Given a positive integer **N**, the task is to represent the fraction **2 / N** as the sum of three distinct positive integers of the form **1 / m** i.e. **(2 / N) = (1 / x) + (1 / y) + (1 / z)** and print **x**, **y** and **z**.

**Examples:**

Input:N = 3

Output:3 4 12

(1 / 3) + (1 / 4) + (1 / 12) = ((4 + 3 + 1) / 12)

= (8 / 12) = (2 / 3) i.e. 2 / N

Input:N = 28

Output:28 29 812

**Approach:** It can be easily inferred that for **N = 1**, there will be no solution. For **N > 1**, **(2 / N)** can be represented as **(1 / N) + (1 / N)** and the problem gets reduced to representing it as a sum of two fractions. Now, find the difference between **(1 / N)** and **1 / (N + 1)** and get the fraction **1 / (N * (N + 1))**. Therefore, the solution is **(2 / N) = (1 / N) + (1 / (N + 1)) + (1 / (N * (N + 1)))** where **x = N**, **y = N + 1** and **z = N * (N + 1)**.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to find the required fractions ` `void` `find_numbers(` `int` `N) ` `{ ` ` ` `// Base condition ` ` ` `if` `(N == 1) { ` ` ` `cout << -1; ` ` ` `} ` ` ` ` ` `// For N > 1 ` ` ` `else` `{ ` ` ` `cout << N << ` `" "` `<< N + 1 << ` `" "` ` ` `<< N * (N + 1); ` ` ` `} ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `N = 5; ` ` ` ` ` `find_numbers(N); ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of the approach ` `class` `GFG ` `{ ` ` ` `// Function to find the required fractions ` `static` `void` `find_numbers(` `int` `N) ` `{ ` ` ` `// Base condition ` ` ` `if` `(N == ` `1` `) ` ` ` `{ ` ` ` `System.out.print(-` `1` `); ` ` ` `} ` ` ` ` ` `// For N > 1 ` ` ` `else` ` ` `{ ` ` ` `System.out.print(N + ` `" "` `+ (N + ` `1` `) + ` ` ` `" "` `+ (N * (N + ` `1` `))); ` ` ` `} ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String []args) ` `{ ` ` ` `int` `N = ` `5` `; ` ` ` ` ` `find_numbers(N); ` `} ` `} ` ` ` `// This code is contributed by Rajput-Ji ` |

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## Python3

`# Python3 implementation of the approach ` ` ` `# Function to find the required fractions ` `def` `find_numbers(N) : ` ` ` ` ` `# Base condition ` ` ` `if` `(N ` `=` `=` `1` `) : ` ` ` `print` `(` `-` `1` `, end ` `=` `""); ` ` ` ` ` `# For N > 1 ` ` ` `else` `: ` ` ` `print` `(N, N ` `+` `1` `, N ` `*` `(N ` `+` `1` `)); ` ` ` `# Driver code ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` ` ` `N ` `=` `5` `; ` ` ` ` ` `find_numbers(N); ` ` ` `# This code is contributed by AnkitRai01 ` |

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## C#

`// C# implementation of the approach ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `// Function to find the required fractions ` `static` `void` `find_numbers(` `int` `N) ` `{ ` ` ` `// Base condition ` ` ` `if` `(N == 1) ` ` ` `{ ` ` ` `Console.Write(-1); ` ` ` `} ` ` ` ` ` `// For N > 1 ` ` ` `else` ` ` `{ ` ` ` `Console.Write(N + ` `" "` `+ (N + 1) + ` ` ` `" "` `+ (N * (N + 1))); ` ` ` `} ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main(String []args) ` `{ ` ` ` `int` `N = 5; ` ` ` ` ` `find_numbers(N); ` `} ` `} ` ` ` `// This code is contributed by PrinciRaj1992 ` |

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**Output:**

5 6 30

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