**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Seating in a Bus
time_limit_ms: 2000
memory_limit_mb: 256
Description: In Berland, a bus consists of a row of $n$ seats numbered from $1$ to $n$. Passengers are advised to always board the bus following these rules:

  * If there are no occupied seats in the bus, a passenger can sit in any free seat;   * Otherwise, a passenger should sit in any free seat that has at least one occupied neighboring seat. In other words, a passenger can sit in a seat with index $i$ ($1 \le i \le n$) only if at least one of the seats with indices $i-1$ or $i+1$ is occupied.

Today, $n$ passengers boarded the bus. The array $a$ chronologically records the seat numbers they occupied. That is, $a_1$ contains the seat number where the first passenger sat, $a_2$ — the seat number where the second passenger sat, and so on.

You know the contents of the array $a$. Determine whether all passengers followed the recommendations.

For example, if $n = 5$, and $a$ = [$5, 4, 2, 1, 3$], then the recommendations were not followed, as the $3$-rd passenger sat in seat number $2$, while the neighboring seats with numbers $1$ and $3$ were free.

Input: The first line of input contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

The following describes the input test cases.

The first line of each test case contains exactly one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of seats in the bus and the number of passengers who boarded the bus.

The second line of each test case contains $n$ distinct integers $a_i$ ($1 \le a_i \le n$) — the seats that the passengers occupied in chronological order.

It is guaranteed that the sum of $n$ values across all test cases does not exceed $2 \cdot 10^5$, and that no passenger sits in an already occupied seat.

Output: For each test case, output on a separate line:

  * "YES", if all passengers followed the recommendations;    * "NO" otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
4
5
5 4 2 1 3
3
2 3 1
4
2 3 1 4
5
1 2 3 5 4
output:
NO
YES
YES
NO

Note: The first test case is explained in the problem statement.

Buggy Code:
T = int(input())

for i in range(T):
    n = int(input())
    arr = list(map(int,input().split()))
    cover = [arr[0]]
    take=0
    for  j in range(1,n):
        if(not (arr[j]+1 in cover or arr[j]-1 in cover)):
            take=1
            break
        cover.append(arr[j])
    if(take):
        print("No")
    else:
        print("Yes")

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Seating in a Bus
time_limit_ms: 2000
memory_limit_mb: 256
Description: In Berland, a bus consists of a row of $n$ seats numbered from $1$ to $n$. Passengers are advised to always board the bus following these rules:

  * If there are no occupied seats in the bus, a passenger can sit in any free seat;   * Otherwise, a passenger should sit in any free seat that has at least one occupied neighboring seat. In other words, a passenger can sit in a seat with index $i$ ($1 \le i \le n$) only if at least one of the seats with indices $i-1$ or $i+1$ is occupied.

Today, $n$ passengers boarded the bus. The array $a$ chronologically records the seat numbers they occupied. That is, $a_1$ contains the seat number where the first passenger sat, $a_2$ — the seat number where the second passenger sat, and so on.

You know the contents of the array $a$. Determine whether all passengers followed the recommendations.

For example, if $n = 5$, and $a$ = [$5, 4, 2, 1, 3$], then the recommendations were not followed, as the $3$-rd passenger sat in seat number $2$, while the neighboring seats with numbers $1$ and $3$ were free.

Input: The first line of input contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

The following describes the input test cases.

The first line of each test case contains exactly one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of seats in the bus and the number of passengers who boarded the bus.

The second line of each test case contains $n$ distinct integers $a_i$ ($1 \le a_i \le n$) — the seats that the passengers occupied in chronological order.

It is guaranteed that the sum of $n$ values across all test cases does not exceed $2 \cdot 10^5$, and that no passenger sits in an already occupied seat.

Output: For each test case, output on a separate line:

  * "YES", if all passengers followed the recommendations;    * "NO" otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
4
5
5 4 2 1 3
3
2 3 1
4
2 3 1 4
5
1 2 3 5 4
output:
NO
YES
YES
NO

Note: The first test case is explained in the problem statement.

Buggy Code:
t=int(input())
for i in range(t):
    n=int(input())
    a=list(map(int,input().split()))
    k=0
    for i in range(n):
        for j in range(i):
            if abs(a[i]-a[j])==1:
                k+=1
                break
        if k!=i:
            break
    if i==k:
        print("Yes")
    else:
        print("No")

**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 where the algorithm might fail. The key issue in the buggy code is that it does not correctly handle the scenario where a passenger sits in a seat that has no occupied neighbors, even though there are other occupied seats in the bus.

### Step-by-Step Analysis:
1. **Initial Seat Occupation**: The first passenger can sit in any seat.
2. **Subsequent Seat Occupation**: Each subsequent passenger must sit in a seat that has at least one occupied neighbor.

The buggy code fails to correctly check if a passenger is sitting in a seat with no occupied neighbors. It only increments `k` if a neighbor is found, but it does not ensure that `k` is incremented for every valid seat.

### Challenging Test Input:
Let's create a test case where the first few passengers sit in seats that are not adjacent, and then a passenger sits in a seat that has no occupied neighbors. This should trigger the bug in the code.

#### Test Case:
```plaintext
1
6
1 3 5 2 4 6
```

### Explanation:
- **n = 6**: There are 6 seats and 6 passengers.
- **a = [1, 3, 5, 2, 4, 6]**:
  - The first passenger sits in seat 1.
  - The second passenger sits in seat 3.
  - The third passenger sits in seat 5.
  - The fourth passenger sits in seat 2, which is adjacent to seat 1.
  - The fifth passenger sits in seat 4, which is adjacent to seat 3.
  - The sixth passenger sits in seat 6, which is not adjacent to any occupied seat (5 is the only occupied seat nearby, but not adjacent).

This test case should expose the bug because the sixth passenger sits in a seat with no occupied neighbors, which should result in a "NO" output. However, the buggy code might incorrectly output "Yes" because it does not correctly check the adjacency condition.

### Expected Output:
```plaintext
NO
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Seating in a Bus
time_limit_ms: 2000
memory_limit_mb: 256
Description: In Berland, a bus consists of a row of $n$ seats numbered from $1$ to $n$. Passengers are advised to always board the bus following these rules:

  * If there are no occupied seats in the bus, a passenger can sit in any free seat;   * Otherwise, a passenger should sit in any free seat that has at least one occupied neighboring seat. In other words, a passenger can sit in a seat with index $i$ ($1 \le i \le n$) only if at least one of the seats with indices $i-1$ or $i+1$ is occupied.

Today, $n$ passengers boarded the bus. The array $a$ chronologically records the seat numbers they occupied. That is, $a_1$ contains the seat number where the first passenger sat, $a_2$ — the seat number where the second passenger sat, and so on.

You know the contents of the array $a$. Determine whether all passengers followed the recommendations.

For example, if $n = 5$, and $a$ = [$5, 4, 2, 1, 3$], then the recommendations were not followed, as the $3$-rd passenger sat in seat number $2$, while the neighboring seats with numbers $1$ and $3$ were free.

Input: The first line of input contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

The following describes the input test cases.

The first line of each test case contains exactly one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of seats in the bus and the number of passengers who boarded the bus.

The second line of each test case contains $n$ distinct integers $a_i$ ($1 \le a_i \le n$) — the seats that the passengers occupied in chronological order.

It is guaranteed that the sum of $n$ values across all test cases does not exceed $2 \cdot 10^5$, and that no passenger sits in an already occupied seat.

Output: For each test case, output on a separate line:

  * "YES", if all passengers followed the recommendations;    * "NO" otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
4
5
5 4 2 1 3
3
2 3 1
4
2 3 1 4
5
1 2 3 5 4
output:
NO
YES
YES
NO

Note: The first test case is explained in the problem statement.

Buggy Code:
import java.util.ArrayList;
import java.util.Scanner;

public class busSeat {
    static String checkSeat(int[] a,int n){
        int i=1;
        ArrayList<Integer> seat = new ArrayList<>();
        seat.add(a[0]);
        while (i < n) {
            if (seat.contains(a[i] + 1) || seat.contains(a[i] - 1)) {
                seat.add(a[i]);
                i++;
            } else {
               return "No";
            }
        }
        if (i == n) {
            return "Yes";
        } else {
            return "No";
        }

    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
            }
           String value= checkSeat(a, n);
            System.out.println(value);
        }
    }
}

**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
3 1 4 2
```

