**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ±1
time_limit_ms: 2000
memory_limit_mb: 256
Description: Bob has a grid with $3$ rows and $n$ columns, each of which contains either $a_i$ or $-a_i$ for some integer $1 \leq i \leq n$. For example, one possible grid for $n=4$ is shown below:

$$\begin{bmatrix} a_1 & -a_2 & -a_3 & -a_2 \\\ -a_4 & a_4 & -a_1 & -a_3 \\\ a_1 & a_2 & -a_2 & a_4 \end{bmatrix}$$

Alice and Bob play a game as follows:

  * Bob shows Alice his grid.    * Alice gives Bob an array $a_1, a_2, \dots, a_n$ of her choosing, whose elements are all $\mathbf{-1}$ or $\mathbf{1}$.    * Bob substitutes these values into his grid to make a grid of $-1$s and $1$s.    * Bob sorts the elements of each column in non-decreasing order.    * Alice wins if all the elements in the middle row are $1$; otherwise, Bob wins.

For example, suppose Alice gives Bob the array $[1, -1, -1, 1]$ for the grid above. Then the following will happen (colors are added for clarity):

$$\begin{bmatrix} \color{red}{a_1} & \color{green}{-a_2} & \color{blue}{-a_3} & \color{green}{-a_2} \\\ -a_4 & a_4 & \color{red}{-a_1} & \color{blue}{-a_3} \\\ \color{red}{a_1} & \color{green}{a_2} & \color{green}{-a_2} & a_4 \end{bmatrix} \xrightarrow{[\color{red}{1},\color{green}{-1},\color{blue}{-1},1]} \begin{bmatrix} \color{red}{1} & \color{green}{1} & \color{blue}{1} & \color{green}{1} \\\ -1 & 1 & \color{red}{-1} & \color{blue}{1} \\\ \color{red}{1} & \color{green}{-1} & \color{green}{1} & 1 \end{bmatrix} \xrightarrow{\text{sort each column}} \begin{bmatrix} -1 & -1 & -1 & 1 \\\ \mathbf{1} & \mathbf{1} & \mathbf{1} & \mathbf{1} \\\ 1 & 1 & 1 & 1 \\\ \end{bmatrix}\,. $$ Since the middle row is all $1$, Alice wins.

Given Bob's grid, determine whether or not Alice can choose the array $a$ to win the game.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 500$) — the number of columns of Bob's grid.

The next three lines each contain $n$ integers, the $i$-th of which contains $g_{i,1}, g_{i,2}, \dots, g_{i,n}$ ($-n \leq g_{i,j} \leq n$, $g_{i,j} \neq 0$), representing Bob's grid.

If cell $x > 0$ is in the input, that cell in Bob's grid should contain $a_x$; if $x < 0$ is in the input, that cell in Bob's grid should contain $-a_{-x}$. See the sample input and notes for a better understanding.

Output: For each test case, output "YES" (without quotes) if Alice can win, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yEs", "yes", and "Yes" will be recognized as a positive response).

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

Note: The first test case is described in the statement.

In the second test case, Bob's grid is as follows:

$$\begin{bmatrix} a_1 & a_2 \\\ -a_1 & -a_2 \\\ a_2 & -a_2 \end{bmatrix}$$

For the last column to have $1$ in the middle row when sorted, Alice must pick $a_2 = -1$. However, it is then impossible to choose $a_1$ such that the first column has $1$ in the middle when sorted. Thus, Alice cannot win.

In the third test case, Alice can pick $a = [1,1,1,1,1]$.

Buggy Code:
from math import *
from collections import *
from bisect import bisect_left, bisect_right, bisect
from itertools import *

# BOOTSTRAP
from types import GeneratorType
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        to = f(*args, **kwargs)
        if stack:
            return to
        else:
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        return to
                    to = stack[-1].send(to)
    return wrappedfunc

R=lambda:map(int,input().split())
S=lambda:map(str,input().split())

MOD = int(1e9 + 7)
# MOD = 998244353
# MOD = 100000000669

t, = R()
# t = 1

mark = []
valid = True

def answer():
    n, = R()

    m = []
    for i in range(3):
        m.append(list(R()))

    g = [[0] * (2 * n) for i in range(2 * n)]

    def adj(x):
        if x > 0:
            return x - 1

        return n + (-x - 1)

    for i in range(n):
        for p in permutations([m[0][i], m[1][i], m[2][i]]):
            g[adj(-p[0])][adj(p[1])] = 1
            g[adj(-p[0])][adj(p[2])] = 1

    @bootstrap
    def dfs(u, p):
        # if u < n:
        #     print("dfs", u)
        # else:
        #     print("dfs", f"~{u - n}")
        global mark
        global valid

        if not valid: yield

        for v in range(2 * n):
            if v == p: continue
            if g[u][v] == 0: continue

            vs = v
            inv = 0
            if vs >= n:
                vs -= n
                inv = 1

            start = False

            if mark[vs] == -1:
                start = True
                mark[vs] = inv
            else:
                if (mark[vs] + inv) % 2 == 1:
                    # print("vertex", vs % n, "conflict")
                    valid = False

            if not valid:
                break

            if start:
                yield dfs(v, u)

        yield

    gvalid = True
    for i in range(n):
        # print("-------- ", i)

        cnt = 0

        for v in [i, i + n]:
            # print()
            # print("att")
            # print()
            global mark
            global valid

            valid = True
            mark = [-1] * n

            if v == i:
                mark[i] = 0
            else:
                mark[i] = 1

            dfs(v, -1)

            if not valid:
                cnt += 1

        if cnt == 2:
            # print("vertex", i, "broken")
            gvalid = False
            break

    # print("g")
    # for i in range(2 * n):
    #     an = []
    #     for j in range(2 * n):
    #         if g[i][j]:
    #             if j < n:
    #                 an.append(f"{j + 1}")
    #             else:
    #                 an.append(f"{-(j - n) - 1}")

    #     if i < n:
    #         print("v: ", i + 1, an)
    #     else:
    #         print("v: ", -(i - n) - 1 , an)

    # print("--")

    if gvalid:
        return 'YES'

    return "NO"

for i in range(t):
    print(answer())

**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
3
1 2 3
-1 -2 -3
3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ±1
time_limit_ms: 2000
memory_limit_mb: 256
Description: Bob has a grid with $3$ rows and $n$ columns, each of which contains either $a_i$ or $-a_i$ for some integer $1 \leq i \leq n$. For example, one possible grid for $n=4$ is shown below:

$$\begin{bmatrix} a_1 & -a_2 & -a_3 & -a_2 \\\ -a_4 & a_4 & -a_1 & -a_3 \\\ a_1 & a_2 & -a_2 & a_4 \end{bmatrix}$$

Alice and Bob play a game as follows:

  * Bob shows Alice his grid.    * Alice gives Bob an array $a_1, a_2, \dots, a_n$ of her choosing, whose elements are all $\mathbf{-1}$ or $\mathbf{1}$.    * Bob substitutes these values into his grid to make a grid of $-1$s and $1$s.    * Bob sorts the elements of each column in non-decreasing order.    * Alice wins if all the elements in the middle row are $1$; otherwise, Bob wins.

For example, suppose Alice gives Bob the array $[1, -1, -1, 1]$ for the grid above. Then the following will happen (colors are added for clarity):

$$\begin{bmatrix} \color{red}{a_1} & \color{green}{-a_2} & \color{blue}{-a_3} & \color{green}{-a_2} \\\ -a_4 & a_4 & \color{red}{-a_1} & \color{blue}{-a_3} \\\ \color{red}{a_1} & \color{green}{a_2} & \color{green}{-a_2} & a_4 \end{bmatrix} \xrightarrow{[\color{red}{1},\color{green}{-1},\color{blue}{-1},1]} \begin{bmatrix} \color{red}{1} & \color{green}{1} & \color{blue}{1} & \color{green}{1} \\\ -1 & 1 & \color{red}{-1} & \color{blue}{1} \\\ \color{red}{1} & \color{green}{-1} & \color{green}{1} & 1 \end{bmatrix} \xrightarrow{\text{sort each column}} \begin{bmatrix} -1 & -1 & -1 & 1 \\\ \mathbf{1} & \mathbf{1} & \mathbf{1} & \mathbf{1} \\\ 1 & 1 & 1 & 1 \\\ \end{bmatrix}\,. $$ Since the middle row is all $1$, Alice wins.

Given Bob's grid, determine whether or not Alice can choose the array $a$ to win the game.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 500$) — the number of columns of Bob's grid.

The next three lines each contain $n$ integers, the $i$-th of which contains $g_{i,1}, g_{i,2}, \dots, g_{i,n}$ ($-n \leq g_{i,j} \leq n$, $g_{i,j} \neq 0$), representing Bob's grid.

If cell $x > 0$ is in the input, that cell in Bob's grid should contain $a_x$; if $x < 0$ is in the input, that cell in Bob's grid should contain $-a_{-x}$. See the sample input and notes for a better understanding.

Output: For each test case, output "YES" (without quotes) if Alice can win, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yEs", "yes", and "Yes" will be recognized as a positive response).

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

Note: The first test case is described in the statement.

In the second test case, Bob's grid is as follows:

$$\begin{bmatrix} a_1 & a_2 \\\ -a_1 & -a_2 \\\ a_2 & -a_2 \end{bmatrix}$$

For the last column to have $1$ in the middle row when sorted, Alice must pick $a_2 = -1$. However, it is then impossible to choose $a_1$ such that the first column has $1$ in the middle when sorted. Thus, Alice cannot win.

In the third test case, Alice can pick $a = [1,1,1,1,1]$.

