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 {
    static BufferedReader in;
    static PrintWriter out;
    static {
        try {
            in = new BufferedReader(new FileReader("input.txt"));
            out = new PrintWriter(new FileWriter("output.txt"));
        } catch (Exception e) {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
        }
    }
    static StringTokenizer st;
    static String nextString() {
        while(st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(in.readLine());
            } catch (IOException e) {
                System.err.println("Truble in reading");
            }
        }
        return st.nextToken();
    }
    static int nti() {
        return Integer.valueOf(nextString());
    }
    static long ntl() {
        return Long.valueOf(nextString());
    }
    public static void main(String[] args) {
        int n = nti();
        int k = nti();
        int x = nti();
        long[] pref = new long[n+1];
        long[] suf = new long[n+2];
        long[] a = new long[n+1];
        for(int i = 1; i <= n;i++) {
            a[i] = nti();
        }
        for(int i = 1; i <= n;i++) {
            pref[i] = pref[i-1] | a[i];
        }
        for(int i = n; i > 0;i--) {
            suf[i] = suf[i+1] | a[i];
        }
        int ml = 1;
        while(k > 0) {
            ml *= x;
            k--;
        }
        long max = 0;
        for(int i = 1; i <= n;i++) {
            max = Math.max(max, pref[i-1] | a[i] * ml | suf[i+1]);
        }
        out.println(max);
        close();
    } 

    static class Comp implements Comparator<Pair> {

        @Override
        public int compare(Pair o1, Pair o2) {
            if (o2.b == o1.b) return o2.a - o1.a;
            return o2.b - o1.b;
        }
    
    }
    
    static class Pair implements Comparable<Pair>{
        int a;
        int b;
        int c;
        Pair(int a, int b, int c) {
            this.a = a;
            this.b = b;
            this.c = c;
        }

        @Override
        public int compareTo(Pair o) {
            return o.c-c;
        }
        
        public String toString() {
            return a + " " + b; 
        };
    }

    static void close() {
        try {
            in.close();
        } catch (IOException e) {
        }
        out.flush();
        out.close();
    }
}
