<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 RMQ:
    def __init__(self, arr):
        self.arr = arr
        self.n = len(arr)
        self.k = self.n.bit_length() + 1
        self.st = [[None for _ in range(self.n)] for _ in range(self.k)]

        for i in range(self.n):
            self.st[0][i] = self.arr[i]
        
        for p in range(1, self.k):
            for i in range(self.n):
                nxt = i + (1 << (p - 1))
                if nxt < self.n:
                    self.st[p][i] = max(self.st[p - 1][i], self.st[p - 1][nxt])
                else:
                    self.st[p][i] = self.st[p - 1][i]
    def query(self, l, r):
        p = (r - l + 1).bit_length() - 1
        len_ = 1 << p
        return max(self.st[p][l], self.st[p][r - len_ + 1])

class Solution:
    def maximumSumQueries(self, nums1: List[int], nums2: List[int], queries: List[List[int]]) -> List[int]:

        yx = list(zip(nums2, nums1))
        xsum = [(x, x + y) for y, x in yx]
        xsum.sort()
        xs = [x for x, _ in xsum]
        sums = [s for _, s in xsum]
        
        rmq = RMQ(sums)

        yx.sort()
        ys = [y for y, _ in yx]
        sufmaxxs = list(reversed(list(accumulate([x for y, x in reversed(yx)],
                                             max))))
        
        ans = []
        for xq, yq in queries:
            indxy = bisect.bisect_left(ys, yq)

            if indxy == len(ys):
                ans.append(-1)
                continue
            maxx = sufmaxxs[indxy]
            if maxx < xq:
                ans.append(-1)
                continue
            rindxx = bisect.bisect_right(xs, maxx) - 1
            lindxx = bisect.bisect_left(xs, xq)
            if lindxx == len(xs):
                ans.append(-1)
                continue
            ans.append(rmq.query(lindxx, rindxx))
        
        return ans
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