**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Easy Version)
time_limit_ms: 6000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the easy version of the problem. In this version, the constraints on $n$ and the time limit are lower. 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 $n$ ($1 \le n \le 2 \cdot 10^4$) — 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 $2 \cdot 10^4$.

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:
def max_complex_subset_size(n, segments):
    # Sort the segments based on their right endpoints
    segments.sort(key=lambda x: x[1])
    g = 0  # Number of non-overlapping intervals
    last_end = -float('inf')
    for l, r in segments:
        if l > last_end:
            g += 1
            last_end = r
    k = n // g  # Group size
    m = g * k   # Maximum size of complex subset
    return m

t = int(input())
for _ in range(t):
    n = int(input())
    l = list(map(int, input().split()))
    r = list(map(int, input().split()))
    segments = list(zip(l, r))
    print(max_complex_subset_size(n, segments))

**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
4
1 2 3 4
4 3 2 1
3
1 3 5
2 4 6
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 (Easy Version)
time_limit_ms: 6000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the easy version of the problem. In this version, the constraints on $n$ and the time limit are lower. 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 $n$ ($1 \le n \le 2 \cdot 10^4$) — 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 $2 \cdot 10^4$.

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
from collections import defaultdict

def main():
    import sys
    input = sys.stdin.read
    data = input().split()
    ptr = 0
    t = int(data[ptr])
    ptr += 1
    for _ in range(t):
        n = int(data[ptr])
        ptr += 1
        l = list(map(int, data[ptr:ptr+n]))
        ptr += n
        r = list(map(int, data[ptr:ptr+n]))
        ptr += n
        # Adjust indices to be 1-based for simplicity
        for i in range(n):
            l[i] -= 1
            r[i] -= 1
        # Find frequency of each point
        freq = [0] * (2*n + 1)
        for i in range(n):
            for p in range(l[i], r[i]+1):
                freq[p] += 1
        # Find maximal cliques by frequency
        max_cliques = {}
        for p in range(2*n + 1):
            if freq[p] > 0:
                # Get the segments that contain p
                clique = [i for i in range(n) if l[i] <= p <= r[i]]
                max_cliques[p] = clique
        # Now, for each possible r, find the maximum number of non-overlapping r-cliques
        max_segments = 0
        r_counts = defaultdict(list)
        for p, clique in max_cliques.items():
            r = len(clique)
            r_counts[r].append((p, clique))
        for r in r_counts:
            cliques = r_counts[r]
            # Sort by the highest segment index used
            cliques.sort(key=lambda x: max(x[1]))
            selected = 0
            current_highest = -1
            for p, clique in cliques:
                if min(clique) > current_highest:
                    selected += 1
                    current_highest = max(clique)
            total_segments = selected * r
            if total_segments > max_segments:
                max_segments = total_segments
        print(max_segments)

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:
```plaintext
1
4
1 2 3 4
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 (Easy Version)
time_limit_ms: 6000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the easy version of the problem. In this version, the constraints on $n$ and the time limit are lower. 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 $n$ ($1 \le n \le 2 \cdot 10^4$) — 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 $2 \cdot 10^4$.

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:
def main():
    import sys
    input = sys.stdin.read
    data = input().split()
    idx = 0
    T = int(data[idx])
    idx += 1
    results = []
    for _ in range(T):
        n = int(data[idx])
        idx += 1
        L = list(map(int, data[idx:idx+n]))
        idx += n
        R = list(map(int, data[idx:idx+n]))
        idx += n
        # Sort segments by right endpoints
        segments = sorted(zip(L, R), key=lambda x: x[1])
        # Find maximum clique size
        max_c = 0
        active = []
        for seg in segments:
            while active and active[0][1] < seg[0]:
                del active[0]
            active.append(seg)
            max_c = max(max_c, len(active))
        # Find maximum matching
        matching = 0
        last_right = -1
        for seg in segments:
            if seg[0] > last_right:
                matching += 1
                last_right = seg[1]
        # Calculate m
        m1 = max_c
        m2 = 2 * matching
        m = min(n, max(m1, m2))
        results.append(str(m))
    print('\n'.join(results))

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 where the algorithm's assumptions might fail. Specifically, the buggy code seems to rely on sorting segments by their right endpoints and then finding a maximum matching, which may not always correctly identify the maximum size of a complex subset.

### Step-by-Step Analysis

1. **Understanding the Bug**:
   - The code sorts segments by their right endpoints.
   - It then attempts to find the maximum clique size and a maximum matching.
   - The issue is that sorting by right endpoints and then finding a maximum matching may not always correctly identify the maximum size of a complex subset, especially when segments have overlapping or nested structures.

2. **Edge Cases**:
   - **Overlapping Segments**: Segments that overlap in a way that they can form a complex subset but are not identified by the sorting and matching approach.
   - **Nested Segments**: Segments where one segment is entirely contained within another, which can lead to incorrect matching.

3. **Test Input Design**:
   - We need to create a set of segments that form a complex subset but are not correctly handled by the sorting and matching approach.
   - Consider a set of segments where each segment overlaps with others in a specific pattern that forms a complex subset but is not easily identified by the sorting method.

### Challenging Test Input

Let's create a test case with 5 segments where the segments are designed to form a complex subset but are not correctly handled by the buggy code.

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

### Explanation

- **Segments**:
  - `[1, 10]`
  - `[3, 6]`
  - `[5, 8]`
  - `[7, 10]`
  - `[9, 10]`

- **Complex Subset**:
  - The segments can be partitioned into two subsets:
    - Subset 1: `[1, 10]`, `[7, 10]`, `[9, 10]`
    - Subset 2: `[3, 6]`, `[5, 8]`
  - Each subset has the same size (3 and 2, respectively).
  - All pairs of segments in the same subset intersect, and no pairs of segments in different subsets intersect.

### Expected Output

The expected output for this test case is `3`, as the maximum size of a complex subset is 3.

### Why This Input is Challenging

- **Sorting by Right Endpoints**:
  - Sorting by right endpoints will not necessarily group the segments correctly to identify the complex subset.
  - The sorting will place `[1, 10]` first, followed by `[7, 10]` and `[9, 10]`, which may not correctly identify the maximum clique size or matching.

- **Maximum Clique and Matching**:
  - The maximum clique size and matching will not correctly identify the complex subset due to the nested and overlapping nature of the segments.

This test input should expose the bug in the buggy code by challenging its assumptions and sorting logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Easy Version)
time_limit_ms: 6000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the easy version of the problem. In this version, the constraints on $n$ and the time limit are lower. 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 $n$ ($1 \le n \le 2 \cdot 10^4$) — 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 $2 \cdot 10^4$.

