**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Swap Dilemma
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given two arrays of distinct positive integers $a$ and $b$ of length $n$, we would like to make both the arrays the same. Two arrays $x$ and $y$ of length $k$ are said to be the same when for all $1 \le i \le k$, $x_i = y_i$.

Now in one move, you can choose some index $l$ and $r$ in $a$ ($l \le r$) and swap $a_l$ and $a_r$, then choose some $p$ and $q$ ($p \le q$) in $b$ such that $r-l=q-p$ and swap $b_p$ and $b_q$.

Is it possible to make both arrays the same?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 2 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 10^5$) — the length of the arrays $a$ and $b$.

The second line of each test case contains $n$ distinct integers $a_1,a_2,a_3,\ldots,a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the integers in the array $a$.

The third line of each test case contains $n$ distinct integers $b_1,b_2,b_3,\ldots,b_n$ ($1 \le b_i \le 2 \cdot 10^5$) — the integers in the array $b$.

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

Output: For each testcase, print "YES" if the arrays $a$ and $b$ can be made the same. Otherwise, print "NO". can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
output:
YES
NO
YES
NO
NO
NO

Note: In the first testcase, you don't need to perform any operations since the arrays are same.

In the second testcase, it can be proven there exists no way to make the arrays same.

In the third testcase, one of the ways to make the arrays same is to first choose $l=1$, $r=3$, $p=1$, $q=3$ then choose $l=1$, $r=2$, $p=3$, $q=4$.

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

public class D_Swap_Dilemma {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

    static FastReader vt = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        try {
            int testCases = vt.nextInt();
            // int testCases = 1;
            while (testCases-- > 0) {
                // write code here
                solve();
            }
            out.flush();
        } catch (Exception e) {
            return;
        }
    }

    static void solve() {
        int n = vt.nextInt();
        int[] ar = new int[n];
        for (int i = 0; i < n; i++) {
            ar[i] = vt.nextInt();
        }
        int[] br = new int[n];
        for (int i = 0; i < n; i++) {
            br[i] = vt.nextInt();
        }
        HashMap<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < n; i++) {
            map.put(ar[i], map.getOrDefault(ar[i], 0) + 1);
        }
        for (int i = 0; i < n; i++) {
            if (!map.containsKey(br[i])) {
                out.println("NO");
                return;
            }
            map.put(br[i], map.getOrDefault(br[i], 0) - 1);
            if (map.get(br[i]) == 0) {
                map.remove(br[i]);
            }
        }
        HashMap<Integer, Integer> mapA = new HashMap<>();
        for (int i = 0; i < n; i++) {
            mapA.put(ar[i], i);
        }
        HashMap<Integer, Integer> mapB = new HashMap<>();
        for (int i = 0; i < n; i++) {
            mapB.put(br[i], i);
        }
        int[] vis = new int[n];
        ArrayList<Integer> cycle = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            if (vis[i] == 0 && ar[i] != br[i]) {
                int j = mapA.get(br[i]);
                int count = 1;
                vis[i] = 1;
                while (j != i) {
                    count++;
                    vis[j] = 1;
                    j = mapA.get(br[j]);
                }
                cycle.add(count);
            }
        }
        if (cycle.size() == 1) {
            out.println("NO");
            return;
        }
        out.println("YES");
    }

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

    static void sort2d(int arr[][], int l, int r, int index) {
        if (l < r) {
            int m = l + (r - l) / 2;
            sort2d(arr, l, m, index);
            sort2d(arr, m + 1, r, index);
            merge2d(arr, l, m, r, index);
        }
    }

    static void merge2d(int arr[][], int l, int m, int r, int index) {
        int n1 = m - l + 1;
        int n2 = r - m;
        int m1 = arr[0].length;
        int L[][] = new int[n1][m1];
        int R[][] = new int[n2][m1];
        for (int i = 0; i < n1; ++i)
            for (int j = 0; j < m1; j++)
                L[i][j] = arr[l + i][j];

        for (int i = 0; i < n2; ++i)
            for (int j = 0; j < m1; j++)
                R[i][j] = arr[m + 1 + i][j];

        int i = 0, j = 0;
        int k = l;
        while (i < n1 && j < n2) {
            if (L[i][index] <= R[j][index]) {
                for (int t = 0; t < m1; t++)
                    arr[k][t] = L[i][t];
                i++;
            } else {
                for (int t = 0; t < m1; t++)
                    arr[k][t] = R[j][t];
                j++;
            }
            k++;
        }
        while (i < n1) {
            for (int t = 0; t < m1; t++)
                arr[k][t] = L[i][t];
            i++;
            k++;
        }
        while (j < n2) {
            for (int t = 0; t < m1; t++)
                arr[k][t] = R[j][t];
            j++;
            k++;
        }
    }
}

**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
3
1 2 3
3 2 1
4
1 2 3 4
4 3 2 1
5
1 2 3 4 5
5 4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Swap Dilemma
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given two arrays of distinct positive integers $a$ and $b$ of length $n$, we would like to make both the arrays the same. Two arrays $x$ and $y$ of length $k$ are said to be the same when for all $1 \le i \le k$, $x_i = y_i$.

Now in one move, you can choose some index $l$ and $r$ in $a$ ($l \le r$) and swap $a_l$ and $a_r$, then choose some $p$ and $q$ ($p \le q$) in $b$ such that $r-l=q-p$ and swap $b_p$ and $b_q$.

Is it possible to make both arrays the same?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 2 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 10^5$) — the length of the arrays $a$ and $b$.

The second line of each test case contains $n$ distinct integers $a_1,a_2,a_3,\ldots,a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the integers in the array $a$.

The third line of each test case contains $n$ distinct integers $b_1,b_2,b_3,\ldots,b_n$ ($1 \le b_i \le 2 \cdot 10^5$) — the integers in the array $b$.

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

Output: For each testcase, print "YES" if the arrays $a$ and $b$ can be made the same. Otherwise, print "NO". can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
output:
YES
NO
YES
NO
NO
NO

Note: In the first testcase, you don't need to perform any operations since the arrays are same.

In the second testcase, it can be proven there exists no way to make the arrays same.

In the third testcase, one of the ways to make the arrays same is to first choose $l=1$, $r=3$, $p=1$, $q=3$ then choose $l=1$, $r=2$, $p=3$, $q=4$.

Buggy Code:
#include <iostream>
using namespace std;
#include <set>
#include <algorithm>
#include <cmath>
#include <map>
#include <cstdio>
#include <string>
#include <cstring>
#include <string.h>
#include <stdlib.h>
#include <iomanip>
#include <fstream>
#include <stdio.h>
#include <stack>
#include <queue>
#include <ctype.h>
#include <vector>
#include <random>
#define ll long long
#define ull unsigned long long
#define pb push_back
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define pii pair<int, int>
#define pli pair<ll, int>
#define pil pair<int, ll>
#define pll pair<ll, ll>
#define endl '\n'
const double pai = acos(-1);
ll extend_gcd(ll a, ll b, ll &x, ll &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    ll d = extend_gcd(b, a % b, y, x);
    y -= a / b * x;
    return d;
}
ll fastpow(ll a, ll n, ll mod)
{
    ll ans = 1;
    a %= mod;
    while (n)
    {
        if (n & 1)
            ans = (ans * a) % mod; //% mod
        a = (a * a) % mod;         //% mod
        n >>= 1;
    }
    return ans;
}
int dir[4][2] =
    {
        {0, 1}, {0, -1}, {1, 0}, {-1, 0}}; // d a w s

const ll inf = 1000000000000000000ll;
const ll mod = 1e9 + 7, P1 = 13331;
const double eps = 1e-7;
const int N = 1e5 + 10, M = 1e6 + 10;

int t, n;
map<int ,int > ma;
int posa[N],posb[N];
int a[N];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    //freopen("ain.txt", "r", stdin); freopen("aout.txt", "w", stdout);
    cin >> t;
    while (t--)
    {
        ma.clear();
        cin>>n;
        rep(i,1,n)
        {
            int temp;
            cin>>temp;
            a[i]=temp;
            ma[temp]=1;
            posa[temp]=i;
        }
        bool flag=1;
        rep(i,1,n)
        {
            int temp;
            cin>>temp;
            ma[temp]--;
            if(ma[temp]!=0)
            {
                flag=0;
            }
            posb[temp]=i;
        }
        if(flag==0)
        {
            cout<<"NO\n";
            continue;
        }
        sort(a+1,a+n+1);
        ll suma=0,sumb=0;
        rep(i,1,n)
        {
            suma+=1ll*abs(posa[i]-i);
            sumb+=1ll*abs(posb[i]-i);
        }
        if(suma/2%2==sumb/2%2)
        {
            cout<<"YES\n";
        }
        else
        {
            cout<<"NO\n";
        }

    }

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on scenarios where the code's logic might fail. The code attempts to determine if two arrays can be made the same by swapping elements. The key points to consider are:

1. **Distinct Integers**: Both arrays must contain distinct integers.
2. **Swapping Logic**: The code calculates the sum of absolute differences between the positions of elements and their sorted positions. It then checks if these sums are such that their parity (even or odd) matches.

### Steps to Create a Challenging Test Input

1. **Edge Case 1: Arrays with the Same Elements in Different Orders**:
   - This should be a straightforward "YES" case, but we need to ensure it works for different lengths and specific values.

2. **Edge Case 2: Arrays with Different Elements**:
   - This should be a straightforward "NO" case. The code should correctly identify that the arrays cannot be made the same.

3. **Edge Case 3: Arrays with Elements That Require Specific Swaps**:
   - This case should test the code's ability to handle specific swaps and ensure the parity check is correct.

