#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int N = 1e3 + 5;
const long long inf = 1e18 + 1;
const long double eps = pow(2, -52);
int main() {
  long long n, cnt = 0;
  cin >> n;
  if (n == 1) {
    cout << -1;
    return 0;
  }
  for (long long x = 1; x < 10005; x++) {
    for (long long y = 1; y < 10005; y++) {
      long long temp1 = x * y;
      long long temp2 = 2 * x * y - n * y - n * x;
      if (temp2 != 0 && temp2 > 0) {
        if (temp1 % temp2 == 0) {
          if (n * temp1 / temp2 <= 1000000000) {
            cout << x << " " << y << " " << n * temp1 / temp2;
            return 0;
          }
        }
      }
    }
  }
  cout << -1;
  return 0;
}

### Prompt

Your task is to create 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;
const int MOD = 1e9 + 7;
const int N = 1e3 + 5;
const long long inf = 1e18 + 1;
const long double eps = pow(2, -52);
int main() {
  long long n, cnt = 0;
  cin >> n;
  if (n == 1) {
    cout << -1;
    return 0;
  }
  for (long long x = 1; x < 10005; x++) {
    for (long long y = 1; y < 10005; y++) {
      long long temp1 = x * y;
      long long temp2 = 2 * x * y - n * y - n * x;
      if (temp2 != 0 && temp2 > 0) {
        if (temp1 % temp2 == 0) {
          if (n * temp1 / temp2 <= 1000000000) {
            cout << x << " " << y << " " << n * temp1 / temp2;
            return 0;
          }
        }
      }
    }
  }
  cout << -1;
  return 0;
}

```