<p>You are given a <strong>0-indexed</strong> array <code>heights</code> of positive integers, where <code>heights[i]</code> represents the height of the <code>i<sup>th</sup></code> building.</p>

<p>If a person is in building <code>i</code>, they can move to any other building <code>j</code> if and only if <code>i &lt; j</code> and <code>heights[i] &lt; heights[j]</code>.</p>

<p>You are also given another array <code>queries</code> where <code>queries[i] = [a<sub>i</sub>, b<sub>i</sub>]</code>. On the <code>i<sup>th</sup></code> query, Alice is in building <code>a<sub>i</sub></code> while Bob is in building <code>b<sub>i</sub></code>.</p>

<p>Return <em>an array</em> <code>ans</code> <em>where</em> <code>ans[i]</code> <em>is <strong>the index of the leftmost building</strong> where Alice and Bob can meet on the</em> <code>i<sup>th</sup></code> <em>query</em>. <em>If Alice and Bob cannot move to a common building on query</em> <code>i</code>, <em>set</em> <code>ans[i]</code> <em>to</em> <code>-1</code>.</p>

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

<pre>
<strong>Input:</strong> heights = [6,4,8,5,2,7], queries = [[0,1],[0,3],[2,4],[3,4],[2,2]]
<strong>Output:</strong> [2,5,-1,5,2]
<strong>Explanation:</strong> In the first query, Alice and Bob can move to building 2 since heights[0] &lt; heights[2] and heights[1] &lt; heights[2]. 
In the second query, Alice and Bob can move to building 5 since heights[0] &lt; heights[5] and heights[3] &lt; heights[5]. 
In the third query, Alice cannot meet Bob since Alice cannot move to any other building.
In the fourth query, Alice and Bob can move to building 5 since heights[3] &lt; heights[5] and heights[4] &lt; heights[5].
In the fifth query, Alice and Bob are already in the same building.  
For ans[i] != -1, It can be shown that ans[i] is the leftmost building where Alice and Bob can meet.
For ans[i] == -1, It can be shown that there is no building where Alice and Bob can meet.
</pre>

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

<pre>
<strong>Input:</strong> heights = [5,3,8,2,6,1,4,6], queries = [[0,7],[3,5],[5,2],[3,0],[1,6]]
<strong>Output:</strong> [7,6,-1,4,6]
<strong>Explanation:</strong> In the first query, Alice can directly move to Bob&#39;s building since heights[0] &lt; heights[7].
In the second query, Alice and Bob can move to building 6 since heights[3] &lt; heights[6] and heights[5] &lt; heights[6].
In the third query, Alice cannot meet Bob since Bob cannot move to any other building.
In the fourth query, Alice and Bob can move to building 4 since heights[3] &lt; heights[4] and heights[0] &lt; heights[4].
In the fifth query, Alice can directly move to Bob&#39;s building since heights[1] &lt; heights[6].
For ans[i] != -1, It can be shown that ans[i] is the leftmost building where Alice and Bob can meet.
For ans[i] == -1, It can be shown that there is no building where Alice and Bob can meet.

</pre>

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

<ul>
	<li><code>1 &lt;= heights.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= heights[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= queries.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>queries[i] = [a<sub>i</sub>, b<sub>i</sub>]</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt;= heights.length - 1</code></li>
</ul>

class TreeNode:
    def __init__(self, start, end, val, left, right):
        self.start = start
        self.end = end
        self.val = val
        self.left = left
        self.right = right
class SegmentTee:
    def __init__(self, nums):
        def build(start, end):
            if start == end:
                return TreeNode(start, end, nums[start], None, None)
            mid = start + (end - start) // 2
            left = build(start, mid)
            right = build(mid+1, end)
            val = max(left.val, right.val)
            return TreeNode(start, end, val, left, right)
        self.root = build(0, len(nums)-1)
    def query(self, start, end):
        def doQuery(node, start, end):
            if node.start == start and node.end == end:
                return node.val
            mid = node.start + (node.end - node.start) // 2
            if mid >= end:
                return doQuery(node.left, start, end)
            if mid < start:
                return doQuery(node.right, start, end)
            return max(doQuery(node.left, start, mid), doQuery(node.right, mid+1, end))
        return doQuery(self.root, start, end)
class Solution:
    def leftmostBuildingQueries(self, heights: List[int], queries: List[List[int]]) -> List[int]:
        res = []
        N = len(heights)
        tree = SegmentTee(heights)

        @cache
        def queryab(a, b):
            if a == b:
                return a
            a, b = min(a, b), max(a, b)
            if heights[a] < heights[b]:
                return b

            left, right = b+1, N-1
            if left < N and tree.query(left, right) <= heights[a]:
                return -1

            res = -1
            while left < right:
                mid = left + (right - left)//2
                if tree.query(b+1, mid) > heights[a]:
                    right = mid
                else:
                    left = mid + 1

            if left < N and heights[left] > heights[a]:
                res = left
            return res

        for a, b in queries:
            res.append(queryab(a, b))
        return res



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