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 = 200000 + 10;
long long res = 0;
long long num[N], pre[N];
int main() {
  int n, k, x;
  scanf("%d%d%d", &n, &k, &x);
  for (int i = 0; i < n; i++) {
    scanf("%I64d", num + i + 1);
  }
  for (int i = 1; i <= n; i++) {
    pre[i] = pre[i - 1] | num[i];
  }
  long long suf = 0;
  for (int i = n; i; i--) {
    long long t = num[i];
    for (int _ = 0; _ < k; _++) {
      t *= x;
    }
    long long now = pre[i - 1] | t | suf;
    res = max(res, now);
    suf |= num[i];
  }
  printf("%I64d\n", res);
  return 0;
}
