**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Hard Version)
time_limit_ms: 13000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the hard version of the problem. In this version, the constraints on $n$ and the time limit are higher. You can make hacks only if both versions of the problem are solved.

A set of (closed) segments is complex if it can be partitioned into some subsets such that

  * all the subsets have the same size; and    * a pair of segments intersects if and only if the two segments are in the same subset.

You are given $n$ segments $[l_1, r_1], [l_2, r_2], \ldots, [l_n, r_n]$. Find the maximum size of a complex subset of these segments.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the number of segments.

The second line of each test case contains $n$ integers $l_1, l_2, \ldots, l_n$ ($1 \le l_i \le 2n$) — the left endpoints of the segments.

The third line of each test case contains $n$ integers $r_1, r_2, \ldots, r_n$ ($l_i \leq r_i \le 2n$) — the right endpoints of the segments.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer: the maximum size of a complex subset of the given segments.

Examples:
input:
3
3
1 2 3
5 4 6
5
1 2 3 6 8
5 4 7 9 10
5
3 1 4 1 5
7 2 6 5 10
output:
3
4
4

Note: In the first test case, all pairs of segments intersect, therefore it is optimal to form a single group containing all of the three segments.

In the second test case, there is no valid partition for all of the five segments. A valid partition with four segments is the following: $\\{\\{ [1, 5], [2, 4] \\}, \\{ [6, 9], [8, 10] \\}\\}$.

In the third test case, it is optimal to make a single group containing all the segments except the second.

Buggy Code:
import sys

def main():

    t = int(sys.stdin.readline())
    for _ in range(t):
        n = int(sys.stdin.readline())
        l = list(map(int, sys.stdin.readline().split()))
        r = list(map(int, sys.stdin.readline().split()))
        events = []
        for i in range(n):
            events.append((l[i], 1))
            events.append((r[i], -1))
        # Sort by position ascending, for same position, +1 before -1
        events.sort(key=lambda x: (x[0], -x[1]))
        counter = 0
        C = 0
        for pos, typ in events:
            if typ ==1:
                counter +=1
                if counter > C:
                    C = counter
            else:
                counter -=1
        max_total = 0
        if C >=2:
            for S in range(2, C+1):
                K = n // S
                total = S * K
                if total > max_total:
                    max_total = total
        # Else, max_total remains 0
        print(max_total)

if __name__ == '__main__':
    main()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause the algorithm to fail. The key issues in the buggy code include:

1. **Incorrect handling of segment intersections**: The code does not correctly determine the maximum size of a complex subset by only considering the maximum number of overlapping segments.
2. **Incorrect partitioning logic**: The code attempts to partition the segments into subsets of equal size but does not ensure that the segments in the same subset intersect and those in different subsets do not intersect.

Given these issues, we can create a test case that:

- Has segments that intersect in a way that does not form a valid complex subset.
- Has a large number of segments to ensure the code's logic is thoroughly tested.
- Ensures that the maximum size of a complex subset is not simply the maximum number of overlapping segments.

Here is a small-scale test input that meets these criteria:

```plaintext
3
4
1 2 3 4
5 4 6 7
6
1 2 3 4 5 6
7 6 5 4 3 2
5
1 2 3 4 5
6 5 4 3 2
```

### Explanation:

1. **Test Case 1:**
   - Segments: `[1, 5]`, `[2, 4]`, `[3, 6]`, `[4, 7]`
   - Intersections:
     - `[1, 5]` intersects with `[2, 4]`, `[3, 6]`, `[4, 7]`
     - `[2, 4]` intersects with `[3, 6]`, `[4, 7]`
     - `[3, 6]` intersects with `[4, 7]`
   - The maximum size of a complex subset is 2 (e.g., `{[1, 5], [2, 4]}` and `{[3, 6], [4, 7]}`).

2. **Test Case 2:**
   - Segments: `[1, 7]`, `[2, 6]`, `[3, 5]`, `[4, 4]`, `[5, 3]`, `[6, 2]`
   - Intersections:
     - `[1, 7]` intersects with all other segments.
     - `[2, 6]` intersects with `[3, 5]`, `[4, 4]`, `[5, 3]`, `[6, 2]`
     - `[3, 5]` intersects with `[4, 4]`, `[5, 3]`
     - `[4, 4]` does not intersect with `[5, 3]`
   - The maximum size of a complex subset is 2 (e.g., `{[1, 7], [2, 6]}` and `{[3, 5], [4, 4]}`).

3. **Test Case 3:**
   - Segments: `[1, 6]`, `[2, 5]`, `[3, 4]`, `[4, 3]`, `[5, 2]`
   - Intersections:
     - `[1, 6]` intersects with all other segments.
     - `[2, 5]` intersects with `[3, 4]`, `[4, 3]`, `[5, 2]`
     - `[3, 4]` intersects with `[4, 3]`
     - `[4, 3]` does not intersect with `[5, 2]`
   - The maximum size of a complex subset is 2 (e.g., `{[1, 6], [2, 5]}` and `{[3, 4], [4, 3]}`).

This test input should expose the bug in the buggy code by demonstrating that the maximum size of a complex subset is not always the maximum number of overlapping segments.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Hard Version)
time_limit_ms: 13000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the hard version of the problem. In this version, the constraints on $n$ and the time limit are higher. You can make hacks only if both versions of the problem are solved.

A set of (closed) segments is complex if it can be partitioned into some subsets such that

  * all the subsets have the same size; and    * a pair of segments intersects if and only if the two segments are in the same subset.

You are given $n$ segments $[l_1, r_1], [l_2, r_2], \ldots, [l_n, r_n]$. Find the maximum size of a complex subset of these segments.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the number of segments.

The second line of each test case contains $n$ integers $l_1, l_2, \ldots, l_n$ ($1 \le l_i \le 2n$) — the left endpoints of the segments.

The third line of each test case contains $n$ integers $r_1, r_2, \ldots, r_n$ ($l_i \leq r_i \le 2n$) — the right endpoints of the segments.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer: the maximum size of a complex subset of the given segments.

Examples:
input:
3
3
1 2 3
5 4 6
5
1 2 3 6 8
5 4 7 9 10
5
3 1 4 1 5
7 2 6 5 10
output:
3
4
4

Note: In the first test case, all pairs of segments intersect, therefore it is optimal to form a single group containing all of the three segments.

In the second test case, there is no valid partition for all of the five segments. A valid partition with four segments is the following: $\\{\\{ [1, 5], [2, 4] \\}, \\{ [6, 9], [8, 10] \\}\\}$.

In the third test case, it is optimal to make a single group containing all the segments except the second.

Buggy Code:
import sys

def main():
    input = sys.stdin.readline
    t_line = input()
    while t_line.strip() == '':
        t_line = input()
    t = int(t_line)
    for _ in range(t):
        n_line = input()
        while n_line.strip() == '':
            n_line = input()
        n = int(n_line)
        if n == 0:
            print(0)
            continue
        l = []
        while len(l) < n:
            l_line = input()
            if l_line.strip() == '':
                continue
            l += list(map(int, l_line.strip().split()))
        r = []
        while len(r) < n:
            r_line = input()
            if r_line.strip() == '':
                continue
            r += list(map(int, r_line.strip().split()))
        events = []
        for i in range(n):
            events.append((l[i], 1))  # start
            events.append((r[i] + 1, -1))  # end
        events.sort()
        k_max = 0
        current = 0
        for pos, delta in events:
            current += delta
            if current > k_max:
                k_max = current
        if k_max >=2:
            m_max = 0
            for k in range(2, k_max + 1):
                s = n // k
                m = k * s
                if m > m_max:
                    m_max = m
            print(m_max)
        else:
            print(n)

if __name__ == '__main__':
    main()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To expose the bug in the given code, we need to craft a test case that specifically targets the logic where the code might fail. The code's main issue is in how it calculates the maximum size of a complex subset. It seems to be overly simplistic in its approach to finding the maximum intersection count and then deriving the maximum subset size.

### Key Points to Consider:
1. **Intersection Logic**: The code calculates the maximum number of intersecting segments at any point and then tries to form groups based on this maximum.
2. **Group Formation**: The code assumes that if the maximum intersection count is `k_max`, then the maximum size of a complex subset is the largest multiple of `k` that fits into `n`.

### Strategy:
- **Create a Test Case with Multiple Intersecting Segments**: We need to create a test case where the segments intersect in a way that the code's logic fails to correctly identify the maximum complex subset.
- **Edge Case**: Consider a scenario where the segments are not uniformly distributed, and the maximum intersection count does not directly translate to the correct group size.

### Example Test Case:
Let's create a test case with 5 segments where the segments intersect in a specific pattern that the code might misinterpret.

```plaintext
1
5
1 2 3 4 5
3 4 5 6 7
```

### Explanation:
- **Segments**:
  - `[1, 3]`
  - `[2, 4]`
  - `[3, 5]`
  - `[4, 6]`
  - `[5, 7]`

- **Intersections**:
  - `[1, 3]` intersects with `[2, 4]` and `[3, 5]`
  - `[2, 4]` intersects with `[1, 3]`, `[3, 5]`, and `[4, 6]`
  - `[3, 5]` intersects with `[1, 3]`, `[2, 4]`, and `[4, 6]`
  - `[4, 6]` intersects with `[2, 4]`, `[3, 5]`, and `[5, 7]`
  - `[5, 7]` intersects with `[4, 6]`

- **Correct Answer**:
  - The maximum size of a complex subset is 2. The segments can be grouped into two subsets: `{[1, 3], [4, 6]}` and `{[2, 4], [5, 7]}`.

### Why This Test Case is Challenging:
- The code might incorrectly calculate the maximum intersection count as 3 (since `[2, 4]` intersects with three other segments).
- However, the correct maximum size of a complex subset is 2, not 3.

