<p>You are given a <strong>0-indexed </strong>integer array <code>nums</code>.</p>

<p>The <strong>distinct count</strong> of a subarray of <code>nums</code> is defined as:</p>

<ul>
	<li>Let <code>nums[i..j]</code> be a subarray of <code>nums</code> consisting of all the indices from <code>i</code> to <code>j</code> such that <code>0 &lt;= i &lt;= j &lt; nums.length</code>. Then the number of distinct values in <code>nums[i..j]</code> is called the distinct count of <code>nums[i..j]</code>.</li>
</ul>

<p>Return <em>the sum of the <strong>squares</strong> of <strong>distinct counts</strong> of all subarrays of </em><code>nums</code>.</p>

<p>Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>A subarray is a contiguous <strong>non-empty</strong> sequence of elements within an array.</p>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,1]
<strong>Output:</strong> 15
<strong>Explanation:</strong> Six possible subarrays are:
[1]: 1 distinct value
[2]: 1 distinct value
[1]: 1 distinct value
[1,2]: 2 distinct values
[2,1]: 2 distinct values
[1,2,1]: 2 distinct values
The sum of the squares of the distinct counts in all subarrays is equal to 1<sup>2</sup> + 1<sup>2</sup> + 1<sup>2</sup> + 2<sup>2</sup> + 2<sup>2</sup> + 2<sup>2</sup> = 15.
</pre>

<p><strong>Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,2]
<strong>Output:</strong> 3
<strong>Explanation:</strong> Three possible subarrays are:
[2]: 1 distinct value
[2]: 1 distinct value
[2,2]: 1 distinct value
The sum of the squares of the distinct counts in all subarrays is equal to 1<sup>2</sup> + 1<sup>2</sup> + 1<sup>2</sup> = 3.</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 Solution:
    def sumCounts(self, nums: List[int]) -> int:
        n = len(nums)
        arr = [0] * (4 * n)
        todo = [0] * (4 * n)

        def update(left, right, l, r, idx):
            if left <= l and r <= right:
                arr[idx] += r - l + 1
                todo[idx] += 1
                return
            mid = (l + r) // 2
            if todo[idx] != 0:
                arr[idx * 2] += todo[idx] * (mid - l + 1)
                arr[idx * 2 + 1] += todo[idx] * (r - mid)
                todo[idx * 2] += todo[idx]
                todo[idx * 2 + 1] += todo[idx]
                todo[idx] = 0
            if left <= mid:
                update(left, right, l, mid, idx * 2)
            if right > mid:
                update(left, right, mid + 1, r, idx * 2 + 1)
            arr[idx] = arr[idx * 2] + arr[idx * 2 + 1]
            return
        
        def query(left, right, l, r, idx):
            if left <= l and r <= right:
                return arr[idx]
            mid = (l + r) // 2
            if todo[idx] != 0:
                arr[idx * 2] += todo[idx] * (mid - l + 1)
                arr[idx * 2 + 1] += todo[idx] * (r - mid)
                todo[idx * 2] += todo[idx]
                todo[idx * 2 + 1] += todo[idx]
                todo[idx] = 0
            ans = 0
            if left <= mid:
                ans += query(left, right, l, mid, idx * 2)
            if right > mid:
                ans += query(left, right, mid + 1, r, idx * 2 + 1)
            return ans
        
        table = {}
        ans = 0
        pre = 0
        n = len(nums)
        for idx, val in enumerate(nums):
            v = table.get(val, -1)
            update(v + 1, idx, 0, n - 1, 1)
            pre += 2 * query(v + 1, idx, 0, n - 1, 1) - (idx - v)
            ans += pre
            table[val] = idx
        return ans % (10 ** 9 + 7)

                
        
5068