**Explanation:**
- The test case is designed to challenge the buggy code by providing a sequence where the first passenger sits in seat 3, and the subsequent passengers sit in seats 1, 4, and 2.
- According to the rules, the first passenger can sit in any seat, so seat 3 is valid.
- The second passenger sits in seat 1, which is not adjacent to seat 3, so this should be marked as "NO" because the second passenger did not follow the recommendation.
- This test case will expose the bug in the code, as the code might incorrectly return "YES" if it does not properly check the initial condition for the first passenger.


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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Seating in a Bus
time_limit_ms: 2000
memory_limit_mb: 256
Description: In Berland, a bus consists of a row of $n$ seats numbered from $1$ to $n$. Passengers are advised to always board the bus following these rules:

  * If there are no occupied seats in the bus, a passenger can sit in any free seat;   * Otherwise, a passenger should sit in any free seat that has at least one occupied neighboring seat. In other words, a passenger can sit in a seat with index $i$ ($1 \le i \le n$) only if at least one of the seats with indices $i-1$ or $i+1$ is occupied.

Today, $n$ passengers boarded the bus. The array $a$ chronologically records the seat numbers they occupied. That is, $a_1$ contains the seat number where the first passenger sat, $a_2$ — the seat number where the second passenger sat, and so on.

You know the contents of the array $a$. Determine whether all passengers followed the recommendations.

For example, if $n = 5$, and $a$ = [$5, 4, 2, 1, 3$], then the recommendations were not followed, as the $3$-rd passenger sat in seat number $2$, while the neighboring seats with numbers $1$ and $3$ were free.

Input: The first line of input contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

The following describes the input test cases.

The first line of each test case contains exactly one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of seats in the bus and the number of passengers who boarded the bus.

The second line of each test case contains $n$ distinct integers $a_i$ ($1 \le a_i \le n$) — the seats that the passengers occupied in chronological order.

It is guaranteed that the sum of $n$ values across all test cases does not exceed $2 \cdot 10^5$, and that no passenger sits in an already occupied seat.

Output: For each test case, output on a separate line:

  * "YES", if all passengers followed the recommendations;    * "NO" otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
4
5
5 4 2 1 3
3
2 3 1
4
2 3 1 4
5
1 2 3 5 4
output:
NO
YES
YES
NO

Note: The first test case is explained in the problem statement.

Buggy Code:
#include <vector>
#include <iostream>

void gain(int n, std::vector<int> passenger){

    for(int i = 1; i<n;i++){

        bool hit = 0;

    for(int j = 0; j<i;j++){
        int next1; int next2;
        if(passenger[i] == 1){
        next1 = 2; next2 = 2;
        }
        else if(passenger[i] == n){
        next1 = n-1; next2 = n-1;
        }else{
            next1 = passenger[i]-1; next2 = passenger[i]+1;
        }

        if(passenger[j] == next1 || passenger[j] == next2){
            hit = 1;
            break;
        }
    }
    if(hit == 0){
        std::cout<<"no"<<std::endl;
        return;
    }
    }
    std::cout<<"yes"<<std::endl;
    return;

}

int main(){
    int t; int n;
    std::cin>>t;
    for(int i = 0;i<t;i++){
        std::cin>>n;
        std::vector<int> passenger(n,0);
        for(int i=0;i<n;i++){
            std::cin>>passenger[i];
        }
    gain(n, passenger);
    }

}

**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 where the code might fail. The buggy code has a few potential issues:

1. **Edge Cases for Seat 1 and Seat n**: The code incorrectly handles the edge cases for seat 1 and seat n by setting `next1` and `next2` to the same value, which is incorrect.
2. **Order of Occupied Seats**: The code might not correctly handle the order in which seats are occupied, especially when the first few seats are occupied in a way that violates the rules.

Let's construct a test input that highlights these issues:

1. **Test Case with Seat 1 and Seat n**: Create a scenario where the first passenger sits in seat 1 and the next passenger sits in seat n, and then a passenger tries to sit in a seat that is not next to an occupied seat.
2. **Test Case with Incorrect Order**: Create a scenario where the first few passengers sit in a way that violates the rules, but the code might not catch it.

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

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

Explanation:
1. **First Test Case (5 seats, 5 passengers)**:
   - Passenger 1 sits in seat 1.
   - Passenger 2 sits in seat 5.
   - Passenger 3 sits in seat 3, which is not next to an occupied seat.
   - This should output "NO".

2. **Second Test Case (4 seats, 4 passengers)**:
   - Passenger 1 sits in seat 1.
   - Passenger 2 sits in seat 4.
   - Passenger 3 sits in seat 2, which is not next to an occupied seat.
   - This should output "NO".

These test cases should help expose the bug in the buggy code by highlighting the incorrect handling of edge cases and the order of occupied seats.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Seating in a Bus
time_limit_ms: 2000
memory_limit_mb: 256
Description: In Berland, a bus consists of a row of $n$ seats numbered from $1$ to $n$. Passengers are advised to always board the bus following these rules:

  * If there are no occupied seats in the bus, a passenger can sit in any free seat;   * Otherwise, a passenger should sit in any free seat that has at least one occupied neighboring seat. In other words, a passenger can sit in a seat with index $i$ ($1 \le i \le n$) only if at least one of the seats with indices $i-1$ or $i+1$ is occupied.

Today, $n$ passengers boarded the bus. The array $a$ chronologically records the seat numbers they occupied. That is, $a_1$ contains the seat number where the first passenger sat, $a_2$ — the seat number where the second passenger sat, and so on.

You know the contents of the array $a$. Determine whether all passengers followed the recommendations.

For example, if $n = 5$, and $a$ = [$5, 4, 2, 1, 3$], then the recommendations were not followed, as the $3$-rd passenger sat in seat number $2$, while the neighboring seats with numbers $1$ and $3$ were free.

Input: The first line of input contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

The following describes the input test cases.

The first line of each test case contains exactly one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of seats in the bus and the number of passengers who boarded the bus.

The second line of each test case contains $n$ distinct integers $a_i$ ($1 \le a_i \le n$) — the seats that the passengers occupied in chronological order.

It is guaranteed that the sum of $n$ values across all test cases does not exceed $2 \cdot 10^5$, and that no passenger sits in an already occupied seat.