### Output:
```plaintext
1
5
1 2 3 4 5
3 4 5 6 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Hard Version)
time_limit_ms: 13000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the hard version of the problem. In this version, the constraints on $n$ and the time limit are higher. You can make hacks only if both versions of the problem are solved.

A set of (closed) segments is complex if it can be partitioned into some subsets such that

  * all the subsets have the same size; and    * a pair of segments intersects if and only if the two segments are in the same subset.

You are given $n$ segments $[l_1, r_1], [l_2, r_2], \ldots, [l_n, r_n]$. Find the maximum size of a complex subset of these segments.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the number of segments.

The second line of each test case contains $n$ integers $l_1, l_2, \ldots, l_n$ ($1 \le l_i \le 2n$) — the left endpoints of the segments.

The third line of each test case contains $n$ integers $r_1, r_2, \ldots, r_n$ ($l_i \leq r_i \le 2n$) — the right endpoints of the segments.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer: the maximum size of a complex subset of the given segments.

Examples:
input:
3
3
1 2 3
5 4 6
5
1 2 3 6 8
5 4 7 9 10
5
3 1 4 1 5
7 2 6 5 10
output:
3
4
4

Note: In the first test case, all pairs of segments intersect, therefore it is optimal to form a single group containing all of the three segments.

In the second test case, there is no valid partition for all of the five segments. A valid partition with four segments is the following: $\\{\\{ [1, 5], [2, 4] \\}, \\{ [6, 9], [8, 10] \\}\\}$.

In the third test case, it is optimal to make a single group containing all the segments except the second.

Buggy Code:
import sys

def process_test_case(n, l, r):
    events = []
    for i in range(n):
        events.append((l[i], 0))  # Start event
        events.append((r[i], 1))  # End event
    events.sort()
    m = 0
    m_max = 0
    for x, typ in events:
        if typ == 0:
            m += 1
            if m > m_max:
                m_max = m
        else:
            m -= 1
    if m_max == 1:
        return n
    else:
        def f(c):
            return c * (n // c)
        left = 2
        right = m_max
        while left < right:
            mid = left + (right - left) // 2
            mid2 = mid + 1
            if f(mid) < f(mid2):
                left = mid2
            else:
                right = mid
        c = left
        return c * (n // c)

def main():
    input = sys.stdin.readline
    t = int(input())
    for _ in range(t):
        n_line = ''
        while n_line.strip() == '':
            n_line = input()
        n = int(n_line)
        if n >=1:
            l = []
            while len(l) < n:
                l_line = ''
                while l_line.strip() == '':
                    l_line = input()
                l += list(map(int, l_line.strip().split()))
            r = []
            while len(r) < n:
                r_line = ''
                while r_line.strip() == '':
                    r_line = input()
                r += list(map(int, r_line.strip().split()))
            s_max = process_test_case(n, l, r)
        else:
            s_max=0
        print(s_max)

if __name__ == '__main__':
    main()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To expose the bug in the provided code, we need to craft a test case that highlights the issues in the logic. The code uses a binary search to find the maximum size of a complex subset, but it might fail in scenarios where the segments have overlapping or non-overlapping properties that are not correctly handled.

Let's consider the following aspects:
1. **Overlapping Segments**: Ensure there are segments that overlap in a way that could confuse the algorithm.
2. **Non-Overlapping Segments**: Ensure there are segments that do not overlap, which might also confuse the algorithm.
3. **Edge Cases**: Test with segments that have the same start or end points.

A good test case would be one where the segments can form a complex set, but the algorithm might incorrectly partition them due to its logic.

Here's a small-scale test input that should expose the bug:

```plaintext
3
6
1 2 3 4 5 6
6 6 6 6 6 6
4
1 2 3 4
4 3 2 1
5
1 2 3 4 5
2 3 4 5 6
```

### Explanation:
1. **First Test Case (6 segments)**:
   - All segments are `[1, 6]`, `[2, 6]`, `[3, 6]`, `[4, 6]`, `[5, 6]`, `[6, 6]`.
   - All segments overlap with each other, forming a single complex subset of size 6.

2. **Second Test Case (4 segments)**:
   - Segments are `[1, 4]`, `[2, 3]`, `[3, 2]`, `[4, 1]`.
   - These segments do not overlap in a way that can form a complex set, but the algorithm might incorrectly partition them.

3. **Third Test Case (5 segments)**:
   - Segments are `[1, 2]`, `[2, 3]`, `[3, 4]`, `[4, 5]`, `[5, 6]`.
   - These segments form a chain where each segment intersects only with its adjacent segments, forming a complex set of size 2 (e.g., `{[1, 2], [2, 3]}`, `{[3, 4], [4, 5]}`, and `[5, 6]`).

This test input should help identify issues in the binary search and segment intersection logic of the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Hard Version)
time_limit_ms: 13000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the hard version of the problem. In this version, the constraints on $n$ and the time limit are higher. You can make hacks only if both versions of the problem are solved.

A set of (closed) segments is complex if it can be partitioned into some subsets such that

  * all the subsets have the same size; and    * a pair of segments intersects if and only if the two segments are in the same subset.

You are given $n$ segments $[l_1, r_1], [l_2, r_2], \ldots, [l_n, r_n]$. Find the maximum size of a complex subset of these segments.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the number of segments.

The second line of each test case contains $n$ integers $l_1, l_2, \ldots, l_n$ ($1 \le l_i \le 2n$) — the left endpoints of the segments.

The third line of each test case contains $n$ integers $r_1, r_2, \ldots, r_n$ ($l_i \leq r_i \le 2n$) — the right endpoints of the segments.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer: the maximum size of a complex subset of the given segments.

Examples:
input:
3
3
1 2 3
5 4 6
5
1 2 3 6 8
5 4 7 9 10
5
3 1 4 1 5
7 2 6 5 10
output:
3
4
4

Note: In the first test case, all pairs of segments intersect, therefore it is optimal to form a single group containing all of the three segments.

In the second test case, there is no valid partition for all of the five segments. A valid partition with four segments is the following: $\\{\\{ [1, 5], [2, 4] \\}, \\{ [6, 9], [8, 10] \\}\\}$.

In the third test case, it is optimal to make a single group containing all the segments except the second.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 300005;
int l[MAXN], r[MAXN], buc[MAXN << 1], p[MAXN << 1], px[MAXN << 1];
namespace dsu {
	int fa[MAXN << 1], siz[MAXN << 1], mn[MAXN << 1];
	void init(int n) {for (int i = 0; i <= n; i++) fa[i] = mn[i] = i, siz[i] = 1;}
	int findfa(int a) {return fa[a] == a ? a : fa[a] = findfa(fa[a]);}
	void merge(int u, int v) {
		u = findfa(u); v = findfa(v); if (u == v) return;
		if (siz[u] < siz[v]) swap(u, v);
		fa[v] = u; siz[u] += siz[v]; mn[u] = min(mn[u], mn[v]);
	}
}
int solve(int n, int m) {
	dsu::init(n << 1); int ret = 0, lat = 0, cnt = 0;
	for (int i = 1; i <= n << 1; i++) {
		if (px[i]) {
			int x = dsu::mn[dsu::findfa(l[p[i]] - 1)];
			if (x) dsu::merge(x, x - 1); else cnt++;
			if (cnt == m) {
				ret++; while (lat < i) {
					if (px[++lat] == 0) {
						x = dsu::mn[dsu::findfa(r[p[lat]])];
						if (x) dsu::merge(x, x - 1), cnt--;
					}
				}
			}
		}
		else dsu::merge(i, i - 1);
	}
	return ret;
}
int dc(int n, int l, int r, int x, int y) {
	if (l == r) return l * solve(n, l);
	int mid = (l + r) >> 1, ret = solve(n, mid);
	int ans = ret * mid;
	if (x > ret && l < mid) ans = max(ans, dc(n, l, mid - 1, x, ret));
	if (ret > y && mid < r) ans = max(ans, dc(n, mid + 1, r, ret, y));
	return ans;
}
int main() {
	int t; scanf("%d", &t); while (t--) {
		int n; scanf("%d", &n);
		for (int i = 1; i <= n; i++) scanf("%d", &l[i]);
		for (int i = 1; i <= n; i++) scanf("%d", &r[i]);
		for (int i = 1; i <= n << 1; i++) buc[i] = 0;
		for (int i = 1; i <= n; i++) buc[l[i]]++, buc[r[i]]++;
		for (int i = 2; i <= n << 1; i++) buc[i] += buc[i - 1];
		for (int i = 1; i <= n; i++) r[i] = buc[r[i]]--;
		for (int i = 1; i <= n; i++) l[i] = buc[l[i]]--;
		// for (int i = 1; i <= n; i++) printf("%d %d\n", l[i], r[i]);
		for (int i = 1; i <= n; i++) p[l[i]] = p[r[i]] = i, px[l[i]] = 0, px[r[i]] = 1;
		// for (int i = 1; i <= n; i++) printf("%d%c", solve(n, i), " \n"[i == n]);
		printf("%d\n", dc(n, 1, n, n, 0));
	}
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that are likely to cause issues. Here are the steps to think through the problem:

1. **Understanding the Bug**: The buggy code involves a complex algorithm with union-find and dynamic programming. The main issue could be in how the segments are processed, especially in the union-find operations and the dynamic programming part.

2. **Edge Cases**:
   - **All segments intersect**: This is a simple case but can help in testing the basic functionality.
   - **No segments intersect**: This can test the boundary conditions and the logic for non-intersecting segments.
   - **Segments with overlapping ranges**: This can test the union-find operations and the merging logic.
   - **Segments with minimal and maximal ranges**: This can test the limits of the algorithm and the input constraints.

3. **Challenging Scenarios**:
   - **Large number of segments with complex intersections**: This can test the performance and correctness of the union-find operations.
   - **Segments with identical endpoints**: This can test the edge cases in the union-find operations and the dynamic programming logic.

4. **Small-Scale Test Input**:
   - We need to create a small-scale test input that is complete and valid, but still challenging enough to expose the bug.

Here is a small-scale test input that includes a mix of the above scenarios:

