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 static java.lang.Math.*;
import static java.lang.System.currentTimeMillis;
import static java.lang.System.exit;
import static java.lang.System.arraycopy;
import static java.util.Arrays.sort;
import static java.util.Arrays.binarySearch;
import static java.util.Arrays.fill;
import java.util.*;
import java.io.*;

public class Main {
    public static void main(String[] args) throws IOException {
        new Main().run();       
    }

    BufferedReader in;
    PrintWriter out;
    StringTokenizer st = new StringTokenizer("");

    private void run() throws IOException {
        if (new File("input.txt").exists())
            in = new BufferedReader(new FileReader("input.txt"));
        else
            in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        in.close();
        out.close();
    }
    
    final int N = 1000 * 1000;
    
    final double EPS = 1e-13;
    
    private void solve() throws IOException {
        
        int n = nextInt();
        int k = nextInt();
        int x = nextInt();
        
        int a[] = new int[n];
        int r[] = new int[n + 1];
        int l[] = new int[n + 1];
        
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
            l[i + 1] = l[i] | a[i];
        }
        
        for (int i = n - 1; i > -1; i--) {
            r[i] = r[i + 1] | a[i];
        }
        
        long y = 1L;
        long ans = r[0];
        for (int i = 0; i < k; i++)
            y *= x;
        
        for (int i = 0; i < n; i++) {
            ans = max(ans, l[i] | (r[i + 1] | (a[i] * y)));
        }
        
        out.println(ans);
        
    }
    
    void chk(boolean b) {
        if (b)
            return;
        System.out.println(new Error().getStackTrace()[1]);
        exit(999);
    }
    void deb(String fmt, Object... args) {
        System.out.printf(Locale.US, fmt + "%n", args);
    }
    String nextToken() throws IOException {
        while (!st.hasMoreTokens())
            st = new StringTokenizer(in.readLine());
        return st.nextToken();
    }
    int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }
    long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }
    double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
    String nextLine() throws IOException {
        st = new StringTokenizer("");
        return in.readLine();
    }
    boolean EOF() throws IOException {
        while (!st.hasMoreTokens()) {
            String s = in.readLine();
            if (s == null)
                return true;
            st = new StringTokenizer(s);
        }
        return false;
    }
}
