**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DIY
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a list of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You need to pick $$$8$$$ elements from the list and use them as coordinates of four points. These four points should be corners of a rectangle which has its sides parallel to the coordinate axes. Your task is to pick coordinates in such a way that the resulting rectangle has the maximum possible area. The rectangle can be degenerate, i. e. its area can be $$$0$$$. Each integer can be used as many times as it occurs in the list (or less).

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 25\,000$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$8 \le n \le 2 \cdot 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print the answer as follows:

- if it is impossible to construct a rectangle which meets the constraints from the statement, print a single line containing the word NO (case-insensitive);
- otherwise, in the first line, print YES (case-insensitive). In the second line, print $$$8$$$ integers $$$x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4$$$ — the coordinates of the corners of the rectangle. You can print the corners in any order.

Examples:
input:
3
16
-5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10
8
0 0 -1 2 2 1 1 3
8
0 0 0 0 0 5 0 5
output:
YES
1 2 1 7 6 2 6 7
NO
YES
0 0 0 5 0 0 0 5

Note:

Buggy Code:
for _ in range(int(input())):
    n = int(input())
    a = list(map(int, input().split()))
    d = dict()
    for i in a:
        if i not in d:
            d[i] = 1
        else:
            d[i] += 1
    nums = []
    sch = 0
    for k, v in d.items():
        if v >= 2:
            nums.append((k, v - (v % 2)))
            sch += (v - (v % 2))
    if sch < 8:
        print('NO')
    else:
        nums.sort()
        mx1, mx2, mn1, mn2 = 0, 0, 0, 0
        for j in range(len(nums)):
            if mn1 == 0:
                if nums[j][1] == 2:
                    mn1 = nums[j][0]
                else:
                    mn2 = nums[j][0]
                    mn1 = nums[j][0]
                    break
            else:
                mn2 = nums[j][0]
                break
        nums.reverse()
        for j in range(len(nums)):
            if mx1 == 0:
                if nums[j][1] == 2:
                    mx1 = nums[j][0]
                else:
                    mx2 = nums[j][0]
                    mx1 = nums[j][0]
                    break
            else:
                mx2 = nums[j][0]
                break
        print('YES')
        print(mn1, mn2, mn1, mx1, mx2, mx1, mx2, mn2)

**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
16
-5 -5 1 1 2 2 3 3 4 4 5 5 6 6 7 10
8
0 0 0 0 0 0 0 0
8
1 2 3 4 5 6 7 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DIY
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a list of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You need to pick $$$8$$$ elements from the list and use them as coordinates of four points. These four points should be corners of a rectangle which has its sides parallel to the coordinate axes. Your task is to pick coordinates in such a way that the resulting rectangle has the maximum possible area. The rectangle can be degenerate, i. e. its area can be $$$0$$$. Each integer can be used as many times as it occurs in the list (or less).

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 25\,000$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$8 \le n \le 2 \cdot 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print the answer as follows:

- if it is impossible to construct a rectangle which meets the constraints from the statement, print a single line containing the word NO (case-insensitive);
- otherwise, in the first line, print YES (case-insensitive). In the second line, print $$$8$$$ integers $$$x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4$$$ — the coordinates of the corners of the rectangle. You can print the corners in any order.

Examples:
input:
3
16
-5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10
8
0 0 -1 2 2 1 1 3
8
0 0 0 0 0 5 0 5
output:
YES
1 2 1 7 6 2 6 7
NO
YES
0 0 0 5 0 0 0 5

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
#define pb push_back
#define mp make_pair
void solve(){
    int n;
    cin >> n;
    vector<int> arr(n);
    for(int i = 0; i < n; i++)  cin >> arr[i];
    sort(arr.begin(), arr.end());
    vector<int> ava;
    for(int i = 1; i < n; i++){
        if(arr[i] == arr[i-1]){
            ava.pb(arr[i]);
            i ++;
        }
    }
    if(ava.size() >= 4){
        cout << "YES\n";
        ll sz = ava.size()-1;
        ll kvadrat1 = 1LL * (ava[sz] - ava[0]) * (ava[sz-1] * ava[1]);
        ll kvadrat2 = 1LL * (ava[sz] - ava[1]) * (ava[sz-1] * ava[0]);
        if(kvadrat1 > kvadrat2){
            cout << ava[1] << " " << ava[0] << " " << ava[1] << " " << ava[sz-1] << " " << ava[sz] << " " << ava[0] << " " << ava[sz] << " " << ava[sz-1] << endl;
         }
        else{
            cout << ava[0] << " " << ava[1] << " " << ava[0] << " " << ava[sz-1] << " " << ava[sz] << " " << ava[1] << " " << ava[sz] << " " << ava[sz-1] << endl;
        }
    }
    else    cout << "NO\n";
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin>>t;
    while(t--)  solve();
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
8
1 1 1 1 2 2 2 2
8
1 1 1 1 1 1 1 1
8
-1 -1 -1 -1 1 1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DIY
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a list of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You need to pick $$$8$$$ elements from the list and use them as coordinates of four points. These four points should be corners of a rectangle which has its sides parallel to the coordinate axes. Your task is to pick coordinates in such a way that the resulting rectangle has the maximum possible area. The rectangle can be degenerate, i. e. its area can be $$$0$$$. Each integer can be used as many times as it occurs in the list (or less).

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 25\,000$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$8 \le n \le 2 \cdot 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print the answer as follows:

- if it is impossible to construct a rectangle which meets the constraints from the statement, print a single line containing the word NO (case-insensitive);
- otherwise, in the first line, print YES (case-insensitive). In the second line, print $$$8$$$ integers $$$x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4$$$ — the coordinates of the corners of the rectangle. You can print the corners in any order.

Examples:
input:
3
16
-5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10
8
0 0 -1 2 2 1 1 3
8
0 0 0 0 0 5 0 5
output:
YES
1 2 1 7 6 2 6 7
NO
YES
0 0 0 5 0 0 0 5

Note:

Buggy Code:
#include <chrono>
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <random>
using namespace std;
using namespace __gnu_pbds;
#pragma GCC optimize("O3,Ofast,unroll-loops")
//#pragma GCC target("popcnt")
#define int long long
#define ll long long
#define ld long double
#define endl '\n'
#define rep(i,k,n) for(int i=k;i<n;i++)
#define pii pair<int,int>
#define debug(...) printf(__VA_ARGS__)
//#define debug(...)
#define trace(x) cout<<#x<<"="<<x<<endl;
//#define trace(x)
#define vi vector<int>
#define popcnt(x) (int)__builtin_popcountll(x)
#define vpii vector<pii >
#define vaii vector<array<int,2>>
#define vvi vector<vector<int> >
#define umap unordered_map
#define uset unordered_set
#define mk make_pair
#define pb push_back
#define pob pop_back
#define _ <<' '<<
#define all(x) x.begin(),x.end()
#define allr(x) x.rbegin(),x.rend()
#define read(x) for (auto& zu: (x)) cin >> zu;
#define write(x) for (auto& zu: (x))cout << zu<<' '; cout<<endl;
static inline void print(const int &x) {
    if (x > 9)print(x / 10);
    putchar('0' + x % 10);
}
//template<typename... T>void read(T&... args) {((for (auto &zu: (args)) { cin >> zu; }),...);}
#define F first
#define S second
template<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<class T> using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define mustawa ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie();
#define saudi main
const int inf=1e9+7;
void fa(bool ok){ cout << (ok ? "YES": "NO") << endl;}
int pw(ll a, ll b, ll mod= 1e9 + 7){
    a %= mod;
    ll res = 1;
    while (b > 0) {
        if (b % 2 == 1) {
            res = res * a%mod;
        }b /= 2;
        a = a * a%mod;
    }
    return res;
}
inline void usaco(string s) {
    freopen((s + ".in").c_str(), "r", stdin);
    freopen((s + ".out").c_str(), "w", stdout);
}inline void usaco(string s1,string s2) {
    freopen((s1 + ".txt").c_str(), "r", stdin);
    freopen((s2 + ".txt").c_str(), "w", stdout);
    freopen("err.txt", "w", stderr);
}int inve(int x,int mod=inf){
    return pow(x,mod-2);
}
int gcd(int a,int b){
    if (b == 0)return a;
    return gcd(b, a % b);
}

int lcm(int a, int b){
    return (a / gcd(a, b)) * b;
}
//long double log(int base,int x){
//    return log2(x)/log(base);
//}
int log(int base,int x){
    return (int)(log2(x)/log2(base));
}
int sum(int l,int r) {
    if (l > r)swap(l, r);
    int a = ((l - 1) * (l)) / 2;
    int b = (r * (r + 1)) / 2;
    return b - a;
}
void solve() {
    int n;
    cin>>n;
    set<int> s;
    vi v;
    for(int i=0;i<n;i++) {
        int x;
        cin>>x;
        if (s.count(x)) {
            s.erase(x);
            v.pb(x);
        }
        else s.insert(x);
    }
    /*
        0{v[0],v[1]}-------1{v[0],v[end-1]}
        |                  |
        |                  |
        2{v[end-2],v[1]}-------3{v[end-2],v[end-1]}

     */
    if (v.size()<4) return fa(0);
    fa(1);
    vi g=v;
    pii p[4];
    p[0]={v[0],v[1]};
    p[1]={v[0],v[v.size()-1]};
    p[2]={v[v.size()-2],v[1]};
    p[3]={v[v.size()-2],v[v.size()-1]};
    for (int i=0;i<4;i++) {
        cout<<p[i].F<<' '<<p[i].S<<' ';
    }cout<<endl;

}
signed saudi() {
    mustawa;
    // usaco("pow");
// #ifndef ONLINE_JUDGE
//     usaco("input","output");
// #endif
    int tescases=1;
    cin>>tescases;
    for (int i = 1; i <= tescases; i++) {
        //cout<<"Case #"<<i<<":"<<' ';
        solve();
    }
}

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

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

Think step by step.

