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>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
long long n, k, x, a[1000006], b[1000006], p[100];
map<int, bool> m;
int main() {
  cin >> n >> k >> x;
  long long y = x;
  for (int i = 2; i <= k; i++) x *= y;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    b[i] = a[n - i + 1];
  }
  long long res = 0;
  long long ans = 0;
  for (int i = 0; i <= 100; i++) a[i] = 0;
  for (int i = 1; i <= n; i++) {
    ans = ans | b[i];
    res = b[i];
    int j = 0;
    while (res > 0) {
      a[j] += res % 2;
      res /= 2;
      j++;
    }
  }
  p[0] = 1;
  for (int i = 1; i <= 40; i++) p[i] = p[i - 1] * 2;
  res = 0;
  for (int i = 1; i <= n; i++) {
    long long z = b[i], y = b[i], mda = ans;
    z *= x;
    int j = 0;
    while (y > 0) {
      if (y % 2 == 1 && a[j] == 1) mda -= p[j];
      y /= 2;
      j++;
    }
    mda = mda | z;
    if (mda > res) res = mda;
  }
  cout << res << endl;
  return 0;
}
