<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:
        # there are 2^10 possible products of a square-free integer (assuming just numbers below 30 in the array)
        # because there are 10 prime integers below 30
        # match each product with a bitmask indicating what prime factors multiply into the product
            #bit mask corresponds to a number
        
        primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29]

        #dp[i][bitmask] = number of square free subsets using nums[:i+1] and where the product has a bitmask of "bitmask"
            # each bit of the bitmask indicating the prime factors
        dp = [[0 for _ in range(1024)] for _ in nums]

        dp = [0 for _ in range(1024)]

        for i in range(len(nums)):
            prevdp = dp
            if i > 0:
                dp = prevdp.copy()
            else:
                dp = [0 for _ in range(1024)]
            
            n = nums[i]
            bitmask = 0
            for j in range(len(primes)):
                p = primes[j]
                if n % p == 0:
                    bitmask += pow(2,j)
                    n = n // p
            
            if n != 1:
                continue
            
            # the subset with just nums[i]
            dp[bitmask] += 1

            if i > 0:
                for prevBitmask in range(1024):
                    if prevBitmask & bitmask == 0:
                        dp[prevBitmask | bitmask] += prevdp[prevBitmask]
                        dp[prevBitmask | bitmask] = dp[prevBitmask | bitmask] % (pow(10, 9) + 7)
        

        #for row in dp:
        #    print(row)
        return sum(dp) % (pow(10, 9) + 7)


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