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

1. **Edge Cases**:
   - The input list might contain fewer than 4 unique values, making it impossible to form a rectangle.
   - The input list might contain repeated values that are not sufficient to form a rectangle.
   - The input list might contain values that are too close to each other, leading to a degenerate rectangle with zero area.

2. **Faulty Logic**:
   - The code uses a set to remove duplicates and then pairs the values to form the rectangle. If the set does not have enough unique values, it will fail to form a rectangle.

3. **Small-Scale Test Input**:
   - We need a small input that can be processed within the constraints but is still challenging enough to expose the bug.

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

- **Test Case 1**: A list with fewer than 4 unique values.
- **Test Case 2**: A list with repeated values that are not sufficient to form a rectangle.
- **Test Case 3**: A list with values that can form a degenerate rectangle.

Here is the test input:

```plaintext
3
8
1 1 1 1 1 1 1 1
8
1 1 1 1 2 2 2 2
16
-5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10
```

### Explanation:
1. **Test Case 1**:
   - Input: `8 1 1 1 1 1 1 1 1`
   - Output: `NO`
   - Explanation: There are not enough unique values to form a rectangle.

2. **Test Case 2**:
   - Input: `8 1 1 1 1 2 2 2 2`
   - Output: `NO`
   - Explanation: There are not enough unique values to form a rectangle.

3. **Test Case 3**:
   - Input: `16 -5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10`
   - Output: `YES`
   - Explanation: The values can form a rectangle with sides parallel to the coordinate axes. The coordinates can be chosen as follows:
     - `(1, 2)`, `(1, 7)`, `(6, 2)`, `(6, 7)`

This test input should effectively expose the bug in the provided code by testing edge cases and ensuring that the code handles them correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DIY
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a list of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You need to pick $$$8$$$ elements from the list and use them as coordinates of four points. These four points should be corners of a rectangle which has its sides parallel to the coordinate axes. Your task is to pick coordinates in such a way that the resulting rectangle has the maximum possible area. The rectangle can be degenerate, i. e. its area can be $$$0$$$. Each integer can be used as many times as it occurs in the list (or less).

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 25\,000$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$8 \le n \le 2 \cdot 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print the answer as follows:

- if it is impossible to construct a rectangle which meets the constraints from the statement, print a single line containing the word NO (case-insensitive);
- otherwise, in the first line, print YES (case-insensitive). In the second line, print $$$8$$$ integers $$$x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4$$$ — the coordinates of the corners of the rectangle. You can print the corners in any order.

Examples:
input:
3
16
-5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10
8
0 0 -1 2 2 1 1 3
8
0 0 0 0 0 5 0 5
output:
YES
1 2 1 7 6 2 6 7
NO
YES
0 0 0 5 0 0 0 5

Note:

Buggy Code:
def solve():
	n=int(input())
	arr=list(map(int,input().split()))
	hashmap={}
	for val in arr:
		hashmap[val]=hashmap.get(val,0)+1
	points=[val for val in hashmap for _ in range(hashmap[val] // 2)]
	if len(points)<4:
		print('NO')
	else:
		print("YES")
		p0=points[0]
		p1=points[1]
		p2=points[-1]
		p3=points[-2]
		print(f'{p0} {p1} {p0} {p2} {p3} {p1} {p3} {p2}')

if __name__ == '__main__':
	t=int(input())
	while t>0:
		t-=1
		solve()

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

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

Think step by step.

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

1. **Edge Cases**:
   - Cases where the list contains fewer than 4 unique values.
   - Cases where the list contains many duplicate values.
   - Cases where the list contains values that are very far apart to ensure the algorithm handles large numbers correctly.

2. **Scenarios to Maximize Failure**:
   - Ensure the code correctly handles cases where the list has fewer than 4 unique values.
   - Ensure the code correctly handles cases where the list has many duplicates and the required points cannot be formed.
   - Ensure the code correctly handles cases where the list has values that are very far apart.

Given these considerations, a small-scale test input that can expose the bug in the buggy code is:

```plaintext
3
8
1 1 1 1 1 1 1 1
8
1 2 3 4 5 6 7 8
16
-1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000
```

### Explanation:
1. **Test Case 1**:
   - List: `1 1 1 1 1 1 1 1`
   - This list contains only one unique value, making it impossible to form a rectangle. The expected output is `NO`.

2. **Test Case 2**:
   - List: `1 2 3 4 5 6 7 8`
   - This list contains 8 unique values. The code should be able to form a rectangle, but the specific coordinates chosen might not be correct if the algorithm is faulty. The expected output is `YES` followed by the coordinates of the rectangle.

3. **Test Case 3**:
   - List: `[-1000000000, 1000000000, -1000000000, 1000000000, -1000000000, 1000000000, -1000000000, 1000000000, -1000000000, 1000000000, -1000000000, 1000000000, -1000000000, 1000000000, -1000000000, 1000000000]`
   - This list contains large values and many duplicates. The code should be able to form a rectangle, but the specific coordinates chosen might not be correct if the algorithm is faulty. The expected output is `YES` followed by the coordinates of the rectangle.

This input will help identify issues with the buggy code, especially in handling edge cases and large values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DIY
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a list of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You need to pick $$$8$$$ elements from the list and use them as coordinates of four points. These four points should be corners of a rectangle which has its sides parallel to the coordinate axes. Your task is to pick coordinates in such a way that the resulting rectangle has the maximum possible area. The rectangle can be degenerate, i. e. its area can be $$$0$$$. Each integer can be used as many times as it occurs in the list (or less).

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 25\,000$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$8 \le n \le 2 \cdot 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print the answer as follows:

- if it is impossible to construct a rectangle which meets the constraints from the statement, print a single line containing the word NO (case-insensitive);
- otherwise, in the first line, print YES (case-insensitive). In the second line, print $$$8$$$ integers $$$x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4$$$ — the coordinates of the corners of the rectangle. You can print the corners in any order.

Examples:
input:
3
16
-5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10
8
0 0 -1 2 2 1 1 3
8
0 0 0 0 0 5 0 5
output:
YES
1 2 1 7 6 2 6 7
NO
YES
0 0 0 5 0 0 0 5

Note:

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

import static java.lang.System.out;
import static java.util.Arrays.sort;

public class cf2 {

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

    }

    public static void shuffle(int[] a) {
        Random r = new Random();

        for (int i = 0; i <= a.length - 2; i++) {
            int j = i + r.nextInt(a.length - i);

            swap(a, i, j);
        }

        sort(a);
    }

    public static void swap(int[] a, int i, int j) {
        int temp = a[i];
        a[i] = a[j];
        a[j] = temp;
    }
    public static int N;
    public static void main(String[] args) {
        FastReader in = new FastReader();
        PrintWriter out = new PrintWriter(System.out);
        int test = in.nextInt();

        while (test-- > 0) {
            int n = in.nextInt();
            int[] a = new int[n];
            for( int i =0 ; i < n ; i++ ) a[i] = in.nextInt();
            HashMap<Integer, Integer> freq = new HashMap<>();
            for( int i = 0; i < n ; i++ ) {
                freq.put(a[i], freq.getOrDefault(a[i], 0)+1);
            }
            ArrayList<Integer> freq1 = new ArrayList<>();
            for( int ele : freq.keySet()) {
                if( freq.get(ele) >= 2 ) freq1.add(ele);
            }
            int m = freq1.size();
            int[] freq2 = new int[m];
            for( int i = 0; i < m ; i++ ) freq2[i] = freq1.get(i);
            Collections.sort(freq1);
            int totalFreq = 0;
            for( int ele : freq2 ) {
                totalFreq += freq.get(ele)%2 == 0 ? freq.get(ele) : freq.get(ele)-1;
            }
            if( totalFreq < 8 ) System.out.println("NO");
            else {
//                if( freq.get(freq2[0]))
                int min1, min2, max1, max2;
                min1 = freq2[0];
                if( freq.get(freq2[0])>= 4 ) min2 = min1;
                else min2 = freq2[1];

                max1 = freq2[m-1];
                if(freq.get(freq2[m-1]) >= 4 ) max2 = max1;
                else max2 = freq2[m-2];
                out.println("YES");
                out.println(min1 + " " + min2 + " " + min1 + " " + max1 + " " + max2 + " " + min2 + " " + max2 + " " + max1 );
            }
            out.flush();
        }
        out.close();
    }
    // 9 2
    // 5,
    // 1 11
    // 6, 1,5 and 7, 11
    //6, 3 and 9
    // 8919, 64
    // 4460, 1, 4459 and 4461, 8919
    // 55, 13
    // 28, 1, 27 and 29, 55
    // 28, 14, 1, 13 and 15, 27, and 42, 29, 41 and 43, 55
    // 28 + 14 + 42 + 7 + 21 + 35 + 49
    public static long segment(int l, int r, int a) {
        if( r-l+1 < a ) {
            return 0;
        }
        int m = (l+r)/2;
        if(( (r-l+1) % 2) == 0 ) {
            return 2*segment(1, (r-l+1)/2, a);
        } else if( r-l+1 >= a ) {
            int k = N/((r-l+1)/2);
            return (long)k*((r-l+1)/2) + N/m*segment(l, (((r-l+1)/2)-1), a) + N/m*segment((((r-l+1)/2)+1), r, a)  ;//+ segment(m+1, r, a);
        } else return 0;
    }
    public static String findFromBegAndEnd( String s ) {
        int n = s.length(); int max = 0; String maxStr = "";
        for( int st1 = 0; st1 < 1; st1++ ) {
            for (int end1 = 0; end1 < n; end1++) {
                String a = s.substring(st1, end1 + 1);
                if(isPalindrome(a) && max < a.length() ) {
                    maxStr = a; max = a.length();
                }
            }
        }
        for( int st1 = 0; st1 <= n-1; st1++ ) {
            for (int end1 = n-1; end1 < n; end1++) {
                String a = s.substring(st1, end1 + 1);
                if (isPalindrome(a) && max < a.length() ) {
                    maxStr = a;
                    max = a.length();
                }
            }
        }
        return maxStr;
    }
    public static String findExtraFromBeginning( int i, int j, String s ) {
        // i to some index k (k <= j)
        for( int k = j; k >= i ; k-- ) {
            String pal = s.substring(i, k+1);
            if(isPalindrome(pal)) return pal;
        }
        return "";
    }

