<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]]:
        n = len(nums)

        def solve(index):
            r = {0}
            for i in range(index+1, n):
                if nums[index] > nums[i]:
                    continue
                r |= solve(i)
            r = set([ (1 << index) + l for l in r ]) | r
            return r
        
        result = set()
        for i in range(n):
            result |= solve(i)
        
        # print(result)
        final = set()
        for l in result:
            _l = []
            index = 0
            while l != 0:
                if l % 2 == 1:
                    _l.append(nums[index])
                index += 1
                l //= 2
            # print(_l)
            if len(_l) >= 2:
                final.add(tuple(_l))
        return [*final]
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