<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>

from typing import List
class Solution:
    def findSubsequences(self, nums: List[int]) -> List[List[int]]:
        self.res = set()


        self.backtracking(nums, [] , 0)

        return list(self.res)
        

    def backtracking(self, nums, curr, idx):
        if len(curr) > 1:
            self.res.add(tuple(curr))

        used = set()
        for i in range(idx, len(nums)):
            if nums[i] in used:
                continue
            if not curr or nums[i] >= curr[-1]:
                used.add(nums[i])
                self.backtracking(nums, curr + [nums[i]], i + 1)

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