```plaintext
3
5
1 2 3 4 5
5 4 6 7 8
5
1 3 5 7 9
2 4 6 8 10
5
1 1 1 1 1
5 5 5 5 5
```

Explanation:
- **Test Case 1**: All segments intersect with each other. This tests the basic functionality of the algorithm.
- **Test Case 2**: No segments intersect. This tests the boundary conditions and the logic for non-intersecting segments.
- **Test Case 3**: All segments are identical. This tests the edge cases in the union-find operations and the dynamic programming logic.

This test input should help in exposing any potential bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Hard Version)
time_limit_ms: 13000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the hard version of the problem. In this version, the constraints on $n$ and the time limit are higher. You can make hacks only if both versions of the problem are solved.

A set of (closed) segments is complex if it can be partitioned into some subsets such that

  * all the subsets have the same size; and    * a pair of segments intersects if and only if the two segments are in the same subset.

You are given $n$ segments $[l_1, r_1], [l_2, r_2], \ldots, [l_n, r_n]$. Find the maximum size of a complex subset of these segments.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the number of segments.

The second line of each test case contains $n$ integers $l_1, l_2, \ldots, l_n$ ($1 \le l_i \le 2n$) — the left endpoints of the segments.

The third line of each test case contains $n$ integers $r_1, r_2, \ldots, r_n$ ($l_i \leq r_i \le 2n$) — the right endpoints of the segments.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer: the maximum size of a complex subset of the given segments.

Examples:
input:
3
3
1 2 3
5 4 6
5
1 2 3 6 8
5 4 7 9 10
5
3 1 4 1 5
7 2 6 5 10
output:
3
4
4

Note: In the first test case, all pairs of segments intersect, therefore it is optimal to form a single group containing all of the three segments.

In the second test case, there is no valid partition for all of the five segments. A valid partition with four segments is the following: $\\{\\{ [1, 5], [2, 4] \\}, \\{ [6, 9], [8, 10] \\}\\}$.

In the third test case, it is optimal to make a single group containing all the segments except the second.

Buggy Code:
#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
#include<ext/pb_ds/tree_policy.hpp>
#include<ext/pb_ds/assoc_container.hpp>
#ifdef ONLINE_JUDGE
	#define getchar()(p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
	char buf[1<<21],*p1=buf,*p2=buf;
#endif
using namespace std;using namespace __gnu_pbds;typedef pair<int,int> pii;typedef tree<pii,null_type,less<pii>,rb_tree_tag,tree_order_statistics_node_update> tr;int read(){int num=0,f=1;char c;while(!isdigit(c=getchar()))if(c=='-')f=-1;while(isdigit(c))num=num*10+(c&15),c=getchar();return num*f;}void write(int x,char ch=' '){int F[20],cnt=0;if(!x)putchar('0');if(x<0)putchar('-'),x=-x;while(x)F[cnt++]=x%10+'0',x/=10;while(cnt)putchar(F[--cnt]);putchar(ch);}
namespace Main
{
    //nBlogn nlog^2n
    //get pos of n and can solve this!
    const int N=600010,B=500,inf=1e9;int n,m,q,L[N/B+10],R[N/B+10],bl[N],x,y,z,xx,yy,zz,anss,ans[N];set<pii>e;namespace sgt{int v[N<<2],tag[N<<2];void U(int k,int z){v[k]+=z,tag[k]+=z;}void pushdown(int k){if(tag[k])U(k<<1,tag[k]),U(k<<1|1,tag[k]),tag[k]=0;}void pushup(int k){v[k]=max(v[k<<1],v[k<<1|1]);}void build(int k,int l,int r){v[k]=tag[k]=0;if(l<r){int mid=(l+r)>>1;build(k<<1,l,mid),build(k<<1|1,mid+1,r);}}
    void upd(int k,int l,int r,int L,int R,int z){if(L<=l&&r<=R)U(k,z);else{int mid=(l+r)>>1;pushdown(k);if(L<=mid)upd(k<<1,l,mid,L,R,z);if(mid<R)upd(k<<1|1,mid+1,r,L,R,z);pushup(k);}}int qry(int k,int l,int r,int L,int R,int z){if(v[k]<z)return m+1;if(l==r)return l;int mid=(l+r)>>1,zz;pushdown(k);if(L<=mid&&(zz=qry(k<<1,l,mid,L,R,z))<=m)return zz;return (mid<R?qry(k<<1|1,mid+1,r,L,R,z):m+1);}}
    namespace sgt2{tr v[N],w[N/B+10];void init(){for(int i=1;i<=q;++i)w[i].clear();for(int i=1;i<=m;++i)v[i].clear();}void ins(int x,int y,pii z){if(bl[x]==bl[y])for(int i=x;i<=y;++i)v[i].insert(z);else{for(int i=x;i<=R[bl[x]];++i)v[i].insert(z);for(int i=L[bl[y]];i<=y;++i)v[i].insert(z);for(int i=bl[x]+1;i<bl[y];++i)w[i].insert(z);}}void del(int x,int y,pii z){if(bl[x]==bl[y])for(int i=x;i<=y;++i)v[i].erase(z);else{for(int i=x;i<=R[bl[x]];++i)v[i].erase(z);for(int i=L[bl[y]];i<=y;++i)v[i].erase(z);for(int i=bl[x]+1;i<bl[y];++i)w[i].erase(z);}}
    int Q(int x,int k){int l=1,r=m,mid;for(;l<r;){mid=(l+r)>>1;if((int)(w[bl[x]].order_of_key({mid+1,-inf})+v[x].order_of_key({mid+1,-inf}))>=k)r=mid;else l=mid+1;}return l;}}
    struct nd{int x,y;}a[N];bool cmpx(nd&x,nd&y){return x.x<y.x;}void ins(int i){sgt::upd(1,1,m,a[i].x,a[i].y,1),sgt2::ins(a[i].x,a[i].y,make_pair(a[i].y,i));}void del(int i){sgt::upd(1,1,m,a[i].x,a[i].y,-1),sgt2::del(a[i].x,a[i].y,make_pair(a[i].y,i));}
    void main()
    {
        // c.insert({1,2}),c.insert({3,4}),c.insert({5,6}),c.insert({7,8}),write(c.order_of_key({2,3}));exit(0); 1
        n=read(),m=n<<1,q=anss=0,memset(ans,0,(n+1)<<2),e.clear(),sgt::build(1,1,m);for(int i=1;i<=m;i+=B){L[++q]=i,R[q]=min(i+B-1,m);for(int j=L[q];j<=R[q];++j)bl[j]=q;}sgt2::init();for(int i=1;i<=n;++i)a[i].x=read();for(int i=1;i<=n;++i)a[i].y=read();sort(a+1,a+n+1,cmpx);
        for(int i=1;i<=n;++i)ins(i),e.insert(make_pair(1,i));for(int i=1,j=1;i<=m;++i){for(;(y=(*e.begin()).first)==i;){x=(*e.begin()).second,e.erase(e.begin());if(y<=m&&(z=sgt::qry(1,1,m,y,m,x))<=m)++ans[x],y=sgt2::Q(z,x),e.insert({y+1,x});}
        for(;j<=n&&a[j].x==i;++j)del(j);}for(int i=1;i<=n;++i)anss=max(anss,ans[i]*i)/*,write(ans[i]," \n"[i==n])*/;write(anss,'\n');
    }
};
int main()
{
    // freopen("E.in","r",stdin);
    // freopen(".in","r",stdin),freopen(".out","w",stdout);
    const bool base=1,IO=1;int T;if(!base)T=1;else if(IO)T=read();else ios::sync_with_stdio(0),cin>>T;for(;T--;)Main::main();
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
5
1 2 3 4 5
6 7 8 9 10
5
1 2 3 4 5
5 6 7 8 10
5
1 2 3 4 5
5 6 7 8 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Hard Version)
time_limit_ms: 13000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the hard version of the problem. In this version, the constraints on $n$ and the time limit are higher. You can make hacks only if both versions of the problem are solved.

A set of (closed) segments is complex if it can be partitioned into some subsets such that

  * all the subsets have the same size; and    * a pair of segments intersects if and only if the two segments are in the same subset.

You are given $n$ segments $[l_1, r_1], [l_2, r_2], \ldots, [l_n, r_n]$. Find the maximum size of a complex subset of these segments.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the number of segments.

The second line of each test case contains $n$ integers $l_1, l_2, \ldots, l_n$ ($1 \le l_i \le 2n$) — the left endpoints of the segments.

The third line of each test case contains $n$ integers $r_1, r_2, \ldots, r_n$ ($l_i \leq r_i \le 2n$) — the right endpoints of the segments.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer: the maximum size of a complex subset of the given segments.

Examples:
input:
3
3
1 2 3
5 4 6
5
1 2 3 6 8
5 4 7 9 10
5
3 1 4 1 5
7 2 6 5 10
output:
3
4
4

Note: In the first test case, all pairs of segments intersect, therefore it is optimal to form a single group containing all of the three segments.

In the second test case, there is no valid partition for all of the five segments. A valid partition with four segments is the following: $\\{\\{ [1, 5], [2, 4] \\}, \\{ [6, 9], [8, 10] \\}\\}$.

In the third test case, it is optimal to make a single group containing all the segments except the second.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;

#define rng(i,a,b) for(int i=int(a);i<=int(b);i++)
#define rep(i,b) rng(i,0,b-1)
#define gnr(i,b,a) for(int i=int(b);i>=int(a);i--)
#define per(i,b) gnr(i,b-1,0)
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define bg begin()
#define ed end()
#define all(x) x.bg,x.ed
#define si(x) int(x.size())
template<class t> using vc=vector<t>;
template<class t> using vvc=vc<vc<t>>;
typedef long long ll;
using pii=pair<int,int>;
using vi=vc<int>;
using uint=unsigned;
using ull=unsigned long long;
using pil=pair<int,ll>;
using pli=pair<ll,int>;
using pll=pair<ll,ll>;
using t3=tuple<int,int,int>;

ll rand_int(ll l, ll r) { //[l, r]
	#ifdef LOCAL
	static mt19937_64 gen;
	#else
    static mt19937_64 gen(chrono::steady_clock::now().time_since_epoch().count());
    #endif
    return uniform_int_distribution<ll>(l, r)(gen);
}

template <uint MD> struct ModInt {
    using M = ModInt;
    const static M G;
    uint v;
    ModInt(ll _v = 0) { set_v(_v % MD + MD); }
    M& set_v(uint _v) {
        v = (_v < MD) ? _v : _v - MD;
        return *this;
    }
    explicit operator bool() const { return v != 0; }
    M operator-() const { return M() - *this; }
    M operator+(const M& r) const { return M().set_v(v + r.v); }
    M operator-(const M& r) const { return M().set_v(v + MD - r.v); }
    M operator*(const M& r) const { return M().set_v(ull(v) * r.v % MD); }
    M operator/(const M& r) const { return *this * r.inv(); }
    M& operator+=(const M& r) { return *this = *this + r; }
    M& operator-=(const M& r) { return *this = *this - r; }
    M& operator*=(const M& r) { return *this = *this * r; }
    M& operator/=(const M& r) { return *this = *this / r; }
    bool operator==(const M& r) const { return v == r.v; }
    M pow(ll n) const {
        M x = *this, r = 1;
        while (n) {
            if (n & 1) r *= x;
            x *= x;
            n >>= 1;
        }
        return r;
    }
    M inv() const { return pow(MD - 2); }
    friend ostream& operator<<(ostream& os, const M& r) { return os << r.v; }
};
using Mint = ModInt<998244353>;
template<> const Mint Mint::G = Mint(3);
using a3 = array<int,3>;
#define N_ 40100
const int SZ = (1<<16);

struct Node{
    int s,l,r,m;
}IT[SZ+SZ];

int n, R[N_];

void UDT(Node &cur, Node &c1, Node &c2){
    cur.s = c1.s+c2.s;
    cur.l = max(c1.l,c1.s+c2.l);
    cur.r = max(c2.r,c2.s+c1.r);
    cur.m = max({c1.m,c2.m,c1.r+c2.l});
}
void Add(int a, int b){
    a+=SZ;
    IT[a].s += b;
    IT[a].m=IT[a].r=IT[a].l=max(IT[a].s,0);
    while(a!=1){
        a>>=1;
        UDT(IT[a], IT[a*2],IT[a*2+1]);
    }
}
vc<pii>w;
int TP[SZ], ans;
set<pii>Set;
int Get(int m){
    if(R[m]!=-1)return R[m];
    int pv=0, res=0, Mn = -1, c=0;
    rep(i,n){
        auto [b,e] = w[i];
        if(b <= Mn){
            continue;
        }
        TP[c++] = i;
        Add(b,1);
        Add(e+1,-1);
        if(IT[1].m>=m){
            Mn = e;
            rep(j,c){
                auto [b,e] = w[TP[j]];
                Add(b,-1);Add(e+1,1);
            }
            c=0;
            res++;
        }
    }
    rep(j,c){
        auto [b,e] = w[TP[j]];
        Add(b,-1);Add(e+1,1);
    }
    R[m]=res;
    Set.insert({R[m],m});
    ans = max(ans,R[m]*m);
    return res;
}
void Solve(){
    Set.clear();
    scanf("%d",&n);
    w.resize(n);
    rng(i,1,n)R[i]=-1;
    rep(i,n)scanf("%d",&w[i].se);
    rep(i,n)scanf("%d",&w[i].fi);
    sort(all(w));
    rep(i,n)swap(w[i].fi,w[i].se);
    int B = sqrt(n*(log2(n)+1));
    B=min(B,n);
    ans=0;
    rng(i,1,B){
        Get(i);
    }

    rng(K,1,n/B){
        auto it = Set.lower_bound(pii(K,0));
        if(it==Set.end())continue;
        int b = it->se, e=n/K;
        if(it!=Set.begin()){
            e = prev(it)->se;
        }
        int r=0;
        while(b<=e){
            int m = (b+e)/2;
            if(Get(m)>=K){
                r=m;
                b=m+1;
            }
            else{
                e=m-1;
            }
        }
    }
    printf("%d\n",ans);
}
int main(){
    int TC;
    scanf("%d",&TC);
    while(TC--){
        Solve();
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that might cause the algorithm to fail. Here are some steps to think through the process:

1. **Understand the Bug**: The provided code seems to have issues with handling overlapping segments and partitioning them correctly. The use of a segment tree and binary search suggests that the algorithm is trying to find the maximum size of a complex subset by partitioning the segments into groups.

2. **Edge Cases**:
   - **All segments overlap**: This should form a single complex subset.
   - **No segments overlap**: This should form multiple complex subsets of size 1.
   - **Partially overlapping segments**: This should form multiple complex subsets of varying sizes.
   - **Segments with the same endpoints**: This should form a single complex subset if they overlap.
   - **Segments with different lengths**: This should form complex subsets based on the intersection criteria.

3. **Small-Scale Test Input**:
   - We will create a test case with a small number of segments to ensure it is complete and valid.
   - The test case should be designed to challenge the algorithm's ability to correctly partition the segments.

Given these considerations, let's create a test input:

