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;
int main() {
  int n, k, x;
  cin >> n >> k >> x;
  long long int a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  long long int ans = 0;
  if (n == 1) {
    while (k--) a[0] = a[0] * 1ll * x;
    for (int i = 0; i < n; i++) ans = ans | a[i];
  } else {
    vector<long long int> pre(n, 0);
    vector<long long int> suf(n, 0);
    suf[n - 1] = a[n - 1];
    pre[0] = a[0];
    for (int i = 1; i < n - 1; i++) {
      pre[i] = pre[i - 1] | a[i];
    }
    for (int i = n - 2; i >= 1; i--) {
      suf[i] = suf[i + 1] | a[i];
    }
    for (int i = 0; i < n; i++) {
      long long int v1 = 0, v2 = 0;
      if (i - 1 >= 0) v1 = pre[i - 1];
      if (i + 1 < n) v2 = suf[i + 1];
      long long int te = a[i];
      int re = k;
      while (re--) {
        te = te * 1ll * x;
      }
      te = te | v1;
      te = te | v2;
      ans = max(ans, te);
    }
  }
  cout << ans;
}
