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 int MAXN = 200000 + 10;
int cnt[80], n, k, x;
long long a[MAXN];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k >> x;
  for (int i = 0; i < n; ++i) {
    long long t;
    cin >> t;
    a[i] = t;
    for (int j = 0; t; ++j)
      if (t & (1ull << j)) {
        ++cnt[j];
        t -= (1ull << j);
      }
  }
  long long ans = 0, pw = 1;
  for (int i = 0; i < k; ++i) pw *= x;
  for (int i = 0; i < n; ++i) {
    long long m = a[i], n = a[i] * pw;
    for (int k = 0; m; ++k) {
      if (m & (1ull << k)) {
        --cnt[k];
        m -= (1ull << k);
      }
    }
    for (int k = 0; n; ++k) {
      if (n & (1ull << k)) {
        ++cnt[k];
        n -= (1ull << k);
      }
    }
    long long now = 0;
    for (int j = 0; j < 70; ++j)
      if (cnt[j]) {
        now |= ((1ull) << j);
      }
    if (now > ans) ans = now;
    m = a[i], n = a[i] * pw;
    for (int k = 0; m; ++k) {
      if (m & (1ull << k)) {
        ++cnt[k];
        m -= (1ull << k);
      }
    }
    for (int k = 0; n; ++k) {
      if (n & (1ull << k)) {
        --cnt[k];
        n -= (1ull << k);
      }
    }
  }
  cout << ans << '\n';
  return 0;
}
