<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:
        offset = {2: 0, 3: 1, 5: 2, 7: 3, 11: 4, 13: 5, 17: 6, 19: 7, 23: 8, 29: 9}

        def get_mask(num: int) -> int:
            if num == 1:
                return 1 << 10
            mask = 0
            for prime in offset:
                if num % prime == 0:
                    if num % (prime ** 2) == 0: return 0
                    mask = mask | (1 << offset[prime])
            return mask

        nums_masked = list(filter(lambda x: x > 0, list(map(get_mask, nums))))
        nums_tally = list(Counter(nums_masked).items())
        # print(nums_tally)

        @cache
        def recur(idx: int, acc_mask: int):
            if idx >= len(nums_tally):
                return 1 if acc_mask != 0 else 0

            # dont include
            ret = recur(idx+1, acc_mask)

            # include
            curr_mask, freq = nums_tally[idx]
            if acc_mask & curr_mask == 0:
                if curr_mask == 1024:
                    ret += ((1 << freq) - 1) * recur(idx + 1, acc_mask | curr_mask)
                else:
                    ret += freq * recur(idx + 1, acc_mask | curr_mask)

            return ret % (10**9 + 7)

        return recur(0, 0) 
        
26800