#include <bits/stdc++.h>
using namespace std;
int numSet[40];
vector<long long> nums;
void add(long long num) {
  for (long long i = 0; i < 33; i++) {
    numSet[i] += (num & (1LL << i)) ? 1 : 0;
  }
}
long long rem(long long num) {
  long long res = 0;
  for (long long i = 0; i < 33; i++) {
    if (numSet[i] == 1 && (num & (1LL << i))) {
      continue;
    }
    if (numSet[i] > 0) res |= (1LL << i);
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  long long n, k, x;
  cin >> n >> k >> x;
  nums.resize(n);
  long long acc;
  for (int i = 0; i < n; i++) {
    cin >> nums[i];
    add(nums[i]);
  }
  long long res = nums[0];
  for (int i = 0; i < n; i++) {
    long long without = rem(nums[i]);
    for (int j = 0; j < k; j++) nums[i] *= x;
    res = max(res, without | nums[i]);
  }
  cout << res;
  return 0;
}

### Prompt

Please formulate 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;
int numSet[40];
vector<long long> nums;
void add(long long num) {
  for (long long i = 0; i < 33; i++) {
    numSet[i] += (num & (1LL << i)) ? 1 : 0;
  }
}
long long rem(long long num) {
  long long res = 0;
  for (long long i = 0; i < 33; i++) {
    if (numSet[i] == 1 && (num & (1LL << i))) {
      continue;
    }
    if (numSet[i] > 0) res |= (1LL << i);
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  long long n, k, x;
  cin >> n >> k >> x;
  nums.resize(n);
  long long acc;
  for (int i = 0; i < n; i++) {
    cin >> nums[i];
    add(nums[i]);
  }
  long long res = nums[0];
  for (int i = 0; i < n; i++) {
    long long without = rem(nums[i]);
    for (int j = 0; j < k; j++) nums[i] *= x;
    res = max(res, without | nums[i]);
  }
  cout << res;
  return 0;
}

```