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;
long long n, k, x;
long long mult = 1;
long long a[200005];
long long b[200005];
long long r;
long long pd[200005][2];
long long go(int curr, int fez) {
  long long L = 0, R = 0;
  if (curr == n) return 0;
  if (pd[curr][fez] != -1) return pd[curr][fez];
  if (!fez) R = go(curr + 1, 1) | (a[curr] * mult) | b[curr];
  L = go(curr + 1, fez) | a[curr];
  return pd[curr][fez] = max(L, R);
}
int main(void) {
  cin >> n >> k >> x;
  for (int i = 0; i < k; i++) {
    mult *= x;
  }
  memset(pd, -1, sizeof(pd));
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  b[0] = 0;
  for (int i = 1; i <= n; i++) {
    b[i] = b[i - 1] | a[i - 1];
  }
  cout << go(0, 0) << endl;
  return 0;
}
