<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]]:
         
        sequences=[]
        seen=set()
        def dfs(nums,seq_so_far):
            if len(nums)==0:
                if len(seq_so_far)>=2:
                    if tuple(seq_so_far) not in seen:
                        seen.add(tuple(seq_so_far))
                        sequences.append(seq_so_far)
                return
            if len(seq_so_far)==0:
                seq_so_far_n=seq_so_far+[nums[0]]
            else:
                if seq_so_far[-1]<=nums[0]:
                    seq_so_far_n=seq_so_far+[nums[0]]
                else:
                    seq_so_far_n=seq_so_far
            dfs(nums[1::],seq_so_far_n)
            dfs(nums[1::],seq_so_far)
        dfs(nums,[])
        return sequences
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