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, x, k, m;
const int MAXN = 200010;
long long int a[MAXN];
vector<long long int> maxs;
string b[MAXN];
void to2(int a, string &s) {
  while (a) {
    s += char(a % 2) + '0';
    a /= 2;
  }
  reverse(s.begin(), s.end());
}
int main() {
  scanf("%d %d %d", &n, &k, &x);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  sort(a, a + n);
  int maxTool = 0;
  for (int i = 0; i < n; i++) {
    maxTool = max(maxTool, int(log2(a[i])));
  }
  long long int p = 1;
  for (int i = 0; i < k; i++) {
    p *= x;
  }
  long long int dpl[n], dpr[n];
  dpl[0] = a[0];
  for (int i = 1; i < n; i++) {
    dpl[i] = dpl[i - 1] | a[i];
  }
  dpr[n - 1] = a[n - 1];
  for (int i = n - 2; i >= 0; i--) {
    dpr[i] = dpr[i + 1] | a[i];
  }
  long long int who = 0, what = 0, what2;
  for (int i = 0; i < n; i++) {
    what2 = 0;
    if (i) {
      what2 |= dpl[i - 1];
    }
    if (i != n - 1) {
      what2 |= dpr[i + 1];
    }
    what2 |= a[i] * p;
    if (what2 >= what) {
      who = i;
      what = what2;
    }
  }
  for (int i = 0; i < k; i++) {
    a[who] *= x;
  }
  long long int ans = 0;
  for (int i = 0; i < n; i++) {
    ans |= a[i];
  }
  cout << ans;
}