```plaintext
1
5
1 2 3 4 5
5 4 6 7 8
```

### Explanation:
- **Test Case**:
  - **Number of test cases**: 1
  - **Number of segments**: 5
  - **Left endpoints**: 1, 2, 3, 4, 5
  - **Right endpoints**: 5, 4, 6, 7, 8

- **Expected Output**: 3
  - **Reason**: The segments [1, 5], [2, 4], and [3, 6] form a complex subset because they all intersect. The remaining segments [4, 7] and [5, 8] do not form a complex subset with the others, so the maximum size of a complex subset is 3.

This test input should expose any issues in the algorithm's ability to correctly partition and count the segments.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Hard Version)
time_limit_ms: 13000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the hard version of the problem. In this version, the constraints on $n$ and the time limit are higher. You can make hacks only if both versions of the problem are solved.

A set of (closed) segments is complex if it can be partitioned into some subsets such that

  * all the subsets have the same size; and    * a pair of segments intersects if and only if the two segments are in the same subset.

You are given $n$ segments $[l_1, r_1], [l_2, r_2], \ldots, [l_n, r_n]$. Find the maximum size of a complex subset of these segments.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the number of segments.

The second line of each test case contains $n$ integers $l_1, l_2, \ldots, l_n$ ($1 \le l_i \le 2n$) — the left endpoints of the segments.

The third line of each test case contains $n$ integers $r_1, r_2, \ldots, r_n$ ($l_i \leq r_i \le 2n$) — the right endpoints of the segments.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer: the maximum size of a complex subset of the given segments.

Examples:
input:
3
3
1 2 3
5 4 6
5
1 2 3 6 8
5 4 7 9 10
5
3 1 4 1 5
7 2 6 5 10
output:
3
4
4

Note: In the first test case, all pairs of segments intersect, therefore it is optimal to form a single group containing all of the three segments.

In the second test case, there is no valid partition for all of the five segments. A valid partition with four segments is the following: $\\{\\{ [1, 5], [2, 4] \\}, \\{ [6, 9], [8, 10] \\}\\}$.

In the third test case, it is optimal to make a single group containing all the segments except the second.

Buggy Code:
#include <iostream>
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const ll maxn = 5*1e5+5, INF = 4e18+9;
struct segment{
    int l, r;
};
// supports: point modify, range apply, range query, walk to find first/last with some precedent
// you are to implement the 2 structs Tag and Info
// for the walks, pass a lambda that takes in Info and return true iff the node with that Info will contain the desired element

template<class Info, class Tag>
struct LazySegmentTree {
    int n;
    vector<Info> info;
    vector<Tag> tag;
    LazySegmentTree() : n(0) {}
    LazySegmentTree(int n_, Info v_ = Info()) {
        init(n_, v_);
    }
    template<class T>
    LazySegmentTree(vector<T> init_) {
        init(init_);
    }
    void init(int n_, Info v_ = Info()) {
        init(vector<Info>(n_, v_));
    }
    template<class T>
    void init(vector<T> init_) {
        n = init_.size();
        info.assign(4 << __lg(n), Info());
        tag.assign(4 << __lg(n), Tag());
        function<void(int, int, int)> build = [&](int p, int l, int r) {
            if (r - l == 1) {
                info[p] = init_[l];
                return;
            }
            int m = (l + r) / 2;
            build(2 * p, l, m);
            build(2 * p + 1, m, r);
            pull(p);
        };
        build(1, 0, n);
    }
    void pull(int p) {
        info[p] = info[2 * p] + info[2 * p + 1];
    }
    void apply(int p, const Tag &v) {
        info[p].apply(v);
        tag[p].apply(v);
    }
    void push(int p) {
        apply(2 * p, tag[p]);
        apply(2 * p + 1, tag[p]);
        tag[p] = Tag();
    }
    void modify(int p, int l, int r, int x, const Info &v) {
        if (r - l == 1) {
            info[p] = v;
            return;
        }
        int m = (l + r) / 2;
        push(p);
        if (x < m) {
            modify(2 * p, l, m, x, v);
        } else {
            modify(2 * p + 1, m, r, x, v);
        }
        pull(p);
    }
    void modify(int p, const Info &v) {
        modify(1, 0, n, p, v);
    }
    Info rangeQuery(int p, int l, int r, int x, int y) {
        if (l >= y || r <= x) {
            return Info();
        }
        if (l >= x && r <= y) {
            return info[p];
        }
        int m = (l + r) / 2;
        push(p);
        return rangeQuery(2 * p, l, m, x, y) + rangeQuery(2 * p + 1, m, r, x, y);
    }
    Info rangeQuery(int l, int r) {
        return rangeQuery(1, 0, n, l, r);
    }
    void rangeApply(int p, int l, int r, int x, int y, const Tag &v) {
        if (l >= y || r <= x) {
            return;
        }
        if (l >= x && r <= y) {
            apply(p, v);
            return;
        }
        int m = (l + r) / 2;
        push(p);
        rangeApply(2 * p, l, m, x, y, v);
        rangeApply(2 * p + 1, m, r, x, y, v);
        pull(p);
    }
    void rangeApply(int l, int r, const Tag &v) {
        return rangeApply(1, 0, n, l, r, v);
    }
    template<class F>
    int findFirst(int p, int l, int r, int x, int y, F &&pred) {
        if (l >= y || r <= x) {
            return -1;
        }
        if (l >= x && r <= y && !pred(info[p])) {
            return -1;
        }
        if (r - l == 1) {
            return l;
        }
        int m = (l + r) / 2;
        push(p);
        int res = findFirst(2 * p, l, m, x, y, pred);
        if (res == -1) {
            res = findFirst(2 * p + 1, m, r, x, y, pred);
        }
        return res;
    }
    template<class F>
    int findFirst(int l, int r, F &&pred) {
        return findFirst(1, 0, n, l, r, pred);
    }
    template<class F>
    int findLast(int p, int l, int r, int x, int y, F &&pred) {
        if (l >= y || r <= x) {
            return -1;
        }
        if (l >= x && r <= y && !pred(info[p])) {
            return -1;
        }
        if (r - l == 1) {
            return l;
        }
        int m = (l + r) / 2;
        push(p);
        int res = findLast(2 * p + 1, m, r, x, y, pred);
        if (res == -1) {
            res = findLast(2 * p, l, m, x, y, pred);
        }
        return res;
    }
    template<class F>
    int findLast(int l, int r, F &&pred) {
        return findLast(1, 0, n, l, r, pred);
    }
};

struct Tag {
    int add = 0;
    void apply(const Tag &t) & {
        add += t.add;
    }
};

struct Info {
    int mx = 0;
    void apply(const Tag &t) & {
        mx += t.add;
    }
    Info operator+(const Info &b) {
        return {max(mx, b.mx)};
    }
};
void solve(){
    int n;
    cin >> n;
    vector<segment> a(n+1);
    for(int i = 1; i <= n; i++){
        cin >> a[i].l;
    }
    for(int i = 1; i <= n; i++){
        cin >> a[i].r;
    }
    sort(a.begin()+1, a.end(), [&](segment a, segment b){
        return a.r < b.r;
    });
    auto maxk = [&](int m) -> int{
        LazySegmentTree<Info, Tag> T(2*n+1);
        int R = 0, cnt = 0;
        for(int i = 1; i <= n; i++){
            if(a[i].l <= R){
                continue;
            }
            T.rangeApply(a[i].l, a[i].r+1, {1});
            if(T.rangeQuery(a[i].l, a[i].r+1).mx >= m){
                cnt++;
                R = a[i].r;
            }
        }
        return cnt;
    };
    vector<int> b(n+2);
    b[0] = n+1, b[n+1] = 0;
    auto dnc = [&](auto dnc, int l, int r) -> void{
        if(l > r){
            return;
        }
        int mid = (l+r)/2;
        if(b[l-1] == b[r+1]){
            for(int i = l; i <= r; i++){
                b[i] = b[l-1];
            }
            return;
        }
        b[mid] = maxk(mid);
        dnc(dnc, l, mid-1);
        dnc(dnc, mid+1, r);
    };
    dnc(dnc, 1, n);
    int ans = 0;
    for(int i = 1; i <= n; i++){
        ans = max(ans, b[i]*i);
    }
    cout << ans;
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while(t--){
        solve();
        cout << "\n";
    }
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a test input that exposes the bug in the given buggy code, let's analyze the problem and the code's logic. The main goal is to create a scenario where the code's assumptions or logic fail, particularly focusing on edge cases or complex segment configurations.

### Key Points to Consider:
1. **Segment Overlaps**: The code assumes that if a segment starts after the end of the last added segment (`a[i].l <= R`), it can be added to the segment tree. This assumption might fail in certain edge cases where segments have complex overlaps.
2. **Segment Tree Operations**: The segment tree is used to track the maximum size of complex subsets. If the segment tree's operations are not correctly handled, it could lead to incorrect results.
3. **Sorting and Partitioning**: The code sorts the segments by their right endpoints and then tries to form complex subsets. This sorting and partitioning logic needs to be robust against various segment configurations.

### Test Input:
Let's create a test case where the segments have complex overlaps and the sorting and partitioning logic is challenged.

