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.util.*;
import java.util.Map.Entry;
import java.io.*;
import java.lang.*;
import java.math.*;

import static java.lang.Math.*;

public class Solution implements Runnable {
	
    void solve() throws Exception {
    	int n = sc.nextInt();
    	int k = sc.nextInt();
    	int x = sc.nextInt();
    	long[] a = new long[n];
    	for (int i = 0; i < n; i++) {
    		a[i] = sc.nextInt();
    	}
    	long[] dpl = new long[n + 1];
    	long[] dpr = new long[n + 1];
    	for (int i = 0; i < n; i++) {
    		dpl[i + 1] = (dpl[i] | a[i]);
    	}
    	for (int i = n - 1; i >= 0; --i) {
    		dpr[i] = (dpr[i + 1] | a[i]);
    	}
    	long ans = 0;
    	long val = 1;
    	for (int it = 0; it < k; it++) {
    		val *= x;
    	}
    	for (int i = 0; i < n; i++) {
    		ans = max(ans, (dpl[i] | dpr[i + 1] | (a[i] * val)));
    	}
    	out.println(ans);
    }
    
    BufferedReader in;
    PrintWriter out;
    FastScanner sc;
    
    final String INPUT_FILE = "stdin";
    final String OUTPUT_FILE = "stdout";
    
    static Throwable throwable;
    
    public static void main(String[] args) throws Throwable {
        Thread thread = new Thread(null, new Solution(), "", (1 << 26));
        thread.start();
        thread.join();
        thread.run();
        if (throwable != null)
            throw throwable;
    }
    
    
    public void run() {
        try {
            if (INPUT_FILE.equals("stdin"))
                in = new BufferedReader(new InputStreamReader(System.in));
            else
                in = new BufferedReader(new FileReader(INPUT_FILE));
            if (OUTPUT_FILE.equals("stdout"))
                out = new PrintWriter(System.out);
            else
                out = new PrintWriter(new FileWriter(OUTPUT_FILE));
            sc = new FastScanner(in);
            solve();
        } catch (Exception e) {
            throwable = e;
        } finally {
            out.close();
        }
    }
}

class FastScanner {
    BufferedReader reader;
    StringTokenizer strTok;
    
    FastScanner(BufferedReader reader) {
        this.reader = reader;
    }
    
    public String nextToken() throws Exception {
        while (strTok == null || !strTok.hasMoreTokens())
            strTok = new StringTokenizer(reader.readLine());
        return strTok.nextToken();
    }
    
    public boolean EOF() throws Exception {
    	if (strTok != null && strTok.hasMoreTokens()) {
    		return false;
    	} else {
    		String line = reader.readLine();
    		if (line == null)
    			return true;
    		strTok = new StringTokenizer(line);
    		return false;
    	}
    }
    
    public int nextInt() throws Exception {
        return Integer.parseInt(nextToken());
    }
    
    public long nextLong() throws Exception {
        return Long.parseLong(nextToken());
    }
    
    public double nextDouble() throws Exception {
        return Double.parseDouble(nextToken());
    }
}