<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:
    set<vector<int>>fin;
    void fun(vector<int>&nums,int ind, int prev,vector<int>&ans){
        int n=nums.size();
        if(ans.size()>=2)fin.insert(ans);
        if(ind>=n)return;
        
        if(nums[ind]>=prev){
            ans.push_back(nums[ind]);
            fun(nums,ind+1,nums[ind],ans);
            ans.pop_back();
        }
        fun(nums,ind+1,prev,ans);
    }
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        vector<int>ans;
        fun(nums,0,-102,ans);
        vector<vector<int>>anss;
        for(auto x:fin){
            vector<int>temp;
            for(auto y:x)temp.push_back(y);
            anss.push_back(temp);
        }
        return anss;
    }
};
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