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<cmath>
#include<cstring>
#define N 3000111
using namespace std;
int total1=0,n,T;
int father[N],left[N],getans[N],add[N],a[N],b[N],f[N],Next[N],head[N],edge[N];
bool bz[N];
void insert(int x,int y)
{
    total1++;
    Next[total1]=head[x];
    head[x]=total1;
    edge[total1]=y;
}
int max(int x,int y)
{
    if (x<y) return y;
    return x;
}
int getfather(int x)
{
    if (father[x]==x) return x;
    father[x]=getfather(father[x]);
    return father[x];
}
int find(int x)
{
    for (int i=1;i<=n*2;i++) father[i]=i,left[i]=i,add[i]=0;
    if (getans[x]!=0) return getans[x];
    int curl=1,cnt=0;
    int curr=1;
    int first=-1;
    for (int i=1;i<=n*2;i++)
    {
        if (i==5)
        {
            n++;n--;
        }
        for (int j=head[i];j;j=Next[j])
        {
            int y=edge[j];
            if (y<curl) continue;
            add[i+1]--;
            int p=getfather(y);
            add[p]++;
            while (add[p]>0)
            {
                if (left[p]<=curl) break;
                int q=getfather(left[p]-1);
                left[p]=left[q];
                add[p]+=add[q];
                add[q]=0;
                father[q]=p;
                if (q==first) first=p;
            }
            if (first==-1&&add[p]>0) first=p;
            if (p<first)
                add[first]+=add[p];
        }
        if (left[i]==i&&i!=curl&&add[i]>=0)
        {
            left[i]=left[i-1];
            father[i-1]=i;
            add[i]+=add[i-1];
            if (i-1==first) first=i;
        }
        if (first!=-1&&add[first]>=x)
            curl=i+1,cnt++,first=-1,add[i+1]=0;
    }
    getans[x]=cnt;
    return cnt;
}
int main()
{
    memset(bz,false,sizeof(bz));
    scanf("%d",&T);
    while (T)
    {
        T--;
        scanf("%d",&n);
        for (int i=1;i<=n;i++) getans[i]=0;
        for (int i=1;i<=n;i++) scanf("%d",&a[i]);
        for (int i=1;i<=n;i++) scanf("%d",&b[i]);
        total1=0;
        for (int i=1;i<=2*n;i++) head[i]=0;
        for (int i=1;i<=n;i++) insert(b[i],a[i]);
        int s=sqrt(n);
        if (s>n) s=n;
        int ans=find(3)*3;
        for (int i=1;i<=s;i++)
        {
            ans=max(ans,find(i)*i);
        }
        int nex=1;
        for (int i=1;i<=n/s;i++)
        {
            int l=ans/i+1,r=n/i;
            int pans=0;
            while (l<=r)
            {
                int mid=(l+r)/2;
                int t=find(mid);
                ans=max(ans,t*mid);
                if (t<i)
                    r=mid-1;
                else    l=mid+1;
            }

        }
        printf("%d\n",ans);
    }
}

**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 can maximize the failure probability. The code is complex and involves several steps, including segment intersection checks and union-find operations. Here are some potential scenarios to consider:

1. **All segments intersecting**: This can cause issues in the union-find and partitioning logic.
2. **Non-intersecting segments**: This can test the logic for partitioning into subsets.
3. **Mixed intersecting and non-intersecting segments**: This can test the robustness of the partitioning logic.
4. **Segments with identical endpoints**: This can cause edge cases in the union-find and partitioning logic.
5. **Segments with large gaps**: This can test the efficiency and correctness of the intersection checks.

Given the constraints and the need for a small-scale test input, let's create a test case that includes a mix of intersecting and non-intersecting segments, with some identical endpoints and large gaps.

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

Explanation of the test cases:
1. **4 segments**:
   - [1, 5], [2, 4], [3, 7], [4, 6]
   - The segments [1, 5] and [2, 4] intersect, [3, 7] and [4, 6] intersect, but [1, 5] and [3, 7] do not intersect, and [2, 4] and [4, 6] do not intersect.
   - This tests the logic for partitioning into subsets with different sizes.
2. **3 segments**:
   - [1, 5], [2, 4], [3, 6]
   - All segments intersect with each other.
   - This tests the logic for a single complex subset.
3. **5 segments**:
   - [1, 5], [2, 4], [3, 7], [6, 9], [8, 10]
   - The segments [1, 5] and [2, 4] intersect, [3, 7] does not intersect with any other segment, and [6, 9] and [8, 10] intersect.
   - This tests the logic for partitioning into multiple subsets with different sizes and non-intersecting segments.

This test input should help expose any issues in the buggy code by covering a variety of scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Easy Version)
time_limit_ms: 6000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the easy version of the problem. In this version, the constraints on $n$ and the time limit are lower. 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 $n$ ($1 \le n \le 2 \cdot 10^4$) — 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 $2 \cdot 10^4$.

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:
/**
 * author: ClHg2
 * created: 2024-09-27 22:55:55
 */

#include <bits/stdc++.h>

using namespace std;

#define rep(i, l, r) for (int i = (l); i <= (r); ++i)
#define per(i, l, r) for (int i = (r); i >= (l); --i)
#define sz(x) int((x).size())

using ll = long long;
using db = long double;

template <typename T>
inline bool ckmin(T& x, const T& y) {
  return y < x && (x = y, true);
}

template <typename T>
inline bool ckmax(T& x, const T& y) {
  return x < y && (x = y, true);
}

constexpr int N = 2e4 + 5;

int n;
array<int, N> minR, maxR, f;
array<int, 2 * N> buc;

struct S {
  int l, r;
  friend bool operator<(S x, S y) { return x.l < y.l; }
};

array<S, N> a;

void proc() {
  cin >> n;
  rep (i, 1, n) cin >> a[i].l;
  rep (i, 1, n) cin >> a[i].r;
  sort(&a[1], &a[n] + 1);
  fill_n(&minR[1], n, 2 * n);
  fill_n(&maxR[1], n, 0);
  int ans = 0;
  rep (d, 1, n) {
    rep (i, 1, n - d + 1)
      ckmin(minR[i], a[i + d - 1].r), ckmax(maxR[i], a[i + d - 1].r);
    fill_n(&buc[1], 2 * n, 0);
    for (int i = d, j = 1, val = 0; i <= n; ++i) {
      if (a[i].l > minR[i - d + 1]) {
        f[i] = 0;
        continue;
      }
      while (j < a[i - d + 1].l) ckmax(val, buc[j++]);
      f[i] = val + 1;
      ckmax(ans, f[i] * d);
      ckmax(maxR[i - d + 1] < j ? val : buc[maxR[i - d + 1]], f[i]);
    }
  }
  cout << ans << "\n";
}

int main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  cin.exceptions(cin.failbit);

  int t;
  cin >> t;
  rep (i, 1, t) proc();
}

**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 given buggy code, we need to focus on edge cases and scenarios that are likely to cause the algorithm to fail. The key points to consider are:

1. **Intersecting Segments**: Ensure that some segments intersect in a way that should be detected but might not be handled correctly by the buggy code.
2. **Non-Intersecting Segments**: Create scenarios where segments do not intersect, which should be handled correctly.
3. **Edge Cases**: Consider the boundaries of the input constraints, such as the smallest and largest possible values for \( n \) and the segment endpoints.

Given the constraints and the nature of the problem, a small-scale test input that is complete and valid can be constructed as follows:

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

### Explanation:
1. **First Test Case**:
   - Segments: \([1, 5]\), \([2, 4]\), \([3, 6]\)
   - All segments intersect, so the maximum size of a complex subset is 3.

2. **Second Test Case**:
   - Segments: \([1, 5]\), \([2, 4]\), \([3, 7]\), \([6, 9]\), \([8, 10]\)
   - The segments can be partitioned into two subsets: \(\{[1, 5], [2, 4]\}\) and \(\{[6, 9], [8, 10]\}\), with the segment \([3, 7]\) not fitting into any subset. The maximum size of a complex subset is 4.

3. **Third Test Case**:
   - Segments: \([1, 2]\), \([2, 3]\), \([3, 4]\), \([4, 5]\)
   - Each segment intersects with the next one, but no segment intersects with more than one other segment. The maximum size of a complex subset is 2 (e.g., \(\{[1, 2], [3, 4]\}\) and \(\{[2, 3], [4, 5]\}\)).

This test input is designed to challenge the buggy code by including both intersecting and non-intersecting segments, as well as edge cases where the segments are tightly packed. The expected outputs for the test cases are provided in the comments to ensure the correctness of the solution.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Easy Version)
time_limit_ms: 6000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the easy version of the problem. In this version, the constraints on $n$ and the time limit are lower. 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 $n$ ($1 \le n \le 2 \cdot 10^4$) — 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 $2 \cdot 10^4$.

