<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 SegTree:

    def __init__(self, nums: List[int]):
        self.n = 1<<len(nums).bit_length()
        # Tracks 1-based indexes of ranges of 1's
        # e.g. [3, 5] means ones are at index 3 and 4 (but not 5)
        self.idxs = [0] * (self.n*2)
        # Tracks sums of ranges of 1's
        self.sums = [0] * (self.n*2)

        prev = 0
        for i,num in enumerate(nums+[0],self.n):
            if num^prev:
                self.idxs[i] = i - self.n + 1
                prev ^= 1

        for i in range(self.n-1,0,-1):
            l, r = i*2, i*2+1
            
            # 4 cases
            # a) Both are indexes
            # b) Both are sums
            # c) Left is index only
            # d) Right is index only
            l_is_idx = self.idxs[l] > 0
            r_is_idx = self.idxs[r] > 0
            if l_is_idx and r_is_idx:
                # Case A) Take the difference and save in sums
                self.sums[i] = self.idxs[r] - self.idxs[l]
                self.idxs[i] = 0
            elif not (l_is_idx or r_is_idx):
                # Case B) Add sums up
                self.sums[i] = self.sums[l] + self.sums[r]
                self.idxs[i] = 0
            elif l_is_idx:
                # Case C) "Shift" left idx "up" by number of 1's found in sums
                self.idxs[i] = self.idxs[l] + self.sums[r]
            else:
                # Case D) "Shift" right idx "down" by number of 1's found in sums
                self.idxs[i] = self.idxs[r] - self.sums[l]

    def update(self, j: int):
        j += self.n
        if self.idxs[j] > 0:
            self.idxs[j] = 0
        else:
            self.idxs[j] = j - self.n + 1

        while j > 1:
            if j%2 == 1:    l, r = j-1, j
            else:           l, r = j, j+1
            j //= 2
            
            # 4 cases
            # a) Both are indexes
            # b) Both are sums
            # c) Left is index only
            # d) Right is index only
            l_is_idx = self.idxs[l] > 0
            r_is_idx = self.idxs[r] > 0
            if l_is_idx and r_is_idx:
                # Case A) Take the difference and save in sums
                self.sums[j] = self.idxs[r] - self.idxs[l]
                self.idxs[j] = 0
            elif not (l_is_idx or r_is_idx):
                # Case B) Add sums up
                self.sums[j] = self.sums[l] + self.sums[r]
                self.idxs[j] = 0
            elif l_is_idx:
                # Case C) "Shift" left idx "up" by number of 1's found in sums
                self.idxs[j] = self.idxs[l] + self.sums[r]
            else:
                # Case D) "Shift" right idx "down" by number of 1's found in sums
                self.idxs[j] = self.idxs[r] - self.sums[l]
    
    def query(self) -> int:
        return self.sums[1]


class Solution:
    def handleQuery(self, nums1: List[int], nums2: List[int], queries: List[List[int]]) -> List[int]:
        st = SegTree(nums1)
        sums = sum(nums2)
        results = []
        n = len(nums1)

        for qtype, arg1, arg2 in queries:
            match qtype:
                case 1: # flip segment between arg1 and arg2
                    st.update(arg1)
                    st.update(arg2 + 1)
                    continue
                case 2: # add arg1 * sum(nums1) to sums
                    sums += arg1 * st.query()
                    continue
                case 3: # report
                    results.append(sums)
                    continue
        
        return results

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