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;
template <class T>
T Bitcnt(T a) {
  int sum = 0;
  while (a) {
    if (a & 1) sum++;
    a /= 2;
  }
  return sum;
}
template <class T>
T Max3(T a, T b, T c) {
  return max(a, max(b, c));
}
template <class T>
T Lcm(T a, T b) {
  T tmp = __gcd(a, b);
  return (a / tmp) * b;
}
template <class T>
T Pow(T a, T b) {
  T ans = 1;
  T base = a;
  while (b) {
    if (b & 1) ans = (ans * base);
    base = (base * base);
    b /= 2;
  }
  return ans;
}
long long Bigmod(long long a, long long b) {
  long long res = 1;
  long long pw = a % 1000000007LL;
  while (b > 0) {
    if (b & 1) res = (res * pw) % 1000000007LL;
    pw = (pw * pw) % 1000000007LL;
    b /= 2;
  }
  return res;
}
int a_x[] = {1, -1, 0, 0};
int a_y[] = {0, 0, 1, -1};
long long X, Y;
void extend_euclid(long long a, long long b) {
  if (b == 0) {
    X = 1;
    Y = 0;
    return;
  }
  extend_euclid(b, a % b);
  long long x, y;
  x = Y;
  y = X - (a / b) * Y;
  X = x;
  Y = y;
}
long long inverse_modulo(long long a, long long b) {
  extend_euclid(a, b);
  return (X + 1000000007LL) % 1000000007LL;
}
int main() {
  int n;
  cin >> n;
  if (n == 1)
    cout << -1;
  else
    cout << n << " " << (n + 1) << " " << n * (n + 1) << endl;
}
