Codeforces Round #384 (Div. 2)

题目链接：Vladik and fractions

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 \(\frac{2}{n}\) as a sum of three distinct positive fractions in form \(\frac{1}{m}\).

Help Vladik with that, i.e for a given \(n\) find three distinct positive integers \(x, y\) and \(z\) such that \(\frac{2}{n} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\). Because Chloe can't check Vladik's answer if the numbers are large, he asks you to print numbers not exceeding $10^9$.

If there is no such answer, print \(-1\).

Input

The single line contains single integer \(n (1 \le  n \le  10^4)\).

Output

If the answer exists, print 3 distinct numbers \(x, y\) and \(z (1 \le  x, y, z \le  10^9, x \neq y, x \neq  z, y \neq  z)\). 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

Solution

题意

给定一个正整数 \(n\)，求正整数 \(x,y,z\) 满足 \(\frac{2}{n} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\)。

其中 \(x \neq y,\ x \neq z,\ y \neq z\)。若无解输出 \(-1\)。

题解

构造

\(\frac{1}{n} - \frac{1}{n + 1} = \frac{1}{n(n+1)}\)

\(\frac{1}{n} = \frac{1}{n + 1} + \frac{1}{n(n+1)}\)

\(\frac{2}{n} = \frac{1}{n + 1} + \frac{1}{n(n+1)} + \frac{1}{n}\)

当 \(n=1\) 时，\(\frac{2}{n}=2\)。而 \((\frac{1}{x}+\frac{1}{y}+\frac{1}{z})_{max} = \frac{1}{1}+\frac{1}{2}+\frac{1}{3} < 2\)，所以当 \(n=1\) 时无解。

Code

#include <bits/stdc++.h>

using namespace std;

int main() {

    ios::sync_with_stdio(false);

    cin.tie(0);

    int n;

    cin >> n;

    if(n == 1) cout << -1 << endl;

    else cout << (n + 1) << " " << (n * (n + 1)) << " " << n << endl;

    return 0;

}