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;
struct code {
  int x, y;
};
bool comp(code a, code b) {}
int64_t bin(int a, int n) {
  int64_t p = 1;
  while (n) {
    if (n % 2 == 0) {
      a = (a * a) % 10;
      n /= 2;
    } else
      p = (p * a) % 10, n--;
  }
  return p;
}
const int MX = 1e6 + 10;
vector<int> result;
int n, m, k, ans;
int a[MX];
bool used[MX];
void go(int x, int len) {
  used[x] = 1;
  if (a[x] == x) {
    m = 1e9;
    return;
  }
  if (used[a[x]]) {
    m = len + 1;
    return;
  }
  go(a[x], len + 1);
}
int solver(int n) {
  int64_t x;
  for (int i = n - 1; i <= 10000; i++) {
    for (int j = i + 1; j <= 100000; j++) {
      if (2ll * i * j - 1ll * n * (i + j) <= 0) continue;
      x = (1ll * i * j * n) / (2ll * i * j - 1ll * n * (i + j));
      if (x > 0 and
          2ll * i * j * x == 1ll * n * (i + j) * x + 1ll * i * j * n and
          i != x and j != x) {
        cout << i << " " << j << " " << x << endl;
        return 0;
      }
    }
  }
  return 1;
}
int main() {
  int n;
  cin >> n;
  if (n == 1) {
    cout << -1;
    return 0;
  }
  solver(n);
  return 0;
  for (int i = 1e4; i >= 1; i--) {
    if (solver(i)) {
      cout << i << " "
           << "YOU HACK";
      return 0;
    }
    cout << i << endl;
  }
}