```plaintext
3
5
1 2 3 4 5
5 4 3 2 1
5
1 2 3 4 5
5 5 5 5 5
5
1 3 5 7 9
2 4 6 8 10
```

### Explanation:
1. **Test Case 1**:
   - Segments: `[1, 5]`, `[2, 4]`, `[3, 3]`, `[4, 2]`, `[5, 1]`
   - All segments overlap with each other, but the code should correctly identify that the maximum size of a complex subset is 1 (since no two segments can form a valid complex subset without violating the intersection rule).

2. **Test Case 2**:
   - Segments: `[1, 5]`, `[2, 5]`, `[3, 5]`, `[4, 5]`, `[5, 5]`
   - All segments end at 5, and all segments overlap with each other. The code should correctly identify that the maximum size of a complex subset is 1.

3. **Test Case 3**:
   - Segments: `[1, 2]`, `[3, 4]`, `[5, 6]`, `[7, 8]`, `[9, 10]`
   - No segments overlap with each other. The code should correctly identify that the maximum size of a complex subset is 5 (since all segments can form a valid complex subset).

This test input is designed to challenge the code's ability to handle complex overlaps, edge cases, and correct partitioning.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Hard Version)
time_limit_ms: 13000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the hard version of the problem. In this version, the constraints on $n$ and the time limit are higher. You can make hacks only if both versions of the problem are solved.

A set of (closed) segments is complex if it can be partitioned into some subsets such that

  * all the subsets have the same size; and    * a pair of segments intersects if and only if the two segments are in the same subset.

You are given $n$ segments $[l_1, r_1], [l_2, r_2], \ldots, [l_n, r_n]$. Find the maximum size of a complex subset of these segments.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the number of segments.

The second line of each test case contains $n$ integers $l_1, l_2, \ldots, l_n$ ($1 \le l_i \le 2n$) — the left endpoints of the segments.

The third line of each test case contains $n$ integers $r_1, r_2, \ldots, r_n$ ($l_i \leq r_i \le 2n$) — the right endpoints of the segments.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer: the maximum size of a complex subset of the given segments.

Examples:
input:
3
3
1 2 3
5 4 6
5
1 2 3 6 8
5 4 7 9 10
5
3 1 4 1 5
7 2 6 5 10
output:
3
4
4

Note: In the first test case, all pairs of segments intersect, therefore it is optimal to form a single group containing all of the three segments.

In the second test case, there is no valid partition for all of the five segments. A valid partition with four segments is the following: $\\{\\{ [1, 5], [2, 4] \\}, \\{ [6, 9], [8, 10] \\}\\}$.

In the third test case, it is optimal to make a single group containing all the segments except the second.

Buggy Code:
import sys
import math
from bisect import bisect_left

INF = 2 * 10**18
INFi = 10**9
N = 5 * 10**5 + 5

dsu = [0] * N
val = [0] * N

def root(v):
    if v == dsu[v]:
        return v
    else:
        dsu[v] = root(dsu[v])
        return dsu[v]

def unite(u, v):
    u = root(u)
    v = root(v)
    if u != v:
        dsu[v] = u

def solve():
    n = int(input())
    L = list(map(int, input().split()))
    R = list(map(int, input().split()))

    xx = sorted(set(R))
    sz = len(xx)
    who = [[] for _ in range(sz + 1)]

    for i in range(n):
        l = bisect_left(xx, L[i]) + 1
        r = bisect_left(xx, R[i]) + 1
        who[r].append(l)

    ans = 0

    def internal_solve(one):
        for i in range(sz + 1):
            dsu[i] = i
            val[i] = 0
        l = 1
        val[l - 1] = n + 2
        cnt = 0

        for r in range(l, sz + 1):
            for j in who[r]:
                if j < l:
                    continue
                val[r] += 1
                pos = root(j - 1)
                val[pos] -= 1
                if val[pos] == 0:
                    unite(pos - 1, pos)

            cost = n + 2 - val[l - 1]
            if cost >= one:
                cnt += 1
                l = r + 1
                val[l - 1] = n + 2
            elif val[r] == 0:
                unite(r - 1, r)

        nonlocal ans
        ans = max(ans, cnt * one)
        return cnt

    def rec(l, r, low, high):
        if low == high:
            return
        while r > l and min(high, n // (r - 1)) * (r - 1) <= ans:
            r -= 1
        if l >= r:
            return
        mid = (l + r) // 2
        mid_val = internal_solve(mid)
        rec(l, mid, mid_val, high)
        rec(mid + 1, r, low, mid_val)

    rec(1, n + 1, 0, n)
    print(ans)

def main():
    t = int(input())
    for _ in range(t):
        solve()

if __name__ == "__main__":
    sys.setrecursionlimit(10**6)
    main()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
2
5
1 2 3 4 5
5 4 3 2 1
3
1 1 1
2 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Hard Version)
time_limit_ms: 13000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the hard version of the problem. In this version, the constraints on $n$ and the time limit are higher. You can make hacks only if both versions of the problem are solved.

A set of (closed) segments is complex if it can be partitioned into some subsets such that

  * all the subsets have the same size; and    * a pair of segments intersects if and only if the two segments are in the same subset.

You are given $n$ segments $[l_1, r_1], [l_2, r_2], \ldots, [l_n, r_n]$. Find the maximum size of a complex subset of these segments.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the number of segments.

The second line of each test case contains $n$ integers $l_1, l_2, \ldots, l_n$ ($1 \le l_i \le 2n$) — the left endpoints of the segments.

The third line of each test case contains $n$ integers $r_1, r_2, \ldots, r_n$ ($l_i \leq r_i \le 2n$) — the right endpoints of the segments.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer: the maximum size of a complex subset of the given segments.

Examples:
input:
3
3
1 2 3
5 4 6
5
1 2 3 6 8
5 4 7 9 10
5
3 1 4 1 5
7 2 6 5 10
output:
3
4
4

Note: In the first test case, all pairs of segments intersect, therefore it is optimal to form a single group containing all of the three segments.

In the second test case, there is no valid partition for all of the five segments. A valid partition with four segments is the following: $\\{\\{ [1, 5], [2, 4] \\}, \\{ [6, 9], [8, 10] \\}\\}$.

In the third test case, it is optimal to make a single group containing all the segments except the second.

Buggy Code:
import java.util.*;

public class Main {
    static final int MAXN = 600005;
    static int n, t, res;
    static int[] c = new int[MAXN], f = new int[MAXN];
    static Pair[] p = new Pair[MAXN];

    static class Pair {
        int first, second;
        Pair(int first, int second) {
            this.first = first;
            this.second = second;
        }
    }

    static int find(int x) {
        if (f[x] == x) return x;
        return f[x] = find(f[x]);
    }

    static int calc(int m) {
        int cnt = 0;
        int cur = 0, lim = 0, lst = 0;

        for (int i = 1; i <= n; i++) {
            int l = p[i].first, r = p[i].second;
            if (l <= lim) continue;

            for (int j = lst + 1; j < r; j++) {
                f[j] = lst;
            }

            cur++;
            lst = f[r] = r;
            int x = find(l);

            if (x > lim) {
                cur--;
                f[x] = x - 1;
            }

            if (cur == m) {
                cnt++;
                cur = 0;
                lim = r;
            }
        }
        res = Math.max(res, cnt * m);
        return cnt;
    }

    static void solve(int l, int r, int L, int R) {
        if (l > r || L == R) return;

        int mid = (l + r) / 2;
        int val = calc(mid);
        solve(l, mid - 1, L, val);
        solve(mid + 1, r, val, R);
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        t = scanner.nextInt();

        while (t-- > 0) {
            n = scanner.nextInt();
            Arrays.fill(c, 0, 8 * n, 0);
            res = 0;

            for (int i = 1; i <= n; i++) {
                int first = scanner.nextInt();
                p[i] = new Pair(first, 0);
                c[first]++;
            }

            for (int i = 1; i <= n; i++) {
                int second = scanner.nextInt();
                p[i].second = second;
                c[second]++;
            }

            for (int i = 1; i <= 2 * n; i++) {
                c[i] += c[i - 1];
            }

            for (int i = 1; i <= n; i++) {
                p[i].second = c[p[i].second]--;
            }

            for (int i = 1; i <= n; i++) {
                p[i].first = c[p[i].first]--;
            }

            Arrays.sort(p, 1, n + 1, Comparator.comparingInt(a -> a.second));

            solve(1, n, n + 1, 0);

            System.out.println(res);
        }

        scanner.close();
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that are likely to cause issues in the algorithm. Here are the steps to think about:

1. **Edge Cases**:
   - Segments that are all the same.
   - Segments that do not intersect at all.
   - Segments that form a complex structure with multiple valid subsets.

2. **Maximizing Failure Probability**:
   - Ensure that the input is within the constraints but is designed to stress the algorithm.
   - Use a mix of overlapping and non-overlapping segments to test the partitioning logic.

3. **Small-Scale Test Input**:
   - Generate a test case with a small number of segments to ensure the input is manageable and valid.

Given these considerations, let's create a test input:

