<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:
    def findSubsequences(self, nums: List[int]) -> List[List[int]]:
        # Use recursive
        ans = []
        n = len(nums)
        
        visited = set()
        for i in range(1, n):
            # Use a DP solving
            tmp = []
            # First consider the new added things
            for j in range(i):
                if nums[j] <= nums[i]:
                    myarray = [nums[j], nums[i]]
                    
                    if tuple(myarray) not in visited:
                        visited.add(tuple(myarray))
                        tmp.append(myarray)
            # Then consider the new things with tmp
            for oldarray in ans:
                if oldarray[-1] <= nums[i]:
                    myarray = oldarray + [nums[i]]
                    if tuple(myarray) not in visited:
                        visited.add(tuple(myarray))
                        tmp.append(myarray)
            
            ans = ans + tmp
        
        print(ans)
        return ans
                    
                    
                    
        
        
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