#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
int n, k, x;
long long a[maxn], b[maxn];
long long fr[maxn], be[maxn];
int main() {
  while (~scanf("%d%d%d", &n, &k, &x)) {
    fr[0] = 0;
    long long cnt = 0;
    long long xx = 1;
    while (k--) xx *= x;
    for (int i = 0; i < n; i++) {
      scanf("%lld", &a[i]);
      b[i] = a[i] * xx;
      if (i == 0) continue;
      cnt = cnt | a[i - 1];
      fr[i] = cnt;
    }
    cnt = 0;
    be[n - 1] = 0;
    for (int i = n - 1; i >= 0; i--) {
      if (i == n - 1) continue;
      cnt = cnt | a[i + 1];
      be[i] = cnt;
    }
    long long maxn = 0;
    for (int i = 0; i < n; i++) {
      long long num = fr[i] | b[i] | be[i];
      maxn = max(maxn, num);
    }
    printf("%lld\n", maxn);
  }
  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 = 2e5 + 10;
int n, k, x;
long long a[maxn], b[maxn];
long long fr[maxn], be[maxn];
int main() {
  while (~scanf("%d%d%d", &n, &k, &x)) {
    fr[0] = 0;
    long long cnt = 0;
    long long xx = 1;
    while (k--) xx *= x;
    for (int i = 0; i < n; i++) {
      scanf("%lld", &a[i]);
      b[i] = a[i] * xx;
      if (i == 0) continue;
      cnt = cnt | a[i - 1];
      fr[i] = cnt;
    }
    cnt = 0;
    be[n - 1] = 0;
    for (int i = n - 1; i >= 0; i--) {
      if (i == n - 1) continue;
      cnt = cnt | a[i + 1];
      be[i] = cnt;
    }
    long long maxn = 0;
    for (int i = 0; i < n; i++) {
      long long num = fr[i] | b[i] | be[i];
      maxn = max(maxn, num);
    }
    printf("%lld\n", maxn);
  }
  return 0;
}

```