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;
struct tour {
	int sum, p;
};
int l [20010]; int r [20010]; int n;
vector < pair <int, int> > in;
const int OFF=(1<<20); tour t [4*OFF+20];
int gr1=500; int gr2=20000/gr1+10;
void prop (int x, int lo, int hi){
	if (x>=OFF || t [x].p==0) return ;
	int le=(hi-lo+1)/2;
	t [x*2].sum+=t [x].p; t [x*2+1].sum+=t [x].p;
	t [x*2].p+=t [x].p; t [x*2+1].p+=t [x].p;
	t [x].p=0;
	return ;
}
void add (int le, int ri, int val, int x=1, int lo=0, int hi=OFF-1){
	if (le>hi || lo>ri) return ;
	if (lo>=le && hi<=ri){
		t [x].sum+=val;
		t [x].p+=val;
		return ;
	}
	int mid=(lo+hi)/2;
	prop (x, lo, hi);
	add (le, ri, val, x*2, lo, mid);
	add (le, ri, val, x*2+1, mid+1, hi);
	t [x].sum=max (t [x*2].sum, t [x*2+1].sum);
	return ;
}
int q (int le, int ri, int x=1, int lo=0, int hi=OFF-1){
	if (le>hi || lo>ri) return 0;
	if (lo>=le && hi<=ri) return t [x].sum;
	int mid=(lo+hi)/2;
	prop (x, lo, hi);
	return max (q (le, ri, x*2, lo, mid), q (le, ri, x*2+1, mid+1, hi));
}
int maxi (int sz){
	int zad=0; int cou=0;
	for (int i=1; i<n+1; i++){
		if (in [i].second<=in [zad].first) continue;
		add (in [i].second, in [i].first+1, 1);
		if (q (1, 2*n)>=sz){
			cou++;
			for (int j=zad+1; j<=i; j++) if (!(in [j].second<=in [zad].first)) add (in [j].second, in [j].first+1, -1);
			zad=i;
		}
	}
	for (int i=zad+1; i<n+1; i++) if (!(in [i].second<=in [zad].first)) add (in [i].second, in [i].first+1, -1);
	return cou;
}
int main(){
	int t; cin >> t;
	while (t--){
		cin >> n; in.clear ();
		for (int i=0; i<n; i++) cin >> l [i];
		for (int j=0; j<n; j++) cin >> r [j];
		for (int i=0; i<n; i++) in.push_back ({r [i], l [i]});
		in.push_back ({-1, -1});
		sort (in.begin(), in.end());
		/*for (int i=1; i<n+1; i++){
			cout << maxi (i) << endl;
		}*/
		int ans=0;
		for (int i=1; i<min (n+1, gr1); i++) ans=max (ans, i*maxi (i));
		for (int i=1; i<min (n+1, gr2); i++){
			int lo=0; int hi=n;
			while (lo!=hi){
				int mid=(lo+hi+1)/2;
				if (maxi (mid)>=i) lo=mid;
				else hi=mid-1;
			}
			ans=max (ans, lo*i);
		}
		cout << ans << endl;
	}
	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
3
1 2 3
3 2 1
5
1 3 5 7 9
9 7 5 3 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Easy Version)
time_limit_ms: 6000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the easy version of the problem. In this version, the constraints on $n$ and the time limit are lower. 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 $n$ ($1 \le n \le 2 \cdot 10^4$) — 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 $2 \cdot 10^4$.

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 fi first
#define se second
using namespace std;
typedef pair<int,int> pii;

const int N=4e5+5;
int n,a[N];
pii p[N];

struct seg_tree
{
    int mx[N*4],tag[N*4];

    void init(int n) {for(int i=1;i<=n*8;i++) mx[i]=tag[i]=0;}

    void pushdown(int x)
    {
        int T=tag[x]; tag[x]=0;
        mx[x<<1]+=T,tag[x<<1]+=T;
        mx[x<<1|1]+=T,tag[x<<1|1]+=T;
    }

    void modify(int x,int l,int r,int L,int R,int val)
    {
        if(L<=l&&r<=R)
        {
            mx[x]+=val,tag[x]+=val;
            return;
        }
        int mid=(l+r)>>1;
        pushdown(x);
        if(L<=mid) modify(x<<1,l,mid,L,R,val);
        if(R>mid) modify(x<<1|1,mid+1,r,L,R,val);
        mx[x]=max(mx[x<<1],mx[x<<1|1]);
    }
    void modify(int L,int R,int val) {modify(1,1,n*2,L,R,val);}
}TR;

bool cmp(pii a,pii b) {return a.se<b.se;}

void input()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&p[i].fi);
    for(int i=1;i<=n;i++) scanf("%d",&p[i].se);
    sort(p+1,p+n+1,cmp);
}

int docount(int m)
{
    // cerr<<'@'<<m<<endl;
    int res=0,L=0;
    TR.init(n);
    for(int i=1;i<=n;i++) if(p[i].fi>L)
    {
        auto[l,r]=p[i];
        // cerr<<' '<<l<<' '<<r<<endl;
        TR.modify(l,r,1);
        if(TR.mx[1]>=m) TR.modify(l,r,-1),L=r,res++;
    }
    return res;
}

int findsize(int cnt)
{
    int L=1,R=n/cnt,res=0;
    while(L<=R)
    {
        int mid=(L+R)>>1;
        if(docount(mid)>=cnt) res=mid,L=mid+1;
        else R=mid-1;
    }
    return res;
}

void solve()
{
    input();
    int B=sqrt(n*log(n)),res=0;
    //cerr<<B<<endl;
    for(int m=1;m*B<=n;m++) res=max(res,m*docount(m));
    for(int cnt=1;cnt<=B;cnt++) res=max(res,cnt*findsize(cnt));
    printf("%d\n",res);
}

int main()
{
    int T=1;
    scanf("%d",&T);
    while(T--) 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 focus on edge cases and scenarios that are likely to cause issues. Here are some key points to consider:

1. **Overlapping Segments**: Segments that overlap in complex ways can cause issues in the algorithm's logic for partitioning and counting.
2. **Edge Cases**: Segments that have the same endpoints, or segments that are very short, can cause edge cases.
3. **Large and Small Values**: Test cases with both large and small values for the endpoints can help identify issues with the algorithm's handling of different scales.
4. **Special Patterns**: Segments that form specific patterns (e.g., all segments intersecting, no segments intersecting, or a mix) can help expose bugs.

Given the constraints, let's create a small-scale test input that includes a mix of these elements:

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

### Explanation:
1. **Test Case 1**:
   - Segments: `[1, 5], [2, 5], [3, 5], [4, 5], [5, 5]`
   - All segments intersect with each other, so the maximum size of a complex subset is 5.

2. **Test Case 2**:
   - Segments: `[1, 2], [1, 2], [1, 2], [1, 2]`
   - All segments intersect with each other, so the maximum size of a complex subset is 4.

3. **Test Case 3**:
   - Segments: `[1, 6], [2, 5], [3, 4], [4, 3], [5, 2], [6, 1]`
   - These segments form a pattern where each segment intersects with its mirror segment, but not with others. The maximum size of a complex subset is 3 (e.g., `{[1, 6], [2, 5], [3, 4]}`).

This test input is designed to challenge the algorithm by including various edge cases and complex segment interactions, which should help in exposing 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 (Easy Version)
time_limit_ms: 6000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the easy version of the problem. In this version, the constraints on $n$ and the time limit are lower. 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 $n$ ($1 \le n \le 2 \cdot 10^4$) — 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 $2 \cdot 10^4$.