Buggy Code:
import time

nn = int(input())

for u in range(nn) :

    n = int(input())
    aa = input().split()
    a = [int(aa[i])for i in range(n)]
    a1 = []
    bb = input().split()
    b = [int(bb[i])for i in range(n)]
    b1 = []
    cc = input().split()
    c = [int(cc[i])for i in range(n)]
    c1 = []

    for i in range(n):
        a1.append(-1 if a[i] < 0 else 1)
        b1.append(-1 if b[i] < 0 else 1)
        c1.append(-1 if c[i] < 0 else 1)
        a[i] = abs(a[i])
        b[i] = abs(b[i])
        c[i] = abs(c[i])

    win = "no"

    from itertools import product

    # Define the possible outcomes
    outcomes = [-1,1]  # 'h' for heads, 't' for tails

    # Generate all combinations of outcomes for 5 coins
    coin_combinations = list(product(outcomes, repeat=n))

    # Create the list of lists with each combination
    kk = [list(combination) for combination in coin_combinations]

   # print(kk)
    for r in kk :
        lose = 0
        for j in range(n):
            tar = [a1[j]*r[a[j]-1],b1[j]*r[b[j]-1],c1[j]*r[c[j]-1]]
            tar.sort()
            if tar[1] == -1 :
                lose = 1
                break
        if lose == 0 :
            win = "yes"
            break
    print(win)

"""
    for i1 in range(-1,2,2):
        for i2 in range(-1,2,2):
            for i3 in range(-1,2,2):
                for i4 in range(-1,2,2):
                    r = [i1,i2,i3,i4]
                    lose = 0
                    for j in range(n):

                        tar = [a1[j]*r[a[j]-1],b1[j]*r[b[j]-1],c1[j]*r[c[j]-1]]
                        print(tar)
                        tar.sort()
                        if tar[1] == -1 :
                            lose = 1
                            break
                    if lose == 0 :
                        win = "yes"
                        break
                if win == "yes" :
                    break
            if win == "yes" :
                break
        if win == "yes" :
            break

    print(win)
"""

"""
1 -2 -3 -2
-4 4 -1 -3
1 2 -2 4

5
1 2 3 4 5
-2 3 -4 -5 -1
3 -5 1 2 2

print(a[j])#a1[j]*r[a[j]-1])
                        print(b[j])#b1[j]*r[b[j]-1])
                        print(c[j])#c1[j]*r[c[j]-1])
                        print()
                        time.sleep(0.5)
                    print("##############")
"""

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ±1
time_limit_ms: 2000
memory_limit_mb: 256
Description: Bob has a grid with $3$ rows and $n$ columns, each of which contains either $a_i$ or $-a_i$ for some integer $1 \leq i \leq n$. For example, one possible grid for $n=4$ is shown below:

$$\begin{bmatrix} a_1 & -a_2 & -a_3 & -a_2 \\\ -a_4 & a_4 & -a_1 & -a_3 \\\ a_1 & a_2 & -a_2 & a_4 \end{bmatrix}$$

Alice and Bob play a game as follows:

  * Bob shows Alice his grid.    * Alice gives Bob an array $a_1, a_2, \dots, a_n$ of her choosing, whose elements are all $\mathbf{-1}$ or $\mathbf{1}$.    * Bob substitutes these values into his grid to make a grid of $-1$s and $1$s.    * Bob sorts the elements of each column in non-decreasing order.    * Alice wins if all the elements in the middle row are $1$; otherwise, Bob wins.

For example, suppose Alice gives Bob the array $[1, -1, -1, 1]$ for the grid above. Then the following will happen (colors are added for clarity):

$$\begin{bmatrix} \color{red}{a_1} & \color{green}{-a_2} & \color{blue}{-a_3} & \color{green}{-a_2} \\\ -a_4 & a_4 & \color{red}{-a_1} & \color{blue}{-a_3} \\\ \color{red}{a_1} & \color{green}{a_2} & \color{green}{-a_2} & a_4 \end{bmatrix} \xrightarrow{[\color{red}{1},\color{green}{-1},\color{blue}{-1},1]} \begin{bmatrix} \color{red}{1} & \color{green}{1} & \color{blue}{1} & \color{green}{1} \\\ -1 & 1 & \color{red}{-1} & \color{blue}{1} \\\ \color{red}{1} & \color{green}{-1} & \color{green}{1} & 1 \end{bmatrix} \xrightarrow{\text{sort each column}} \begin{bmatrix} -1 & -1 & -1 & 1 \\\ \mathbf{1} & \mathbf{1} & \mathbf{1} & \mathbf{1} \\\ 1 & 1 & 1 & 1 \\\ \end{bmatrix}\,. $$ Since the middle row is all $1$, Alice wins.

Given Bob's grid, determine whether or not Alice can choose the array $a$ to win the game.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 500$) — the number of columns of Bob's grid.

The next three lines each contain $n$ integers, the $i$-th of which contains $g_{i,1}, g_{i,2}, \dots, g_{i,n}$ ($-n \leq g_{i,j} \leq n$, $g_{i,j} \neq 0$), representing Bob's grid.

If cell $x > 0$ is in the input, that cell in Bob's grid should contain $a_x$; if $x < 0$ is in the input, that cell in Bob's grid should contain $-a_{-x}$. See the sample input and notes for a better understanding.

Output: For each test case, output "YES" (without quotes) if Alice can win, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yEs", "yes", and "Yes" will be recognized as a positive response).

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

Note: The first test case is described in the statement.

In the second test case, Bob's grid is as follows:

$$\begin{bmatrix} a_1 & a_2 \\\ -a_1 & -a_2 \\\ a_2 & -a_2 \end{bmatrix}$$

For the last column to have $1$ in the middle row when sorted, Alice must pick $a_2 = -1$. However, it is then impossible to choose $a_1$ such that the first column has $1$ in the middle when sorted. Thus, Alice cannot win.

In the third test case, Alice can pick $a = [1,1,1,1,1]$.

Buggy Code:
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.concurrent.PriorityBlockingQueue;

public class Main {
	static kattio sc = new kattio();
	static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

	public static void main(String[] args) {
		//System.out.println(1<<25);
		//System.out.println(1000000);

		int t = sc.nextint();
		while (t-- > 0) {
			p=false;
			if (t==1000-91) {
				p=true;
			}
			sovle();
		}
		out.close();
	}
	static boolean p;
	static Map<Integer, List<Integer>>map;
	static void sovle() {
		int n=sc.nextint();
		int [][]arr=new int[n][3];
		int []z=new int[n+1];
		int []f=new int[n+1];
		for (int i = 0; i < 3; i++) {
			for (int j = 0; j < arr.length; j++) {
				arr[j][i]=sc.nextint();

			}
		}
		if (p) {
			for (int i = 0; i < arr.length; i++) {
				//System.out.println(Arrays.toString(arr[i]));
			}
		}
		int []va=new int[n+1];
		for (int i = 0; i < arr.length; i++) {
			int a=Math.abs(arr[i][0]);
			int b=Math.abs(arr[i][1]);
			int c=Math.abs(arr[i][2]);
			if (a==c) {
				int a1=arr[i][2];arr[i][2]=arr[i][1];arr[i][1]=a1;
				b=Math.abs(arr[i][1]);
				c=Math.abs(arr[i][2]);
			}else if (b==c) {
				int a1=arr[i][2];arr[i][2]=arr[i][0];arr[i][0]=a1;
				a=Math.abs(arr[i][0]);
				c=Math.abs(arr[i][2]);
			}
			if (a==b) {
				if (b==c) {
					int sum=arr[i][0]+arr[i][1]+arr[i][2];
					int v=0;
					if (sum>=a) {
						v=1;
					}else {
						v=-1;
					}
					if (va[a]==0) {
						va[a]=v;
					}else {
						if (va[a]!=v) {
							out.println("NO");return;
						}
					}
				}else {
					int sum=arr[i][0]+arr[i][1];
					if (sum==0) {
						if (va[c]==0) {
							va[c]=arr[i][2]/c;
						}else {
							if (va[c]!=arr[i][2]/c) {
								out.println("NO");return;
							}
						}
					}else {
						if (va[a]==0) {
							va[a]=arr[i][0]/a;
						}else {
							if (va[a]!=arr[i][0]/a) {
								out.println("NO");return;
							}
						}
					}
				}
			}
			//System.out.println(Arrays.toString(arr[i]));
		}

		for (int i = 0; i < arr.length; i++) {
			int a=Math.abs(arr[i][0]);
			int b=Math.abs(arr[i][1]);
			int c=Math.abs(arr[i][2]);
			if (a!=b&&a!=c&&b!=c) {
				int sum=0;
				for (int j = 0; j < 3; j++) {
					sum+=va[Math.abs(arr[i][j])]*arr[i][j]/Math.abs(arr[i][j]);
				}

				if (sum<=-2) {
					out.println("NO");return;
				}
				for (int j = 0; j < 3; j++) {
					if (arr[i][j]>0) {
						z[arr[i][j]]++;
					}else {
						f[-arr[i][j]]++;
					}
				}
			}

		}
		//System.out.println(Arrays.toString(va));

		int ge=0;int sum=0;
		for (int i = 1; i < va.length; i++) {
			if (va[i]==0) {
				ge++;
				sum+=Math.min(z[i], f[i]);
				//System.out.println(i+" "+Math.min(z[i], f[i]));
			}
		}
		//System.out.println(ge+" "+sum);
		if (ge<sum) {
			out.println("NO");return;
		}
		//System.out.println(Arrays.toString(va));
		out.println("YES");
	}

	static long pow(long a,long b,long mod) {
		if (b==0) {
			return 1;
		}else {
			long c=pow(a, b/2, mod);
			if (b%2==0) {
				return c*c%mod;
			}else {
				return c*c%mod*a%mod;
			}
		}
	}
	static 	int gcd(int a,int b) {
		return b>0 ? gcd(b,a%b):a;
	}

	static long mod = 998244353;
	static long[] c;

	static int lowbit(int x) {
		return x & -x;
	}

	static void add(int x, long v) {
		while (x < c.length) {
			// System.out.println(x);
			c[x] += v;
			x += lowbit(x);
		}
	}

	static long sum(int x) {
		long ans = 0;
		while (x > 0) {
			ans += c[x];
			x -= lowbit(x);
		}
		return ans;
	}
}

class kattio extends PrintWriter {
	private BufferedReader r;
	private StringTokenizer st;

	public kattio() {
		this(System.in, System.out);
	}

	public kattio(InputStream i, OutputStream o) {
		super(o);
		r = new BufferedReader(new InputStreamReader(i));
	}

	public kattio(String in, String o) throws IOException {
		super(o);
		r = new BufferedReader(new FileReader(in));
	}

	public String next() {
		try {
			while (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(r.readLine());
			}
			return st.nextToken();
		} catch (Exception e) {
			// TODO: handle exception
			return null;
		}
	}

	public int nextint() {
		return Integer.parseInt(next());
	}
	public long nextlong() {
		return Long.parseLong(next());
	}

}

**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 buggy code, we need to focus on edge cases or scenarios that maximize the failure probability. Here are the steps to think through this:

1. **Understand the Bug**: The buggy code seems to have issues with handling certain configurations of the grid, especially when the values in the grid are not distinct or when the sum of values in a column is zero.

2. **Edge Cases**:
   - **All Columns with the Same Value**: If all columns have the same value, the code should handle this correctly.
   - **Columns with Zero Sum**: If a column has a sum of zero, the code should correctly determine if Alice can win.
   - **Distinct Values in Columns**: Ensure that the code handles columns with distinct values correctly.

3. **Small-Scale Test Input**:
   - We will create a small grid (e.g., 3 columns) with values that are likely to expose the bug.
   - We will include columns with the same value, columns with zero sum, and columns with distinct values.

