<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>> ans;
    vector<vector<int>> a;
    int n;

    void rec(int id, vector<int> curr, vector<int>& nums){
        if(id>=n){
            return;
        }
        int l = curr.size();
        rec(id+1, curr, nums);

        if(l==0 || curr[l-1]<=nums[id]){
            curr.push_back(nums[id]);
            if(l>0 && ans.find(curr)==ans.end()) {a.push_back(curr); ans.insert(curr);}
            rec(id+1, curr, nums);
        }
    }

    vector<vector<int>> findSubsequences(vector<int>& nums) {
        ans.clear();
        n = nums.size();
        vector<int> curr;
        rec(0, curr, nums);
        return a;
    }
};
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