Maximum Sum Queries
<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>
Stack; Binary Indexed Tree; Segment Tree; Array; Binary Search; Sorting; Monotonic Stack
type BinaryIndexedTree struct {
	n int
	c []int
}

func NewBinaryIndexedTree(n int) BinaryIndexedTree {
	c := make([]int, n+1)
	for i := range c {
		c[i] = -1
	}
	return BinaryIndexedTree{n: n, c: c}
}

func (bit *BinaryIndexedTree) update(x, v int) {
	for x <= bit.n {
		bit.c[x] = max(bit.c[x], v)
		x += x & -x
	}
}

func (bit *BinaryIndexedTree) query(x int) int {
	mx := -1
	for x > 0 {
		mx = max(mx, bit.c[x])
		x -= x & -x
	}
	return mx
}

func maximumSumQueries(nums1 []int, nums2 []int, queries [][]int) []int {
	n, m := len(nums1), len(queries)
	nums := make([][2]int, n)
	for i := range nums {
		nums[i] = [2]int{nums1[i], nums2[i]}
	}
	sort.Slice(nums, func(i, j int) bool { return nums[j][0] < nums[i][0] })
	sort.Ints(nums2)
	idx := make([]int, m)
	for i := range idx {
		idx[i] = i
	}
	sort.Slice(idx, func(i, j int) bool { return queries[idx[j]][0] < queries[idx[i]][0] })
	tree := NewBinaryIndexedTree(n)
	ans := make([]int, m)
	j := 0
	for _, i := range idx {
		x, y := queries[i][0], queries[i][1]
		for ; j < n && nums[j][0] >= x; j++ {
			k := n - sort.SearchInts(nums2, nums[j][1])
			tree.update(k, nums[j][0]+nums[j][1])
		}
		k := n - sort.SearchInts(nums2, y)
		ans[i] = tree.query(k)
	}
	return ans
}