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;
int INT_MAX_VAL = (int)0x3F3F3F3F;
int INT_MIN_VAL = (int)-0x3F3F3F3F;
long long LONG_MAX_VAL = (long long)0x3F3F3F3F3F3F3F3F;
long long LONG_MIN_VAL = (long long)-0x3F3F3F3F3F3F3F3F;
long long vs[500006];
long long prf[500006];
long long sfx[500006];
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  long long k, x;
  cin >> k >> x;
  for (long long(i) = 0LL; (i) < (n); (i)++) cin >> vs[i];
  prf[0] = vs[0];
  for (int i = 1; i < n; ++i) {
    prf[i] = prf[i - 1] | vs[i];
  }
  sfx[n - 1] = vs[n - 1];
  for (int i = n - 2; i >= 0; --i) {
    sfx[i] = sfx[i + 1] | vs[i];
  }
  long long res = 0;
  for (int i = 0; i < n; ++i) {
    long long cur = vs[i];
    for (long long(i) = 0LL; (i) < (k); (i)++) cur *= x;
    if (i > 0) cur |= prf[i - 1];
    if (i + 1 < n) cur |= sfx[i + 1];
    res = max(res, cur);
  }
  cout << res << '\n';
  return 0;
}
