Number theory question playing with divisors

#1

You are given an integer array A of size x denoting the prime powers of an integer N.
Ai denotes the power of ith prime in the prime factorization of N.
To make it more clear, A1 will denote the power of 2 in the prime factorization of N A2 will denote the power of 3 in the prime factorization of N and so on. Consider a number P.
P equals to the product of all the divisors of N You have to find the number of divisors of P.
Output it modulo 109+7

Sir can you some hint to how to do this because power is in the <= 10^9?

#2

There’s a tutorial video attached for this problem. You can refer that for better understanding. Although the logic is first calculate the number of divisors of given number. Now The product of all the divisors of a number N is N^(x/2) where x is the number of divisors of N. The number of divisors is given by product of prime powers+1 for N.
Also mod inverse will also be used in solving this. It is used when you are performing divisions under the modulo operation.

#3

Sir where is the tutorial link for this problem?
can you give me the link?

#4

There isn’t any dedicated video for this, you can see about modulo operation. Rest follow up this code

#5

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