<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]]:
        subsequences = []
        self._backtracking(nums, 0, [], subsequences)
        return subsequences
    
    def _backtracking(
        self,
        nums: list[int],
        start: int,
        subsequence: list[int],
        subsequences: list[list[int]]
    ):
        if len(subsequence) >= 2:
            subsequences.append(list(subsequence))
        visited = set()
        for i in range(start, len(nums)):
            if subsequence and subsequence[-1] > nums[i]:
                continue
            if nums[i] in visited:
                continue
            subsequence.append(nums[i])
            visited.add(nums[i])
            self._backtracking(nums, i + 1, subsequence, subsequences)
            subsequence.pop()
        
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