<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;
    set<vector<int>>st;
    void f(int i,int n,vector<int>&arr,int c,vector<int>v)
    {
        if(i >= n)
        {
            if(c >= 2)
            st.insert(v);
            return;
        }
        if(c >= 2)
        {
            st.insert(v);
        }
        f(i+1,n,arr,c,v);
        if(v.empty() || v.back() <= arr[i])
        {
            v.push_back(arr[i]);
            f(i+1,n,arr,c+1,v);
            v.pop_back();
        }
    }
    vector<vector<int>> findSubsequences(vector<int>& arr) {
        int n = arr.size();
        f(0,n,arr,0,{});
        for(auto i : st) ans.push_back(i);
        return ans;
    }
};
#pragma GCC optimize("Ofast")
static auto _ = [](){
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    return nullptr;
}();
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