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 Set(int N, int pos) { return N = N | (1 << pos); }
int reset(int N, int pos) { return N = N & ~(1 << pos); }
bool check(int N, int pos) { return (bool)(N & (1 << pos)); }
int cas = 1;
int dxx[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dyy[] = {-1, 0, 1, -1, 1, -1, 0, 1};
int dx[] = {-1, 0, 0, 1};
int dy[] = {0, -1, 1, 0};
long long ans, i, j, l, n, k, x, y, ai[200006], bi[200006], ci[200006];
int main() {
  scanf("%lld%lld%lld", &n, &k, &x);
  for (i = 0; i < n; i++) scanf("%lld", &ai[i]);
  y = 1;
  for (i = 0; i < k; i++) y *= x;
  ans = 0;
  bi[0] = 0;
  for (i = 0; i < n; i++) {
    bi[i + 1] = bi[i] | ai[i];
  }
  ci[n] = 0;
  for (i = n; i >= 0; i--) {
    ci[i - 1] = ci[i] | ai[i - 1];
  }
  ans = 0;
  for (i = 0; i < n; i++) {
    ans = max(ans, bi[i] | ci[i + 1] | ai[i] * y);
  }
  printf("%lld\n", ans);
}
