#include <bits/stdc++.h>
using namespace std;
long long arr[200005], lef[200005], righ[200005];
long long power(long long base, long long exp) {
  long long res = 1;
  while (exp != 0) {
    if (exp % 2 == 1) res = (res * base) % 1000000000000000007;
    base = (base * base) % 1000000000000000007;
    exp = exp >> 1;
  }
  return res;
}
int main() {
  long long n, k, x;
  scanf("%I64d", &n);
  scanf("%I64d", &k);
  scanf("%I64d", &x);
  x = power(x, k);
  long long i, xo = 0, j;
  for (i = 1; i <= n; i++) {
    scanf("%I64d", &arr[i]);
    lef[i] = (lef[i - 1]) | (arr[i]);
  }
  for (i = n; i >= 1; i--) {
    righ[i] = righ[i + 1] | arr[i];
  }
  long long orr = 0;
  k = 1;
  for (i = 1; i <= k; i++) {
    long long in = 0;
    for (j = 1; j <= n; j++) {
      long long y = x * arr[j];
      long long t = lef[j - 1] | righ[j + 1] | y;
      if (t > orr) {
        orr = t;
        in = j;
      }
    }
    if (in != 0) {
      arr[in] *= x;
      for (j = n; j >= 1; j--) {
        righ[j] = righ[j + 1] | arr[j];
      }
      for (j = 1; j <= n; j++) {
        lef[j] = lef[j - 1] | arr[j];
      }
    }
  }
  cout << orr;
  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;
long long arr[200005], lef[200005], righ[200005];
long long power(long long base, long long exp) {
  long long res = 1;
  while (exp != 0) {
    if (exp % 2 == 1) res = (res * base) % 1000000000000000007;
    base = (base * base) % 1000000000000000007;
    exp = exp >> 1;
  }
  return res;
}
int main() {
  long long n, k, x;
  scanf("%I64d", &n);
  scanf("%I64d", &k);
  scanf("%I64d", &x);
  x = power(x, k);
  long long i, xo = 0, j;
  for (i = 1; i <= n; i++) {
    scanf("%I64d", &arr[i]);
    lef[i] = (lef[i - 1]) | (arr[i]);
  }
  for (i = n; i >= 1; i--) {
    righ[i] = righ[i + 1] | arr[i];
  }
  long long orr = 0;
  k = 1;
  for (i = 1; i <= k; i++) {
    long long in = 0;
    for (j = 1; j <= n; j++) {
      long long y = x * arr[j];
      long long t = lef[j - 1] | righ[j + 1] | y;
      if (t > orr) {
        orr = t;
        in = j;
      }
    }
    if (in != 0) {
      arr[in] *= x;
      for (j = n; j >= 1; j--) {
        righ[j] = righ[j + 1] | arr[j];
      }
      for (j = 1; j <= n; j++) {
        lef[j] = lef[j - 1] | arr[j];
      }
    }
  }
  cout << orr;
  return 0;
}

```