When can we get to see the test cases?

Isn't your submission O(Nlog^2N). Aren't you basically using a modified sieve with exponentiation at every step?

A very naive way to think of it is this:

You can calculate all divisors of all numbers from 1 to n using a sieve whose complexity is O(nlogn). Also, in each step, you calculate exactly one divisor of a number. So, there can only be O(nlogn), The sieve can be visualized as:

for(int i = 2; i <= n; i ++) {
   for(int j = i; j <= n; j += i) {
      //now we know that i is a divisor of j
   }
}

I did! 3 penalties + 20 minutes wasted because of that! Would have been in top-500 otherwise :(