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;
long long t, d, ct, q, ans, l, m, r, n, mx, mn, i, j, k, x, y, z, s;
char ch;
string ss = "", str = "";
long long a[200003], b[200003];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  s = k = d = l = mx = m = mn = r = q = i = j = z = x = y = ct = ans = 0;
  cin >> n >> k >> x;
  for (i = 1; i <= n; i++) cin >> a[i];
  long long dp[n + 1][2 * k + 2];
  m = 50;
  while (m--) {
    for (i = 0; i <= n; i++)
      for (j = 0; j <= k; j++) dp[i][j] = 0;
    random_shuffle(a + 1, a + n + 1);
    for (i = 1; i <= n; i++) {
      double cur = clock() * 1.0 / CLOCKS_PER_SEC;
      if (cur > 1.75) break;
      y = a[i];
      for (j = 0; j <= k; j++) {
        for (l = 0; l <= k - j; l++)
          dp[i][j + l] = max(dp[i][j + l], y | dp[i - 1][l]);
        y *= x;
      }
    }
    for (i = 0; i < k + 1; i++) ans = max(ans, dp[n][i]);
  }
  cout << ans;
  return 0;
}
