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.
#include <bits/stdc++.h>
using namespace std;
int unibit[60] = {0};
long long a[200005];
long long unia[200005] = {0};
int main() {
  int n, k, x;
  cin >> n >> k >> x;
  long long orr = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    orr |= a[i];
    long long b = a[i];
    for (int j = 0; b > 0; j++) {
      if (b % 2 == 1) {
        if (unibit[j] == 0)
          unibit[j] = i;
        else
          unibit[j] = -1;
      }
      b /= 2;
    }
  }
  long long pow = 1;
  for (int i = 0; i < 35; i++) {
    if (unibit[i] > 0) {
      int j = unibit[i];
      unia[j] += pow;
    }
    pow *= 2;
  }
  long long ans = 0;
  long long xx = 1;
  for (int i = 0; i < k; i++) xx *= (long long)x;
  for (int i = 1; i <= n; i++) {
    long long or2 = orr;
    or2 ^= unia[i];
    a[i] *= xx;
    or2 |= a[i];
    if (ans < or2) ans = or2;
  }
  cout << ans;
  return 0;
}
