<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:
        MOD = (10 ** 9 + 7)
        numsc = Counter(nums)
        ones = 1 << numsc[1] # get possible combinations of ones
        
        primefactors = [self.prime_factors(i) for i in numsc.keys()] # get prime factors
        primefactors = list(map(tuple, filter(lambda x: x != -1, primefactors))) # filter ones and square products

        if not primefactors:
            return (ones - 1) % MOD
        primefactorscount = {}
        for primefac in primefactors: # map prime factors to count
            primefactorscount[primefac] = numsc[prod(primefac)]
        ans = 0
        if not ones:
            ones = 1
        else:
            ans = ones - 1
        del numsc
        
        for i in range(1, len(primefactors) + 1):  # iterate possible lengths
            for subset in combinations(primefactors, i): # iterate all possible combinations
                total = list(chain(*subset))
                if len(total) == len(set(total)):  # if square-free subset
                    plus = ones  # can use any possible combination of ones
                    for primelist in subset:
                        plus *= primefactorscount[primelist] # multiply by how many instances of this number exists (can choose any one)
                    ans += plus
        return ans % (MOD)


    def prime_factors(self, n):
        primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29]
        factors = []
        for i in primes:
            if i * i > n:
                break
            if n % i == 0:
                n //= i
                if n % i == 0: # has square factor
                    return -1
                factors.append(i)
        if n > 1:
            factors.append(n)
        if not factors:
            return -1 # passed in 1
        return factors
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