<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 collections import Counter
import math

class Solution:
    def waysToFillArray(self, queries):
        def prime_factors(n):
            i = 2
            factors = []
            while i * i <= n:
                if n % i:
                    i += 1
                else:
                    n //= i
                    factors.append(i)
            if n > 1:
                factors.append(n)
            return factors
        
        def binomial_coefficient(n, k, mod):
            if k < 0 or k > n:
                return 0
            if k == 0 or k == n:
                return 1
            numerator = 1
            denominator = 1
            for i in range(1, k+1):
                numerator = (numerator * (n - i + 1)) % mod
                denominator = (denominator * i) % mod
            return (numerator * pow(denominator, mod - 2, mod)) % mod
        
        mod = 10**9 + 7
        answer = []
        
        for ni, ki in queries:
            factors = prime_factors(ki)
            count = Counter(factors)
            ways = 1
            for power in count.values():
                ways = (ways * binomial_coefficient(ni + power - 1, power, mod)) % mod
            answer.append(ways)
        
        return answer

# Example usage:
solution = Solution()
queries = [[2,6],[5,1],[73,660]]
print(solution.waysToFillArray(queries))  # Output: [4, 1, 50734910]
22600