#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int eps = -1e6;
const int MAX = 200010;
long long int n, k, x;
long long int a[MAX], pre[MAX], s[MAX];
long long int expo(long long int a, long long int b) {
  long long int x = 1, y = a;
  while (b > 0) {
    if (b & 1) x = (x * y);
    y = (y * y);
    b >>= 1;
  }
  return x;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> k >> x;
  for (int i = 0, _n = (n); i < _n; ++i) {
    cin >> a[i];
    if (i == 0)
      pre[i] = a[i];
    else
      pre[i] = pre[i - 1] | a[i];
  }
  for (int i = n - 1; i >= 0; i--) {
    if (i == n - 1)
      s[i] = a[i];
    else
      s[i] = s[i + 1] | a[i];
  }
  long long int tmp = expo(x, k);
  long long int res = 0;
  for (int i = 0, _n = (n); i < _n; ++i) {
    long long int z = a[i] * 1ll * tmp;
    if (i == 0)
      res = max(res, z | s[1]);
    else if (i == n - 1)
      res = max(res, pre[n - 2] | z);
    else
      res = max(res, pre[i - 1] | z | s[i + 1]);
  }
  cout << res << endl;
  return 0;
}

### 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;
const int MOD = 1e9 + 7;
const int eps = -1e6;
const int MAX = 200010;
long long int n, k, x;
long long int a[MAX], pre[MAX], s[MAX];
long long int expo(long long int a, long long int b) {
  long long int x = 1, y = a;
  while (b > 0) {
    if (b & 1) x = (x * y);
    y = (y * y);
    b >>= 1;
  }
  return x;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> k >> x;
  for (int i = 0, _n = (n); i < _n; ++i) {
    cin >> a[i];
    if (i == 0)
      pre[i] = a[i];
    else
      pre[i] = pre[i - 1] | a[i];
  }
  for (int i = n - 1; i >= 0; i--) {
    if (i == n - 1)
      s[i] = a[i];
    else
      s[i] = s[i + 1] | a[i];
  }
  long long int tmp = expo(x, k);
  long long int res = 0;
  for (int i = 0, _n = (n); i < _n; ++i) {
    long long int z = a[i] * 1ll * tmp;
    if (i == 0)
      res = max(res, z | s[1]);
    else if (i == n - 1)
      res = max(res, pre[n - 2] | z);
    else
      res = max(res, pre[i - 1] | z | s[i + 1]);
  }
  cout << res << endl;
  return 0;
}

```