#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, k;
long long x, a[200050], bin[32], cnt[32];
int main() {
  bin[0] = 1;
  for (int i = 1; i <= 31; i++) bin[i] = bin[i - 1] << 1;
  n = read(), k = read(), x = read();
  for (int i = 1, t = x; i < k; i++) x *= t;
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 32; j++)
      if (bin[j] & a[i]) cnt[j]++;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long res = 0;
    for (int j = 0; j < 32; j++)
      if (bin[j] & a[i]) cnt[j]--;
    for (int j = 0; j < 32; j++)
      if (cnt[j]) res |= bin[j];
    ans = max(ans, res | ((long long)x * a[i]));
    for (int j = 0; j < 32; j++)
      if (bin[j] & a[i]) cnt[j]++;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Generate 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;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, k;
long long x, a[200050], bin[32], cnt[32];
int main() {
  bin[0] = 1;
  for (int i = 1; i <= 31; i++) bin[i] = bin[i - 1] << 1;
  n = read(), k = read(), x = read();
  for (int i = 1, t = x; i < k; i++) x *= t;
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 32; j++)
      if (bin[j] & a[i]) cnt[j]++;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long res = 0;
    for (int j = 0; j < 32; j++)
      if (bin[j] & a[i]) cnt[j]--;
    for (int j = 0; j < 32; j++)
      if (cnt[j]) res |= bin[j];
    ans = max(ans, res | ((long long)x * a[i]));
    for (int j = 0; j < 32; j++)
      if (bin[j] & a[i]) cnt[j]++;
  }
  cout << ans << endl;
  return 0;
}

```