#include <bits/stdc++.h>
using namespace std;
void Fast() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
void File() {}
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
int dx[] = {1, 0, 0, -1};
int dy[] = {0, 1, -1, 0};
int n, m;
bool valid(int i, int j) { return i >= 0 && i < n && j >= 0 && j < m; }
int main() {
  Fast(), File();
  long long n;
  cin >> n;
  if ((n * (n + 1)) / gcd(n, n + 1) > 1e9 ||
      n == (n * (n + 1)) / gcd(n, n + 1) ||
      n + 1 == (n * (n + 1)) / gcd(n, n + 1))
    cout << -1 << "\n";
  else
    cout << n << " " << n + 1 << " " << (n * (n + 1)) / gcd(n, n + 1) << "\n";
}

### Prompt

In CPP, your task is to solve 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;
void Fast() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
void File() {}
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
int dx[] = {1, 0, 0, -1};
int dy[] = {0, 1, -1, 0};
int n, m;
bool valid(int i, int j) { return i >= 0 && i < n && j >= 0 && j < m; }
int main() {
  Fast(), File();
  long long n;
  cin >> n;
  if ((n * (n + 1)) / gcd(n, n + 1) > 1e9 ||
      n == (n * (n + 1)) / gcd(n, n + 1) ||
      n + 1 == (n * (n + 1)) / gcd(n, n + 1))
    cout << -1 << "\n";
  else
    cout << n << " " << n + 1 << " " << (n * (n + 1)) / gcd(n, n + 1) << "\n";
}

```