4. **Edge Case 4: Arrays with Elements That Have Different Parities**:
   - This case should ensure the code correctly identifies when the parity of the sums does not match.

### Challenging Test Input

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

### Explanation

1. **Test Case 1**:
   - Arrays: `[1, 2, 3, 4]` and `[4, 3, 2, 1]`
   - Expected Output: "YES"
   - Reason: The arrays can be made the same by swapping elements.

2. **Test Case 2**:
   - Arrays: `[1, 3, 4, 2, 5]` and `[7, 1, 2, 5, 4]`
   - Expected Output: "NO"
   - Reason: The arrays contain different elements, making it impossible to make them the same.

3. **Test Case 3**:
   - Arrays: `[1, 2, 3]` and `[1, 3, 2]`
   - Expected Output: "NO"
   - Reason: The arrays cannot be made the same by the allowed swaps.

4. **Test Case 4**:
   - Arrays: `[1, 2, 3, 4, 5, 6]` and `[6, 5, 4, 3, 2, 1]`
   - Expected Output: "YES"
   - Reason: The arrays can be made the same by swapping elements, and the parity of the sums should match.

This test input should help expose any issues in the provided code, particularly in handling different lengths and specific values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Swap Dilemma
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given two arrays of distinct positive integers $a$ and $b$ of length $n$, we would like to make both the arrays the same. Two arrays $x$ and $y$ of length $k$ are said to be the same when for all $1 \le i \le k$, $x_i = y_i$.

Now in one move, you can choose some index $l$ and $r$ in $a$ ($l \le r$) and swap $a_l$ and $a_r$, then choose some $p$ and $q$ ($p \le q$) in $b$ such that $r-l=q-p$ and swap $b_p$ and $b_q$.

Is it possible to make both arrays the same?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 2 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 10^5$) — the length of the arrays $a$ and $b$.

The second line of each test case contains $n$ distinct integers $a_1,a_2,a_3,\ldots,a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the integers in the array $a$.

The third line of each test case contains $n$ distinct integers $b_1,b_2,b_3,\ldots,b_n$ ($1 \le b_i \le 2 \cdot 10^5$) — the integers in the array $b$.

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

Output: For each testcase, print "YES" if the arrays $a$ and $b$ can be made the same. Otherwise, print "NO". can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
output:
YES
NO
YES
NO
NO
NO

Note: In the first testcase, you don't need to perform any operations since the arrays are same.

In the second testcase, it can be proven there exists no way to make the arrays same.

In the third testcase, one of the ways to make the arrays same is to first choose $l=1$, $r=3$, $p=1$, $q=3$ then choose $l=1$, $r=2$, $p=3$, $q=4$.

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

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while(t--){
        int n;
        cin >> n;
        vector<int> a(n), b(n);
        for(auto &x: a) cin >> x;
        for(auto &x: b) cin >> x;
        // Check if both have the same set of elements
        vector<int> a_sorted = a, b_sorted = b;
        sort(a_sorted.begin(), a_sorted.end());
        sort(b_sorted.begin(), b_sorted.end());
        if(a_sorted != b_sorted){
            cout << "NO\n";
            continue;
        }
        // Check if a equals b
        if(a == b){
            cout << "YES\n";
            continue;
        }
        // Check if a equals reversed b
        vector<int> b_reversed = b;
        reverse(b_reversed.begin(), b_reversed.end());
        if(a == b_reversed){
            cout << "YES\n";
            continue;
        }
        // Otherwise, it's not possible
        cout << "NO\n";
    }
}

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

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

Think step by step.