```plaintext
3
3
1 2 3
5 4 6
5
1 2 3 6 8
5 4 7 9 10
5
3 1 4 1 5
7 2 6 5 10
```

This input includes:
1. A simple case where all segments intersect.
2. A case where segments form two valid subsets.
3. A case where one segment is excluded from the complex subset.

This test input is designed to cover different scenarios and should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Hard Version)
time_limit_ms: 13000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the hard version of the problem. In this version, the constraints on $n$ and the time limit are higher. You can make hacks only if both versions of the problem are solved.

A set of (closed) segments is complex if it can be partitioned into some subsets such that

  * all the subsets have the same size; and    * a pair of segments intersects if and only if the two segments are in the same subset.

You are given $n$ segments $[l_1, r_1], [l_2, r_2], \ldots, [l_n, r_n]$. Find the maximum size of a complex subset of these segments.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the number of segments.

The second line of each test case contains $n$ integers $l_1, l_2, \ldots, l_n$ ($1 \le l_i \le 2n$) — the left endpoints of the segments.

The third line of each test case contains $n$ integers $r_1, r_2, \ldots, r_n$ ($l_i \leq r_i \le 2n$) — the right endpoints of the segments.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer: the maximum size of a complex subset of the given segments.

Examples:
input:
3
3
1 2 3
5 4 6
5
1 2 3 6 8
5 4 7 9 10
5
3 1 4 1 5
7 2 6 5 10
output:
3
4
4

Note: In the first test case, all pairs of segments intersect, therefore it is optimal to form a single group containing all of the three segments.

In the second test case, there is no valid partition for all of the five segments. A valid partition with four segments is the following: $\\{\\{ [1, 5], [2, 4] \\}, \\{ [6, 9], [8, 10] \\}\\}$.

In the third test case, it is optimal to make a single group containing all the segments except the second.

Buggy Code:
#include <bits/stdc++.h>
#define ls rt<<1
#define rs rt<<1|1
using namespace std;
const int N=2e4+10;
struct node{
    int l,r;
}a[N];
int cmp(node x,node y){
    return x.r<y.r;
}
int n,fa[N<<1];
int find(int x){
    if(x==fa[x])return x;
    return fa[x]=find(fa[x]);
}
int las,c;
void del(int x){
    if(!x)return;
    fa[x]=find(x-1);
    if(x>las)c--;
}
int check(int sz){
    int cnt=0;c=0,las=0;
    for(int i=1;i<=n*2;i++)fa[i]=i;fa[0]=0;
    for(int i=1,R=0;i<=n;i++){
        int l=a[i].l,r=a[i].r;
        if(l<=las)continue;
        for(int j=R+1;j<a[i].r;j++)fa[j]=R;
        R=r;
        del(find(l-1));
        c++;
        // for(int j=1;j<=r;j++)cout<<find(j)<<" ";
        // cout<<"\n";
        // cout<<i<<" "<<c<<"\n";
        if(c==sz)cnt++,c=0,las=r;

    }
    return cnt;
}
int get(int num){
    int l=1,r=n/num,p=0;
    while(l<=r){
        int mid=l+r>>1;
        if(check(mid)>=num)p=mid,l=mid+1;
        else r=mid-1;
    }
    return p*num;
}
vector<int>le[N<<1],ri[N<<1];
void solve(){
    scanf("%d",&n);for(int i=1;i<=n*2;i++)le[i].clear(),ri[i].clear();
    for(int i=1;i<=n;i++)scanf("%d",&a[i].l),le[a[i].l].push_back(i);
    for(int i=1;i<=n;i++)scanf("%d",&a[i].r),ri[a[i].r].push_back(i);
    if(n==1){
        printf("1\n");
        return;
    }
    int tot=0;
    for(int i=1;i<=n*2;i++){
        for(int x:le[i])a[x].l=++tot;
        for(int x:ri[i])a[x].r=++tot;

    }
    sort(a+1,a+1+n,cmp);
    // for(int i=1;i<=n;i++)cout<<a[i].l<<" "<<a[i].r<<"\n";
    int ans=0,B=min(n,1000);
    // cout<<check(2)<<"\n";
    for(int i=1;i<=B;i++)ans=max(ans,get(i));
    for(int i=ceil(n*1.0/B);i>0;i--)ans=max(ans,check(i)*i);
    printf("%d\n",ans);
}
int main() {
    int t;scanf("%d",&t);
    while(t--)solve();
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. The key points to consider are:

1. **Overlapping Segments**: Segments that overlap in complex ways can cause issues in the partitioning logic.
2. **Edge Cases**: Segments with very specific endpoints that might not be handled correctly.
3. **Large Test Cases**: While the test input should be small-scale, it should still be complex enough to challenge the algorithm.

Let's construct a test input that includes overlapping segments and edge cases:

