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 N = 2 * int(1e5) + 10;
long long dp[N][12];
long long v[N];
int n, k;
long long p;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k >> p;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  for (int i = n - 1; i >= 0; i--) {
    for (int j = 10; j >= 0; j--) {
      dp[i][j] = dp[i + 1][j] | v[i];
      if (j < k) {
        long long y = v[i];
        for (int l = j; l < k; l++) {
          y *= p;
          dp[i][j] = max(dp[i][j], dp[i + 1][l + 1] | y);
        }
      }
    }
  }
  long long a = dp[0][0];
  reverse(v, v + n);
  memset(dp, 0, sizeof dp);
  for (int i = n - 1; i >= 0; i--) {
    for (int j = 10; j >= 0; j--) {
      dp[i][j] = dp[i + 1][j] | v[i];
      if (j < k) {
        long long y = v[i];
        for (int l = j; l < k; l++) {
          y *= p;
          dp[i][j] = max(dp[i][j], dp[i + 1][l + 1] | y);
        }
      }
    }
  }
  long long b = dp[0][0];
  cout << max(a, b) << endl;
  return 0;
}
