#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long A[200003], B[200003], C[200003];
long long count2, r, c;
string S[50001];
int main() {
  long long i, j, k, s, a, b, c, x, y, z, m, n, t, p, q, T, sum, ans, pos, val,
      count;
  scanf("%lld%lld%lld", &n, &k, &x);
  p = 1;
  for (i = 1; i <= k; i++) p *= x;
  B[0] = 0;
  for (i = 1; i <= n; i++) {
    scanf("%lld", &A[i]);
    B[i] = B[i - 1] | A[i];
  }
  C[n + 1] = 0;
  for (i = n; i >= 1; i--) {
    C[i] = C[i + 1] | A[i];
  }
  ans = 0;
  for (i = 1; i <= n; i++) {
    ans = max(ans, B[i - 1] | C[i + 1] | (A[i] * p));
  }
  printf("%lld\n", 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 long long mod = 1e9 + 7;
long long A[200003], B[200003], C[200003];
long long count2, r, c;
string S[50001];
int main() {
  long long i, j, k, s, a, b, c, x, y, z, m, n, t, p, q, T, sum, ans, pos, val,
      count;
  scanf("%lld%lld%lld", &n, &k, &x);
  p = 1;
  for (i = 1; i <= k; i++) p *= x;
  B[0] = 0;
  for (i = 1; i <= n; i++) {
    scanf("%lld", &A[i]);
    B[i] = B[i - 1] | A[i];
  }
  C[n + 1] = 0;
  for (i = n; i >= 1; i--) {
    C[i] = C[i + 1] | A[i];
  }
  ans = 0;
  for (i = 1; i <= n; i++) {
    ans = max(ans, B[i - 1] | C[i + 1] | (A[i] * p));
  }
  printf("%lld\n", ans);
  return 0;
}

```