<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>

MOD = 1_000_000_007


class SegmentTreeNode:
    def __init__(self, lo, hi) -> None:
        self.lo = lo
        self.hi = hi
        self.val = 0
        self.lazy = 0
        if lo + 1 < hi:
            mid = (lo + hi) // 2
            self.left = SegmentTreeNode(lo, mid)
            self.right = SegmentTreeNode(mid, hi)

    def update(self, val):
        self.lazy += val
        self.val += val * (self.hi - self.lo)

    def query(self, lo, hi):
        if self.lo >= lo and self.hi <= hi:
            res = self.val
            self.update(1)
            return res
        mid = (self.lo + self.hi) // 2
        res = 0
        self.left.update(self.lazy)
        self.right.update(self.lazy)
        self.lazy = 0
        if lo < mid:
            res += self.left.query(lo, hi)
        if hi > mid:
            res += self.right.query(lo, hi)
        self.val = self.left.val + self.right.val
        return res


class Solution:
    def sumCounts(self, nums: list[int]) -> int:
        last_idx_dict = {}
        segment_tree = SegmentTreeNode(0, len(nums))
        res = 0
        curr_res = 0
        for i, num in enumerate(nums):
            last_idx = last_idx_dict.get(num, -1) + 1
            curr_res = (
                curr_res + i - last_idx + 1 + 2 * segment_tree.query(last_idx, i + 1)
            ) % MOD
            res += curr_res
            last_idx_dict[num] = i
        return res % MOD
5163