<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:
        p=10**9+7
        dct={}
        for i, n in enumerate(nums):
            if n in dct:
                dct[n].append(i)
            else:
                dct[n]=[-1, i]
        curs={k:0 for k in dct.keys()}
        s2=0
        pw=1
        l=len(nums)
        while pw<l:
            pw<<=1
        
        
        tree=[0, pw, None, None, 0]
        def rec(t):
            if t[1]-t[0]>1:
                mid=(t[1]+t[0])>>1
                t[2]=[t[0], mid, None, None, 0]
                t[3]=[mid, t[1], None, None, 0]
                rec(t[2])
                rec(t[3])
        rec(tree)
        
        def up(t, a, b):
            if a==t[0] and b==t[1]:
                t[4]+=b-a
                return t[4]
            else:
                carry=(t[4]-t[2][4]-t[3][4])>>1
                t[2][4]+=carry
                t[3][4]+=carry
                mid=(t[1]+t[0])>>1
                tmp=0
                if a<mid:
                    tmp+=up(t[2], a, min(mid, b))
                if b>mid:
                    tmp+=up(t[3], max(mid, a), b)
                t[4]+=b-a
                return tmp
                    
        ans=0
        for n in nums:
            a=dct[n][curs[n]]+1
            curs[n]+=1
            b=dct[n][curs[n]]+1
            s2+=2*up(tree, a, b)-(b-a)
            s2%=p
            ans+=s2
        
        return ans%p
5593