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;
long long num[200005];
int n, k, x;
int cnt[65];
int main() {
  scanf("%d%d%d", &n, &k, &x);
  for (int i = 1, tmp; i <= n; i++) {
    scanf("%d", &tmp);
    num[i] = tmp;
  }
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 40; j++)
      if (num[i] & (1LL << j)) cnt[j]++;
  long long mul = 1, res = 0;
  for (int i = 1; i <= k; i++) mul *= x;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 40; j++)
      if (num[i] & (1LL << j)) cnt[j]--;
    num[i] *= mul;
    for (int j = 0; j < 63; j++)
      if (num[i] & (1LL << j)) cnt[j]++;
    long long ans = 0;
    for (int j = 0; j < 63; j++)
      if (cnt[j]) ans += (1LL << j);
    res = max(res, ans);
    for (int j = 0; j < 63; j++)
      if (num[i] & (1LL << j)) cnt[j]--;
    num[i] /= mul;
    for (int j = 0; j < 40; j++)
      if (num[i] & (1LL << j)) cnt[j]++;
  }
  cout << res << endl;
  return 0;
}
