<p>You are given a positive integer <strong>0-indexed</strong>&nbsp;array <code>nums</code>.</p>

<p>A subset of the array <code>nums</code> is <strong>square-free</strong> if the product of its elements is a <strong>square-free integer</strong>.</p>

<p>A <strong>square-free integer</strong> is an integer that is divisible by no square number other than <code>1</code>.</p>

<p>Return <em>the number of square-free non-empty subsets of the array</em> <strong>nums</strong>. Since the answer may be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>A <strong>non-empty</strong>&nbsp;<strong>subset</strong> of <code>nums</code> is an array that can be obtained by deleting some (possibly none but not all) elements from <code>nums</code>. Two subsets are different if and only if the chosen indices to delete are different.</p>

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

<pre>
<strong>Input:</strong> nums = [3,4,4,5]
<strong>Output:</strong> 3
<strong>Explanation:</strong> There are 3 square-free subsets in this example:
- The subset consisting of the 0<sup>th</sup> element [3]. The product of its elements is 3, which is a square-free integer.
- The subset consisting of the 3<sup>rd</sup> element [5]. The product of its elements is 5, which is a square-free integer.
- The subset consisting of 0<sup>th</sup> and 3<sup>rd</sup> elements [3,5]. The product of its elements is 15, which is a square-free integer.
It can be proven that there are no more than 3 square-free subsets in the given array.</pre>

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

<pre>
<strong>Input:</strong> nums = [1]
<strong>Output:</strong> 1
<strong>Explanation:</strong> There is 1 square-free subset in this example:
- The subset consisting of the 0<sup>th</sup> element [1]. The product of its elements is 1, which is a square-free integer.
It can be proven that there is no more than 1 square-free subset in the given array.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length&nbsp;&lt;= 1000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 30</code></li>
</ul>

class Solution:
    def squareFreeSubsets(self, nums: List[int]) -> int:
        M = 10**9 + 7
        primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29]
        nums = [n for n in nums if all(n % p**2 != 0 for p in primes)]
        primes = [p for p in primes if any(n % p == 0 for n in nums)]
        freqs = Counter(nums)
        counts = Counter()
        total = (pow(2, freqs[1], M) - 1) % M

        def factorial_mod(n):
            f = 1
            for i in range(2, n + 1):
                f = (f * i) % M
            return f

        for size in range(1, len(primes) + 1):
            for subset in combinations(primes, size):
                prod = reduce(lambda x, y: x * y, subset)
                counts[prod] = Counter()
                counts[prod][1] = freqs[prod]
                for n in freqs:
                    if prod % n != 0 or n == 1 or n == prod:
                        continue
                    m = prod // n
                    for factors in counts[m]:
                        count = (freqs[n] * counts[m][factors]) % M
                        counts[prod][factors + 1] = (counts[prod][factors + 1] + count) % M
                for factors in counts[prod]:
                    factorial = factorial_mod(factors)
                    break
                    counts[prod][factors] = (counts[prod][factors] * pow(factorial, -1, M)) % M
                    
        for prod in counts:
            for factors in counts[prod]:
                factorial = factorial_mod(factors)
                count = (counts[prod][factors] * pow(factorial, -1, M)) % M
                count = (count * pow(2, freqs[1], M) % M) % M
                total = (total + count) % M
        
        return total
17700