Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction <image> as a sum of three distinct positive fractions in form <image>.

Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that <image>. Because Chloe can't check Vladik's answer if the numbers are large, he asks you to print numbers not exceeding 109.

If there is no such answer, print -1.

Input

The single line contains single integer n (1 ≤ n ≤ 104).

Output

If the answer exists, print 3 distinct numbers x, y and z (1 ≤ x, y, z ≤ 109, x ≠ y, x ≠ z, y ≠ z). Otherwise print -1.

If there are multiple answers, print any of them.

Examples

Input

3


Output

2 7 42


Input

7


Output

7 8 56
#include <bits/stdc++.h>
using namespace std;
template <typename T>
T prod(const T &a, const T &b, const T &mod) {
  return ((a % mod) * (b % mod)) % mod;
}
template <typename T>
T pow_(const T &a, const T &b) {
  if (!b) return 1;
  T p = pow_(a, b / 2);
  p *= p;
  return (b % 2) ? (p * a) : p;
}
template <typename T>
T modpow(const T &a, const T &b, const T &mod) {
  if (!b) return 1;
  T p = modpow(a, b / 2);
  p = prod(p, p);
  return (b % 2) ? (prod(p, a)) : p;
}
template <typename T>
T max(const T &a, const T &b, const T &c) {
  return max(a, max(b, c));
}
template <typename T>
T min(const T &a, const T &b, const T &c) {
  return min(a, min(b, c));
}
template <typename T>
void dbg(const vector<T> &a, int l, int r) {
  for (long long int i = l; i <= r; i++) cout << a[i] << " ";
  cout << endl;
}
template <typename T>
void dbg(const vector<T> &a) {
  for (long long int i = 0; i <= a.size() - 1; i++) cout << a[i] << " ";
  cout << endl;
}
void main_() {
  long long int n;
  cin >> n;
  if (n == 1)
    cout << "-1\n";
  else
    cout << n << " " << n + 1 << " " << n * (n + 1) << endl;
}
int main() {
  { main_(); }
  return 0;
}
