#include <bits/stdc++.h>
using namespace std;
long double PI = acosl(-1);
bool compare_int(int a, int b) { return (a > b); }
bool compare_string(string a, string b) { return a.size() < b.size(); }
bool compare_pair(const pair<int, int> &a, const pair<int, int> &b) {
  if (a.second == b.second)
    return a.first < b.first;
  else
    return (a.second > b.second);
}
bool cmp(pair<string, int> x, pair<string, int> y) {
  return (x.second < y.second);
}
void NA() {
  printf("NO\n");
  exit(0);
}
void YA() {
  printf("YES\n");
  exit(0);
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  if (n == 1)
    cout << -1 << endl;
  else
    cout << n << ' ' << n + 1 << ' ' << n * (n + 1) << endl;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long double PI = acosl(-1);
bool compare_int(int a, int b) { return (a > b); }
bool compare_string(string a, string b) { return a.size() < b.size(); }
bool compare_pair(const pair<int, int> &a, const pair<int, int> &b) {
  if (a.second == b.second)
    return a.first < b.first;
  else
    return (a.second > b.second);
}
bool cmp(pair<string, int> x, pair<string, int> y) {
  return (x.second < y.second);
}
void NA() {
  printf("NO\n");
  exit(0);
}
void YA() {
  printf("YES\n");
  exit(0);
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  if (n == 1)
    cout << -1 << endl;
  else
    cout << n << ' ' << n + 1 << ' ' << n * (n + 1) << endl;
}

```