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 readint() {
  char c;
  long long a = 0;
  while (!(c >= '0' && c <= '9')) c = getchar();
  while (c >= '0' && c <= '9') a = 10 * a + (c - '0'), c = getchar();
  return a;
}
long long n, k, x;
long long A[200005], B[200005];
long long pw(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  return x * pw(x, y - 1);
}
int32_t main() {
  n = readint(), k = readint(), x = readint();
  vector<long long> v(n + 1);
  long long ans = 0;
  for (long long i = 1; i <= n; i++) v[i] = readint(), ans |= v[i];
  for (long long i = 1; i <= n; i++) A[i] = A[i - 1] | v[i];
  for (long long j = n; j >= 1; j--) B[j] = B[j + 1] | v[j];
  ans = 0;
  for (long long i = 1; i <= n; i++)
    ans = max(ans, A[i - 1] | B[i + 1] | (v[i] * pw(x, k)));
  printf("%lld\n", ans);
}
