**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $$$n$$$, $$$m$$$.

Calculate the number of ordered pairs $$$(a, b)$$$ satisfying the following conditions:

- $$$1\le a\le n$$$, $$$1\le b\le m$$$;
- $$$b \cdot \gcd(a,b)$$$ is a multiple of $$$a+b$$$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1\le t\le 10^4$$$). The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$, $$$m$$$ ($$$1\le n,m\le 2 \cdot 10^6$$$).

It is guaranteed that neither the sum of $$$n$$$ nor the sum of $$$m$$$ over all test cases exceeds $$$2 \cdot 10^6$$$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $$$(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$$$ satisfy the conditions.

Buggy Code:
#include<bits/stdc++.h>
#pragma GCC optimize("O3")
#define int long long
#define se second
#define fi first
using namespace std;
const int md = 998244353;
const int N = 4e5 + 10;
int us[N] , a[N] , ans;
vector<int> g[N] , fuck;
void dfs(int x){
    us[x] = 1;
    for(auto to : g[x]){
        if(! us[to] and a[to] == 0){
            dfs(to);
        }
        if(a[to] != 0){
            fuck.push_back(a[to]);
        }
    }
}
void solve(){
    int n , m;
    cin >> n >> m;
    int ans = 0;
    for(int i = 1;i * i <= m; ++ i){
        for(int j = 1;(j + i) * i <= n and (i + j) * j <= m; ++ j){
            if(__gcd(i , j) == 1){
               ans += min(n / ((i + j) * i) , m / ((i + j) * j));
            }
        }
    }
    cout << ans;
}
main()
{
	ios_base::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
    int T = 1;
    cin >> T;
    while(T --)
    {
        solve();
        cout << '\n';
    }
}
//test
/*
1
7
0 1 2 3 1 1 1
1 2
1 3
1 4
2 5
3 6
4 7
// ٩ (◕‿◕｡) ۶
*/

**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
2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $$$n$$$, $$$m$$$.

Calculate the number of ordered pairs $$$(a, b)$$$ satisfying the following conditions:

- $$$1\le a\le n$$$, $$$1\le b\le m$$$;
- $$$b \cdot \gcd(a,b)$$$ is a multiple of $$$a+b$$$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1\le t\le 10^4$$$). The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$, $$$m$$$ ($$$1\le n,m\le 2 \cdot 10^6$$$).

It is guaranteed that neither the sum of $$$n$$$ nor the sum of $$$m$$$ over all test cases exceeds $$$2 \cdot 10^6$$$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $$$(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$$$ satisfy the conditions.

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

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vll;
#define umap unordered_map
#define uset unordered_set
#define forn(i, n) for (int i = 0; i < (n); ++i)
#define forn1(i, n) for (int i = 1; i <= (n); ++i)
#define rforn(i, n) for (int i = (n) - 1; i >= 0; --i)
#define rforn1(i, n) for (int i = (n); i > 0; --i)
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define len(x) (int) x.size()
#define LSB(x) ((x) & -(x))
inline int MSB(int x) { return x ? 1 << (31 - __builtin_clz(x)) : 0; }
#define pb push_back
#define eb emplace_back
#define fi first
#define se second

template <typename... Ts>
inline void print(const Ts&... args) {
    const char* sep = "";
    (((cout << sep << args), sep = " "), ...);
    cout << "\n";
}
template <typename T>
inline void print(const vector<T>& v) {
    for (auto x : v) cout << x << " ";
    cout << "\n";
}
template <typename... Ts>
inline void printerr(const Ts&... args) {
    const char* sep = "";
    cerr << "\u001B[31m";
    (((cerr << sep << args), sep = " "), ...);
    cerr << "\u001B[0m" << endl;
}
template <typename T>
inline void printerr(const vector<T>& v) {
    cerr << "\u001B[31m";
    for (auto x : v) cerr << x << " ";
    cerr << "\u001B[0m" << endl;
}

const double EPS = 1e-9;
const int INF = 1e9;
const int MAXN = 3e5;

void solve() {
    int n, m; cin >> n >> m;
    ll t = min(n, m) / 2;
    forn1(p, min(n, 2002)) {
        forn1(q, p-1) if (gcd(p, q) == 1) {
            int r = p + q;
            t += min(n / (p*r), m / (q*r));
            t += min(n / (q*r), m / (p*r));
        }
    }
    print(t);
}

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

    int tc; cin >> tc;
    while (tc--) {
        solve();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
2002 2002
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $$$n$$$, $$$m$$$.

Calculate the number of ordered pairs $$$(a, b)$$$ satisfying the following conditions:

- $$$1\le a\le n$$$, $$$1\le b\le m$$$;
- $$$b \cdot \gcd(a,b)$$$ is a multiple of $$$a+b$$$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1\le t\le 10^4$$$). The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$, $$$m$$$ ($$$1\le n,m\le 2 \cdot 10^6$$$).

It is guaranteed that neither the sum of $$$n$$$ nor the sum of $$$m$$$ over all test cases exceeds $$$2 \cdot 10^6$$$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $$$(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$$$ satisfy the conditions.

Buggy Code:
from sys import *
input = stdin.readline
inputlist = lambda: map(int, stdin.readline().split())
import math

for _ in range(int(input())):
    n, m = inputlist()
    ans = 0
    for i in range(2, max(n, m)):
        maxa = n//i
        minb = max(1, i - maxa)
        maxb = min(m//i, i-1)
        for b in range(minb, maxb+1):
            a = i-b
            if math.gcd(a,b) != 1: continue
            ans += min(n//(i*a), m//(i*b))
    print(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 maximize the failure probability. The bug in the code seems to be related to the way it handles the calculation of valid pairs \((a, b)\) and the conditions under which it skips certain pairs.

Here are the steps to think through the problem:

1. **Edge Cases**:
   - Small values of \(n\) and \(m\) to ensure the test input is small-scale.
   - Values of \(n\) and \(m\) that are equal to each other.
   - Values of \(n\) and \(m\) that are prime numbers.
   - Values of \(n\) and \(m\) that are powers of small integers.

2. **Critical Conditions**:
   - Pairs \((a, b)\) where \(a\) and \(b\) are both small and their GCD is 1.
   - Pairs \((a, b)\) where \(a\) and \(b\) are both large and their GCD is greater than 1.
   - Pairs \((a, b)\) where \(a + b\) is a multiple of \(b \cdot \gcd(a, b)\).

3. **Input Construction**:
   - Use a small number of test cases to ensure the input is small-scale.
   - Include a mix of the above edge cases.

Given these considerations, a challenging test input that exposes the bug in the buggy code could be:

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

This input includes small values of \(n\) and \(m\) that are equal to each other, which can help expose issues with the way the code handles pairs and their GCDs. The pairs \((2, 2)\), \((3, 3)\), \((4, 4)\), and \((5, 5)\) are particularly interesting because they involve small numbers where the GCD conditions are more likely to be met.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $$$n$$$, $$$m$$$.

Calculate the number of ordered pairs $$$(a, b)$$$ satisfying the following conditions:

- $$$1\le a\le n$$$, $$$1\le b\le m$$$;
- $$$b \cdot \gcd(a,b)$$$ is a multiple of $$$a+b$$$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1\le t\le 10^4$$$). The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$, $$$m$$$ ($$$1\le n,m\le 2 \cdot 10^6$$$).

It is guaranteed that neither the sum of $$$n$$$ nor the sum of $$$m$$$ over all test cases exceeds $$$2 \cdot 10^6$$$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $$$(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$$$ satisfy the conditions.

Buggy Code:
import java.util.Scanner;
import static java.lang.Math.*;

public class Main {
    static Scanner in = new Scanner(System.in);

    static int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }

    static int N = (int) sqrt(2e6);
    static boolean[][] a = new boolean[N + 1][N + 1];

    static void sol() {
        int n, m;
        n = in.nextInt();
        m = in.nextInt();
        long ans = 0;
        for (int p = 1; p < sqrt(2e6); p++) {
            for (int q = 1; q < sqrt(2e6); q++)
                if (a[p][q]) {
                    ans += min(n / p, m / q) / (p + q);
                }
        }
        System.out.println(ans);
    }

    public static void main(String[] args) {
        int t = in.nextInt();
        for (int p = 1; p < sqrt(2e6); p++)
            for (int q = 1; q < sqrt(2e6); q++)
                a[p][q] = gcd(p, q) == 1;
        while (t-- > 0)
            sol();
    }
}

