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
#include <bits/stdc++.h>
using namespace std;
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
bool isitPrime(long long int n) {
  long long int j = 0;
  for (j = 2; j * j <= n; j++)
    if (n % j == 0) break;
  if (j * j > n)
    return true;
  else
    return false;
}
long long int gcd(long long int a, long long int b) {
  long long int c = a % b;
  while (c != 0) {
    a = b;
    b = c;
    c = a % b;
  }
  return b;
}
long long int Com(long long int n, long long int k) {
  if (k > n) return 0;
  long long int r = 1;
  for (long long int d = 1; d <= k; ++d) {
    r *= n--;
    r /= d;
  }
  return r;
}
long long modexpo(long long a, long long b, int n) {
  long long d = 1;
  while (b) {
    if (b % 2) {
      d = (d * a) % n;
    }
    b >>= 1;
    a = (a * a) % n;
  }
  return d;
}
int main() {
  int n;
  cin >> n;
  if (n == 1)
    cout << -1;
  else {
    cout << n << " " << n + 1 << " " << n * (n + 1);
  }
  return 0;
}
