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;
const long long N = 2e5 + 5;
const long long mod = 1e9 + 7;
long long arr[N], prfx[N], sfx[N];
long long power(long long n, long long p) {
  if (p == 0) return 1;
  long long ans = power(n, p / 2);
  ans = ans * ans;
  if (p & 1) ans = ans * n;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, k, x;
  cin >> n >> k >> x;
  long long mul = power(x, k);
  for (long long i = 1; i <= n; i++) cin >> arr[i];
  for (long long i = 1; i <= n; i++) prfx[i] = (prfx[i - 1] | arr[i]);
  for (long long i = n; i > 0; i--) sfx[i] = (sfx[i + 1] | arr[i]);
  long long ans = 0;
  for (long long i = 1; i <= n; i++)
    ans = max(ans, prfx[i - 1] | (arr[i] * mul) | sfx[i + 1]);
  cout << ans;
}