Output: For each test case, output on a separate line:

  * "YES", if all passengers followed the recommendations;    * "NO" otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
4
5
5 4 2 1 3
3
2 3 1
4
2 3 1 4
5
1 2 3 5 4
output:
NO
YES
YES
NO

Note: The first test case is explained in the problem statement.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
    ll t, n;
    cin>>t;
    while(t--)
    {
        cin>>n;
         vector <ll> a(n);
        for(ll i=0;i<n;i++)
        {
            cin>>a[i];
        }
        int man=1;
        for(ll i=1; i<n;i++)
        {
            int giao=0;
            for(ll j=0;j<i;j++)
            {
                if(a[i]-a[j]==1||a[j]-a[i]==1)
                {
                    giao=1;
                    break;
                }
            }
            if(giao==0)
            {
                man=0;
                break;
            }
        }
        if(man!=0) cout<<"yes"<<endl;
        else cout<<"no"<<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
2
1 2
3
1 3 2
```

**Explanation:**

1. **Test Case 1:**
   - **Input:** `1 1`
   - **Output:** `yes`
   - **Explanation:** With only one seat and one passenger, the recommendation is trivially followed.

2. **Test Case 2:**
   - **Input:** `2 1 2`
   - **Output:** `yes`
   - **Explanation:** The first passenger sits in seat 1, and the second passenger sits in seat 2, which is a valid move as seat 1 is a neighboring seat of seat 2.

3. **Test Case 3:**
   - **Input:** `3 1 3 2`
   - **Output:** `no`
   - **Explanation:** The first passenger sits in seat 1, and the second passenger sits in seat 3. When the third passenger sits in seat 2, neither of the neighboring seats (1 and 3) are occupied, which violates the recommendation.

These test cases are designed to cover different scenarios, including the edge case of a single seat, a small bus with two seats, and a bus with three seats where the recommendation is violated. This should help expose the bug in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is an easy version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

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 two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le \frac{n}{2}$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval.

Additional constraint on the input in this version: $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$.

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

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
6
2 0
2 1
1 2
5 1
2 4
8 2
1 4
6 8
7 2
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
21
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 8, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2] = [1, 7]$.

In the fifth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$.

In the sixth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
// ﷽
// https://codeforces.com/contest/2003/problem/E1

#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;

#ifdef SAWALHY
#include "debug.hpp"
#else
#define debug(...) 0
#define debug_itr(...) 0
#define debug_bits(...) 0
#endif

#define ll long long
#define int long long
#define pii pair<int, int>
#define sz(v) (int) (v).size()
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define minit(v, x...) v = min({v, x})
#define maxit(v, x...) v = max({v, x})

template<class T>
using rpq = priority_queue<T, vector<T>, greater<T>>;

const int N = 5e3 + 3;

int dp[N][N];

void solve() {
    int n, m;
    cin >> n >> m;
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= n; j++) {
            dp[i][j] = -1e18;
        }
    }

    int l[m], r[m];
    for (int i = 0; i < m; i++) cin >> l[i] >> r[i], l[i]--, r[i]--;

    dp[0][0] = 0;
    int k = 0;
    for (int i = 0; i < n; i++) {
        while (k < m && r[k] < i) k++;
        if (l[k] == i) {
            for (int h = 1; h < r[k] - l[k] + 1; h++) {
                int hh = r[k] - l[k] + 1 - h;
                for (int j = 0; j <= i; j++) {
                    maxit(dp[r[k] + 1][j + hh], dp[i][j] + j * h);
                }
            }
            i = r[k];
        } else {
            for (int j = 0; j <= i; j++) {
                // put 1
                maxit(dp[i + 1][j + 1], dp[i][j]);
                // put 0
                maxit(dp[i + 1][j], dp[i][j] + j);
            }
        }
    }

    int ans = 0;
    debug(n, m);
    for (int j = 0; j <= n; j++) {
        int x = j;
        int y = n - j;
        int z = dp[n][j];
        debug(x, y, z);
        maxit(ans, x * (x - 1) / 2 + y * (y - 1) / 2 + z);
    }

    cout << ans << endl;
}

int32_t main(int32_t argc, char **argv) {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL), cout.tie(NULL);

    int t = 1;
    cin >> t;
    while (t--)
        solve();

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that are likely to cause issues. The code handles intervals and calculates the maximum number of inversions in interesting permutations. Here are the steps to think through the test case:

1. **Understand the Constraints and Conditions:**
   - The intervals are such that \( r_i < l_{i+1} \) for each \( i \) from \( 1 \) to \( m - 1 \).
   - The code should handle cases where intervals are present and where they are not.

2. **Identify Potential Bugs:**
   - The code might have issues with overlapping or non-overlapping intervals.
   - The code might not correctly handle the maximum number of inversions for specific interval configurations.

3. **Create a Test Case:**
   - We should create a test case with intervals that are non-overlapping but close to each other.
   - We should also consider the edge case where the intervals are at the beginning or end of the permutation.

4. **Test Case Example:**
   - Let's create a test case with \( n = 8 \) and \( m = 2 \) intervals.
   - The intervals should be such that they are close to each other but still non-overlapping.

Here is the test input:

