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;
void sc(int& a) { scanf("%d", &a); }
void sc(long long int& a) { scanf("%lld", &a); }
void sc(int& a, int& b) {
  sc(a);
  sc(b);
}
void sc(long long int& a, long long int& b) {
  sc(a);
  sc(b);
}
void sc(int& a, int& b, int& c) {
  sc(a, b);
  sc(c);
}
void sc(long long int& a, long long int& b, long long int& c) {
  sc(a, b);
  sc(c);
}
void prl(int a) { printf("%d\n", a); }
void prl(long long int a) { printf("%lld\n", a); }
void prs(int a) { printf("%d ", a); }
void prs(long long int a) { printf("%lld ", a); }
int mod = 1000000007;
long long int modpow(long long int a, long long int b, long long int mod) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b = b / 2;
  }
  return res % mod;
}
long long int l[200005], r[200005], b[200005], a[200005];
int main() {
  int n, k, x;
  sc(n, k, x);
  for (int i = 1; i <= n; i++) {
    sc(a[i]);
    b[i] = (long long int)a[i] * ceil(pow(x, k));
    l[i] = l[i - 1] | a[i];
  }
  for (int i = n; i >= 1; i--) {
    r[i] = r[i + 1] | a[i];
  }
  long long int ans = 0;
  for (int i = 1; i <= n; i++) {
    ans = max(ans, l[i - 1] | r[i + 1] | b[i]);
  }
  prl(ans);
}
