You are given the array `nums`

consisting of `n`

positive integers. You computed the sum of all non-empty continuous subarrays from the array and then sorted them in non-decreasing order, creating a new array of `n * (n + 1) / 2`

numbers.

*Return the sum of the numbers from index *`left`

* to index *`right`

(**indexed from 1**)*, inclusive, in the new array. *Since the answer can be a huge number return it modulo `10`

.^{9} + 7

**Example 1:**

Input:nums = [1,2,3,4], n = 4, left = 1, right = 5Output:13Explanation:All subarray sums are 1, 3, 6, 10, 2, 5, 9, 3, 7, 4. After sorting them in non-decreasing order we have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 1 to ri = 5 is 1 + 2 + 3 + 3 + 4 = 13.

**Example 2:**

Input:nums = [1,2,3,4], n = 4, left = 3, right = 4Output:6Explanation:The given array is the same as example 1. We have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 3 to ri = 4 is 3 + 3 = 6.

**Example 3:**

Input:nums = [1,2,3,4], n = 4, left = 1, right = 10Output:50

**Constraints:**

`n == nums.length`

`1 <= nums.length <= 1000`

`1 <= nums[i] <= 100`

`1 <= left <= right <= n * (n + 1) / 2`

class Solution {
public int rangeSum(int[] nums, int n, int left, int right) {
}
}