```plaintext
3
5
1 2 3 4 5
5 4 6 7 8
5
1 2 3 4 5
6 5 4 3 2
6
1 2 3 4 5 6
6 5 4 3 2 1
```

### Explanation:

1. **Test Case 1**:
   - Segments: `[1, 5], [2, 4], [3, 6], [4, 7], [5, 8]`
   - These segments overlap in a way that makes it difficult to partition them into complex subsets. The expected output is `3` because the segments `[1, 5], [2, 4], [3, 6]` can form a complex subset.

2. **Test Case 2**:
   - Segments: `[1, 6], [2, 5], [3, 4], [4, 3], [5, 2]`
   - These segments are designed to test the partitioning logic where segments are in reverse order. The expected output is `4` because the segments `[1, 6], [2, 5], [3, 4], [4, 3]` can form a complex subset.

3. **Test Case 3**:
   - Segments: `[1, 6], [2, 5], [3, 4], [4, 3], [5, 2], [6, 1]`
   - This test case includes a larger number of segments and tests the ability to handle a more complex partitioning. The expected output is `4` because the segments `[1, 6], [2, 5], [3, 4], [4, 3]` can form a complex subset.

This test input should expose any issues in the buggy code by challenging the partitioning logic and handling of overlapping segments.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Hard Version)
time_limit_ms: 13000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the hard version of the problem. In this version, the constraints on $n$ and the time limit are higher. You can make hacks only if both versions of the problem are solved.

A set of (closed) segments is complex if it can be partitioned into some subsets such that

  * all the subsets have the same size; and    * a pair of segments intersects if and only if the two segments are in the same subset.

You are given $n$ segments $[l_1, r_1], [l_2, r_2], \ldots, [l_n, r_n]$. Find the maximum size of a complex subset of these segments.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the number of segments.

The second line of each test case contains $n$ integers $l_1, l_2, \ldots, l_n$ ($1 \le l_i \le 2n$) — the left endpoints of the segments.

The third line of each test case contains $n$ integers $r_1, r_2, \ldots, r_n$ ($l_i \leq r_i \le 2n$) — the right endpoints of the segments.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer: the maximum size of a complex subset of the given segments.

Examples:
input:
3
3
1 2 3
5 4 6
5
1 2 3 6 8
5 4 7 9 10
5
3 1 4 1 5
7 2 6 5 10
output:
3
4
4

Note: In the first test case, all pairs of segments intersect, therefore it is optimal to form a single group containing all of the three segments.

In the second test case, there is no valid partition for all of the five segments. A valid partition with four segments is the following: $\\{\\{ [1, 5], [2, 4] \\}, \\{ [6, 9], [8, 10] \\}\\}$.

In the third test case, it is optimal to make a single group containing all the segments except the second.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")

// #define int long long
#define ll long long
#define lb(x) ((x) & (-x))
#define double long double
#define ff first
#define ss second
#define mr make_pair
#define l(p) t[p].l
#define r(p) t[p].r
#define lz(p) t[p].lz
#define ans(p) t[p].ans
#define res(p) t[p].res
const int mod = 998244353 ;
const int P = 1e9 + 7 ;
// const int modu[10] = {1000000007,1000000009,1000000021,
// 1000000033,1000000087,1000000093,1000000097} ;
void cmin(auto &x, auto y) {
    x = min(x, y) ;
}
void cmax(int &x, int y) {
    x = max(x, y) ;
}
int read() {
    char ch ;int s = 0 ;int w = 1;
    while((ch = getchar()) >'9' || ch < '0' )if(ch == '-')w = -1 ;
    while(ch >= '0' && ch <= '9')s = (s << 1) + (s << 3) + (ch ^ 48) , ch = getchar() ;
    return s * w ;
}
void print(int x) {
    if(x < 0) putchar('-'), x = -x ;
    if(x > 9)print(x / 10) ;
    putchar(x % 10 + '0') ;
}
void prn(int x) {
    print(x) ;
    putchar('\n') ;
}
int pows(int x, int k) {
    long long bace = x, ans = 1 ;
    while(k) {
        if(k & 1) ans *= bace, ans %= mod ;
        bace *= bace, bace %= mod ;
        k >>= 1 ;
    }
    return ans ;
}
mt19937 rnd(time(0)^clock()) ;
int TT ;
int n, m ;
int k, Q ;
pair<int, int> p[300005] ;
struct node {
	int l, r ;
	int maxn ;
	int lz ;
} t[600005<<2] ;
void pup(int p) {t[p].maxn=max(t[p<<1].maxn,t[p<<1|1].maxn) ;}
void build(int l, int r, int p) {
	t[p].maxn=t[p].lz=0;
	l(p) = l, r(p) = r ;
	if(l==r) {
		return ;
	}
	int mid = l + r >> 1 ;
	build(l,mid,p<<1) ;
	build(mid+1,r,p<<1|1) ;
}
void down(int p) {
	if(!lz(p)) return ;
	lz(p<<1)+=lz(p) ;
	lz(p<<1|1)+=lz(p) ;
	t[p<<1].maxn += lz(p) ;
	t[p<<1|1].maxn+=lz(p) ;
	lz(p)=0 ;
}
void change(int l, int r, int p) {
	if(l(p) >= l && r(p) <= r) {
		t[p].maxn++ ;
		lz(p)++ ;
		return ;
	}
	if(lz(p)) down(p) ;
	int mid = l(p) + r(p) >> 1 ;
	if(l<=mid) change(l,r,p<<1) ;
	if(r>mid) change(l,r,p<<1|1) ;
	t[p].maxn=max(t[p<<1].maxn,t[p<<1|1].maxn) ;
}
int ask(int l, int r, int p) {
	if(l(p) >= l && r(p) <= r) return t[p].maxn ;
	if(lz(p)) down(p) ;
	int mid = l(p) + r(p) >> 1 ;
	int res = 0 ;
	if(l <= mid) cmax(res, ask(l, r, p << 1)) ;
	if(r > mid) cmax(res, ask(l,r,p<<1|1)) ;
	return res ;
}
bool cmp(pair<int, int> a, pair<int, int> b) {
	return a.ss< b.ss ;
}
int cal(int now) {
	int R = 0, cnt = 0 ;
	// build(1,2*n,1) ;
	for(int i = 1 ;i <= n<<3 ; i++) t[i].lz=t[i].maxn=0;
	for(int i = 1;  i <= n ; i++) {
		if(p[i].ff <= R) continue ;
		change(p[i].ff, p[i].ss, 1) ;
		// if(now==3)cout<<"!! "<<now<< ' '<<p[i].ff<<' '<<p[i].ss<<' '<<ask(p[i].ff,p[i].ss,1)<<endl;
		if(ask(p[i].ff, p[i].ss, 1) >= now) {
			cnt++ ;
			R = p[i].ss ;
		}
	}
	return cnt ;
}
int ans;
int f[1000005] ;
//小写：每段几个
//大写：分几段
void solve(int l, int r, int L, int R) {
	if(l>r) return ;
	R=min(R,n/l);
	if (L>R) return;
	//如果你能分的最多的段，乘以每段最多的值，都不行，则G
	if ((long long)R*r<=ans) return;
	if(L==R) {
		// cout << l << ' ' << r << ' ' << L << ' ' << R << endl ;
		for(int i = l ; i <= r ; i++)
		{
			f[i] = L ;
			cmax(ans,f[i]*i);
		}
		return ;
	}
	int mid = l + r >> 1 ;
	int val = cal(mid) ;
	f[mid] = val ;
	// cout << mid << ' ' << val << endl ;
	solve(l, mid-1, val, R) ;
	solve(mid + 1, r, L, val) ;
}
void work() {
	n = read() ;
	for (int i=1;i<=n;i++) f[i]=0;
	int m = pow(n * log(n)/log(2), 0.5) ;
	cmax(m,1);
	// prn(m) ;
	build(1,2*n,1);
	for(int i = 1 ; i <= n ; i++) p[i].ff = read() ;
	for(int i = 1 ; i <= n ; i++) p[i].ss = read() ;
	sort(p + 1, p + n + 1, cmp) ;
	ans=0;
	solve(1, n, 0, n) ;
	for(int i = 1 ; i <= n ; i++) cmax(ans, f[i]*i) ;
	cout << ans <<endl;
}
signed main(){
    // freopen("data.in", "r", stdin) ;
    // freopen("ans.out", "w", stdout) ;
	TT = read() ;
    // TT = read() ;
    while(TT--) {

        work() ;
    }
	return 0;
}
/*

*/

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
1
1
1
5
1 2 3 4 5
1 2 3 4 5
10
1 2 3 4 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Hard Version)
time_limit_ms: 13000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the hard version of the problem. In this version, the constraints on $n$ and the time limit are higher. You can make hacks only if both versions of the problem are solved.

A set of (closed) segments is complex if it can be partitioned into some subsets such that

  * all the subsets have the same size; and    * a pair of segments intersects if and only if the two segments are in the same subset.

You are given $n$ segments $[l_1, r_1], [l_2, r_2], \ldots, [l_n, r_n]$. Find the maximum size of a complex subset of these segments.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the number of segments.

The second line of each test case contains $n$ integers $l_1, l_2, \ldots, l_n$ ($1 \le l_i \le 2n$) — the left endpoints of the segments.

The third line of each test case contains $n$ integers $r_1, r_2, \ldots, r_n$ ($l_i \leq r_i \le 2n$) — the right endpoints of the segments.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer: the maximum size of a complex subset of the given segments.

Examples:
input:
3
3
1 2 3
5 4 6
5
1 2 3 6 8
5 4 7 9 10
5
3 1 4 1 5
7 2 6 5 10
output:
3
4
4

Note: In the first test case, all pairs of segments intersect, therefore it is optimal to form a single group containing all of the three segments.

In the second test case, there is no valid partition for all of the five segments. A valid partition with four segments is the following: $\\{\\{ [1, 5], [2, 4] \\}, \\{ [6, 9], [8, 10] \\}\\}$.

In the third test case, it is optimal to make a single group containing all the segments except the second.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define ii pair<int,int>
#define iii tuple<int,int,int>
#define fi first
#define se second
#define endl '\n'
#define debug(x) cout << #x << ": " << x << endl

#define pub push_back
#define pob pop_back
#define puf push_front
#define pof pop_front
#define lb lower_bound
#define ub upper_bound

#define rep(x,start,end) for(int x=start;x<end;x++)
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()

mt19937 rng(chrono::system_clock::now().time_since_epoch().count());

struct node{ //iterative segment tree for range increment queries and range maximum queries
	int BUF;
	int val[1200005],lazy[1200005];

	void reset(int n){
		BUF=n;
		rep(x,0,2*n) val[x]=lazy[x]=0;
	}

	void get(int i){ val[i]=max(val[i<<1],val[i<<1|1])+lazy[i]; }

	void update(int i,int j,int k){ //[i,j]
		i+=BUF-1,j+=BUF+1; //(i,j)
		while (i|j){
			if (j-i>1){
				if (~i&1) val[i^1]+=k,lazy[i^1]+=k;
				if (j&1) val[j^1]+=k,lazy[j^1]+=k;
			}
			i>>=1,j>>=1;
			get(i),get(j);
		}
	}
} root;

int n;
ii arr[300005];

int F[300005];
int B[300005];

int solve(int b){ //find the maximum groups of b vertices that satisfies conditions
	if (F[b]==1e9) return 0;
	if (B[b]<=F[b]) return 1;

	root.reset(2*n+5);
	int L=F[b]+1; //arr[x].fi must be greater than this
	int ans=1;

	rep(x,1,n+1){
		if (L<=arr[x].fi) root.update(arr[x].fi,arr[x].se,1);
		if (arr[x].se>=B[b]) return ans+1;
		if (root.val[1]==b){ //we found a suitable grouping!
			ans++;
			root.update(arr[x].fi,arr[x].se,-1);
			L=arr[x].se+1;
		}
	}
}

int ans[300005];

void dnc(int l,int r){
	if (ans[l-1]==ans[r+1]){
		rep(x,l,r+1) ans[x]=ans[l-1];
		return;
	}

	int m=l+r>>1;
	ans[m]=solve(m);
	if (l<m) dnc(l,m-1);
	if (m<r) dnc(m+1,r);
}

signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin.exceptions(ios::badbit | ios::failbit);

	int TC;
	cin>>TC;
	while (TC--){
		cin>>n;
		rep(x,1,n+1) cin>>arr[x].fi;
		rep(x,1,n+1) cin>>arr[x].se;

		rep(x,1,n+1) F[x]=1e9,B[x]=-1e9;

		sort(arr+1,arr+n+1,[](ii i,ii j){
			return i.fi>j.fi;
		});

		root.reset(2*n+5);
		rep(x,1,n+1){
			root.update(arr[x].fi,arr[x].se,1);
			int v=root.val[1];
			if (B[v]==-1e9) B[v]=arr[x].fi;
		}

		sort(arr+1,arr+n+1,[](ii i,ii j){
			return i.se<j.se;
		});

		root.reset(2*n+5);
		rep(x,1,n+1){
			root.update(arr[x].fi,arr[x].se,1);
			int v=root.val[1];
			if (F[v]==1e9) F[v]=arr[x].se;
		}

		ans[0]=n+1;
		ans[n+1]=0;
		dnc(1,n);

		int fin=0;
		rep(x,1,n+1) fin=max(fin,ans[x]*x);
		cout<<fin<<endl;
	}
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
5
1 2 3 4 5
5 4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Hard Version)
time_limit_ms: 13000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the hard version of the problem. In this version, the constraints on $n$ and the time limit are higher. You can make hacks only if both versions of the problem are solved.

A set of (closed) segments is complex if it can be partitioned into some subsets such that

  * all the subsets have the same size; and    * a pair of segments intersects if and only if the two segments are in the same subset.

You are given $n$ segments $[l_1, r_1], [l_2, r_2], \ldots, [l_n, r_n]$. Find the maximum size of a complex subset of these segments.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the number of segments.

The second line of each test case contains $n$ integers $l_1, l_2, \ldots, l_n$ ($1 \le l_i \le 2n$) — the left endpoints of the segments.

The third line of each test case contains $n$ integers $r_1, r_2, \ldots, r_n$ ($l_i \leq r_i \le 2n$) — the right endpoints of the segments.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer: the maximum size of a complex subset of the given segments.

Examples:
input:
3
3
1 2 3
5 4 6
5
1 2 3 6 8
5 4 7 9 10
5
3 1 4 1 5
7 2 6 5 10
output:
3
4
4

Note: In the first test case, all pairs of segments intersect, therefore it is optimal to form a single group containing all of the three segments.

In the second test case, there is no valid partition for all of the five segments. A valid partition with four segments is the following: $\\{\\{ [1, 5], [2, 4] \\}, \\{ [6, 9], [8, 10] \\}\\}$.

