#include <bits/stdc++.h>
using namespace std;
const int N = 1e9 + 7;
int pow(int x, int y) {
  int res = 1;
  while (y > 0) {
    if (y & 1) res = ((res) * (x));
    res %= N;
    y = y >> 1;
    x = ((x) * (x));
  }
  return res;
}
long long int powm(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void codeit() {
  long long int n;
  cin >> n;
  if (n == 1)
    cout << -1;
  else
    cout << n << " " << n + 1 << " " << n * n + n;
  return;
}
int main() {
  for (int i = 0; i < 1; i++) codeit();
}

### Prompt

Develop a solution in cpp 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 int N = 1e9 + 7;
int pow(int x, int y) {
  int res = 1;
  while (y > 0) {
    if (y & 1) res = ((res) * (x));
    res %= N;
    y = y >> 1;
    x = ((x) * (x));
  }
  return res;
}
long long int powm(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void codeit() {
  long long int n;
  cin >> n;
  if (n == 1)
    cout << -1;
  else
    cout << n << " " << n + 1 << " " << n * n + n;
  return;
}
int main() {
  for (int i = 0; i < 1; i++) codeit();
}

```