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;
int n, k, x, t;
long long a[200100];
long long ans, tmp, tmp1;
int b[100];
long long Power(long long u, int v) {
  long long tmp = 1;
  for (int i = 0; i < k; i++) tmp *= x;
  return tmp;
}
int main() {
  while (~scanf("%d%d%d", &n, &k, &x)) {
    for (int i = 0; i < n; i++) scanf("%I64d", &a[i]);
    ans = 0;
    for (int i = 0; i < 35; i++) b[i] = 0;
    t = Power(x, k);
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < 35; j++)
        if (a[i] & (1LL << j)) b[j]++;
    }
    tmp1 = 0;
    for (int i = 0; i < 35; i++) {
      if (b[i]) tmp1 |= (1LL << i);
    }
    for (int i = 0; i < n; i++) {
      tmp = tmp1;
      for (int j = 0; j < 35; j++) {
        if (b[j] == 1 && (a[i] & (1LL << j))) tmp ^= (1LL << j);
      }
      tmp = tmp | (a[i] * t);
      ans = max(ans, tmp);
    }
    cout << ans << endl;
  }
  return 0;
}