**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
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
```

### Explanation:
1. **Edge Case: (1, 1)**
   - This is the smallest possible input. No pairs satisfy the condition, so the expected output is 0.

2. **Small Input: (2, 3)**
   - This is a small input to check basic functionality. The only valid pair is (2, 3), so the expected output is 1.

3. **Small Input: (3, 5)**
   - Another small input to ensure the algorithm handles different values. The only valid pair is (3, 5), so the expected output is 1.

4. **Medium Input: (10, 8)**
   - A medium-sized input to test a larger range of values. The valid pairs are (2, 2), (3, 6), (4, 4), (6, 3), (6, 6), and (8, 8), so the expected output is 6.

5. **Large Input: (100, 1233)**
   - A larger input to test the algorithm's performance and correctness. The expected output is 423, which is a non-trivial number of valid pairs.

6. **Very Large Input: (1000000, 1145141)**
   - A very large input to test the algorithm's performance and correctness under extreme conditions. The expected output is 5933961, which is a large number of valid pairs.

### Why These Inputs:
- **Edge Cases**: The first test case (1, 1) is an edge case where no pairs satisfy the condition.
- **Small Inputs**: The second and third test cases (2, 3) and (3, 5) are small inputs to ensure basic functionality.
- **Medium Input**: The fourth test case (10, 8) is a medium-sized input to test a larger range of values.
- **Large Inputs**: The fifth and sixth test cases (100, 1233) and (1000000, 1145141) are large inputs to test performance and correctness under more challenging conditions.

These inputs are designed to expose any bugs in the algorithm, especially in handling edge cases and large inputs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Field Trip
time_limit_ms: 2000
memory_limit_mb: 256
Description: Even in university, students need to relax. That is why Sakurakos teacher decided to go on a field trip. It is known that all of the students will be walking in one line. The student with index $i$ has some topic of interest which is described as $a_i$. As a teacher, you want to minimise the disturbance of the line of students.

The disturbance of the line is defined as the number of neighbouring people with the same topic of interest. In other words, disturbance is the number of indices $j$ ($1 \le j < n$) such that $a_j = a_{j + 1}$.

In order to do this, you can choose index $i$ ($1\le i\le n$) and swap students at positions $i$ and $n-i+1$. You can perform any number of swaps.

Your task is to determine the minimal amount of disturbance that you can achieve by doing the operation described above any number of times.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($2 \le n \le 10^5$) — the length of the line of students.    * The second line contains $n$ integers $a_i$ ($1\le a_i\le n$) — the topics of interest of students in line.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output the minimal possible disturbance of the line that you can achieve.

Examples:
input:
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3
output:
1
2
1
0
0
1
1
0
2

Note: In the first example, it is necessary to apply the operation to $i=2$, thus the array will become $[1, \textbf{2}, 1, \textbf{1}, 3]$, with the bold elements indicating those that have swapped places. The disturbance of this array is equal to $1$.

In the fourth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[2, 1, \textbf{2}, \textbf{1}, 2, 4]$. The disturbance of this array is equal to $0$.

In the eighth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[1, 4, \textbf{1}, 5, \textbf{3}, 1, 3]$. The disturbance of this array is equal to $0$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define all(a) a.begin(), a.end()
#define nline '\n'
#define debug(x) cout  << #x << ":" << x << nline;
/* ordered set / mutliset pbds  */
/*comparators less<int>, greater<int>, less_equal<int>, etc*/
/*functions like find_by_order, order_of_key, etc*/
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
vector<bool>seive(int N) { vector<bool>vec(N+1,true); for(int i = 2; i <= N; i++) { if(vec[i]) { for(int j = 2; j*i <= N; j++){ vec[i*j]= false; } } } return vec; }
const int mx = 1e5;

int help(int a, int b)
{
    if(a == b)
    {
        return 1;
    }
    return 0;
}

void solve(){
    int n; cin >> n;
    vector<int>a(n+1);
    for(int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    vector<vector<int>>dp(n+1, vector<int>(2,0));
    // dp[i][0] --> dont dwap a[i];  ans when
    //dp[i][1] --> swap a[i]; ans when

    int s;
    if(n%2==1)
    {
        dp[n/2 + 1][0] = 0;
        dp[n/2 + 1][1] = 0;
        s = n/2 + 1;
    }
    else
    {
        s = n/2;
        if(a[n/2] != a[n/2 + 1])
        {
            dp[n/2][0] = 0;
            dp[n/2][1] = 0;
        }
        else
        {
            dp[n/2][0] = 1;
            dp[n/2][1] = 1;
        }
    }

    for(int i = s-1; i >=1; i--)
    {
        // if swap a[i], a[n-i+1]
        // swap a[i+1], a[n - i];
        int a1 = a[i], a2 = a[i+1], a4 = a[n-i+1], a3 = a[n-i];
        int ans1 = help(a2,a4) + help(a1,a3) + dp[i+1][0];
        int ans2 = help(a1, a2) + help(a3,a4) + dp[i+1][0];
        int ans3 = help(a1,a3) + help(a2,a4) + dp[i+1][1];
        int ans4 = help(a2, a4) + help(a1,a3) + dp[i+1][1];
        dp[i][1] = min(ans2,ans3);
        dp[i][0] = min(ans1,ans4);
    }

    cout << min(dp[1][0], dp[1][1]) << nline;

};

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);cout.tie(0);
    int t; cin >> t;
    while(t--) {
        solve();
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5
1 1 1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Field Trip
time_limit_ms: 2000
memory_limit_mb: 256
Description: Even in university, students need to relax. That is why Sakurakos teacher decided to go on a field trip. It is known that all of the students will be walking in one line. The student with index $i$ has some topic of interest which is described as $a_i$. As a teacher, you want to minimise the disturbance of the line of students.

The disturbance of the line is defined as the number of neighbouring people with the same topic of interest. In other words, disturbance is the number of indices $j$ ($1 \le j < n$) such that $a_j = a_{j + 1}$.

In order to do this, you can choose index $i$ ($1\le i\le n$) and swap students at positions $i$ and $n-i+1$. You can perform any number of swaps.

Your task is to determine the minimal amount of disturbance that you can achieve by doing the operation described above any number of times.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($2 \le n \le 10^5$) — the length of the line of students.    * The second line contains $n$ integers $a_i$ ($1\le a_i\le n$) — the topics of interest of students in line.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output the minimal possible disturbance of the line that you can achieve.

Examples:
input:
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3
output:
1
2
1
0
0
1
1
0
2

Note: In the first example, it is necessary to apply the operation to $i=2$, thus the array will become $[1, \textbf{2}, 1, \textbf{1}, 3]$, with the bold elements indicating those that have swapped places. The disturbance of this array is equal to $1$.

In the fourth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[2, 1, \textbf{2}, \textbf{1}, 2, 4]$. The disturbance of this array is equal to $0$.

In the eighth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[1, 4, \textbf{1}, 5, \textbf{3}, 1, 3]$. The disturbance of this array is equal to $0$.

Buggy Code:
#include <iostream>
#include <bits/stdc++.h>
// #include <sys/resource.h>
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
using namespace std;
// using namespace chrono;
// using namespace __gnu_pbds;

// def
#define Ankit 1

//Speed
#define Code ios_base::sync_with_stdio(false);
#define By cin.tie(NULL);
#define Asquare cout.tie(NULL);

//Aliases
using ll= long long;
using lld= long double;
using ull= unsigned long long;

//Constants
const lld pi= 3.141592653589793238;
const ll INF= LONG_LONG_MAX;
const ll mod=1e9+7;

//TypeDEf
typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef vector<pll> vpll;
typedef vector<string> vs;
typedef unordered_map<ll,ll> umll;
typedef map<ll,ll> mll;

// Macros
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define fl(i,n) for(int i=0;i<n;i++)
#define rl(i,m,n) for(int i=n;i>=m;i--)
#define py cout<<"YES\n";
#define pm cout<<"-1\n";
#define pn cout<<"NO\n";
#define all(v) v.begin(),v.end()
#define rv(v) v.end(),v.begin()

//Debug
#ifdef Ankit
#define debug(x) cerr << #x<<" "; cerr<<x<<" "; cerr << endl;
#else
#define debug(x);
#endif

// Operator overloads
template<typename T1, typename T2> // cin >> pair<T1, T2>
istream& operator>>(istream &istream, pair<T1, T2> &p) { return (istream >> p.first >> p.second); }
template<typename T> // cin >> vector<T>
istream& operator>>(istream &istream, vector<T> &v){for (auto &it : v)cin >> it;return istream;}
template<typename T1, typename T2> // cout << pair<T1, T2>
ostream& operator<<(ostream &ostream, const pair<T1, T2> &p) { return (ostream << p.first << " " << p.second); }
template<typename T> // cout << vector<T>
ostream& operator<<(ostream &ostream, const vector<T> &c) { for (auto &it : c) cout << it << " "; return ostream; }

// Utility functions
template <typename T>
void print(T &&t)  { cout << t << "\n"; }
void printarr(ll arr[], ll n){fl(i,n) cout << arr[i] << " ";cout << "\n";}
template<typename T>
void printvec(vector<T>v){ll n=v.size();fl(i,n)cout<<v[i]<<" ";cout<<"\n";}
template<typename T>
ll sumvec(vector<T>v){ll n=v.size();ll s=0;fl(i,n)s+=v[i];return s;}

// Mathematical functions
ll gcd(ll a, ll b){if (b == 0)return a;return gcd(b, a % b);} //__gcd
ll lcm(ll a, ll b){return (a/gcd(a,b)*b);}
ll moduloMultiplication(ll a,ll b,ll mod){ll res = 0;a %= mod;while (b){if (b & 1)res = (res + a) % mod;b >>= 1;}return res;}
ll powermod(ll x, ll y, ll p){ll res = 1;x = x % p;if (x == 0) return 0;while (y > 0){if (y & 1)res = (res*x) % p;y = y>>1;x = (x*x) % p;}return res;}

//Graph-dfs
// bool gone[MN];
// vector<int> adj[MN];
// void dfs(int loc){
//     gone[loc]=true;
//     for(auto x:adj[loc])if(!gone[x])dfs(x);
// }

//Sorting
bool sorta(const pair<int,int> &a,const pair<int,int> &b){return (a.second < b.second);}
bool sortd(const pair<int,int> &a,const pair<int,int> &b){return (a.second > b.second);}

//Bits
string decToBinary(int n){string s="";int i = 0;while (n > 0) {s =to_string(n % 2)+s;n = n / 2;i++;}return s;}
ll binaryToDecimal(string n){string num = n;ll dec_value = 0;int base = 1;int len = num.length();for(int i = len - 1; i >= 0; i--){if (num[i] == '1')dec_value += base;base = base * 2;}return dec_value;}

//Check
bool isPrime(ll n){if(n<=1)return false;if(n<=3)return true;if(n%2==0||n%3==0)return false;for(int i=5;i*i<=n;i=i+6)if(n%i==0||n%(i+2)==0)return false;return true;}
bool isPowerOfTwo(int n){if(n==0)return false;return (ceil(log2(n)) == floor(log2(n)));}
bool isPerfectSquare(ll x){if (x >= 0) {ll sr = sqrt(x);return (sr * sr == x);}return false;}

//Code
bool check(int idx,vector<int> v){
    int midx=v.size()-idx-1;
    // cout<<"idx"<<idx<<endl;
    // cout<<1<<endl;
    if(idx-1>=0 && v[idx-1]==v[midx]) return false;
    // cout<<2<<endl;
    if(midx!=(v.size())/2 && midx-1>=0 && v[midx-1]==v[idx]) return false;
    // cout<<3<<endl;
    if(midx!=(v.size())/2 && idx+1<v.size() && v[idx+1]==v[midx]) return false;
    // cout<<4<<endl;
    if(midx+1<v.size() && v[midx+1]==v[idx]) return false;
    // cout<<5<<endl;

    return true;

}
void asquare()
{
    ll n;
    cin>>n;
    vector<int> v(n);
    cin>>v;
    for(int i=0;i<n-1;i++){
        if(v[i]==v[i+1]){
            bool doswap=check(i,v);
            if(doswap){
                swap(v[i],v[v.size()-i-1]);
                // cout<<i<<endl;
            }else{
                bool doswap1= check(i+1,v);
                if(doswap1){
                    swap(v[i+1],v[v.size()-(i+1)-1]);
                }
            }
        }
        // for(auto ele: v){
        //     cout<<ele<<" ";
        // }
        // cout<<endl;
    }
    ll ans=0;
    for(int i=0;i<n-1;i++){
        if(v[i]==v[i+1]){
            ans++;
        }
    }
    cout<<ans<<endl;

}

//Main
int main()
{
    // Code By Asquare
    ll t;
    t=1;
    cin>>t; //comment if only 1 input
    fl(i,t)
    {
        asquare();
    }
    // asquare();
    // fl(i,t) //Kickstart
    // {
    //     cout<<"Case #"<<i+1<<": ";
    //     asquare();
    //     cout<<"\n";
    // }
    return 0;
}
//End

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases that can cause the code to fail. Here are the steps to think through this:

1. **Understand the Bug**: The code attempts to minimize the disturbance by swapping elements, but it might not always choose the optimal swap. Specifically, the `check` function might not correctly determine whether a swap will reduce the disturbance.

2. **Identify Edge Cases**:
   - **Small Arrays**: Test with small arrays where the number of elements is just above the minimum (2).
   - **Arrays with All Elements the Same**: This can cause the `check` function to fail because it might not find any valid swaps.
   - **Arrays with Alternating Elements**: These can be tricky because the disturbance is already low, and the code might not find any valid swaps.
   - **Arrays with a Single Element Repeated Multiple Times**: This can also cause issues because the `check` function might not correctly handle multiple occurrences of the same element.

3. **Create the Test Input**:
   - We will create a test case with an array that has a single element repeated multiple times, which can cause the `check` function to fail in finding a valid swap.

Here is the test input:

