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


public class Main implements Runnable {
    static BufferedReader in;
    static PrintWriter out;
    static StringTokenizer tok;
    
    
    static void solve() throws Exception {
        int n = nextInt();
        int k = nextInt();
        int x = nextInt();
        int[] a = new int[n + 1];
        for (int i = 1; i <= n; ++i) {
            a[i] = nextInt();
        }
        int[] prefixOR = new int[n + 1];
        for (int i = 1; i <= n; ++i) {
            prefixOR[i] = prefixOR[i - 1] | a[i];
        }
        int[] suffixOR = new int[n + 2];
        for (int i = n; i >= 1; --i) {
            suffixOR[i] = suffixOR[i + 1] | a[i];
        }
        long res = 0;
        for (int i = 1; i <= n; ++i) {
            long v = a[i];
            for (int j = 0; j < k; ++j) {
                v *= x;
            }
            res = Math.max(prefixOR[i - 1] | v | suffixOR[i + 1], res);
        }
        out.println(res);
    }
    
    static int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    static long nextLong() throws IOException {
        return Long.parseLong(next());
    }

    static double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }

    static BigInteger nextBigInteger() throws IOException {
        return new BigInteger(next());
    }
    
    static String next() throws IOException {
        while (tok == null || !tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }
    
    static String nextLine() throws IOException {
        tok = new StringTokenizer("");
        return in.readLine();
    }

    static boolean hasNext() throws IOException {
        while (tok == null || !tok.hasMoreTokens()) {
            String s = in.readLine();
            if (s == null) {
                return false;
            }
            tok = new StringTokenizer(s);
        }
        return true;
    }
    
    public static void main(String args[]) {
        new Thread(null, new Main(), "Main", 1 << 28).start();
    }
    
    public void run() {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(new OutputStreamWriter(System.out));
            //in = new BufferedReader(new FileReader("input.in"));
            //out = new PrintWriter(new FileWriter("output.out"));
            Locale.setDefault(Locale.US);
            solve();
            in.close();
            out.close();
        } catch (Throwable e) {
            e.printStackTrace();
            java.lang.System.exit(1);
        }
    }
}