Response:
```plaintext
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Swap Dilemma
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given two arrays of distinct positive integers $a$ and $b$ of length $n$, we would like to make both the arrays the same. Two arrays $x$ and $y$ of length $k$ are said to be the same when for all $1 \le i \le k$, $x_i = y_i$.

Now in one move, you can choose some index $l$ and $r$ in $a$ ($l \le r$) and swap $a_l$ and $a_r$, then choose some $p$ and $q$ ($p \le q$) in $b$ such that $r-l=q-p$ and swap $b_p$ and $b_q$.

Is it possible to make both arrays the same?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 2 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 10^5$) — the length of the arrays $a$ and $b$.

The second line of each test case contains $n$ distinct integers $a_1,a_2,a_3,\ldots,a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the integers in the array $a$.

The third line of each test case contains $n$ distinct integers $b_1,b_2,b_3,\ldots,b_n$ ($1 \le b_i \le 2 \cdot 10^5$) — the integers in the array $b$.

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

Output: For each testcase, print "YES" if the arrays $a$ and $b$ can be made the same. Otherwise, print "NO". can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
output:
YES
NO
YES
NO
NO
NO

Note: In the first testcase, you don't need to perform any operations since the arrays are same.

In the second testcase, it can be proven there exists no way to make the arrays same.

In the third testcase, one of the ways to make the arrays same is to first choose $l=1$, $r=3$, $p=1$, $q=3$ then choose $l=1$, $r=2$, $p=3$, $q=4$.

Buggy Code:
def compare(l1, l2, n):
    for i in range(n):
        if l1[i] != l2[i]:
            return False
    return True

for _ in range(int(input())):
    n = int(input())
    l1 = list(map(int, input().split()))
    l2 = list(map(int, input().split()))

    l3 = sorted(l1)
    l4 = sorted(l2)

    if l3 != l4:
        print("NO")
    else:
        if compare(l1, l2, n):
            print("YES")
        else:
            flag = 0
            if n%2==0:
                flag=1
            else:
                flag=0
            if flag == 1:
                print("YES")
            else:
                print("NO")

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases that can highlight the incorrect logic. The buggy code checks if the sorted versions of the arrays are the same and then uses an incorrect condition to determine if the arrays can be made the same. Specifically, the condition `if n % 2 == 0` is not a valid criterion for determining whether the arrays can be made the same.

Let's consider a test case where the arrays can be made the same, but the length of the arrays is odd. This will expose the bug because the code will incorrectly print "NO" for such cases.

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

### Explanation
- **Test Case:**
  - `n = 3`
  - `a = [1, 2, 3]`
  - `b = [3, 2, 1]`

- **Expected Output:**
  - "YES" because the arrays can be made the same by swapping the elements.

- **Buggy Code Behavior:**
  - The sorted versions of both arrays are `[1, 2, 3]`.
  - The initial comparison `l3 != l4` will be `False` (since they are the same).
  - The direct comparison `compare(l1, l2, n)` will be `False` (since the arrays are not the same initially).
  - The code then checks `if n % 2 == 0`, which is `False` for `n = 3`.
  - The code will print "NO", which is incorrect.

This test case will expose the bug in the buggy code by showing that it incorrectly determines that the arrays cannot be made the same when they actually can.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Swap Dilemma
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given two arrays of distinct positive integers $a$ and $b$ of length $n$, we would like to make both the arrays the same. Two arrays $x$ and $y$ of length $k$ are said to be the same when for all $1 \le i \le k$, $x_i = y_i$.

Now in one move, you can choose some index $l$ and $r$ in $a$ ($l \le r$) and swap $a_l$ and $a_r$, then choose some $p$ and $q$ ($p \le q$) in $b$ such that $r-l=q-p$ and swap $b_p$ and $b_q$.

Is it possible to make both arrays the same?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 2 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 10^5$) — the length of the arrays $a$ and $b$.

The second line of each test case contains $n$ distinct integers $a_1,a_2,a_3,\ldots,a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the integers in the array $a$.

The third line of each test case contains $n$ distinct integers $b_1,b_2,b_3,\ldots,b_n$ ($1 \le b_i \le 2 \cdot 10^5$) — the integers in the array $b$.

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

Output: For each testcase, print "YES" if the arrays $a$ and $b$ can be made the same. Otherwise, print "NO". can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
output:
YES
NO
YES
NO
NO
NO

Note: In the first testcase, you don't need to perform any operations since the arrays are same.

In the second testcase, it can be proven there exists no way to make the arrays same.

In the third testcase, one of the ways to make the arrays same is to first choose $l=1$, $r=3$, $p=1$, $q=3$ then choose $l=1$, $r=2$, $p=3$, $q=4$.

Buggy Code:
def swap(ll,i,j):
    ll[i],ll[j]=ll[j],ll[i]
whereA={}
WhereB={}
t=int(input())
for _ in range(t):
    n=int(input())
    a=list(map(int,input().split()))
    b=list(map(int,input().split()))
    for i in range(n):
        whereA[a[i]]=i
        WhereB[b[i]]=i
    bb=set(b)
    for i in a:
        if i not in bb:
            print("NO")
            break
    else:
        for i in range(n-2):
            aindex=WhereB[a[i]]
            if aindex==i:
                continue
            swap(b,i+1,aindex)
            swap(b,i,i+1)
            WhereB[b[i]]=i
            WhereB[b[i+1]]=i+1
            WhereB[b[aindex]]=aindex
            swap(a,i+1,aindex)
            swap(a,i+1,i+2)

        if a[-2]==b[-2]:
            print("YES")
        else:
            print("NO")

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Swap Dilemma
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given two arrays of distinct positive integers $a$ and $b$ of length $n$, we would like to make both the arrays the same. Two arrays $x$ and $y$ of length $k$ are said to be the same when for all $1 \le i \le k$, $x_i = y_i$.

Now in one move, you can choose some index $l$ and $r$ in $a$ ($l \le r$) and swap $a_l$ and $a_r$, then choose some $p$ and $q$ ($p \le q$) in $b$ such that $r-l=q-p$ and swap $b_p$ and $b_q$.

Is it possible to make both arrays the same?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 2 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 10^5$) — the length of the arrays $a$ and $b$.

The second line of each test case contains $n$ distinct integers $a_1,a_2,a_3,\ldots,a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the integers in the array $a$.

The third line of each test case contains $n$ distinct integers $b_1,b_2,b_3,\ldots,b_n$ ($1 \le b_i \le 2 \cdot 10^5$) — the integers in the array $b$.

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

Output: For each testcase, print "YES" if the arrays $a$ and $b$ can be made the same. Otherwise, print "NO". can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
output:
YES
NO
YES
NO
NO
NO

Note: In the first testcase, you don't need to perform any operations since the arrays are same.

In the second testcase, it can be proven there exists no way to make the arrays same.

In the third testcase, one of the ways to make the arrays same is to first choose $l=1$, $r=3$, $p=1$, $q=3$ then choose $l=1$, $r=2$, $p=3$, $q=4$.

Buggy Code:
import java.io.*;
import java.util.*;
public class D_Swap_Dilemma {
    public static long count_inversions(int n, int[] arr){
        int[][] mat = new int[n][2];
        for(int i = 0; i < n; i++){
            mat[i][0] = arr[i];
            mat[i][1] = i;
        }
        Arrays.sort(mat, (a, b) -> {
            return a[0] - b[0];
        });

        // System.out.println("Printing mat: ");
        // for(int i = 0; i < n; i++) System.out.println(mat[i][0] + "..." + mat[i][1] + "..." + i);

        long ans = 0;
        for(int i = 0; i < n; i++){
            ans += Math.abs(mat[i][1] - i);
        }
        return ans / 2;
    }
    public static String fun(int n, int[] arr, int[] brr) {
        HashMap<Integer, Integer> map1 = new HashMap<>();
        for(int x: arr) map1.put(x, map1.getOrDefault(x, 0) + 1);

        HashMap<Integer, Integer> map2 = new HashMap<>();
        for(int x: brr) map2.put(x, map2.getOrDefault(x, 0) + 1);

        for(Map.Entry<Integer, Integer> entry: map1.entrySet()){
            int key = entry.getKey(), val = entry.getValue();
            if(!map2.containsKey(key) || map2.get(key) != val) return "NO";
        }
        for(Map.Entry<Integer, Integer> entry: map2.entrySet()){
            int key = entry.getKey(), val = entry.getValue();
            if(!map1.containsKey(key) || map1.get(key) != val) return "NO";
        }

        long n1 = count_inversions(n, arr);
        long n2 = count_inversions(n, brr);

        // System.out.println("n1: " + n1);
        // System.out.println("n2: " + n2);

        return (n1 % 2 == n2 % 2) ? "YES" : "NO";
    }
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int t = Integer.parseInt(st.nextToken());
        while (t-- > 0) {
            st = new StringTokenizer(br.readLine());
            int n = Integer.parseInt(st.nextToken());
            st = new StringTokenizer(br.readLine());
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = Integer.parseInt(st.nextToken());
            }
            st = new StringTokenizer(br.readLine());
            int[] brr = new int[n];
            for (int i = 0; i < n; i++) {
                brr[i] = Integer.parseInt(st.nextToken());
            }

            bw.write(fun(n, arr, brr) + "\n");
        }
        br.close();
        bw.flush();
        bw.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
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Swap Dilemma
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given two arrays of distinct positive integers $a$ and $b$ of length $n$, we would like to make both the arrays the same. Two arrays $x$ and $y$ of length $k$ are said to be the same when for all $1 \le i \le k$, $x_i = y_i$.

Now in one move, you can choose some index $l$ and $r$ in $a$ ($l \le r$) and swap $a_l$ and $a_r$, then choose some $p$ and $q$ ($p \le q$) in $b$ such that $r-l=q-p$ and swap $b_p$ and $b_q$.

Is it possible to make both arrays the same?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 2 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 10^5$) — the length of the arrays $a$ and $b$.

The second line of each test case contains $n$ distinct integers $a_1,a_2,a_3,\ldots,a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the integers in the array $a$.

The third line of each test case contains $n$ distinct integers $b_1,b_2,b_3,\ldots,b_n$ ($1 \le b_i \le 2 \cdot 10^5$) — the integers in the array $b$.

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

Output: For each testcase, print "YES" if the arrays $a$ and $b$ can be made the same. Otherwise, print "NO". can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
output:
YES
NO
YES
NO
NO
NO

Note: In the first testcase, you don't need to perform any operations since the arrays are same.

In the second testcase, it can be proven there exists no way to make the arrays same.

In the third testcase, one of the ways to make the arrays same is to first choose $l=1$, $r=3$, $p=1$, $q=3$ then choose $l=1$, $r=2$, $p=3$, $q=4$.

Buggy Code:
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;
import java.util.StringTokenizer;

public class D_Swap_Dilemma {

    public static void main(String[] args) throws Exception {
        BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(infile.readLine());
        int T = Integer.parseInt(st.nextToken());

        // System.out.println(T);

        StringBuilder sb = new StringBuilder();

        while (T-- > 0) {
            // read input1: read int
            st = new StringTokenizer(infile.readLine());
            int N = Integer.parseInt(st.nextToken());
            // System.out.println("N: " + N);

            // read input2: read arr
            int[] arr1 = readArr(N, infile, st);
            // System.out.println(Arrays.toString(arr1));

            int[] arr2 = readArr(N, infile, st);
            // System.out.println(Arrays.toString(arr2));

            boolean res = true;

            Map<Integer, Integer> cnt = new HashMap<>();
            boolean isValid = true;
            long ops = 0L;
            for (int i = 0; i < arr2.length; i++) {
                cnt.put(arr2[i], i);
            }

            for (int num : arr1) {
                if (!cnt.containsKey(num)) {
                    isValid = false;
                    break;
                }
