<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: List[int], nums2: List[int], queries: List[List[int]]) -> List[int]:
        n = len(nums1)
        tree = ([0] * (4 * n))
        lazy = ([0] * (4 * n))

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

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

        def query(node, start, end):
            if lazy[node]:
                tree[node] = (((end - start) + 1) - tree[node])
                if (start != end):
                    lazy[(2 * node)] ^= 1
                    lazy[((2 * node) + 1)] ^= 1
                lazy[node] = 0
            return tree[node]
        build(1, 0, (n - 1))
        sum_nums2 = sum(nums2)
        result = []
        for q in queries:
            if (q[0] == 1):
                update(1, 0, (n - 1), q[1], q[2])
            elif (q[0] == 2):
                sum_nums2 += (query(1, 0, (n - 1)) * q[1])
            else:
                result.append(sum_nums2)
        return result

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