<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:
        def prime_factors(n):
            if n == 1:
                return {1}
            factors = set()
            while not n % 2:
                if 2 in factors:
                    return 
                factors.add(2)
                n //= 2
            for i in range(3, int(n**0.5) + 1, 2):
                while not n % i:
                    if i in factors:
                        return 
                    factors.add(i)
                    n //= i
            if n > 2:
                if n in factors:
                    return 
                factors.add(n)
            return factors

        d = {}
        c = Counter(nums)
        for n in sorted(c):
            if n not in d and (p := prime_factors(n)):
                d[n] = p

        dp = Counter({(0,) : 1})                  
        mod = 10 ** 9 + 7
        for k, v in sorted(c.items()):
            if k not in d or k == 1:
                continue
            next_dp = dp.copy()
            for a, b in dp.items():
                if not d[k] & set(a):
                    next_dp[(*a, *d[k])] =  (next_dp[(*a, *d[k])] + b * v) % mod
            dp = next_dp

        k = sum(dp.values()) - 1
        w = sum(comb(c[1], i + 1) for i in range(c[1]))
        return (w + w * k + k) % mod


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