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

```plaintext
1
3
1 2 -3
-1 -2 3
1 2 -3
```

### Explanation:
- **Column 1**: `[1, -1, 1]` - This column has a sum of 1, which should be sorted to `[-1, 1, 1]`. The middle row is 1, so Alice can win.
- **Column 2**: `[2, -2, 2]` - This column has a sum of 2, which should be sorted to `[-2, 2, 2]`. The middle row is 2, so Alice can win.
- **Column 3**: `[-3, 3, -3]` - This column has a sum of -3, which should be sorted to `[-3, -3, 3]`. The middle row is -3, so Alice cannot win.

### Expected Output:
- The expected output should be "NO" because the third column cannot be won by Alice.

This test input should help expose the bug in the buggy code, especially in handling columns with specific sums and values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ±1
time_limit_ms: 2000
memory_limit_mb: 256
Description: Bob has a grid with $3$ rows and $n$ columns, each of which contains either $a_i$ or $-a_i$ for some integer $1 \leq i \leq n$. For example, one possible grid for $n=4$ is shown below:

$$\begin{bmatrix} a_1 & -a_2 & -a_3 & -a_2 \\\ -a_4 & a_4 & -a_1 & -a_3 \\\ a_1 & a_2 & -a_2 & a_4 \end{bmatrix}$$

Alice and Bob play a game as follows:

  * Bob shows Alice his grid.    * Alice gives Bob an array $a_1, a_2, \dots, a_n$ of her choosing, whose elements are all $\mathbf{-1}$ or $\mathbf{1}$.    * Bob substitutes these values into his grid to make a grid of $-1$s and $1$s.    * Bob sorts the elements of each column in non-decreasing order.    * Alice wins if all the elements in the middle row are $1$; otherwise, Bob wins.

For example, suppose Alice gives Bob the array $[1, -1, -1, 1]$ for the grid above. Then the following will happen (colors are added for clarity):

$$\begin{bmatrix} \color{red}{a_1} & \color{green}{-a_2} & \color{blue}{-a_3} & \color{green}{-a_2} \\\ -a_4 & a_4 & \color{red}{-a_1} & \color{blue}{-a_3} \\\ \color{red}{a_1} & \color{green}{a_2} & \color{green}{-a_2} & a_4 \end{bmatrix} \xrightarrow{[\color{red}{1},\color{green}{-1},\color{blue}{-1},1]} \begin{bmatrix} \color{red}{1} & \color{green}{1} & \color{blue}{1} & \color{green}{1} \\\ -1 & 1 & \color{red}{-1} & \color{blue}{1} \\\ \color{red}{1} & \color{green}{-1} & \color{green}{1} & 1 \end{bmatrix} \xrightarrow{\text{sort each column}} \begin{bmatrix} -1 & -1 & -1 & 1 \\\ \mathbf{1} & \mathbf{1} & \mathbf{1} & \mathbf{1} \\\ 1 & 1 & 1 & 1 \\\ \end{bmatrix}\,. $$ Since the middle row is all $1$, Alice wins.

Given Bob's grid, determine whether or not Alice can choose the array $a$ to win the game.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 500$) — the number of columns of Bob's grid.

The next three lines each contain $n$ integers, the $i$-th of which contains $g_{i,1}, g_{i,2}, \dots, g_{i,n}$ ($-n \leq g_{i,j} \leq n$, $g_{i,j} \neq 0$), representing Bob's grid.

If cell $x > 0$ is in the input, that cell in Bob's grid should contain $a_x$; if $x < 0$ is in the input, that cell in Bob's grid should contain $-a_{-x}$. See the sample input and notes for a better understanding.

Output: For each test case, output "YES" (without quotes) if Alice can win, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yEs", "yes", and "Yes" will be recognized as a positive response).

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

Note: The first test case is described in the statement.

In the second test case, Bob's grid is as follows:

$$\begin{bmatrix} a_1 & a_2 \\\ -a_1 & -a_2 \\\ a_2 & -a_2 \end{bmatrix}$$

For the last column to have $1$ in the middle row when sorted, Alice must pick $a_2 = -1$. However, it is then impossible to choose $a_1$ such that the first column has $1$ in the middle when sorted. Thus, Alice cannot win.

In the third test case, Alice can pick $a = [1,1,1,1,1]$.

Buggy Code:
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.concurrent.PriorityBlockingQueue;

public class Main {
	static kattio sc = new kattio();
	static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

	public static void main(String[] args) {
		//System.out.println(1<<25);
		//System.out.println(1000000);

		int t = sc.nextint();
		while (t-- > 0) {
			p=false;
			if (t==2711-1703) {
				p=true;
			}
			sovle();
		}
		out.close();
	}
	static boolean p;
	static Map<Integer, List<Integer>>map;
	static void sovle() {
		int n=sc.nextint();
		int [][]arr=new int[n][3];
		for (int i = 0; i < 3; i++) {
			for (int j = 0; j < arr.length; j++) {
				arr[j][i]=sc.nextint();
			}
		}
		int []va=new int[n+1];
		for (int i = 0; i < arr.length; i++) {
			int a=Math.abs(arr[i][0]);
			int b=Math.abs(arr[i][1]);
			int c=Math.abs(arr[i][2]);
			if (a==c) {
				int a1=arr[i][2];arr[i][2]=arr[i][1];arr[i][1]=a1;
				b=Math.abs(arr[i][1]);
				c=Math.abs(arr[i][2]);
			}else if (b==c) {
				int a1=arr[i][2];arr[i][2]=arr[i][0];arr[i][0]=a1;
				a=Math.abs(arr[i][0]);
				c=Math.abs(arr[i][2]);
			}
			if (a==b) {
				if (b==c) {
					int sum=arr[i][0]+arr[i][1]+arr[i][2];
					int v=0;
					if (sum>=a) {
						v=1;
					}else {
						v=-1;
					}
					if (va[a]==0) {
						va[a]=v;
					}else {
						if (va[a]!=v) {
							out.println("NO");return;
						}
					}
				}else {
					int sum=arr[i][0]+arr[i][1];
					if (sum==0) {
						if (va[c]==0) {
							va[c]=arr[i][2]/c;
						}else {
							if (va[c]!=arr[i][2]/c) {
								out.println("NO");return;
							}
						}
					}else {
						if (va[a]==0) {
							va[a]=arr[i][0]/a;
						}else {
							if (va[a]!=arr[i][0]/a) {
								out.println("NO");return;
							}
						}
					}
				}
			}
			//System.out.println(Arrays.toString(arr[i]));
		}
		for (int i = 0; i < arr.length; i++) {
			int a=Math.abs(arr[i][0]);
			int b=Math.abs(arr[i][1]);
			int c=Math.abs(arr[i][2]);
			if (a!=b&&a!=c&&b!=c) {
				int sum=0;
				for (int j = 0; j < 3; j++) {
					sum+=va[Math.abs(arr[i][j])]*arr[i][j]/Math.abs(arr[i][j]);
				}

				if (sum<=-2) {
					out.println("NO");return;
				}
			}

		}
		//System.out.println(Arrays.toString(va));
		out.println("YES");
	}

	static long pow(long a,long b,long mod) {
		if (b==0) {
			return 1;
		}else {
			long c=pow(a, b/2, mod);
			if (b%2==0) {
				return c*c%mod;
			}else {
				return c*c%mod*a%mod;
			}
		}
	}
	static 	int gcd(int a,int b) {
		return b>0 ? gcd(b,a%b):a;
	}

	static long mod = 998244353;
	static long[] c;

	static int lowbit(int x) {
		return x & -x;
	}

	static void add(int x, long v) {
		while (x < c.length) {
			// System.out.println(x);
			c[x] += v;
			x += lowbit(x);
		}
	}

	static long sum(int x) {
		long ans = 0;
		while (x > 0) {
			ans += c[x];
			x -= lowbit(x);
		}
		return ans;
	}
}

class kattio extends PrintWriter {
	private BufferedReader r;
	private StringTokenizer st;

	public kattio() {
		this(System.in, System.out);
	}

	public kattio(InputStream i, OutputStream o) {
		super(o);
		r = new BufferedReader(new InputStreamReader(i));
	}

	public kattio(String in, String o) throws IOException {
		super(o);
		r = new BufferedReader(new FileReader(in));
	}

	public String next() {
		try {
			while (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(r.readLine());
			}
			return st.nextToken();
		} catch (Exception e) {
			// TODO: handle exception
			return null;
		}
	}

	public int nextint() {
		return Integer.parseInt(next());
	}
	public long nextlong() {
		return Long.parseLong(next());
	}

}

**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 edge cases and scenarios that are likely to cause the code to fail. Here are the steps to think through this:

1. **Understand the Buggy Code's Logic:**
   - The code processes the grid and tries to determine if Alice can win by ensuring the middle row is all 1s after sorting each column.
   - It has a specific logic for handling cases where the values in a column are the same or different.

2. **Identify Potential Bugs:**
   - The code might fail when handling columns with specific patterns of values.
   - It might not correctly handle the sorting and value assignment in certain edge cases.

3. **Construct a Challenging Test Case:**
   - We need a small grid that can expose the potential bugs in the code.
   - Consider a grid where the values in the columns are designed to challenge the sorting and value assignment logic.

4. **Generate the Test Input:**
   - We will create a grid with a small number of columns (e.g., 3) and values that are likely to cause issues in the code's logic.

Here is a challenging test input:

```plaintext
1
3
1 2 3
-2 -1 -3
3 1 -2
```

### Explanation:
- **Grid:**
  - Column 1: [1, -2, 3]
  - Column 2: [2, -1, 1]
  - Column 3: [3, -3, -2]

- **Expected Behavior:**
  - For Alice to win, the middle row of the sorted columns must be all 1s.
  - Sorting each column:
    - Column 1: [-2, 1, 3] (middle is 1)
    - Column 2: [-1, 1, 2] (middle is 1)
    - Column 3: [-3, -2, 3] (middle is -2)

- **Outcome:**
  - The middle row after sorting is [1, 1, -2], which means Alice cannot win because the middle row is not all 1s.

This test case should expose the bug in the code's logic for handling the values and ensuring the middle row is all 1s after sorting.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ±1
time_limit_ms: 2000
memory_limit_mb: 256
Description: Bob has a grid with $3$ rows and $n$ columns, each of which contains either $a_i$ or $-a_i$ for some integer $1 \leq i \leq n$. For example, one possible grid for $n=4$ is shown below:

$$\begin{bmatrix} a_1 & -a_2 & -a_3 & -a_2 \\\ -a_4 & a_4 & -a_1 & -a_3 \\\ a_1 & a_2 & -a_2 & a_4 \end{bmatrix}$$

Alice and Bob play a game as follows:

  * Bob shows Alice his grid.    * Alice gives Bob an array $a_1, a_2, \dots, a_n$ of her choosing, whose elements are all $\mathbf{-1}$ or $\mathbf{1}$.    * Bob substitutes these values into his grid to make a grid of $-1$s and $1$s.    * Bob sorts the elements of each column in non-decreasing order.    * Alice wins if all the elements in the middle row are $1$; otherwise, Bob wins.

