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;
long long s, t, p, ok = 0;
long long a[200001];
int b[200001][101];
int ans[101];
int main() {
  scanf("%d%d%d", &n, &k, &x);
  s = 1;
  for (int i = 1; i <= k; i++) s = s * x;
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    t = a[i];
    int j = 0;
    while (t) {
      j++;
      b[i][j] = t % 2;
      ans[j] += b[i][j];
      t /= 2;
    }
  }
  int maxi = 100;
  while (maxi > 0 && ans[maxi] == 0) maxi--;
  if (maxi == 0) {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= maxi; j++) ans[j] -= b[i][j];
    p = 0;
    for (int j = 1; j <= maxi; j++) {
      if (j == 1)
        t = 1;
      else
        t = t * 2;
      if (ans[j]) p = p + t;
    }
    t = p | ((long long)a[i] * s);
    if (t > ok) ok = t;
    for (int j = 1; j <= maxi; j++) ans[j] += b[i][j];
  }
  cout << ok << endl;
  return 0;
}
