B. Different Divisors

//Day 4 Problem 1
// The key to the question is that we have to find the number with at least 4 divisors so to make it simple we assume exactly 4 divisors, 2 of them being 1 and the number itself 
//and we have to find the other two such that difference between all the divisors is minimum d(given). 
//Point to be noted is that for making the divisors difference minimum d, we consider only prime numbers as if we consider the composite numbers then the differnce might be 
//disturbed. Eg: if the number is 24, thn divisors 1,4,7,24 that gives a minimum differnce of 3 but wait a minute!!!!!  What about the factors of 4 i.e. 2 so the difference will
//boil down to 1 so prime nos. are used.
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
int A[100005];
//First we generate first 10^5 prime nos.
void prime()
{
    int index=0;
    // Declare the variables
    int a=0, b=100001, i, j;
 
 
    // Explicitly handling the cases when a is less than 2
    // since 0 and 1 are not prime numbers
    if (a <= 2) {
        a = 2;
        if (b >= 2) {
 
                A[index]=a;
                index++;
            a++;
        }
    }
 
    // MAKING SURE THAT a IS ODD BEFORE WE BEGIN
    // THE LOOP
    if (a % 2 == 0)
        a++;
 
    // NOTE : WE TRAVERSE THROUGH ODD NUMBERS ONLY
    for (i = a; i <= b; i = i + 2) {
 
        // flag variable to tell
        // if i is prime or not
        bool flag = 1;
 
        // WE TRAVERSE TILL SQUARE ROOT OF j only.
        // (LARGEST POSSIBLE VALUE OF A PRIME FACTOR)
        for (j = 2; j * j <= i; ++j) {
            if (i % j == 0) {
                flag = 0;
                break;
            }
        }
 
        // flag = 1 means i is prime
        // and flag = 0 means i is not prime
        if (flag == 1)
        {
            
            A[index]= i;
            index++;
        }
    }
 
 
}
 
void tests() {
int d;
cin>>d;
int ans=1;
//Then we iterate over the loop, and as soon as we find the difference of d, we return as we have to find the smallest ans.
for(int i=0;i<100005;i++)
{
    if(A[i]-1>=d)
    {
       ans=ans*A[i];
       break;
    }
  
}
  //cout<<ans<<endl;
for(int i=0;i<100005;i++)
{
    if(A[i]-ans>=d)
    {
        ans=ans*A[i];
        break;
    }
}
 
cout<<ans<<endl;
 
 
}
int32_t main( int argc , char ** argv )
{
	ios_base::sync_with_stdio(false) ; 
	cin.tie(NULL) ; 
	prime();
	int t = 1;
	cin>>t;
	while(t--){
		tests();
	}
    return 0 ; 
}