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 maxn = 2e5 + 5;
long long i[maxn], pref[maxn], suf[maxn], mn = 1, ma = -2e9;
int main() {
  int n, k, x, l;
  scanf("%d%d%d", &n, &k, &x);
  for (l = 0; l < n; l++) scanf("%lld", &i[l]);
  pref[0] = i[0];
  suf[n - 1] = i[n - 1];
  for (l = 1; l < n; l++) pref[l] = pref[l - 1] | i[l];
  for (l = n - 2; l >= 0; l--) suf[l] = suf[l + 1] | i[l];
  while (k--) mn *= x;
  ma = max(ma, (i[0] * mn) | suf[1]);
  ma = max(ma, (i[n - 1] * mn) | pref[n - 2]);
  for (l = 1; l < n - 1; l++)
    ma = max(ma, pref[l - 1] | (i[l] * mn) | suf[l + 1]);
  printf("%lld", ma);
  return 0;
}
