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.
#include <bits/stdc++.h>
using namespace std;
long long n, k, x, sum[500000], sq[500000], sh[500000], maxr;
int main() {
  while (~scanf("%ld%ld%ld", &n, &k, &x)) {
    for (int a = 0; a < n; a++) {
      scanf("%ld", &sum[a]);
      a ? maxr = maxr | sum[a] : maxr = sum[a];
      sq[a] = maxr;
    }
    for (int a = n - 1; a >= 0; a--)
      a == n - 1 ? sh[a] = sum[a] : sh[a] = sh[a + 1] | sum[a];
    for (int b = 1; b <= k; b++)
      for (int a = 0; a < n; a++) {
        if (!a)
          maxr = max(maxr, sh[a + 1] | sum[a] * (long long)pow(x, b));
        else if (a == n - 1)
          maxr = max(maxr, sq[a - 1] | sum[a] * (long long)pow(x, b));
        else
          maxr =
              max(maxr, sq[a - 1] | sh[a + 1] | sum[a] * (long long)pow(x, b));
      }
    cout << maxr << endl;
  }
  return 0;
}
