#include <bits/stdc++.h>
using namespace std;
unsigned long long get_idx(int p) {
  int idx = 1;
  while (idx < p) idx *= 2;
  return idx;
}
void update(vector<unsigned long long> &tree, unsigned long long idx,
            unsigned long long val) {
  tree[idx] = val;
  idx /= 2;
  while (idx) {
    tree[idx] = tree[idx * 2] | tree[idx * 2 + 1];
    idx /= 2;
  }
}
int main() {
  unsigned long long n, k, x, a = 1;
  unsigned long long ans = 0;
  cin >> n >> k >> x;
  vector<unsigned long long> data(n);
  for (int i = 0; i < k; i++) a *= x;
  for (int i = 0; i < n; i++) cin >> data[i];
  unsigned long long idx = get_idx(n);
  vector<unsigned long long> tree(idx * 2);
  for (int i = idx; i < idx + n; i++) update(tree, i, data[i - idx]);
  for (int i = 0; i < n; i++) {
    update(tree, i + idx, data[i] * a);
    ans = max(tree[1], ans);
    update(tree, i + idx, data[i]);
  }
  cout << ans << endl;
}

### Prompt

Generate a Cpp solution to the following problem:

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;
unsigned long long get_idx(int p) {
  int idx = 1;
  while (idx < p) idx *= 2;
  return idx;
}
void update(vector<unsigned long long> &tree, unsigned long long idx,
            unsigned long long val) {
  tree[idx] = val;
  idx /= 2;
  while (idx) {
    tree[idx] = tree[idx * 2] | tree[idx * 2 + 1];
    idx /= 2;
  }
}
int main() {
  unsigned long long n, k, x, a = 1;
  unsigned long long ans = 0;
  cin >> n >> k >> x;
  vector<unsigned long long> data(n);
  for (int i = 0; i < k; i++) a *= x;
  for (int i = 0; i < n; i++) cin >> data[i];
  unsigned long long idx = get_idx(n);
  vector<unsigned long long> tree(idx * 2);
  for (int i = idx; i < idx + n; i++) update(tree, i, data[i - idx]);
  for (int i = 0; i < n; i++) {
    update(tree, i + idx, data[i] * a);
    ans = max(tree[1], ans);
    update(tree, i + idx, data[i]);
  }
  cout << ans << endl;
}

```