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 MaxN = 200010;
int N, K, X;
long long S;
long long a[MaxN], b[MaxN];
long long L[MaxN], R[MaxN];
long long Ans;
int main() {
  scanf("%d%d%d", &N, &K, &X);
  S = 1LL;
  for (int i = 1; i <= K; ++i) S *= X;
  for (int i = 1; i <= N; ++i) {
    scanf("%I64d", &a[i]);
    b[i] = a[i] * S;
    L[i] = L[i - 1] | a[i];
  }
  for (int i = N; i; --i) R[i] = R[i + 1] | a[i];
  for (int i = 1; i <= N; ++i) Ans = max(Ans, L[i - 1] | b[i] | R[i + 1]);
  printf("%I64d\n", Ans);
}
