**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Job Interview
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp is opening his own IT company. He wants to hire $n$ programmers and $m$ testers.

There are $n+m+1$ candidates, numbered from $1$ to $n+m+1$ in chronological order of their arriving time. The $i$-th candidate has programming skill $a_i$ and testing skill $b_i$ (a person's programming skill is different from their testing skill). The skill of the team is the sum of the programming skills of all candidates hired as programmers, and the sum of the testing skills of all candidates hired as testers.

When a candidate arrives to interview, Monocarp tries to assign them to the most suitable position for them (if their programming skill is higher, then he hires them as a programmer, otherwise as a tester). If all slots for that position are filled, Monocarp assigns them to the other position.

Your task is, for each candidate, calculate the skill of the team if everyone except them comes to interview. Note that it means that exactly $n+m$ candidates will arrive, so all $n+m$ positions in the company will be filled.

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

Each test case consists of three lines:

  * the first line contains two integers $n$ and $m$ ($0 \le n, m \le 2 \cdot 10^5$; $2 \le n + m + 1 \le 2 \cdot 10^5$) — the number of programmers and the number of testers Monocarp wants to hire, respectively;    * the second line contains $n + m + 1$ integers $a_1, a_2, \dots, a_{n+m+1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the programming skill of the $i$-th candidate;    * the third line contains $n + m + 1$ integers $b_1, b_2, \dots, b_{n+m+1}$ ($1 \le b_i \le 10^9$; $b_i \ne a_i$), where $b_i$ is the testing skill of the $i$-th candidate.

Additional constraint on the input: the sum of $(n + m + 1)$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print $n + m + 1$ integers, where the $i$-th integer should be equal to the skill of the team if everyone except the $i$-th candidate comes to interview.

Examples:
input:
4
1 0
2 1
1 2
0 2
4 5 5
5 4 1
1 2
2 1 5 4
5 2 3 1
3 1
4 3 3 4 1
5 5 4 5 2
output:
1 2
5 6 9
8 11 11 12
13 13 13 12 15

Note: Let's consider the third test case of the example:

  * if the $1$-st candidate does not arrive, the $2$-nd candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer, the $4$-th candidate gets hired as a tester. The total skill of the team will be $2 + 5 + 1 = 8$;    * if the $2$-nd candidate does not arrive, the $1$-st candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer, the $4$-th candidate gets hired as a tester. The total skill of the team will be $5 + 5 + 1 = 11$;    * if the $3$-rd candidate does not arrive, the $1$-st candidate gets hired as a tester, the $2$-nd candidate gets hired as a tester, the $4$-th candidate gets hired as a programmer. The total skill of the team will be $5 + 2 + 4 = 11$;    * if the $4$-th candidate does not arrive, the $1$-st candidate gets hired as a tester, the $2$-nd candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer. The total skill of the team will be $5 + 2 + 5 = 12$.

Buggy Code:
import sys
sys.setrecursionlimit(2*10**5)
input = sys.stdin.readline

def solve():
    n, m = map(int,input().split())
    #hire n programmer, m tester
    #"chornological order"
    #hire for programmer or tester, with higher skill he has.

    #may use DP

    ps = list(map(int,input().split()))
    ts = list(map(int,input().split()))
    #DP[programmer][tester]
    dp = [[0]*(m+1) for _ in range((n+1))]  # after without
    # max mem: 10**10 * 2 = 100MB * 2

    before = []
    #everyone come
    P, T = 0, 0
    summ = 0
    for i in range(n+m):
        if P==n:
            T+=1
            summ+=ts[i]
        elif T==m:
            P+=1
            summ+=ps[i]

        else:
            if ps[i]>ts[i]:
                P+=1
                summ+=ps[i]
            else:
                T+=1
                summ+=ts[i]
        before.append((P,T,summ))

    def cwa(idx,pro_h,test_h):
        if idx==n+m+1: return 0
        if dp[pro_h][test_h]: return dp[pro_h][test_h]
        else:
            answer = 0
            if pro_h == n: #tester 강제
                answer = cwa(idx+1,pro_h,test_h+1) + ts[idx]
            elif test_h == m:
                answer = cwa(idx+1,pro_h+1,test_h) + ps[idx]
            else:
                if ps[idx]>ts[idx]:
                    answer = cwa(idx+1,pro_h+1,test_h) + ps[idx]
                else:
                    answer = cwa(idx+1,pro_h,test_h+1) + ts[idx]

            dp[pro_h][test_h] = answer
            return answer

    retval = [cwa(1,0,0)]
    for i in range(1,n+m+1):
        P, T, s = before[i-1]

        retval.append(s+cwa(i+1,P,T))

    return retval

if __name__ == "__main__":
    t = int(input())
    results = []
    for _ in range(t):
        results.append(solve())
    for res in results:
        for r in res:
            print(r,end=" ")
        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:
To generate a challenging test input that exposes the bug in the provided buggy code, let's consider the following edge cases and scenarios:

1. **Small Number of Candidates**: Ensure that the test case is small enough to be easily verifiable but still complex enough to expose bugs.
2. **Candidates with Similar Skills**: Include candidates with programming and testing skills that are close to each other to test the logic for assigning positions.
3. **Edge Cases for Hiring Limits**: Test scenarios where the number of programmers or testers to be hired is zero or very close to the total number of candidates.
4. **Candidates with High and Low Skills**: Include candidates with a wide range of skills to test the robustness of the algorithm.

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

```plaintext
1
2 1
10 20 30
5 25 35
```

**Explanation:**
- **t = 1**: One test case.
- **n = 2, m = 1**: Hiring 2 programmers and 1 tester.
- **a = [10, 20, 30]**: Programming skills of the candidates.
- **b = [5, 25, 35]**: Testing skills of the candidates.