```plaintext
1
4
1 1 1 1
```

Explanation:
- **t = 1**: One test case.
- **n = 4**: The length of the array.
- **a = [1, 1, 1, 1]**: The array with all elements the same.

This test case is designed to expose the bug in the `check` function, as it will not be able to find any valid swaps to reduce the disturbance. The expected output for this test case should be 3, but the buggy code might not handle it correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Field Trip
time_limit_ms: 2000
memory_limit_mb: 256
Description: Even in university, students need to relax. That is why Sakurakos teacher decided to go on a field trip. It is known that all of the students will be walking in one line. The student with index $i$ has some topic of interest which is described as $a_i$. As a teacher, you want to minimise the disturbance of the line of students.

The disturbance of the line is defined as the number of neighbouring people with the same topic of interest. In other words, disturbance is the number of indices $j$ ($1 \le j < n$) such that $a_j = a_{j + 1}$.

In order to do this, you can choose index $i$ ($1\le i\le n$) and swap students at positions $i$ and $n-i+1$. You can perform any number of swaps.

Your task is to determine the minimal amount of disturbance that you can achieve by doing the operation described above any number of times.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($2 \le n \le 10^5$) — the length of the line of students.    * The second line contains $n$ integers $a_i$ ($1\le a_i\le n$) — the topics of interest of students in line.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output the minimal possible disturbance of the line that you can achieve.

Examples:
input:
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3
output:
1
2
1
0
0
1
1
0
2

Note: In the first example, it is necessary to apply the operation to $i=2$, thus the array will become $[1, \textbf{2}, 1, \textbf{1}, 3]$, with the bold elements indicating those that have swapped places. The disturbance of this array is equal to $1$.

In the fourth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[2, 1, \textbf{2}, \textbf{1}, 2, 4]$. The disturbance of this array is equal to $0$.

In the eighth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[1, 4, \textbf{1}, 5, \textbf{3}, 1, 3]$. The disturbance of this array is equal to $0$.

Buggy Code:
#include <bits/stdc++.h>
#include <algorithm>
typedef long long ll;
using namespace std;
const ll MOD = 1000000007;
int gcd(int a, int b)
{
    // Euclids algo....
    if (a < b)
    {
        swap(a, b);
    }
    if (b == 0)
    {
        return a;
    }
    return gcd(b, a % b);
}

bool isSquare(int x)
{
    int y = sqrt(x);
    return y * y == x;
}
int SmallestOfXdivedeByY(int x, int y)
{
    return x / y + 1;
}
bool is_true(const pair<ll, ll> &a, const pair<ll, ll> &b)
{
    return (a.first + b.first) < (b.first + b.second);
}
void solve()
{

    ll n;
    cin >> n;
    ll a[n + 1];
    a[0] = 0;
    for (int i = 1; i < n + 1; i++)
    {
        cin >> a[i];
    }
    for (int i = 1; i < n; i++)
    {
        if (a[i] == a[i + 1])
        {
            swap(a[i +1], a[n - i]);
        }
    }
    ll ans = 0;
    if (a[n - 1] == a[n] && a[n] != a[1])
    {
        swap(a[1], a[n]);
    }
    if (a[1] == a[2] && a[1] != a[n])
    {
        swap(a[1], a[n]);
    }
    for (int i = 1; i <= n; i++)
    {
        if (a[i] == a[i + 1])
        {
            ans++;
        }
        // cout<<a[i]<<" ";
    }
    // cout<<"\n";
    cout << ans << "\n";
}
int main()
{
    int T;
    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 expose the bug in the provided buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. The main issue in the code is that it only attempts to swap elements when it encounters a pair of adjacent elements that are the same, without considering the overall impact on the entire array. This can lead to suboptimal results, especially when the array has multiple pairs of adjacent elements that are the same.

Let's consider a few scenarios:

1. **All elements are the same**: This is a straightforward case where the disturbance is maximized, and the code should ideally reduce it to the minimum possible.
2. **Alternating elements**: This is another simple case where the disturbance is zero, and the code should not change anything.
3. **Multiple pairs of adjacent elements**: This is a more complex case where the code might not handle all pairs correctly, leading to suboptimal results.

### Test Input
We will create a test input that includes a mix of these scenarios to ensure the code is thoroughly tested.

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

### Explanation
1. **Test Case 1**: All elements are the same.
   - Input: `1 1 1 1 1`
   - Expected Output: `4`
   - The code should ideally reduce the disturbance to `0` by swapping elements, but it fails to do so correctly.

2. **Test Case 2**: Alternating elements.
   - Input: `1 2 1 2 1 2`
   - Expected Output: `0`
   - The code should not change anything since the disturbance is already `0`.

3. **Test Case 3**: Multiple pairs of adjacent elements.
   - Input: `1 1 2 2 3 3 4 4`
   - Expected Output: `0`
   - The code should be able to reduce the disturbance to `0` by swapping elements, but it might not handle all pairs correctly.

This test input should help expose the bug 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: Sakurako's Field Trip
time_limit_ms: 2000
memory_limit_mb: 256
Description: Even in university, students need to relax. That is why Sakurakos teacher decided to go on a field trip. It is known that all of the students will be walking in one line. The student with index $i$ has some topic of interest which is described as $a_i$. As a teacher, you want to minimise the disturbance of the line of students.

The disturbance of the line is defined as the number of neighbouring people with the same topic of interest. In other words, disturbance is the number of indices $j$ ($1 \le j < n$) such that $a_j = a_{j + 1}$.

In order to do this, you can choose index $i$ ($1\le i\le n$) and swap students at positions $i$ and $n-i+1$. You can perform any number of swaps.

Your task is to determine the minimal amount of disturbance that you can achieve by doing the operation described above any number of times.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($2 \le n \le 10^5$) — the length of the line of students.    * The second line contains $n$ integers $a_i$ ($1\le a_i\le n$) — the topics of interest of students in line.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output the minimal possible disturbance of the line that you can achieve.

Examples:
input:
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3
output:
1
2
1
0
0
1
1
0
2

Note: In the first example, it is necessary to apply the operation to $i=2$, thus the array will become $[1, \textbf{2}, 1, \textbf{1}, 3]$, with the bold elements indicating those that have swapped places. The disturbance of this array is equal to $1$.

In the fourth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[2, 1, \textbf{2}, \textbf{1}, 2, 4]$. The disturbance of this array is equal to $0$.

In the eighth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[1, 4, \textbf{1}, 5, \textbf{3}, 1, 3]$. The disturbance of this array is equal to $0$.

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int,input().split()))
    #print(a)
    already = {}
    for i in range(n-1):
        #print(i,n-i-1)
        if a[i]==a[i+1]:
            #print(i,'II',a[i])
            if a[i]!=a[n-i-1]:
                if i not in already:
                    already[i]=True
                    already[n-i-1]=True
                    a[i],a[n-i-1]=a[n-i-1],a[i]

    dist1 = 0
    #print(a)
    for i in range(n-1):
        if a[i]==a[i+1]:
            dist1+=1
    #print(a)
    already = {}
    for i in range(n-1):
        #print(i,n-i-1)
        if a[i]==a[i+1]:
            #print(i,'II',a[i])
            if a[i]!=a[n-i-1]:
                if i not in already:
                    already[i]=True
                    already[n-i-1]=True
                    a[i],a[n-i-1]=a[n-i-1],a[i]

    dist = 0
    #print(a)
    for i in range(n-1):
        if a[i]==a[i+1]:
            dist+=1
    print(min(dist,dist1))

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

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

Think step by step.

Response:
```plaintext
1
5
1 1 2 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Field Trip
time_limit_ms: 2000
memory_limit_mb: 256
Description: Even in university, students need to relax. That is why Sakurakos teacher decided to go on a field trip. It is known that all of the students will be walking in one line. The student with index $i$ has some topic of interest which is described as $a_i$. As a teacher, you want to minimise the disturbance of the line of students.

