<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:
        newNums = set()
        exclusion = {4, 8, 12, 16, 20, 24, 28, 9, 18, 27, 25}
        count = {}
        count[1] = 0
        m = 1000000007

        for i in range(len(nums)):
            if nums[i] == 1:
                count[1] += 1
            elif nums[i] not in exclusion:
                if nums[i] not in count:
                    count[nums[i]] = 0
                newNums.add(nums[i])
                count[nums[i]] += 1
        
        nums = list(newNums)

        mask = 1 << len(nums)
        ans = 0
        for i in range(1, mask):
            product = total = 1
            for j in range(len(nums)):
                if i & (1 << j):
                    if gcd(product, nums[j]) == 1:
                        product *= nums[j]
                        total = (total * count[nums[j]]) % m
                    else:
                        total = 0
                        break
            ans = (ans + total) % m
        
        ans = ((ans * pow(2, count[1], m)) % m + (pow(2, count[1], m)) - 1 + m) % m
        return ans
16500