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>
long long int arr[210000];
long long int count[210000];
long long int sum[210000];
long long int temp[210000];
long long int bin[210000][64];
long long int spc[210000];
int main() {
  long long int i, j, p, s;
  long long int mul = 1;
  long long int n, k, x;
  scanf("%I64d%I64d%I64d", &n, &k, &x);
  for (i = 1; i <= k; i++) mul *= x;
  for (i = 1; i <= n; i++) scanf("%I64d", &arr[i]);
  for (i = 1; i <= n; i++) {
    for (j = 0, p = 1; j < 64; j++, p *= 2) {
      if (((arr[i]) & (p)) == (p)) {
        bin[i][j] = 1;
        sum[j] += 1;
      } else
        bin[i][j] = 0;
      if (arr[i] < p) break;
    }
  }
  long long int t;
  long long int max = 0;
  long long int summ;
  for (i = 1; i <= n; i++) {
    for (s = 0; s < 64; s++) {
      count[s] = sum[s];
      if (count[s] - bin[i][s] == 0) count[s] = 0;
    }
    t = arr[i] * mul;
    summ = 0;
    for (j = 0, p = 1; j < 64; j++, p *= 2) {
      if (((t) & (p)) == (p)) count[j] = 1;
      if (count[j] >= 1) summ += p;
    }
    if (summ > max) max = summ;
  }
  printf("%I64d", max);
  return 0;
}
