#include <bits/stdc++.h>
const double PI = acos(-1), EPS = 1e-7;
using namespace std;
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
long long gcd(long long x, long long y) { return (!y) ? x : gcd(y, x % y); }
long long lcm(long long x, long long y) { return ((x / gcd(x, y)) * y); }
const long long mod = 998244353;
long long power(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  long long r = power(x, y / 2) % mod;
  return r % mod * r % mod * power(x, y % 2) % mod;
}
void file() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int main() {
  file();
  int n;
  cin >> n;
  if (n == 1)
    cout << "-1" << '\n';
  else
    cout << n << " " << n + 1 << " " << n * (n + 1) << '\n';
}

### Prompt

Please create a solution in cpp 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>
const double PI = acos(-1), EPS = 1e-7;
using namespace std;
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
long long gcd(long long x, long long y) { return (!y) ? x : gcd(y, x % y); }
long long lcm(long long x, long long y) { return ((x / gcd(x, y)) * y); }
const long long mod = 998244353;
long long power(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  long long r = power(x, y / 2) % mod;
  return r % mod * r % mod * power(x, y % 2) % mod;
}
void file() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int main() {
  file();
  int n;
  cin >> n;
  if (n == 1)
    cout << "-1" << '\n';
  else
    cout << n << " " << n + 1 << " " << n * (n + 1) << '\n';
}

```