```plaintext
1
8 2
1 4
5 8
```

This test case has:
- \( n = 8 \)
- \( m = 2 \) intervals: \([1, 4]\) and \([5, 8]\)

This test case is designed to check if the code can correctly handle non-overlapping intervals and calculate the maximum number of inversions for the interesting permutations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is an easy version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

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 two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le \frac{n}{2}$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval.

Additional constraint on the input in this version: $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$.

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

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
6
2 0
2 1
1 2
5 1
2 4
8 2
1 4
6 8
7 2
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
21
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 8, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2] = [1, 7]$.

In the fifth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$.

In the sixth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<int, int> vii;
typedef vector<ll, ll> vll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const unsigned int MOD = 1000000007;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif

    // code begins here:-
    int t;
    cin >> t;
    while(t--)
    {
        int n, m;
        cin >> n  >> m;

        if(m > 0)
        {
            int l[m];
            int r[m];

            for(int i = 0; i < m ; i++)
            {
                cin >> l[i] >> r[i];
            }

            bool fr[n]= {false};

            for(int i = l[0]+1; i <= r[0]; i++)
            {
                fr[i-1] = true;
            }

            for(int i = 1; i < m ; i++)
            {
                fr[r[i]-1] = true;
            }

            int ans[n];

            int k = n;

            for(int i = 0; i < l[0]-1; i++)
            {
                ans[i] = k;
                k--;
            }

            for(int i = l[0]-1; i < n; i++)
            {
                if(fr[i])
                {
                    ans[i] = k;
                    k--;
                }
            }

            for(int i = l[0]-1; i < n; i++)
            {
                if(!fr[i])
                {
                    ans[i] = k;
                    k--;
                }
            }

            // for(int i = 0; i < n; i++)
            // {
            //     cout << ans[i] << " ";
            // }

            ll inv = 0;

            for(int i = 0; i < n ; i++)
            {
                for(int j = i+1; j < n ; j++)
                {
                    if(ans[i] > ans[j])
                    {
                        inv++;
                    }
                }
            }

            cout << inv <<  endl;
        }
        else
        {
            cout << (n*(n-1))/2 << endl;
        }

    }
    // code ends here.
    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 are likely to cause issues. Here are the steps to think through this:

1. **Understand the Bug**:
   - The code initializes the `fr` array to mark certain positions as "fixed" (i.e., positions where the permutation must have specific values to satisfy the interval conditions).
   - The code then tries to construct the permutation by placing the largest values in the non-fixed positions first, followed by the fixed positions.
   - The bug might be in how the `fr` array is set up or how the permutation is constructed.

2. **Identify Edge Cases**:
   - **Single Interval**: Test with a single interval to see if the permutation is constructed correctly.
   - **Multiple Intervals**: Test with multiple intervals to ensure the intervals are handled correctly.
   - **Overlapping Intervals**: Although the problem constraints state that intervals do not overlap, we can still test the code to ensure it handles this case correctly.
   - **No Intervals**: Test with no intervals to ensure the code handles this case correctly.
   - **Maximum Number of Intervals**: Test with the maximum number of intervals allowed (i.e., `m = n/2`).

3. **Construct the Test Case**:
   - Choose a small value for `n` and `m` to keep the test case small.
   - Ensure the intervals are valid and follow the constraints `r_i < l_{i+1}`.
   - Use intervals that might cause the code to construct the permutation incorrectly.

Based on these considerations, here is a challenging test input:

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

Explanation:
- **Test Case 1**: `n = 5, m = 1` with interval `[2, 4]`.
  - The permutation should have the maximum value in the first position, and the next largest values in the non-interval positions.
  - The correct permutation is `[5, 2, 4, 3, 1]` with 8 inversions.
- **Test Case 2**: `n = 6, m = 2` with intervals `[1, 3]` and `[4, 6]`.
  - The permutation should have the maximum value in the first position, and the next largest values in the non-interval positions.
  - The correct permutation is `[6, 4, 3, 5, 2, 1]` with 13 inversions.
- **Test Case 3**: `n = 8, m = 2` with intervals `[1, 4]` and `[6, 8]`.
  - The permutation should have the maximum value in the first position, and the next largest values in the non-interval positions.
  - The correct permutation is `[8, 6, 5, 4, 7, 3, 2, 1]` with 21 inversions.

This test input should help expose any issues 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: Turtle and Inversions (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is an easy version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

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 two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le \frac{n}{2}$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval.

Additional constraint on the input in this version: $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$.

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

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
6
2 0
2 1
1 2
5 1
2 4
8 2
1 4
6 8
7 2
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
21
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 8, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2] = [1, 7]$.

In the fifth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$.

In the sixth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
#include <bits/stdc++.h>
#define int long long
#define yes "YES"
#define no "NO"
using ll = long long;
using namespace std;
const int mod=1e9+7;
const int len=2e5+5;

int T;

void solve(){
    int n, m;
    cin >> n >> m;
    vector<int> p(n, -1);
    int now = n;
    int l, r;
    int pre = -1;
    int ok = 1;
    for (int i = 0; i < m; i++) {
        cin >> l >> r;
        l--;
        r--;
        if (ok) {
            for (int j = 0; j < l; j++) {
                p[j] = now--;
            }
            for (int j = l + 1; j <= r; j++) {
                p[j] = now--;
            }
            ok = 0;
            pre = r;
        } else {
            for (int i = pre + 1; i < l; i++) p[i] = now--;
            pre = r;
            p[r] = now--;
        }
    }
    if (ok) {
        cout << n * (n - 1) / 2 << "\n";
        return;
    }
    now = 1;
    for (int i = n - 1; i >= 0; i--) {
        if (p[i] != -1) continue;
        p[i] = now++;
    }
    int ans = 0;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            ans += p[i] > p[j];
        }
    }
    cout << ans << "\n";

}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>T;
    while(T--){
       solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The key issues in the buggy code include:

1. **Improper Handling of Intervals**: The code attempts to fill the permutation array `p` based on the intervals, but it might not correctly handle the intervals, especially when there are multiple intervals.
2. **Inversion Calculation**: The code calculates the number of inversions in the final permutation, but it might not correctly ensure that the permutation is interesting.

Let's consider a small-scale test input that can expose these issues:

1. **Test Case with No Intervals**: This should output the maximum number of inversions possible, which is `n * (n - 1) / 2`.
2. **Test Case with One Interval**: This should test the handling of a single interval.
3. **Test Case with Multiple Intervals**: This should test the handling of multiple intervals and the correctness of the permutation.

Here is the test input:

```plaintext
3
2 0
5 1
2 4
7 2
1 4
6 8
```

### Explanation:
1. **Test Case 1:**
   - **Input**: `2 0`
   - **Output**: `1`
   - **Explanation**: With no intervals, the maximum number of inversions is 1 (the only possible permutation is [2, 1]).

2. **Test Case 2:**
   - **Input**: `5 1
                 2 4`
   - **Output**: `8`
   - **Explanation**: The interesting permutation with the maximum number of inversions is [5, 2, 4, 3, 1]. The inversions are (5, 2), (5, 4), (5, 3), (5, 1), (2, 1), (4, 3), (4, 1), (3, 1), which totals to 8.

3. **Test Case 3:**
   - **Input**: `7 2
                 1 4
                 6 8`
   - **Output**: `15`
   - **Explanation**: The interesting permutation with the maximum number of inversions is [4, 7, 3, 6, 2, 5, 1]. The inversions are (7, 4), (7, 3), (7, 6), (7, 2), (7, 5), (7, 1), (6, 4), (6, 3), (6, 2), (6, 5), (6, 1), (4, 3), (4, 2), (4, 1), (3, 2), (3, 1), (2, 1), which totals to 15.

This test input should help expose any issues in the handling of intervals and the calculation of inversions in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is an easy version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

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 two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le \frac{n}{2}$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval.

Additional constraint on the input in this version: $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$.

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

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
6
2 0
2 1
1 2
5 1
2 4
8 2
1 4
6 8
7 2
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
21
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 8, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2] = [1, 7]$.

In the fifth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$.

In the sixth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
import sys
from random import getrandbits
input = sys.stdin.readline
RANDOM = getrandbits(32)

class Wrapper(int):
    def __init__(self, x):
        int.__init__(x)
    def __hash__(self):
        return super(Wrapper, self).__hash__() ^ RANDOM

def chkInv(p):
    res, n = 0, len(p)-1
    for i in range(1, n+1):
        for j in range(i+1, n+1):
            if p[i] > p[j]:
                res += 1
    return res

for _ in range(int(input())):
    n, m = map(int, input().split())

    if m == 0:
        p = [0] + [i for i in range(n, 0, -1)]
    else:
        p = [0]*(n+1)
        local = [tuple(map(int, input().split())) for _ in range(m)]
        ls = set()
        rs = []
        for i in range(m):
            ls.add(Wrapper(local[i][0]))
            rs.append(local[i][1])

        ptr = n
        for i in range(1, local[-1][-1]+1):
            if Wrapper(i) in ls:
                continue
            p[i] = ptr
            ptr -= 1

        for i in range(1, n+1):
            if p[i] == 0:
                p[i] = ptr
                ptr -= 1

    res = chkInv(p)

    if m and local[-1][1] == n:
        p = [0]*(n+1)
        ptr = n
        l, r = local[-1]
        for i in range(1, n+1):
            if Wrapper(i) in ls:
                continue
            if l <= i < r:
                continue
            p[i] = ptr
            ptr -= 1

        for i in range(1, n+1):
            if p[i] == 0:
                p[i] = ptr
                ptr -= 1

    ans = chkInv(p)

    print(max(res, 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 given buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The key aspects to consider are:

1. **Intervals that cover the entire permutation**:
   - This can cause issues if the code does not handle the case where the last interval ends at `n`.

2. **Intervals that have overlapping constraints**:
   - Although the problem constraints state that `r_i < l_{i + 1}` for each `i` from `1` to `m - 1`, we should still consider how the code handles such constraints.

3. **Intervals that are very close to each other**:
   - This can cause issues in the permutation generation logic.

4. **Intervals that are at the boundaries**:
   - Intervals starting at `1` or ending at `n` can be particularly tricky.

Given these considerations, let's create a test case that includes these edge cases:

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

### Explanation:
1. **Test Case 1**:
   - `n = 5`, `m = 1`, Interval: `[1, 5]`
   - This interval covers the entire permutation, which can cause issues in the permutation generation logic.

2. **Test Case 2**:
   - `n = 5`, `m = 2`, Intervals: `[1, 3]` and `[4, 5]`
   - These intervals are close to each other and cover the entire permutation, which can expose issues in the permutation generation logic.

3. **Test Case 3**:
   - `n = 8`, `m = 3`, Intervals: `[1, 4]`, `[5, 7]`, and `[7, 8]`
   - These intervals are close to each other and cover the entire permutation, which can expose issues in the permutation generation logic.

This test input should help in identifying the bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is an easy version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

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 two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le \frac{n}{2}$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval.

Additional constraint on the input in this version: $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$.

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

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
6
2 0
2 1
1 2
5 1
2 4
8 2
1 4
6 8
7 2
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
21
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 8, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2] = [1, 7]$.

In the fifth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$.

In the sixth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
import math

for _ in range(int(input())):
    n, m = map(int, input().split())

    a = [list(map(int, input().split()))for _ in range(m)]

    highSpots = set()
    lowSpots = set()

    for l, r in a:
        highSpots.add(r)
        lowSpots.add(l)

        disBeginHigh = r + 1
        disEndLow = n - l + 2

        if disBeginHigh < disEndLow:
            highSpots.update(range(l + 1, r))
        else:
            lowSpots.update(range(l + 1, r))

    exclude = set(range(1, n + 1)) - lowSpots - highSpots

    for i in exclude:
        if i > n // 2:
            lowSpots.add(i)
        else:
            highSpots.add(i)

    result = [0 for _ in range(n)]

    count = 0
    for i in sorted(lowSpots, reverse=True):
        result[i-1] = count
        count += 1

    for i in sorted(highSpots, reverse=True):
        result[i-1] = count
        count += 1

    # print(result)

    segTreeSize = math.ceil(math.log2(n)) + 1
    segTreeInternalSize = 2**segTreeSize
    segTree = [[0 for k in range(2**i)] for i in range(segTreeSize)]

    # print(segTree)

    total = 0
    for i in range(n-1, -1, -1):
        val = result[i]

        left = val
        tempTotal = 0
        for k in range(segTreeSize-1, -1, -1):
            if left % 2 > 0:
                tempTotal += segTree[k][left - 1]
            left = left // 2

        total += tempTotal

        for k in range(segTreeSize):
            segTree[k][val//2**(segTreeSize-k-1)] += 1

        # print(val, tempTotal, segTree)

    print(total)

**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 code, let's consider the following aspects:

1. **Edge Cases**: The code might fail in edge cases where the intervals are very tight or where the intervals do not cover the entire range of the permutation.
2. **Overlapping Intervals**: The code might not handle overlapping intervals correctly, but since the problem constraints specify that $r_i < l_{i + 1}$, we need to focus on other aspects.
3. **Maximizing Inversions**: The code might not correctly calculate the maximum number of inversions, especially when the intervals are positioned in a way that maximizes the inversions.

Given these considerations, let's create a test case where the intervals are positioned to maximize the inversions and to test the code's ability to handle such cases correctly.