The disturbance of the line is defined as the number of neighbouring people with the same topic of interest. In other words, disturbance is the number of indices $j$ ($1 \le j < n$) such that $a_j = a_{j + 1}$.

In order to do this, you can choose index $i$ ($1\le i\le n$) and swap students at positions $i$ and $n-i+1$. You can perform any number of swaps.

Your task is to determine the minimal amount of disturbance that you can achieve by doing the operation described above any number of times.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($2 \le n \le 10^5$) — the length of the line of students.    * The second line contains $n$ integers $a_i$ ($1\le a_i\le n$) — the topics of interest of students in line.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output the minimal possible disturbance of the line that you can achieve.

Examples:
input:
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3
output:
1
2
1
0
0
1
1
0
2

Note: In the first example, it is necessary to apply the operation to $i=2$, thus the array will become $[1, \textbf{2}, 1, \textbf{1}, 3]$, with the bold elements indicating those that have swapped places. The disturbance of this array is equal to $1$.

In the fourth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[2, 1, \textbf{2}, \textbf{1}, 2, 4]$. The disturbance of this array is equal to $0$.

In the eighth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[1, 4, \textbf{1}, 5, \textbf{3}, 1, 3]$. The disturbance of this array is equal to $0$.

Buggy Code:
# Author: Colin Rolance D

from sys import stdin
from collections import *
from heapq import *
from bisect import *
from math import *

def input():
    return stdin.readline().strip()

def main():
    t = int(input())
    for _ in range(t):
        n = int(input())
        arr = list(map(int,input().split()))
        ans = res = 0
        for i in range(n//2):
            mini = min(arr[i],arr[n-i-1])
            maxi = max(arr[i],arr[n-i-1])
            arr[i] = mini
            arr[n-i-1] = maxi
        # print(arr)
        mark = [0] * n
        for i in range(1,n):
            if arr[i]==arr[i-1]:
                if not mark[i]:
                    arr[i],arr[n-i-1] = arr[n-i-1],arr[i]
                    mark[i] = mark[n-i-1] = 1
        for i in range(1,n):
            if arr[i]==arr[i-1]:
                ans += 1
        print(ans)

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider the following:

1. **Edge Cases**: The code should handle small and large inputs, including the minimum and maximum allowed values.
2. **Symmetric Arrays**: Arrays where elements are symmetrically placed can challenge the swapping logic.
3. **Uniform Arrays**: Arrays with all elements being the same can test the disturbance calculation.
4. **Alternating Arrays**: Arrays with alternating elements can test the disturbance reduction logic.
5. **Randomized Arrays**: Randomly generated arrays can help identify general issues.

Given the constraints and the nature of the problem, let's create a test input that includes a mix of these cases.

### Test Input
```plaintext
4
2
1 1
3
1 2 1
4
1 2 2 1
6
1 2 2 1 2 1
```

### Explanation
1. **Test Case 1**: `2 1 1`
   - This is the smallest possible input where both elements are the same. The minimal disturbance is 1, and the code should handle this correctly.

2. **Test Case 2**: `3 1 2 1`
   - This is a small symmetric array. The minimal disturbance is 1, and the code should be able to reduce it to 0 by swapping the middle element with itself.

3. **Test Case 3**: `4 1 2 2 1`
   - This is a symmetric array with a pair of identical elements in the middle. The minimal disturbance is 1, and the code should be able to reduce it to 0 by swapping the middle elements.

4. **Test Case 4**: `6 1 2 2 1 2 1`
   - This is a larger symmetric array with multiple pairs of identical elements. The minimal disturbance is 2, and the code should be able to reduce it to 0 by appropriate swaps.

This test input should help in identifying if the buggy code fails to handle these specific scenarios correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Field Trip
time_limit_ms: 2000
memory_limit_mb: 256
Description: Even in university, students need to relax. That is why Sakurakos teacher decided to go on a field trip. It is known that all of the students will be walking in one line. The student with index $i$ has some topic of interest which is described as $a_i$. As a teacher, you want to minimise the disturbance of the line of students.

The disturbance of the line is defined as the number of neighbouring people with the same topic of interest. In other words, disturbance is the number of indices $j$ ($1 \le j < n$) such that $a_j = a_{j + 1}$.

In order to do this, you can choose index $i$ ($1\le i\le n$) and swap students at positions $i$ and $n-i+1$. You can perform any number of swaps.

Your task is to determine the minimal amount of disturbance that you can achieve by doing the operation described above any number of times.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($2 \le n \le 10^5$) — the length of the line of students.    * The second line contains $n$ integers $a_i$ ($1\le a_i\le n$) — the topics of interest of students in line.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output the minimal possible disturbance of the line that you can achieve.

Examples:
input:
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3
output:
1
2
1
0
0
1
1
0
2

Note: In the first example, it is necessary to apply the operation to $i=2$, thus the array will become $[1, \textbf{2}, 1, \textbf{1}, 3]$, with the bold elements indicating those that have swapped places. The disturbance of this array is equal to $1$.

In the fourth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[2, 1, \textbf{2}, \textbf{1}, 2, 4]$. The disturbance of this array is equal to $0$.

In the eighth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[1, 4, \textbf{1}, 5, \textbf{3}, 1, 3]$. The disturbance of this array is equal to $0$.

Buggy Code:
# Problem: C. Sakurako's Field Trip
# Contest: Codeforces - Codeforces Round 981 (Div. 3)
# URL: https://codeforces.com/contest/2033/problem/C
# Memory Limit: 256 MB
# Time Limit: 2000 ms
#
# Powered by CP Editor (https://cpeditor.org)

def bro(a):
    c = 0
    for j in range(len(a) - 1):
        if a[j] == a[j + 1]:
            c += 1
    return c

t = int(input())

for _ in range(t):
    n = int(input())
    a = list(map(int, input().split()))

    m = bro(a)

    for i in range(n // 2):
        a[i], a[n - i - 1] = a[n - i - 1], a[i]
        m = min(m, bro(a))

    print(m)

**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
1 1
4
1 1 2 2
5
1 2 2 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Field Trip
time_limit_ms: 2000
memory_limit_mb: 256
Description: Even in university, students need to relax. That is why Sakurakos teacher decided to go on a field trip. It is known that all of the students will be walking in one line. The student with index $i$ has some topic of interest which is described as $a_i$. As a teacher, you want to minimise the disturbance of the line of students.

The disturbance of the line is defined as the number of neighbouring people with the same topic of interest. In other words, disturbance is the number of indices $j$ ($1 \le j < n$) such that $a_j = a_{j + 1}$.

In order to do this, you can choose index $i$ ($1\le i\le n$) and swap students at positions $i$ and $n-i+1$. You can perform any number of swaps.

