You are given an array nums consisting of integers. You are also given a 2D array queries, where queries[i] = [posi, xi].
For query i, we first set nums[posi] equal to xi, then we calculate the answer to query i which is the maximum sum of a subsequence of nums where no two adjacent elements are selected.
Return the sum of the answers to all queries.
\n\nSince the final answer may be very large, return it modulo 109 + 7.
A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.
\n\n\n
Example 1:
\n\nInput: nums = [3,5,9], queries = [[1,-2],[0,-3]]
\n\nOutput: 21
\n\nExplanation:
\nAfter the 1st query, nums = [3,-2,9] and the maximum sum of a subsequence with non-adjacent elements is 3 + 9 = 12.
\nAfter the 2nd query, nums = [-3,-2,9] and the maximum sum of a subsequence with non-adjacent elements is 9.
Example 2:
\n\nInput: nums = [0,-1], queries = [[0,-5]]
\n\nOutput: 0
\n\nExplanation:
\nAfter the 1st query, nums = [-5,-1] and the maximum sum of a subsequence with non-adjacent elements is 0 (choosing an empty subsequence).
\n
Constraints:
\n\n1 <= nums.length <= 5 * 104-105 <= nums[i] <= 1051 <= queries.length <= 5 * 104queries[i] == [posi, xi]0 <= posi <= nums.length - 1-105 <= xi <= 105给你一个整数数组 nums 和一个二维数组 queries，其中 queries[i] = [posi, xi]。
对于每个查询 i，首先将 nums[posi] 设置为 xi，然后计算查询 i 的答案，该答案为 nums 中 不包含相邻元素 的 子序列 的 最大 和。
返回所有查询的答案之和。
\n\n由于最终答案可能非常大，返回其对 109 + 7 取余 的结果。
子序列 是指从另一个数组中删除一些或不删除元素而不改变剩余元素顺序得到的数组。
\n\n\n\n
示例 1：
\n\n输入：nums = [3,5,9], queries = [[1,-2],[0,-3]]
\n\n输出：21
\n\n解释：
\n执行第 1 个查询后，nums = [3,-2,9]，不包含相邻元素的子序列的最大和为 3 + 9 = 12。
\n执行第 2 个查询后，nums = [-3,-2,9]，不包含相邻元素的子序列的最大和为 9 。
示例 2：
\n\n输入：nums = [0,-1], queries = [[0,-5]]
\n\n输出：0
\n\n解释：
\n执行第 1 个查询后，nums = [-5,-1]，不包含相邻元素的子序列的最大和为 0（选择空子序列）。
\n\n
提示：
\n\n1 <= nums.length <= 5 * 104-105 <= nums[i] <= 1051 <= queries.length <= 5 * 104queries[i] == [posi, xi]0 <= posi <= nums.length - 1-105 <= xi <= 105O(nums.length) with dynamic programming?",
"In order to optimize, we will use segment tree where each node contains the maximum value of (front element has been chosen or not, back element has been chosen or not)."
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