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;
long long prr[] = {2,   3,   5,   7,   11,  13,  17,  19,  23,  29,  31,  37,
                   41,  43,  47,  53,  59,  61,  67,  71,  73,  79,  83,  89,
                   97,  101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151,
                   157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223,
                   227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281,
                   283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359,
                   367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433,
                   439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499};
long long powmod(long long a, long long b, long long modulo) {
  if (b == 0 || a == 1) return 1;
  long long half = powmod(a, (b / 2), modulo) % modulo;
  long long full = (half * half) % modulo;
  if (b % 2) return (full * a) % modulo;
  return full % modulo;
}
long long invmod(long long a, long long modulo) {
  long long check = powmod(a, modulo - 2, modulo) % modulo;
  return check;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
bool isPrime[100001];
int prime[100001];
void sieve(long long N) {
  for (long long i = 0; i <= N; ++i) {
    isPrime[i] = true;
  }
  isPrime[0] = false;
  isPrime[1] = false;
  for (int i = 2; i * i <= N; ++i) {
    if (isPrime[i] == true) {
      for (int j = i * i; j <= N; j += i) isPrime[j] = false;
    }
  }
  int j = 0;
  for (int i = 2; i <= N; i++) {
    if (isPrime[i]) {
      prime[j] = i;
      j++;
    }
  }
}
bool palindrome(string s) {
  for (long long i = 0; i < s.length() / 2; i++) {
    if (s[i] != s[s.length() - i - 1]) return 0;
  }
  return 1;
}
void solve() {
  int x;
  cin >> x;
  int a = x;
  int b = x + 1;
  int c = a * b;
  if (a == b || b == c || c == a)
    cout << -1 << "\n";
  else
    cout << a << " " << b << " " << c << "\n";
}
int main() {
  clock_t start, end;
  start = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  while (t--) {
    solve();
  }
  end = clock();
  double time = double(end - start) / (double)(CLOCKS_PER_SEC);
  return 0;
}