    public static String findExtraFromEnd( int i, int j, String s ) {
        //  some index k to j (i <= k)
        for( int k = i; k <= j ; k++ ) {
            String pal = s.substring(k, j+1);
            if(isPalindrome(pal)) return pal;
        }
        return "";
    }
    public static boolean isPalindrome( String s ) {

        char[] ch = s.toCharArray();
        int n = s.length();
        boolean isNot = false;
        for( int i = 0; i < s.length()/2; i++ ) {
            if( ch[i] != ch[n-1-i]) isNot = true;
        }
        return !isNot;
    }
    public static int findMin( int q, int[] a ) {
        int low = 0; int high = a.length-1;
        int res = -1;
        while( low <= high ) {
            int mid = (low+high)/2;
            if( a[mid] > q ) {
                high = mid-1;
            }
            else {
                res = mid;
                low = mid+1 ;
            }
        }
        if( res == -1 || a[res] > q ) return -1;
        return res+1;
    }
    public static int findMax( int q, int[] a ) {
        int low = 0; int high = a.length-1;
        int res = -1;
        while( low <= high ) {
            int mid = (low+high)/2;
            if( a[mid] >= q ) {
                res = mid; high = mid-1;
            }
            else {
                low = mid+1 ;
            }
        }

        if( res == -1 || a[res] < q ) return -1;
        return res+1;
    }
    public static int findGroup(int target, int[] a ) {
        // find larget i such that a[i] <= target
        int n = a.length;
        int low = 0; int high = n-1; int res = -1;
        while( low <= high && low >= 0 && high <= a.length-1 ) {
            int mid = (low+high)/2;

//            out.println(low + " " + high + " " + mid);
            if( a[mid] <= target ) {
                res = mid;
                low = mid+1;
            }
            else high = mid-1;

        }
        return res;
    }
    public static int findSubstrings(HashSet<String> set, String str) {
        for(int i = 0; i < str.length(); i++) {
            for (int j = i + 1; j <= str.length(); j++) {
                set.add(str.substring(i, j));
            }
        }
        return set.size();
    }
    public static int findIndex(int[][] a, int lMax ) {
        int low = 0;
        int j = -1;
        int high = a.length-1;
        while( low <= high ) {
            int mid = (low+high)/2;
            if( a[mid][0] <= lMax ) {
                j = mid;
                low = mid+1;
            } else {
                high = mid-1;
            }
        }
        return j;
    }
    public static void reverse2DArray(int[][] array) {
        int n = array.length;
        for (int i = 0; i < n / 2; i++) {
            int[] temp = new int[]{array[i][0], array[i][1]};
            array[i][0] = array[n-1-i][0]; array[i][1] = array[n-1-i][1];
            array[n-1-i][0] = temp[0]; array[n-1-i][1] = temp[1];
        }
    }
    static boolean findIfInOrder(int i, int[] a, HashMap<Integer, Integer> map ) {
        int n = a.length; int oper = 0;
        int st = i+1; int end = n-(i+1)+1;
//                out.println(st + end );
        int last = -1;
        boolean perform = true;
        for( int val = st; i != n/2 && val <= end; val++ ) {
            if( map.get(val) > last ) {
                last = map.get(val); //continue;
            }
            else perform = false;
        }
//        if( perform && map.get(i) < map.get(st) && map.get(n-i+1) > map.get(end)) {perform = false;}
        return perform;
    }
    static final int MOD = (int)(1e9 + 7);
    static long[] fac = new long[200001];

    static long power(long x, long y, int p) {
        long res = 1;
        x %= p;

        while (y > 0) {
            if ((y & 1) != 0)
                res = (res * x) % p;
            y >>= 1;
            x = (x * x) % p;
        }
        return res;
    }

    static long modInverse(long n, int p) {
        return power(n, p - 2, p);
    }

    static long nCr(int n, int r, int p) {
        if (n < r)
            return 0;

        if (r == 0)
            return 1;

        return fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p;
    }

    static void computeFactorials() {
        fac[0] = 1;
        for (int i = 1; i < fac.length; i++)
            fac[i] = fac[i - 1] * i % MOD;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases where the code might fail. Here are the steps to think through the problem:

1. **Understand the Bug**: The buggy code has issues with handling cases where the frequency of elements is not sufficient to form a valid rectangle. It also has logic flaws in selecting the minimum and maximum values for the coordinates.

2. **Edge Cases**:
   - **Insufficient Frequency**: Test cases where the frequency of elements is not enough to form a rectangle.
   - **Equal Elements**: Test cases where all elements are the same, which should form a degenerate rectangle.
   - **Mixed Elements**: Test cases with a mix of elements where the frequency is just enough to form a rectangle.
   - **Large Range**: Test cases with a large range of values to ensure the code handles large integers correctly.

3. **Construct the Test Case**:
   - **Test Case 1**: Insufficient frequency to form a rectangle.
   - **Test Case 2**: All elements are the same.
   - **Test Case 3**: Mixed elements with just enough frequency to form a rectangle.
   - **Test Case 4**: Large range of values.

Given the constraints, we will focus on a small-scale test input that is complete and valid.

### Test Case 1: Insufficient Frequency
```plaintext
1
10
1 2 3 4 5 6 7 8 9 10
```
This test case should output "NO" because there are no elements that appear at least twice.

### Test Case 2: All Elements are the Same
```plaintext
1
8
1 1 1 1 1 1 1 1
```
This test case should output "YES" with coordinates that form a degenerate rectangle.

### Test Case 3: Mixed Elements with Just Enough Frequency
```plaintext
1
12
1 1 2 2 3 3 4 4 5 6 7 8
```
This test case should output "YES" with coordinates that form a valid rectangle.

### Test Case 4: Large Range of Values
```plaintext
1
16
-1000000000 1000000000 1 2 3 4 5 6 7 8 9 10 11 12 13 14
```
This test case should output "NO" because there are no elements that appear at least twice.

### Final Test Input
```plaintext
4
10
1 2 3 4 5 6 7 8 9 10
8
1 1 1 1 1 1 1 1
12
1 1 2 2 3 3 4 4 5 6 7 8
16
-1000000000 1000000000 1 2 3 4 5 6 7 8 9 10 11 12 13 14
```

This test input covers a variety of edge cases and should help expose any bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DIY
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a list of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You need to pick $$$8$$$ elements from the list and use them as coordinates of four points. These four points should be corners of a rectangle which has its sides parallel to the coordinate axes. Your task is to pick coordinates in such a way that the resulting rectangle has the maximum possible area. The rectangle can be degenerate, i. e. its area can be $$$0$$$. Each integer can be used as many times as it occurs in the list (or less).

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 25\,000$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$8 \le n \le 2 \cdot 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print the answer as follows:

- if it is impossible to construct a rectangle which meets the constraints from the statement, print a single line containing the word NO (case-insensitive);
- otherwise, in the first line, print YES (case-insensitive). In the second line, print $$$8$$$ integers $$$x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4$$$ — the coordinates of the corners of the rectangle. You can print the corners in any order.

Examples:
input:
3
16
-5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10
8
0 0 -1 2 2 1 1 3
8
0 0 0 0 0 5 0 5
output:
YES
1 2 1 7 6 2 6 7
NO
YES
0 0 0 5 0 0 0 5

Note:

Buggy Code:
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Map;

public class CCC {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static String [] s ;

    public static void main(String [] args) throws Exception {
        int op = Integer.parseInt(br.readLine());
        while(op-->0){
            text();
        }
    }

    public static void text() throws Exception{
        int n = Integer.parseInt(br.readLine());
        s = br.readLine().split(" ");
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < n; i++) {
            int a = Integer.parseInt(s[i]);
            if(map.containsKey(a)){
                map.put(a,map.get(a)+1);
            }else{
                map.put(a,1);
            }
        }
        ArrayList<Integer> list = new ArrayList<>();
        map.forEach((k,v)->{
            for (int i = 0; i < v/2 ; i++) {
                list.add(k);
            }
        });
        if(list.size()<4){
            System.out.println("NO");
        }else{
            System.out.println("YES");
            int x1 = list.get(0);
            int x2 = list.get(1);
            int y1 = list.get(list.size()-2);
            int y2 = list.get(list.size()-1);
            System.out.println(x1+" "+x2+" "+y1+" "+y2+ " " +x1+" "+y2+" "+y1+" "+x2);

        }

    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on the following aspects:

1. **Edge Cases**: The code should handle cases where the input list has very few unique values or where the values are repeated in a way that makes it impossible to form a rectangle.
2. **Boundary Conditions**: Test the limits of the input constraints, such as the minimum and maximum values for `n` and the elements in the list.
3. **Invalid Rectangle Formation**: Create a test case where it is impossible to form a rectangle, but the code might incorrectly output a valid rectangle.

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

1. **Test Case 1**: A list with only one unique value repeated multiple times. This should result in a degenerate rectangle (area 0).
2. **Test Case 2**: A list with values that cannot form a valid rectangle (i.e., less than 4 unique values with sufficient occurrences).
3. **Test Case 3**: A list with values that can form a valid rectangle, but the code might incorrectly handle the coordinates.