```plaintext
6
2 0
2 1
1 2
5 1
2 4
8 2
1 4
6 8
7 2
1 3
4 7
7 3
1 2
3 4
5 6
```

This test input includes:
1. A simple case with no intervals.
2. A case with one interval that covers the entire permutation.
3. A case with one interval that is positioned to maximize inversions.
4. A case with two intervals that are positioned to maximize inversions.
5. A case with multiple intervals that are positioned to maximize inversions.
6. A case with three intervals that are positioned to maximize inversions.

These cases will help in exposing any issues in the code's handling of interval positions and the calculation of maximum inversions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is an easy version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

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 two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le \frac{n}{2}$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval.

Additional constraint on the input in this version: $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$.

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

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
6
2 0
2 1
1 2
5 1
2 4
8 2
1 4
6 8
7 2
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
21
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 8, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2] = [1, 7]$.

In the fifth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$.

In the sixth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
import heapq
import os, sys
from io import BytesIO, IOBase

from collections import defaultdict, deque, Counter
from bisect import bisect_left, bisect_right
from heapq import heappush, heappop
from functools import lru_cache, reduce
from itertools import accumulate
from operator import xor
import random
import math
import sys

# sys.setrecursionlimit(10000)
# Fast IO Region
BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
sys.setrecursionlimit(10000)

mag = random.getrandbits(20)

class D:
    def __init__(self, type=int):
        self.d = defaultdict(type)
        self.mag = random.getrandbits(20)

    def __setitem__(self, key, value):
        self.d[key + self.mag] = value

    def __getitem__(self, key):
        return self.d[key + self.mag]

    def __delitem__(self, key):
        del self.d[key + self.mag]

    def __contains__(self, key):
        return key + self.mag in self.d

    def __iter__(self):
        self.keys = iter(self.d.keys())
        return self

    def __next__(self):
        return next(self.keys) - self.mag

    def __len__(self):
        return len(self.d)

ii = lambda: int(input())
mii = lambda: map(int, input().split())
fii = lambda: map(float, input().split())
lmii = lambda: list(map(int, input().split()))
i2c = lambda n: chr(ord('a') + n)
c2i = lambda c: ord(c) - ord('a')

# mod = 998244353
mod = 10 ** 9 + 7

def solve():
    n, m = mii()
    T = []
    tot = 0
    for _ in range(m):
        x, y = mii()
        x, y = x - 1, y - 1
        T.append([x, y])
        tot += y - x + 1
    y_min = len(T)
    y_max = tot - len(T)
    res = 0
    for tot_y in range(y_min, y_max + 1):
        tot_x = tot - tot_y
        r_y = tot_y
        cx, cy = 0, 0
        r = 0
        L = []
        for i, (a, b) in enumerate(T):
            need_y = len(T) - i
            if r_y == need_y:
                L.append([b - a, 1])
            elif r_y - need_y < b - a:
                L.append([b - a - r_y + need_y, r_y - need_y + 1])
                r_y = 1
            else:
                L.append([1, b - a])
                r_y -= b - a
        left = 0
        cz = 0
        for i in range(len(L)):
            x, y = L[i]
            a, b = T[i]
            z = a - left
            r += z * (z - 1) // 2 + z * (cy + tot_x - cx) + z * max(cx, tot_y - cy) + z * cz
            cz += z
            cx += x
            cy += y
            r += x * (x - 1) // 2 + y * (y - 1) // 2 + x * (tot_x - cx) + y * (tot_x - cx + tot_y - cy)
            left = b + 1
        z = n - left
        r += z * (z - 1) // 2 + z * (cy + tot_x - cx) + z * max(cx, tot_y - cy) + z * cz
        res = max(res, r)

    print(res)

for _ in range(ii()):
    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:
```plaintext
3
2 0
2 1
1 2
5 2
1 3
4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is an easy version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

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 two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le \frac{n}{2}$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval.

Additional constraint on the input in this version: $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$.

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

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
6
2 0
2 1
1 2
5 1
2 4
8 2
1 4
6 8
7 2
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
21
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 8, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2] = [1, 7]$.

In the fifth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$.

In the sixth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
import java.io.*;
import java.util.*;

public class E {

	static IOHandler sc = new IOHandler();
	static PrintWriter out = new PrintWriter(System.out);

	public static void main(String[] args) throws IOException {

        new Thread(null, new Solution(), "solution", 1 << 26 ).start();
    }

	private static class Solution implements Runnable {

		public void run() {
			// TODO Auto-generated method stub
			int testCases = sc.nextInt();

			for (int i = 1; i <= testCases; ++i) {
				solve(i);
			}

			out.flush();
		}

		private void advice() {
			// Read question and Test cases thoroughly
			// Check for overflow , Avoid Integer ==
			// Handle invalid cases
			// Think before solving
			// Search for concepts
			// Double check Constraints
			// expected alg like seg tree, trie, binSearch, sqrt decom ...

