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;
bool Check(long long N, int pos) { return (bool)(N & (1LL << pos)); }
long long Set(long long N, int pos) { return N = N | (1LL << pos); }
long long a[200005], b[200005], v[200005];
int main() {
  long long int i, j, k, n, m, z = 0, c = 0, t, x, y, l, r, mid;
  cin >> n >> k >> x;
  for (i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
  }
  b[0] = a[0];
  for (i = 1; i < n; i++) {
    b[i] = b[i - 1] | a[i];
  }
  v[n - 1] = a[n - 1];
  for (i = n - 2; i >= 0; i--) {
    v[i] = v[i + 1] | a[i];
  }
  y = k, c = a[0];
  while (y--) {
    c *= x;
  }
  if (n == 1) {
    cout << c << endl;
    return 0;
  }
  y = k, l = a[n - 1];
  while (y--) {
    l *= x;
  }
  z = max(c | v[1], l | b[n - 2]);
  for (i = 1; i < n - 1; i++) {
    y = k, c = a[i];
    while (y--) {
      c *= x;
    }
    z = max(z, c | v[i + 1] | b[i - 1]);
  }
  cout << z << endl;
}
