<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]]:
        Solution.answer = []
        Solution.set = set()
        self.backtrack(nums, 0, [])
        return Solution.answer
        
    def backtrack(self, nums, index, comb):
        comb_str = ",".join([str(n) for n in comb])
        if len(comb) > 1 and comb_str not in Solution.set:
            Solution.answer.append(comb[:])
            Solution.set.add(comb_str)

        for i in range(index, len(nums)):
            if not len(comb) or nums[i] >= comb[-1]:
                comb.append(nums[i])
                self.backtrack(nums, i+1, comb)
                comb.pop()
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