count*=(freqOfPrimeFact[i]+1)
why did we do +1 in the above?
Editorial doubt
If x=a^m∗b^n∗c^p…
where a, b, c, … are the prime factors of x, then the number of factors of x is
(m+1)(n+1)(p+1)…
This is because any factor can be made by selecting 0 to m numbers of a (in m+1 ways), 0 to n numbers of b (in n+1 ways) and so on.