<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>

def increment(bases):
    bases[0] += 1
    for x in range(len(bases) - 1):
        if bases[x] > 1:
            bases[x] = 0
            bases[x + 1] += 1

def all_increasing(nums):
    for x in range(len(nums) - 1):
        if nums[x] > nums[x + 1]:
            return False

    return True

class Solution:
    def findSubsequences(self, nums: List[int]) -> List[List[int]]:
        bases = [0] * len(nums)
        ret = set()

        for x in range(2 ** len(nums)):
            
            if sum(bases) > 1:
                # print(bases)
                temp = [num for choose, num in zip(bases, nums) if choose == 1]
                # print(temp)

                if all_increasing(temp):
                    ret.add(tuple(temp))

            if x == 2 ** len(nums) - 1:
                break

            increment(bases)

        return list(map(list, ret))
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