Here is the test input:

```plaintext
3
8
1 1 1 1 1 1 1 1
8
1 2 3 4 5 6 7 8
16
-5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10
```

Explanation:
1. **Test Case 1**: All elements are 1. This should result in a degenerate rectangle (area 0).
2. **Test Case 2**: The elements are all unique and cannot form a rectangle with sides parallel to the coordinate axes.
3. **Test Case 3**: The elements can form a valid rectangle, but the code might incorrectly handle the coordinates.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DIY
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a list of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You need to pick $$$8$$$ elements from the list and use them as coordinates of four points. These four points should be corners of a rectangle which has its sides parallel to the coordinate axes. Your task is to pick coordinates in such a way that the resulting rectangle has the maximum possible area. The rectangle can be degenerate, i. e. its area can be $$$0$$$. Each integer can be used as many times as it occurs in the list (or less).

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 25\,000$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$8 \le n \le 2 \cdot 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print the answer as follows:

- if it is impossible to construct a rectangle which meets the constraints from the statement, print a single line containing the word NO (case-insensitive);
- otherwise, in the first line, print YES (case-insensitive). In the second line, print $$$8$$$ integers $$$x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4$$$ — the coordinates of the corners of the rectangle. You can print the corners in any order.

Examples:
input:
3
16
-5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10
8
0 0 -1 2 2 1 1 3
8
0 0 0 0 0 5 0 5
output:
YES
1 2 1 7 6 2 6 7
NO
YES
0 0 0 5 0 0 0 5

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
bool cmp(pair <int, int> a, pair <int, int> b){
    return a.second>b.second;
}
struct Cmp{
    bool operator()(const string& lhs, const string& rhs) const{
        return stol(lhs)<stol(rhs);
    }
};
bool isprime(int n){
    if(n<=1) return false;
    for(int i=2;i*i<=n;i++){
        if(n%i==0) return false;
    }
    return true;
}
int tc=1;
void s(int cn){

    ll n;
    cin>>n;

    map <string, ll, Cmp> mp;
    for(ll i=0;i<n;i++){
        string temp;
        cin>>temp;
        mp[temp]++;
    }
    // if(tc==14272 && cn!=5248){
    //     return;
    // }
    // if(tc==14272 && cn==5248){
    //     for(auto [first, second]: mp){
    //         cout<<first<<" "<<second<<endl;
    //     }
    //     return;
    // }
    ll cnt=0, amount=0;
    vector <pair<string, ll>> v;
    // for(auto itr=mp.begin();itr!=mp.end();itr++){
    //     if(itr->second>=2) cout<<itr->first<<" "<<itr->second<<endl;
    // }cout<<endl;
    for(auto it=mp.begin();it!=mp.end();){
        if(mp.size()==0 || amount>=2) break;
        if(it->second>=2){
            cnt+=(it->second - it->second%2);
            amount++;
            v.push_back({it->first, (it->second)});
            mp.erase(it);
            it=mp.begin();
        }
        else it++;
    }
    // for(auto [first, second]: v) cout<<first<<" ";cout<<endl;
    // cout<<"cnt: "<<cnt<<endl;
    amount=0;
    for(auto it=mp.rbegin();it!=mp.rend() && cnt<=8;){
        if(amount>=2) break;
        if(it->second>=2){
            // cout<<"it->fi - it->second%2rst: "<<it->first<<endl;
            cnt+=it->second;
            amount++;
            v.push_back({it->first, it->second});
            mp.erase(it->first);
            it=mp.rbegin();
        }
        else it++;
    }
    if(cnt<8) {
        cout<<"NO"<<endl;
        return;
    }
    cout<<"YES"<<endl;
    sort(v.begin(), v.end());
    vector <ll> V(9);
    V[0]=0;
    ll index=1;
    for(auto [first, second]: v) {
        for(ll i=1;i<=second;i++){
            V[index++]=stoi(first);
        }
    }
    sort(V.begin()+1, V.end());
    cout<<V[1]<<" "<<V[3]<<" "<<V[2]<<" "<<V[8]<<" "<<V[5]<<" "<<V[4]<<" "<<V[6]<<" "<<V[7]<<" "<<endl;

}
int main(int argc, char const *argv[]){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    #ifndef ONLINE_JUDGE
        freopen("in", "r", stdin);
        freopen("out","w", stdout);
    #endif

    cin>>tc;
    int cn=1;
    for(int i=1;i<=tc;i++){
        s(cn++);
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
8
0 0 0 0 0 0 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DIY
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a list of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You need to pick $$$8$$$ elements from the list and use them as coordinates of four points. These four points should be corners of a rectangle which has its sides parallel to the coordinate axes. Your task is to pick coordinates in such a way that the resulting rectangle has the maximum possible area. The rectangle can be degenerate, i. e. its area can be $$$0$$$. Each integer can be used as many times as it occurs in the list (or less).

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 25\,000$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$8 \le n \le 2 \cdot 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print the answer as follows:

- if it is impossible to construct a rectangle which meets the constraints from the statement, print a single line containing the word NO (case-insensitive);
- otherwise, in the first line, print YES (case-insensitive). In the second line, print $$$8$$$ integers $$$x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4$$$ — the coordinates of the corners of the rectangle. You can print the corners in any order.

Examples:
input:
3
16
-5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10
8
0 0 -1 2 2 1 1 3
8
0 0 0 0 0 5 0 5
output:
YES
1 2 1 7 6 2 6 7
NO
YES
0 0 0 5 0 0 0 5

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
bool cmp(pair <int, int> a, pair <int, int> b){
    return a.second>b.second;
}
struct Cmp{
    bool operator()(const std::string& lhs, const std::string& rhs) const{
        return stol(lhs)<stol(rhs);
    }
};
bool isprime(int n){
    if(n<=1) return false;
    for(int i=2;i*i<=n;i++){
        if(n%i==0) return false;
    }
    return true;
}
void s(){
    ll n;
    cin>>n;
    map <string, ll, Cmp> mp;
    for(ll i=0;i<n;i++){
        string temp;
        cin>>temp;
        mp[temp]++;
    }
    ll cnt=0, amount=0;
    vector <pair<string, ll>> v;
    // for(auto itr=mp.begin();itr!=mp.end();itr++){
    //     cout<<itr->first<<" ";
    // }cout<<endl;
    for(auto it=mp.begin();it!=mp.end();){
        if(mp.size()==0 || amount>=2) break;
        if((it->second)%2==0){
            cnt+=(it->second);
            amount++;
            v.push_back({it->first, (it->second)});
            mp.erase(it);
            it=mp.begin();
        }
        else it++;
    }
    // for(auto [first, second]: v) cout<<first<<" ";cout<<endl;
    // cout<<"cnt: "<<cnt<<endl;
    amount=0;
    for(auto it=mp.rbegin();it!=mp.rend() && cnt<=8;){
        if(amount>=2) break;
        if((it->second%2)==0){
            // cout<<"it->first: "<<it->first<<endl;
            cnt+=it->second;
            amount++;
            v.push_back({it->first, it->second});
            mp.erase(it->first);
            it=mp.rbegin();
        }
        else it++;
    }
    if(cnt<8) {
        cout<<"NO"<<endl;
        return;
    }
    cout<<"YES"<<endl;
    sort(v.begin(), v.end());
    vector <ll> V(9);
    V[0]=0;
    ll index=1;
    for(auto [first, second]: v) {
        for(ll i=1;i<=second;i++){
            V[index++]=stoi(first);
        }
    }
    sort(V.begin()+1, V.end());
    cout<<V[1]<<" "<<V[3]<<" "<<V[2]<<" "<<V[8]<<" "<<V[5]<<" "<<V[4]<<" "<<V[6]<<" "<<V[7]<<" "<<endl;

}
int main(int argc, char const *argv[]){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    #ifndef ONLINE_JUDGE
        freopen("in", "r", stdin);
        freopen("out","w", stdout);
    #endif
    int tc=1;
    cin>>tc;
    while(tc--){
        s();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
8
0 0 0 0 0 0 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DIY
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a list of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You need to pick $$$8$$$ elements from the list and use them as coordinates of four points. These four points should be corners of a rectangle which has its sides parallel to the coordinate axes. Your task is to pick coordinates in such a way that the resulting rectangle has the maximum possible area. The rectangle can be degenerate, i. e. its area can be $$$0$$$. Each integer can be used as many times as it occurs in the list (or less).

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 25\,000$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$8 \le n \le 2 \cdot 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print the answer as follows:

- if it is impossible to construct a rectangle which meets the constraints from the statement, print a single line containing the word NO (case-insensitive);
- otherwise, in the first line, print YES (case-insensitive). In the second line, print $$$8$$$ integers $$$x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4$$$ — the coordinates of the corners of the rectangle. You can print the corners in any order.

