<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
        is_primes = [True] * 35
        is_primes[0] = False
        is_primes[1] = False
        for i in range(2, len(is_primes)):
            for j in range(i * i, len(is_primes), i):
                is_primes[j] = False    
        primes = [num for num, is_prime in enumerate(is_primes) if is_prime]

        def get_mask(num):
            mask = 0
            for i, prime in enumerate(primes):
                if num % prime != 0:
                    continue
                mask |= (1 << i)
                num = num // prime
                if num % prime == 0:
                    return None
            return mask

        num_of_ones = 0

        masked_nums = []
        masked_counter = collections.Counter()
        for num in nums:
            if num == 1:
                num_of_ones += 1
                continue
            mask = get_mask(num)
            if mask is None:
                continue
            if mask not in masked_counter:
                masked_nums.append(mask)
            masked_counter[mask] += 1
        
        @cache
        def dp(index, mask):
            if index >= len(masked_nums):
                return 1
            ans = dp(index + 1, mask)
            if mask & masked_nums[index] == 0:
                ans += (dp(index + 1, mask | masked_nums[index]) * masked_counter[masked_nums[index]])
            return ans % MOD

        return (((dp(0, 0) * pow(2, num_of_ones, MOD)) % MOD - 1) + MOD) % MOD


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