You are given a 0-indexed integer array nums.
The distinct count of a subarray of nums is defined as:
nums[i..j] be a subarray of nums consisting of all the indices from i to j such that 0 <= i <= j < nums.length. Then the number of distinct values in nums[i..j] is called the distinct count of nums[i..j].Return the sum of the squares of distinct counts of all subarrays of nums.
Since the answer may be very large, return it modulo 109 + 7.
A subarray is a contiguous non-empty sequence of elements within an array.
\n\n\n
Example 1:
\n\n\nInput: nums = [1,2,1]\nOutput: 15\nExplanation: Six possible subarrays are:\n[1]: 1 distinct value\n[2]: 1 distinct value\n[1]: 1 distinct value\n[1,2]: 2 distinct values\n[2,1]: 2 distinct values\n[1,2,1]: 2 distinct values\nThe sum of the squares of the distinct counts in all subarrays is equal to 12 + 12 + 12 + 22 + 22 + 22 = 15.\n\n\n
Example 2:
\n\n\nInput: nums = [2,2]\nOutput: 3\nExplanation: Three possible subarrays are:\n[2]: 1 distinct value\n[2]: 1 distinct value\n[2,2]: 1 distinct value\nThe sum of the squares of the distinct counts in all subarrays is equal to 12 + 12 + 12 = 3.\n\n
\n
Constraints:
\n\n1 <= nums.length <= 1051 <= nums[i] <= 105给你一个下标从 0 开始的整数数组 nums 。
定义 nums 一个子数组的 不同计数 值如下：
nums[i..j] 表示 nums 中所有下标在 i 到 j 范围内的元素构成的子数组（满足 0 <= i <= j < nums.length ），那么我们称子数组 nums[i..j] 中不同值的数目为 nums[i..j] 的不同计数。请你返回 nums 中所有子数组的 不同计数 的 平方 和。
由于答案可能会很大，请你将它对 109 + 7 取余 后返回。
子数组指的是一个数组里面一段连续 非空 的元素序列。
\n\n\n\n
示例 1：
\n\n\n输入：nums = [1,2,1]\n输出：15\n解释：六个子数组分别为：\n[1]: 1 个互不相同的元素。\n[2]: 1 个互不相同的元素。\n[1]: 1 个互不相同的元素。\n[1,2]: 2 个互不相同的元素。\n[2,1]: 2 个互不相同的元素。\n[1,2,1]: 2 个互不相同的元素。\n所有不同计数的平方和为 12 + 12 + 12 + 22 + 22 + 22 = 15 。\n\n\n
示例 2：
\n\n\n输入：nums = [2,2]\n输出：3\n解释：三个子数组分别为：\n[2]: 1 个互不相同的元素。\n[2]: 1 个互不相同的元素。\n[2,2]: 1 个互不相同的元素。\n所有不同计数的平方和为 12 + 12 + 12 = 3 。\n\n\n
\n\n
提示：
\n\n1 <= nums.length <= 1051 <= nums[i] <= 105i, let’s call it sum. Now if you need the sum of all subarrays ending with index i + 1 think how it can be related to sum and what extra will be needed to add to this.",
"You can find that extra sum using the segment tree."
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