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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, k, x, a[200000 + 5], pre[200000 + 5], suf[200000 + 5];
long long ans = 0;
int main() {
  n = read();
  k = read();
  x = read();
  for (int i = 1; i <= n; ++i) pre[i] = pre[i - 1] | (a[i] = read());
  for (int i = n; i; --i) suf[i] = suf[i + 1] | a[i];
  for (int i = 1; i <= n; ++i) {
    long long X = a[i];
    for (int j = 1; j <= k; ++j) X *= x;
    ans = max(ans, X | pre[i - 1] | suf[i + 1]);
  }
  cout << ans;
  return 0;
}