            }

            if (!isValid) sb.append("NO" + "\n");
            else {
                for (int i = 0; i < arr1.length - 1; i++) {
                    int num = arr1[i];
                    int curInd = cnt.get(num);
                    if (curInd == i) continue;

                    int temp = arr2[i];
                    arr2[i] = num;
                    arr2[curInd] = temp;

                    cnt.put(num, i);
                    cnt.put(temp, curInd);

                    ops = ops + curInd - i + curInd - i - 1;
                }

                if (ops % 2 == 0) {
                    if (arr1[N-1] != arr2[N-1] || arr1[N-2] != arr2[N-2]) res = false;
                } else {
                    if (arr1[N-1] != arr2[N-2] || arr1[N-2] != arr2[N-1]) res = false;
                }

                if (isValid && res) sb.append("YES" + "\n");
                else sb.append("NO" + "\n");
            }

        }

        System.out.println(sb.toString());
    }

    public static int[] readArr(int N, BufferedReader infile, StringTokenizer st) throws Exception {
        int[] arr = new int[N];
        st = new StringTokenizer(infile.readLine());
        for(int i=0; i < N; i++)
            arr[i] = Integer.parseInt(st.nextToken());
        return arr;
    }

    public static long[] readArr2(int N, BufferedReader infile, StringTokenizer st) throws Exception {
        long[] arr = new long[N];
        st = new StringTokenizer(infile.readLine());
        for(int i=0; i < N; i++)
            arr[i] = Long.parseLong(st.nextToken());
        return arr;
    }

    public static void print(int[] arr) {
        //for debugging only
        for(int x: arr)
            System.out.print(x+" ");
        System.out.println();
    }

    private static long quickPower(long a, int p, int MOD) {
        long res = 1L;
        int mask = 1;
        long base = a;

        base = base % MOD;

        for (int i = 0; i < 31; i++) {
            if ((mask & p) != 0) {
                res = (res * base) % MOD;
            }
            base = (base * base) % MOD;
            mask <<= 1;
        }
        return res;
    }

    private static void buildStringHash(String s, long[] hash, long[] power, int MOD, int P) {
        power[0] = 1L;
        for (int i = 1; i < s.length()+1; i++) {
            int ch = s.charAt(i-1) - 'a' + 1;
            hash[i] = (hash[i-1] * P + ch) % MOD;
            power[i] = (power[i-1] * P) % MOD;
        }
    }

    private static long getStringHash(int l, int r, long[] hash, long[] power, int MOD) {
        return (hash[r+1] - (hash[l] * power[r-l+1]) % MOD + MOD) % MOD;
    }

/*     // to be checked
    private static int kmp(String text, String pattern) {
        int[] match = calcMaxMatch(pattern);
        int lenP = pattern.length();
        int matchCnt = 0;
        int c = 0;
        for (int i = 0; i < text.length(); i++) {
            char v = text.charAt(i);
            while (c > 0 && pattern.charAt(c) != v) {
                c = match[c - 1];
            }
            if (pattern.charAt(c) == v) {
                c++;
            }
            if (c == lenP) {
                matchCnt++;
                c = match[c - 1];
            }
        }
        return matchCnt;
    }
    private static int[] calcMaxMatch(String s) {
        int[] match = new int[s.length()];
        int c = 0;
        for (int i = 1; i < s.length(); i++) {
            char v = s.charAt(i);
            while (c > 0 && s.charAt(c) != v) {
                c = match[c - 1];
            }
            if (s.charAt(c) == v) {
                c++;
            }
            match[i] = c;
        }
        return match;
    } */

}

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

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

Think step by step.

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

1. **Incorrect Handling of Swaps**: The code attempts to swap elements in `arr2` to match `arr1`, but it does not correctly handle the constraints of the problem.
2. **Incorrect Check for Validity**: The final checks for `ops % 2 == 0` and the specific comparisons for the last two elements are not robust enough to ensure the arrays are the same.

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

- **Has distinct elements in both arrays**.
- **Ensures that the arrays can be made the same with a specific number of swaps**.
- **Triggers the incorrect handling of swaps or the final checks**.

Let's create a test case where the arrays can be made the same, but the code's logic for swapping and checking validity fails.

### Test Case
```plaintext
1
4
1 3 2 4
2 1 4 3
```

### Explanation
- **Test Case Count**: 1
- **Array Length**: 4
- **Array a**: [1, 3, 2, 4]
- **Array b**: [2, 1, 4, 3]

### Why This Test Case?
1. **Distinct Elements**: Both arrays have distinct elements.
2. **Possible to Make Arrays the Same**: The arrays can be made the same with the following swaps:
   - Swap `a[0]` and `a[1]` to get [3, 1, 2, 4]
   - Swap `a[1]` and `a[2]` to get [3, 2, 1, 4]
   - Swap `a[2]` and `a[3]` to get [3, 2, 4, 1]
   - Swap `a[0]` and `a[3]` to get [1, 2, 4, 3]
   - Swap `a[2]` and `a[3]` to get [1, 2, 3, 4]

3. **Triggers Incorrect Handling**: The code's logic for swapping and checking the final validity might fail because it does not correctly handle the sequence of swaps and the final checks for the last two elements.

### Expected Output
- **Correct Output**: YES
- **Buggy Code Output**: NO

This test case should expose the bug in the buggy code by triggering its incorrect handling of swaps and final checks.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Swap Dilemma
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given two arrays of distinct positive integers $a$ and $b$ of length $n$, we would like to make both the arrays the same. Two arrays $x$ and $y$ of length $k$ are said to be the same when for all $1 \le i \le k$, $x_i = y_i$.

Now in one move, you can choose some index $l$ and $r$ in $a$ ($l \le r$) and swap $a_l$ and $a_r$, then choose some $p$ and $q$ ($p \le q$) in $b$ such that $r-l=q-p$ and swap $b_p$ and $b_q$.

Is it possible to make both arrays the same?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 2 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 10^5$) — the length of the arrays $a$ and $b$.

The second line of each test case contains $n$ distinct integers $a_1,a_2,a_3,\ldots,a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the integers in the array $a$.

The third line of each test case contains $n$ distinct integers $b_1,b_2,b_3,\ldots,b_n$ ($1 \le b_i \le 2 \cdot 10^5$) — the integers in the array $b$.

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

Output: For each testcase, print "YES" if the arrays $a$ and $b$ can be made the same. Otherwise, print "NO". can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
output:
YES
NO
YES
NO
NO
NO

Note: In the first testcase, you don't need to perform any operations since the arrays are same.

In the second testcase, it can be proven there exists no way to make the arrays same.

In the third testcase, one of the ways to make the arrays same is to first choose $l=1$, $r=3$, $p=1$, $q=3$ then choose $l=1$, $r=2$, $p=3$, $q=4$.

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

int func(vector<int> &arr){
    int n = arr.size();

    int cnt = 0;

    int i, j;
    bool swapped;
    for (i = 0; i < n - 1; i++) {
        swapped = false;
        for (j = 0; j < n - i - 1; j++) {
            if (arr[j] > arr[j + 1]) {
                cnt++;
                swap(arr[j], arr[j + 1]);
                swapped = true;
            }
        }

        if (swapped == false)
            break;
    }

    return cnt;
}

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

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

        vector<int> a(n),b(n);

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

        int anum = func(a);
        int bnum = func(b);

        if(a!=b){
            cout<<"NO"<<endl;
            continue;
        }

        if(abs(anum-bnum)%2==0){
            cout<<"YES"<<endl;
        }
        else{
            cout<<"NO"<<endl;
        }

    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Swap Dilemma
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given two arrays of distinct positive integers $a$ and $b$ of length $n$, we would like to make both the arrays the same. Two arrays $x$ and $y$ of length $k$ are said to be the same when for all $1 \le i \le k$, $x_i = y_i$.

Now in one move, you can choose some index $l$ and $r$ in $a$ ($l \le r$) and swap $a_l$ and $a_r$, then choose some $p$ and $q$ ($p \le q$) in $b$ such that $r-l=q-p$ and swap $b_p$ and $b_q$.

Is it possible to make both arrays the same?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 2 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 10^5$) — the length of the arrays $a$ and $b$.

The second line of each test case contains $n$ distinct integers $a_1,a_2,a_3,\ldots,a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the integers in the array $a$.

The third line of each test case contains $n$ distinct integers $b_1,b_2,b_3,\ldots,b_n$ ($1 \le b_i \le 2 \cdot 10^5$) — the integers in the array $b$.

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

Output: For each testcase, print "YES" if the arrays $a$ and $b$ can be made the same. Otherwise, print "NO". can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
output:
YES
NO
YES
NO
NO
NO

Note: In the first testcase, you don't need to perform any operations since the arrays are same.

In the second testcase, it can be proven there exists no way to make the arrays same.

In the third testcase, one of the ways to make the arrays same is to first choose $l=1$, $r=3$, $p=1$, $q=3$ then choose $l=1$, $r=2$, $p=3$, $q=4$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define F first
#define S second
#define pb push_back
#define endl '\n'
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
const int N = 2e5 + 5;
const int M = 1e9 + 7;
const int INF = 1e18;
int a[N], b[N], c[N];
void scan(int arr[], int n){
    for(int i = 0; i<n; i++) cin >> arr[i];
 }
void print(int arr[], int n){
    for(int i = 0; i<n; i++) cerr << arr[i] << " ";
    cerr << '\n';
}
void print(vector<int> &vec){
    for(auto it : vec) cerr << it << " ";
    cerr << endl;
}
void print(vector<pair<int, int>> &vec){
      for(auto it : vec) cerr << it.first << " " << it.second << endl;
 }
 void print(vector<vector<int>> &vec){

      for(auto v : vec) {
           for(auto it : v){
               cerr << it << " ";
           }
           cerr << endl;
      }
 }
 int inv(vector<int>& A) {
    int N = A.size();
    if (N <= 1) {
        return 0;
    }

    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> sortList;
    int result = 0;

    // Heapsort, O(N*log(N))
    for (int i = 0; i < N; i++) {
        sortList.push(make_pair(A[i], i));
    }

    // Create a sorted list of indexes
    vector<int> x;
    while (!sortList.empty()) {

        // O(log(N))
        int v = sortList.top().first;
        int i = sortList.top().second;
        sortList.pop();

        // Find the current minimum's index
        // the index y can represent how many minimums on the left
        int y = upper_bound(x.begin(), x.end(), i) - x.begin();

        // i can represent how many elements on the left
        // i - y can find how many bigger nums on the left
        result += i - y;

        x.insert(upper_bound(x.begin(), x.end(), i), i);
    }

    return result;
}

 int n;
 void solve(){
     cin >> n;
     vector<int> v1(n), v2(n);
     for(int i = 0; i<n; i++) cin >> v1[i];
     for(int i = 0; i<n; i++) cin >> v2[i];
     int l = inv(v1), r = inv(v2);
     sort(all(v1));
     sort(all(v2));
     for(int i = 0; i<n; i++) {
            if(v1[i] != v2[i]) {
                  cout << "NO\n";
                  return;
            }
     }
     if(l%2 == r%2) cout << "YES\n";
     else cout << "NO\n";

 }

signed main(){
      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
2
5
1 2 3 4 5
5 4 3 2 1
4
1 2 3 4
4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Swap Dilemma
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given two arrays of distinct positive integers $a$ and $b$ of length $n$, we would like to make both the arrays the same. Two arrays $x$ and $y$ of length $k$ are said to be the same when for all $1 \le i \le k$, $x_i = y_i$.

Now in one move, you can choose some index $l$ and $r$ in $a$ ($l \le r$) and swap $a_l$ and $a_r$, then choose some $p$ and $q$ ($p \le q$) in $b$ such that $r-l=q-p$ and swap $b_p$ and $b_q$.

Is it possible to make both arrays the same?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 2 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 10^5$) — the length of the arrays $a$ and $b$.

The second line of each test case contains $n$ distinct integers $a_1,a_2,a_3,\ldots,a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the integers in the array $a$.

