Sum of Prime

Problem Statement: Given a number (greater than 2), print two prime numbers whose sum will be equal to given number, else print -1 if no such number exists.

NOTE: A solution will always exist if the number is even.

If [a, b] is one solution with a <= b, and [c, d] is another solution with c <= d, and a < c then print [a, b] only
and not all possible solutions.

Input:
The first line contains an integer T, depicting total number of test cases. Then following T lines contains an integer N.

Output:
Print the two prime numbers in a single line with space in between.

Constraints:
1 ≤ T ≤ 50
2 ≤ N ≤ 1000000

Example:
Input:
2
8
3
Output
3 5
-1

Explanation:
Testcase 1: two prime numbers from 1 to 8 are 3 and 5 whose sum is 8.

This problem is available in Geeksforgeeks, the link of which is mentioned below.

https://practice.geeksforgeeks.org/problems/sum-of-prime/0

Solution:

CPP Solution

#include<iostream>

using namespace std;

//Function to determine whether a number is prime or not

bool isprime(int n)   

{

    if(n<=1)

    {

        return false;

    }

    for(int i=2;i*i<=n;i++)

    {

        if(n%i==0)

        {

            return false;

        }

    }

    return true;

}

int main()

{

    int t;

    cin>>t;

    while(t--)

    {

        int n;

        cin>>n;

        int flag=0;

        for(int i=1;i<n;i++)

        {

            if(isprime(i)==true && isprime(n-i)==true)

            {

                cout<<i<<" "<<n-i<<endl;

                flag=1;

                break;

            }

        }

        if(flag==0)

        {

            cout<<-1<<endl;

        }

    }

return 0;

}

*Please comment if you have any confusion in this solution*