#include <bits/stdc++.h>
using namespace std;
long long n, k, x;
long long m[200010];
long long answer;
struct node {
  long long pre;
  long long next;
} p[200010];
int main() {
  long long a;
  while (~scanf("%I64d %I64d %I64d", &n, &k, &x)) {
    answer = p[1].pre = p[n].next = 0;
    a = x;
    for (long long i = k - 1; i > 0; i--) a *= x;
    for (long long i = 1; i <= n; i++) {
      scanf("%d", &m[i]);
      p[i].pre = (m[i - 1] | p[i - 1].pre);
    }
    for (long long i = n; i >= 1; i--) {
      if (i < n) p[i].next = (p[i + 1].next | m[i + 1]);
      answer = max(answer, (p[i].pre | p[i].next | m[i] * a));
    }
    printf("%I64d\n", answer);
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
long long m[200010];
long long answer;
struct node {
  long long pre;
  long long next;
} p[200010];
int main() {
  long long a;
  while (~scanf("%I64d %I64d %I64d", &n, &k, &x)) {
    answer = p[1].pre = p[n].next = 0;
    a = x;
    for (long long i = k - 1; i > 0; i--) a *= x;
    for (long long i = 1; i <= n; i++) {
      scanf("%d", &m[i]);
      p[i].pre = (m[i - 1] | p[i - 1].pre);
    }
    for (long long i = n; i >= 1; i--) {
      if (i < n) p[i].next = (p[i + 1].next | m[i + 1]);
      answer = max(answer, (p[i].pre | p[i].next | m[i] * a));
    }
    printf("%I64d\n", answer);
  }
  return 0;
}

```