<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]:    
        sum3 = []
        sum2 = sum(nums2)
        n = len(nums1)
        psums1 = self.getPsums(nums1)
        flipIdxs = []
        for i1, i2, i3 in queries:
            if i1 == 1:
                for flipIdx in i2, i3 + 1:
                    flipIdxs.insert(
                        bisect.bisect_left(flipIdxs, flipIdx),
                        flipIdx)
                if len(flipIdxs) * len(flipIdxs) > n * 4:
                    self.flushFlipIdxs(nums1, flipIdxs)
                    psums1 = self.getPsums(nums1)
                    flipIdxs = []
            elif i1 == 2:
                sum2 += self.getSum(psums1, flipIdxs) * i2
            else:
                sum3.append(sum2)
        return sum3
    
    @classmethod    
    def getPsums(cls, nums):
        psums = [0] + nums
        for i in range(len(nums)):
            psums[i + 1] += psums[i]
        return psums
    
    @classmethod
    def getSum(cls, psums, flipIdxs):
        sum1 = psums[-1]
        for i, flipIdx in enumerate(flipIdxs):
            isLo = i % 2 == 0
            if isLo:
                sum1 += psums[flipIdx] * 2 - flipIdx
            else:
                sum1 -= psums[flipIdx] * 2 - flipIdx
        return sum1    
    
    @classmethod
    def flushFlipIdxs(cls, nums, flipIdxs):
        for i in range(0, len(flipIdxs), 2):
            flipOnIdx = flipIdxs[i]
            flipOffIdx = flipIdxs[i + 1]
            for j in range(flipOnIdx, flipOffIdx):
                nums[j] = 1 - nums[j]
            
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