<p>You are given two <strong>0-indexed</strong> integer arrays <code>nums1</code> and <code>nums2</code>, each of length <code>n</code>, and a <strong>1-indexed 2D array</strong> <code>queries</code> where <code>queries[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>.</p>

<p>For the <code>i<sup>th</sup></code> query, find the <strong>maximum value</strong> of <code>nums1[j] + nums2[j]</code> among all indices <code>j</code> <code>(0 &lt;= j &lt; n)</code>, where <code>nums1[j] &gt;= x<sub>i</sub></code> and <code>nums2[j] &gt;= y<sub>i</sub></code>, or <strong>-1</strong> if there is no <code>j</code> satisfying the constraints.</p>

<p>Return <em>an array </em><code>answer</code><em> where </em><code>answer[i]</code><em> is the answer to the </em><code>i<sup>th</sup></code><em> query.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [4,3,1,2], nums2 = [2,4,9,5], queries = [[4,1],[1,3],[2,5]]
<strong>Output:</strong> [6,10,7]
<strong>Explanation:</strong> 
For the 1st query <code node="[object Object]">x<sub>i</sub> = 4</code>&nbsp;and&nbsp;<code node="[object Object]">y<sub>i</sub> = 1</code>, we can select index&nbsp;<code node="[object Object]">j = 0</code>&nbsp;since&nbsp;<code node="[object Object]">nums1[j] &gt;= 4</code>&nbsp;and&nbsp;<code node="[object Object]">nums2[j] &gt;= 1</code>. The sum&nbsp;<code node="[object Object]">nums1[j] + nums2[j]</code>&nbsp;is 6, and we can show that 6 is the maximum we can obtain.

For the 2nd query <code node="[object Object]">x<sub>i</sub> = 1</code>&nbsp;and&nbsp;<code node="[object Object]">y<sub>i</sub> = 3</code>, we can select index&nbsp;<code node="[object Object]">j = 2</code>&nbsp;since&nbsp;<code node="[object Object]">nums1[j] &gt;= 1</code>&nbsp;and&nbsp;<code node="[object Object]">nums2[j] &gt;= 3</code>. The sum&nbsp;<code node="[object Object]">nums1[j] + nums2[j]</code>&nbsp;is 10, and we can show that 10 is the maximum we can obtain. 

For the 3rd query <code node="[object Object]">x<sub>i</sub> = 2</code>&nbsp;and&nbsp;<code node="[object Object]">y<sub>i</sub> = 5</code>, we can select index&nbsp;<code node="[object Object]">j = 3</code>&nbsp;since&nbsp;<code node="[object Object]">nums1[j] &gt;= 2</code>&nbsp;and&nbsp;<code node="[object Object]">nums2[j] &gt;= 5</code>. The sum&nbsp;<code node="[object Object]">nums1[j] + nums2[j]</code>&nbsp;is 7, and we can show that 7 is the maximum we can obtain.

Therefore, we return&nbsp;<code node="[object Object]">[6,10,7]</code>.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [3,2,5], nums2 = [2,3,4], queries = [[4,4],[3,2],[1,1]]
<strong>Output:</strong> [9,9,9]
<strong>Explanation:</strong> For this example, we can use index&nbsp;<code node="[object Object]">j = 2</code>&nbsp;for all the queries since it satisfies the constraints for each query.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [2,1], nums2 = [2,3], queries = [[3,3]]
<strong>Output:</strong> [-1]
<strong>Explanation:</strong> There is one query in this example with <code node="[object Object]">x<sub>i</sub></code> = 3 and <code node="[object Object]">y<sub>i</sub></code> = 3. For every index, j, either nums1[j] &lt; <code node="[object Object]">x<sub>i</sub></code> or nums2[j] &lt; <code node="[object Object]">y<sub>i</sub></code>. Hence, there is no solution. 
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>nums1.length == nums2.length</code>&nbsp;</li>
	<li><code>n ==&nbsp;nums1.length&nbsp;</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums1[i], nums2[i] &lt;= 10<sup>9</sup>&nbsp;</code></li>
	<li><code>1 &lt;= queries.length &lt;= 10<sup>5</sup></code></li>
	<li><code>queries[i].length ==&nbsp;2</code></li>
	<li><code>x<sub>i</sub>&nbsp;== queries[i][1]</code></li>
	<li><code>y<sub>i</sub> == queries[i][2]</code></li>
	<li><code>1 &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
</ul>

import bisect


class SegNode:

    def __init__(self, start: int, end: int):
        self.start = start
        self.end = end
        self.left = None
        self.right = None
        self.m = -float("inf")

    @classmethod
    def build(cls, start: int, end: int):
        if start > end:
            return None

        node = cls(start, end)
        if start == end:
            return node

        mid = (start + end) >> 1
        node.left = cls.build(start, mid)
        node.right = cls.build(mid + 1, end)
        return node

    def update(self, index: int, m: int):
        if index < self.start:
            return

        if index > self.end:
            return

        if self.start <= index <= self.end:
            self.m = max(self.m, m)

        if self.left:
            self.left.update(index, m)

        if self.right:
            self.right.update(index, m)

    def max(self, start: int, end: int):
        if end < self.start:
            return -float("inf")

        if self.end < start:
            return -float("inf")

        if start <= self.start and self.end <= end:
            return self.m

        m = -float("inf")
        if self.left:
            m = max(m, self.left.max(start, end))
        if self.right:
            m = max(m, self.right.max(start, end))
        return m


class Solution:
    def maximumSumQueries(self, nums1: List[int], nums2: List[int], queries: List[List[int]]) -> List[int]:
        indexes = sorted(range(len(nums1)), key=lambda idx: nums1[idx])

        max_val = [0] * len(indexes)
        max_sum = [0] * len(indexes)
        max_val[-1] = nums2[indexes[-1]]
        for idx in reversed(range(len(indexes) - 1)):
            index = indexes[idx]
            max_val[idx] = max(nums2[index], max_val[idx + 1])

        ans = [0] * len(queries)
        nums1.sort()

        max_sum = [n + m for n, m in zip(nums1, max_val)]

        node = SegNode.build(0, len(max_sum) - 1)
        for idx, s in enumerate(max_sum):
            node.update(idx, s)

        for idx, (x, y) in enumerate(queries):
            i1 = bisect.bisect_left(nums1, x)
            if i1 >= len(nums1):
                ans[idx] = -1
                continue

            if y > max_val[i1]:
                ans[idx] = -1
                continue

            left = i1 + 1
            right = len(max_val)
            while left < right:
                mid = (left + right) >> 1
                if max_val[mid] >= y:
                    left = mid + 1
                else:
                    right = mid

            ans[idx] = node.max(i1, left - 1)
            # ans[idx] = max(max_sum[i1: left])

        return ans

81900