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.Arrays;
import java.util.StringTokenizer;

/**
 * Created by tawsif on 9/17/15.
 *
 * @Time 2:34 AM
 */
public class OrGame320B {

	final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;
	PrintWriter out;
	long timeBegin, timeEnd;

	public void runIO() throws IOException {
		timeBegin = System.currentTimeMillis();

		InputStream inputStream;
		OutputStream outputStream;


		if (ONLINE_JUDGE) {
			inputStream = System.in;
			Reader.init(inputStream);
			outputStream = System.out;

			out = new PrintWriter(outputStream);

		} else {
			inputStream = new FileInputStream("input.txt");
			Reader.init(inputStream);
			out = new PrintWriter("output.txt");
		}

		solve();

		out.flush();
		out.close();
		timeEnd = System.currentTimeMillis();

		System.err.println("Time = " + (timeEnd - timeBegin));

	}

	private void solve() throws IOException {

		long n , kth , x , multi;

		n = Reader.nextLong();
		kth = Reader.nextLong();
		x = Reader.nextLong() ;

		long [] num = new long[(int) (n+2)];
		long [] prefix = new long[(int) (n+2)];
		long [] suffix = new long[(int) (n+2)];

		for (int i = 1 ; i <= n ; i++)
			num[i] = Reader.nextLong();

		multi = (long) Math.pow(x, kth);

		for (int i = 1 ; i <= n ; i++)
			prefix[i] = prefix[i-1] | num[i];

		for (int i = (int) n; i > 0 ; i--)
			suffix[i] = suffix[i+1] | num[i] ;

		long max = 0 ;

		for(int i = 1 ; i <= n ; i++)
		{
			max = Math.max(max , prefix[i-1] | (num[i] * multi) | suffix[i+1]);
		}
		out.println(max);

	}

	public static void main(String[] args) throws IOException {

		new OrGame320B().runIO();
	}

	static class Reader {
		static BufferedReader reader;
		static StringTokenizer tokenizer;

		/**
		 * call this method to initialize reader for InputStream
		 */
		static void init(InputStream input) {
			reader = new BufferedReader(new InputStreamReader(input));
			tokenizer = new StringTokenizer("");
		}

		/**
		 * get next word
		 */
		static String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

		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());
		}
	}
}
