#include <bits/stdc++.h>
using namespace std;
int a[200010], bit[35];
int main() {
  memset(a, 0, sizeof(a));
  int n, k, x;
  scanf("%d%d%d", &n, &k, &x);
  for (int i = 0, k, cur; i < n; i++) {
    scanf("%d", &a[i]);
    k = a[i];
    cur = 0;
    while (k) {
      bit[cur++] += (k & 1);
      k >>= 1;
    }
  }
  long long ans = 0, st = 1;
  for (int i = 0; i < k; i++) st *= x;
  for (int i = 0, k = a[i], cur; i < n; i++) {
    long long tmp = 0;
    k = a[i];
    cur = 0;
    while (k) {
      bit[cur++] -= (k & 1);
      k >>= 1;
    }
    for (int j = 0; j <= 30; j++)
      if (bit[j]) tmp |= (1 << j);
    tmp |= a[i] * st;
    if (ans < tmp) ans = tmp;
    k = a[i];
    cur = 0;
    while (k) {
      bit[cur++] += (k & 1);
      k >>= 1;
    }
  }
  printf("%I64d\n", ans);
  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 a[200010], bit[35];
int main() {
  memset(a, 0, sizeof(a));
  int n, k, x;
  scanf("%d%d%d", &n, &k, &x);
  for (int i = 0, k, cur; i < n; i++) {
    scanf("%d", &a[i]);
    k = a[i];
    cur = 0;
    while (k) {
      bit[cur++] += (k & 1);
      k >>= 1;
    }
  }
  long long ans = 0, st = 1;
  for (int i = 0; i < k; i++) st *= x;
  for (int i = 0, k = a[i], cur; i < n; i++) {
    long long tmp = 0;
    k = a[i];
    cur = 0;
    while (k) {
      bit[cur++] -= (k & 1);
      k >>= 1;
    }
    for (int j = 0; j <= 30; j++)
      if (bit[j]) tmp |= (1 << j);
    tmp |= a[i] * st;
    if (ans < tmp) ans = tmp;
    k = a[i];
    cur = 0;
    while (k) {
      bit[cur++] += (k & 1);
      k >>= 1;
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

```