#include <bits/stdc++.h>
using namespace std;
const long long INF = 9e18L;
const long long mod = 1e9 + 7;
long long power(long long x, long long y) {
  long long res = 1;
  int mod = 1e9 + 7;
  while (y > 0) {
    if (y & 1) (res = res * x) %= mod;
    (x = x * x) %= mod;
    y = y >> 1;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int x = 1; x < 20001; x++) {
    if (n != x) {
      if ((n * x) % (x - n) == 0 && (n * x) / (x - n) > 0 &&
          (n * x) / (x - n) != x && (n * x) / (x - n) != n) {
        cout << n << " " << x << " " << (n * x) / (x - n) << "\n";
        return 0;
      }
    }
  }
  cout << -1 << '\n';
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
const long long INF = 9e18L;
const long long mod = 1e9 + 7;
long long power(long long x, long long y) {
  long long res = 1;
  int mod = 1e9 + 7;
  while (y > 0) {
    if (y & 1) (res = res * x) %= mod;
    (x = x * x) %= mod;
    y = y >> 1;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int x = 1; x < 20001; x++) {
    if (n != x) {
      if ((n * x) % (x - n) == 0 && (n * x) / (x - n) > 0 &&
          (n * x) / (x - n) != x && (n * x) / (x - n) != n) {
        cout << n << " " << x << " " << (n * x) / (x - n) << "\n";
        return 0;
      }
    }
  }
  cout << -1 << '\n';
}

```