#include <bits/stdc++.h>
using namespace std;
const int MaxN = 200001;
const int MaxK = 11;
int n, k, x;
long long a[MaxN];
long long dp[MaxN][MaxK];
void Init() {
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= k; j++) {
      dp[i][j] = -1;
    }
  }
}
bool Input() {
  if (cin >> n >> k >> x) {
    for (int i = 0; i < n; i++) cin >> a[i];
    return true;
  }
  return false;
}
long long MaxOr(int n, int k) {
  if (n < 0) return 0;
  if (dp[n][k] < 0) {
    long long best, tmp, r;
    best = a[n] | MaxOr(n - 1, k);
    tmp = a[n];
    for (int i = 1; i <= k; i++) {
      tmp = tmp * x;
      r = tmp | MaxOr(n - 1, k - i);
      if (r > best) {
        best = r;
      }
    }
    dp[n][k] = best;
  }
  return dp[n][k];
}
int main() {
  while (Input()) {
    long long r1;
    Init();
    sort(a, a + n, greater<long long>());
    r1 = MaxOr(n - 1, k);
    long long r2;
    Init();
    sort(a, a + n, less<long long>());
    r2 = MaxOr(n - 1, k);
    cout << max(r1, r2) << endl;
  }
  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 MaxN = 200001;
const int MaxK = 11;
int n, k, x;
long long a[MaxN];
long long dp[MaxN][MaxK];
void Init() {
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= k; j++) {
      dp[i][j] = -1;
    }
  }
}
bool Input() {
  if (cin >> n >> k >> x) {
    for (int i = 0; i < n; i++) cin >> a[i];
    return true;
  }
  return false;
}
long long MaxOr(int n, int k) {
  if (n < 0) return 0;
  if (dp[n][k] < 0) {
    long long best, tmp, r;
    best = a[n] | MaxOr(n - 1, k);
    tmp = a[n];
    for (int i = 1; i <= k; i++) {
      tmp = tmp * x;
      r = tmp | MaxOr(n - 1, k - i);
      if (r > best) {
        best = r;
      }
    }
    dp[n][k] = best;
  }
  return dp[n][k];
}
int main() {
  while (Input()) {
    long long r1;
    Init();
    sort(a, a + n, greater<long long>());
    r1 = MaxOr(n - 1, k);
    long long r2;
    Init();
    sort(a, a + n, less<long long>());
    r2 = MaxOr(n - 1, k);
    cout << max(r1, r2) << endl;
  }
  return 0;
}

```