<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 Solution:
    def handleQuery(self, nums1, nums2, queries):
        n = len(nums1)
        m = len(queries)
        seg = [0] * (4 * n)
        lazy = [0] * (4 * n)
        ans = []
        total = sum(nums2)

        def build(node, start, end):
            if start == end:
                seg[node] = nums1[start]
            else:
                mid = (start + end) // 2
                build(2 * node, start, mid)
                build(2 * node + 1, mid + 1, end)
                seg[node] = seg[2 * node] + seg[2 * node + 1]

        def update(node, start, end, l, r):
            if lazy[node]:
                seg[node] = (end - start + 1) - seg[node]
                if start != end:
                    lazy[2 * node] ^= 1
                    lazy[2 * node + 1] ^= 1
                lazy[node] = 0
            if start > end or start > r or end < l:
                return
            if start >= l and end <= r:
                seg[node] = (end - start + 1) - seg[node]
                if start != end:
                    lazy[2 * node] ^= 1
                    lazy[2 * node + 1] ^= 1
                return
            mid = (start + end) // 2
            update(2 * node, start, mid, l, r)
            update(2 * node + 1, mid + 1, end, l, r)
            seg[node] = seg[2 * node] + seg[2 * node + 1]

        def query(node, start, end, l, r):
            if start > end or start > r or end < l:
                return 0
            if lazy[node]:
                seg[node] = (end - start + 1) - seg[node]
                if start != end:
                    lazy[2 * node] ^= 1
                    lazy[2 * node + 1] ^= 1
                lazy[node] = 0
            if start >= l and end <= r:
                return seg[node]
            mid = (start + end) // 2
            p1 = query(2 * node, start, mid, l, r)
            p2 = query(2 * node + 1, mid + 1, end, l, r)
            return p1 + p2

        build(1, 0, n - 1)
        for q in queries:
            if q[0] == 1:
                update(1, 0, n - 1, q[1], q[2])
            elif q[0] == 2:
                total += query(1, 0, n - 1, 0, n - 1) * q[1]
            else:
                ans.append(total)
        return ans
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