			// Sample Intuitions
			// Hashing, Inout Red, binary partition

			// Source of Error
			// Int overflow, Not returning after special case
			// Not taking mod properly
			// sorting input groups after reading certain values
			// Not using the right vairable eg idx in method for i in loop
			// Accounting for special case in general logic. eg special in dp
			// for BIT. Do not use tree maps with update query
			// Assuming Constraints on an unbound variable
			// Not properly calculating time complexity
		}

		private void solve(int t) {
			int n = sc.nextInt();

			int m = sc.nextInt();

			int [][] arr = new int [m][];

			boolean [] visited = new boolean [n + 1];

			boolean [] skipLeft = new boolean [n + 1];
			boolean [] skipRight = new boolean [n + 1];

			for (int i = 0; i < m; ++i) {
				arr[i] = sc.readArray(2);

				// System.out.println(Arrays.toString(arr[i]));
				skipRight[arr[i][1]] = true;
				skipLeft[arr[i][0]] = true;
			}

			long result = 0;

			for (int i = n + 1; i > 0; --i) {
				result = Math.max(result, solve(n, skipLeft, skipRight, i));
			}

			out.println(result);
		}

		private long solve(int n, boolean [] left, boolean [] right, int stop) {

			int low = 1;
			int high = n;

			int [] arr = new int [n + 1];

			for (int i = n; i >= stop; --i) {
				if (right[i])
					continue;

				arr[i] = low++;
			}

			for (int i = 1; i <= n; ++i) {
				if (arr[i] != 0)
					continue;
				else if (left[i])
					arr[i] = low++;
				else
					arr[i] = high--;
			}

			BIT bit = new BIT(n + 2);

			long result = 0;

			for (int i = 1; i <= n; ++i) {
				result += bit.get(arr[i]);
				bit.update(arr[i], 1);
			}

			return result;
		}

	}

	private static class BIT {

		long [] arr;

		BIT(int n){
			arr = new long [n + 1];
		}

		void update (int idx, long val) {
			for (int i = idx; i > 0; i -= (i & -i)) {
				arr[i] += val;
			}
		}

		long get(int idx) {
			long result = 0;

			for (int i = idx; i < arr.length; i += (i & -i)) {
				result += arr[i];
			}

			return result;
		}
	}

	private static class IOHandler {
      BufferedReader br;
      StringTokenizer st;

      public IOHandler() {
         br = new BufferedReader(new InputStreamReader(System.in));
      }

      String next() {
          while (st == null || !st.hasMoreElements()) {
              try {
                  st = new StringTokenizer(br.readLine());
              } catch (IOException e) {
                  e.printStackTrace();
              }
          }
          return st.nextToken();
      }

      int nextInt() {
          return Integer.parseInt(next());
      }

      int [] readArray(int n) {
    	  int [] res = new int [n];

    	  for (int i = 0; i < n; ++i)
    		  res[i] = nextInt();

    	  return res;
      }

      int [] readArray2(int n) {
    	  int [] res = new int [n + 1];

    	  for (int i = 1; i <= n; ++i)
    		  res[i] = nextInt();

    	  return res;
      }

      int [][] readGraph(int n, int m, int c) {
    	  int [][] graph = new int [n + 1][];

    	  int [] count = new int [n + 1];
    	  int [][] arr = new int [m][1 + c];

    	  int a, b;

    	  for (int i = 0; i < m; ++i) {
    		  a = sc.nextInt();
    		  b = sc.nextInt();

    		  arr[i][0] = a;
    		  arr[i][1] = b;

    		  ++count[a];
    		  ++count[b];

    		  for (int j = 2; j <= c; ++j) {
    			  arr[i][j] = sc.nextInt();
    		  }
    	  }

    	  for (int i = 0; i <= n; ++i) {
    		  graph[i] = new int [count[i] * c];
    		  count[i] = 0;
    	  }
    	  int swap;

    	  for (int [] tArr : arr) {
    		  for (int j = 1; j < tArr.length; ++j) {
    			  graph[tArr[0]][count[tArr[0]]++] = tArr[j];
    		  }

    		  swap = tArr[0];
    		  tArr[0] = tArr[1];
    		  tArr[1] = swap;

    		  for (int j = 1; j < tArr.length; ++j) {
    			  graph[tArr[0]][count[tArr[0]]++] = tArr[j];
    		  }
    	  }

    	  return graph;
      }

      long nextLong() {
          return Long.parseLong(next());
      }

      double nextDouble() {
          return Double.parseDouble(next());
      }

      String nextLine(){
          String str = "";
		  try {
		     str = br.readLine();
		  } catch (IOException e) {
		     e.printStackTrace();
		  }
		  return str;
      }

   }

}

**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 code, we need to focus on edge cases and scenarios that can highlight issues in the logic. The code seems to be handling the construction of permutations and calculating the maximum number of inversions, but there are potential issues in how it manages the intervals and the placement of elements in the permutation.

### Step-by-Step Analysis

1. **Constraints and Edge Cases:**
   - The code should handle cases where the number of intervals \( m \) is zero.
   - The code should handle cases where the intervals are very small or very large.
   - The code should handle cases where the intervals are tightly packed or widely spaced.

2. **Key Points to Test:**
   - **No Intervals:** The code should correctly handle the case where \( m = 0 \).
   - **Single Interval:** The code should handle a single interval correctly.
   - **Multiple Intervals:** The code should handle multiple intervals, especially when they are tightly packed.
   - **Boundary Conditions:** The code should handle boundary conditions for \( l_i \) and \( r_i \).

3. **Challenging Scenarios:**
   - **Tightly Packed Intervals:** Intervals that are very close to each other.
   - **Widely Spaced Intervals:** Intervals that are far apart.
   - **Intervals at Boundaries:** Intervals that start or end at the boundaries of the permutation.

### Challenging Test Input

Let's create a test input that includes a mix of the above scenarios:

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

### Explanation

1. **Test Case 1:**
   - \( n = 5 \)
   - \( m = 2 \)
   - Intervals: \([1, 3]\) and \([4, 5]\)
   - This test case checks how the code handles tightly packed intervals.

2. **Test Case 2:**
   - \( n = 6 \)
   - \( m = 1 \)
   - Interval: \([1, 6]\)
   - This test case checks how the code handles a single interval that spans the entire permutation.

3. **Test Case 3:**
   - \( n = 8 \)
   - \( m = 3 \)
   - Intervals: \([1, 2]\), \([3, 4]\), and \([6, 8]\)
   - This test case checks how the code handles multiple intervals, including some that are widely spaced.

This test input is designed to cover a range of scenarios that can help expose potential bugs in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is an easy version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

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 two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le \frac{n}{2}$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval.

Additional constraint on the input in this version: $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$.

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

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
6
2 0
2 1
1 2
5 1
2 4
8 2
1 4
6 8
7 2
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
21
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 8, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2] = [1, 7]$.

In the fifth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$.

