<p>You are given two <strong>0-indexed</strong> arrays <code>nums1</code> and <code>nums2</code> and a 2D array <code>queries</code> of queries. There are three types of queries:</p>

<ol>
	<li>For a query of type 1, <code>queries[i]&nbsp;= [1, l, r]</code>. Flip the values from <code>0</code> to <code>1</code> and from <code>1</code> to <code>0</code> in <code>nums1</code>&nbsp;from index <code>l</code> to index <code>r</code>. Both <code>l</code> and <code>r</code> are <strong>0-indexed</strong>.</li>
	<li>For a query of type 2, <code>queries[i]&nbsp;= [2, p, 0]</code>. For every index <code>0 &lt;= i &lt; n</code>, set&nbsp;<code>nums2[i] =&nbsp;nums2[i]&nbsp;+ nums1[i]&nbsp;* p</code>.</li>
	<li>For a query of type 3, <code>queries[i]&nbsp;= [3, 0, 0]</code>. Find the sum of the elements in <code>nums2</code>.</li>
</ol>

<p>Return <em>an array containing all the answers to the third type&nbsp;queries.</em></p>

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

<pre>
<strong>Input:</strong> nums1 = [1,0,1], nums2 = [0,0,0], queries = [[1,1,1],[2,1,0],[3,0,0]]
<strong>Output:</strong> [3]
<strong>Explanation:</strong> After the first query nums1 becomes [1,1,1]. After the second query, nums2 becomes [1,1,1], so the answer to the third query is 3. Thus, [3] is returned.
</pre>

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

<pre>
<strong>Input:</strong> nums1 = [1], nums2 = [5], queries = [[2,0,0],[3,0,0]]
<strong>Output:</strong> [5]
<strong>Explanation:</strong> After the first query, nums2 remains [5], so the answer to the second query is 5. Thus, [5] is returned.
</pre>

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

<ul>
	<li><code>1 &lt;= nums1.length,nums2.length &lt;= 10<sup>5</sup></code></li>
	<li><code>nums1.length = nums2.length</code></li>
	<li><code>1 &lt;= queries.length &lt;= 10<sup>5</sup></code></li>
	<li><code><font face="monospace">queries[i].length = 3</font></code></li>
	<li><code><font face="monospace">0 &lt;= l &lt;= r &lt;= nums1.length - 1</font></code></li>
	<li><code><font face="monospace">0 &lt;= p &lt;= 10<sup>6</sup></font></code></li>
	<li><code>0 &lt;= nums1[i] &lt;= 1</code></li>
	<li><code>0 &lt;= nums2[i] &lt;= 10<sup>9</sup></code></li>
</ul>

# high-memory implementation: explicit tree

class Node:

    def __init__(self, l, r, val, left=None, right=None):
        self.l = l
        self.r = r
        self.val = val
        self.left = left
        self.right = right
        self.flipped = False

    @classmethod
    def build(cls, nums, start, end):
        if start >= end:
            return None
        if start == end - 1:
            return Node(start, end - 1, nums[start])
        left = cls.build(nums, start, start + (end - start) // 2)
        right = cls.build(nums, start + (end - start) // 2, end)
        return Node(start, end - 1, left.val + right.val, left=left, right=right)

    def sum(self, l, r):
        if l == self.l and r == root.r:
            return self.val
        if r < self.right.l:
            return self.left.sum(l, r)
        if l > self.left.r:
            return self.right.sum(l, r)
        return self.left.sum(l, self.left.r) + self.right.sum(self.right.l, r)

    def flip(self, l, r):
        if l == self.l and r == self.r:
            self.flipped = not self.flipped
            self.val = r - l + 1 - self.val
            return self.val
        if self.flipped:
            self.val = self.left.flip(self.left.l, self.left.r) + self.right.flip(self.right.l, self.right.r)
            self.flipped = False
        if r < self.right.l:
            self.val = self.left.flip(l, r) + self.right.val
        elif l > self.left.r:
            self.val = self.left.val + self.right.flip(l, r)
        else:
            self.val = self.left.flip(l, self.left.r) + self.right.flip(self.right.l, r)
        return self.val


class Solution:
    def handleQuery(self, nums1: List[int], nums2: List[int], queries: List[List[int]]) -> List[int]:
        ans = []
        tot = sum(nums2)
        ones = Node.build(nums1, 0, len(nums1))
        for qt, q1, q2 in queries:
            if qt == 3:
                ans.append(tot)
            elif qt == 2:
                tot += ones.val * q1
            else:
                ones.flip(q1, q2)
        return ans
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