<p>You are given a 2D integer array, <code>queries</code>. For each <code>queries[i]</code>, where <code>queries[i] = [n<sub>i</sub>, k<sub>i</sub>]</code>, find the number of different ways you can place positive integers into an array of size <code>n<sub>i</sub></code> such that the product of the integers is <code>k<sub>i</sub></code>. As the number of ways may be too large, the answer to the <code>i<sup>th</sup></code> query is the number of ways <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>Return <em>an integer array </em><code>answer</code><em> where </em><code>answer.length == queries.length</code><em>, and </em><code>answer[i]</code><em> is the answer to the </em><code>i<sup>th</sup></code><em> query.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> queries = [[2,6],[5,1],[73,660]]
<strong>Output:</strong> [4,1,50734910]
<strong>Explanation:</strong>&nbsp;Each query is independent.
[2,6]: There are 4 ways to fill an array of size 2 that multiply to 6: [1,6], [2,3], [3,2], [6,1].
[5,1]: There is 1 way to fill an array of size 5 that multiply to 1: [1,1,1,1,1].
[73,660]: There are 1050734917 ways to fill an array of size 73 that multiply to 660. 1050734917 modulo 10<sup>9</sup> + 7 = 50734910.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> queries = [[1,1],[2,2],[3,3],[4,4],[5,5]]
<strong>Output:</strong> [1,2,3,10,5]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= queries.length &lt;= 10<sup>4</sup> </code></li>
	<li><code>1 &lt;= n<sub>i</sub>, k<sub>i</sub> &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution:
    # set up mod, primes, and a memo for expensive calculation repeat prevention 
    def __init__(self) : 
        self.mod = 10**9 + 7 
        self.primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59,
                       61, 67, 71, 73, 79, 83, 89, 97]
        self.memo = {}

    # solution progression 
    def waysToFillArray(self, queries: List[List[int]]) -> List[int]:
        # set up answer 
        answer = [0] * len(queries)
        # enumerate queries 
        for index, (n_i, k_i) in enumerate(queries) : 
            # if state is stored, mark it again 
            if (n_i, k_i) in self.memo : 
                answer[index] = self.memo[(n_i, k_i)]
            else : 
                # otherwise, preserve initial for proper marking 
                k_i_o = k_i 
                n_i_o = n_i
                # set up result and decrement n_i 
                result = 1 
                n_i -= 1 
                # loop over primes 
                for p in self.primes : 
                    # if k_i is a prime factor 
                    if k_i % p == 0 : 
                        # get count of prime factors 
                        c = 1 
                        k_i //= p 
                        while k_i % p == 0 : 
                            c += 1 
                            k_i //= p 
                        # and increment result accordingly by combinatorial logic 
                        result = (result * math.comb(n_i + c, c))% self.mod 
                # when done if k_i is gt 1 update result 
                if k_i > 1 : 
                    result = (result * (n_i_o)) % self.mod 
                else : 
                    result *= 1
                # then store result 
                answer[index] = result 
                self.memo[(n_i_o, k_i_o)] = result 
        # return when done 
        return answer 
23300