<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
        def factorize(n):
            
            basis = 2
            
            factors = []
            
            while basis <= n:
                
                while n % basis == 0:
                    factors.append(basis)
                    n //= basis
                basis += 1
                
            return factors
        
        
        nums = sorted(list(nums))
        
        
        nums = [i for i in nums if len(factorize(i)) ==len(list(set(factorize(i))))]
        
        
        
        toots = 0
        
        
        hazschecorpsys = defaultdict(int)
        
        def generateHash(basis):
            basil = 0
            
            
            for ea in basis:
                basil |= (1 << ea)
            return basil
        
        def collision(hash1, hash2):
            
            if hash1 & hash2:
                return True
            return False
        
        def combine(h1, h2):
            return h1 | h2
        
        
        
        ans = 0
        for i in range(len(nums)):
            
            keys = list(hazschecorpsys.keys())
            
            
            
            anoo = defaultdict(int)
            basis = 0
            for ea in keys:
                if not collision(generateHash(factorize(nums[i])), ea):
                    thesis = ea | generateHash(factorize(nums[i]))
                    anoo[thesis] += hazschecorpsys[ea]
                    basis += hazschecorpsys[ea]
            basis += 1
            
            for key in list(anoo.keys()):
                hazschecorpsys[key] += anoo[key]
            hazschecorpsys[generateHash(factorize(nums[i]))] += 1
            ans += basis
            ans %= MOD
            
                    
            
            
            
            
            
            
            
            
        
        return ans
                
        
17100