#include <bits/stdc++.h>
using namespace std;
inline int init() {
  int now = 0, ju = 1;
  char c;
  bool flag = false;
  while (1) {
    c = getchar();
    if (c == '-')
      ju = -1;
    else if (c >= '0' && c <= '9') {
      now = now * 10 + c - '0';
      flag = true;
    } else if (flag)
      return now * ju;
  }
}
inline long long llinit() {
  long long now = 0, ju = 1;
  char c;
  bool flag = false;
  while (1) {
    c = getchar();
    if (c == '-')
      ju = -1;
    else if (c >= '0' && c <= '9') {
      now = now * 10 + c - '0';
      flag = true;
    } else if (flag)
      return now * ju;
  }
}
long long a, b, c;
int main() {
  int n;
  n = init();
  if (n == 1) {
    printf("-1\n");
    return 0;
  }
  a = n;
  b = n + 1;
  c = n * (n + 1);
  cout << a << " " << b << " " << c << endl;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
inline int init() {
  int now = 0, ju = 1;
  char c;
  bool flag = false;
  while (1) {
    c = getchar();
    if (c == '-')
      ju = -1;
    else if (c >= '0' && c <= '9') {
      now = now * 10 + c - '0';
      flag = true;
    } else if (flag)
      return now * ju;
  }
}
inline long long llinit() {
  long long now = 0, ju = 1;
  char c;
  bool flag = false;
  while (1) {
    c = getchar();
    if (c == '-')
      ju = -1;
    else if (c >= '0' && c <= '9') {
      now = now * 10 + c - '0';
      flag = true;
    } else if (flag)
      return now * ju;
  }
}
long long a, b, c;
int main() {
  int n;
  n = init();
  if (n == 1) {
    printf("-1\n");
    return 0;
  }
  a = n;
  b = n + 1;
  c = n * (n + 1);
  cout << a << " " << b << " " << c << endl;
  return 0;
}

```