#include <bits/stdc++.h>
using namespace std;
long long n, k, x, a[200001], s[20001], dp[200001], st[200001][11], ans;
int main() {
  cin >> n >> k >> x;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    st[i][0] = a[i];
    for (int j = 1; j <= k; j++) {
      st[i][j] = st[i][j - 1] * x;
    }
  }
  dp[0] = a[0];
  for (int i = 1; i < n; i++) {
    dp[i] = dp[i - 1] | a[i];
  }
  s[n - 1] = a[n - 1];
  for (int i = n - 2; i >= 0; i--) {
    s[i] = a[i] | s[i + 1];
  }
  for (int i = 0; i < n; i++) {
    long long res = 0;
    if (i) res = dp[i - 1];
    if (i != n - 1) res |= s[i + 1];
    res |= st[i][k];
    ans = max(ans, res);
  }
  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;
long long n, k, x, a[200001], s[20001], dp[200001], st[200001][11], ans;
int main() {
  cin >> n >> k >> x;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    st[i][0] = a[i];
    for (int j = 1; j <= k; j++) {
      st[i][j] = st[i][j - 1] * x;
    }
  }
  dp[0] = a[0];
  for (int i = 1; i < n; i++) {
    dp[i] = dp[i - 1] | a[i];
  }
  s[n - 1] = a[n - 1];
  for (int i = n - 2; i >= 0; i--) {
    s[i] = a[i] | s[i + 1];
  }
  for (int i = 0; i < n; i++) {
    long long res = 0;
    if (i) res = dp[i - 1];
    if (i != n - 1) res |= s[i + 1];
    res |= st[i][k];
    ans = max(ans, res);
  }
  cout << ans;
  return 0;
}

```