Given a <strong>0-indexed</strong> integer array <code>nums</code> of length <code>n</code> and an integer <code>k</code>, return <em>the <strong>number of pairs</strong></em> <code>(i, j)</code> <em>where</em> <code>0 &lt;= i &lt; j &lt; n</code>, <em>such that</em> <code>nums[i] == nums[j]</code> <em>and</em> <code>(i * j)</code> <em>is divisible by</em> <code>k</code>.
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,1,2,2,2,1,3], k = 2
<strong>Output:</strong> 4
<strong>Explanation:</strong>
There are 4 pairs that meet all the requirements:
- nums[0] == nums[6], and 0 * 6 == 0, which is divisible by 2.
- nums[2] == nums[3], and 2 * 3 == 6, which is divisible by 2.
- nums[2] == nums[4], and 2 * 4 == 8, which is divisible by 2.
- nums[3] == nums[4], and 3 * 4 == 12, which is divisible by 2.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4], k = 1
<strong>Output:</strong> 0
<strong>Explanation:</strong> Since no value in nums is repeated, there are no pairs (i,j) that meet all the requirements.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
	<li><code>1 &lt;= nums[i], k &lt;= 100</code></li>
</ul>

class Solution:
    def countPairs(self, nums: List[int], k: int) -> int:
        idx = {}

        for i,x in enumerate(nums):
            if x not in idx:
                idx[x] = []

            idx[x].append(i)

        ans = 0
        for x in idx.values():
            n = len(x)
            for i in range(n):
                for j in range(i+1, n):
                    if (x[i] * x[j]) % k == 0:
                        ans += 1

        return ans
56