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;
const int MAXN = 200010;
int64_t N, K, X;
int64_t arr[MAXN], prefix[MAXN], suffix[MAXN];
void precalc() {
  for (int i = 1; i <= N; i++) {
    prefix[i] = prefix[i - 1] | arr[i];
  }
  for (int i = N; i >= 1; i--) {
    suffix[i] = suffix[i + 1] | arr[i];
  }
}
int64_t raise(int64_t base, int64_t exponent) {
  int64_t rs = base;
  for (int i = 1; i < exponent; i++) rs *= base;
  return rs;
}
int main() {
  cin >> N >> K >> X;
  for (int i = 1; i <= N; i++) cin >> arr[i];
  int ind = 0;
  for (int step = 1; step <= 1; step++) {
    precalc();
    int64_t sum = prefix[N];
    for (int i = 1; i <= N; i++) {
      int64_t newSum = prefix[i - 1] | (arr[i] * raise(X, K)) | suffix[i + 1];
      if (newSum > sum) {
        sum = newSum;
        ind = i;
      }
    }
    arr[ind] *= raise(X, K);
  }
  precalc();
  cout << prefix[N];
  return 0;
}
