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;
const int Nmax = 2 * 1e5;
int a[Nmax], n, k, x;
long long lef[Nmax], rgt[Nmax];
long long mPow(long long a, long long b) {
  long long res = 1;
  for (int i = 0; i < b; i++) {
    res *= a;
  }
  return res;
}
int main() {
  scanf("%d %d %d", &n, &k, &x);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  lef[0] = a[0];
  rgt[n - 1] = a[n - 1];
  for (int i = 1; i < n; i++) {
    lef[i] = lef[i - 1] | a[i];
  }
  for (int i = n - 2; i >= 0; i--) {
    rgt[i] = rgt[i + 1] | a[i];
  }
  long long Max = -1;
  if (n == 1) {
    cout << mPow(x, k) * a[0];
  } else {
    for (int i = 0; i < n; i++) {
      if (i == 0) {
        Max = max((mPow(x, k) * a[0]) | rgt[1], Max);
      } else if (i == n - 1) {
        Max = max((mPow(x, k) * a[n - 1]) | lef[n - 2], Max);
      } else {
        Max = max((mPow(x, k) * a[i]) | lef[i - 1] | rgt[i + 1], Max);
      }
    }
    cout << Max;
  }
  int j = 1;
}
