<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>, and you are allowed to <strong>traverse</strong> between its indices. You can traverse between index <code>i</code> and index <code>j</code>, <code>i != j</code>, if and only if <code>gcd(nums[i], nums[j]) &gt; 1</code>, where <code>gcd</code> is the <strong>greatest common divisor</strong>.</p>

<p>Your task is to determine if for <strong>every pair</strong> of indices <code>i</code> and <code>j</code> in nums, where <code>i &lt; j</code>, there exists a <strong>sequence of traversals</strong> that can take us from <code>i</code> to <code>j</code>.</p>

<p>Return <code>true</code><em> if it is possible to traverse between all such pairs of indices,</em><em> or </em><code>false</code><em> otherwise.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,3,6]
<strong>Output:</strong> true
<strong>Explanation:</strong> In this example, there are 3 possible pairs of indices: (0, 1), (0, 2), and (1, 2).
To go from index 0 to index 1, we can use the sequence of traversals 0 -&gt; 2 -&gt; 1, where we move from index 0 to index 2 because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 &gt; 1, and then move from index 2 to index 1 because gcd(nums[2], nums[1]) = gcd(6, 3) = 3 &gt; 1.
To go from index 0 to index 2, we can just go directly because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 &gt; 1. Likewise, to go from index 1 to index 2, we can just go directly because gcd(nums[1], nums[2]) = gcd(3, 6) = 3 &gt; 1.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,9,5]
<strong>Output:</strong> false
<strong>Explanation:</strong> No sequence of traversals can take us from index 0 to index 2 in this example. So, we return false.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,3,12,8]
<strong>Output:</strong> true
<strong>Explanation:</strong> There are 6 possible pairs of indices to traverse between: (0, 1), (0, 2), (0, 3), (1, 2), (1, 3), and (2, 3). A valid sequence of traversals exists for each pair, so we return true.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

# # class DSU:

# #     def __init__(self, N):
# #         self.dsu = list(range(N + 1))
# #         self.size = [1] * (N + 1)

# #     def find(self, x):
# #         return x if self.dsu[x] == x else self.find(self.dsu[x])

# #     def merge(self, x, y):
# #         fx = self.find(x)
# #         fy = self.find(y)
# #         if fx == fy:
# #             return
# #         if self.size[fx] > self.size[fy]:
# #             fx, fy = fy, fx
# #         self.dsu[fx] = fy
# #         self.size[fy] += self.size[fx]

class Solution:
    @functools.lru_cache()
    def gcd(self, left, right):
        if right == 0:
            return left
        else:
            return gcd(right, left % right)

    def is_invalid(self, left, right):
        return self.gcd(left, right) - 1

    def canTraverseAllPairs(self, nums: List[int]) -> bool:
        
        if len(nums) == 1:
            return True

        nums = set(nums)
        sz = len(nums)

        if 1 in nums:
            return False
        if sz == 1:
            return True

        nums = sorted(list(nums), reverse=True)

        for i in range(sz - 1):
            
            has_bad_route = False

            for j in range(i + 1, sz):
                if self.is_invalid(nums[i], nums[j]):
                    nums[j] *= nums[i]
                    has_bad_route = True
                    break

            if not has_bad_route:
                return False

        return True


#         # sieve = [0 for i in range(max_val + 1)]
#         # for d in range(2, max_val + 1):
#         #     if sieve[d] == 0:
#         #         for v in range(d, max_val + 1, d)
#         #             sieve[v] = d
        

# class Solution:
#     def canTraverseAllPairs(self, nums: List[int]) -> bool:
#         if len(nums)==1:return True
#         nums = set(nums)
#         if 1 in nums:return False
#         if len(nums)==1:return True
        
#         nums = sorted(nums,reverse=True)

#         for i in range(len(nums)-1):
#             for j in range(i+1,len(nums)):
#                 if gcd(nums[i],nums[j])-1:
#                     nums[j]*=nums[i]
#                     break
#             else:
#                 return False
#         return True
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