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;
const int MOD = (int)1e9 + 7;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
long long pow(long long n, long long k) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n;
    n = n * n;
  }
  return r;
}
template <class T>
void setmin(T& a, T val) {
  if (a > val) a = val;
}
template <class T>
void setmax(T& a, T val) {
  if (a < val) a = val;
}
void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
const int N = 200005;
long long a[N], pref[N], suff[N];
long long solve(int n, int k, int x) {
  long long val = pow(x, k);
  long long ans = 0;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  long long lim = a[n] / 2, pos = 0, sum = 0;
  for (int i = 1; a[i] <= lim; i++) {
    sum = sum | a[i];
    pos = i;
  }
  for (int i = n; i > pos; i--) suff[i] = suff[i + 1] | a[i];
  for (int i = pos + 1; i <= n; i++) pref[i] = pref[i - 1] | a[i];
  for (int i = pos + 1; i <= n; i++) {
    long long value = a[i] * val;
    long long fans = pref[i - 1] | value | suff[i + 1];
    ans = max(ans, sum | fans);
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(NULL);
  int n, k, x;
  cin >> n >> k >> x;
  cout << solve(n, k, x) << "\n";
  return 0;
}