Examples:
input:
3
16
-5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10
8
0 0 -1 2 2 1 1 3
8
0 0 0 0 0 5 0 5
output:
YES
1 2 1 7 6 2 6 7
NO
YES
0 0 0 5 0 0 0 5

Note:

Buggy Code:
from collections import Counter
import math

for s in [*open(0)][2::2]:
    a = list(map(int, s.split()))
    b = []
    res = Counter(a)
    st = set(a)
    chk = 0
    for i in st:
        if res[i] >= 2:
            b.append(i)
            chk += res[i]
    lb = len(b)
    if lb < 4 and chk < 8:
        print('NO')

    else:
        b.sort()

        if lb == 1:
            print('YES')
            print(b[0], b[0], b[0], b[0], b[0], b[0], b[0], b[0])

        elif lb == 2:
            if res[b[0]] == res[b[1]]:
                print('YES')
                print(b[0], b[0], b[0], b[1], b[1], b[0], b[1], b[1])
            elif res[b[0]] == 6:
                print('YES')
                print(b[0], b[0], b[0], b[1], b[0], b[0], b[0], b[1])
            elif res[b[1]] == 6:
                print('YES')
                print(b[1], b[1], b[1], b[0], b[1], b[1], b[1], b[0])

        elif max(res[b[0]], res[b[-1]], res[b[1]], res[b[-2]]) == 2:
            b.sort()
            s1 = (b[0]-b[-1]) * (b[1]-b[-2])
            s2 = (b[0]-b[-2]) * (b[1]-b[-1])
            print('YES')
            if s1 > s2:
                print(b[0], b[1], b[0], b[-2], b[-1], b[1], b[-1], b[-2])
            else:
                print(b[0], b[1], b[0], b[-1], b[-2], b[1], b[-2], b[-1])

        else:
            b.sort()
            print('YES')

            if res[b[0]] and res[b[-1]] >= 4:
                print(b[0],b[0],b[0],b[-1],b[-1],b[0],b[-1],b[-1])

            elif res[b[0]] and res[b[-2]] >= 4:
                s1 = (b[0] - b[-2]) ** 2
                s2 = (b[0] - b[-1]) * (b[0] - b[-2])
                if s1 > s2:
                    print(b[0],b[0],b[0],b[-2],b[-2],b[0],b[-2],b[-2])
                else :
                    print(b[0],b[0],b[0],b[-2],b[-1],b[0],b[-1],b[-2])
            else:
                s1 = (b[0]-b[-1]) * (b[1]-b[-2])
            s2 = (b[0]-b[-2]) * (b[1]-b[-1])
            print('YES')
            if s1 > s2:
                print(b[0], b[1], b[0], b[-2], b[-1], b[1], b[-1], b[-2])
            else:
                print(b[0], b[1], b[0], b[-1], b[-2], b[1], b[-2], b[-1])

**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
8
1 1 1 1 2 2 2 2
8
1 1 1 1 1 1 2 2
8
1 1 1 1 1 1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DIY
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a list of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You need to pick $$$8$$$ elements from the list and use them as coordinates of four points. These four points should be corners of a rectangle which has its sides parallel to the coordinate axes. Your task is to pick coordinates in such a way that the resulting rectangle has the maximum possible area. The rectangle can be degenerate, i. e. its area can be $$$0$$$. Each integer can be used as many times as it occurs in the list (or less).

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 25\,000$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$8 \le n \le 2 \cdot 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print the answer as follows:

- if it is impossible to construct a rectangle which meets the constraints from the statement, print a single line containing the word NO (case-insensitive);
- otherwise, in the first line, print YES (case-insensitive). In the second line, print $$$8$$$ integers $$$x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4$$$ — the coordinates of the corners of the rectangle. You can print the corners in any order.

Examples:
input:
3
16
-5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10
8
0 0 -1 2 2 1 1 3
8
0 0 0 0 0 5 0 5
output:
YES
1 2 1 7 6 2 6 7
NO
YES
0 0 0 5 0 0 0 5

Note:

Buggy Code:
# https://codeforces.com/problemset/problem/2038/C

def solve():
    n = int(input())
    s = list(map(int, input().split()))
    if len(set(s)) > n-4 or n < 8: return "NO"
    print("YES")
    coord = []
    s.sort()
    for i in range(n):
        if s[i] not in coord:
            coord += [s[i]]*(s.count(s[i])//2)

    coord = [str(coord[0]), str(coord[1]), str(coord[0]), str(coord[-1]), str(coord[-2]), str(coord[1]), str(coord[-2]), str(coord[-1])]
    return " ".join(coord)

for _ in range(int(input())): print(solve())

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

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

Think step by step.

Response:
```plaintext
1
8
1 1 1 1 1 1 1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DIY
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a list of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You need to pick $$$8$$$ elements from the list and use them as coordinates of four points. These four points should be corners of a rectangle which has its sides parallel to the coordinate axes. Your task is to pick coordinates in such a way that the resulting rectangle has the maximum possible area. The rectangle can be degenerate, i. e. its area can be $$$0$$$. Each integer can be used as many times as it occurs in the list (or less).

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 25\,000$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$8 \le n \le 2 \cdot 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print the answer as follows:

- if it is impossible to construct a rectangle which meets the constraints from the statement, print a single line containing the word NO (case-insensitive);
- otherwise, in the first line, print YES (case-insensitive). In the second line, print $$$8$$$ integers $$$x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4$$$ — the coordinates of the corners of the rectangle. You can print the corners in any order.

Examples:
input:
3
16
-5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10
8
0 0 -1 2 2 1 1 3
8
0 0 0 0 0 5 0 5
output:
YES
1 2 1 7 6 2 6 7
NO
YES
0 0 0 5 0 0 0 5

Note:

Buggy Code:
import java.math.BigInteger;
import java.util.*;

public class Solution {
    public static class StringHashing {
        private static final long MOD = (long)1e9 + 7;
        private static final int BASE1 = 43;  // First prime base
        private static final int BASE2 = 57;  // Second prime base

        private final String s;
        private final List<Long> hash1;
        private final List<Long> hash2;
        private final List<Long> power1;
        private final List<Long> power2;

        public StringHashing(String s) {
            this.s = s;
            int n = s.length();
            hash1 = new ArrayList<>(n + 1);
            hash2 = new ArrayList<>(n + 1);
            power1 = new ArrayList<>(n + 1);
            power2 = new ArrayList<>(n + 1);

            precomputeHashes();
        }

        private void precomputeHashes() {
            int n = s.length();

            hash1.add(0L);
            hash2.add(0L);
            power1.add(1L);
            power2.add(1L);

            for (int i = 0; i < n; i++) {
                char c = s.charAt(i);
                long newHash1 = (hash1.get(i) * BASE1 + c) % MOD;
                hash1.add(newHash1);
                long newHash2 = (hash2.get(i) * BASE2 + c) % MOD;
                hash2.add(newHash2);
                long newPower1 = (power1.get(i) * BASE1) % MOD;
                power1.add(newPower1);
                long newPower2 = (power2.get(i) * BASE2) % MOD;
                power2.add(newPower2);
            }
        }

        public Pair<Long, Long> getHash(int left, int right) {
            long hashtemp1 = (hash1.get(right + 1) - (hash1.get(left) * power1.get(right - left + 1)) % MOD + MOD) % MOD;
            long hashtemp2 = (hash2.get(right + 1) - (hash2.get(left) * power2.get(right - left + 1)) % MOD + MOD) % MOD;
            return new Pair<>(hashtemp1, hashtemp2);
        }
    }

    public static class Pair<T1, T2> {
        private T1 first;
        private T2 second;

        public Pair(T1 first, T2 second) {
            this.first = first;
            this.second = second;
        }

        public T1 getFirst() {
            return first;
        }

        public T2 getSecond() {
            return second;
        }

        @Override
        public String toString() {
            return "Pair{" + "first=" + first + ", second=" + second + '}';
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Pair<?, ?> pair = (Pair<?, ?>) o;
            return first.equals(pair.first) && second.equals(pair.second);
        }

        @Override
        public int hashCode() {
            return 31 * first.hashCode() + second.hashCode();
        }
    }

    public static int phi(int n) {
       int x=n;
        for (long i = 2; i * i <= n; i++) {
            if (n % i == 0) {

               n/=i;
               x-=n/i;
            }
        }
        if(n>1)x-=x/n;
        return x;
    }

    public static ArrayList<Long> factors(long n) {
        ArrayList<Long> x = new ArrayList<>();
        for (long i = 2; i * i <= n; i++) {
            if (n % i == 0) {
                x.add(i);
                if (n / i != i) {
                    x.add(n / i);
                }
            }
        }
        if(n!=1){
            x.add((long)1);
            x.add(n);
        }
        else{
            x.add(n);
        }

        return x;
    }

    public static ArrayList<Pair<Long, Integer>> primeFactorization(long n) {
        ArrayList<Pair<Long, Integer>> x = new ArrayList<>();
        for (long i = 2; i * i <= n; i++) {
            if (n % i == 0) {
                int count = 0;
                while (n % i == 0) {
                    count++;
                    n /= i;
                }
                x.add(new Pair<>(i, count));
            }
        }
        if (n > 1) {
            x.add(new Pair<>(n, 1));
        }
        return x;
    }

    public static boolean[] sieve(int n) {
        boolean[] isprime = new boolean[n + 1];
        Arrays.fill(isprime, true);
        isprime[0] = false;
        isprime[1] = false;
        for (int i = 2; i * i <= n; i++) {
            if (isprime[i]) {
                for (int j = i * i; j <= n; j += i) {
                    isprime[j] = false;
                }
            }
        }
        return isprime;
    }

    public static long gcd(long a, long b) {
        while (b != 0) {
            long temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }

    public static int setBits(int num, int bit) {
        int cyc = (num + 1) / (1 << (bit + 1));
        int remain = (num + 1) % (1 << (bit + 1));
        return cyc * (1 << bit) + Math.max(0, remain - (1 << bit));
    }

    public static int flip(int l, int r) {
        int tot = r - l + 1;
        int max1 = 0;
        for (int i = 0; i < 31; i++) {
            int l1 = setBits(l - 1, i);
            int r1 = setBits(r, i);
            max1 = Math.max(r1 - l1, max1);
        }
        return tot - max1;
    }

    public static boolean predicate(int mid,String k,String l,StringHashing a,StringHashing b) {
        Pair<Long,Long>p1=a.getHash(0, mid-1);
        Pair<Long,Long>p2=b.getHash(k.length()-mid, k.length()-1);
        Pair<Long,Long>p3=b.getHash(0, mid-1);
       Pair<Long,Long>p4=a.getHash(k.length()-mid, k.length()-1);
        System.out.println(p1+" "+p2);
       if(p1.equals(p2)||p3.equals(p4))return true;

       return false;

    }

// Helper function to find modular inverse
public static long modInverse(long a, long mod) {
    return power(a, mod - 2, mod);
}

public static long power(long x, long y, long mod) {
    long res = 1;
    x = x % mod;
    while (y > 0) {
        if ((y & 1) == 1)
            res = (res * x) % mod;
        y = y >> 1;
        x = (x * x) % mod;
    }
    return res;
}

  public static void main(String[] args)  {
        Solution solution = new Solution();
       Scanner sc=new Scanner(System.in);
//  int max_n=(int)(2e6+2);
//         long MOD=(long)( 998244353)
//         ;

//         long fact[]=new long[max_n+1];
//         long ifact[]=new long[max_n+1];
//         fact[0] = 1;
//         for (int i = 1; i <= max_n; i++) {
//             fact[i] = (fact[i - 1] * i) % MOD;
//         }

//         ifact[max_n] = modInverse(fact[max_n], MOD);
//         for (int i = max_n - 1; i >= 0; i--) {
//             ifact[i] = (ifact[i + 1] * (i + 1)) % MOD;
//         }

int t=sc.nextInt();
while(t-->0){
    int n=sc.nextInt();
    long arr[]=new long[n];
    TreeMap<Long,Integer> tm=new TreeMap<>();
    for(int i=0;i<n;i++){
        arr[i]=sc.nextLong();
        tm.put(arr[i],tm.getOrDefault(arr[i],0)+1);
    }
    ArrayList<Long> x=new ArrayList<>();
    for(long l:tm.keySet()){
if(tm.get(l)>=2){
    int k=tm.get(l);
    while(k%2==0&&k!=0){
        x.add(l);
       k-=2;
    }
}
    }
    if(x.size()<4){
        System.out.println("NO");
    }
    else{
        Collections.sort(x);
        System.out.println("YES");
        System.out.println(x.get(0)+" "+x.get(1)+" "+x.get(0)+" "+x.get(x.size()-1)+" "+x.get(x.size()-2)+" "+x.get(1)+" "+x.get(x.size()-2)+" "+x.get(x.size()-1));
    }
}
}

}

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

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

Think step by step.

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

1. **Identify the Bug**: The buggy code has a logical error in the way it selects the coordinates for the rectangle. It only considers the elements that appear at least twice and tries to form a rectangle using these elements. However, it does not correctly handle cases where there are not enough unique elements to form a valid rectangle.

2. **Edge Cases**:
   - **Case 1**: All elements are unique, making it impossible to form a rectangle.
   - **Case 2**: There are not enough elements that appear at least twice to form a valid rectangle.
   - **Case 3**: There are exactly enough elements that appear at least twice to form a valid rectangle, but they are not in the correct order.

3. **Generate Test Input**:
   - **Case 1**: All elements are unique.
   - **Case 2**: Only a few elements appear more than once, but not enough to form a valid rectangle.
   - **Case 3**: Exactly four elements appear at least twice, but in a way that makes it difficult to form a valid rectangle.

Here is a small-scale test input that includes these edge cases:

