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 = 200000 + 5;
int a[maxn], num[40];
long long q[40];
int main() {
  q[0] = 1;
  for (int i = 1; i < 40; ++i) {
    q[i] = q[i - 1] << 1;
  }
  int n, k, x;
  while (~scanf("%d%d%d", &n, &k, &x)) {
    memset(num, 0, sizeof(num));
    long long temp = 0, ans = 0;
    for (int i = 0; i < n; ++i) {
      scanf("%d", a + i);
      temp |= a[i];
      for (int j = 0; j < 32; ++j) {
        if (a[i] >> j & 1) {
          ++num[j];
        }
      }
    }
    long long p = 1;
    ans = temp;
    for (int l = 0; l < k; ++l) {
      p *= x;
    }
    for (int i = 0; i < n; ++i) {
      long long s = temp;
      for (int j = 0; j < 32; ++j) {
        if ((a[i] >> j & 1) && num[j] == 1) {
          s ^= q[j];
        }
      }
      ans = max(ans, s | (a[i] * p));
    }
    printf("%I64d\n", ans);
  }
  return 0;
}