Your task is to determine the minimal amount of disturbance that you can achieve by doing the operation described above any number of times.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($2 \le n \le 10^5$) — the length of the line of students.    * The second line contains $n$ integers $a_i$ ($1\le a_i\le n$) — the topics of interest of students in line.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output the minimal possible disturbance of the line that you can achieve.

Examples:
input:
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3
output:
1
2
1
0
0
1
1
0
2

Note: In the first example, it is necessary to apply the operation to $i=2$, thus the array will become $[1, \textbf{2}, 1, \textbf{1}, 3]$, with the bold elements indicating those that have swapped places. The disturbance of this array is equal to $1$.

In the fourth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[2, 1, \textbf{2}, \textbf{1}, 2, 4]$. The disturbance of this array is equal to $0$.

In the eighth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[1, 4, \textbf{1}, 5, \textbf{3}, 1, 3]$. The disturbance of this array is equal to $0$.

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

public class Main {
    public static void swap(int arr[],int i,int j){
        int temp=arr[i];
        arr[i]=arr[j];
        arr[j]=temp;
    }
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int tt = s.nextInt();
        while (tt-- > 0) {
            int n=s.nextInt();
            int arr[]=new int[n];
            for(int i=0;i<n;i++)arr[i]=s.nextInt();

            int a=0,b=0;
            if(arr[0]==arr[1])a++;
            if(arr[0]==arr[n-2])b++;

            if(arr[n-1]==arr[n-2])a++;
            if(arr[n-1]==arr[1])b++;

            if(b<a){
                swap(arr,0,n-1);
            }

            for(int i=1;i<=(n-1)/2;i++){
                a=0;b=0;
                if(arr[i-1]==arr[i])a++;
                if(arr[i+1]==arr[i])a++;
                if(arr[n-1-i]==arr[n-i])a++;
                if(arr[n-1-i]==arr[n-i-2])a++;

                if(arr[n-i-2]==arr[i])b++;
                if(arr[n-i]==arr[i])b++;
                if(arr[n-1-i]==arr[i+1])b++;
                if(arr[n-1-i]==arr[i-1])b++;
                if(b<a){
                    swap(arr,i,n-i-1);
                }
            }

            a=0;b=0;
            if(arr[0]==arr[1])a++;
            if(arr[0]==arr[n-2])b++;

            if(arr[n-1]==arr[n-2])a++;
            if(arr[n-1]==arr[1])b++;

            int sum=0;
            if(b<a){
                swap(arr,0,n-1);
            }

            for(int i=0;i<n-1;i++){
                if(arr[i+1]==arr[i])sum++;
            }

            // for(int i=0;i<n;i++){
            //     System.out.print(arr[i]+" ");
            // }

            System.out.println(sum);
        }
        s.close();
    }
}

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

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

Think step by step.

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

1. **Small and Valid Input**: The input should be small enough to fit within the constraints but still be challenging.
2. **Edge Cases**: Consider cases where the array has a very specific pattern that could trip up the algorithm, such as all elements being the same, or a specific pattern that the swapping logic might not handle correctly.
3. **Odd and Even Lengths**: Test both odd and even lengths of the array to ensure the algorithm handles them correctly.
4. **Specific Patterns**: Create patterns that might cause the algorithm to make incorrect swaps or fail to minimize the disturbance.

Let's construct a test input that fits these criteria:

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

### Explanation:
1. **Test Case 1**: `3 1 1 1`
   - All elements are the same, which is a challenging case for the algorithm to handle.
   - Expected disturbance: 2 (since all elements are the same, no swaps can reduce the disturbance).

2. **Test Case 2**: `4 1 2 2 1`
   - This pattern has a middle section with repeated elements, which could cause the algorithm to make incorrect swaps.
   - Expected disturbance: 0 (by swapping the middle elements, the disturbance can be reduced to 0).

3. **Test Case 3**: `5 1 2 3 2 1`
   - This pattern is symmetric, which could be challenging for the algorithm to handle correctly.
   - Expected disturbance: 1 (the middle element cannot be swapped to reduce the disturbance further).

This test input is small, valid, and designed to expose potential issues in the algorithm's logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Field Trip
time_limit_ms: 2000
memory_limit_mb: 256
Description: Even in university, students need to relax. That is why Sakurakos teacher decided to go on a field trip. It is known that all of the students will be walking in one line. The student with index $i$ has some topic of interest which is described as $a_i$. As a teacher, you want to minimise the disturbance of the line of students.

The disturbance of the line is defined as the number of neighbouring people with the same topic of interest. In other words, disturbance is the number of indices $j$ ($1 \le j < n$) such that $a_j = a_{j + 1}$.

In order to do this, you can choose index $i$ ($1\le i\le n$) and swap students at positions $i$ and $n-i+1$. You can perform any number of swaps.

Your task is to determine the minimal amount of disturbance that you can achieve by doing the operation described above any number of times.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($2 \le n \le 10^5$) — the length of the line of students.    * The second line contains $n$ integers $a_i$ ($1\le a_i\le n$) — the topics of interest of students in line.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output the minimal possible disturbance of the line that you can achieve.

Examples:
input:
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3
output:
1
2
1
0
0
1
1
0
2

Note: In the first example, it is necessary to apply the operation to $i=2$, thus the array will become $[1, \textbf{2}, 1, \textbf{1}, 3]$, with the bold elements indicating those that have swapped places. The disturbance of this array is equal to $1$.

In the fourth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[2, 1, \textbf{2}, \textbf{1}, 2, 4]$. The disturbance of this array is equal to $0$.

In the eighth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[1, 4, \textbf{1}, 5, \textbf{3}, 1, 3]$. The disturbance of this array is equal to $0$.

Buggy Code:
import java.util.*;

public class C {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t1 = sc.nextInt();
		StringBuilder sb = new StringBuilder();

		while (t1-- > 0) {
			int n = sc.nextInt();
			int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
            }
            int arr[] = new int[n];
            for(int i=0;i<n;i++)
            	arr[i] = a[i];
            int d = 0;
            for (int i = 0; i < n - 1; i++) {
                if (a[i] == a[i + 1]) {
                    d++;
                }
            }
            for(int i=1;i<n/2;i++) {
            	if(a[i]==a[i-1] && a[i] == a[i+1] && a[i]!=a[n-i-1]) {
            		int temp = a[i];
            		a[i] = a[n-i-1];
            		a[n-i-1] = temp;
            	}
            	else if(a[i]==a[i-1] && a[i]!=a[n-i-1]) {
            		int temp = a[i];
            		a[i] = a[n-i-1];
            		a[n-i-1] = temp;
            	}
//            	res = Math.min(res, d);
            }
            int res1=0;
            for (int i = 0; i < n - 1; i++) {
                if (a[i] == a[i + 1]) {
                    res1++;
                }
            }
            int temp1 = n/2;
            for(int i=temp1+1;i<n-1;i++) {
            	if(a[i]==a[i-1] && a[i] == a[i+1] && a[i]!=a[n-i-1]) {
            		int temp = a[i];
            		a[i] = a[n-i-1];
            		a[n-i-1] = temp;
            	}
            	else if(a[i]==a[i-1] && a[i]!=a[n-i-1]) {
            		int temp = a[i];
            		a[i] = a[n-i-1];
            		a[n-i-1] = temp;
            	}
//            	res = Math.min(res, d);
            }
            int res =0;
            for (int i = 0; i < n - 1; i++) {
                if (a[i] == a[i + 1]) {
                    res++;
                }
            }
            for(int i=1;i<n/2;i++) {
            	if(arr[i]==arr[i+1] && arr[i] == arr[i-1] && arr[i]!=arr[n-i-1]) {
            		int temp = arr[i];
            		arr[i] = arr[n-i-1];
            		arr[n-i-1] = temp;
            	}
            	else if(arr[i]==arr[i+1] && arr[i]!=arr[n-i-1]) {
            		int temp = arr[i];
            		arr[i] = arr[n-i-1];
            		arr[n-i-1] = temp;
            	}
//            	res = Math.min(res, d);
            }
            int res2=0;
            for (int i = 0; i < n - 1; i++) {
                if (arr[i] == arr[i + 1]) {
                    res2++;
                }
            }
            temp1 = n/2;
            for(int i=temp1+1;i<n-1;i++) {
            	if(arr[i]==arr[i+1] && arr[i] == arr[i-1] && arr[i]!=arr[n-i-1]) {
            		int temp = arr[i];
            		arr[i] = arr[n-i-1];
            		arr[n-i-1] = temp;
            	}
            	else if(arr[i]==arr[i+1] && arr[i]!=arr[n-i-1]) {
            		int temp = arr[i];
            		arr[i] = arr[n-i-1];
            		arr[n-i-1] = temp;
            	}
//            	res = Math.min(res, d);
            }
            int res3 =0;
            for (int i = 0; i < n - 1; i++) {
                if (arr[i] == arr[i + 1]) {
                    res3++;
                }
            }
            int tempo = Math.min(res2, res3);
            sb.append(Math.min(Math.min(tempo, d), Math.min(res, res1))+"\n");
		}

		System.out.print(sb.toString());
		sc.close();
	}
}

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

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

Think step by step.