```plaintext
3
8
1 2 3 4 5 6 7 8
12
1 1 2 2 3 4 5 6 7 8 9 10
16
1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4
```

### Explanation:
1. **First Test Case**:
   - **Input**: `8 1 2 3 4 5 6 7 8`
   - **Expected Output**: `NO`
   - **Reason**: All elements are unique, making it impossible to form a rectangle.

2. **Second Test Case**:
   - **Input**: `12 1 1 2 2 3 4 5 6 7 8 9 10`
   - **Expected Output**: `NO`
   - **Reason**: Only two elements (`1` and `2`) appear more than once, which is not enough to form a valid rectangle.

3. **Third Test Case**:
   - **Input**: `16 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4`
   - **Expected Output**: `YES 1 1 1 4 4 1 4 4`
   - **Reason**: There are exactly four unique elements, each appearing at least twice, making it possible to form a valid rectangle.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DIY
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a list of $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. You need to pick $$$8$$$ elements from the list and use them as coordinates of four points. These four points should be corners of a rectangle which has its sides parallel to the coordinate axes. Your task is to pick coordinates in such a way that the resulting rectangle has the maximum possible area. The rectangle can be degenerate, i. e. its area can be $$$0$$$. Each integer can be used as many times as it occurs in the list (or less).

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 25\,000$$$) — the number of test cases.

The first line of each test case contains one integer $$$n$$$ ($$$8 \le n \le 2 \cdot 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print the answer as follows:

- if it is impossible to construct a rectangle which meets the constraints from the statement, print a single line containing the word NO (case-insensitive);
- otherwise, in the first line, print YES (case-insensitive). In the second line, print $$$8$$$ integers $$$x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4$$$ — the coordinates of the corners of the rectangle. You can print the corners in any order.

Examples:
input:
3
16
-5 1 1 2 2 3 3 4 4 5 5 6 6 7 7 10
8
0 0 -1 2 2 1 1 3
8
0 0 0 0 0 5 0 5
output:
YES
1 2 1 7 6 2 6 7
NO
YES
0 0 0 5 0 0 0 5

Note:

Buggy Code:
import java.util.*;

public class contest2 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = Integer.parseInt(sc.nextLine());
        while(t>0){
            t--;
            int n = sc.nextInt();
            int[] arr = new int[n];
            for(int i=0; i<n; i++){
                arr[i] = sc.nextInt();
            }
            Arrays.sort(arr);
            Stack<Integer> st = new Stack<>();
            ArrayList<Integer> ar = new ArrayList<>();
            int count = 0;
            int temp = arr[0];
            for(int i=0; i<n; i++){
                if(arr[i]==temp) count++;
                else count=1;
                if(count==2) {
                    count=0;
                    st.push(arr[i]);
                    if(ar.size()==0 || ar.get(ar.size()-1)!=arr[i]){
                        ar.add(arr[i]);
                    }
                }
                temp=arr[i];
            }
            if(st.size()<4) System.out.println("NO");
            else{
                System.out.println("Yes");
                if(ar.size()>3){
                    int a1 = ar.get(0);
                    int a2 = ar.get(1);
                    int a3 = ar.get(ar.size()-1);
                    int a4 = ar.get(ar.size()-2);
                    if((a3-a1)*(a4-a2)>=(a4-a1)*(a3-a2)){
                        System.out.println(a1+" "+a2+" "+a1+" "+a4+" "+a3+" "+a2+" "+a3+" "+a4);
                    }
                    else System.out.println(a1+" "+a2+" "+a1+" "+a3+" "+a4+" "+a2+" "+a4+" "+a3);
                }
                else{
                    int a1 = st.pop();
                    int a2 = st.pop();
                    int a3 = st.pop();
                    int a4 = st.pop();
                    System.out.println(a1+" "+a2+" "+a1+" "+a4+" "+a3+" "+a2+" "+a3+" "+a4);
                }
            }
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
8
0 0 0 0 1 1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Annual Ants' Gathering
time_limit_ms: 2000
memory_limit_mb: 256
Description: Deep within a forest lies an ancient tree, home to $$$n$$$ ants living in $$$n$$$ tiny houses, indexed from $$$1$$$ to $$$n$$$, connected by the branches of the tree.

Once a year, all the ants need to gather to watch the EUC. For this, all ants move along the $$$n-1$$$ branches of the tree they live on to meet at the home of one ant.

However, this year the ants could not agree on where to meet and need your help to gather up. You can tell all the ants currently at house $$$u$$$ to move to house $$$v$$$ if there is a branch directly connecting those two houses. However, the ants ignore your command if there are fewer ants gathered in house $$$v$$$ than in house $$$u$$$, i.e., if it would be easier for the ants from house $$$v$$$ to move. This even holds true if no ant at all is currently in house $$$v$$$. You can give this kind of commands as many times as you want.

Is it possible for you to gather all the ants in a single house?

Input: The first line contains one integer $$$n$$$ ($$$1\leq n\leq 200\,000$$$) — the number of ant homes.

Each of the following $$$n-1$$$ lines contains two integers $$$u$$$ and $$$v$$$ ($$$1\leq u, v\leq n$$$) — there is a branch directly connecting the house $$$u$$$ and house $$$v$$$. It is guaranteed that every ant can reach the house of any other ant just by following the branches of the tree.

Output: Print $$$\texttt{YES}$$$ if it is possible to gather all the ants in a single house. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
7
5 1
3 2
4 6
3 6
7 1
1 3
output:
YES

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

input:
6
4 5
5 6
6 1
2 6
3 2
output:
YES

Note: In the first sample, you can gather all the ants at house $$$3$$$ as follows:

- You tell to the ant at house $$$4$$$ to move to house $$$6$$$.
- You tell to the ant at house $$$2$$$ to move to house $$$3$$$.
- You tell to the two ants at house $$$6$$$ to move to house $$$3$$$ (which already contains two ants).
- You tell to the ant at house $$$5$$$ to move to house $$$1$$$.
- You tell to the ant at house $$$7$$$ to move to house $$$1$$$ (which already contains two ants).
- You tell to the three ants at house $$$1$$$ to move to house $$$3$$$ (which already contains four ants).

In the second sample, it is impossible to gather all the ants in a single house.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
long long int t,i,j,l,n,m,a,b,c,x,y,p,aux,minn,maxx,ok,cnt,ras,frec[400008],aux2,val[400008],k,unu,doi,ok2,ultnod,viz[400008];
char ch;
struct element
{
    long long x,y,r;
};
set<long long>dat[400008];
vector<long long>G[400008];
bool verificare(long long nod)
{
    aux=1;
    for(auto j:dat[nod])
    {
        if(aux<j)
            return 0;
        aux+=j;
    }
    return 1;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin>>n;
    for(i=1; i<n; i++)
    {
        cin>>a>>b;
        G[a].push_back(b);
        G[b].push_back(a);
        frec[a]++;
        frec[b]++;
    }
    queue<long long>Q;
    for(i=1; i<=n; i++)
    {
        if(frec[i]==1)
        {
            Q.push(i);
            frec[i]=0;
        }
        val[i]=1;
    }
    ok=1;
    while(!Q.empty())
    {
        long long nod=Q.front();
        Q.pop();
        if(val[nod]<n/2)
        {
            viz[nod]=1;
            for(auto i:G[nod])
            {
                if(frec[i]!=0)
                {
                    frec[i]--;
                    dat[i].insert(val[nod]);
                    val[i]+=val[nod];
                    if(frec[i]==1)
                    {
                        Q.push(i);
                        frec[i]=0;
                    }
                }
            }
        }
    }
    for(i=1;i<=n;i++)
    {
        if(verificare(i)==0)
        {
            ok=0;
            break;
        }
    }
    if(ok==1)
        cout<<"YES";
    else
        cout<<"NO";
    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
1 2
2 3
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Annual Ants' Gathering
time_limit_ms: 2000
memory_limit_mb: 256
Description: Deep within a forest lies an ancient tree, home to $$$n$$$ ants living in $$$n$$$ tiny houses, indexed from $$$1$$$ to $$$n$$$, connected by the branches of the tree.

Once a year, all the ants need to gather to watch the EUC. For this, all ants move along the $$$n-1$$$ branches of the tree they live on to meet at the home of one ant.

However, this year the ants could not agree on where to meet and need your help to gather up. You can tell all the ants currently at house $$$u$$$ to move to house $$$v$$$ if there is a branch directly connecting those two houses. However, the ants ignore your command if there are fewer ants gathered in house $$$v$$$ than in house $$$u$$$, i.e., if it would be easier for the ants from house $$$v$$$ to move. This even holds true if no ant at all is currently in house $$$v$$$. You can give this kind of commands as many times as you want.

Is it possible for you to gather all the ants in a single house?

Input: The first line contains one integer $$$n$$$ ($$$1\leq n\leq 200\,000$$$) — the number of ant homes.

Each of the following $$$n-1$$$ lines contains two integers $$$u$$$ and $$$v$$$ ($$$1\leq u, v\leq n$$$) — there is a branch directly connecting the house $$$u$$$ and house $$$v$$$. It is guaranteed that every ant can reach the house of any other ant just by following the branches of the tree.

Output: Print $$$\texttt{YES}$$$ if it is possible to gather all the ants in a single house. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
7
5 1
3 2