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.read().split()
    t = int(input[0])
    index =1
    for _ in range(t):
        n = int(input[index])
        index +=1
        l = []
        r = []
        for _ in range(n):
            l.append(int(input[index]))
            index +=1
        for _ in range(n):
            r.append(int(input[index]))
            index +=1
        segments = list(zip(l, r))
        # Compute D
        events = []
        for a, b in segments:
            if a > b:
                # Invalid segment, skip or adjust
                pass
            events.append((a, 1))
            events.append((b, -1))
        events.sort()
        D =0
        current =0
        for x, delta in events:
            current += delta
            D = max(D, current)
        max_n =0
        for s in range(D,0,-1):
            if s ==1:
                if D ==1:
                    k = n
                    n_prime =1 *k
                    if n_prime > max_n:
                        max_n =n_prime
                continue
            n_prime = get_n(s, sorted(segments, key=lambda x: x[1]))
            if n_prime > max_n:
                max_n = n_prime
        print(max_n)

def get_n(s, segments):
    subsets = []
    k=0
    for start, end in segments:
        for subset in subsets.copy():
            if subset['size'] <s and subset['last_end'] >= start:
                subset['size'] +=1
                if subset['size'] ==s:
                    subsets.remove(subset)
                    k +=1
                break
        else:
            new_subset = {'size':1, 'last_end':end}
            if new_subset['size'] ==s:
                k +=1
            else:
                subsets.append(new_subset)
    return s *k

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 focus on edge cases that might cause issues in the logic. The code has several potential points of failure, including how it handles invalid segments, the sorting of events, and the logic for forming subsets.

### Key Points to Consider:
1. **Invalid Segments**: The code skips invalid segments (where `a > b`), but it doesn't handle them properly, which could lead to incorrect results.
2. **Event Handling**: The code uses events to compute the maximum number of overlapping segments (`D`). If the segments are not properly handled, this could lead to incorrect `D` values.
3. **Subsets Formation**: The logic for forming subsets and calculating `n_prime` might not correctly handle all cases, especially when segments overlap in complex ways.

### Test Case to Expose the Bug:
Let's create a test case with a small number of segments that have complex overlaps and include invalid segments to see how the code handles them.

### Test Input:
```plaintext
3
4
1 2 3 4
2 1 4 3
3
1 1 1
2 2 2
5
1 2 3 4 5
5 4 3 2 1
```

### Explanation:
1. **First Test Case**:
   - Segments: `[1, 2]`, `[2, 1]`, `[3, 4]`, `[4, 3]`
   - The segment `[2, 1]` is invalid and should be skipped. The valid segments are `[1, 2]`, `[3, 4]`, and `[4, 3]`.
   - The maximum size of a complex subset is 2 (either `[1, 2]` and `[3, 4]` or `[1, 2]` and `[4, 3]`).

2. **Second Test Case**:
   - Segments: `[1, 1]`, `[1, 1]`, `[1, 1]`
   - All segments are the same and intersect with each other. The maximum size of a complex subset is 3.

3. **Third Test Case**:
   - Segments: `[1, 5]`, `[2, 4]`, `[3, 3]`, `[4, 2]`, `[5, 1]`
   - The segments `[4, 2]` and `[5, 1]` are invalid and should be skipped. The valid segments are `[1, 5]`, `[2, 4]`, and `[3, 3]`.
   - The maximum size of a complex subset is 3 (either `[1, 5]`, `[2, 4]`, and `[3, 3]`).

### Expected Output:
```plaintext
2
3
3
```

This test case should help expose issues in the code's handling of invalid segments and the formation of complex subsets.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Easy Version)
time_limit_ms: 6000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the easy version of the problem. In this version, the constraints on $n$ and the time limit are lower. 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 $n$ ($1 \le n \le 2 \cdot 10^4$) — 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 $2 \cdot 10^4$.

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>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define N 1000010
#define int long long
#define db long double
#define pii pair<int,int>
#define st first
#define ed second
#define mkp make_pair
#define pb push_back
#define eps 1e-9
#define mod 998244353
#define mod2 1000000007
#define bs 13131
#define bs2 131
#define INF 0x3f3f3f3f3f3f3f3f
#define il inline
#define vi vector<int>
#define ins insert
#define umap unordered_map
#define uset unordered_set
#define R(x) x.begin(),x.end()
#define B(x) x.begin()
#define E(x) x.end()
#define lb lower_bound
#define ub upper_bound
#define prq priority_queue
il int read(){
	int x=0,f=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){
		if(ch=='-')f=-1;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9'){
		x=(x<<1)+(x<<3)+(ch^48);
		ch=getchar();
	}
	return x*f;
}
int T=1,n,m,q,k;
int s[N];
char c[N];
vector<int> vl[N],vr[N];
int B;
il int solvem(int w){
	int cnt=0,R=0,ans=0;
	for(int i=1;i<=2*n;i++){
		i=max(i,R+1);
		cnt+=vl[i].size();
		int num=0;
		for(auto x:vr[i])if(x>R)num++;
		if(num&&cnt>=w){
			R=-1;
			int num=0;
			for(int j=i;j<=2*n;j++){
				for(auto x:vr[j])
					if(x<=i)num++;
				if(num>=w){
					R=j;
					break;
				}
			}
			if(R==-1)break;
			ans++;
			cnt=0;
			continue;
		}
		cnt-=num;
	}
	return ans;
}
il int solvek(int w){
	int l=1,r=n,ans=0;
	while(l<=r){
		int mid=(l+r)>>1;
		if(solvem(mid)>=w)ans=max(ans,mid),l=mid+1;
		else r=mid-1;
	}
	return ans;
}
int l[N],r[N];
int fl;
il void solve(){
	n=read();
	if(T==999&&n==7)fl=1;
	for(int i=1;i<=n*2;i++)
		vl[i].clear(),vr[i].clear();
	for(int i=1;i<=n;i++)
		l[i]=read();
	for(int i=1;i<=n;i++)
		r[i]=read();
	for(int i=1;i<=n;i++)
		vl[l[i]].pb(r[i]),vr[r[i]].pb(l[i]);
	B=sqrt(n);
	int ans=0;
	for(int i=1;i<=B;i++)ans=max(ans,i*solvem(i));
	for(int i=1;i<=B;i++)ans=max(ans,i*solvek(i));
//	for(int i=1;i<=n;i++)printf("%lld ",solvem(i));
//	puts("");
	if(!fl)printf("%lld\n",ans);
	if(fl&&T==538){
		puts("!!!!!");
		printf("%lld\n",n);
		for(int i=1;i<=n;i++)
			printf("%lld ",l[i]);puts("");
		for(int i=1;i<=n;i++)
			printf("%lld ",r[i]);puts("");
	}
}
signed main(){
	T=read();
	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 given buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are the steps to think through this:

1. **Understand the Problem**: The problem requires finding the maximum size of a complex subset of segments. A complex subset is one where all subsets have the same size, and pairs of segments intersect if and only if they are in the same subset.

2. **Identify Potential Bugs**: The buggy code might have issues with:
   - Handling large numbers of segments.
   - Correctly identifying and partitioning intersecting segments.
   - Efficiently managing the dynamic programming or binary search logic.

3. **Create a Challenging Test Case**:
   - **Edge Case 1**: Segments that are highly overlapping.
   - **Edge Case 2**: Segments that are disjoint.
   - **Edge Case 3**: Segments that form a complex pattern of intersections.

4. **Generate the Test Input**:
   - We will create a test case with a small number of segments to ensure it fits the constraints and is valid.
   - The segments will be designed to form a complex pattern of intersections to challenge the algorithm.

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

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

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

2. **Test Case 2**:
   - Segments: `[1, 5], [2, 4], [3, 3], [4, 2], [5, 1]`
   - This is a mirrored version of the first test case, where the segments are still highly overlapping but in a different order.

3. **Test Case 3**:
   - Segments: `[1, 2], [2, 3], [3, 4], [4, 5], [5, 6]`
   - These segments form a chain where each segment intersects with the next one, but not with segments further apart. This should form a complex subset of size 2.

This test input should help in identifying any issues in the buggy code, especially in handling complex patterns of intersections and efficiently partitioning 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 (Easy Version)
time_limit_ms: 6000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the easy version of the problem. In this version, the constraints on $n$ and the time limit are lower. 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 $n$ ($1 \le n \le 2 \cdot 10^4$) — 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 $2 \cdot 10^4$.

