We have an array of integers, `nums`

, and an array of `requests`

where `requests[i] = [start`

. The _{i}, end_{i}]`i`

request asks for the sum of ^{th}`nums[start`

. Both _{i}] + nums[start_{i} + 1] + ... + nums[end_{i} - 1] + nums[end_{i}]`start`

and _{i}`end`

are _{i}*0-indexed*.

Return *the maximum total sum of all requests among all permutations of*

`nums`

.Since the answer may be too large, return it **modulo** `10`

.^{9} + 7

**Example 1:**

Input:nums = [1,2,3,4,5], requests = [[1,3],[0,1]]Output:19Explanation:One permutation of nums is [2,1,3,4,5] with the following result: requests[0] -> nums[1] + nums[2] + nums[3] = 1 + 3 + 4 = 8 requests[1] -> nums[0] + nums[1] = 2 + 1 = 3 Total sum: 8 + 3 = 11. A permutation with a higher total sum is [3,5,4,2,1] with the following result: requests[0] -> nums[1] + nums[2] + nums[3] = 5 + 4 + 2 = 11 requests[1] -> nums[0] + nums[1] = 3 + 5 = 8 Total sum: 11 + 8 = 19, which is the best that you can do.

**Example 2:**

Input:nums = [1,2,3,4,5,6], requests = [[0,1]]Output:11Explanation:A permutation with the max total sum is [6,5,4,3,2,1] with request sums [11].

**Example 3:**

Input:nums = [1,2,3,4,5,10], requests = [[0,2],[1,3],[1,1]]Output:47Explanation:A permutation with the max total sum is [4,10,5,3,2,1] with request sums [19,18,10].

**Constraints:**

`n == nums.length`

`1 <= n <= 10`

^{5}`0 <= nums[i] <= 10`

^{5}`1 <= requests.length <= 10`

^{5}`requests[i].length == 2`

`0 <= start`

_{i}<= end_{i}< n

class Solution {
public int maxSumRangeQuery(int[] nums, int[][] requests) {
}
}