<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>

class BinaryIndexedTree:
    __slots__ = ["n", "c"]

    def __init__(self, n: int):
        self.n = n
        self.c = [-1] * (n + 1)

    def update(self, x: int, v: int):
        while x <= self.n:
            self.c[x] = max(self.c[x], v)
            x += x & -x

    def query(self, x: int) -> int:
        mx = -1
        while x:
            mx = max(mx, self.c[x])
            x -= x & -x
        return mx


class Solution:
    def maximumSumQueries(
        self, nums1: List[int], nums2: List[int], queries: List[List[int]]
    ) -> List[int]:
        nums = sorted(zip(nums1, nums2), key=lambda x: -x[0])
        nums2.sort()
        n, m = len(nums1), len(queries)
        ans = [-1] * m
        j = 0
        tree = BinaryIndexedTree(n)
        for i in sorted(range(m), key=lambda i: -queries[i][0]):
            x, y = queries[i]
            while j < n and nums[j][0] >= x:
                k = n - bisect_left(nums2, nums[j][1])
                tree.update(k, nums[j][0] + nums[j][1])
                j += 1
            k = n - bisect_left(nums2, y)
            ans[i] = tree.query(k)
        return ans
69700