<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>>vec;
set<vector<int>>st;
    vector<vector<int>> findSubsequences(vector<int>& nums){
        vector<int>curr;
        for(int i=0;i<nums.size()-1;i++){
            curr.push_back(nums[i]);
            nondecrease(i+1,curr,nums);
            curr.pop_back();
        }
        for(auto x : st) vec.push_back(x);
        return vec;
    } 
    void nondecrease(int i,vector<int>curr,vector<int>&nums){ 
      if(i==nums.size()){
        return;
      }
       for(int j=i;j<nums.size();j++){
           if(nums[j]>=curr.back()){
            curr.push_back(nums[j]);
            nondecrease(j+1,curr,nums);
            st.insert(curr);
            curr.pop_back();
           } 
        }
       }   
};
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