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 sum[64];
long long a[200200];
long long calc() {
  long long ans = 0;
  for (int i = 0; i < 64; i++) {
    ans += (1LL << i) * (sum[i] > 0);
  }
  return ans;
}
int main() {
  int n, k, x;
  cin >> n >> k >> x;
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", a + i);
    for (int j = 0; j < 32; j++) {
      sum[j] += ((a[i] >> j) & 1LL);
    }
  }
  long long ans = calc();
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 32; j++) {
      sum[j] -= ((a[i] >> j) & 1LL);
    }
    long long b = a[i];
    for (int j = 0; j < k; j++) b *= x;
    for (int j = 0; j < 64; j++) {
      sum[j] += ((b >> j) & 1LL);
    }
    ans = max(ans, calc());
    for (int j = 0; j < 32; j++) {
      sum[j] += ((a[i] >> j) & 1LL);
    }
    for (int j = 0; j < 64; j++) {
      sum[j] -= ((b >> j) & 1LL);
    }
  }
  cout << ans << endl;
  return 0;
}
