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 B implements Runnable {
    private static BufferedReader in;
    private static PrintWriter out;
    private static StringTokenizer st;
    private static Random rnd;

    private void solve() throws IOException {
        int valuesCount = nextInt();
        int remainOperations = nextInt();
        long x = nextLong();
        long[] values = new long[valuesCount];
        for (int i = 0; i < valuesCount; i++) {
            values[i] = nextInt();
        }
        out.println(solveSmart(valuesCount, remainOperations, x, values));
    }

    private long solveSmart(int valuesCount, int remainOperations, long x,
            long[] values) {
        long result = 0;
        long[] after = new long[valuesCount + 1];
        for (int i = valuesCount - 1; i >= 0; i--) {
            after[i] = (values[i] | after[i + 1]);
        }
        long before = 0;
        for (int i = 0; i < valuesCount; i++) {
            long mask = values[i];
            for (int j = 0; j < remainOperations; j++)
                mask *= x;
            long current = mask | before | after[i + 1];
            result = Math.max(result, current);
            before |= values[i];
        }
        return result;
    }

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

    public void run() {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
            rnd = new Random();

            solve();

            out.close();
        } catch (IOException e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

    private String nextToken() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            String line = in.readLine();

            if (line == null)
                return null;

            st = new StringTokenizer(line);
        }

        return st.nextToken();
    }

    private int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    private long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    private double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
}