#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using i32 = int32_t;
using i64 = int64_t;
using ui32 = uint32_t;
using ui64 = uint64_t;
using pii = pair<int, int>;
using pdd = pair<double, double>;
using vpii = vector<pii>;
using vpdd = vector<pdd>;
int64_t fuse(int i1, int i2) { return (((int64_t)i1) << 32) + i2; }
pii split(int64_t l) { return {(int)(l >> 32), (int)(l & 0xFFFFFFFF)}; }
static int constexpr dbg = 0;
int main(int argc, char *argv[]) {
  if (!dbg) {
    std::ios::sync_with_stdio(false);
  }
  ll n;
  cin >> n;
  if (n == 1) {
    cout << -1 << endl;
  } else {
    cout << n << " " << (n + 1) << " " << (n * (n + 1)) << endl;
  }
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using i32 = int32_t;
using i64 = int64_t;
using ui32 = uint32_t;
using ui64 = uint64_t;
using pii = pair<int, int>;
using pdd = pair<double, double>;
using vpii = vector<pii>;
using vpdd = vector<pdd>;
int64_t fuse(int i1, int i2) { return (((int64_t)i1) << 32) + i2; }
pii split(int64_t l) { return {(int)(l >> 32), (int)(l & 0xFFFFFFFF)}; }
static int constexpr dbg = 0;
int main(int argc, char *argv[]) {
  if (!dbg) {
    std::ios::sync_with_stdio(false);
  }
  ll n;
  cin >> n;
  if (n == 1) {
    cout << -1 << endl;
  } else {
    cout << n << " " << (n + 1) << " " << (n * (n + 1)) << endl;
  }
  return 0;
}

```