For example, suppose Alice gives Bob the array $[1, -1, -1, 1]$ for the grid above. Then the following will happen (colors are added for clarity):

$$\begin{bmatrix} \color{red}{a_1} & \color{green}{-a_2} & \color{blue}{-a_3} & \color{green}{-a_2} \\\ -a_4 & a_4 & \color{red}{-a_1} & \color{blue}{-a_3} \\\ \color{red}{a_1} & \color{green}{a_2} & \color{green}{-a_2} & a_4 \end{bmatrix} \xrightarrow{[\color{red}{1},\color{green}{-1},\color{blue}{-1},1]} \begin{bmatrix} \color{red}{1} & \color{green}{1} & \color{blue}{1} & \color{green}{1} \\\ -1 & 1 & \color{red}{-1} & \color{blue}{1} \\\ \color{red}{1} & \color{green}{-1} & \color{green}{1} & 1 \end{bmatrix} \xrightarrow{\text{sort each column}} \begin{bmatrix} -1 & -1 & -1 & 1 \\\ \mathbf{1} & \mathbf{1} & \mathbf{1} & \mathbf{1} \\\ 1 & 1 & 1 & 1 \\\ \end{bmatrix}\,. $$ Since the middle row is all $1$, Alice wins.

Given Bob's grid, determine whether or not Alice can choose the array $a$ to win the game.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 500$) — the number of columns of Bob's grid.

The next three lines each contain $n$ integers, the $i$-th of which contains $g_{i,1}, g_{i,2}, \dots, g_{i,n}$ ($-n \leq g_{i,j} \leq n$, $g_{i,j} \neq 0$), representing Bob's grid.

If cell $x > 0$ is in the input, that cell in Bob's grid should contain $a_x$; if $x < 0$ is in the input, that cell in Bob's grid should contain $-a_{-x}$. See the sample input and notes for a better understanding.

Output: For each test case, output "YES" (without quotes) if Alice can win, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yEs", "yes", and "Yes" will be recognized as a positive response).

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

Note: The first test case is described in the statement.

In the second test case, Bob's grid is as follows:

$$\begin{bmatrix} a_1 & a_2 \\\ -a_1 & -a_2 \\\ a_2 & -a_2 \end{bmatrix}$$

For the last column to have $1$ in the middle row when sorted, Alice must pick $a_2 = -1$. However, it is then impossible to choose $a_1$ such that the first column has $1$ in the middle when sorted. Thus, Alice cannot win.

In the third test case, Alice can pick $a = [1,1,1,1,1]$.

Buggy Code:
//
// Created by JMKov on 6/21/2024.
//
#include <iostream>
#include <set>
#include <vector>
#include <algorithm>

using namespace std;

int n;
int arr[504][3];
set<int> adj[1008];
set<int> radj[1008];
bool explored[1008];
int scc[1008];
int cnt = 1;
vector<int> ord;

void addEdges(int a, int b, int c) {
    int aInd = (a < 0) ? a * -1 - 1 : a + n - 1;
    int bInd = (b < 0) ? b * -1 + n - 1 : b - 1;
    int cInd = (c < 0) ? c * -1 + n - 1 : c - 1;
    if (aInd != bInd) { // dont know if necessary if statement
        adj[aInd].insert(bInd);
        radj[bInd].insert(aInd);
    }
    if (aInd != cInd) {
        adj[aInd].insert(cInd);
        radj[cInd].insert(aInd);
    }
}

void dfs1(int v) {
    explored[v] = true;
    for (int e : adj[v]) {
        if (explored[e]) continue;
        dfs1(e);
    }
    ord.push_back(v);
}

void dfs2(int v) {
    scc[v] = cnt;
    for (int e : radj[v]) {
        if (scc[e] != 0) continue;
        dfs2(e);
    }
}

void kosaraju() {
    for (int i = 0; i < 2 * n; i++) {
        if (!explored[i]) dfs1(i);
    }
    reverse(ord.begin(), ord.end());

    for (int i : ord) {
        if (scc[i] == 0) {
            dfs2(i);
            cnt++;
        }
    }
}

bool works() {
    for (int i = 0; i < n; i++) {
        if (scc[i] == scc[i + n]) return false;
    }
    return true;
}

void clean() {
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j < 3; j++) arr[i][j] = 0;
    }
    for (int i = 0; i <= 2 * n; i++) {
        adj[i].clear();
        radj[i].clear();
        explored[i] = false;
        scc[i] = 0;
    }
    cnt = 1;
    ord.clear();
}

void solve() {
    cin >> n;
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < n; j++) {
            cin >> arr[j][i];
        }
    }
    for (int i = 0; i < n; i++) {
        addEdges(arr[i][0], arr[i][1], arr[i][2]);
        addEdges(arr[i][1], arr[i][0], arr[i][2]);
        addEdges(arr[i][2], arr[i][1], arr[i][0]);
    }

    kosaraju();
//    for (int i = 0; i < 2 * n; i++) {
//        for (int j : adj[i]) cout << i << " " << j << endl;
//        cout << endl;
//    }
    bool ans = works();

    cout << (ans ? "YES" : "NO") << endl;
    clean();
}

int main() {
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}
// 1 2 3 4 5
// 1 1 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
2
1 2
-1 -2
2 -2
4
1 2 3 4
-1 -2 -3 -4
1 -2 3 -4
5
1 2 3 4 5
-1 -2 -3 -4 -5
1 -2 3 -4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ±1
time_limit_ms: 2000
memory_limit_mb: 256
Description: Bob has a grid with $3$ rows and $n$ columns, each of which contains either $a_i$ or $-a_i$ for some integer $1 \leq i \leq n$. For example, one possible grid for $n=4$ is shown below:

$$\begin{bmatrix} a_1 & -a_2 & -a_3 & -a_2 \\\ -a_4 & a_4 & -a_1 & -a_3 \\\ a_1 & a_2 & -a_2 & a_4 \end{bmatrix}$$

Alice and Bob play a game as follows:

  * Bob shows Alice his grid.    * Alice gives Bob an array $a_1, a_2, \dots, a_n$ of her choosing, whose elements are all $\mathbf{-1}$ or $\mathbf{1}$.    * Bob substitutes these values into his grid to make a grid of $-1$s and $1$s.    * Bob sorts the elements of each column in non-decreasing order.    * Alice wins if all the elements in the middle row are $1$; otherwise, Bob wins.

For example, suppose Alice gives Bob the array $[1, -1, -1, 1]$ for the grid above. Then the following will happen (colors are added for clarity):

$$\begin{bmatrix} \color{red}{a_1} & \color{green}{-a_2} & \color{blue}{-a_3} & \color{green}{-a_2} \\\ -a_4 & a_4 & \color{red}{-a_1} & \color{blue}{-a_3} \\\ \color{red}{a_1} & \color{green}{a_2} & \color{green}{-a_2} & a_4 \end{bmatrix} \xrightarrow{[\color{red}{1},\color{green}{-1},\color{blue}{-1},1]} \begin{bmatrix} \color{red}{1} & \color{green}{1} & \color{blue}{1} & \color{green}{1} \\\ -1 & 1 & \color{red}{-1} & \color{blue}{1} \\\ \color{red}{1} & \color{green}{-1} & \color{green}{1} & 1 \end{bmatrix} \xrightarrow{\text{sort each column}} \begin{bmatrix} -1 & -1 & -1 & 1 \\\ \mathbf{1} & \mathbf{1} & \mathbf{1} & \mathbf{1} \\\ 1 & 1 & 1 & 1 \\\ \end{bmatrix}\,. $$ Since the middle row is all $1$, Alice wins.

Given Bob's grid, determine whether or not Alice can choose the array $a$ to win the game.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 500$) — the number of columns of Bob's grid.

The next three lines each contain $n$ integers, the $i$-th of which contains $g_{i,1}, g_{i,2}, \dots, g_{i,n}$ ($-n \leq g_{i,j} \leq n$, $g_{i,j} \neq 0$), representing Bob's grid.

If cell $x > 0$ is in the input, that cell in Bob's grid should contain $a_x$; if $x < 0$ is in the input, that cell in Bob's grid should contain $-a_{-x}$. See the sample input and notes for a better understanding.

Output: For each test case, output "YES" (without quotes) if Alice can win, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yEs", "yes", and "Yes" will be recognized as a positive response).

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

Note: The first test case is described in the statement.

In the second test case, Bob's grid is as follows:

$$\begin{bmatrix} a_1 & a_2 \\\ -a_1 & -a_2 \\\ a_2 & -a_2 \end{bmatrix}$$

For the last column to have $1$ in the middle row when sorted, Alice must pick $a_2 = -1$. However, it is then impossible to choose $a_1$ such that the first column has $1$ in the middle when sorted. Thus, Alice cannot win.

In the third test case, Alice can pick $a = [1,1,1,1,1]$.

