<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 SegTreeLazy: 
    
    def __init__(self, arr: List[int]): 
        """Build the segmentation tree."""
        self.n = n = len(arr)
        self.tree = [0]*(4*n)
        self.lazy = [0]*(4*n)
        
        def build(k: int, lo: int, hi: int) -> None: 
            """Build segment tree from array."""
            if lo+1 == hi: self.tree[k] = arr[lo]
            else: 
                mid = lo + hi >> 1
                build(2*k+1, lo, mid)
                build(2*k+2, mid, hi)
                self.tree[k] = self.tree[2*k+1] + self.tree[2*k+2]
        
        build(0, 0, n)


    def update(self, qlo: int, qhi: int, k: int = 0, lo: int = 0, hi: int = 0) -> None:
        """Update segment tree when value in [qlo, qhi) is flipped."""
        if not hi: hi = self.n
        if self.lazy[k]: 
            self.tree[k] = (hi - lo) - self.tree[k]
            if lo+1 < hi: 
                self.lazy[2*k+1] ^= 1
                self.lazy[2*k+2] ^= 1 
            self.lazy[k] = 0
        if lo < hi and qlo < hi and lo < qhi: 
            if qlo <= lo and hi <= qhi: # total overlap
                self.tree[k] = (hi - lo) - self.tree[k]
                if lo+1 < hi: 
                    self.lazy[2*k+1] ^= 1
                    self.lazy[2*k+2] ^= 1
                return 
            mid = lo + hi >> 1
            self.update(qlo, qhi, 2*k+1, lo, mid) 
            self.update(qlo, qhi, 2*k+2, mid, hi)
            self.tree[k] = self.tree[2*k+1] + self.tree[2*k+2]
            

class Solution:
    def handleQuery(self, nums1: List[int], nums2: List[int], queries: List[List[int]]) -> List[int]:
        tree = SegTreeLazy(nums1)
        ans = []
        val = sum(nums2)
        for x, y, z in queries: 
            if x == 1: tree.update(y, z+1)
            elif x == 2: val += y * tree.tree[0]
            else: ans.append(val)
        return ans
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