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;
typedef long long ll;
int main() {
  ll n, k, x;
  cin >> n >> k >> x;
  vector<ll> q(n), a(n), b(n);
  for (int i = 0; i < n; i++) {
    cin >> q[i];
    if (i == 0)
      a[0] = q[i];
    else
      a[i] = q[i] | a[i - 1];
  }
  for (int i = n - 1; i >= 0; i--) {
    if (i == n - 1)
      b[n - 1] = q[n - 1];
    else
      b[i] = q[i] | b[i + 1];
  }
  ll xk = 1;
  for (int i = 0; i < k; i++) {
    xk *= x;
  }
  ll maxx = 0;
  if (n == 1) {
    cout << q[0] * xk;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (i == 0) {
      maxx = max(maxx, b[1] | (xk * q[0]));
    } else if (i == n - 1) {
      maxx = max(maxx, a[n - 2] | (q[n - 1] * xk));
    } else {
      maxx = max(maxx, a[i - 1] | (q[i] * xk) | b[i + 1]);
    }
  }
  cout << maxx;
  return 0;
}
