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;
const long long mod = 998244353;
const long long N = 2e5;
const long long inf = 1e18;
template <typename... T>
void read(T&... args) {
  ((cin >> args), ...);
}
template <typename... T>
void write(T&&... args) {
  ((cout << args << " "), ...);
}
template <typename... T>
void writeline(T&&... args) {
  ((cout << args << " "), ...);
  cout << "\n";
}
long long add(long long a, long long b) {
  long long res = (a + b) % mod;
  return (res < 0) ? res + mod : res;
}
long long mul(long long a, long long b) {
  long long res = (a * 1LL * b) % mod;
  return (res < 0) ? res + mod : res;
}
string binary(long long x) {
  string res = "";
  for (long long i = 0; i < 32; i++) res += (1 << i & x) ? '1' : '0';
  return res;
}
long long in(long long x, long long y, long long z) {
  if (y > z) swap(y, z);
  return (x >= y && x <= z);
}
long long valid(long long x, long long sum) {
  long long s = 0;
  while (x) {
    s += x % 10;
    x /= 10;
  }
  return s == sum;
}
void solve() {
  long long n;
  cin >> n;
  if (n == 1) {
    cout << -1;
    return;
  }
  writeline(n, n + 1, n * (n + 1));
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
