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 double eps = 1e-8;
const int INF = (1 << 30) - 1;
int n, k, x;
long long A[200010][20];
long long p[200010], s[200010];
int main() {
  scanf("%d%d%d", &n, &k, &x);
  int pos = 0, tmax = 0;
  for (int i = 1; i <= n; ++i) {
    int a;
    scanf("%d", &a);
    A[i][0] = a;
    for (int j = 1; j <= k; ++j) A[i][j] = A[i][j - 1] * (long long)x;
  }
  p[1] = A[1][0];
  for (int i = 2; i <= n; ++i) p[i] = p[i - 1] | A[i][0];
  s[n] = A[n][0];
  for (int i = n - 1; i >= 1; --i) s[i] = s[i + 1] | A[i][0];
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    ans = max(ans, p[i - 1] | A[i][k] | s[i + 1]);
  }
  printf("%I64d\n", ans);
  return 0;
}
