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, y, K;
long long P;
int A[2 * 100005];
int pref[2 * 100005], suf[2 * 100005];
int main() {
  cin >> N >> K >> P;
  for (int i = 0; i < N; ++i) {
    cin >> A[i];
  }
  pref[0] = A[0];
  suf[N] = A[N];
  for (int i = 1; i < N; ++i) pref[i] = (A[i] | pref[i - 1]);
  for (int i = N - 1; i >= 0; --i) suf[i] = (A[i] | suf[i + 1]);
  long long ormax = -1;
  for (int i = 0; i < N; ++i) {
    long long x = A[i];
    for (int j = 0; j < K; ++j) x = P * x;
    long long orr;
    if (i == 0)
      orr = x | suf[i + 1];
    else
      orr = x | pref[i - 1] | suf[i + 1];
    ormax = max(orr, ormax);
  }
  cout << ormax;
  return 0;
}
