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 int modpow(long long int a, long long int n, long long int temp) {
  long long int res = 1, y = a;
  while (n > 0) {
    if (n & 1) res = (res * y) % temp;
    y = (y * y) % temp;
    n /= 2;
  }
  return res % temp;
}
long long int a[200005];
long long int fr[200005];
long long int x, tot;
void rec(long long int rem, long long int ans, long long int st) {
  if (st < 0) {
    tot = ((ans) > (tot) ? (ans) : (tot));
    return;
  }
  if (rem == 0) {
    tot = ((ans | fr[st]) > (tot) ? (ans | fr[st]) : (tot));
    return;
  }
  long long int i, j, cr = 1;
  rec(rem, ans | (a[st] * cr), st - 1);
  rec(0, ans | (a[st] * x), st - 1);
}
int main() {
  long long int i, j, n, t, k;
  scanf("%lld", &n);
  scanf("%lld", &k);
  scanf("%lld", &x);
  long long int mx = 0;
  for (i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
  }
  sort(a, a + n);
  long long int ck = k;
  long long int cx = 1;
  while (ck != 0) {
    cx *= x;
    ck--;
  }
  x = cx;
  for (i = 0; i < n; i++) {
    fr[i] = fr[i - 1] | a[i];
  }
  rec(k, 0, n - 1);
  cout << tot << endl;
  return 0;
}
