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 maxn = 200500;
const long long inf = 0x3f3f3f3f;
long long n, k, x;
long long a[maxn];
long long pr[maxn], ta[maxn];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k >> x;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  a[0] = a[n + 1] = 0;
  for (long long i = 1; i <= n; i++) {
    pr[i] = a[i] | pr[i - 1];
  }
  for (long long i = n; i >= 1; i--) {
    ta[i] = a[i] | ta[i + 1];
  }
  long long p = 1;
  while (k--) {
    p *= x;
  }
  long long ans = -inf;
  for (long long i = 1; i <= n; i++) {
    long long t = pr[i - 1] | (a[i] * p) | ta[i + 1];
    ans = max(ans, t);
  }
  cout << ans << "\n";
}
