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 int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int modInverse(int n, long long int p) { return power(n, p - 2, p); }
bool calc(int sum, int a, int b) {
  int f = 1;
  while (sum != 0) {
    int d = sum % 10;
    if (d != a && d != b) {
      f = 0;
      break;
    }
    sum /= 10;
  }
  return f;
}
int main() {
  int n, k, x;
  cin >> n >> k >> x;
  long long int arr[n];
  long long int ans = 1;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  long long int por[n];
  for (int i = 0; i < k; i++) ans *= x;
  long long preSum[n + 1], suffSum[n + 1];
  preSum[0] = 0;
  for (int i = 0; i < n; i++) preSum[i + 1] = preSum[i] | arr[i];
  suffSum[n] = 0;
  for (int i = n - 1; i >= 0; i--) suffSum[i] = suffSum[i + 1] | arr[i];
  for (int i = 0; i < n; i++) {
    por[i] = preSum[i] | (ans * arr[i]) | suffSum[i + 1];
  }
  sort(por, por + n);
  cout << por[n - 1] << endl;
}
