<p>Given an integer array <code>nums</code>, return <em>all the different possible non-decreasing subsequences of the given array with at least two elements</em>. You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,6,7,7]
<strong>Output:</strong> [[4,6],[4,6,7],[4,6,7,7],[4,7],[4,7,7],[6,7],[6,7,7],[7,7]]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,4,3,2,1]
<strong>Output:</strong> [[4,4]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 15</code></li>
	<li><code>-100 &lt;= nums[i] &lt;= 100</code></li>
</ul>

class Solution:
    def findSubsequences(self, nums: List[int]) -> List[List[int]]:
        def findSubseqAux(nums):
            if len(nums) == 0:
                return [tuple()]
            if len(nums) == 1:
                return [tuple(), (nums[0],)]
            a = nums[0]
            i = 1
            while i < len(nums) and nums[i] == a:
                i += 1
            old_res, res = findSubseqAux(nums[i:]), set()
            for r in old_res:
                res.add(r)
                if len(r) == 0 or r[0] >= a:
                    for j in range(1, i+1):
                        res.add(tuple([a] * j) + r)
            return res
            
        return [list(x) for x in findSubseqAux(nums) if len(x) > 1]
        
        
        
        

        

        
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