human
Give me a the Python code you requested given the following:
Solve in Python:
We have an array of integers, nums, and an array of requests where requests[i] = [starti, endi]. The ith request asks for the sum of nums[starti] + nums[starti + 1] + ... + nums[endi - 1] + nums[endi]. Both starti and endi are 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 109 + 7.
 
Example 1:
Input: nums = [1,2,3,4,5], requests = [[1,3],[0,1]]
Output: 19
Explanation: 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: 11
Explanation: 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: 47
Explanation: 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 <= 105
0 <= nums[i] <= 105
1 <= requests.length <= 105
requests[i].length == 2
0 <= starti <= endi < n
assistant
#Here is the Python code you requested.

class Solution:
    def maxSumRangeQuery(self, nums: List[int], requests: List[List[int]]) -> int:
        
        frequency = [0 for i in range(len(nums))]
        
        
        for u, v in requests:
            
            frequency[u] += 1
            if v + 1 < len(nums): frequency[v + 1] -= 1
                
        for i in range(1, len(frequency)):
            
            frequency[i] += frequency[i - 1]
            
        frequency.sort()
        nums.sort()
        
        Sum = 0
        while nums:
            
            Sum += nums.pop()*frequency.pop()
            
        return Sum%(10**9 + 7)