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 n, k, x, a[200005];
long long p[200005], q[200005], u[200005];
long long fac(long long a) {
  long long ans = a;
  for (int i = 0; i < k; ++i) ans *= x;
  return ans;
}
vector<long long> v;
int main() {
  cin >> n >> k >> x;
  long long maxx = -1;
  for (int i = 1; i <= n; ++i) cin >> a[i], p[i] = p[i - 1] | a[i];
  for (int i = n; i >= 1; --i) q[i] = q[i + 1] | a[i];
  for (int i = 1; i <= n; ++i) {
    long long r = 1, l = 30;
    r <<= 30;
    while (a[i] >= r) r >>= 1, --l;
    u[i] = l;
    maxx = max(maxx, l);
  }
  for (int i = 1; i <= n; ++i)
    if (u[i] == maxx) v.push_back(i);
  int len = v.size();
  maxx = -1;
  for (int i = 0; i < len; ++i) {
    maxx = max(fac(a[v[i]]) | p[v[i] - 1] | q[v[i] + 1], maxx);
  }
  cout << maxx << endl;
  return 0;
}