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

using namespace std;
using ll = long long;

ll N;

ll opposite(ll num){
    return (num < N)? num+N : num-N;
}

void addClause(vector<set<ll>> &graph, ll n, ll a, ll b){
    ll negA = opposite(a), negB = opposite(b);
    if(a == opposite(b)) return;
    graph[negA].insert(b);
    graph[negB].insert(a);
}

bool stronglyConnected(vector<set<ll>> graph, vector<bool> &visited, set<ll> avoid, set<ll> avoidFr, ll curr){
    if(avoidFr.find(curr) != avoidFr.end()) return true;
    if(avoid.find(curr) != avoid.end()) {
        avoidFr.insert(opposite(curr));
        visited[opposite(curr)] = false;
    }
    avoid.insert(opposite(curr));
    visited[curr] = true;
    auto it = graph[curr].begin();
    while (it != graph[curr].end())
    {
        if(!visited[*it]){
            if(stronglyConnected(graph, visited, avoid, avoidFr, *it)) return true;
        }
        it++;
    }
    return false;
}

void solve(){
    ll n; cin >> n;
    N = n;

    vector<set<ll>> graph(2*n);
    ll mt[3][n];

    for (int i = 0; i < 3; i++){
        for (int j = 0; j < n; j++) cin >> mt[i][j];
    }
    for (int i = 0; i < n; i++)
    {
        vector<ll> vx;
        for (int j = 0; j < 3; j++) {
            if(mt[j][i] > 0){
                vx.push_back(mt[j][i]);
            }else{
                vx.push_back(abs(mt[j][i])+n);
            }
        }
        addClause(graph, n, vx[0]-1, vx[1]-1);
        addClause(graph, n, vx[0]-1, vx[2]-1);
        addClause(graph, n, vx[1]-1, vx[2]-1);
    }

    /*for (int i = 0; i < 2*n; i++)
    {
        cout << "\033[31;1m" << i+1 << ":\033[0m ";
        for (int j = 0; j < graph[i].size(); j++)
        {
            cout << graph[i][j]+1 << ' ';
        }
        cout << '\n';
    }*/

    vector<bool> visited (2*n, false);
    for (int i = 0; i < n; i++)
    {
        set<ll> avoid, avoidFr;
        if(stronglyConnected(graph, visited, avoid, avoidFr, i)) {
            cout << "NO\n";
            return;
        }
    }
    cout << "YES\n";
}

int main(){
    ll t; cin >> t; while(t--) solve();
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ±1
time_limit_ms: 2000
memory_limit_mb: 256
Description: Bob has a grid with $3$ rows and $n$ columns, each of which contains either $a_i$ or $-a_i$ for some integer $1 \leq i \leq n$. For example, one possible grid for $n=4$ is shown below:

$$\begin{bmatrix} a_1 & -a_2 & -a_3 & -a_2 \\\ -a_4 & a_4 & -a_1 & -a_3 \\\ a_1 & a_2 & -a_2 & a_4 \end{bmatrix}$$

Alice and Bob play a game as follows:

  * Bob shows Alice his grid.    * Alice gives Bob an array $a_1, a_2, \dots, a_n$ of her choosing, whose elements are all $\mathbf{-1}$ or $\mathbf{1}$.    * Bob substitutes these values into his grid to make a grid of $-1$s and $1$s.    * Bob sorts the elements of each column in non-decreasing order.    * Alice wins if all the elements in the middle row are $1$; otherwise, Bob wins.

For example, suppose Alice gives Bob the array $[1, -1, -1, 1]$ for the grid above. Then the following will happen (colors are added for clarity):

$$\begin{bmatrix} \color{red}{a_1} & \color{green}{-a_2} & \color{blue}{-a_3} & \color{green}{-a_2} \\\ -a_4 & a_4 & \color{red}{-a_1} & \color{blue}{-a_3} \\\ \color{red}{a_1} & \color{green}{a_2} & \color{green}{-a_2} & a_4 \end{bmatrix} \xrightarrow{[\color{red}{1},\color{green}{-1},\color{blue}{-1},1]} \begin{bmatrix} \color{red}{1} & \color{green}{1} & \color{blue}{1} & \color{green}{1} \\\ -1 & 1 & \color{red}{-1} & \color{blue}{1} \\\ \color{red}{1} & \color{green}{-1} & \color{green}{1} & 1 \end{bmatrix} \xrightarrow{\text{sort each column}} \begin{bmatrix} -1 & -1 & -1 & 1 \\\ \mathbf{1} & \mathbf{1} & \mathbf{1} & \mathbf{1} \\\ 1 & 1 & 1 & 1 \\\ \end{bmatrix}\,. $$ Since the middle row is all $1$, Alice wins.

Given Bob's grid, determine whether or not Alice can choose the array $a$ to win the game.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 500$) — the number of columns of Bob's grid.

The next three lines each contain $n$ integers, the $i$-th of which contains $g_{i,1}, g_{i,2}, \dots, g_{i,n}$ ($-n \leq g_{i,j} \leq n$, $g_{i,j} \neq 0$), representing Bob's grid.

If cell $x > 0$ is in the input, that cell in Bob's grid should contain $a_x$; if $x < 0$ is in the input, that cell in Bob's grid should contain $-a_{-x}$. See the sample input and notes for a better understanding.

Output: For each test case, output "YES" (without quotes) if Alice can win, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yEs", "yes", and "Yes" will be recognized as a positive response).

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

Note: The first test case is described in the statement.

In the second test case, Bob's grid is as follows:

$$\begin{bmatrix} a_1 & a_2 \\\ -a_1 & -a_2 \\\ a_2 & -a_2 \end{bmatrix}$$

For the last column to have $1$ in the middle row when sorted, Alice must pick $a_2 = -1$. However, it is then impossible to choose $a_1$ such that the first column has $1$ in the middle when sorted. Thus, Alice cannot win.

In the third test case, Alice can pick $a = [1,1,1,1,1]$.

Buggy Code:
import sys
input = sys.stdin.read

def tarjan(u):
    global tot, scc, top
    dfn[u] = low[u] = tot = tot + 1  # 初始化dfn和low数组
    stk.append(u)  # 将节点u入栈
    mk[u] = 1  # 标记节点u在栈中
    for v in g[u]:  # 遍历u的所有邻接节点
        if dfn[v] == 0:  # 如果v未被访问过
            tarjan(v)  # 递归访问v
            low[u] = min(low[u], low[v])  # 更新low值
        elif mk[v] == 1:  # 如果v在栈中
            low[u] = min(low[u], dfn[v])  # 更新low值
    if dfn[u] == low[u]:  # 如果u是强连通分量的根节点
        scc += 1  # 强连通分量数量加1
        while stk[-1] != u:  # 将栈中u及其之后的节点出栈
            mk[stk[-1]] = 0  # 标记节点不在栈中
            f[stk.pop()] = scc  # 记录节点所属的强连通分量编号
        mk[stk[-1]] = 0  # 标记节点不在栈中
        f[stk.pop()] = scc  # 记录节点所属的强连通分量编号

data = input().split()  # 读取所有输入数据
T = int(data[0])  # 读取测试用例数量
index = 1

for _ in range(T):
    n = int(data[index])  # 读取节点数量
    index += 1
    a = [[0] * (n + 1) for _ in range(4)]  # 初始化输入数据数组
    for i in range(1, 4):
        for j in range(1, n + 1):
            a[i][j] = int(data[index])  # 读取输入数据
            index += 1

    g = [[] for _ in range(2 * n + 1)]  # 初始化邻接表
    for i in range(1, n + 1):
        for j in range(1, 4):
            for k in range(1, 4):
                if j == k:
                    continue  # 跳过相同的情况
                if a[j][i] < 0 and a[k][i] < 0:
                    g[-a[j][i] + n].append(-a[k][i])  # 构建图
                if a[j][i] < 0 and a[k][i] > 0:
                    g[-a[j][i] + n].append(a[k][i] + n)
                if a[j][i] > 0 and a[k][i] < 0:
                    g[a[j][i]].append(-a[k][i])
                if a[j][i] > 0 and a[k][i] > 0:
                    g[a[j][i]].append(a[k][i] + n)

    dfn = [0] * (2 * n + 1)  # 初始化dfn数组
    low = [0] * (2 * n + 1)  # 初始化low数组
    mk = [0] * (2 * n + 1)  # 初始化mk数组
    f = [0] * (2 * n + 1)  # 初始化f数组
    stk = []  # 初始化栈
    tot = 0  # 初始化tot
    scc = 0  # 初始化scc
    top = 0  # 初始化top

    for i in range(1, 2 * n + 1):
        if dfn[i] == 0:
            tarjan(i)  # 对每个未访问的节点执行Tarjan算法

    flag = 1
    for i in range(1, n + 1):
        if f[i] == f[i + n]:  # 检查是否存在矛盾
            flag = 0
            break

    if flag:
        print("YES")  # 输出结果
    else:
        print("NO")

    for i in range(1, 2 * n + 1):
        g[i].clear()  # 清空邻接表
        dfn[i] = low[i] = mk[i] = f[i] = 0  # 清空数组

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ±1
time_limit_ms: 2000
memory_limit_mb: 256
Description: Bob has a grid with $3$ rows and $n$ columns, each of which contains either $a_i$ or $-a_i$ for some integer $1 \leq i \leq n$. For example, one possible grid for $n=4$ is shown below:

$$\begin{bmatrix} a_1 & -a_2 & -a_3 & -a_2 \\\ -a_4 & a_4 & -a_1 & -a_3 \\\ a_1 & a_2 & -a_2 & a_4 \end{bmatrix}$$

Alice and Bob play a game as follows:

  * Bob shows Alice his grid.    * Alice gives Bob an array $a_1, a_2, \dots, a_n$ of her choosing, whose elements are all $\mathbf{-1}$ or $\mathbf{1}$.    * Bob substitutes these values into his grid to make a grid of $-1$s and $1$s.    * Bob sorts the elements of each column in non-decreasing order.    * Alice wins if all the elements in the middle row are $1$; otherwise, Bob wins.

For example, suppose Alice gives Bob the array $[1, -1, -1, 1]$ for the grid above. Then the following will happen (colors are added for clarity):

$$\begin{bmatrix} \color{red}{a_1} & \color{green}{-a_2} & \color{blue}{-a_3} & \color{green}{-a_2} \\\ -a_4 & a_4 & \color{red}{-a_1} & \color{blue}{-a_3} \\\ \color{red}{a_1} & \color{green}{a_2} & \color{green}{-a_2} & a_4 \end{bmatrix} \xrightarrow{[\color{red}{1},\color{green}{-1},\color{blue}{-1},1]} \begin{bmatrix} \color{red}{1} & \color{green}{1} & \color{blue}{1} & \color{green}{1} \\\ -1 & 1 & \color{red}{-1} & \color{blue}{1} \\\ \color{red}{1} & \color{green}{-1} & \color{green}{1} & 1 \end{bmatrix} \xrightarrow{\text{sort each column}} \begin{bmatrix} -1 & -1 & -1 & 1 \\\ \mathbf{1} & \mathbf{1} & \mathbf{1} & \mathbf{1} \\\ 1 & 1 & 1 & 1 \\\ \end{bmatrix}\,. $$ Since the middle row is all $1$, Alice wins.

Given Bob's grid, determine whether or not Alice can choose the array $a$ to win the game.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 500$) — the number of columns of Bob's grid.

The next three lines each contain $n$ integers, the $i$-th of which contains $g_{i,1}, g_{i,2}, \dots, g_{i,n}$ ($-n \leq g_{i,j} \leq n$, $g_{i,j} \neq 0$), representing Bob's grid.

If cell $x > 0$ is in the input, that cell in Bob's grid should contain $a_x$; if $x < 0$ is in the input, that cell in Bob's grid should contain $-a_{-x}$. See the sample input and notes for a better understanding.

Output: For each test case, output "YES" (without quotes) if Alice can win, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yEs", "yes", and "Yes" will be recognized as a positive response).

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

Note: The first test case is described in the statement.

In the second test case, Bob's grid is as follows:

$$\begin{bmatrix} a_1 & a_2 \\\ -a_1 & -a_2 \\\ a_2 & -a_2 \end{bmatrix}$$

For the last column to have $1$ in the middle row when sorted, Alice must pick $a_2 = -1$. However, it is then impossible to choose $a_1$ such that the first column has $1$ in the middle when sorted. Thus, Alice cannot win.