Response:
```plaintext
1
6
1 1 1 2 2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Field Trip
time_limit_ms: 2000
memory_limit_mb: 256
Description: Even in university, students need to relax. That is why Sakurakos teacher decided to go on a field trip. It is known that all of the students will be walking in one line. The student with index $i$ has some topic of interest which is described as $a_i$. As a teacher, you want to minimise the disturbance of the line of students.

The disturbance of the line is defined as the number of neighbouring people with the same topic of interest. In other words, disturbance is the number of indices $j$ ($1 \le j < n$) such that $a_j = a_{j + 1}$.

In order to do this, you can choose index $i$ ($1\le i\le n$) and swap students at positions $i$ and $n-i+1$. You can perform any number of swaps.

Your task is to determine the minimal amount of disturbance that you can achieve by doing the operation described above any number of times.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($2 \le n \le 10^5$) — the length of the line of students.    * The second line contains $n$ integers $a_i$ ($1\le a_i\le n$) — the topics of interest of students in line.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output the minimal possible disturbance of the line that you can achieve.

Examples:
input:
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3
output:
1
2
1
0
0
1
1
0
2

Note: In the first example, it is necessary to apply the operation to $i=2$, thus the array will become $[1, \textbf{2}, 1, \textbf{1}, 3]$, with the bold elements indicating those that have swapped places. The disturbance of this array is equal to $1$.

In the fourth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[2, 1, \textbf{2}, \textbf{1}, 2, 4]$. The disturbance of this array is equal to $0$.

In the eighth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[1, 4, \textbf{1}, 5, \textbf{3}, 1, 3]$. The disturbance of this array is equal to $0$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int dx[4]={0,0,-1,1};
int dy[4]={-1,1,0,0};
#define int long long
typedef pair<int,int> pii;
const int mod=998244353;

const int N=100010;

int n;
int a[N];

void solve()
{
	cin>>n;
	for(int i=1;i<=n;i++)cin>>a[i];
	int cnt=0;
	for(int i=1;i<n;i++)
	{
		if(a[i]==a[i+1])
		{
			cnt++;
		}
	}
	if(cnt==0)
	{
		cout<<"0"<<endl;
		return;
	}
	for(int i=1;i<n;i++)
	{
		if(a[i]==a[i+1])
		{
			if(i==1)
			{
				int cnt1=0;
				if(a[1]==a[2])cnt1++;
				if(a[n]==a[n-1])cnt1++;
				int cnt2=0;
				swap(a[1],a[n]);
				if(a[1]==a[2])cnt2++;
				if(a[n]==a[n-1])cnt2++;
				if(cnt2<cnt1)continue;
				swap(a[1],a[n]);
			}else{
				int l=i-1,r=i+1;
				int l1=n+i,r1=n+2-i;
				int cnt1=0;
				if(a[i]==a[i-1])cnt1++;
				if(a[i]==a[i+1])cnt1++;
				if(a[n+1-i]==a[l1])cnt1++;
				if(a[n+1-i]==a[r1])cnt1++;
				int cnt2=0;
				swap(a[i],a[n+1-i]);
				if(a[i]==a[i-1])cnt2++;
				if(a[i]==a[i+1])cnt2++;
				if(a[n+1-i]==a[l1])cnt2++;
				if(a[n+1-i]==a[r1])cnt2++;

			//	cout<<i<<" "<<cnt1<<" "<<cnt2<<endl;
				if(cnt2<cnt1)continue;
				swap(a[i],a[n+1-i]);
			}
		}
	}
	int ans=0;
	//for(int i=1;i<=n;i++)cout<<a[i]<<" ";
	//cout<<endl;
	for(int i=1;i<n;i++)
	{
		if(a[i]==a[i+1])ans++;
	}
	cout<<ans<<endl;
	cout<<endl;
}
signed main()
{
    int good_luck_to_you;
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    good_luck_to_you=1;
    cin>>good_luck_to_you;
    while(good_luck_to_you--)
    {
       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
1 1
3
1 2 1
4
1 2 1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Field Trip
time_limit_ms: 2000
memory_limit_mb: 256
Description: Even in university, students need to relax. That is why Sakurakos teacher decided to go on a field trip. It is known that all of the students will be walking in one line. The student with index $i$ has some topic of interest which is described as $a_i$. As a teacher, you want to minimise the disturbance of the line of students.

The disturbance of the line is defined as the number of neighbouring people with the same topic of interest. In other words, disturbance is the number of indices $j$ ($1 \le j < n$) such that $a_j = a_{j + 1}$.

In order to do this, you can choose index $i$ ($1\le i\le n$) and swap students at positions $i$ and $n-i+1$. You can perform any number of swaps.

Your task is to determine the minimal amount of disturbance that you can achieve by doing the operation described above any number of times.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($2 \le n \le 10^5$) — the length of the line of students.    * The second line contains $n$ integers $a_i$ ($1\le a_i\le n$) — the topics of interest of students in line.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output the minimal possible disturbance of the line that you can achieve.

Examples:
input:
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3
output:
1
2
1
0
0
1
1
0
2

Note: In the first example, it is necessary to apply the operation to $i=2$, thus the array will become $[1, \textbf{2}, 1, \textbf{1}, 3]$, with the bold elements indicating those that have swapped places. The disturbance of this array is equal to $1$.

In the fourth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[2, 1, \textbf{2}, \textbf{1}, 2, 4]$. The disturbance of this array is equal to $0$.

In the eighth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[1, 4, \textbf{1}, 5, \textbf{3}, 1, 3]$. The disturbance of this array is equal to $0$.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int d(vector<int>& a, int n) {
    int cnt = 0;
    for (int i = 0; i < n - 1; ++i) {
        if (a[i] == a[i + 1]) cnt++;
    }
    return cnt;
}

int minD(vector<int>& a, int n) {
    int curD = d(a, n);
    int minD = curD;

    for (int i = 0; i < n / 2; ++i) {
        if (a[i] != a[n - i - 1]) {
            swap(a[i], a[n - i - 1]);
        }
        int newD = d(a, n);
        minD = min(minD, newD);
    }

    return minD;
}

int main() {
    int t;
    cin >> t;

    while (t--) {
        int n;
        cin >> n;
        vector<int> a(n);

        for (int i = 0; i < n; ++i) {
            cin >> a[i];
        }

        cout << minD(a, n) << 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
1
6
1 2 3 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Field Trip
time_limit_ms: 2000
memory_limit_mb: 256
Description: Even in university, students need to relax. That is why Sakurakos teacher decided to go on a field trip. It is known that all of the students will be walking in one line. The student with index $i$ has some topic of interest which is described as $a_i$. As a teacher, you want to minimise the disturbance of the line of students.

The disturbance of the line is defined as the number of neighbouring people with the same topic of interest. In other words, disturbance is the number of indices $j$ ($1 \le j < n$) such that $a_j = a_{j + 1}$.

In order to do this, you can choose index $i$ ($1\le i\le n$) and swap students at positions $i$ and $n-i+1$. You can perform any number of swaps.

Your task is to determine the minimal amount of disturbance that you can achieve by doing the operation described above any number of times.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($2 \le n \le 10^5$) — the length of the line of students.    * The second line contains $n$ integers $a_i$ ($1\le a_i\le n$) — the topics of interest of students in line.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output the minimal possible disturbance of the line that you can achieve.

Examples:
input:
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3
output:
1
2
1
0
0
1
1
0
2

Note: In the first example, it is necessary to apply the operation to $i=2$, thus the array will become $[1, \textbf{2}, 1, \textbf{1}, 3]$, with the bold elements indicating those that have swapped places. The disturbance of this array is equal to $1$.

In the fourth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[2, 1, \textbf{2}, \textbf{1}, 2, 4]$. The disturbance of this array is equal to $0$.

In the eighth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[1, 4, \textbf{1}, 5, \textbf{3}, 1, 3]$. The disturbance of this array is equal to $0$.

Buggy Code:
"""
@File        :    2033C.py
@Time        :    2024/10/24 23:20:41
@Author      :    Bobo
@Description :    None
"""

import sys
import math
import heapq
import bisect

from collections import deque
from collections import defaultdict

from itertools import accumulate

from functools import lru_cache

MAX_NUM = 10000000
INF = int(1e9)
MOD_NUM = int(1e9) + 7

def input():
    return sys.stdin.readline()

def main():
    case_num = int(input())
    for _ in range(case_num):
        n = int(input())
        a = list(map(int, input().split()))

        for i in range(n - 1):
            if a[i] == a[i + 1]:
                if (
                    a[i] != a[n - 2 - i]
                    and a[n - 1 - i] != a[i + 1]
                    and (
                        i == 0 or a[i - 1] != a[n - 1 - i] and a[i] != a[n - i]
                    )
                ):
                    a[i], a[n - 1 - i] = a[n - 1 - i], a[i]
                elif a[i + 1] != a[n - 1 - i] and a[i] != a[n - 2 - i]:
                    a[i + 1], a[n - 2 - i] = a[n - 2 - i], a[i + 1]

        # print(a)
        ans = 0
        for i in range(n - 1):
            if a[i] == a[i + 1]:
                ans += 1

        print(ans)

# sys.stdin = open("in.txt", "r")
# sys.stdout = open("out.txt", "w")
main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that maximize the failure probability. The bug in the code is in the way it handles swaps to minimize the disturbance. The code might fail in cases where the swap logic does not correctly account for all possible pairs that can reduce the disturbance.

### Step-by-Step Reasoning:
1. **Edge Cases:**
   - **Single Element Arrays:** Not applicable since the minimum length is 2.
   - **Arrays with All Elements the Same:** This is a good candidate as it will have the maximum initial disturbance.
   - **Arrays with Alternating Elements:** This can also be a good candidate as it might not require any swaps.
   - **Arrays with a Few Repeated Elements:** This can expose issues in the swap logic.

2. **Specific Scenarios:**
   - **Symmetric Arrays:** Arrays where the elements are symmetric can be tricky because the swap logic might not correctly handle the symmetry.
   - **Arrays with Repeated Elements at the Ends:** This can expose issues in the swap logic for edge elements.

### Chosen Test Input:
Let's choose an array with repeated elements at the ends and a few alternating elements in the middle. This will test the swap logic and the disturbance calculation.

### Test Input:
```plaintext
1
6
1 2 2 1 2 1
```

### Explanation:
- **Array:** `1 2 2 1 2 1`
- **Initial Disturbance:** 2 (at indices 2 and 4)
- **Expected Output:** 1

### Why This Input is Challenging:
- The initial disturbance is 2, and the array has repeated elements at the ends.
- The optimal solution is to swap elements at indices 2 and 5, resulting in the array `1 1 2 2 1 2`, which has a disturbance of 1.
- The given code might fail to correctly identify this optimal swap, leading to a higher disturbance.

This test input should effectively expose the bug in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Field Trip
time_limit_ms: 2000
memory_limit_mb: 256
Description: Even in university, students need to relax. That is why Sakurakos teacher decided to go on a field trip. It is known that all of the students will be walking in one line. The student with index $i$ has some topic of interest which is described as $a_i$. As a teacher, you want to minimise the disturbance of the line of students.

The disturbance of the line is defined as the number of neighbouring people with the same topic of interest. In other words, disturbance is the number of indices $j$ ($1 \le j < n$) such that $a_j = a_{j + 1}$.

In order to do this, you can choose index $i$ ($1\le i\le n$) and swap students at positions $i$ and $n-i+1$. You can perform any number of swaps.

Your task is to determine the minimal amount of disturbance that you can achieve by doing the operation described above any number of times.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($2 \le n \le 10^5$) — the length of the line of students.    * The second line contains $n$ integers $a_i$ ($1\le a_i\le n$) — the topics of interest of students in line.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output the minimal possible disturbance of the line that you can achieve.

Examples:
input:
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3
output:
1
2
1
0
0
1
1
0
2

Note: In the first example, it is necessary to apply the operation to $i=2$, thus the array will become $[1, \textbf{2}, 1, \textbf{1}, 3]$, with the bold elements indicating those that have swapped places. The disturbance of this array is equal to $1$.

In the fourth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[2, 1, \textbf{2}, \textbf{1}, 2, 4]$. The disturbance of this array is equal to $0$.

In the eighth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[1, 4, \textbf{1}, 5, \textbf{3}, 1, 3]$. The disturbance of this array is equal to $0$.

Buggy Code:
t=int(input())
ans=''
for x in range(t):
    n=int(input())
    a=input().split()
    for j in range(n//2):
        for i in range(1,n):
            yo=int()
            coun=0
            if a[i-1]==a[i]:
                temp=a[i]
                a[i-1]=a[((n-i-1)+1)]
                a[((n-i-1)+1)]=temp
    for i in range(1,n):
        if a[i-1]==a[i]:
            if yo==a[i] and yo==a[i-1]:
                continue
            yo=a[i]
            coun+=1
    ans+=f'{coun}\n'
print(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:
```plaintext
3
5
1 2 3 4 5
6
1 1 2 2 3 3
7
1 2 2 3 3 4 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Field Trip
time_limit_ms: 2000
memory_limit_mb: 256
Description: Even in university, students need to relax. That is why Sakurakos teacher decided to go on a field trip. It is known that all of the students will be walking in one line. The student with index $i$ has some topic of interest which is described as $a_i$. As a teacher, you want to minimise the disturbance of the line of students.

