<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:
  def waysToFillArray(self, queries: list[list[int]]) -> list[int]:
    kMod = 1_000_000_007
    kMax = 10_000
    minPrimeFactors = self._sieveEratosthenes(kMax + 1)

    @functools.lru_cache(None)
    def fact(i: int) -> int:
      return 1 if i <= 1 else i * fact(i - 1) % kMod

    @functools.lru_cache(None)
    def inv(i: int) -> int:
      return pow(i, kMod - 2, kMod)

    @functools.lru_cache(None)
    def nCk(n: int, k: int) -> int:
      return fact(n) * inv(fact(k)) * inv(fact(n - k)) % kMod

    ans = []

    for n, k in queries:
      res = 1
      for freq in self._getPrimeFactorsCount(k, minPrimeFactors).values():
        res = res * nCk(n - 1 + freq, freq) % kMod
      ans.append(res)

    return ans

  def _sieveEratosthenes(self, n: int) -> list[int]:
    """Gets the minimum prime factor of i, where 1 < i <= n."""
    minPrimeFactors = [i for i in range(n + 1)]
    for i in range(2, int(n**0.5) + 1):
      if minPrimeFactors[i] == i:  # `i` is prime.
        for j in range(i * i, n, i):
          minPrimeFactors[j] = min(minPrimeFactors[j], i)
    return minPrimeFactors

  def _getPrimeFactorsCount(self, num: int, minPrimeFactors: list[int]) -> dict[int, int]:
    count = collections.Counter()
    while num > 1:
      divisor = minPrimeFactors[num]
      while num % divisor == 0:
        num //= divisor
        count[divisor] += 1
    return count
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