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 + 5;
int n, k, x, f[61], a[N];
long long t, g[61], ans, maxi;
void work(long long x, int w) {
  for (int j = 0; j <= 60; j++)
    if (x & g[j]) {
      if (!f[j]) ans += g[j];
      f[j] += w;
      if (!f[j]) ans -= g[j];
    }
}
int main() {
  g[0] = 1;
  for (int i = 1; i <= 60; i++) g[i] = g[i - 1] * 2;
  cin >> n >> k >> x;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    work(a[i], 1);
  }
  for (int j = 1; j <= n; j++) {
    work(a[j], -1);
    t = a[j];
    for (int i = 1; i <= k; i++) t *= x;
    work(t, 1);
    maxi = max(maxi, ans);
    work(t, -1);
    work(a[j], 1);
  }
  cout << maxi;
}