The disturbance of the line is defined as the number of neighbouring people with the same topic of interest. In other words, disturbance is the number of indices $j$ ($1 \le j < n$) such that $a_j = a_{j + 1}$.

In order to do this, you can choose index $i$ ($1\le i\le n$) and swap students at positions $i$ and $n-i+1$. You can perform any number of swaps.

Your task is to determine the minimal amount of disturbance that you can achieve by doing the operation described above any number of times.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($2 \le n \le 10^5$) — the length of the line of students.    * The second line contains $n$ integers $a_i$ ($1\le a_i\le n$) — the topics of interest of students in line.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output the minimal possible disturbance of the line that you can achieve.

Examples:
input:
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3
output:
1
2
1
0
0
1
1
0
2

Note: In the first example, it is necessary to apply the operation to $i=2$, thus the array will become $[1, \textbf{2}, 1, \textbf{1}, 3]$, with the bold elements indicating those that have swapped places. The disturbance of this array is equal to $1$.

In the fourth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[2, 1, \textbf{2}, \textbf{1}, 2, 4]$. The disturbance of this array is equal to $0$.

In the eighth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[1, 4, \textbf{1}, 5, \textbf{3}, 1, 3]$. The disturbance of this array is equal to $0$.

Buggy Code:
import java.util.*;
public class Main
{

	public static void main(String[] args) {
	    Scanner sc = new Scanner(System.in);
	    int t = sc.nextInt();
	    while(t -- > 0) {
	        int n = sc.nextInt();
	        int arr[] = new int[n];
	        for(int i = 0; i<n; i++) {
	            arr[i] = sc.nextInt();
	        }
	       // int l = 1, r = arr.length-2;
	       // int count = 0;
	       // while(l < r) {
	       //     if(arr[l] == arr[r] || (arr[r-1] == arr[l] && arr[r+1] == arr[l]) || (arr[l-1] == arr[r] && arr[l+1] == arr[r])) {
	       //         l++;
	       //         r--;
	       //         continue;
	       //     }
	       //     int c1 = 0;
	       //     if(arr[l] == arr[l-1]) c1++;
	       //     if(arr[l] == arr[l+1]) c1++;
	       //     if(arr[r] == arr[r-1]) c1++;
	       //     if(arr[r] == arr[r+1]) c1++;
	       //     int c2 = 0;
	       //     if(arr[l] == arr[r-1]) c2++;
	       //     if(arr[l] == arr[r+1]) c2++;
	       //     if(arr[r] == arr[l-1]) c2++;
	       //     if(arr[r] == arr[l+1]) c2++;
	       //     if(c1 > c2) {
	       //         int temp = arr[l];
	       //         arr[l] = arr[r];
	       //         arr[r] = temp;
	       //     }
	       //     l++;
	       //     r--;
	       // }
	       // for(int i = 1; i<n; i++) {
	       //     count += arr[i] == arr[i-1] ? 1 : 0;
	       // }
	       // System.out.println(Arrays.toString(arr));
    	    System.out.println(util(arr, 1, n-2, new Integer[n][n]));
	    }
	}
	private static int util(int arr[], int l, int r, Integer dp[][]) {
	    if(l >= r) {
	        int count = 0;
	        for(int i = 1; i<arr.length; i++) {
	            count += arr[i] == arr[i-1] ? 1 : 0;
	        }
	        return count;
	    }
	    if(dp[l][r] != null) return dp[l][r];
	    int count = util(arr, l+1, r-1, dp);
	   // if(arr[l] == arr[r] || (arr[r-1] == arr[l] && arr[r+1] == arr[l]) || (arr[l-1] == arr[r] && arr[l+1] == arr[r])) {
    //         return count;
    //     }
    //     int c1 = 0;
    //     if(arr[l] == arr[l-1]) c1++;
    //     if(arr[l] == arr[l+1]) c1++;
    //     if(arr[r] == arr[r-1]) c1++;
    //     if(arr[r] == arr[r+1]) c1++;
    //     int c2 = 0;
    //     if(arr[l] == arr[r-1]) c2++;
    //     if(arr[l] == arr[r+1]) c2++;
    //     if(arr[r] == arr[l-1]) c2++;
    //     if(arr[r] == arr[l+1]) c2++;
    //     if(c1 > c2) {
            int temp = arr[l];
            arr[l] = arr[r];
            arr[r] = temp;
            count = Math.min(util(arr, l+1, r-1, dp), count);
            temp = arr[l];
            arr[l] = arr[r];
            arr[r] = temp;
        // }
        return dp[l][r] = count;
	}
}

**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. Here are the steps to think through this:

1. **Edge Cases**:
   - Small arrays (e.g., n = 2, 3, 4).
   - Arrays with all elements the same.
   - Arrays with alternating elements.

2. **Scenarios**:
   - Arrays where the first and last elements are the same.
   - Arrays where the first and last elements are different.
   - Arrays where the middle elements are the same or different.

3. **Test Case Generation**:
   - Create a small-scale test input that covers these scenarios.

Let's generate a test input:

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

### Explanation:
1. **Test Case 1**: `n = 2, [1, 1]`
   - This is the smallest possible array where both elements are the same.
   - The disturbance is 1, and it cannot be reduced by any swap.

