<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>>ans;
     vector<vector<int>>dp;
     void f(int i,int prev,vector<int>v,vector<int>& nums,int n){
        if(i>=n){
            if(v.size()>=2){
                ans.push_back(v);
            }
            return;
        }
        if(prev==-1 || nums[i]>=nums[prev]){
           v.push_back(nums[i]);
           f(i+1,i,v,nums,n);
           v.pop_back();
        }
        f(i+1,prev,v,nums,n);
     }
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        int n=nums.size();vector<int>l;
         f(0,-1,l,nums,n);
       set<vector<int>>st;
         for(auto c:ans)st.insert(c);
         vector<vector<int>> g;
         for(auto x:st)g.push_back(x);
        return g;
    }
};
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