<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>

MOD = int(1e9 + 7)

class Solution:
    def __init__(self):
        self.primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29]
        self.dp = []

    def getMask(self, num):
        mask = 0
        for i in range(10):
            tim = 0
            while num % self.primes[i] == 0:
                tim += 1
                num //= self.primes[i]
            if tim > 1:
                return -1  # If a number is divisible by a prime more than once, return -1
            if tim == 1:
                mask |= (1 << i)  
        return mask

    def dfs(self, ind, prodmask, nums):
        if ind >= len(nums): return 0

        if self.dp[ind][prodmask] != -1:
            return self.dp[ind][prodmask]
        
        mask = self.getMask(nums[ind])
        ans = self.dfs(ind + 1, prodmask, nums)
        
        if mask != -1 and (prodmask & mask) == 0:
            ans = (ans + self.dfs(ind + 1, prodmask | mask, nums) + 1) % MOD
        
        self.dp[ind][prodmask] = ans
        return ans

    def squareFreeSubsets(self, nums):
        self.dp = [[-1] * (1 << 11) for _ in range(len(nums) + 1)]
        return self.dfs(0, 0, nums)
40400