In the third test case, Alice can pick $a = [1,1,1,1,1]$.

Buggy Code:
import sys
from collections import deque, Counter
import heapq
from bisect import bisect_left
from itertools import combinations
import math
from typing import Callable

def I():
	return [int(i) for i in sys.stdin.readline().split()]

def II():
	return int(sys.stdin.readline())

def S():
	return sys.stdin.readline().split()

def IS():
	return sys.stdin.readline().replace('\n', '')

def main():
	n = II()
	m = [I(), I(), I()]
	info = [0] * n
	for i in range(n):
		v = [abs(m[j][i]) - 1 for j in range(3)]
		z = [-1 if m[j][i] < 0 else 1 for j in range(3)]
		sv = set(v)
		if len(sv) == 1:
			if z.count(-1) < 2:
				if info[v[0]] == -1:
					print('NO')
					return
				info[v[0]] = 1
			else:
				if info[v[0]] == 1:
					print('NO')
					return
				info[v[0]] = -1
		elif len(sv) == 2:
			c = Counter(v)
			target = v[0]
			for vi in v:
				if c[vi] == 2:
					target = vi
					break
			tz = []
			_tz, _v = 0, 0
			for j in range(3):
				if v[j] == target:
					tz.append(z[j])
				else:
					_tz = z[j]
					_v = v[j]

			if tz[0] == tz[1]:
				if info[target] == -tz[0]:
					print('NO')
					return
				info[target] = tz[0]
			else:
				if info[_v] == -_tz:
					print('NO')
					return
				info[_v] = _tz

	for _ in range(10):
		if _ % 2 == 0:
			start, end, step = 0, n, 1
		else:
			start, end, step = n - 1, -1, -1
		for i in range(start, end, step):
			v = [abs(m[j][i]) - 1 for j in range(3)]
			z = [-1 if m[j][i] < 0 else 1 for j in range(3)]
			total = []
			o = []
			oz = []
			for j in range(3):
				if info[v[j]] != 0:
					total.append(info[v[j]] * z[j])
				else:
					o.append(v[j])
					oz.append(z[j])
			if len(total) == 3:
				if total.count(-1) > 1:
					print('NO')
					return
			elif len(total) == 2:
				if total.count(-1) == 2:
					print('NO')
					return
				if total.count(-1) == 1:
					if info[o[0]] == -oz[0]:
						print('NO')
						return
					else:
						info[o[0]] = oz[0]
			elif len(total) == 1:
				if total.count(-1) == 1:
					if info[o[0]] == -oz[0]:
						print('NO')
						return
					info[o[0]] = oz[0]
					if info[o[1]] == -oz[1]:
						print('NO')
						return
					info[o[1]] = oz[1]
	print('YES')

if __name__ == '__main__':
	for _ in range(II()):
		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 create a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases that might cause the code to fail. Here are some steps to consider:

1. **Columns with all the same absolute values:**
   - If a column has the same absolute value in all three rows, the code should handle this correctly. However, we need to ensure that the sign logic is correct.

2. **Columns with two different absolute values:**
   - If a column has two of the same absolute values and one different, the code should handle the sign logic correctly.

3. **Columns with all different absolute values:**
   - If a column has all different absolute values, the code should handle the sign logic correctly.

4. **Test cases where the middle row cannot be all 1s:**
   - Create a test case where it is impossible for the middle row to be all 1s after sorting.

Given these considerations, let's create a small-scale test input:

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

### Explanation:

- **First Column:**
  - Values: `1, -1, -1`
  - After sorting: `[-1, -1, 1]`
  - Middle row: `-1` (Alice cannot win if this column is included)

- **Second Column:**
  - Values: `-2, 2, 2`
  - After sorting: `[-2, 2, 2]`
  - Middle row: `2` (Alice can win if this column is included)

- **Third Column:**
  - Values: `-3, 3, -3`
  - After sorting: `[-3, -3, 3]`
  - Middle row: `-3` (Alice cannot win if this column is included)

- **Fourth Column:**
  - Values: `-2, -3, 2`
  - After sorting: `[-3, -2, 2]`
  - Middle row: `-2` (Alice cannot win if this column is included)

This test case is designed to have a mix of columns that are impossible to make the middle row all 1s, which should trigger the "NO" output. The buggy code should fail to handle this correctly, exposing the bug.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ±1
time_limit_ms: 2000
memory_limit_mb: 256
Description: Bob has a grid with $3$ rows and $n$ columns, each of which contains either $a_i$ or $-a_i$ for some integer $1 \leq i \leq n$. For example, one possible grid for $n=4$ is shown below:

$$\begin{bmatrix} a_1 & -a_2 & -a_3 & -a_2 \\\ -a_4 & a_4 & -a_1 & -a_3 \\\ a_1 & a_2 & -a_2 & a_4 \end{bmatrix}$$

Alice and Bob play a game as follows:

  * Bob shows Alice his grid.    * Alice gives Bob an array $a_1, a_2, \dots, a_n$ of her choosing, whose elements are all $\mathbf{-1}$ or $\mathbf{1}$.    * Bob substitutes these values into his grid to make a grid of $-1$s and $1$s.    * Bob sorts the elements of each column in non-decreasing order.    * Alice wins if all the elements in the middle row are $1$; otherwise, Bob wins.

For example, suppose Alice gives Bob the array $[1, -1, -1, 1]$ for the grid above. Then the following will happen (colors are added for clarity):

$$\begin{bmatrix} \color{red}{a_1} & \color{green}{-a_2} & \color{blue}{-a_3} & \color{green}{-a_2} \\\ -a_4 & a_4 & \color{red}{-a_1} & \color{blue}{-a_3} \\\ \color{red}{a_1} & \color{green}{a_2} & \color{green}{-a_2} & a_4 \end{bmatrix} \xrightarrow{[\color{red}{1},\color{green}{-1},\color{blue}{-1},1]} \begin{bmatrix} \color{red}{1} & \color{green}{1} & \color{blue}{1} & \color{green}{1} \\\ -1 & 1 & \color{red}{-1} & \color{blue}{1} \\\ \color{red}{1} & \color{green}{-1} & \color{green}{1} & 1 \end{bmatrix} \xrightarrow{\text{sort each column}} \begin{bmatrix} -1 & -1 & -1 & 1 \\\ \mathbf{1} & \mathbf{1} & \mathbf{1} & \mathbf{1} \\\ 1 & 1 & 1 & 1 \\\ \end{bmatrix}\,. $$ Since the middle row is all $1$, Alice wins.

Given Bob's grid, determine whether or not Alice can choose the array $a$ to win the game.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 500$) — the number of columns of Bob's grid.

The next three lines each contain $n$ integers, the $i$-th of which contains $g_{i,1}, g_{i,2}, \dots, g_{i,n}$ ($-n \leq g_{i,j} \leq n$, $g_{i,j} \neq 0$), representing Bob's grid.

If cell $x > 0$ is in the input, that cell in Bob's grid should contain $a_x$; if $x < 0$ is in the input, that cell in Bob's grid should contain $-a_{-x}$. See the sample input and notes for a better understanding.

Output: For each test case, output "YES" (without quotes) if Alice can win, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yEs", "yes", and "Yes" will be recognized as a positive response).

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

Note: The first test case is described in the statement.

In the second test case, Bob's grid is as follows:

$$\begin{bmatrix} a_1 & a_2 \\\ -a_1 & -a_2 \\\ a_2 & -a_2 \end{bmatrix}$$

For the last column to have $1$ in the middle row when sorted, Alice must pick $a_2 = -1$. However, it is then impossible to choose $a_1$ such that the first column has $1$ in the middle when sorted. Thus, Alice cannot win.

In the third test case, Alice can pick $a = [1,1,1,1,1]$.

Buggy Code:
import static java.lang.Integer.parseInt;
import static java.lang.Math.abs;
import static java.lang.Math.min;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.UncheckedIOException;
import java.util.ArrayList;
import java.util.Collection;
import java.util.HashMap;
import java.util.Iterator;
import java.util.List;
import java.util.NoSuchElementException;
import java.util.Stack;
import java.util.concurrent.atomic.AtomicInteger;
import java.util.function.Consumer;
import java.util.function.Function;
import java.util.function.Predicate;

public class PlusMinusOneGraph {

  public static final IllegalStateException EX = new IllegalStateException("Nope");

  public static void main(final String[] args) {
    final Iterator<String> inputIterator = readLinesFromStdIn();

    final int n = parseInt(inputIterator.next());

    for (int i = 0; i < n; i++) {
      final int nCols = parseInt(inputIterator.next());
      final String[] row1 = inputIterator.next().split(" ");
      final String[] row2 = inputIterator.next().split(" ");
      final String[] row3 = inputIterator.next().split(" ");

      if (isWinnable(nCols, row1, row2, row3)) {
        System.out.println("YES");
      } else {
        System.out.println("NO");
      }
    }
  }

  private static boolean isWinnable(final int n, final String[] row1, final String[] row2,
      final String[] row3) {
    final Graph graph = new Graph(2 * n + 1);

    for (int i = 0; i < n; i++) {
      final int i1 = parseInt(row1[i]);
      final int i2 = parseInt(row2[i]);
      final int i3 = parseInt(row3[i]);
      graph.addEdge(-i1 + n, i2 + n);
      graph.addEdge(-i2 + n, i1 + n);

      graph.addEdge(-i2 + n, i3 + n);
      graph.addEdge(-i3 + n, i2 + n);

      graph.addEdge(-i3 + n, i1 + n);
      graph.addEdge(-i1 + n, i3 + n);
    }

    try {
      graph.findComponents(scc -> {
        if (scc.size() > 1) {
          final HashMap<Integer, Integer> contradictions = new HashMap<>(scc.size());
          for (final int var : scc) {
            final int realValue = var - n;
            final int absVar = abs(realValue);
            final Integer otherValue = contradictions.get(absVar);
            if (otherValue == null) {
              contradictions.put(absVar, realValue);
            } else if (realValue == -otherValue) {
              throw EX;
            }
          }
        }
      });
    } catch (final Exception e) {
      if (e == EX) {
        return false;
      } else {
        throw e;
      }
    }

    return true;
  }

  static class Graph {

    private final Vertex[] vertices;

    public Graph(final int n) {
      this.vertices = new Vertex[n];
    }

    public void addEdge(final int from, final int to) {
      final Vertex v;
      if ((v = vertices[from]) != null) {
        v.add(to);
      } else {
        vertices[from] = new Vertex(from, new ArrayList<>() {{
          add(to);
        }});
      }

      if (vertices[to] == null) {
        vertices[to] = new Vertex(to, new ArrayList<>());
      }