The third line of each test case contains $n$ distinct integers $b_1,b_2,b_3,\ldots,b_n$ ($1 \le b_i \le 2 \cdot 10^5$) — the integers in the array $b$.

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

Output: For each testcase, print "YES" if the arrays $a$ and $b$ can be made the same. Otherwise, print "NO". can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
output:
YES
NO
YES
NO
NO
NO

Note: In the first testcase, you don't need to perform any operations since the arrays are same.

In the second testcase, it can be proven there exists no way to make the arrays same.

In the third testcase, one of the ways to make the arrays same is to first choose $l=1$, $r=3$, $p=1$, $q=3$ then choose $l=1$, $r=2$, $p=3$, $q=4$.

Buggy Code:
def count_inversions(arr):
    # Helper function to count inversions using a modified merge sort
    def merge_count_split_inv(arr, temp_arr, left, right):
        if left >= right:
            return 0
        mid = (left + right) // 2
        inv_count = 0
        inv_count += merge_count_split_inv(arr, temp_arr, left, mid)
        inv_count += merge_count_split_inv(arr, temp_arr, mid + 1, right)
        inv_count += merge_and_count(arr, temp_arr, left, mid, right)
        return inv_count

    def merge_and_count(arr, temp_arr, left, mid, right):
        i = left    # Starting index for left subarray
        j = mid + 1 # Starting index for right subarray
        k = left    # Starting index to be sorted
        inv_count = 0

        while i <= mid and j <= right:
            if arr[i] <= arr[j]:
                temp_arr[k] = arr[i]
                i += 1
            else:
                temp_arr[k] = arr[j]
                inv_count += (mid - i + 1)
                j += 1
            k += 1

        while i <= mid:
            temp_arr[k] = arr[i]
            i += 1
            k += 1

        while j <= right:
            temp_arr[k] = arr[j]
            j += 1
            k += 1

        for i in range(left, right + 1):
            arr[i] = temp_arr[i]

        return inv_count

    n = len(arr)
    temp_arr = [0] * n
    return merge_count_split_inv(arr, temp_arr, 0, n - 1)

def can_make_arrays_same(a, b):
    if sorted(a) != sorted(b):
        return False

    # Count inversions in both arrays
    inversions_a = count_inversions(a[:])
    inversions_b = count_inversions(b[:])

    # Check if both have the same parity
    return (inversions_a % 2) == (inversions_b % 2)

def main():
    import sys
    input = sys.stdin.read
    data = input().split()

    index = 0
    t = int(data[index])
    index += 1
    results = []

    for _ in range(t):
        n = int(data[index])
        index += 1
        a = list(map(int, data[index:index + n]))
        index += n
        b = list(map(int, data[index:index + n]))
        index += n

        if can_make_arrays_same(a, b):
            results.append("YES")
        else:
            results.append("NO")

    print("\n".join(results))

if __name__ == "__main__":
    main()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Swap Dilemma
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given two arrays of distinct positive integers $a$ and $b$ of length $n$, we would like to make both the arrays the same. Two arrays $x$ and $y$ of length $k$ are said to be the same when for all $1 \le i \le k$, $x_i = y_i$.

Now in one move, you can choose some index $l$ and $r$ in $a$ ($l \le r$) and swap $a_l$ and $a_r$, then choose some $p$ and $q$ ($p \le q$) in $b$ such that $r-l=q-p$ and swap $b_p$ and $b_q$.

Is it possible to make both arrays the same?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 2 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 10^5$) — the length of the arrays $a$ and $b$.

The second line of each test case contains $n$ distinct integers $a_1,a_2,a_3,\ldots,a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the integers in the array $a$.

The third line of each test case contains $n$ distinct integers $b_1,b_2,b_3,\ldots,b_n$ ($1 \le b_i \le 2 \cdot 10^5$) — the integers in the array $b$.

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

Output: For each testcase, print "YES" if the arrays $a$ and $b$ can be made the same. Otherwise, print "NO". can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
output:
YES
NO
YES
NO
NO
NO

Note: In the first testcase, you don't need to perform any operations since the arrays are same.

In the second testcase, it can be proven there exists no way to make the arrays same.

In the third testcase, one of the ways to make the arrays same is to first choose $l=1$, $r=3$, $p=1$, $q=3$ then choose $l=1$, $r=2$, $p=3$, $q=4$.

Buggy Code:
#定义一个逆序数函数即可
def tau(l):
    t=0
    leng = len(l)
    for i in range(leng) :
        for j in l[i+1:]:
            if j<l[i]:t+=1
    return(t)

t=int(input())
for _ in range(t):
    n=int(input())
    a = list(map(int,input().split()))
    b = list(map(int,input().split()))
    #test3 搞了一个巨长的b，所以防一手
    if len(a)!=len(b):
        print("NO")
    elif sorted(a) != sorted(b):
        print("NO")
    else:
        if tau(a)%2==tau(b)%2:
            print("YES")
        else :
            print("NO")

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Swap Dilemma
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given two arrays of distinct positive integers $a$ and $b$ of length $n$, we would like to make both the arrays the same. Two arrays $x$ and $y$ of length $k$ are said to be the same when for all $1 \le i \le k$, $x_i = y_i$.

Now in one move, you can choose some index $l$ and $r$ in $a$ ($l \le r$) and swap $a_l$ and $a_r$, then choose some $p$ and $q$ ($p \le q$) in $b$ such that $r-l=q-p$ and swap $b_p$ and $b_q$.

Is it possible to make both arrays the same?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 2 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 10^5$) — the length of the arrays $a$ and $b$.

The second line of each test case contains $n$ distinct integers $a_1,a_2,a_3,\ldots,a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the integers in the array $a$.

The third line of each test case contains $n$ distinct integers $b_1,b_2,b_3,\ldots,b_n$ ($1 \le b_i \le 2 \cdot 10^5$) — the integers in the array $b$.

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

Output: For each testcase, print "YES" if the arrays $a$ and $b$ can be made the same. Otherwise, print "NO". can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
output:
YES
NO
YES
NO
NO
NO

Note: In the first testcase, you don't need to perform any operations since the arrays are same.

In the second testcase, it can be proven there exists no way to make the arrays same.

In the third testcase, one of the ways to make the arrays same is to first choose $l=1$, $r=3$, $p=1$, $q=3$ then choose $l=1$, $r=2$, $p=3$, $q=4$.

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

public class Main {
    public static int INF = 0x3f3f3f3f, mod = 1000000007, mod9 = 998244353;
    public static void main(String args[]){
        try {
            PrintWriter o = new PrintWriter(System.out);
            boolean multiTest = true;
            // init
            if(multiTest) {
                int t = nextInt(), loop = 0;
                while (loop < t) {loop++;solve(o);}
            } else solve(o);
            o.close();
        } catch (Exception e) {e.printStackTrace();}
    }
    static void solve(PrintWriter o) {
        try {
            int n = nextInt();
            int[] a = new int[n];
            int[] b = new int[n];
            List<Integer> li1 = new ArrayList<>();
            List<Integer> li2 = new ArrayList<>();
            for(int i=0;i<n;i++) {
                a[i] = nextInt();
                li1.add(a[i]);
            }
            for(int i=0;i<n;i++) {
                b[i] = nextInt();
                li2.add(b[i]);
            }
            Collections.sort(li1);
            Collections.sort(li2);
            boolean ok = true;
            for(int i=0;i<n;i++) if(li1.get(i) != li2.get(i)) ok = false;
            if(!ok) {
                o.println("NO");
                return;
            }
            int m = li1.size();
            FenWick[] fen = new FenWick[2];
            for(int i=0;i<2;i++) fen[i] = new FenWick(m);
            long inv1 = 0l, inv2 = 0l;
            for(int i=n-1;i>=0;i--) {
                int id1 = lower_bound(li1, a[i])+1;
                int id2 = lower_bound(li2, b[i])+1;
                inv1 += fen[0].query(id1);
                fen[0].add(id1, 1);
                inv2 += fen[1].query(id2);
                fen[1].add(id2, 1);
            }
            if((inv1%2) == (inv2%2)) o.println("YES");
            else o.println("NO");
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
    public static int lower_bound(List<Integer> li, int val){
        int l = 0, r = li.size();
        while(l < r){
            int mid = l + (r - l) / 2;
            if(li.get(mid) < val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int lower_bound(int[] a, int val){
        int l = 0, r = a.length;
        while(l < r){
            int mid = l + (r - l) / 2;
            if(a[mid] < val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int lower_bound(long[] a, long val){
        int l = 0, r = a.length;
        while(l < r){
            int mid = l + (r - l) / 2;
            if(a[mid] < val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a%b);
    }
    public static long[] extgcd(long a, long b) {
        if(b == 0) return new long[]{1, 0};
        long[] it = extgcd(b, a%b);
        long x = it[1], y = it[0];
        y -= a/b*x;
        return new long[]{x, y};
    }
    public static long lcm(long a, long b){
        return a / gcd(a,b)*b;
    }
    public static long qpow(long a, long n, int md){
        a %= md;
        long ret = 1l;
        while(n > 0){
            if((n & 1) == 1) ret = ret * a % md;
            n >>= 1;
            a = a * a % md;
        }
        return ret;
    }
    public static class FenWick {
        int n;
        long[] a;
        long[] tree;
        public FenWick(int n){
            this.n = n;
            a = new long[n+1];
            tree = new long[n+1];
        }
        private void add(int x, long val){
            while(x <= n){
                tree[x] += val;
                x += x&-x;
            }
        }
        private void addMx(int x, long val) {
            a[x] += val;
            tree[x] = a[x];
            while(x <= n) {
                for(int i=1;i<(x&-x);i<<=1) tree[x] = Math.max(tree[x], tree[x-i]);
                x += x&-x;
            }
        }
        private long query(int x){
            long ret = 0l;
            while(x > 0){
                ret += tree[x];
                x -= x&-x;
            }
            return ret;
        }
        private long queryMx(int l, int r) {
            long res = 0l;
            while(l <= r) {
                if(r-(r&-r) >= l) {
                    res = Math.max(res, tree[r]);
                    r -= r&-r;
                }
                else {
                    res = Math.max(res, a[r]);
                    r--;
                }
            }
            return res;
        }
    }
    static class DSU {
        int[] parent;
        public DSU(int n) {
            parent = new int[n];
            for(int i=0;i<n;i++) parent[i] = i;
        }
        public int find(int x) {
            if(parent[x] != x) parent[x] = find(parent[x]);
            return parent[x];
        }
        public void union(int x, int y) {
            int root_x = find(x);
            int root_y = find(y);
            if(root_x == root_y) return;
            parent[root_x] = root_y;
        }
    }
    private static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    private static StringTokenizer tokenizer = new StringTokenizer("");
    private static String next() throws IOException{
        while(!tokenizer.hasMoreTokens()){tokenizer = new StringTokenizer(reader.readLine());}
        return tokenizer.nextToken();
    }
    public static int nextInt() throws IOException {return Integer.parseInt(next());}
    public static Long nextLong() throws IOException {return Long.parseLong(next());}
    public static double nextDouble() throws IOException {return Double.parseDouble(next());}
    public static char nextChar() throws IOException {return next().toCharArray()[0];}
    public static String nextString() throws IOException {return next();}
    public static String nextLine() throws IOException {return reader.readLine();}
}

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

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

Think step by step.

Response:
```plaintext
3
4
1 2 3 4
4 3 2 1
5
1 2 3 4 5
5 4 3 2 1
3
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: Swap Dilemma
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given two arrays of distinct positive integers $a$ and $b$ of length $n$, we would like to make both the arrays the same. Two arrays $x$ and $y$ of length $k$ are said to be the same when for all $1 \le i \le k$, $x_i = y_i$.

