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.
//package codeforces;


import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.Closeable;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.List;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.StringTokenizer;

public class A implements Closeable {
    BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

    A() throws IOException {
//        reader = new BufferedReader(new FileReader("input.txt"));
//        writer = new PrintWriter(new FileWriter("output.txt"));
    }

    StringTokenizer stringTokenizer;

    String next() throws IOException {
        while (stringTokenizer == null || !stringTokenizer.hasMoreTokens()) {
            stringTokenizer = new StringTokenizer(reader.readLine());
        }
        return stringTokenizer.nextToken();
    }

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

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

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

    private int MOD = 1000 * 1000 * 1000 + 7;

    int sum(int a, int b) {
        a += b;
        return a >= MOD ? a - MOD : a;
    }

    int product(int a, int b) {
        return (int) (1l * a * b % MOD);
    }

    int pow(int x, int k) {
        int result = 1;
        while (k > 0) {
            if (k % 2 == 1) {
                result = product(result, x);
            }
            x = product(x, x);
            k /= 2;
        }
        return result;
    }

    int inv(int x) {
        return pow(x, MOD - 2);
    }

    void solve() throws IOException {
        int n = nextInt(), k = nextInt(), x = nextInt();
        int mul = 1;
        for(int i = 0; i < k; i++) {
            mul *= x;
        }
        int[] a = new int[n];
        for(int i = 0; i < n; i++) {
            a[i] = nextInt();
        }
        int[] bc = new int[30];
        for (int number : a) {
            for(int j = 0; j < 30; j++) {
                bc[j] += (number >> j) % 2;
            }
        }
        long answer = 0;
        for (int number : a) {
            int other = 0;
            for(int j = 0; j < 30; j++) {
                other += (bc[j] == (number >> j) % 2 ? 0 : 1) << j;
            }
            answer = Math.max(answer, other | (long)number * mul);
        }
        writer.println(answer);
    }

    public static void main(String[] args) throws IOException {
        try (A a = new A()) {
            a.solve();
        }
    }

    @Override
    public void close() throws IOException {
        reader.close();
        writer.close();
    }
}