<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 {
public:
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        set<vector<int>> result;
        vector<int> subsequence;
        recursiveSubsequences(nums, 0, subsequence, result);
        return vector<vector<int>>(result.begin(), result.end());
    }
    
private:
    void recursiveSubsequences(vector<int>& nums, int index, vector<int>& subsequence, set<vector<int>>& result) {
        if (subsequence.size() >= 2) {
            result.insert(subsequence);
        }
        for (int i = index; i < nums.size(); ++i) {
            if (subsequence.empty() || nums[i] >= subsequence.back()) {
                subsequence.push_back(nums[i]);
                recursiveSubsequences(nums, i + 1, subsequence, result);
                subsequence.pop_back();
            }
        }
    }
};
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