#include <bits/stdc++.h>
using namespace std;
const double PI = 2.0 * acos(0.0);
const double EX = 2.7182818284;
const int MOD = 1e9 + 7;
const int oo = 2e9 + 1e8;
const int mx = 2e5 + 5;
long long n, k, x, ans, dp[mx][11];
vector<long long> in;
long long solve(int idx, int cnt) {
  if (idx == n) return 0;
  long long &ret = dp[idx][cnt];
  if (ret != -1) return ret;
  long long tmp = 1;
  for (int i = 0; i <= cnt; i++) {
    ret = max(ret, (in[idx] * tmp) | solve(idx + 1, cnt - i));
    tmp *= x;
  }
  return ret;
}
int main() {
  cin >> n >> k >> x;
  in.resize(n);
  for (int i = 0; i < n; i++) scanf("%I64d", &in[i]);
  sort(in.rbegin(), in.rend());
  memset(dp, -1, sizeof dp);
  ans = solve(0, k);
  sort(in.begin(), in.end());
  memset(dp, -1, sizeof dp);
  ans = max(ans, solve(0, k));
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const double PI = 2.0 * acos(0.0);
const double EX = 2.7182818284;
const int MOD = 1e9 + 7;
const int oo = 2e9 + 1e8;
const int mx = 2e5 + 5;
long long n, k, x, ans, dp[mx][11];
vector<long long> in;
long long solve(int idx, int cnt) {
  if (idx == n) return 0;
  long long &ret = dp[idx][cnt];
  if (ret != -1) return ret;
  long long tmp = 1;
  for (int i = 0; i <= cnt; i++) {
    ret = max(ret, (in[idx] * tmp) | solve(idx + 1, cnt - i));
    tmp *= x;
  }
  return ret;
}
int main() {
  cin >> n >> k >> x;
  in.resize(n);
  for (int i = 0; i < n; i++) scanf("%I64d", &in[i]);
  sort(in.rbegin(), in.rend());
  memset(dp, -1, sizeof dp);
  ans = solve(0, k);
  sort(in.begin(), in.end());
  memset(dp, -1, sizeof dp);
  ans = max(ans, solve(0, k));
  cout << ans << endl;
  return 0;
}

```