#include <bits/stdc++.h>
using namespace std;
long long rem(long long a, long long b) { return a % b; }
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long power(long long a, long long b, long long m) {
  long long ans = 1;
  while (b > 0) {
    if (b % 2 != 0) {
      ans = (ans * a) % m;
    }
    a = (a * a) % m;
    b >>= 1;
  }
  return ans;
}
inline void sf(long long *a) {
  char c = 0;
  while (c < 33) c = getc(stdin);
  *a = 0;
  while (c > 33) {
    *a = (*a) * 10 + c - '0';
    c = getc(stdin);
  }
}
long long n;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  if (n == 1) {
    cout << "-1" << endl;
  } else
    cout << n << " " << n + 1 << " " << n * (n + 1) << endl;
  return 0;
}

### 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;
long long rem(long long a, long long b) { return a % b; }
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long power(long long a, long long b, long long m) {
  long long ans = 1;
  while (b > 0) {
    if (b % 2 != 0) {
      ans = (ans * a) % m;
    }
    a = (a * a) % m;
    b >>= 1;
  }
  return ans;
}
inline void sf(long long *a) {
  char c = 0;
  while (c < 33) c = getc(stdin);
  *a = 0;
  while (c > 33) {
    *a = (*a) * 10 + c - '0';
    c = getc(stdin);
  }
}
long long n;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  if (n == 1) {
    cout << "-1" << endl;
  } else
    cout << n << " " << n + 1 << " " << n * (n + 1) << endl;
  return 0;
}

```