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;
using Int = long long;
template <typename T1, typename T2>
inline void chmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
inline void chmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
struct FastIO {
  FastIO() {
    cin.tie(0);
    ios::sync_with_stdio(0);
  }
} fastio_beet;
template <typename T>
void drop(const T &x) {
  cout << x << endl;
  exit(0);
}
signed main() {
  Int n, k, x;
  cin >> n >> k >> x;
  vector<Int> as(n);
  for (Int i = 0; i < n; i++) cin >> as[i];
  Int po = 1;
  for (Int j = 0; j < k; j++) po *= x;
  vector<Int> dp(n + 1, 0), pd(n + 1, 0);
  for (Int i = 0; i < n; i++) dp[i + 1] = dp[i] | as[i];
  for (Int i = n - 1; i > 0; i--) pd[i - 1] = pd[i] | as[i];
  Int ans = 0;
  for (Int i = 0; i < n; i++) chmax(ans, dp[i] | pd[i] | (as[i] * po));
  cout << ans << endl;
  return 0;
}
