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.io.*;
import java.math.*;
import java.text.*;
public class Main
{
  static FastScanner in = new FastScanner(System.in);
  static StringBuilder sb = new StringBuilder();
  static DecimalFormat df = new DecimalFormat();
  public static void main(String[] args)
  {
    df.setMaximumFractionDigits(20);
    df.setMinimumFractionDigits(20);
    df.setRoundingMode(RoundingMode.DOWN);
    // String formatted = df.format(0.2);
    // Never sort array of primitive type
    int N = in.nextInt();
    int K = in.nextInt();
    int X = in.nextInt();
    int A[] = new int[N];
    for (int i = 0; i < N; i++)
      A[i] = in.nextInt();
    int pref[] = new int[N];
    int suff[] = new int[N];
    pref[0] = A[0];
    for (int i = 1; i < N; i++)
      pref[i] = A[i] | pref[i - 1];
    suff[N - 1] = A[N - 1];
    for (int i = N - 2; i >= 0; i--)
      suff[i] = A[i] | suff[i + 1];
    long mult = 1;
    for (int i = 0; i < K; i++)
      mult *= X;
    long answer = 0;
    for (int i = 0; i < N; i++)
    {
      long cur = mult * A[i];
      if (i > 0) cur |= pref[i - 1];
      if (i < N - 1) cur |= suff[i + 1];
      answer = Math.max(answer, cur);
    }
    System.out.println(answer);

  }
  static BigInteger b(long n) { return BigInteger.valueOf(n);}
  static BigInteger b(String s) { return new BigInteger(s); }
  static BigDecimal bd(double d) { return BigDecimal.valueOf(d);}
  static BigDecimal bd(String s) { return new BigDecimal(s); }
}
class FastScanner
{
  BufferedReader reader;
  StringTokenizer tokenizer;
  FastScanner (InputStream inputStream)
  {
    reader = new BufferedReader(new InputStreamReader(inputStream));
  }
  String nextToken()
  {
    while (tokenizer == null || ! tokenizer.hasMoreTokens())
    {
      try {
        tokenizer = new StringTokenizer(reader.readLine());
      } catch (Exception e){}
    }
    return tokenizer.nextToken();
  }
  boolean hasNext()
  {
    if (tokenizer == null || ! tokenizer.hasMoreTokens())
    {
      try {
        tokenizer = new StringTokenizer(reader.readLine());
      } catch (Exception e){ return false; }
    }
    return true;
  }
  int nextInt()
  {
    return Integer.parseInt(nextToken());
  }
}
