#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 50;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
long long num[maxn * 2], xx;
int n, k, len, a[50];
void get(long long x) {
  len = -1;
  while (x) {
    len++;
    x >>= 1;
  }
}
void deal(long long x) {
  int i = 0;
  while (x) {
    a[i] += x & 1;
    x >>= 1;
    i++;
  }
}
void solve(long long x, int d) {
  int i = 0;
  while (x) {
    a[i] += d * (x & 1);
    x >>= 1;
    i++;
  }
}
long long get_num(long long x) {
  int i;
  long long res = 0;
  for (i = len; i >= 0; i--) {
    res <<= 1;
    if (a[i]) res++;
  }
  return res | x;
}
int main() {
  scanf("%d%d%I64d", &n, &k, &xx);
  int d = xx;
  while (--k) xx *= d;
  for (int i = 0; i < n; i++) scanf("%I64d", num + i);
  sort(num, num + n);
  get(num[n - 1]);
  for (int i = 0; i < n; i++) deal(num[i]);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    long long t = num[i] * xx;
    solve(num[i], -1);
    long long tmp = get_num(t);
    ans = max(tmp, ans);
    solve(num[i], 1);
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Your task is to create 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;
const int maxn = 1e5 + 50;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
long long num[maxn * 2], xx;
int n, k, len, a[50];
void get(long long x) {
  len = -1;
  while (x) {
    len++;
    x >>= 1;
  }
}
void deal(long long x) {
  int i = 0;
  while (x) {
    a[i] += x & 1;
    x >>= 1;
    i++;
  }
}
void solve(long long x, int d) {
  int i = 0;
  while (x) {
    a[i] += d * (x & 1);
    x >>= 1;
    i++;
  }
}
long long get_num(long long x) {
  int i;
  long long res = 0;
  for (i = len; i >= 0; i--) {
    res <<= 1;
    if (a[i]) res++;
  }
  return res | x;
}
int main() {
  scanf("%d%d%I64d", &n, &k, &xx);
  int d = xx;
  while (--k) xx *= d;
  for (int i = 0; i < n; i++) scanf("%I64d", num + i);
  sort(num, num + n);
  get(num[n - 1]);
  for (int i = 0; i < n; i++) deal(num[i]);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    long long t = num[i] * xx;
    solve(num[i], -1);
    long long tmp = get_num(t);
    ans = max(tmp, ans);
    solve(num[i], 1);
  }
  cout << ans << "\n";
  return 0;
}

```