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>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define N 1000010
#define int long long
#define db long double
#define pii pair<int,int>
#define st first
#define ed second
#define mkp make_pair
#define pb push_back
#define eps 1e-9
#define mod 998244353
#define mod2 1000000007
#define bs 13131
#define bs2 131
#define INF 0x3f3f3f3f3f3f3f3f
#define il inline
#define vi vector<int>
#define ins insert
#define umap unordered_map
#define uset unordered_set
#define R(x) x.begin(),x.end()
#define B(x) x.begin()
#define E(x) x.end()
#define lb lower_bound
#define ub upper_bound
#define prq priority_queue
il int read(){
	int x=0,f=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){
		if(ch=='-')f=-1;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9'){
		x=(x<<1)+(x<<3)+(ch^48);
		ch=getchar();
	}
	return x*f;
}
int T=1,n,m,q,k;
int s[N];
char c[N];
vector<int> vl[N],vr[N];
int B;
il int solvem(int w){
	int cnt=0,R=0,ans=0;
	for(int i=1;i<=2*n;i++){
		i=max(i,R+1);
		cnt+=vl[i].size();
//		printf("%lld %lld\n",i,cnt);
		int num=0;
		for(auto x:vr[i])if(x>R)num++;
		if(num&&cnt>=w){
//			printf("!! %lld\n",i);
			int fl=0;
			int num=0;
			for(int j=i;j<=2*n;j++){
				for(auto x:vr[j]){
					if(x<=i&&x>R){
//						printf("add %lld %lld\n",x,j);
						num++;
					}
				}
				if(num>=w){
					R=j;
					fl=1;
					break;
				}
			}
			if(!fl)break;
//			printf("*%lld\n",R);
			ans++;
			cnt=0;
			continue;
		}
		cnt-=num;
	}
	return ans;
}
il int solvek(int w){
	int l=1,r=n,ans=0;
	while(l<=r){
		int mid=(l+r)>>1;
		if(solvem(mid)>=w)ans=max(ans,mid),l=mid+1;
		else r=mid-1;
	}
	return ans;
}
int l[N],r[N];
int fl;
il void solve(){
	n=read();
//	if(T==999&&n==7)fl=1;
	for(int i=1;i<=n*2;i++)
		vl[i].clear(),vr[i].clear();
	for(int i=1;i<=n;i++)
		l[i]=read();
	for(int i=1;i<=n;i++)
		r[i]=read();
	for(int i=1;i<=n;i++)
		vl[l[i]].pb(r[i]),vr[r[i]].pb(l[i]);
	B=sqrt(n);
	int ans=0;
	for(int i=1;i<=B;i++)ans=max(ans,i*solvem(i));
	for(int i=1;i<=B;i++)ans=max(ans,i*solvek(i));
//	for(int i=1;i<=n;i++)printf("%lld\n",solvem(i));
//	puts("");
	if(!fl)printf("%lld\n",ans);
	if(fl&&T==538){
		puts("!!!!!");
		printf("%lld\n",n);
		for(int i=1;i<=n;i++)
			printf("%lld ",l[i]);puts("");
		for(int i=1;i<=n;i++)
			printf("%lld ",r[i]);puts("");
	}
}
signed main(){
	T=read();
	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 given buggy code, we need to focus on edge cases or scenarios that can cause the algorithm to fail. Here are the steps to think through this:

1. **Understand the Problem and Constraints:**
   - The problem involves finding the maximum size of a complex subset of segments.
   - A complex subset is one where all subsets have the same size and a pair of segments intersects if and only if they are in the same subset.
   - The constraints are \(1 \le t \le 10^3\) and \(1 \le n \le 2 \cdot 10^4\).

2. **Identify Potential Bugs:**
   - The buggy code might have issues with handling overlapping segments, especially when segments are very close to each other or when they form complex patterns.
   - The code might also have issues with the partitioning logic, especially when the segments are not uniformly distributed.

3. **Create a Test Case:**
   - We need a test case that is small but complex enough to expose the bug.
   - Consider a scenario where segments are very close to each other and form a complex pattern that is difficult to partition correctly.

4. **Example Test Case:**
   - Let's create a test case with 5 segments where the segments are close to each other and form a complex pattern.

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

**Explanation:**
- This test case has 5 segments with the following endpoints:
  - Segment 1: [1, 5]
  - Segment 2: [2, 4]
  - Segment 3: [3, 6]
  - Segment 4: [4, 7]
  - Segment 5: [5, 8]

- The segments are close to each other and form a complex pattern:
  - Segments 1, 2, and 3 intersect with each other.
  - Segments 4 and 5 intersect with each other but not with the first three segments.

- This test case is designed to challenge the partitioning logic of the buggy code. The correct answer should be 3, as the maximum size of a complex subset is 3 (segments 1, 2, and 3).

This test case should help expose any issues with the partitioning logic in the 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 (Easy Version)
time_limit_ms: 6000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the easy version of the problem. In this version, the constraints on $n$ and the time limit are lower. 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 $n$ ($1 \le n \le 2 \cdot 10^4$) — 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 $2 \cdot 10^4$.

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 N = 300005;
int n;
int ct[N], pos[N];
pair<int, int> a[N];
namespace sgt {
int mx[N << 3], lazy[N << 3];

void clear() {
  int m = n * 8;
  memset(mx, 0, m << 2);
  memset(lazy, 0, m << 2);
}

void modify(int i, int l, int r, int x, int y, int z) {
  if (x <= l && y >= r) {
    mx[i] += z, lazy[i] += z;
    return;
  }

  int mid = (l + r) >> 1;
  if (x <= mid) {
    modify(i << 1, l, mid, x, y, z);
  }
  if (y > mid) {
    modify(i << 1 | 1, mid + 1, r, x, y, z);
  }
  mx[i] = max(mx[i << 1], mx[i << 1 | 1]) + lazy[i];
}
}  // namespace sgt

int main() {
  ios::sync_with_stdio(0), cin.tie(0);

  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i].second;
    }
    for (int i = 1; i <= n; i++) {
      cin >> a[i].first;
    }
    sort(a + 1, a + n + 1);

    int B = sqrt(n + 0.5) + 1;
    B = min(B, n);
    memset(ct, 0, (n + 1) << 2);
    for (int b = 1; b <= B; b++) {
      for (int i = 1, lt = 0; i <= n; i++) {
        if (a[i].second > lt) {
          sgt::modify(1, 1, n * 2, a[i].second, a[i].first, 1);
          if (sgt::mx[1] >= b) {
            sgt::clear();
            lt = a[i].first;
            ct[b]++;
          }
        }
      }
      sgt::clear();
    }

    memset(pos, 0, (n + 1) << 2);
    for (int gu = 1; gu <= B; gu++) {
      priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> q;
      for (int i = B + 1, r = pos[B + 1]; i <= n && pos[i] <= n; i++) {
        while ((!q.empty()) && q.top().first < a[pos[i]].first) {
          int id = q.top().second;
          q.pop();
          sgt::modify(1, 1, 2 * n, a[id].second, a[id].first, -1);
        }
        while (r + 1 <= n && sgt::mx[1] < i) {
          r++;
          if (a[r].second < a[pos[i]].first) {
            continue;
          }
          q.emplace(a[r].second, r);
          sgt::modify(1, 1, 2 * n, a[r].second, a[r].first, 1);
        }
        if (sgt::mx[1] < i) {
          fill(pos + i, pos + n + 1, n + 1);
          break;
        } else {
          pos[i] = r;
          ct[i] = gu;
        }
      }
      sgt::clear();
    }

    int ans = 0;
    for (int i = 1; i <= n; i++) {
      assert(n / i >= ct[i]);
      ans = max(ans, i * ct[i]);
    }
    cout << ans << '\n';
  }
}

**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 provided buggy code, we need to consider edge cases and scenarios that are likely to cause issues. Here are some steps to think through this:

1. **Overlap and Non-Overlap Scenarios**: The code needs to handle both overlapping and non-overlapping segments correctly.
2. **Boundary Conditions**: Test cases with segments that are just touching or have very small overlaps.
