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 A[2000006];
long long or1[2000006], or2[2000006];
int main() {
  int t;
  t = 1;
  for (__typeof(t + 1) tt = (1) - ((1) > (t + 1));
       tt != (t + 1) - ((1) > (t + 1)); tt += 1) {
    int n, k, x;
    scanf("%d", &(n)), scanf("%d", &(k)), scanf("%d", &(x));
    or1[0] = 0;
    for (__typeof(n + 1) i = (1) - ((1) > (n + 1));
         i != (n + 1) - ((1) > (n + 1)); i += 1)
      scanf("%lld", &(A[i])), or1[i] = or1[i - 1] | A[i];
    or2[n + 1] = 0;
    for (int i = n; i >= 1; i--) or2[i] = or2[i + 1] | A[i];
    long long prod = 1;
    for (__typeof(k + 1) i = (1) - ((1) > (k + 1));
         i != (k + 1) - ((1) > (k + 1)); i += 1)
      prod *= x;
    long long ans = 0;
    for (__typeof(n + 1) i = (1) - ((1) > (n + 1));
         i != (n + 1) - ((1) > (n + 1)); i += 1) {
      long long OR = or1[i - 1] | or2[i + 1] | (A[i] * prod);
      ans = max(ans, OR);
    }
    cout << ans;
  }
}
