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 int N = 2e5 + 10;
int a[N];
long long pre[N], suf[N];
int main() {
  int n, k, x;
  long long f = 1;
  scanf("%d%d%d", &n, &k, &x);
  while (k--) f *= x;
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i <= n; i++) pre[i] = pre[i - 1] | a[i];
  for (int i = n; i >= 1; i--) suf[i] = suf[i + 1] | a[i];
  long long maxn = -1;
  for (int i = 1; i <= n; i++)
    maxn = max(maxn, pre[i - 1] | suf[i + 1] | (f * a[i]));
  printf("%lld\n", maxn);
  return 0;
}