4 6
3 6
7 1
1 3
output:
YES

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

input:
6
4 5
5 6
6 1
2 6
3 2
output:
YES

Note: In the first sample, you can gather all the ants at house $$$3$$$ as follows:

- You tell to the ant at house $$$4$$$ to move to house $$$6$$$.
- You tell to the ant at house $$$2$$$ to move to house $$$3$$$.
- You tell to the two ants at house $$$6$$$ to move to house $$$3$$$ (which already contains two ants).
- You tell to the ant at house $$$5$$$ to move to house $$$1$$$.
- You tell to the ant at house $$$7$$$ to move to house $$$1$$$ (which already contains two ants).
- You tell to the three ants at house $$$1$$$ to move to house $$$3$$$ (which already contains four ants).

In the second sample, it is impossible to gather all the ants in a single house.

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

// @author : amaiya

//Main Driver Code
int main(){
    ios_base::sync_with_stdio(false);
    cin. tie(NULL);

    int n;
    cin >> n;

    vector<vector<int>> adj(n);
    int u, v;
    for (int i = 0; i < n-1; i++){
        cin >> u >> v;
        u--; v--;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }

    vector<int> leaves;
    vector<bool> vis(n, false);

    for (int i = 0; i < n; i++){
        if (adj[i].size() == 1) leaves.push_back(i);
    }

    vector<int> people(n, 1);
    bool poss = true;
    int index = 0;
    vector<int> count(n, 0);
    while (leaves.size() - index > 1){
        vector<int> next_leaves;
        for (auto leaf: leaves){
            for (auto neighbour: adj[leaf]){
                if (!vis[neighbour]){
                    if (people[neighbour] >= people[leaf]){
                        // next_leaves.push_back(neighbour);
                        people[neighbour] += people[leaf];
                        people[leaf] = 0;

                        count[neighbour]++;
                        if (adj[neighbour].size() - count[neighbour] <= 1){
                            next_leaves.push_back(neighbour);
                        }
                    }else{
                        poss = false;
                    }
                }
            }
            vis[leaf] = true;
        }
        index = leaves.size();

        leaves = next_leaves;
        if (!poss) break;

        // leaves.clear();
        // for (auto leaf : next_leaves){
        //     int count = adj[leaf].size();
        //     for (auto neighbour: adj[leaf]){
        //         if (vis[neighbour]) count--;
        //     }
        //     if (count <= 1){
        //         leaves.push_back(leaf);
        //     }
        // }
    }

    cout << ((poss)?"YES":"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
4
1 2
2 3
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Annual Ants' Gathering
time_limit_ms: 2000
memory_limit_mb: 256
Description: Deep within a forest lies an ancient tree, home to $$$n$$$ ants living in $$$n$$$ tiny houses, indexed from $$$1$$$ to $$$n$$$, connected by the branches of the tree.

Once a year, all the ants need to gather to watch the EUC. For this, all ants move along the $$$n-1$$$ branches of the tree they live on to meet at the home of one ant.

However, this year the ants could not agree on where to meet and need your help to gather up. You can tell all the ants currently at house $$$u$$$ to move to house $$$v$$$ if there is a branch directly connecting those two houses. However, the ants ignore your command if there are fewer ants gathered in house $$$v$$$ than in house $$$u$$$, i.e., if it would be easier for the ants from house $$$v$$$ to move. This even holds true if no ant at all is currently in house $$$v$$$. You can give this kind of commands as many times as you want.

Is it possible for you to gather all the ants in a single house?

Input: The first line contains one integer $$$n$$$ ($$$1\leq n\leq 200\,000$$$) — the number of ant homes.

Each of the following $$$n-1$$$ lines contains two integers $$$u$$$ and $$$v$$$ ($$$1\leq u, v\leq n$$$) — there is a branch directly connecting the house $$$u$$$ and house $$$v$$$. It is guaranteed that every ant can reach the house of any other ant just by following the branches of the tree.

Output: Print $$$\texttt{YES}$$$ if it is possible to gather all the ants in a single house. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
7
5 1
3 2
4 6
3 6
7 1
1 3
output:
YES

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

input:
6
4 5
5 6
6 1
2 6
3 2
output:
YES

Note: In the first sample, you can gather all the ants at house $$$3$$$ as follows:

- You tell to the ant at house $$$4$$$ to move to house $$$6$$$.
- You tell to the ant at house $$$2$$$ to move to house $$$3$$$.
- You tell to the two ants at house $$$6$$$ to move to house $$$3$$$ (which already contains two ants).
- You tell to the ant at house $$$5$$$ to move to house $$$1$$$.
- You tell to the ant at house $$$7$$$ to move to house $$$1$$$ (which already contains two ants).
- You tell to the three ants at house $$$1$$$ to move to house $$$3$$$ (which already contains four ants).

In the second sample, it is impossible to gather all the ants in a single house.

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

using namespace std;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

int main() {
    fastio;

    long long n;
    cin >> n;

    map<long long, vector<long long>> edges;
    map<long long, long long> degree;
    map<long long, long long> ants;
    int flag = 0;
    for (long long i = 0; i < n - 1; ++i) {
        long long x, y;
        cin >> x >> y;
        if(i == 0 && x == 186416)
            flag = 1;
        edges[x].push_back(y);
        edges[y].push_back(x);
        degree[x]++;
        degree[y]++;
        ants[x] = 1;
        ants[y] = 1;
    }

    while (true) {
        vector<long long> remove;
        vector<long long> decrease;
        for (auto& entry : degree) {
            long long key = entry.first;
            long long val = entry.second;
            if (val == 0 || val == 1) {
                remove.push_back(key);
                for (long long elem : edges[key]) {
                    if (degree.find(elem) != degree.end()) {
                        if (ants[elem] >= ants[key]) {
                            decrease.push_back(elem);
                        }
                    }
                }
            }
        }

        for (size_t i = 0; i < decrease.size(); ++i) {
            long long key = decrease[i];
            degree[key]--;
            ants[key] += ants[remove[i]];
        }

        for (long long key : remove) {
            degree.erase(key);
            edges.erase(key);
            ants.erase(key);
        }

        if (remove.empty()) {
            break;
        }
    }

    if (!degree.empty() && flag) {
        cout << "No" << endl;
    } else {
        cout << "Yes" << 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
7
1 2
2 3
3 4
4 5
5 6
6 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Annual Ants' Gathering
time_limit_ms: 2000
memory_limit_mb: 256
Description: Deep within a forest lies an ancient tree, home to $$$n$$$ ants living in $$$n$$$ tiny houses, indexed from $$$1$$$ to $$$n$$$, connected by the branches of the tree.

Once a year, all the ants need to gather to watch the EUC. For this, all ants move along the $$$n-1$$$ branches of the tree they live on to meet at the home of one ant.

However, this year the ants could not agree on where to meet and need your help to gather up. You can tell all the ants currently at house $$$u$$$ to move to house $$$v$$$ if there is a branch directly connecting those two houses. However, the ants ignore your command if there are fewer ants gathered in house $$$v$$$ than in house $$$u$$$, i.e., if it would be easier for the ants from house $$$v$$$ to move. This even holds true if no ant at all is currently in house $$$v$$$. You can give this kind of commands as many times as you want.

Is it possible for you to gather all the ants in a single house?

Input: The first line contains one integer $$$n$$$ ($$$1\leq n\leq 200\,000$$$) — the number of ant homes.

Each of the following $$$n-1$$$ lines contains two integers $$$u$$$ and $$$v$$$ ($$$1\leq u, v\leq n$$$) — there is a branch directly connecting the house $$$u$$$ and house $$$v$$$. It is guaranteed that every ant can reach the house of any other ant just by following the branches of the tree.

Output: Print $$$\texttt{YES}$$$ if it is possible to gather all the ants in a single house. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
7
5 1
3 2
4 6
3 6
7 1
1 3
output:
YES

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

input:
6
4 5
5 6
6 1
2 6
3 2
output:
YES

Note: In the first sample, you can gather all the ants at house $$$3$$$ as follows:

- You tell to the ant at house $$$4$$$ to move to house $$$6$$$.
- You tell to the ant at house $$$2$$$ to move to house $$$3$$$.
- You tell to the two ants at house $$$6$$$ to move to house $$$3$$$ (which already contains two ants).
- You tell to the ant at house $$$5$$$ to move to house $$$1$$$.
- You tell to the ant at house $$$7$$$ to move to house $$$1$$$ (which already contains two ants).
- You tell to the three ants at house $$$1$$$ to move to house $$$3$$$ (which already contains four ants).

In the second sample, it is impossible to gather all the ants in a single house.

Buggy Code:
def can_gather_ants(n, edges):
    if n == 1:
        return "YES"

    from collections import defaultdict

    degree = defaultdict(int)

    for u, v in edges:
        degree[u] += 1
        degree[v] += 1

    # Check the degrees
    degree_count = defaultdict(int)
    for d in degree.values():
        degree_count[d] += 1

    # If any node has degree > 2, we can gather all ants there
    if any(d > 2 for d in degree.values()):
        return "YES"

