<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>> findSubsequences(vector<int>& nums) {
        dfs(0, INT_MIN, nums);
        return ans;
    }

private:
    vector<vector<int>> ans;
    vector<int> temp;

    void dfs(int cur, int last, vector<int>& nums) {
        if (cur == nums.size()) {
            if (temp.size() >= 2) {
                ans.push_back(vector<int>(temp));
            }
            return;
        }
        if (nums[cur] >= last) {
            temp.push_back(nums[cur]);
            dfs(cur + 1, nums[cur], nums);
            temp.pop_back();
        }
        if (nums[cur] != last) {
            dfs(cur + 1, last, nums);
        }
    }
};
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