Now in one move, you can choose some index $l$ and $r$ in $a$ ($l \le r$) and swap $a_l$ and $a_r$, then choose some $p$ and $q$ ($p \le q$) in $b$ such that $r-l=q-p$ and swap $b_p$ and $b_q$.

Is it possible to make both arrays the same?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 2 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 10^5$) — the length of the arrays $a$ and $b$.

The second line of each test case contains $n$ distinct integers $a_1,a_2,a_3,\ldots,a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the integers in the array $a$.

The third line of each test case contains $n$ distinct integers $b_1,b_2,b_3,\ldots,b_n$ ($1 \le b_i \le 2 \cdot 10^5$) — the integers in the array $b$.

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

Output: For each testcase, print "YES" if the arrays $a$ and $b$ can be made the same. Otherwise, print "NO". can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
6
4
1 2 3 4
1 2 3 4
5
1 3 4 2 5
7 1 2 5 4
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
3
1 4 2
1 3 2
output:
YES
NO
YES
NO
NO
NO

Note: In the first testcase, you don't need to perform any operations since the arrays are same.

In the second testcase, it can be proven there exists no way to make the arrays same.

In the third testcase, one of the ways to make the arrays same is to first choose $l=1$, $r=3$, $p=1$, $q=3$ then choose $l=1$, $r=2$, $p=3$, $q=4$.

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

public class Main {
    public static int INF = 0x3f3f3f3f, mod = 1000000007, mod9 = 998244353;
    public static void main(String args[]){
        try {
            PrintWriter o = new PrintWriter(System.out);
            boolean multiTest = true;
            // init
            if(multiTest) {
                int t = nextInt(), loop = 0;
                while (loop < t) {loop++;solve(o);}
            } else solve(o);
            o.close();
        } catch (Exception e) {e.printStackTrace();}
    }
    static void solve(PrintWriter o) {
        try {
            int n = nextInt();
            int[] a = new int[n];
            int[] b = new int[n];
            List<Integer> li1 = new ArrayList<>();
            List<Integer> li2 = new ArrayList<>();
            for(int i=0;i<n;i++) {
                a[i] = nextInt();
                li1.add(a[i]);
            }
            for(int i=0;i<n;i++) {
                b[i] = nextInt();
                li2.add(b[i]);
            }
            Collections.sort(li1);
            Collections.sort(li2);
            boolean ok = true;
            for(int i=0;i<n;i++) if(li1.get(i) != li2.get(i)) ok = false;
            if(!ok) {
                o.println("NO");
                return;
            }
            int m = li1.size();
            FenWick[] fen = new FenWick[2];
            for(int i=0;i<2;i++) fen[i] = new FenWick(m);
            long inv1 = 0l, inv2 = 0l;
            for(int i=n-1;i>=0;i--) {
                int id1 = lower_bound(li1, a[i])+1;
                int id2 = lower_bound(li2, b[i])+1;
                inv1 += fen[0].query(id1);
                fen[0].add(id1, 1);
                inv2 += fen[1].query(id2);
                fen[1].add(id2, 1);
            }
            if(inv1%2 == inv2%2) o.println("YES");
            else o.println("NO");
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
    public static int lower_bound(List<Integer> li, int val){
        int l = 0, r = li.size();
        while(l < r){
            int mid = l + (r - l) / 2;
            if(li.get(mid) < val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int lower_bound(int[] a, int val){
        int l = 0, r = a.length;
        while(l < r){
            int mid = l + (r - l) / 2;
            if(a[mid] < val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int lower_bound(long[] a, long val){
        int l = 0, r = a.length;
        while(l < r){
            int mid = l + (r - l) / 2;
            if(a[mid] < val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a%b);
    }
    public static long[] extgcd(long a, long b) {
        if(b == 0) return new long[]{1, 0};
        long[] it = extgcd(b, a%b);
        long x = it[1], y = it[0];
        y -= a/b*x;
        return new long[]{x, y};
    }
    public static long lcm(long a, long b){
        return a / gcd(a,b)*b;
    }
    public static long qpow(long a, long n, int md){
        a %= md;
        long ret = 1l;
        while(n > 0){
            if((n & 1) == 1) ret = ret * a % md;
            n >>= 1;
            a = a * a % md;
        }
        return ret;
    }
    public static class FenWick {
        int n;
        long[] a;
        long[] tree;
        public FenWick(int n){
            this.n = n;
            a = new long[n+1];
            tree = new long[n+1];
        }
        private void add(int x, long val){
            while(x <= n){
                tree[x] += val;
                x += x&-x;
            }
        }
        private void addMx(int x, long val) {
            a[x] += val;
            tree[x] = a[x];
            while(x <= n) {
                for(int i=1;i<(x&-x);i<<=1) tree[x] = Math.max(tree[x], tree[x-i]);
                x += x&-x;
            }
        }
        private long query(int x){
            long ret = 0l;
            while(x > 0){
                ret += tree[x];
                x -= x&-x;
            }
            return ret;
        }
        private long queryMx(int l, int r) {
            long res = 0l;
            while(l <= r) {
                if(r-(r&-r) >= l) {
                    res = Math.max(res, tree[r]);
                    r -= r&-r;
                }
                else {
                    res = Math.max(res, a[r]);
                    r--;
                }
            }
            return res;
        }
    }
    static class DSU {
        int[] parent;
        public DSU(int n) {
            parent = new int[n];
            for(int i=0;i<n;i++) parent[i] = i;
        }
        public int find(int x) {
            if(parent[x] != x) parent[x] = find(parent[x]);
            return parent[x];
        }
        public void union(int x, int y) {
            int root_x = find(x);
            int root_y = find(y);
            if(root_x == root_y) return;
            parent[root_x] = root_y;
        }
    }
    private static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    private static StringTokenizer tokenizer = new StringTokenizer("");
    private static String next() throws IOException{
        while(!tokenizer.hasMoreTokens()){tokenizer = new StringTokenizer(reader.readLine());}
        return tokenizer.nextToken();
    }
    public static int nextInt() throws IOException {return Integer.parseInt(next());}
    public static Long nextLong() throws IOException {return Long.parseLong(next());}
    public static double nextDouble() throws IOException {return Double.parseDouble(next());}
    public static char nextChar() throws IOException {return next().toCharArray()[0];}
    public static String nextString() throws IOException {return next();}
    public static String nextLine() throws IOException {return reader.readLine();}
}

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

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

Think step by step.