3. **Complexity in Grouping**: Ensure the test case has a mix of segments that can and cannot form complex subsets.
4. **Edge Cases for Segment Limits**: Test cases with segments that have the minimum and maximum possible values for `l` and `r`.

Given these considerations, let's create a small-scale test input that is complete and valid:

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

### Explanation of the Test Input:

1. **First Test Case**:
   - Segments: `[1, 5], [2, 4], [3, 6], [4, 7], [5, 8]`
   - All pairs of segments intersect, making it a valid complex subset of size 5.

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

3. **Third Test Case**:
   - Segments: `[1, 6], [2, 5], [3, 7], [4, 8], [5, 9], [6, 10]`
   - The segments can be grouped into three complex subsets: `{[1, 6], [2, 5]}`, `{[3, 7], [4, 8]}`, and `{[5, 9], [6, 10]}`. The maximum size of a complex subset is 2.

This test input should help expose any issues in the buggy code, particularly in handling overlapping segments and forming valid complex subsets.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Easy Version)
time_limit_ms: 6000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the easy version of the problem. In this version, the constraints on $n$ and the time limit are lower. 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 $n$ ($1 \le n \le 2 \cdot 10^4$) — 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 $2 \cdot 10^4$.

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 N = 20'000 + 10,
          S = 500;
int n;
pair<int, int> p[N];

namespace IT {
  int sz;
  int f[N << 2], lazy[N << 2];

  void clear() {
    memset(f, 0, ((sz << 2) + 10) * sizeof(int));
    memset(lazy, 0, ((sz << 2) + 10) * sizeof(int));
  }
  inline void push(int s) {
    if (!lazy[s]) return;
    f[s << 1] += lazy[s];     f[s << 1 | 1] += lazy[s];
    lazy[s << 1] += lazy[s];  lazy[s << 1 | 1] += lazy[s];
    lazy[s] = 0;
  }

  void update(int s, int l, int r, int u, int v, int x) {
    if (v < l || u > r) return;
    if (u <= l && r <= v) {
      f[s] += x;
      lazy[s] += x;
      return;
    }
    push(s);
    int mid = (l + r) >> 1;
    update(s << 1, l, mid, u, v, x); update(s << 1 | 1, mid + 1, r, u, v, x);
    f[s] = max(f[s << 1], f[s << 1 | 1]);
  }
}

int32_t main() {
  cin.tie(0)->sync_with_stdio(0);

  int test; cin >> test;
  while (test--) {
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> p[i].first;
    for (int i = 1; i <= n; ++i) cin >> p[i].second;

    sort(p + 1, p + n + 1, [](const auto& a, const auto& b) {
      return make_pair(a.second, a.first) < make_pair(b.second, b.first);
    });

    vector<int> allValue;
    for (int i = 1; i <= n; ++i) allValue.push_back(p[i].first), allValue.push_back(p[i].second);
    sort(allValue.begin(), allValue.end());
    allValue.erase(unique(allValue.begin(), allValue.end()), allValue.end());
    for (int i = 1; i <= n; ++i) {
      auto& [l, r] = p[i];
      l = upper_bound(allValue.begin(), allValue.end(), l) - allValue.begin();
      r = upper_bound(allValue.begin(), allValue.end(), r) - allValue.begin();
    }

    IT::sz = allValue.size();
    vector<pair<int, int>> segments;
    auto cal = [&](int subsetSize) {
      int ret = 0;

      int pre = 0;
      for (int i = 1; i <= n; ++i) {
        const auto& [l, r] = p[i];
        if (l <= pre) continue;
        IT::update(1, 1, IT::sz, l, r, 1);
        segments.push_back({l, r});

        if (IT::f[1] >= subsetSize) {
          ret += 1;
          for (const auto& [lt, rt] : segments) IT::update(1, 1, IT::sz, lt, rt, -1);
          segments.clear();
          pre = r;
        }
      }
      IT::clear();
      segments.clear();

      return ret;
    };

    int answer = 0;
    for (int subsetSize = 1; subsetSize <= S; ++subsetSize)
      answer = max(answer, subsetSize * cal(subsetSize));
    for (int numSubset = 1; numSubset <= n / S; ++numSubset) {
      int le = S + 1, ri = n, subsetSize = 0;
      while (le <= ri) {
        int mid = (le + ri) >> 1;
        if (cal(mid) >= numSubset) le = (subsetSize = mid) + 1;
        else ri = mid - 1;
      }
      answer = max(answer, numSubset * subsetSize);
    }

    cout << answer << "\n";
  }
  cerr << 1.0 * clock() / CLOCKS_PER_SEC << "\n";
}

**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. Here are some potential areas to consider:

1. **Overlap and Intersection**:
   - Segments that overlap in complex ways.
   - Segments that form multiple valid complex subsets.

2. **Edge Cases**:
   - Single segment.
   - Segments with the same endpoints.
   - Segments with minimal and maximal endpoints.

3. **Large and Small Subsets**:
   - Test cases with a large number of segments to ensure the algorithm handles large inputs correctly.
   - Test cases with a small number of segments to ensure the algorithm handles small inputs correctly.

4. **Special Constraints**:
   - Test cases where the number of segments is exactly at the boundary of the constraints (e.g., \( n = 2 \cdot 10^4 \)).

Given the constraints and the need for a small-scale test input, let's create a test input that includes a mix of these cases:

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

### Explanation:
1. **First Test Case**:
   - Segments: \([1, 6], [2, 5], [3, 7], [4, 8], [5, 9]\)
   - This set of segments has overlapping and non-overlapping segments, which can form multiple complex subsets.

2. **Second Test Case**:
   - Segments: \([1, 2], [1, 2], [2, 3], [2, 3]\)
   - This set of segments includes segments with the same endpoints, which should form complex subsets of size 2.

3. **Third Test Case**:
   - Segments: \([1, 2], [2, 3]\)
   - This set of segments has minimal and maximal endpoints, which should form a single complex subset of size 2.

This test input is designed to cover a range of edge cases and scenarios that are likely to expose 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 (Easy Version)
time_limit_ms: 6000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the easy version of the problem. In this version, the constraints on $n$ and the time limit are lower. 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 $n$ ($1 \le n \le 2 \cdot 10^4$) — 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 $2 \cdot 10^4$.

