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 N, K, X;
bitset<32> A[200200];
int bits[64];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> K >> X;
  for (int n = 0; n < N; ++n) {
    unsigned a;
    cin >> a;
    A[n] = bitset<32>(a);
    for (int i = 0; i < 32; ++i)
      if (A[n][i]) ++bits[i];
  }
  unsigned long long Y = 1;
  while (K--) Y *= X;
  unsigned long long ans = 0;
  for (int n = 0; n < N; ++n) {
    for (int i = 0; i < 32; ++i)
      if (A[n][i]) --bits[i];
    bitset<64> B(Y * A[n].to_ulong());
    bitset<64> tmpans;
    for (int i = 0; i < 64; ++i)
      if (bits[i] || B[i]) tmpans[i] = 1;
    ans = max(ans, tmpans.to_ullong());
    for (int i = 0; i < 32; ++i)
      if (A[n][i]) ++bits[i];
  }
  cout << ans << '\n';
}
