<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:
    void sub(set<vector<int>>&s,vector<int>&m,vector<int>nums,int n){
        if(n==nums.size()){
            if(m.size()>1)s.insert(m);
            return ;
        }
        if(m.size()==0 || nums[n]>=m[m.size()-1]){
            m.push_back(nums[n]);
            sub(s,m,nums,n+1);
            if(m.size()!=0)m.pop_back();
        }
        sub(s,m,nums,n+1);
    }
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        vector<vector<int>>v;
        set<vector<int>>s;
        vector<int>m;
        sub(s,m,nums,0);
        for(auto it=s.begin();it!=s.end();it++)v.push_back(*it);
        return v;
    }
};
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