#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
long long dp[200005][15];
long long dp2[200005][15];
long long a[200005];
long long power[25] = {0};
long long x, k, n;
long long func1(int idx, int k) {
  if (idx == n)
    return 0;
  else if (dp[idx][k] != -1)
    return dp[idx][k];
  else {
    long long ret = -1;
    for (long long i = 0; i <= k; i++) {
      ret = max(ret, (a[idx] * power[i]) | func1(idx + 1, k - i));
    }
    dp[idx][k] = ret;
    return dp[idx][k];
  }
}
long long func2(int idx, int k) {
  if (idx == -1)
    return 0;
  else if (dp2[idx][k] != -1)
    return dp2[idx][k];
  else {
    long long ret = -1;
    for (long long i = 0; i <= k; i++) {
      ret = max(ret, (a[idx] * power[i]) | func2(idx - 1, k - i));
    }
    dp2[idx][k] = ret;
    return dp2[idx][k];
  }
}
void test() {
  long long a, b, c;
  for (int i = 0; i < 10; i++) {
    cin >> a >> b >> c;
    cout << a << " " << b << " " << c << " " << (a | b | c) << endl;
  }
}
int main() {
  memset(dp, -1, sizeof dp);
  memset(dp2, -1, sizeof dp2);
  cin >> n >> k >> x;
  power[0] = 1;
  power[1] = x;
  for (int i = 2; i <= k; i++) power[i] = power[i - 1] * x;
  for (int i = 0; i < n; i++) cin >> a[i];
  cout << max(func2(n - 1, k), func1(0, k)) << endl;
}

### Prompt

Create a solution in cpp for 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 gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
long long dp[200005][15];
long long dp2[200005][15];
long long a[200005];
long long power[25] = {0};
long long x, k, n;
long long func1(int idx, int k) {
  if (idx == n)
    return 0;
  else if (dp[idx][k] != -1)
    return dp[idx][k];
  else {
    long long ret = -1;
    for (long long i = 0; i <= k; i++) {
      ret = max(ret, (a[idx] * power[i]) | func1(idx + 1, k - i));
    }
    dp[idx][k] = ret;
    return dp[idx][k];
  }
}
long long func2(int idx, int k) {
  if (idx == -1)
    return 0;
  else if (dp2[idx][k] != -1)
    return dp2[idx][k];
  else {
    long long ret = -1;
    for (long long i = 0; i <= k; i++) {
      ret = max(ret, (a[idx] * power[i]) | func2(idx - 1, k - i));
    }
    dp2[idx][k] = ret;
    return dp2[idx][k];
  }
}
void test() {
  long long a, b, c;
  for (int i = 0; i < 10; i++) {
    cin >> a >> b >> c;
    cout << a << " " << b << " " << c << " " << (a | b | c) << endl;
  }
}
int main() {
  memset(dp, -1, sizeof dp);
  memset(dp2, -1, sizeof dp2);
  cin >> n >> k >> x;
  power[0] = 1;
  power[1] = x;
  for (int i = 2; i <= k; i++) power[i] = power[i - 1] * x;
  for (int i = 0; i < n; i++) cin >> a[i];
  cout << max(func2(n - 1, k), func1(0, k)) << endl;
}

```