<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>

from typing import List
from math import isqrt

MOD = 10**9 + 7

class Solution:
    def waysToFillArray(self, queries: List[List[int]]) -> List[int]:
        max_n = max(q[0] for q in queries)
        max_k = max(q[1] for q in queries)
        
        # Precompute factorials and inverse factorials
        max_factorial = max_n + max_k
        fact = [1] * (max_factorial + 1)
        inv_fact = [1] * (max_factorial + 1)
        
        for i in range(2, max_factorial + 1):
            fact[i] = fact[i - 1] * i % MOD
        
        inv_fact[max_factorial] = pow(fact[max_factorial], MOD - 2, MOD)
        for i in range(max_factorial - 1, 0, -1):
            inv_fact[i] = inv_fact[i + 1] * (i + 1) % MOD
        
        def binomial(n, k):
            if k > n or k < 0:
                return 0
            return fact[n] * inv_fact[k] % MOD * inv_fact[n - k] % MOD
        
        def prime_factors(x):
            factors = {}
            for i in range(2, isqrt(x) + 1):
                while x % i == 0:
                    if i not in factors:
                        factors[i] = 0
                    factors[i] += 1
                    x //= i
            if x > 1:
                factors[x] = 1
            return factors
        
        results = []
        for n, k in queries:
            factors = prime_factors(k)
            ways = 1
            for prime, exponent in factors.items():
                ways = ways * binomial(n + exponent - 1, exponent) % MOD
            results.append(ways)
        
        return results
23200