    }

    public void findComponents(final Consumer<List<Integer>> sccConsumer) {
      final VertexWrapper[] vertexWrappers = new VertexWrapper[vertices.length];
      final AtomicInteger vCount = new AtomicInteger(0);
      final Stack<VertexWrapper> path = new Stack<>();
      final Function<Vertex, VertexWrapper> getWrapper = v -> vertexWrappers[v.id] == null
          ? (vertexWrappers[v.id] = new VertexWrapper(v, vCount.getAndIncrement()))
          : vertexWrappers[v.id];
      final Predicate<Vertex> alreadyVisited = v -> vertexWrappers[v.id] != null;
      for (final Vertex v : vertices) {
        if (v != null && !alreadyVisited.test(v)) {
          findComponents(v, getWrapper, alreadyVisited, path, sccConsumer);
        }
      }
    }

    private void findComponents(final Vertex v, final Function<Vertex, VertexWrapper> getWrapper,
        final Predicate<Vertex> alreadyVisited, final Stack<VertexWrapper> path,
        final Consumer<List<Integer>> sccConsumer) {
      final VertexWrapper wr = getWrapper.apply(v);
      path.push(wr);
      wr.onStack = true;

      for (final int vId : wr.vertex.tos) {
        if (!alreadyVisited.test(vertices[vId])) {
          findComponents(vertices[vId], getWrapper, alreadyVisited, path, sccConsumer);
          final VertexWrapper to = getWrapper.apply(vertices[vId]);
          wr.minIndex = min(wr.minIndex, to.minIndex);
        } else {
          final VertexWrapper to = getWrapper.apply(vertices[vId]);
          if (to.onStack) {
            wr.minIndex = min(wr.minIndex, to.index);
          }
        }
      }

      if (wr.minIndex == wr.index) {
        final List<Integer> stronglyConnectedComponent = new ArrayList<>();
        VertexWrapper cV;
        do {
          cV = path.pop();
          cV.onStack = false;
          stronglyConnectedComponent.add(cV.vertex.id);
        } while (cV != wr);
        sccConsumer.accept(stronglyConnectedComponent);
      }
    }

    private static class VertexWrapper {

      private final Vertex vertex;
      private final int index;
      private int minIndex;
      private boolean onStack;

      public VertexWrapper(final Vertex vertex, final int index) {
        this.vertex = vertex;
        this.index = index;
        this.minIndex = index;
        this.onStack = false;
      }
    }

    static class Vertex {

      private final int id;
      private final Collection<Integer> tos;

      public Vertex(final int form, final Collection<Integer> tos) {
        this.id = form;
        this.tos = tos;
      }

      public void add(final Integer to) {
        if (!tos.contains(to)) {
          tos.add(to);
        }
      }

      @Override
      public String toString() {
        return String.valueOf(id);
      }
    }
  }

  private static Iterator<String> readLinesFromStdIn() {
    final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    return new Iterator<>() {
      String nextLine = null;

      @Override
      public boolean hasNext() {
        if (nextLine != null) {
          return true;
        } else {
          try {
            nextLine = br.readLine();
            return (nextLine != null);
          } catch (final IOException e) {
            throw new UncheckedIOException(e);
          }
        }
      }

      @Override
      public String next() {
        if (nextLine != null || hasNext()) {
          final String line = nextLine;
          nextLine = null;
          return line;
        } else {
          throw new NoSuchElementException();
        }
      }
    };
  }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases that might cause the algorithm to fail. The key aspects to consider are:

1. **Contradictions in the Graph**: The code uses a graph to detect contradictions (i.e., a variable and its negation being in the same strongly connected component). We need to create a grid that leads to such a contradiction.

2. **Small Scale**: The test input should be small to ensure it fits within the constraints and is easy to validate.

3. **Complex Interactions**: The grid should have complex interactions between the rows to increase the likelihood of exposing the bug.

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

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

### Explanation:
- **Grid**:
  ```
  1  2  3
 -2  3 -1
  3 -1  2
  ```

- **Analysis**:
  - For the first column: `1, -2, 3`
    - If Alice picks `a_1 = 1`, `a_2 = 1`, and `a_3 = 1`, the column becomes `[1, -1, 1]` which sorts to `[-1, 1, 1]`.
  - For the second column: `2, 3, -1`
    - If Alice picks `a_1 = 1`, `a_2 = 1`, and `a_3 = 1`, the column becomes `[1, 1, -1]` which sorts to `[-1, 1, 1]`.
  - For the third column: `3, -1, 2`
    - If Alice picks `a_1 = 1`, `a_2 = 1`, and `a_3 = 1`, the column becomes `[1, -1, 1]` which sorts to `[-1, 1, 1]`.

- **Contradiction**:
  - The graph will detect a contradiction because `1` and `-1` are in the same strongly connected component, which should make the algorithm return `NO`.

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

This test input is designed to expose the bug in the graph-based contradiction detection logic. If the code correctly identifies the contradiction, it should return `NO`. If it fails to detect the contradiction, it might return `YES`, indicating a bug.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: ±1
time_limit_ms: 2000
memory_limit_mb: 256
Description: Bob has a grid with $3$ rows and $n$ columns, each of which contains either $a_i$ or $-a_i$ for some integer $1 \leq i \leq n$. For example, one possible grid for $n=4$ is shown below:

$$\begin{bmatrix} a_1 & -a_2 & -a_3 & -a_2 \\\ -a_4 & a_4 & -a_1 & -a_3 \\\ a_1 & a_2 & -a_2 & a_4 \end{bmatrix}$$

Alice and Bob play a game as follows:

  * Bob shows Alice his grid.    * Alice gives Bob an array $a_1, a_2, \dots, a_n$ of her choosing, whose elements are all $\mathbf{-1}$ or $\mathbf{1}$.    * Bob substitutes these values into his grid to make a grid of $-1$s and $1$s.    * Bob sorts the elements of each column in non-decreasing order.    * Alice wins if all the elements in the middle row are $1$; otherwise, Bob wins.

For example, suppose Alice gives Bob the array $[1, -1, -1, 1]$ for the grid above. Then the following will happen (colors are added for clarity):

$$\begin{bmatrix} \color{red}{a_1} & \color{green}{-a_2} & \color{blue}{-a_3} & \color{green}{-a_2} \\\ -a_4 & a_4 & \color{red}{-a_1} & \color{blue}{-a_3} \\\ \color{red}{a_1} & \color{green}{a_2} & \color{green}{-a_2} & a_4 \end{bmatrix} \xrightarrow{[\color{red}{1},\color{green}{-1},\color{blue}{-1},1]} \begin{bmatrix} \color{red}{1} & \color{green}{1} & \color{blue}{1} & \color{green}{1} \\\ -1 & 1 & \color{red}{-1} & \color{blue}{1} \\\ \color{red}{1} & \color{green}{-1} & \color{green}{1} & 1 \end{bmatrix} \xrightarrow{\text{sort each column}} \begin{bmatrix} -1 & -1 & -1 & 1 \\\ \mathbf{1} & \mathbf{1} & \mathbf{1} & \mathbf{1} \\\ 1 & 1 & 1 & 1 \\\ \end{bmatrix}\,. $$ Since the middle row is all $1$, Alice wins.

Given Bob's grid, determine whether or not Alice can choose the array $a$ to win the game.

Input: The first line contains a single integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($2 \leq n \leq 500$) — the number of columns of Bob's grid.

The next three lines each contain $n$ integers, the $i$-th of which contains $g_{i,1}, g_{i,2}, \dots, g_{i,n}$ ($-n \leq g_{i,j} \leq n$, $g_{i,j} \neq 0$), representing Bob's grid.

If cell $x > 0$ is in the input, that cell in Bob's grid should contain $a_x$; if $x < 0$ is in the input, that cell in Bob's grid should contain $-a_{-x}$. See the sample input and notes for a better understanding.

Output: For each test case, output "YES" (without quotes) if Alice can win, and "NO" (without quotes) otherwise.

You can output "YES" and "NO" in any case (for example, strings "yEs", "yes", and "Yes" will be recognized as a positive response).

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

Note: The first test case is described in the statement.

In the second test case, Bob's grid is as follows:

$$\begin{bmatrix} a_1 & a_2 \\\ -a_1 & -a_2 \\\ a_2 & -a_2 \end{bmatrix}$$

For the last column to have $1$ in the middle row when sorted, Alice must pick $a_2 = -1$. However, it is then impossible to choose $a_1$ such that the first column has $1$ in the middle when sorted. Thus, Alice cannot win.

In the third test case, Alice can pick $a = [1,1,1,1,1]$.

Buggy Code:
//package plusminusone;

import static java.lang.Integer.parseInt;
import static java.lang.Math.abs;
import static java.lang.Math.min;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.UncheckedIOException;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Iterator;
import java.util.List;
import java.util.NoSuchElementException;
import java.util.Stack;
import java.util.concurrent.atomic.AtomicInteger;
import java.util.function.Consumer;
import java.util.function.Function;
import java.util.function.Predicate;

public class PlusMinusOneGraph {

  public static final IllegalStateException EX = new IllegalStateException("Nope");

  public static void main(final String[] args) {
    final Iterator<String> inputIterator = readLinesFromStdIn();

    final int n = parseInt(inputIterator.next());

    for (int i = 0; i < n; i++) {
      final int nCols = parseInt(inputIterator.next());
      final String[] row1 = inputIterator.next().split(" ");
      final String[] row2 = inputIterator.next().split(" ");
      final String[] row3 = inputIterator.next().split(" ");

      if (isWinnable(nCols, row1, row2, row3)) {
        System.out.println("YES");
      } else {
        System.out.println("NO");
      }
    }
  }

  private static boolean isWinnable(final int n, final String[] row1, final String[] row2,
      final String[] row3) {
    final Graph graph = new Graph(2 * n + 1);

    for (int i = 0; i < n; i++) {
      final int i1 = parseInt(row1[i]);
      final int i2 = parseInt(row2[i]);
      final int i3 = parseInt(row3[i]);
      graph.addEdge(-i1 + n, i2 + n);
      graph.addEdge(-i2 + n, i1 + n);

      graph.addEdge(-i2 + n, i3 + n);
      graph.addEdge(-i3 + n, i2 + n);

      graph.addEdge(-i3 + n, i1 + n);
      graph.addEdge(-i1 + n, i3 + n);
    }

    try {
      graph.findComponents(scc -> {
        if (scc.size() > 1) {
          final HashMap<Integer, Integer> contradictions = new HashMap<>(scc.size());
          for (final int var : scc) {
            final int realValue = var - n;
            final int absVar = abs(realValue);
            final Integer otherValue = contradictions.get(absVar);
            if (otherValue == null) {
              contradictions.put(absVar, realValue);
            } else if (realValue == -otherValue) {
              throw EX;
            }
          }
        }
      });
    } catch (final Exception e) {
      if (e == EX) {
        return false;
      } else {
        throw e;
      }
    }

    return true;
  }

  static class Graph {

    private final Vertex[] vertices;

    public Graph(final int n) {
      this.vertices = new Vertex[n];
    }

    public void addEdge(final int from, final int to) {
      final Vertex v;
      if ((v = vertices[from]) != null) {
        v.add(to);
      } else {
        vertices[from] = new Vertex(from, new ArrayList<>(10) {{
          add(to);
        }});
      }

      if (vertices[to] == null) {
        vertices[to] = new Vertex(to, new ArrayList<>(10));
      }

