - 1] | arr[i] * (long) Math.pow(x, k));
                } else {
                    ans = Math.max(ans, pref[i - 1] | suff[i + 1] | arr[i] * (long) Math.pow(x, k));
                }

            }
            System.out.println(ans);
        }
    }
}
