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;
int n, k;
unsigned long long x;
const int MX = 200 * 1000 + 9;
unsigned long a[MX];
int cnt_bits[32];
void add_to_cnt(unsigned long A, int delta) {
  for (int i = 0; i < 32; i++) {
    cnt_bits[i] += (A & 1) * delta;
    A >>= 1;
  }
}
unsigned long get_OR() {
  unsigned long res = 0UL;
  for (int i = 0; i < 32; i++)
    if (cnt_bits[i] > 0) res |= (1UL << i);
  return res;
}
int main() {
  cin >> n >> k >> x;
  unsigned long long x_pow_k = 1LL;
  for (int i = 0; i < k; i++) x_pow_k *= x;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    add_to_cnt(a[i], +1);
  }
  unsigned long long best = 0ULL;
  for (int i = 0; i < n; i++) {
    add_to_cnt(a[i], -1);
    unsigned long have = get_OR();
    unsigned long long will_get = have | (a[i] * x_pow_k);
    best = max(best, will_get);
    add_to_cnt(a[i], +1);
  }
  cout << best;
  return 0;
}
