<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]]:
        
        self.output = []
        def dfs(i, sequence):
            # stopping condition 
            if i >= len(nums):
                if len(sequence) > 1 and sequence not in self.output:
                    self.output.append(sequence)
                return 

            # call without current element
            dfs(i +1 , sequence)
            # call with current element only if 
            # empty sequence, or squence[-1] smaller than current
            if not sequence or (sequence and sequence[-1] <= nums[i]) : 
                dfs(i +1 , sequence +[nums[i]])
                

            # add current element if bigger than previous


        
        dfs(0, [])
        return self.output
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