In the third test case, it is optimal to make a single group containing all the segments except the second.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define ii pair<int,int>
#define iii tuple<int,int,int>
#define fi first
#define se second
#define endl '\n'
#define debug(x) cout << #x << ": " << x << endl

#define pub push_back
#define pob pop_back
#define puf push_front
#define pof pop_front
#define lb lower_bound
#define ub upper_bound

#define rep(x,start,end) for(int x=start;x<end;x++)
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()

mt19937 rng(chrono::system_clock::now().time_since_epoch().count());

struct node{ //iterative segment tree for range increment queries and range maximum queries
	int BUF;
	int val[1200005],lazy[1200005];

	void reset(int n){
		BUF=n;
		rep(x,0,2*n) val[x]=lazy[x]=0;
	}

	void get(int i){ val[i]=max(val[i<<1],val[i<<1|1])+lazy[i]; }

	void update(int i,int j,int k){ //[i,j]
		i+=BUF-1,j+=BUF+1; //(i,j)
		while (i|j){
			if (j-i>1){
				if (~i&1) val[i^1]+=k,lazy[i^1]+=k;
				if (j&1) val[j^1]+=k,lazy[j^1]+=k;
			}
			i>>=1,j>>=1;
			get(i),get(j);
		}
	}
} root;

int n;
ii arr[300005];

int F[300005];
int B[300005];

int solve(int b){ //find the maximum groups of b vertices that satisfies conditions
	if (F[b]==1e9) return 0;
	if (B[b]<=F[b]) return 1;

	root.reset(2*n+5);
	int L=F[b]+1; //arr[x].fi must be greater than this
	int ans=1;

	rep(x,1,n+1){
		if (L<=arr[x].fi) root.update(arr[x].fi,arr[x].se,1);
		if (root.val[1]==b){ //we found a suitable grouping!
			ans++;
			if (B[b]<=arr[x].se) return ans;
			root.update(arr[x].fi,arr[x].se,-1);
			L=arr[x].se+1;
		}
	}

	return ans;
}

int ans[300005];

void dnc(int l,int r){
	if (ans[l-1]==ans[r+1]){
		rep(x,l,r+1) ans[x]=ans[l-1];
		return;
	}

	int m=l+r>>1;
	ans[m]=solve(m);
	if (l<m) dnc(l,m-1);
	if (m<r) dnc(m+1,r);
}

signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin.exceptions(ios::badbit | ios::failbit);

	int TC;
	cin>>TC;
	while (TC--){
		cin>>n;
		rep(x,1,n+1) cin>>arr[x].fi;
		rep(x,1,n+1) cin>>arr[x].se;

		rep(x,1,n+1) F[x]=1e9,B[x]=-1e9;

		sort(arr+1,arr+n+1,[](ii i,ii j){
			return i.fi>j.fi;
		});

		root.reset(2*n+5);
		rep(x,1,n+1){
			root.update(arr[x].fi,arr[x].se,1);
			int v=root.val[1];
			if (B[v]==-1e9) B[v]=arr[x].fi;
		}

		sort(arr+1,arr+n+1,[](ii i,ii j){
			return i.se<j.se;
		});

		root.reset(2*n+5);
		rep(x,1,n+1){
			root.update(arr[x].fi,arr[x].se,1);
			int v=root.val[1];
			if (F[v]==1e9) F[v]=arr[x].se;
		}

		ans[0]=n+1;
		ans[n+1]=0;
		dnc(1,n);

		int fin=0;
		rep(x,1,n+1) fin=max(fin,ans[x]*x);
		cout<<fin<<endl;
	}
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
5
1 2 3 4 5
5 4 6 7 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Hard Version)
time_limit_ms: 13000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the hard version of the problem. In this version, the constraints on $n$ and the time limit are higher. You can make hacks only if both versions of the problem are solved.

A set of (closed) segments is complex if it can be partitioned into some subsets such that

  * all the subsets have the same size; and    * a pair of segments intersects if and only if the two segments are in the same subset.

You are given $n$ segments $[l_1, r_1], [l_2, r_2], \ldots, [l_n, r_n]$. Find the maximum size of a complex subset of these segments.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the number of segments.

The second line of each test case contains $n$ integers $l_1, l_2, \ldots, l_n$ ($1 \le l_i \le 2n$) — the left endpoints of the segments.

The third line of each test case contains $n$ integers $r_1, r_2, \ldots, r_n$ ($l_i \leq r_i \le 2n$) — the right endpoints of the segments.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer: the maximum size of a complex subset of the given segments.

Examples:
input:
3
3
1 2 3
5 4 6
5
1 2 3 6 8
5 4 7 9 10
5
3 1 4 1 5
7 2 6 5 10
output:
3
4
4

Note: In the first test case, all pairs of segments intersect, therefore it is optimal to form a single group containing all of the three segments.

In the second test case, there is no valid partition for all of the five segments. A valid partition with four segments is the following: $\\{\\{ [1, 5], [2, 4] \\}, \\{ [6, 9], [8, 10] \\}\\}$.

In the third test case, it is optimal to make a single group containing all the segments except the second.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
using ll=long long;
using pi=pair<int,int>;
#define fi first
#define se second
const int N=3e5+5,B=800;
int n,ans,fa[N*2],F[N];
pi a[N],b[N*2];
int gf(int x){return x==fa[x]?x:fa[x]=gf(fa[x]);}
int calc(int k) {
	if(~F[k])return F[k];
	int as=0;iota(fa+1,fa+n*2+1,1);
	for(int i=1,p=0,j=0,las=0;i<=n;i++)if(a[i].fi>las) {
		int l=a[i].fi,r=a[i].se,t;
		for(int j=p+1;j<r;j++)fa[j]=p;
		j++,p=r,t=gf(l);
		if(t!=las)fa[t]=gf(t-1),j--;
		if(j==k)las=r,j=0,as++;
	}return F[k]=as;
}
void solve() {
	cin>>n,ans=0;
	for(int i=1;i<=n;i++)cin>>b[i].fi,b[i].se=i,F[i]=-1;
	for(int i=1;i<=n;i++)cin>>b[i+n].fi,b[i+n].se=i+n;
	sort(b+1,b+n*2+1);
	for(int i=1;i<=2*n;i++)if(b[i].se>n)a[b[i].se-n].se=i;else a[b[i].se].fi=i;
	sort(a+1,a+n+1,[&](const pi&x,const pi&y){return x.se<y.se;});
	for(int i=1;i<=min(n,B);i++)ans=max(ans,i*calc(i));
	for(int i=1;i<=n/B;i++) {
		int l=ans/i+1,r=n/i,mid,rs=0;
		while(l<=r)mid=l+r>>1,calc(mid)>=i?l=mid+1,rs=mid:r=mid-1;
		ans=max(ans,i*rs);
	}cout<<ans<<'\n';
}
int main() {
    #ifndef ONLINE_JUDGE
        freopen(".in","r",stdin);
        freopen(".out","w",stdout);
    #endif
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	int tt;cin>>tt;
	while(tt--)solve();
	return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
6
1 2 3 4 5 6
6 5 4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Hard Version)
time_limit_ms: 13000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the hard version of the problem. In this version, the constraints on $n$ and the time limit are higher. You can make hacks only if both versions of the problem are solved.

A set of (closed) segments is complex if it can be partitioned into some subsets such that

  * all the subsets have the same size; and    * a pair of segments intersects if and only if the two segments are in the same subset.

You are given $n$ segments $[l_1, r_1], [l_2, r_2], \ldots, [l_n, r_n]$. Find the maximum size of a complex subset of these segments.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^3$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the number of segments.

The second line of each test case contains $n$ integers $l_1, l_2, \ldots, l_n$ ($1 \le l_i \le 2n$) — the left endpoints of the segments.

The third line of each test case contains $n$ integers $r_1, r_2, \ldots, r_n$ ($l_i \leq r_i \le 2n$) — the right endpoints of the segments.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer: the maximum size of a complex subset of the given segments.

Examples:
input:
3
3
1 2 3
5 4 6
5
1 2 3 6 8
5 4 7 9 10
5
3 1 4 1 5
7 2 6 5 10
output:
3
4
4

Note: In the first test case, all pairs of segments intersect, therefore it is optimal to form a single group containing all of the three segments.

In the second test case, there is no valid partition for all of the five segments. A valid partition with four segments is the following: $\\{\\{ [1, 5], [2, 4] \\}, \\{ [6, 9], [8, 10] \\}\\}$.

In the third test case, it is optimal to make a single group containing all the segments except the second.

Buggy Code:
#include <cstdio>
#include <map>
#include <iostream>
#include <algorithm>
#include <bitset>
#include <queue>
#include <stack>
#include <vector>
#include <random>
#include <cstring>
#include <ctime>
#include <cmath>
#include <assert.h>
#include <unordered_map>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#define LL long long
#define pp pair<int,int>
#define mp make_pair
#define ull unsigned long long
namespace IO{
	const int sz=1<<22;
	char a[sz+5],b[sz+5],*p1=a,*p2=a,*t=b,p[105];
	inline char gc(){
	//	return p1==p2?(p2=(p1=a)+fread(a,1,sz,stdin),p1==p2?EOF:*p1++):*p1++;
		return getchar();
	}
	template<class T> void gi(T& x){
		x=0; int f=1;char c=gc();
		if(c=='-')f=-1;
		for(;c<'0'||c>'9';c=gc())if(c=='-')f=-1;
		for(;c>='0'&&c<='9';c=gc())
			x=x*10+(c-'0');
		x=x*f;
	}
	inline void flush(){fwrite(b,1,t-b,stdout),t=b; }
	inline void pc(char x){*t++=x; if(t-b==sz) flush(); }
	template<class T> void pi(T x,char c='\n'){
		if(x<0)pc('-'),x=-x;
		if(x==0) pc('0'); int t=0;
		for(;x;x/=10) p[++t]=x%10+'0';
		for(;t;--t) pc(p[t]); pc(c);
	}
	struct F{~F(){flush();}}f;
}
using IO::gi;
using IO::pi;
using IO::pc;
const int mod=998244353;
inline int add(int x,int y){
	return x+y>=mod?x+y-mod:x+y;
}
inline int dec(int x,int y){
	return x-y<0?x-y+mod:x-y;
}
inline int mul(int x,int y){
	return 1ll*x*y%mod;
}
inline int qkpow(int a,int b){
	if(b<0)return 0;
	int ans=1,base=a%mod;
	while(b){
		if(b&1)ans=1ll*ans*base%mod;
		base=1ll*base*base%mod;
		b>>=1;
	}
	return ans;
}
int fac[1000005],inv[1000005],Invn[600005];
inline int binom(int n,int m){
	if(n<m||m<0)return 0;
	return 1ll*fac[n]*inv[m]%mod*inv[n-m]%mod;
}
void init_C(int n){
	fac[0]=1;
	for(int i=1;i<=n;i++)fac[i]=1ll*fac[i-1]*i%mod;
	inv[0]=1;
	inv[n]=qkpow(fac[n],mod-2);
	for(int i=n-1;i>=1;i--)inv[i]=1ll*inv[i+1]*(i+1)%mod;
	Invn[0]=Invn[1]=1;
	for(int i=1;i<=200000;i++)Invn[i]=(LL)(mod-mod/i)*Invn[mod%i]%mod;
}
int t,n,l[300005],r[300005],po[600005],cnt,f[300005],res,fa[600005],tot;
vector<int>G1[600005],G2[600005];
inline int findSet(int u){
	return fa[u]==u?u:fa[u]=findSet(fa[u]);
}
inline void merge(int u,int v){
	u=findSet(u),v=findSet(v);
	if(u==v)return ;
	if(u>v)swap(u,v);
	fa[v]=u;
	tot++;
}
inline int calc(int m){
	for(int i=0;i<=2*n;i++)fa[i]=i;
	int lst=0,pos=0,ans=0;
	tot=0;
	for(int i=1;i<=2*n;i++){
		if(po[i]!=-1){
			if(po[i]<=lst)continue;
			for(int j=pos+1;j<i;j++)merge(pos,j);
			pos=i;
			int pre=findSet(po[i]-1);
			if(pre)merge(pre-1,pre);
			if(i-tot==m){
				ans++;
				lst=i;
				while(1){
					int pre=findSet(i);
					if(pre)merge(pre-1,pre);
					else break;
				}
			}
		}
	}
	res=max(res,ans*m);
	return ans;
}
inline void solve2(int l,int r){
	if(f[l]==f[r])return ;
	if(r-l<=1)return ;
	if(res>=(r-1)*f[l])return ;
	int mid=(l+r)>>1;
	f[mid]=calc(mid);
	solve2(l,mid);
	solve2(mid,r);
}
inline void solve(){
	gi(n);
	for(int i=1;i<=n;i++)gi(l[i]);
	for(int i=1;i<=n;i++)gi(r[i]);
	for(int i=1;i<=2*n;i++)G1[i].clear(),G2[i].clear();
	for(int i=1;i<=n;i++){
		G1[l[i]].push_back(i);
		G2[r[i]].push_back(i);
	}
	cnt=0;
	for(int i=1;i<=2*n;i++){
		for(auto x:G1[i])l[x]=++cnt;
		for(auto x:G2[i])r[x]=++cnt;
		po[i]=-1;
	}
	for(int i=1;i<=n;i++)po[r[i]]=l[i];
	res=0;
	f[1]=calc(1),f[n]=calc(n);
	solve2(1,n);
	pi(res);
}
signed main(){
	srand(time(0));
	gi(t);
	while(t--)solve();
	return 0;
}
/*
*/

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. Here are the steps to think through this:

1. **Understand the Bug**: The provided code seems to have issues with how it handles the merging and calculating of segments. Specifically, the `calc` function might not correctly handle the merging of segments, especially when dealing with overlapping segments or segments that are close to each other.

2. **Edge Cases**:
   - **Single Segment**: A single segment should be considered a complex set of size 1.
   - **All Segments Overlapping**: All segments overlap with each other, forming a single complex set.
   - **Disjoint Segments**: Segments that do not overlap at all should form separate complex sets.
   - **Mixed Overlapping and Non-overlapping Segments**: A mix of segments that overlap and do not overlap, which can expose issues in the merging logic.

3. **Challenging Scenarios**:
   - **Segments with the Same Endpoints**: This can cause issues if the merging logic does not handle it correctly.
   - **Segments with Small Gaps**: Segments that have very small gaps between them can cause issues in the merging logic.

4. **Generate the Test Input**:
   - We will create a test case with a small number of segments that include a mix of overlapping and non-overlapping segments, and segments with the same endpoints.

Here is the test input:

```plaintext
1
6
1 2 3 4 5 6
2 3 4 5 6 7
```

**Explanation**:
- This test case includes 6 segments: `[1, 2]`, `[2, 3]`, `[3, 4]`, `[4, 5]`, `[5, 6]`, and `[6, 7]`.
- Each segment overlaps with the next segment, forming a chain of overlapping segments.
- This scenario is challenging because it requires the merging logic to correctly handle the overlapping segments and form a single complex set.

This test input should help expose any issues in the merging and calculation logic of the provided buggy code.