    # If all nodes have degree 2 or less, check if it's a path
    if degree_count[1] == 2 and degree_count[2] == n - 2:
        return "YES"

    return "NO"

# Read input
import sys
input = sys.stdin.read
data = input().split()

n = int(data[0])
edges = []

index = 1
for _ in range(n - 1):
    u = int(data[index])
    v = int(data[index + 1])
    edges.append((u, v))
    index += 2

# Output the result
print(can_gather_ants(n, edges))

**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
1 2
2 3
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Annual Ants' Gathering
time_limit_ms: 2000
memory_limit_mb: 256
Description: Deep within a forest lies an ancient tree, home to $$$n$$$ ants living in $$$n$$$ tiny houses, indexed from $$$1$$$ to $$$n$$$, connected by the branches of the tree.

Once a year, all the ants need to gather to watch the EUC. For this, all ants move along the $$$n-1$$$ branches of the tree they live on to meet at the home of one ant.

However, this year the ants could not agree on where to meet and need your help to gather up. You can tell all the ants currently at house $$$u$$$ to move to house $$$v$$$ if there is a branch directly connecting those two houses. However, the ants ignore your command if there are fewer ants gathered in house $$$v$$$ than in house $$$u$$$, i.e., if it would be easier for the ants from house $$$v$$$ to move. This even holds true if no ant at all is currently in house $$$v$$$. You can give this kind of commands as many times as you want.

Is it possible for you to gather all the ants in a single house?

Input: The first line contains one integer $$$n$$$ ($$$1\leq n\leq 200\,000$$$) — the number of ant homes.

Each of the following $$$n-1$$$ lines contains two integers $$$u$$$ and $$$v$$$ ($$$1\leq u, v\leq n$$$) — there is a branch directly connecting the house $$$u$$$ and house $$$v$$$. It is guaranteed that every ant can reach the house of any other ant just by following the branches of the tree.

Output: Print $$$\texttt{YES}$$$ if it is possible to gather all the ants in a single house. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
7
5 1
3 2
4 6
3 6
7 1
1 3
output:
YES

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

input:
6
4 5
5 6
6 1
2 6
3 2
output:
YES

Note: In the first sample, you can gather all the ants at house $$$3$$$ as follows:

- You tell to the ant at house $$$4$$$ to move to house $$$6$$$.
- You tell to the ant at house $$$2$$$ to move to house $$$3$$$.
- You tell to the two ants at house $$$6$$$ to move to house $$$3$$$ (which already contains two ants).
- You tell to the ant at house $$$5$$$ to move to house $$$1$$$.
- You tell to the ant at house $$$7$$$ to move to house $$$1$$$ (which already contains two ants).
- You tell to the three ants at house $$$1$$$ to move to house $$$3$$$ (which already contains four ants).

In the second sample, it is impossible to gather all the ants in a single house.

Buggy Code:
def can_gather_ants(n, edges):
    from collections import defaultdict, deque

    # Create adjacency list
    graph = defaultdict(list)
    for u, v in edges:
        graph[u].append(v)
        graph[v].append(u)

    # Function to simulate ant movement
    def simulate_movement(start):
        visited = set()
        queue = deque([(start, 0)])  # (house, number of ants)
        while queue:
            house, ants = queue.popleft()
            if house in visited:
                continue
            visited.add(house)
            for neighbor in graph[house]:
                if (neighbor, ants) not in visited:
                    queue.append((neighbor, ants))
        return len(visited) == n

    # Check if all ants can gather at any house
    for house in range(1, n+1):
        if simulate_movement(house):
            return True
    return False

# Input reading
n = int(input())
edges = [tuple(map(int, input().split())) for _ in range(n-1)]

# Output the result
print("YES" if can_gather_ants(n, edges) else "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
4
1 2
2 3
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Annual Ants' Gathering
time_limit_ms: 2000
memory_limit_mb: 256
Description: Deep within a forest lies an ancient tree, home to $$$n$$$ ants living in $$$n$$$ tiny houses, indexed from $$$1$$$ to $$$n$$$, connected by the branches of the tree.

Once a year, all the ants need to gather to watch the EUC. For this, all ants move along the $$$n-1$$$ branches of the tree they live on to meet at the home of one ant.

However, this year the ants could not agree on where to meet and need your help to gather up. You can tell all the ants currently at house $$$u$$$ to move to house $$$v$$$ if there is a branch directly connecting those two houses. However, the ants ignore your command if there are fewer ants gathered in house $$$v$$$ than in house $$$u$$$, i.e., if it would be easier for the ants from house $$$v$$$ to move. This even holds true if no ant at all is currently in house $$$v$$$. You can give this kind of commands as many times as you want.

Is it possible for you to gather all the ants in a single house?

Input: The first line contains one integer $$$n$$$ ($$$1\leq n\leq 200\,000$$$) — the number of ant homes.

Each of the following $$$n-1$$$ lines contains two integers $$$u$$$ and $$$v$$$ ($$$1\leq u, v\leq n$$$) — there is a branch directly connecting the house $$$u$$$ and house $$$v$$$. It is guaranteed that every ant can reach the house of any other ant just by following the branches of the tree.

Output: Print $$$\texttt{YES}$$$ if it is possible to gather all the ants in a single house. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
7
5 1
3 2
4 6
3 6
7 1
1 3
output:
YES

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

input:
6
4 5
5 6
6 1
2 6
3 2
output:
YES

Note: In the first sample, you can gather all the ants at house $$$3$$$ as follows:

- You tell to the ant at house $$$4$$$ to move to house $$$6$$$.
- You tell to the ant at house $$$2$$$ to move to house $$$3$$$.
- You tell to the two ants at house $$$6$$$ to move to house $$$3$$$ (which already contains two ants).
- You tell to the ant at house $$$5$$$ to move to house $$$1$$$.
- You tell to the ant at house $$$7$$$ to move to house $$$1$$$ (which already contains two ants).
- You tell to the three ants at house $$$1$$$ to move to house $$$3$$$ (which already contains four ants).

In the second sample, it is impossible to gather all the ants in a single house.

Buggy Code:
import sys
input = sys.stdin.read
sys.setrecursionlimit(200000)

def find_centroid(n, adjacency_list):
    parent = [-1] * n
    subtree_sizes = [0] * n

    def dfs(node):
        subtree_size = 1
        for neighbor in adjacency_list[node]:
            if neighbor != parent[node]:
                parent[neighbor] = node
                subtree_size += dfs(neighbor)
        subtree_sizes[node] = subtree_size
        return subtree_size

    dfs(0)  # Assuming nodes are 0-indexed for internal processing

    # Find the centroid
    for i in range(n):
        is_centroid = True
        majority = (n // 2)
        for neighbor in adjacency_list[i]:
            if neighbor != parent[i]:  # Check child subtree
                if subtree_sizes[neighbor] > majority:
                    is_centroid = False
                    break
        if parent[i] != -1:  # Check the "parent part" of the subtree
            if n - subtree_sizes[i] > majority:
                is_centroid = False
        if is_centroid:
            return True
    return False

def main():
    data = input().split()
    n = int(data[0])
    if n == 1:
        print("YES")
        return

    adjacency_list = [[] for _ in range(n)]
    index = 1
    for _ in range(n - 1):
        u = int(data[index]) - 1
        v = int(data[index + 1]) - 1
        adjacency_list[u].append(v)
        adjacency_list[v].append(u)
        index += 2

    if find_centroid(n, adjacency_list):
        print("YES")
    else:
        print("NO")

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on the logic that determines whether a centroid exists in the tree. The centroid of a tree is a node such that when removed, no subtree has more than half of the nodes of the original tree. The provided code has a bug in the centroid-finding logic, which might fail to correctly identify the centroid in certain edge cases.

### Step-by-Step Analysis:
1. **Identify the Bug**: The bug in the code is in the `find_centroid` function. It checks if a node is a centroid by ensuring that no subtree has more than half of the nodes. However, it might not correctly handle the case where the tree is not balanced, or the tree has a specific structure that makes it difficult to find a centroid.

2. **Create a Test Case**: We need a test case where the tree structure is such that the centroid-finding logic fails to identify a valid centroid. One such scenario is a tree where the nodes form a long chain, and the centroid is not easily identifiable by the given logic.

### Test Case:
Consider a tree with 7 nodes where the nodes form a long chain. In this case, the tree will look like a linear structure, and the centroid should be the middle node. However, the given code might fail to identify this middle node correctly.

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

### Explanation:
- **Input**: The tree has 7 nodes, and the nodes are connected in a linear fashion.
- **Expected Output**: The expected output is `YES` because the middle node (node 4) can be the centroid, and all ants can be gathered there.
- **Why This Test Case**: This test case exposes the bug in the centroid-finding logic. The code might fail to correctly identify the middle node as the centroid, leading to an incorrect output of `NO`.

By using this test case, you can verify if the centroid-finding logic in the given code is working correctly or not.