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>

#ifndef ONLINE_JUDGE

#include "../debug.cpp"

#else
#define debug(...) (void)(0721)
#endif

using namespace std;

using i64 = long long;
using u64 = unsigned long long;

constexpr int inf = 1e9;

struct Max {
    int x;

    Max(int x = 0) : x(x) {}

    Max operator+=(const Max &ano) {
        x = max(x, ano.x);
        return *this;
    }

    friend Max operator+(const Max &l, const Max &r) {
        Max ret = l;
        ret += r;
        return ret;
    }
};

void infoApply(Max &info, int l, int r, const int &tag) {
    info.x += tag;
}

void tagApply(int &son, int l, int r, const int &fa) {
    son += fa;
}

template<class Info, class Tag>
class segTree {
#define ls i << 1
#define rs i << 1 | 1
#define mid ((l + r) >> 1)
#define lson ls, l, mid
#define rson rs, mid + 1, r

    int n;
    vector<Info> info;
    vector<Tag> tag;
public:
    segTree(const vector<Info> &init) : n(init.size()) {
        assert(n > 0);
        info.resize(4 << __lg(n));
        tag.resize(4 << __lg(n));
        auto build = [&](auto dfs, int i, int l, int r) {
            if (l == r) {
                info[i] = init[l];
                return;
            }
            dfs(dfs, lson);
            dfs(dfs, rson);
            pull(i);
        };
        build(build, 1, 0, n - 1);
    }

private:
    void pull(int i) { info[i] = info[ls] + info[rs]; }

    template<class... T>
    void apply(int i, int l, int r, const T &... val) {
        ::infoApply(info[i], l, r, val...);
        ::tagApply(tag[i], l, r, val...);
    }

    void push(int i, int l, int r) {
        if (tag[i] == Tag{}) return;
        apply(lson, tag[i]);
        apply(rson, tag[i]);
        tag[i] = {};
    }

public:
    template<class... T>
    void rangeApply(int ql, int qr, const T &... val) {
        auto dfs = [&](auto dfs, int i, int l, int r) {
            if (qr < l || r < ql) return;
            if (ql <= l && r <= qr) {
                apply(i, l, r, val...);
                return;
            }
            push(i, l, r);
            dfs(dfs, lson);
            dfs(dfs, rson);
            pull(i);
        };
        dfs(dfs, 1, 0, n - 1);
    }

    Info rangeAsk(int ql, int qr) {
        Info res{};
        auto dfs = [&](auto dfs, int i, int l, int r) {
            if (qr < l || r < ql) return;
            if (ql <= l && r <= qr) {
                res = res + info[i];
                return;
            }
            push(i, l, r);
            dfs(dfs, lson);
            dfs(dfs, rson);
        };
        dfs(dfs, 1, 0, n - 1);
        return res;
    }

#undef rson
#undef lson
#undef mid
#undef rs
#undef ls
};

void run(int testCase) {
    int n;
    cin >> n;
    vector<int> l(n), r(n);
    for (int i = 0; i < n; ++i) {
        cin >> l[i];
        l[i]--;
    }
    for (int i = 0; i < n; ++i) {
        cin >> r[i];
        r[i]--;
    }

    auto max_k = [&](int m) {
        vector<vector<int>> ls(2 * n);
        for (int i = 0; i < n; ++i) {
            ls[r[i]].push_back(l[i]);
        }
        segTree<Max, int> t((vector<Max>(2 * n)));
        int i = 0;
        int ret = 0;
        for (int j = 0; j < 2 * n; ++j) {
            for (int L: ls[j]) {
                if (L >= i) {
                    t.rangeApply(L, j, 1);
                }
            }
            if (t.rangeAsk(i, j).x >= m) {
                ret++;
                i = j + 1;
            }
        }
        return ret;
    };

    auto max_m = [&](int k) {
        auto check = [&](int m) {
            return k <= max_k(m);
        };
        return *ranges::partition_point(views::iota(0, n + 1), check) - 1;
    };

    int ans = 0;

    const int B = sqrt(n * (__lg(n) + 1));

    for (int m = 1; m <= B; ++m) {
        int k = max_k(m);
        ans = max(ans, m * k);
    }

    for(int k = 1; k * B <= n; ++k) {
        int m = max_m(k);
        ans = max(ans, m * k);
    }

    cout << ans << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int T = 1;
    cin >> T;
    for (int t = 1; t <= T; ++t) {
        run(t);
    }
    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
2
1 2
2 3
4
1 3 5 7
2 4 6 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Easy Version)
time_limit_ms: 6000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the easy version of the problem. In this version, the constraints on $n$ and the time limit are lower. 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 $n$ ($1 \le n \le 2 \cdot 10^4$) — 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 $2 \cdot 10^4$.

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 = 40005;
    static int n, t, res;
    static int[] c = new int[MAXN];
    static int[] f = new int[MAXN];
    static Pair[] p = new Pair[MAXN];

    static class Pair {
        int fi, se;

        Pair(int fi, int se) {
            this.fi = fi;
            this.se = se;
        }
    }

    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++) {
                p[i] = new Pair(scanner.nextInt(), 0);
                c[p[i].fi]++;
            }
            for (int i = 1; i <= n; i++) {
                p[i].se = scanner.nextInt();
                c[p[i].se]++;
            }
            for (int i = 1; i <= 2 * n; i++) c[i] += c[i - 1];
            for (int i = 1; i <= n; i++) p[i].se = c[p[i].se]--;
            for (int i = 1; i <= n; i++) p[i].fi = c[p[i].fi]--;
            Arrays.sort(p, 1, n + 1, Comparator.comparingInt(x -> x.se));
            solve(1, n, n + 1, 0);
            System.out.println(res);
        }
        scanner.close();
    }

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

    static int calc(int m) {
        int cnt = 0;
        int x = 0, cur = 0, lim = 0, lst = 0;
        for (int i = 1; i <= n; i++) {
            int l = p[i].fi, r = p[i].se;
            if (l <= lim) continue;
            for (int j = lst + 1; j < r; j++) f[j] = lst;
            cur++;
            lst = f[r] = r;
            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) >> 1;
        int val = calc(mid);
        solve(l, mid - 1, L, val);
        solve(mid + 1, r, val, R);
    }
}

**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 given buggy code, let's consider the following points:

1. **Edge Cases**: The code might fail when dealing with segments that have overlapping or non-overlapping endpoints.
2. **Complexity**: The code should handle multiple test cases, and the segments should be designed to test the partitioning logic.
3. **Boundary Conditions**: Ensure the input includes segments that test the boundaries of the constraints.

Given these points, a good test input would include:
- Segments that overlap in complex ways.
- Segments with the same endpoints.
- Segments that do not overlap at all.

Here's a small-scale test input that covers these scenarios:

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

Explanation:
1. **Test Case 1**: Segments [1,5], [2,4], [3,6], [4,7], [5,8]. This tests overlapping and non-overlapping segments.
2. **Test Case 2**: Segments [1,5], [2,4], [3,3], [4,2], [5,1]. This tests segments with decreasing right endpoints.
3. **Test Case 3**: Segments [1,2], [1,2], [1,2], [1,2], [1,2]. This tests segments with the same endpoints.

This input should help expose any issues in 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 (Easy Version)
time_limit_ms: 6000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the easy version of the problem. In this version, the constraints on $n$ and the time limit are lower. 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 $n$ ($1 \le n \le 2 \cdot 10^4$) — 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 $2 \cdot 10^4$.

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 long long long int
#define DEBUG
using namespace std;

// @author: pashka

struct test {
    void solve() {
        int n;
        cin >> n;
        vector<int> ll(n);
        vector<int> rr(n);
        for (int i = 0; i < n; i++) {
            cin >> ll[i];
        }
        for (int i = 0; i < n; i++) {
            cin >> rr[i];
        }
        vector<vector<int>> p(4 * n + 2);
        for (int i = 0; i < n; i++) {
            p[2 * ll[i]].push_back(rr[i]);
            p[2 * rr[i] + 1].push_back(ll[i]);
        }
        for (int i = 0; i < p.size(); i++) {
            if (i % 2 == 0)
                sort(p[i].begin(), p[i].end());
            else
                sort(p[i].rbegin(), p[i].rend());
        }
//        sort(p.begin(), p.end());

        int res = 0;
        for (int k = 1; k <= n; k++) {
            int end = 2 * n;
            int ss = 0;
            while (true) {
                int l = -1;
                int r = end + 1;
                while (r > l + 1) {
                    int m = (l + r) / 2;
                    int mx = 0;
                    int c = 0;
                    for (int x = m; x <= end; x++) {
                        for (int rr: p[2 * x]) {
                            if (rr <= end) {
                                c++;
                            } else {
                                break;
                            }
                        }
                        mx = max(mx, c);
                        for (int ll: p[2 * x + 1]) {
                            if (ll >= m) {
                                c--;
                            } else {
                                break;
                            }
                        }

                    }
                    if (mx >= k) {
                        l = m;
                    } else {
                        r = m;
                    }
                }
                if (l == -1) break;
                ss += k;
                end = l - 1;
            }
//            cout << k << " " << ss << "\n";
            res = max(res, ss);
        }
        cout << res << "\n";
    }
};

int main() {
    ios::sync_with_stdio(false);

    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
        test().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 expose the bug in the provided buggy code, we need to focus on edge cases that might cause the algorithm to fail. The algorithm seems to handle the segments in a way that might not correctly account for the intersections and the formation of complex subsets. Here are some key points to consider:

1. **Overlapping Segments**: The code might not correctly handle cases where segments overlap in a complex manner.
