You are given a number A, convert A to number D by following steps.

Convert A to basic form B Basic form contains exactly single occurence of all prime factors of A. eg. if A=12 -> (2 x 2 x 3) then B = 2x3 = 6 if A=120, then B=2x3x5=30 After converting into basic number, then multiply all it's divisors to get a number C.

Count the number of divisors of C, you get your final answer which is D.

Find D and print is module 1e9+7

Constraints: 2<=A<=10^15

Sample input: 12 Sample output: 9

Explanation: A = 2 x 2 x 3 hence basic form B = 2 x 3 = 6

The divisors of B are 1, 2, 3, 4. Hence C = 1 x 2 x 3 x 6 = 36

The divisors of C are 1, 2, 3, 4, 6, 9, 12, 18, 36. Therefore the D or the count of divisors of C is 9

Sample Input: 71 Sample Output: 2

I calculated prime numbers and divisors in O(sqrt(N)) each but my soln only passed 5/10 test cases. How to solve this question?

Thanks in advance.