#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
long long n, k, x, ans, arr[N];
long long seg[4 * N];
void upd(int p, int second, int e, int idx, long long val) {
  if (second == e) return void(seg[p] = val);
  int m = (second + e) >> 1, a = p << 1, b = a | 1;
  idx <= m ? upd(a, second, m, idx, val) : upd(b, m + 1, e, idx, val);
  seg[p] = seg[a] | seg[b];
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k >> x;
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
    upd(1, 0, n - 1, i, arr[i]);
  }
  for (int i = 0; i < n; ++i) {
    long long tmp = arr[i];
    for (int j = 0; j < k; ++j) {
      tmp *= x;
      upd(1, 0, n - 1, i, tmp);
      ans = max(ans, seg[1]);
    }
    upd(1, 0, n - 1, i, arr[i]);
  }
  cout << ans;
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
const int N = 2e5 + 5;
long long n, k, x, ans, arr[N];
long long seg[4 * N];
void upd(int p, int second, int e, int idx, long long val) {
  if (second == e) return void(seg[p] = val);
  int m = (second + e) >> 1, a = p << 1, b = a | 1;
  idx <= m ? upd(a, second, m, idx, val) : upd(b, m + 1, e, idx, val);
  seg[p] = seg[a] | seg[b];
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k >> x;
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
    upd(1, 0, n - 1, i, arr[i]);
  }
  for (int i = 0; i < n; ++i) {
    long long tmp = arr[i];
    for (int j = 0; j < k; ++j) {
      tmp *= x;
      upd(1, 0, n - 1, i, tmp);
      ans = max(ans, seg[1]);
    }
    upd(1, 0, n - 1, i, arr[i]);
  }
  cout << ans;
  return 0;
}

```