<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>

class SegNode:
    def __init__(self, l, r, val=0, left=None, right=None, lazy=False):
        self.l, self.r, self.val = l, r, val
        self.left, self.right = left, right
        self.lazy = lazy


class SegTree:
    def __init__(self, values):
        self.values = values
        self.root = self.build(0, len(values) - 1)

    def build(self, l, r):
        if l == r:
            return SegNode(l, r, self.values[l])
        m = (l + r) // 2
        left, right = self.build(l, m), self.build(m + 1, r)
        return SegNode(l, r, left.val + right.val, left, right)

    def update(self, l, r, node):
        if node.lazy:
            self.propagate(node)
        if node.r < l or r < node.l:
            return 
        if l <= node.l and r >= node.r:
            node.val = (node.r - node.l + 1) - node.val
            if node.l != node.r:
                node.left.lazy ^= True
                node.right.lazy ^= True
            return
        self.update(l, r, node.left)
        self.update(l, r, node.right)
        node.val = node.left.val + node.right.val
        
    def propagate(self, node):
        node.val = (node.r - node.l + 1) - node.val
        if node.l != node.r:
            node.left.lazy ^= True
            node.right.lazy ^= True
        node.lazy = False

    def query(self, node):
        if node.lazy:
            self.propagate(node)
        return node.val


class Solution:
    def handleQuery(
        self, nums1: List[int], nums2: List[int], queries: List[List[int]]
    ) -> List[int]:
        n = len(nums1)
        seg_tree = SegTree(nums1)
        ans = []
        s = sum(nums2)

        for i, j, k in queries:
            if i == 1:
                seg_tree.update(j, k, seg_tree.root)
            elif i == 2:
                s += seg_tree.query(seg_tree.root) * j
            elif i == 3:
                ans.append(s)

        return ans

68700