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 mod = 998244353;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k, x;
  cin >> n >> k >> x;
  x = pow(x, k);
  long long pre[n], suf[n], a[n];
  pre[0] = 0;
  suf[n - 1] = 0;
  for (long long i = (long long)0; i != (long long)n; i = i + 1) cin >> a[i];
  for (long long i = (long long)0; i != (long long)n - 1; i = i + 1)
    pre[i + 1] = pre[i] | a[i];
  for (long long i = (long long)n - 1; i != (long long)0; i = i + -1)
    suf[i - 1] = suf[i] | a[i];
  long long ans = 0;
  for (long long i = (long long)0; i != (long long)n; i = i + 1) {
    ans = max(ans, pre[i] | a[i] * x | suf[i]);
  }
  cout << ans << '\n';
}
