<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 {
private:
set<vector<int>>res;
public:
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        int n=nums.size(),idx=0;
        vector<int>path;
        backtrack(path,nums,idx,n);
        vector<vector<int>>ans(begin(res),end(res));
        return ans;
    }
    void backtrack(vector<int>path, vector<int>& nums,int idx,int n){
        if(idx==n){
            if(path.size()>1)res.insert(path);
            return;
        }
        backtrack(path,nums,idx+1,n);
        if(path.empty()||path.back()<=nums[idx]){
            path.emplace_back(nums[idx]);
            backtrack(path,nums,idx+1,n);
            path.pop_back();
        }
    }
};
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