In the sixth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<unordered_map>
#include<vector>
#include<bitset>
#include<queue>
#include<set>
#include<map>
#include<ctime>
#include<random>
#include<numeric>
#include<assert.h>
using namespace std;
#define int long long
#define ll long long
#define ull unsigned long long
#define lc (x<<1)
#define rc (x<<1|1)
#define pii pair<int,int>
#define mkp make_pair
#define fi first
#define se second
#define cout (cerr<<">> ",cout)
const int Mx=5005,p=998244353;
int read(){
	char ch=getchar();
	int Alice=0,Aya=1;
	while(ch<'0'||ch>'9'){
		if(ch=='-') Aya=-Aya;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9')
		Alice=(Alice<<3)+(Alice<<1)+(ch^48),ch=getchar();
	return (Aya==1?Alice:-Alice);
}
int n,m;
int c[Mx];
int f[Mx][Mx][2];
void Solve(){
	n=read(),m=read();
	for(int i=1,l,r;i<=m;i++){
		l=read(),r=read();
		for(int j=l;j<=r;j++) c[j]=i;
	}
	int tot=m+1;
	for(int i=1;i<=n;i++) if(!c[i]) c[i]=tot++;
	for(int i=0;i<=n;i++){
		for(int j=0;j<=n;j++){
			f[i][j][0]=f[i][j][1]=-1e18;
		}
	}
	if(c[1]>m) f[1][1][0]=(n-1)*(n-2)/2,f[1][0][1]=n*(n-1)/2;
	else f[1][1][0]=(n-1)*(n-2)/2;
	c[n+1]=0;
	for(int i=2;i<=n;i++){
		for(int j=0;j<=i;j++){
			if(c[i]==c[i-1]&&c[i]==c[i+1]){
				if(j>0) f[i][j][0]=f[i-1][j-1][0]+(i-j)+(j-1)-(n-j);
				f[i][j][1]=max(f[i-1][j][0],(j>0?f[i-1][j][1]:(int)-1e18));
			}
			else if(c[i]!=c[i-1]&&c[i]==c[i+1]){
				if(j>0) f[i][j][0]=max(f[i-1][j-1][0]+(i-j)+(j-1)-(n-j),f[i-1][j-1][1]+(i-j)+(j-1)-(n-j));
			}
			else if(c[i]==c[i-1]&&c[i]!=c[i+1]){
				f[i][j][1]=max(f[i-1][j][0],(j>0?f[i-1][j][1]:(int)-1e18));
			}
			else if(c[i]!=c[i-1]&&c[i]!=c[i+1]){
				if(j>0) f[i][j][0]=max(f[i-1][j-1][0]+(i-j)+(j-1)-(n-j),f[i-1][j-1][1]+(i-j)+(j-1)-(n-j));
				f[i][j][1]=max(f[i-1][j][0],(j>0?f[i-1][j][1]:(int)-1e18));
			}
		}
	}
	int ans=-1e18;
	for(int j=0;j<=n;j++){
		ans=max(ans,max(f[n][j][0],f[n][j][1]));
		c[j]=0;
	}
	cout<<ans<<endl;

}
signed main(){
	int T=read();
	while(T--) Solve();
	return 0;
}
/*
Hello!!
Sample:

-------------------
*/

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is an easy version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

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 two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le \frac{n}{2}$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval.

Additional constraint on the input in this version: $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$.

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

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
6
2 0
2 1
1 2
5 1
2 4
8 2
1 4
6 8
7 2
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
21
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 8, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2] = [1, 7]$.

In the fifth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$.

In the sixth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
#include <bits/stdc++.h>

using namespace std;

int l[2505], r[2505];

int Ans[5005];

int main()
{
	int T;
	scanf("%d", &T);
	while (T --)
	  {
	  	int n, m;
	  	scanf("%d%d", &n, &m);
	  	for (int i = 1 ; i <= m ; ++ i)
	  	  scanf("%d%d", l + i, r + i);
	  	for (int i = 1 ; i <= n ; ++ i)
	  	  Ans[i] = 0;
	  	int x = n;
	  	for (int i = 1 ; i < l[1] ; ++ i)
	  	  Ans[i] = x --;
	  	for (int i = l[1] + 1 ; i <= r[1] ; ++ i)
	  	  Ans[i] = x --;
	  	for (int i = 2 ; i <= m ; ++ i)
	  	  Ans[r[i]] = x --;
	  	for (int i = 1 ; i <= n ; ++ i)
	  	  if (!Ans[i])
	  	    Ans[i] = x --;
		int R = 0;
		for (int i = 2 ; i <= n ; ++ i)
		  for (int j = 1 ; j < i ; ++ j)
		    if (Ans[j] > Ans[i])
		      ++ R;
		printf("%d\n", R);
	  }
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability. Here are the steps to think through this:

1. **Understand the Bug**: The provided code seems to have issues with how it handles the intervals and the permutation generation. Specifically, it might not correctly place elements to ensure the condition \(\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i\) holds.

2. **Edge Cases**:
   - **No Intervals**: The code should handle cases where there are no intervals correctly.
   - **Single Interval**: Test cases with a single interval to see if the code correctly maximizes inversions.
   - **Multiple Disjoint Intervals**: Test cases with multiple intervals to ensure the code handles the placement of elements correctly.
   - **Boundary Conditions**: Test cases with intervals that are at the boundaries of the permutation length.

3. **Challenging Scenarios**:
   - **Overlapping Intervals**: Although the problem specifies that \(r_i < l_{i + 1}\), we can still test the code with intervals that are very close to each other to see if it handles the placement correctly.
   - **Large Permutations**: Test cases with larger \(n\) values to ensure the code doesn't have performance issues.

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

```plaintext
4
2 0
2 1
1 2
5 1
2 4
8 2
1 4
6 8
```

This test input includes:
1. A case with no intervals.
2. A case with a single interval.
3. A case with multiple intervals that are close to each other.
4. A case with multiple intervals to ensure the code handles the placement of elements correctly.

This should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is an easy version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

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 two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le \frac{n}{2}$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval.

Additional constraint on the input in this version: $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$.

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

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
6
2 0
2 1
1 2
5 1
2 4
8 2
1 4
6 8
7 2
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
21
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 8, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2] = [1, 7]$.

In the fifth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$.

In the sixth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

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

void solve() {
    ll ans = 0;
    int n, m;
    cin >> n >> m;
    if (m == 0) {
        cout << (ll)n * (n - 1)/2 << endl;
        return;
    }
    vector<pair<int, int>> a;
    vector<int> ranges;
    for(int i = 0; i < m; i ++) {
        int x, y;
        cin >> x >> y;
        a.emplace_back(x, y);
        ranges.push_back(y - x + 1);
    }
    vector<int> suf(m + 1, 0);
    for(int i = m - 1; i >= 0; i --) {
        suf[i] = suf[i + 1] + ranges[i] - 1;
    }
    for(int i = m; i <= suf[0]; i ++) {
        vector<int> k;
        int remains = i;
        for(int j = 0; j < m; j ++) {
            int x = max(remains - suf[j + 1], 1);
            k.push_back(x);
            remains -= x;
        }
        remains = i;
        ll now = 0;
        int lst = 0;
        for(int j = 0; j < m; j ++) {
            int x, y;
            x = a[j].first - 1 - lst;
            y = n - a[j].first + 1;
            now += (ll)x * y + (ll)x * (x - 1)/2;

            x = k[j];
            y = ranges[j] - x;
            now += (ll)x * (x - 1)/2 + (ll)y * (y - 1)/2;
            now += (ll)y * (n - a[j].second);

            remains -= x;
            now += (ll)x * remains;

            lst = a[j].second;
        }
        int x = n - lst;
        now += (ll)x * (x - 1)/2;
        now += (ll)i * x;
        ans = max(ans, now);
    }

    cout << ans << endl;
}

int main() {
    // freopen("input.txt", "r", stdin);
    ios::sync_with_stdio(false);
    std::cin.tie(nullptr);

    int t; cin >> t;
    for(;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 that can cause the algorithm to fail. Here are the steps to think through this:

1. **Understand the Bug**: The buggy code might have issues with handling edge cases, especially when the intervals are very tight or when there are no intervals at all.

2. **Edge Case 1**: No intervals (i.e., \( m = 0 \)). This should be straightforward and the code should handle it correctly by outputting the maximum number of inversions for a permutation of length \( n \).

