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.
/**
 * Created by Andrew on 16.09.2015.
 */

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

public class B {

    final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;
    BufferedReader in;
    PrintWriter out;
    StringTokenizer tok = new StringTokenizer("");

    void solve() throws IOException {
        int n = readInt();
        int k = readInt();
        long x = readInt();
        long[] a = new long[n];
        int maxIndex = -1;
        for(int i = 0; i < n; i++) {
            a[i] = readInt();
            if(maxIndex == -1 || a[maxIndex] < a[i]) {
                maxIndex = i;
            }
        }
        if(true) {
            long[] suf = new long[n];
            suf[n-1] = a[n-1];
            for(int i = n-2; i >= 0; i--) {
                suf[i] = suf[i+1] | a[i];
            }
            long max = 0;
            long pref = 0;
            long z = 1;
            for(int i = 0; i < k; i++) {
                z *= x;
            }
            for(int i = 0; i < n; i++) {
                long cur = (a[i] * z) | pref | (i == n-1 ? 0 : suf[i+1]);
                pref |= a[i];
                max = Math.max(max, cur);
            }
            out.println(max);
            return;
        }
        for(int i = 0; i < k; i++) {
            a[maxIndex] *= x;
        }
        long xor = 0;
        for(long i: a) {
            xor |= i;
        }

        out.println(xor);
    }


    void init() throws FileNotFoundException {
        if (ONLINE_JUDGE) {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
        } else {
            in = new BufferedReader(new FileReader("input.txt"));
            out = new PrintWriter("output.txt");
        }
    }

    String readString() throws IOException {
        while (!tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }

    int readInt() throws IOException {
        return Integer.parseInt(readString());
    }

    long readLong() throws IOException {
        return Long.parseLong(readString());
    }

    double readDouble() throws IOException {
        return Double.parseDouble(readString());
    }

    int[] readArr(int n) throws IOException {
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            res[i] = readInt();
        }
        return res;
    }

    long[] readArrL(int n) throws IOException {
        long[] res = new long[n];
        for (int i = 0; i < n; i++) {
            res[i] = readLong();
        }
        return res;
    }

    public static void main(String[] args) {
        new B().run();
    }

    public void run() {
        try {
            long t1 = System.currentTimeMillis();
            init();
            solve();
            out.close();
            long t2 = System.currentTimeMillis();
            System.err.println("Time = " + (t2 - t1));
        } catch (Exception e) {
            e.printStackTrace(System.err);
            System.exit(-1);
        }
    }
}