<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:
        def update_lazy(l,r,i,lzy,sum,sqr,mod):
            if l!=r:
                lzy[2*i+1]+=lzy[i]
                lzy[2*i+2]+=lzy[i]
            gap=r-l+1
            new_sum=sum[i]+lzy[i]*gap
            new_sqr=sqr[i] + lzy[i]*sum[i]*2 + lzy[i]*lzy[i]*gap

            sum[i]=new_sum%mod
            sqr[i]=new_sqr%mod
            lzy[i]=0
        def add_one(x,y,l,r,i,lzy,sum,sqr,mod):
            update_lazy(l,r,i,lzy,sum,sqr,mod)
            if l>y or r<x:
                return 
            if l>=x and r<=y:
                lzy[i]=1
                return update_lazy(l,r,i,lzy,sum,sqr,mod)
            m=(l+r)//2
            add_one(x,y,l,m,2*i+1,lzy,sum,sqr,mod)
            add_one(x,y,m+1,r,2*i+2,lzy,sum,sqr,mod)
            sum[i]=(sum[i*2+1]+sum[i*2+2])%mod
            sqr[i]=(sqr[2*i+1]+sqr[2*i+2])%mod

        
        n=len(nums)
        mod=1000000007
        last_pos=[-1]*100001
        lzy=[0]*4*n
        sum=[0]*4*n
        sqr=[0]*4*n
        ans=0

        for i in range(len(nums)):
            st=last_pos[nums[i]]+1
            ed=i
            add_one(st,i,0,n-1,0,lzy,sum,sqr,mod)
            ans=(ans+sqr[0])%mod
            last_pos[nums[i]]=i
        return ans
45100