<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 {
    vector<vector<int>>ans;set<vector<int>>st;

public: 

    vector<vector<int>> findSubsequences(vector<int>& nums) {   
solve(nums,0,{});
for(auto i:st){ans.push_back(i);}
     return ans;   
    }
    void solve(vector<int>&nums,int i,vector<int>temp){
         for(int j=i;j<nums.size();j++){
            if(j>0 && temp.size() && temp.back()>nums[j]){continue;}
            temp.push_back(nums[j]);
           if(temp.size()>=2){ st.insert(temp);}
            solve(nums,j+1,temp);
            temp.pop_back();
         }
    }
};
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