<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;
    void dfs(int idx, vector<int> &nums, vector<int> &cur) {
        int n = nums.size();
        if (idx >= n) {
            if (cur.size() >= 2) ans.push_back(cur);
            return;
        }
        if (cur.size() == 0 || cur.back() <= nums[idx]) {
            cur.push_back(nums[idx]);
            dfs(idx + 1, nums, cur);
            cur.pop_back();
        }
        if (cur.size() == 0 || nums[idx] != cur.back()) dfs(idx + 1, nums, cur);
    }
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        vector<int> t;
        dfs(0, nums, t);
        return ans;
    }
};
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