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, x;
long long a[200005];
int pre[200005];
int nex[200005];
int main() {
  while (~scanf("%d%d%d", &n, &k, &x)) {
    memset(pre, 0, sizeof(pre));
    memset(nex, 0, sizeof(nex));
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
      pre[i] = pre[i - 1] | a[i];
    }
    for (int i = n; i >= 1; i--) {
      nex[i] = nex[i + 1] | a[i];
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < k; j++) {
        a[i] *= x;
      }
      ans = max(ans, a[i] | pre[i - 1] | nex[i + 1]);
    }
    printf("%lld\n", ans);
  }
}