3. **Edge Case 2**: Single interval that covers the entire permutation. This should also be straightforward, but we need to ensure the code handles it correctly.

4. **Edge Case 3**: Multiple intervals that are very tight and close to each other. This can cause issues in the logic that calculates the maximum number of inversions.

5. **Edge Case 4**: Intervals that are very large, covering a significant portion of the permutation. This can also expose issues in the logic.

6. **Edge Case 5**: Intervals that are very small and disjoint. This can test the code's ability to handle small ranges and ensure the correct calculation of inversions.

Given these considerations, let's create a small-scale test input that includes a mix of these edge cases:

```plaintext
4
5 0
5 1
1 5
8 2
1 2
7 8
10 3
1 2
3 4
5 6
```

This test input includes:
- A test case with no intervals (5 0).
- A test case with a single interval that covers the entire permutation (5 1 1 5).
- A test case with two intervals that are tight and close to each other (8 2 1 2 7 8).
- A test case with three small, disjoint intervals (10 3 1 2 3 4 5 6).

This should be a good mix to challenge the buggy code and expose any potential issues.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Inversions (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is an easy version of this problem. The differences between the versions are the constraint on $m$ and $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$ in the easy version. You can make hacks only if both versions of the problem are solved.

Turtle gives you $m$ intervals $[l_1, r_1], [l_2, r_2], \ldots, [l_m, r_m]$. He thinks that a permutation $p$ is interesting if there exists an integer $k_i$ for every interval ($l_i \le k_i < r_i$), and if he lets $a_i = \max\limits_{j = l_i}^{k_i} p_j, b_i = \min\limits_{j = k_i + 1}^{r_i} p_j$ for every integer $i$ from $1$ to $m$, the following condition holds:

$$\max\limits_{i = 1}^m a_i < \min\limits_{i = 1}^m b_i$$

Turtle wants you to calculate the maximum number of inversions of all interesting permutations of length $n$, or tell him if there is no interesting permutation.

An inversion of a permutation $p$ is a pair of integers $(i, j)$ ($1 \le i < j \le n$) such that $p_i > p_j$.

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 two integers $n, m$ ($2 \le n \le 5 \cdot 10^3, 0 \le m \le \frac{n}{2}$) — the length of the permutation and the number of intervals.

The $i$-th of the following $m$ lines contains two integers $l_i, r_i$ ($1 \le l_i < r_i \le n$) — the $i$-th interval.

Additional constraint on the input in this version: $r_i < l_{i + 1}$ holds for each $i$ from $1$ to $m - 1$.

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

Output: For each test case, if there is no interesting permutation, output a single integer $-1$.

Otherwise, output a single integer — the maximum number of inversions.

Examples:
input:
6
2 0
2 1
1 2
5 1
2 4
8 2
1 4
6 8
7 2
1 3
4 7
7 3
1 2
3 4
5 6
output:
1
0
8
21
15
15

Note: In the third test case, the interesting permutation with the maximum number of inversions is $[5, 2, 4, 3, 1]$.

In the fourth test case, the interesting permutation with the maximum number of inversions is $[4, 8, 7, 6, 3, 2, 1, 5]$. In this case, we can let $[k_1, k_2] = [1, 7]$.

In the fifth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 6, 3, 2, 1, 5]$.

In the sixth test case, the interesting permutation with the maximum number of inversions is $[4, 7, 3, 6, 2, 5, 1]$.

Buggy Code:
import heapq
import os, sys
from io import BytesIO, IOBase

from collections import defaultdict, deque, Counter
from bisect import bisect_left, bisect_right
from heapq import heappush, heappop
from functools import lru_cache, reduce
from itertools import accumulate
from operator import xor
import random
import math
import sys

# sys.setrecursionlimit(10000)
# Fast IO Region
BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
sys.setrecursionlimit(10000)

mag = random.getrandbits(20)

class D:
    def __init__(self, type=int):
        self.d = defaultdict(type)
        self.mag = random.getrandbits(20)

    def __setitem__(self, key, value):
        self.d[key + self.mag] = value

    def __getitem__(self, key):
        return self.d[key + self.mag]

    def __delitem__(self, key):
        del self.d[key + self.mag]

    def __contains__(self, key):
        return key + self.mag in self.d

    def __iter__(self):
        self.keys = iter(self.d.keys())
        return self

    def __next__(self):
        return next(self.keys) - self.mag

    def __len__(self):
        return len(self.d)

ii = lambda: int(input())
mii = lambda: map(int, input().split())
fii = lambda: map(float, input().split())
lmii = lambda: list(map(int, input().split()))
i2c = lambda n: chr(ord('a') + n)
c2i = lambda c: ord(c) - ord('a')

# mod = 998244353
mod = 10 ** 9 + 7

def solve():
    n, m = mii()
    T = []
    tot = 0
    for _ in range(m):
        x, y = mii()
        x, y = x - 1, y - 1
        T.append([x, y])
        tot += y - x + 1
    y_min = len(T)
    y_max = tot - len(T)
    res = 0
    if len(T) == 0:
        print(n * (n - 1) // 2)
        return
    for tot_y in range(y_min, y_max + 1):
        tot_x = tot - tot_y
        r_y = tot_y
        cx, cy = 0, 0
        r = 0
        L = []
        for i, (a, b) in enumerate(T):
            need_y = len(T) - i
            if r_y == need_y:
                L.append([b - a, 1])
            elif r_y - need_y < b - a:
                L.append([b - a - r_y + need_y, r_y - need_y + 1])
                r_y = 1
            else:
                L.append([1, b - a])
                r_y -= b - a
        left = 0
        cz = 0
        for i in range(len(L)):
            x, y = L[i]
            a, b = T[i]
            z = a - left
            r += z * (z - 1) // 2 + z * (cy + tot_x - cx) + z * max(cx, tot_y - cy) + z * cz
            cz += z
            cx += x
            cy += y
            r += x * (x - 1) // 2 + y * (y - 1) // 2 + x * (tot_x - cx) + y * (tot_x - cx + tot_y - cy)
            left = b + 1
        z = n - left
        r += z * (z - 1) // 2 + z * (cy + tot_x - cx) + z * max(cx, tot_y - cy) + z * cz
        res = max(res, r)

    print(res)

for _ in range(ii()):
    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 given buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. Here are the steps to think through this:

1. **Understand the Constraints and Requirements:**
   - The intervals must satisfy \( r_i < l_{i + 1} \).
   - The intervals are used to define the interesting permutations.
   - The goal is to maximize the number of inversions in an interesting permutation.

2. **Identify Potential Bugs:**
   - The code might not handle cases where the intervals are very small or overlap in a specific way.
   - The code might have issues with edge cases where the number of intervals \( m \) is zero or very large.
   - The code might not correctly handle the calculation of the maximum number of inversions.

3. **Create a Test Case:**
   - We will create a test case with a small number of intervals that are carefully chosen to maximize the failure probability.
   - We will ensure that the intervals are non-overlapping and satisfy the constraint \( r_i < l_{i + 1} \).

4. **Test Case Generation:**
   - Let's create a test case with \( n = 5 \) and \( m = 2 \).
   - We will choose intervals that are close to each other but non-overlapping.

Here is the test input:

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

This test case has:
- \( n = 5 \) (permutation length)
- \( m = 2 \) (number of intervals)
- Intervals: \([1, 2]\) and \([3, 4]\)

This test case is designed to check if the code can handle a small number of intervals and correctly calculate the maximum number of inversions. The intervals are non-overlapping and satisfy the constraint \( r_i < l_{i + 1} \). The expected output for this test case is 6, as the permutation \([3, 4, 1, 2, 5]\) has 6 inversions.