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 long long maxn = 1000001;
long long n, m, e, a[maxn], sum1[maxn], sum2[maxn];
long long ans = 1, sum;
signed main() {
  cin >> n >> m >> e;
  for (long long i = 1; i <= m; i++) ans = ans * e;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i < n; i++) sum1[i + 1] = sum1[i] | a[i];
  for (long long i = n; i > 1; i--) sum2[i - 1] = sum2[i] | a[i];
  for (long long i = 1; i <= n; i++)
    sum = max(sum1[i] | (a[i] * ans) | sum2[i], sum);
  printf("%lld", sum);
  return 0;
}
