<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 SegmentTreeNode: 
    def __init__(self, low, high) -> None: 
        self.low = low 
        self.high = high 
        self.val = 0 
        self.lazy = 0 

        
        if low + 1 < high: 
            mid = (low + high)//2 
            self.left = SegmentTreeNode(low, mid)   
            self.right = SegmentTreeNode(mid, high)

        
    def update(self, val): 
        self.lazy += val 
        self.val += val * (self.high - self.low)

    def query(self, low, high): 
        if self.low >= low and self.high <= high:
            res = self.val 
            self.update(1)  # update called 

            return res 

        mid = (self.low + self.high)//2 
        res = 0 
        self.left.update(self.lazy)    # update called to update the left node of the current node 
        self.right.update(self.lazy)

        self.lazy = 0 
        if low < mid: 
            res += self.left.query(low, high)     # query calling it's self for the right node of cur node

        if high > mid: 
            res += self.right.query(low, high)
        
        self.val = self.left.val + self.right.val 

        return res 


class Solution:
    def sumCounts(self, nums: List[int]) -> int:

        MOD = 1_000_000_007
        last_index = {}
        segment_tree = SegmentTreeNode(0,len(nums))
        result, current_result = 0, 0 

        for i, num in enumerate(nums): 
            prev_index = last_index.get(num, -1) + 1

            current_result = (current_result + i - prev_index + 1 + 2*segment_tree.query(prev_index, i+1)) % MOD
            result += current_result 
            last_index[num] = i   

        return result % MOD

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