<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:
        n = len(nums)
        primes = [2,3,5,7,11,13,17,19,23,29]
        masks = defaultdict(int)
        exclude = set([1])
        for i in range(n):
            curr = nums[i]
            j = 0
            curr_mask = 0
            if curr not in masks:
                while j < len(primes):
                    if curr % primes[j] == 0:
                        if curr_mask & (1 << j) != 0:
                            exclude.add(nums[i])
                        curr_mask |= (1 << j)
                        curr /= primes[j]
                    else:
                        j += 1
                masks[nums[i]] = curr_mask
        count_ones = sum([1 for num in nums if num == 1])
        nums = [num for num in nums if num not in exclude]
        freq = Counter(nums)
        uniq_vals = list(freq.keys())
        n = len(uniq_vals)
        # print("masks", masks)
        @cache
        def dfs(mask, idx):
            if idx == n:
                if mask != 0:
                    return 1
                return 0
            ans = dfs(mask, idx+1)
            if mask & masks[uniq_vals[idx]] == 0:
                ans += freq[uniq_vals[idx]] * dfs(mask | masks[uniq_vals[idx]], idx + 1)
            return ans
        # print('nums', nums)
        # print("count_ones", count_ones)
        subsets = dfs(0, 0)
        return ( subsets*(2**count_ones) + (2**count_ones -1)) % (10**9 + 7)

            
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