<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)
        dp = [None] * (n+1)
        dp[n] = [()]

        for index in reversed(range(n)):
            result = []
            for s in dp[index+1]:
                if not s:
                    result.append((index,))
                elif nums[index] <= nums[s[0]]:
                    result.append((index, *s))
            dp[index] = dp[index+1] + result

        result = set()
        for s in dp[0]:
            l = [ nums[i] for i in s ]
            if len(l) >= 2:
                result.add(tuple(l))
        
        return result

        
            
            


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