2. **Non-Intersecting Segments**: The code might fail to correctly identify subsets where segments do not intersect.
3. **Boundary Conditions**: The code might have issues with segments that have the same endpoints or very close endpoints.

Given these points, let's create a small-scale test input that includes segments with various overlaps and non-overlapping segments.

### Test Input
```plaintext
3
4
1 2 3 4
5 6 7 8
5
1 2 3 4 5
3 4 5 6 7
6
1 2 3 4 5 6
2 3 4 5 6 7
```

### Explanation
1. **First Test Case**:
   - Segments: `[1, 5]`, `[2, 6]`, `[3, 7]`, `[4, 8]`
   - These segments do not intersect with each other, so the maximum size of a complex subset is 1.

2. **Second Test Case**:
   - Segments: `[1, 3]`, `[2, 4]`, `[3, 5]`, `[4, 6]`, `[5, 7]`
   - These segments have some intersections, but not all. The maximum size of a complex subset is 2 (e.g., `{[1, 3], [2, 4]}` and `{[3, 5], [4, 6]}`).

3. **Third Test Case**:
   - Segments: `[1, 2]`, `[2, 3]`, `[3, 4]`, `[4, 5]`, `[5, 6]`, `[6, 7]`
   - These segments are all adjacent and do not intersect with each other, so the maximum size of a complex subset is 1.

This test input should help expose any issues in the buggy code by covering a variety of segment configurations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Complex Segments (Easy Version)
time_limit_ms: 6000
memory_limit_mb: 256
Description: [Ken Arai - COMPLEX](https://soundcloud.com/diatomichail2/complex)

This is the easy version of the problem. In this version, the constraints on $n$ and the time limit are lower. 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 $n$ ($1 \le n \le 2 \cdot 10^4$) — 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 $2 \cdot 10^4$.

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>

#ifndef ONLINE_JUDGE

#include "../debug.cpp"

#else
#define debug(...) (void)(0721)
#endif

using namespace std;

using i64 = long long;
using u64 = unsigned long long;

constexpr int inf = 1e9;

struct Max {
    int x;

    Max(int x = 0) : x(x) {}

    Max operator+=(const Max &ano) {
        x = max(x, ano.x);
        return *this;
    }

    friend Max operator+(const Max &l, const Max &r) {
        Max ret = l;
        ret += r;
        return ret;
    }
};

void infoApply(Max &info, int l, int r, const int &tag) {
    info.x += tag;
}

void tagApply(int &son, int l, int r, const int &fa) {
    son += fa;
}

template<class Info, class Tag>
class segTree {
#define ls i << 1
#define rs i << 1 | 1
#define mid ((l + r) >> 1)
#define lson ls, l, mid
#define rson rs, mid + 1, r

    int n;
    vector<Info> info;
    vector<Tag> tag;
public:
    segTree(const vector<Info> &init) : n(init.size()) {
        assert(n > 0);
        info.resize(4 << __lg(n));
        tag.resize(4 << __lg(n));
        auto build = [&](auto dfs, int i, int l, int r) {
            if (l == r) {
                info[i] = init[l];
                return;
            }
            dfs(dfs, lson);
            dfs(dfs, rson);
            pull(i);
        };
        build(build, 1, 0, n - 1);
    }

private:
    void pull(int i) { info[i] = info[ls] + info[rs]; }

    template<class... T>
    void apply(int i, int l, int r, const T &... val) {
        ::infoApply(info[i], l, r, val...);
        ::tagApply(tag[i], l, r, val...);
    }

    void push(int i, int l, int r) {
        if (tag[i] == Tag{}) return;
        apply(lson, tag[i]);
        apply(rson, tag[i]);
        tag[i] = {};
    }

public:
    template<class... T>
    void rangeApply(int ql, int qr, const T &... val) {
        auto dfs = [&](auto dfs, int i, int l, int r) {
            if (qr < l || r < ql) return;
            if (ql <= l && r <= qr) {
                apply(i, l, r, val...);
                return;
            }
            push(i, l, r);
            dfs(dfs, lson);
            dfs(dfs, rson);
            pull(i);
        };
        dfs(dfs, 1, 0, n - 1);
    }

    Info rangeAsk(int ql, int qr) {
        Info res{};
        auto dfs = [&](auto dfs, int i, int l, int r) {
            if (qr < l || r < ql) return;
            if (ql <= l && r <= qr) {
                res = res + info[i];
                return;
            }
            push(i, l, r);
            dfs(dfs, lson);
            dfs(dfs, rson);
        };
        dfs(dfs, 1, 0, n - 1);
        return res;
    }

#undef rson
#undef lson
#undef mid
#undef rs
#undef ls
};

void run(int testCase) {
    int n;
    cin >> n;
    vector<int> l(n), r(n);
    for (int i = 0; i < n; ++i) {
        cin >> l[i];
        l[i]--;
    }
    for (int i = 0; i < n; ++i) {
        cin >> r[i];
        r[i]--;
    }

    auto max_k = [&](int m) {
        vector<vector<int>> ls(2 * n);
        for (int i = 0; i < n; ++i) {
            ls[r[i]].push_back(l[i]);
        }
        segTree<Max, int> t((vector<Max>(2 * n)));
        int i = 0;
        int ret = 0;
        for (int j = 0; j < 2 * n; ++j) {
            for (int L: ls[j]) {
                if (L >= i) {
                    t.rangeApply(L, j, 1);
                }
            }
            if (t.rangeAsk(i, j).x >= m) {
                ret++;
                i = j + 1;
            }
        }
        return ret;
    };

    auto max_m = [&](int k) {
        auto check = [&](int m) {
            return k <= max_k(m);
        };
        return *ranges::partition_point(views::iota(0, n + 1), check) - 1;
    };

    int ans = 0;

    const int B = 200;

    for (int m = 1; m <= B and m <= n; ++m) {
        int k = max_k(m);
        ans = max(ans, m * k);
    }

    if (n == 20000) assert(false);

    for (int k = 1; k * B <= n; ++k) {
        int m = max_m(k);
        ans = max(ans, m * k);
    }

    cout << ans << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int T = 1;
    cin >> T;
    for (int t = 1; t <= T; ++t) {
        run(t);
    }
    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 given buggy code, we need to focus on edge cases and scenarios that can highlight the issues in the algorithm. The key points to consider are:

1. **Intersecting Segments**: Ensure that the segments have complex intersection patterns.
2. **Non-Intersecting Segments**: Ensure that some segments do not intersect with others.
3. **Edge Cases**: Test with the smallest and largest possible values for the number of segments and their endpoints.
4. **Special Patterns**: Test with specific patterns that might cause the algorithm to fail, such as all segments being the same, or segments that form a specific complex structure.

Given the constraints and the need for a small-scale test input, let's create a test case that includes a mix of intersecting and non-intersecting segments, and also includes some edge cases.

### 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
```

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

2. **Second Test Case**:
   - Segments: `[1, 5]`, `[2, 4]`, `[3, 7]`, `[6, 9]`, `[8, 10]`
   - Segments `[1, 5]` and `[2, 4]` intersect, and segments `[6, 9]` and `[8, 10]` intersect, but there is no intersection between the two pairs. The optimal solution is to form two subsets of size 2, resulting in a complex subset of size 4.

3. **Third Test Case**:
   - Segments: `[3, 7]`, `[1, 2]`, `[4, 6]`, `[1, 5]`, `[5, 10]`
   - Segments `[3, 7]`, `[4, 6]`, and `[5, 10]` form a complex subset, and segment `[1, 2]` does not intersect with any other segment. The optimal solution is to form a single subset of size 4 by including all segments except `[1, 2]`.

This test input is designed to cover a range of scenarios and should help expose any bugs in the algorithm.