Response:
```plaintext
3
4
1 2 3 4
4 3 2 1
3
1 2 3
1 3 2
5
1 5 7 1000 4
4 1 7 5 1000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Min-Fund Prison (Medium)
time_limit_ms: 1500
memory_limit_mb: 256
Description: In the medium version, $$$2 \leq \sum n \leq 300$$$ and $$$1 \leq \sum m \leq 300$$$

After a worker's strike organized by the Dementors asking for equal rights, the prison of Azkaban has suffered some damage. After settling the spirits, the Ministry of Magic is looking to renovate the prison to ensure that the Dementors are kept in check. The prison consists of $$$n$$$ prison cells and $$$m$$$ bi-directional corridors. The $$$i^{th}$$$ corridor is from cells $$$u_i$$$ to $$$v_i$$$. A subset of these cells $$$S$$$ is called a complex if any cell in $$$S$$$ is reachable from any other cell in $$$S$$$. Formally, a subset of cells $$$S$$$ is a complex if $$$x$$$ and $$$y$$$ are reachable from each other for all $$$x, y \in S$$$, using only cells from $$$S$$$ on the way. The funding required for a complex $$$S$$$ consisting of $$$k$$$ cells is defined as $$$k^2$$$.

As part of your Intro to Magical Interior Design course at Hogwarts, you have been tasked with designing the prison. The Ministry of Magic has asked that you divide the prison into $$$2$$$ complexes with $$$\textbf{exactly one corridor}$$$ connecting them, so that the Dementors can't organize union meetings. For this purpose, you are allowed to build bi-directional corridors. The funding required to build a corridor between any $$$2$$$ cells is $$$c$$$.

Due to budget cuts and the ongoing fight against the Death Eaters, you must find the $$$\textbf{minimum total funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Note: The total funding is the sum of the funding required for the $$$2$$$ complexes and the corridors built. If after the division, the two complexes have $$$x$$$ and $$$y$$$ cells respectively and you have built a total of $$$a$$$ corridors, the total funding will be $$$x^2 + y^2 + c \times a$$$. Note that $$$x+y=n$$$.

Input: The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 300$$$) — the number of test cases. Then $$$t$$$ test cases follow.

The first line of each test case consists of three integers $$$n, m$$$ and $$$c$$$ ($$$2 \leq n \leq 300$$$, $$$1 \leq m \leq 300$$$, $$$1 \leq c \leq 10^9$$$)

$$$m$$$ lines follow, each consisting of 2 integers — $$$u_i, v_i$$$ indicating a corridor is present between cells $$$u_i$$$ and $$$v_i$$$ ($$$1 \leq u_i, v_i \leq n$$$, $$$u_i \neq v_i$$$)

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$300$$$.

It is guaranteed that the sum of $$$m$$$ over all test cases does not exceed $$$300$$$.

It is guaranteed that there exists at most one corridor between any two cells.

Output: Print the $$$\textbf{minimum funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Examples:
input:
4
4 6 5
4 3
2 3
2 4
1 2
4 1
3 1
6 6 2
1 4
2 5
3 6
1 5
3 5
6 5
6 5 7
1 4
2 5
3 6
3 5
6 5
7 5 4
1 4
3 6
3 5
6 5
2 7
output:
-1
20
25
33

Note: In the first test case of the sample input, there is no way to divide the prison according to the Ministry's requirements.

In the second test case, consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{2, 3, 5, 6\}$$$ and $$$\{1, 4\}$$$ cells respectively. There are no new corridors built. The total funding is $$$4^2 + 2^2 = 20$$$. You can verify this is the minimum funding required.

In the third test case, build a corridor between $$$2$$$ and $$$4$$$. Consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{3, 5, 6\}$$$ and $$$\{1, 2, 4\}$$$ cells respectively. The total funding is $$$3^2 + 3^2 + 7 \times 1 = 25$$$. You can verify this is the minimum funding required.

In the fourth test case, build a corridor between $$$2$$$ and $$$4$$$ and between $$$5$$$ and $$$7$$$. Consider the corridor between cells $$$5$$$ and $$$7$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{1, 2, 4, 7\}$$$ and $$$\{3, 5, 6\}$$$ cells respectively. The total funding is $$$4^2 + 3^2 + 4 \times 2 = 33$$$. You can verify this is the minimum funding required.

Note for all test cases that there may be multiple ways to get the same funding but there is no other division which will have a more optimal minimum funding.

Buggy Code:
// LUOGU_RID: 169887899
#include <iostream>
#include <algorithm>
#include <vector>
#include <utility>
#include <unordered_map>
#include <queue>
#include <cmath>
#include <cstring>
#include <set>

using namespace std;

typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pi;

const int maxn=350;

int n,m,c;
int head[maxn],ecnt;
int fa[maxn];
int dfn[maxn],low[maxn],dfscnt;
int siz[maxn];
int bl[maxn],blcnt;
int rt[maxn];

vector<int> cut[maxn];

struct edge{
    int to,next;
    #define to e[i].to
}e[maxn<<1];

void add(int x,int y){
    e[++ecnt]=(edge){y,head[x]};
    head[x]=ecnt;
}

void tarjan(int x){
    bl[x]=blcnt;
    dfn[x]=low[x]=++dfscnt;
    siz[x]=1;
    for(int i=head[x];i;i=e[i].next){
        if(to==fa[x]) continue;
        if(!dfn[to]){
            fa[to]=x;
            tarjan(to);
            low[x]=min(low[x],low[to]);
            siz[x]+=siz[to];
            if(low[to]>dfn[x]){
                //printf("bridge : %d - %d\n",x,to);
                cut[blcnt].push_back(siz[to]);
            }
        }
        else{
            low[x]=min(low[x],dfn[to]);
        }
    }
    // printf("dfn[%d] = %d\n",x,dfn[x]);
    // printf("low[%d] = %d\n",x,low[x]);
}

bool f[2][2][maxn];

void work(){
    cin>>n>>m>>c;

    for(int i=1;i<=blcnt;i++){
        cut[i].clear();
    }
    ecnt=dfscnt=blcnt=0;
    memset(head,0,(sizeof (int))*(n+5));
    memset(fa,0,(sizeof (int))*(n+5));
    memset(bl,0,(sizeof (int))*(n+5));
    memset(dfn,0,(sizeof (int))*(n+5));
    memset(siz,0,(sizeof (int))*(n+5));
    memset(rt,0,(sizeof (int))*(n+5));
    memset(f,0,sizeof f);
    f[0][0][0]=1;
    int totsiz=0;

    for(int i=1,x,y;i<=m;i++){
        cin>>x>>y;
        add(x,y);
        add(y,x);
    }
    for(int i=1;i<=n;i++){
        if(!dfn[i]){
            rt[++blcnt]=i;
            tarjan(i);
        }
    }
    for(int i=1;i<=blcnt;i++){
        //printf("size %d : %d\n",i,siz[rt[i]]);
        int t=siz[rt[i]];
        memset(f[i&1][0],0,(sizeof(bool))*(totsiz+5));
        memset(f[i&1][1],0,(sizeof(bool))*(totsiz+5));
        for(int j=0;j<=totsiz;j++){
            f[i&1][0][j+t]|=f[i&1^1][0][j];
            f[i&1][1][j]|=f[i&1^1][0][j];
            f[i&1][1][j+t]|=f[i&1^1][1][j];
            f[i&1][1][j]|=f[i&1^1][1][j];
        }
        for(auto x:cut[blcnt]){
            //printf(" cut size %d\n",x);
            for(int j=0;j<=totsiz;j++){
                f[i&1][1][j+x]|=f[i&1^1][0][j]|f[i&1^1][1][j];
                f[i&1][1][j+t-x]|=f[i&1^1][0][j]|f[i&1^1][1][j];
            }
        }
        totsiz+=t;
    }
    for(int i=n/2;i>=1;i--){
        if(f[blcnt&1][1][i]|f[blcnt&1][1][n-i]){
            cout<<1LL*i*i+1LL*(n-i)*(n-i)+1LL*(blcnt-1)*c<<'\n';
            return ;
        }
    }
    cout<<"-1\n";
}

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

    int T;
    cin>>T;
    while(T--) work();

    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Min-Fund Prison (Medium)
time_limit_ms: 1500
memory_limit_mb: 256
Description: In the medium version, $$$2 \leq \sum n \leq 300$$$ and $$$1 \leq \sum m \leq 300$$$

After a worker's strike organized by the Dementors asking for equal rights, the prison of Azkaban has suffered some damage. After settling the spirits, the Ministry of Magic is looking to renovate the prison to ensure that the Dementors are kept in check. The prison consists of $$$n$$$ prison cells and $$$m$$$ bi-directional corridors. The $$$i^{th}$$$ corridor is from cells $$$u_i$$$ to $$$v_i$$$. A subset of these cells $$$S$$$ is called a complex if any cell in $$$S$$$ is reachable from any other cell in $$$S$$$. Formally, a subset of cells $$$S$$$ is a complex if $$$x$$$ and $$$y$$$ are reachable from each other for all $$$x, y \in S$$$, using only cells from $$$S$$$ on the way. The funding required for a complex $$$S$$$ consisting of $$$k$$$ cells is defined as $$$k^2$$$.

As part of your Intro to Magical Interior Design course at Hogwarts, you have been tasked with designing the prison. The Ministry of Magic has asked that you divide the prison into $$$2$$$ complexes with $$$\textbf{exactly one corridor}$$$ connecting them, so that the Dementors can't organize union meetings. For this purpose, you are allowed to build bi-directional corridors. The funding required to build a corridor between any $$$2$$$ cells is $$$c$$$.

Due to budget cuts and the ongoing fight against the Death Eaters, you must find the $$$\textbf{minimum total funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Note: The total funding is the sum of the funding required for the $$$2$$$ complexes and the corridors built. If after the division, the two complexes have $$$x$$$ and $$$y$$$ cells respectively and you have built a total of $$$a$$$ corridors, the total funding will be $$$x^2 + y^2 + c \times a$$$. Note that $$$x+y=n$$$.

Input: The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 300$$$) — the number of test cases. Then $$$t$$$ test cases follow.

The first line of each test case consists of three integers $$$n, m$$$ and $$$c$$$ ($$$2 \leq n \leq 300$$$, $$$1 \leq m \leq 300$$$, $$$1 \leq c \leq 10^9$$$)

$$$m$$$ lines follow, each consisting of 2 integers — $$$u_i, v_i$$$ indicating a corridor is present between cells $$$u_i$$$ and $$$v_i$$$ ($$$1 \leq u_i, v_i \leq n$$$, $$$u_i \neq v_i$$$)

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$300$$$.

It is guaranteed that the sum of $$$m$$$ over all test cases does not exceed $$$300$$$.

It is guaranteed that there exists at most one corridor between any two cells.