2. **Test Case 2**: `n = 3, [1, 2, 1]`
   - This is a small array with the first and last elements the same.
   - The disturbance is 1, and it cannot be reduced by any swap.

3. **Test Case 3**: `n = 4, [1, 2, 2, 1]`
   - This is a small array with the first and last elements the same and the middle elements also the same.
   - The disturbance is 2, and it can be reduced to 0 by swapping the second and third elements.

These test cases should help expose the bug in the buggy code by testing different scenarios and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Field Trip
time_limit_ms: 2000
memory_limit_mb: 256
Description: Even in university, students need to relax. That is why Sakurakos teacher decided to go on a field trip. It is known that all of the students will be walking in one line. The student with index $i$ has some topic of interest which is described as $a_i$. As a teacher, you want to minimise the disturbance of the line of students.

The disturbance of the line is defined as the number of neighbouring people with the same topic of interest. In other words, disturbance is the number of indices $j$ ($1 \le j < n$) such that $a_j = a_{j + 1}$.

In order to do this, you can choose index $i$ ($1\le i\le n$) and swap students at positions $i$ and $n-i+1$. You can perform any number of swaps.

Your task is to determine the minimal amount of disturbance that you can achieve by doing the operation described above any number of times.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($2 \le n \le 10^5$) — the length of the line of students.    * The second line contains $n$ integers $a_i$ ($1\le a_i\le n$) — the topics of interest of students in line.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output the minimal possible disturbance of the line that you can achieve.

Examples:
input:
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3
output:
1
2
1
0
0
1
1
0
2

Note: In the first example, it is necessary to apply the operation to $i=2$, thus the array will become $[1, \textbf{2}, 1, \textbf{1}, 3]$, with the bold elements indicating those that have swapped places. The disturbance of this array is equal to $1$.

In the fourth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[2, 1, \textbf{2}, \textbf{1}, 2, 4]$. The disturbance of this array is equal to $0$.

In the eighth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[1, 4, \textbf{1}, 5, \textbf{3}, 1, 3]$. The disturbance of this array is equal to $0$.

Buggy Code:
import java.util.*;

public class C_Sakurako_s_Field_Trip {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        while (T-- > 0) {
            // Your code here
            ////Bunny is topper
            int n=sc.nextInt();
            int arr[]=new int[n];
            for(int i=0;i<n;i++){
                arr[i]=sc.nextInt();
            }
            for(int i=n/2;i>=0;i--){
                int index=n-i-1;
                int now=0;
                int fut=0;
                if(i!=0){
                    if(arr[i]==arr[i-1]){
                        now++;
                    }
                    if(arr[index]==arr[i-1]){
                        fut++;
                    }
                }
                if(arr[i]==arr[i+1]){
                    now++;
                }
                if(arr[index]==arr[i+1]){
                    fut++;
                }
                if(arr[index]==arr[index-1]){
                    now++;
                }
                if(arr[i]==arr[index-1]){
                    fut++;
                }
                if(index!=n-1&&arr[index]==arr[index+1]){
                    now++;

                }
                if(index!=n-1&&arr[i]==arr[index+1]){
                    fut++;
                }
                if(fut<now){
                    int swap=arr[i];
                    arr[i]=arr[index];
                    arr[index]=swap;
                }
            }
            int ans=0;
            // System.out.print(arr[0]+" ");
            for(int i=1;i<n;i++){
                // System.out.print(arr[i]+" ");
                if(arr[i]==arr[i-1]){
                    ans++;
                }
            }
            // System.out.println();
            System.out.println(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:
```plaintext
1
6
1 2 1 2 1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Field Trip
time_limit_ms: 2000
memory_limit_mb: 256
Description: Even in university, students need to relax. That is why Sakurakos teacher decided to go on a field trip. It is known that all of the students will be walking in one line. The student with index $i$ has some topic of interest which is described as $a_i$. As a teacher, you want to minimise the disturbance of the line of students.

The disturbance of the line is defined as the number of neighbouring people with the same topic of interest. In other words, disturbance is the number of indices $j$ ($1 \le j < n$) such that $a_j = a_{j + 1}$.

In order to do this, you can choose index $i$ ($1\le i\le n$) and swap students at positions $i$ and $n-i+1$. You can perform any number of swaps.

Your task is to determine the minimal amount of disturbance that you can achieve by doing the operation described above any number of times.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($2 \le n \le 10^5$) — the length of the line of students.    * The second line contains $n$ integers $a_i$ ($1\le a_i\le n$) — the topics of interest of students in line.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output the minimal possible disturbance of the line that you can achieve.

Examples:
input:
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3
output:
1
2
1
0
0
1
1
0
2

Note: In the first example, it is necessary to apply the operation to $i=2$, thus the array will become $[1, \textbf{2}, 1, \textbf{1}, 3]$, with the bold elements indicating those that have swapped places. The disturbance of this array is equal to $1$.

In the fourth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[2, 1, \textbf{2}, \textbf{1}, 2, 4]$. The disturbance of this array is equal to $0$.

In the eighth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[1, 4, \textbf{1}, 5, \textbf{3}, 1, 3]$. The disturbance of this array is equal to $0$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
using ll = int64_t;
#define rep(i, x) for(int i = 0; i < int(x); i++)
#define all(x) (x).begin(), (x).end()

void run_case();
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int tests; cin >> tests;
    while (tests-- > 0){
        run_case();
    }
    return 0;
}

void run_case() {
    int n;
    cin >> n;
    vector<int> a(n);
    rep(i, n) cin >> a[i];

    auto check = [&](const vector<int>& arr) {
        int same = 0;
        for (int i = 0; i < n - 1; i++) {
            if (arr[i] == arr[i + 1]) {
                same++;
            }
        }
        return same;
    };

    int prev = check(a);
    int now = prev;
    for (int i = n/2; i >= 0; i--) {
        swap(a[i], a[n - i - 1]);
        now = min(now, check(a));
        swap(a[i], a[n - i - 1]);
    }

    cout << now << endl;
}

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

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

Think step by step.

Response:
```plaintext
1
6
1 1 1 1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sakurako's Field Trip
time_limit_ms: 2000
memory_limit_mb: 256
Description: Even in university, students need to relax. That is why Sakurakos teacher decided to go on a field trip. It is known that all of the students will be walking in one line. The student with index $i$ has some topic of interest which is described as $a_i$. As a teacher, you want to minimise the disturbance of the line of students.

The disturbance of the line is defined as the number of neighbouring people with the same topic of interest. In other words, disturbance is the number of indices $j$ ($1 \le j < n$) such that $a_j = a_{j + 1}$.

In order to do this, you can choose index $i$ ($1\le i\le n$) and swap students at positions $i$ and $n-i+1$. You can perform any number of swaps.

Your task is to determine the minimal amount of disturbance that you can achieve by doing the operation described above any number of times.

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

Each test case is described by two lines.

  * The first line contains one integer $n$ ($2 \le n \le 10^5$) — the length of the line of students.    * The second line contains $n$ integers $a_i$ ($1\le a_i\le n$) — the topics of interest of students in line.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2\cdot 10^5$.

Output: For each test case, output the minimal possible disturbance of the line that you can achieve.

Examples:
input:
9
5
1 1 1 2 3
6
2 1 2 2 1 1
4
1 2 1 1
6
2 1 1 2 2 4
4
2 1 2 3
6
1 2 2 1 2 1
5
4 5 5 1 5
7
1 4 3 5 1 1 3
7
3 1 3 2 2 3 3
output:
1
2
1
0
0
1
1
0
2

Note: In the first example, it is necessary to apply the operation to $i=2$, thus the array will become $[1, \textbf{2}, 1, \textbf{1}, 3]$, with the bold elements indicating those that have swapped places. The disturbance of this array is equal to $1$.

In the fourth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[2, 1, \textbf{2}, \textbf{1}, 2, 4]$. The disturbance of this array is equal to $0$.

In the eighth example, it is sufficient to apply the operation to $i=3$, thus the array will become $[1, 4, \textbf{1}, 5, \textbf{3}, 1, 3]$. The disturbance of this array is equal to $0$.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

// Function to count current disturbance in the array
int countDisturbance(const vector<int>& arr) {
    int count = 0;
    for(int i = 0; i < arr.size() - 1; i++) {
        if(arr[i] == arr[i + 1]) {
            count++;
        }
    }
    return count;
}

// Function to simulate swap operation at index i
void performSwap(vector<int>& arr, int i) {
    int n = arr.size();
    swap(arr[i], arr[n - i - 1]);
}

// Function to find minimum disturbance possible
int findMinDisturbance(vector<int> arr) {
    int n = arr.size();
    int minDist = countDisturbance(arr);

    // Try each possible swap position
    for(int i = 0; i < n/2; i++) {
        // Make a copy of original array
        vector<int> temp = arr;
        performSwap(temp, i);

        // Calculate new disturbance
        int newDist = countDisturbance(temp);
        minDist = min(minDist, newDist);

        // If we already found zero disturbance, no need to continue
        if(minDist == 0) break;
    }

    return minDist;
}

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

    int t;
    cin >> t;

    while(t--) {
        int n;
        cin >> n;

        vector<int> arr(n);
        for(int i = 0; i < n; i++) {
            cin >> arr[i];
        }

        cout << findMinDisturbance(arr) << '\n';
    }

    return 0;
}

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

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

Think step by step.

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