    }

    public void findComponents(final Consumer<List<Integer>> sccConsumer) {
      final VertexWrapper[] vertexWrappers = new VertexWrapper[vertices.length];
      final AtomicInteger vCount = new AtomicInteger(0);
      final Stack<VertexWrapper> path = new Stack<>();
      final Function<Vertex, VertexWrapper> getWrapper = v -> vertexWrappers[v.id] == null
          ? (vertexWrappers[v.id] = new VertexWrapper(v, vCount.getAndIncrement()))
          : vertexWrappers[v.id];
      final Predicate<Vertex> alreadyVisited = v -> vertexWrappers[v.id] != null;
      for (final Vertex v : vertices) {
        if (v != null && !alreadyVisited.test(v)) {
          findComponents(v, getWrapper, alreadyVisited, path, sccConsumer);
        }
      }
    }

    private void findComponents(final Vertex v, final Function<Vertex, VertexWrapper> getWrapper,
        final Predicate<Vertex> alreadyVisited, final Stack<VertexWrapper> path,
        final Consumer<List<Integer>> sccConsumer) {
      final VertexWrapper wr = getWrapper.apply(v);
      path.push(wr);
      wr.onStack = true;

      for (final int vId : wr.vertex.tos) {
        if (!alreadyVisited.test(vertices[vId])) {
          findComponents(vertices[vId], getWrapper, alreadyVisited, path, sccConsumer);
          final VertexWrapper to = getWrapper.apply(vertices[vId]);
          wr.minIndex = min(wr.minIndex, to.minIndex);
        } else {
          final VertexWrapper to = getWrapper.apply(vertices[vId]);
          if (to.onStack) {
            wr.minIndex = min(wr.minIndex, to.index);
          }
        }
      }

      if (wr.minIndex == wr.index) {
        final List<Integer> stronglyConnectedComponent = new ArrayList<>();
        VertexWrapper cV;
        do {
          cV = path.pop();
          cV.onStack = false;
          stronglyConnectedComponent.add(cV.vertex.id);
        } while (cV != wr);
        sccConsumer.accept(stronglyConnectedComponent);
      }
    }

    private static class VertexWrapper {

      private final Vertex vertex;
      private final int index;
      private int minIndex;
      private boolean onStack;

      public VertexWrapper(final Vertex vertex, final int index) {
        this.vertex = vertex;
        this.index = index;
        this.minIndex = index;
        this.onStack = false;
      }
    }

    static class Vertex {

      private final int id;
      private final List<Integer> tos;

      public Vertex(final int form, final List<Integer> tos) {
        this.id = form;
        this.tos = tos;
      }

      public void add(final Integer to) {
        if (!tos.contains(to)) {
          tos.add(to);
        }
      }

      @Override
      public String toString() {
        return String.valueOf(id);
      }
    }
  }

  private static Iterator<String> readLinesFromStdIn() {
    final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    return new Iterator<>() {
      String nextLine = null;

      @Override
      public boolean hasNext() {
        if (nextLine != null) {
          return true;
        } else {
          try {
            nextLine = br.readLine();
            return (nextLine != null);
          } catch (final IOException e) {
            throw new UncheckedIOException(e);
          }
        }
      }

      @Override
      public String next() {
        if (nextLine != null || hasNext()) {
          final String line = nextLine;
          nextLine = null;
          return line;
        } else {
          throw new NoSuchElementException();
        }
      }
    };
  }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases that can cause the algorithm to fail. The key points to consider are:

1. **Contradictions in the Graph**: The code uses a graph to detect contradictions. If the graph contains a contradiction (i.e., both a variable and its negation are in the same strongly connected component), it should return "NO". We need to create a grid that forces such a contradiction.

2. **Small Scale**: The test input should be small enough to be manageable but still complex enough to expose the bug.

3. **Valid Input**: The input should be valid according to the problem constraints.

Let's construct a test case step by step:

### Step 1: Define the Grid
We need a grid where the values in the columns will lead to a contradiction when the graph is constructed. For example, we can create a grid where a variable and its negation are forced to be in the same strongly connected component.

### Step 2: Construct the Grid
Consider a grid with 3 columns:
- Column 1: \(a_1, -a_1, a_2\)
- Column 2: \(a_2, -a_2, a_3\)
- Column 3: \(a_3, -a_3, a_1\)

This grid will force a contradiction because each column will have a variable and its negation, and the graph will detect this.

### Step 3: Format the Input
The input should follow the format specified in the problem statement.

```plaintext
1
3
1 -1 2
2 -2 3
3 -3 1
```

### Explanation:
- **t = 1**: One test case.
- **n = 3**: Three columns.
- **Grid**:
  - Column 1: \(a_1, -a_1, a_2\)
  - Column 2: \(a_2, -a_2, a_3\)
  - Column 3: \(a_3, -a_3, a_1\)

This test case should expose the bug in the code because the graph will detect a contradiction (both a variable and its negation in the same strongly connected component), and the code should return "NO".

### Final Test Input:
```plaintext
1
3
1 -1 2
2 -2 3
3 -3 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Triangle Formation
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given $n$ sticks, numbered from $1$ to $n$. The length of the $i$-th stick is $a_i$.

You need to answer $q$ queries. In each query, you are given two integers $l$ and $r$ ($1 \le l < r \le n$, $r - l + 1 \ge 6$). Determine whether it is possible to choose $6$ distinct sticks from the sticks numbered $l$ to $r$, to form $2$ non-degenerate triangles$^{\text{∗}}$.

$^{\text{∗}}$A triangle with side lengths $a$, $b$, and $c$ is called non-degenerate if:

  * $a < b + c$,    * $b < a + c$, and    * $c < a + b$.

Input: The first line contains two integers $n$ and $q$ ($6 \le n \le 10^5$, $1 \le q \le 10^5$) — the number of sticks and the number of queries respectively.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — $a_i$ denotes the length of the $i$-th stick.

Each of the following $q$ lines contains two integers $l$ and $r$ ($1 \le l < r \le n$, $r - l + 1 \ge 6$) — the parameters of each query.

Output: For each query, output "YES" (without quotes) if it is possible to form $2$ triangles, and "NO" (without quotes) otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

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

Note: In the first query, the lengths of the sticks are $[5, 2, 2, 10, 4, 10]$. Two sets of sticks $[2, 4, 5]$ and $[2, 10, 10]$ can be selected to form $2$ non-degenerate triangles.

In the second query, the lengths of the sticks are $[2, 2, 10, 4, 10, 6]$. It can be shown that it is impossible to form $2$ non-degenerate triangles.

In the third query, the lengths of the sticks are $[2, 2, 10, 4, 10, 6, 1]$. Two sets of sticks $[1, 2, 2]$ and $[4, 10, 10]$ can be selected to form $2$ non-degenerate triangles.

In the fourth query, the lengths of the sticks are $[4, 10, 6, 1, 5, 3]$. It can be shown that it is impossible to form $2$ non-degenerate triangles.

In the fifth query, the lengths of the sticks are $[10, 4, 10, 6, 1, 5, 3]$. Two sets of sticks $[1, 10, 10]$ and $[3, 4, 5]$ can be selected to form $2$ non-degenerate triangles.

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

using namespace std;

void solve() {
    int N, Q;
    cin >> N >> Q;
    vector<int> a(N);
    for (int i = 0; i < N; i++) {
        cin >> a[i];
    }
    while (Q--) {
        int l, r;
        cin >> l >> r;
        l--;
        r--;
        if (r - l + 1 >= 200) {
            cout << "YES\n";
        }else {
            vector<int> v;
            for (int i = l; i <= r; i++) {
                v.push_back(a[i]);
            }
            sort(v.begin(), v.end());

            int m = v.size();
            vector<int> cnt(m);
            for (int i = 3; i < v.size(); i++) {
                if (v[i] - v[i - 1] < v[0]) {
                    cnt[i] = 1;
                }
                if (i > 2 && v[i] - v[i - 1] < v[1]) {
                    cnt[i] = 2;
                }
            }
            int ok = 0;
            for (int i = 3; i + 2 < v.size(); i++) {
                cnt[i] = max(cnt[i], cnt[i - 1]);
                if (cnt[i] + cnt[i + 2] >= 3) {
                    ok = 1;
                }
            }
            if (v[0] + v[1] > v[2]) {
                for (int i = 5; i < v.size(); i++) {
                    if (v[i] - v[i - 1] < v[3]) {
                        ok = 1;
                    }
                }
            }
            cout << (ok ? "YES\n" : "NO\n");
        }
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    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
6 1
1 1 2 2 3 3
1 6
```

### Explanation:
This test case is designed to expose potential issues in the buggy code, particularly with the logic for forming two non-degenerate triangles from a set of sticks.

1. **Input Details:**
   - **Number of sticks (n):** 6
   - **Number of queries (q):** 1
   - **Stick lengths:** [1, 1, 2, 2, 3, 3]
   - **Query:** [1, 6]

2. **Expected Behavior:**
   - The query asks if it is possible to form two non-degenerate triangles from the sticks numbered 1 to 6.
   - The lengths of the sticks in the query are [1, 1, 2, 2, 3, 3].

3. **Possible Triangles:**
   - The sticks [1, 2, 2] can form a non-degenerate triangle.
   - The sticks [1, 3, 3] can form a non-degenerate triangle.
   - Therefore, the expected output should be "YES".

4. **Buggy Code Analysis:**
   - The buggy code uses a complex logic to determine if two non-degenerate triangles can be formed.
   - It sorts the sticks and uses a `cnt` array to track potential valid triangles.
   - However, the logic for updating the `cnt` array and checking for valid triangles might not handle all edge cases correctly.

5. **Why This Test Case:**
   - This test case is small and simple, making it easy to verify the correctness of the solution.
   - It includes repeated stick lengths, which can often cause issues in algorithms that assume unique values.
   - The expected output is "YES," which can help identify if the buggy code incorrectly outputs "NO".

By using this test case, you can effectively test the buggy code and identify any logical errors in the triangle formation algorithm.