<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>

# lower-memory implementation: implicit tree

class Solution:
    def handleQuery(self, nums1: List[int], nums2: List[int], queries: List[List[int]]) -> List[int]:
        ones = [[0, False] for _ in range(2 * len(nums2) - 1)]
        def build(v, l, r):
            if l == r:
                ones[v][0] = nums1[l]
            else:
                m = l + (r - l) // 2
                right = v + 2 * (m - l + 1)
                build(v + 1, l, m)
                build(right, m + 1, r)
                ones[v][0] = ones[v + 1][0] + ones[right][0]

        def flip(v, l, r, lpos, rpos):
            if l == lpos and r == rpos:
                ones[v][1] = not ones[v][1]
                ones[v][0] = r - l + 1 - ones[v][0]
                return ones[v][0]
            m = l + (r - l) // 2
            mpos = lpos + (rpos - lpos) // 2
            right = v + 2 * (mpos - lpos + 1)
            if ones[v][1]:
                ones[v][1] = False
                ones[v][0] = (
                    flip(v + 1, lpos, mpos, lpos, mpos) + 
                    flip(right, mpos + 1, rpos, mpos + 1, rpos))
            if r <= mpos:
                ones[v][0] = flip(v + 1, l, r, lpos, mpos) + ones[right][0]
            elif l > mpos:
                ones[v][0] = ones[v + 1][0] + flip(right, l, r, mpos + 1, rpos)
            else:
                ones[v][0] = flip(v + 1, l, mpos, lpos, mpos) + flip(right, mpos + 1, r, mpos + 1, rpos)
            return ones[v][0]

        build(0, 0, len(nums2) - 1)
        ans = []
        tot = sum(nums2)
        r = len(nums2) - 1
        for qt, q1, q2 in queries:
            if qt == 3:
                ans.append(tot)
            elif qt == 2:
                tot += ones[0][0] * q1
            else:
                flip(0, q1, q2, 0, r)
        return ans
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