Output: Print the $$$\textbf{minimum funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Examples:
input:
4
4 6 5
4 3
2 3
2 4
1 2
4 1
3 1
6 6 2
1 4
2 5
3 6
1 5
3 5
6 5
6 5 7
1 4
2 5
3 6
3 5
6 5
7 5 4
1 4
3 6
3 5
6 5
2 7
output:
-1
20
25
33

Note: In the first test case of the sample input, there is no way to divide the prison according to the Ministry's requirements.

In the second test case, consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{2, 3, 5, 6\}$$$ and $$$\{1, 4\}$$$ cells respectively. There are no new corridors built. The total funding is $$$4^2 + 2^2 = 20$$$. You can verify this is the minimum funding required.

In the third test case, build a corridor between $$$2$$$ and $$$4$$$. Consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{3, 5, 6\}$$$ and $$$\{1, 2, 4\}$$$ cells respectively. The total funding is $$$3^2 + 3^2 + 7 \times 1 = 25$$$. You can verify this is the minimum funding required.

In the fourth test case, build a corridor between $$$2$$$ and $$$4$$$ and between $$$5$$$ and $$$7$$$. Consider the corridor between cells $$$5$$$ and $$$7$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{1, 2, 4, 7\}$$$ and $$$\{3, 5, 6\}$$$ cells respectively. The total funding is $$$4^2 + 3^2 + 4 \times 2 = 33$$$. You can verify this is the minimum funding required.

Note for all test cases that there may be multiple ways to get the same funding but there is no other division which will have a more optimal minimum funding.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <deque>

#define ff first
#define ss second

using namespace std;

typedef pair<int, int> pii;

const int maxn = 310;
const int inf = 1e9+10;

int n;
bool mark[maxn];

vector<pii> grafo[maxn];

int dfs(int u, int e)
{
	mark[u] = 1;
	int tot = 1;

	for (auto pp: grafo[u])
		if (!mark[pp.first] and pp.second != e)
			tot += dfs(pp.first, e);
	return tot;
}

bool bridge(int e)
{
	for (int i = 1; i <= n; i++)
		mark[i] = 0;

	int cc = 0;
	for (int i = 1; i <= n; i++)
	{
		if (!mark[i])
		{
			cc++;
			dfs(i, e);
		}
	}

	return cc > 1;
}

int getCost(int e)
{
	for (int i = 1; i <= n; i++)
		mark[i] = 0;

	for (int u = 1; u <= n; u++)
	{
		for (auto pp: grafo[u])
		{
			if (pp.second == e)
			{
				int x = dfs(u, e);
				return x*x + (n-x)*(n-x);
			}
		}
	}

	return inf;
}

int main(void)
{
	int tc;
	cin >> tc;

	while (tc--)
	{
		int m, c;
		cin >> n >> m >> c;

		for (int i = 1; i <= n; i++)
		{
			grafo[i].clear();
			mark[i] = 0;
		}

		for (int i = 1; i <= m; i++)
		{
			int u, v;
			cin >> u >> v;

			grafo[u].push_back({v, i});
			grafo[v].push_back({u, i});
		}

		int cc = 0;
		vector<int> ccHead;

		for (int i = 1; i <= n; i++)
		{
			if (!mark[i])
			{
				cc++;
				dfs(i, -1);
				ccHead.push_back(i);
			}
		}

		int ans = inf;

		if (cc > 1)
		{
			for (int i = 1; i < ccHead.size(); i++)
			{
				int u = ccHead[i-1], v = ccHead[i];
				grafo[u].push_back({v, ++m});
				grafo[v].push_back({u, m});
			}
		}

		for (int u = 1; u <= n; u++)
		{
			for (auto pp: grafo[u])
			{
				int v = pp.ff, e = pp.ss;
				if (bridge(e))
					ans = min(ans, getCost(e) + (cc-1)*c);
			}
		}

		if (ans == inf)
			printf("-1\n");
		else
			printf("%d\n", ans);
	}
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Min-Fund Prison (Medium)
time_limit_ms: 1500
memory_limit_mb: 256
Description: In the medium version, $$$2 \leq \sum n \leq 300$$$ and $$$1 \leq \sum m \leq 300$$$

After a worker's strike organized by the Dementors asking for equal rights, the prison of Azkaban has suffered some damage. After settling the spirits, the Ministry of Magic is looking to renovate the prison to ensure that the Dementors are kept in check. The prison consists of $$$n$$$ prison cells and $$$m$$$ bi-directional corridors. The $$$i^{th}$$$ corridor is from cells $$$u_i$$$ to $$$v_i$$$. A subset of these cells $$$S$$$ is called a complex if any cell in $$$S$$$ is reachable from any other cell in $$$S$$$. Formally, a subset of cells $$$S$$$ is a complex if $$$x$$$ and $$$y$$$ are reachable from each other for all $$$x, y \in S$$$, using only cells from $$$S$$$ on the way. The funding required for a complex $$$S$$$ consisting of $$$k$$$ cells is defined as $$$k^2$$$.

As part of your Intro to Magical Interior Design course at Hogwarts, you have been tasked with designing the prison. The Ministry of Magic has asked that you divide the prison into $$$2$$$ complexes with $$$\textbf{exactly one corridor}$$$ connecting them, so that the Dementors can't organize union meetings. For this purpose, you are allowed to build bi-directional corridors. The funding required to build a corridor between any $$$2$$$ cells is $$$c$$$.

Due to budget cuts and the ongoing fight against the Death Eaters, you must find the $$$\textbf{minimum total funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Note: The total funding is the sum of the funding required for the $$$2$$$ complexes and the corridors built. If after the division, the two complexes have $$$x$$$ and $$$y$$$ cells respectively and you have built a total of $$$a$$$ corridors, the total funding will be $$$x^2 + y^2 + c \times a$$$. Note that $$$x+y=n$$$.

Input: The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 300$$$) — the number of test cases. Then $$$t$$$ test cases follow.

The first line of each test case consists of three integers $$$n, m$$$ and $$$c$$$ ($$$2 \leq n \leq 300$$$, $$$1 \leq m \leq 300$$$, $$$1 \leq c \leq 10^9$$$)

$$$m$$$ lines follow, each consisting of 2 integers — $$$u_i, v_i$$$ indicating a corridor is present between cells $$$u_i$$$ and $$$v_i$$$ ($$$1 \leq u_i, v_i \leq n$$$, $$$u_i \neq v_i$$$)

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$300$$$.

It is guaranteed that the sum of $$$m$$$ over all test cases does not exceed $$$300$$$.

It is guaranteed that there exists at most one corridor between any two cells.

Output: Print the $$$\textbf{minimum funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Examples:
input:
4
4 6 5
4 3
2 3
2 4
1 2
4 1
3 1
6 6 2
1 4
2 5
3 6
1 5
3 5
6 5
6 5 7
1 4
2 5
3 6
3 5
6 5
7 5 4
1 4
3 6
3 5
6 5
2 7
output:
-1
20
25
33

Note: In the first test case of the sample input, there is no way to divide the prison according to the Ministry's requirements.

In the second test case, consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{2, 3, 5, 6\}$$$ and $$$\{1, 4\}$$$ cells respectively. There are no new corridors built. The total funding is $$$4^2 + 2^2 = 20$$$. You can verify this is the minimum funding required.

In the third test case, build a corridor between $$$2$$$ and $$$4$$$. Consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{3, 5, 6\}$$$ and $$$\{1, 2, 4\}$$$ cells respectively. The total funding is $$$3^2 + 3^2 + 7 \times 1 = 25$$$. You can verify this is the minimum funding required.

In the fourth test case, build a corridor between $$$2$$$ and $$$4$$$ and between $$$5$$$ and $$$7$$$. Consider the corridor between cells $$$5$$$ and $$$7$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{1, 2, 4, 7\}$$$ and $$$\{3, 5, 6\}$$$ cells respectively. The total funding is $$$4^2 + 3^2 + 4 \times 2 = 33$$$. You can verify this is the minimum funding required.

Note for all test cases that there may be multiple ways to get the same funding but there is no other division which will have a more optimal minimum funding.

Buggy Code:
// LUOGU_RID: 170169400
#include <bits/stdc++.h>
#define pii pair <int, int>
#define INF 0x3f3f3f3f
using namespace std;

const int N = 300 + 5;
int fa[N], sz[N], cnt;
bool vis[N];
int d[N];
int a[N], b[N];
int dp[N];
int n, m, c;

int find(int x) {
	return (x == fa[x] ? x : fa[x] = find(fa[x]));
}
bool merge(int x, int y) {
	x = find(x), y = find(y);
	if (x == y) return 0;
	fa[x] = y;
	sz[y] += sz[x];
	return 1;
}

void solve() {
	scanf("%d%d%d", &n, &m, &c);
	for (int i = 1; i <= m; i++) {
		scanf("%d%d", &a[i], &b[i]);
	}

	int ans = INF;
	for (int i = 1; i <= n; i++) {
		fa[i] = i;
		sz[i] = 1;
		vis[i] = 0;
	}
	for (int i = 1; i <= m; i++) {
		merge(a[i], b[i]);
	}
	cnt = 0;
	for (int i = 1; i <= n; i++) {
		int x = find(i);
		if (!vis[x]) {
			d[++cnt] = sz[x];
		}
		vis[x] = 1;
	}
	dp[0] = 1;
	int mx = 0;
	for (int j = 1; j <= n / 2; j++) { dp[j] = 0; }
	for (int j = 1; j <= cnt; j++) {
		for (int k = n / 2; k >= d[j]; k--) {
			dp[k] |= dp[k - d[j]];
			mx = dp[k] * k;
		}
	}
	if (mx) {
		ans = min(mx * mx + (n - mx) * (n - mx) + (cnt - 1) * c, ans);
	}
	int tmp = cnt;
	for (int i = 1; i <= m; i++) {
		for (int j = 1; j <= n; j++) {
			fa[j] = j;
			sz[j] = 1;
			vis[j] = 0;
		}
		for (int j = 1; j <= m; j++) {
			if (j == i) continue;
			merge(a[j], b[j]);
		}
		cnt = 0;
		int ax = find(a[i]), bx = find(b[i]);
		if (ax == bx) {
			continue;
		}
		int c1 = 0, c2 = 0;
		for (int j = 1; j <= n; j++) {
			int x = find(j);
			if (x == ax) { c1 = sz[x]; continue; }
			if (x == bx) { c2 = sz[x]; continue; }
			if (!vis[x]) {
				d[++cnt] = sz[x];
				vis[x] = 1;
			}
		}
		for (int j = 1; j <= n / 2; j++) { dp[j] = 0; }
		dp[0] = 1, dp[c1] = 1, dp[c2] = 1;
		for (int j = 1; j <= cnt; j++) {
			for (int k = n / 2; k >= d[j]; k--) {
				dp[k] |= dp[k - d[j]];
			}
		}
		int mx = 0;
		for (int j = 1; j <= n / 2; j++) {
			if (dp[j]) mx = j;
		}
		ans = min(mx * mx + (n - mx) * (n - mx) + (tmp - 1) * c, ans);
	}
	if (ans == INF) puts("-1");
	else printf("%d\n", ans);
}

int main() {
	#ifndef ONLINE_JUDGE
	freopen("test.in", "r", stdin);
	freopen("test.out", "w", stdout);
	#endif

	int T;
	scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}

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

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

Think step by step.

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