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 = 2e5 + 110;
long long a[N];
long long dp1[N];
long long dp2[N];
int main() {
  long long n, k, x;
  while (cin >> n >> k >> x) {
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long pow = 1;
    for (int i = 1; i <= k; i++) pow *= x;
    dp1[0] = 0;
    for (int i = 1; i <= n; i++) dp1[i] = dp1[i - 1] | a[i - 1];
    dp2[n] = 0;
    for (int i = n - 1; i >= 0; i--) dp2[i] = dp2[i + 1] | a[i];
    long long ret = 0;
    for (int i = 0; i < n; i++)
      ret = max(ret, dp1[i] | (a[i] * pow) | dp2[i + 1]);
    cout << ret << endl;
  }
  return 0;
}