This test case is designed to:
- Test the logic for assigning the first candidate (10, 5) when there are still slots for both programmers and testers.
- Test the logic for assigning the second candidate (20, 25) when there are still slots for both programmers and testers.
- Test the logic for assigning the third candidate (30, 35) when there is only one slot left for a tester.

By ensuring that the skills are varied and the number of candidates is small, this test case can help identify any issues in the logic of the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Job Interview
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp is opening his own IT company. He wants to hire $n$ programmers and $m$ testers.

There are $n+m+1$ candidates, numbered from $1$ to $n+m+1$ in chronological order of their arriving time. The $i$-th candidate has programming skill $a_i$ and testing skill $b_i$ (a person's programming skill is different from their testing skill). The skill of the team is the sum of the programming skills of all candidates hired as programmers, and the sum of the testing skills of all candidates hired as testers.

When a candidate arrives to interview, Monocarp tries to assign them to the most suitable position for them (if their programming skill is higher, then he hires them as a programmer, otherwise as a tester). If all slots for that position are filled, Monocarp assigns them to the other position.

Your task is, for each candidate, calculate the skill of the team if everyone except them comes to interview. Note that it means that exactly $n+m$ candidates will arrive, so all $n+m$ positions in the company will be filled.

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

Each test case consists of three lines:

  * the first line contains two integers $n$ and $m$ ($0 \le n, m \le 2 \cdot 10^5$; $2 \le n + m + 1 \le 2 \cdot 10^5$) — the number of programmers and the number of testers Monocarp wants to hire, respectively;    * the second line contains $n + m + 1$ integers $a_1, a_2, \dots, a_{n+m+1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the programming skill of the $i$-th candidate;    * the third line contains $n + m + 1$ integers $b_1, b_2, \dots, b_{n+m+1}$ ($1 \le b_i \le 10^9$; $b_i \ne a_i$), where $b_i$ is the testing skill of the $i$-th candidate.

Additional constraint on the input: the sum of $(n + m + 1)$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print $n + m + 1$ integers, where the $i$-th integer should be equal to the skill of the team if everyone except the $i$-th candidate comes to interview.

Examples:
input:
4
1 0
2 1
1 2
0 2
4 5 5
5 4 1
1 2
2 1 5 4
5 2 3 1
3 1
4 3 3 4 1
5 5 4 5 2
output:
1 2
5 6 9
8 11 11 12
13 13 13 12 15

Note: Let's consider the third test case of the example:

  * if the $1$-st candidate does not arrive, the $2$-nd candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer, the $4$-th candidate gets hired as a tester. The total skill of the team will be $2 + 5 + 1 = 8$;    * if the $2$-nd candidate does not arrive, the $1$-st candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer, the $4$-th candidate gets hired as a tester. The total skill of the team will be $5 + 5 + 1 = 11$;    * if the $3$-rd candidate does not arrive, the $1$-st candidate gets hired as a tester, the $2$-nd candidate gets hired as a tester, the $4$-th candidate gets hired as a programmer. The total skill of the team will be $5 + 2 + 4 = 11$;    * if the $4$-th candidate does not arrive, the $1$-st candidate gets hired as a tester, the $2$-nd candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer. The total skill of the team will be $5 + 2 + 5 = 12$.

Buggy Code:
import java.util.Scanner;

public class PittuCodeForces{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int tc = sc.nextInt();

        while (tc-- > 0) {
            int p = sc.nextInt();
            int t = sc.nextInt();
            int n = p + t + 1;

            int[] prr = new int[n];
            int[] trr = new int[n];

            for (int i = 0; i < n; i++) {
                prr[i] = sc.nextInt();
            }
            for (int i = 0; i < n; i++) {
                trr[i] = sc.nextInt();
            }

            int totalProgrammers = p;
            int totalTesters = t;
            int[] result = new int[n];

            for (int exclude = 0; exclude < n; exclude++) {
                int sum = 0;
                int remainingP = totalProgrammers;
                int remainingT = totalTesters;

                for (int j = 0; j < n; j++) {
                    if (j == exclude) continue;

                    if (remainingP > 0 && remainingT > 0) {
                        if (prr[j] > trr[j]) {
                            sum += prr[j];
                            remainingP--;
                        } else {
                            sum += trr[j];
                            remainingT--;
                        }
                    } else if (remainingP > 0) {
                        sum += prr[j];
                        remainingP--;
                    } else if (remainingT > 0) {
                        sum += trr[j];
                        remainingT--;
                    }
                }
                result[exclude] = sum;
            }

            for (int res : result) {
                System.out.print(res + " ");
            }
            System.out.println();
        }

        sc.close();
    }
}

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

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

Think step by step.

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

1. **Edge Case: All Programmers or All Testers**:
   - If all candidates are better as programmers or testers, the algorithm should handle this correctly. However, the buggy code might not handle the case where all positions are filled with the same type of role.

2. **Edge Case: One Candidate Excluded**:
   - If one candidate is excluded, the algorithm should still be able to fill all positions correctly. The buggy code might fail if the excluded candidate is the only one who can fill a specific role.

3. **Edge Case: Equal Skills**:
   - The problem states that programming and testing skills are different, so this case is not applicable. However, we can still test the algorithm with skills that are very close to each other to see if it handles the decision-making correctly.

4. **Edge Case: Large Skill Differences**:
   - If the skill differences are very large, the algorithm should clearly favor one role over the other. The buggy code might fail if it does not handle large skill differences correctly.

5. **Edge Case: Balanced Skills**:
   - If the skills are balanced, the algorithm should distribute the candidates fairly. The buggy code might fail if it does not handle balanced skills correctly.

Given these considerations, let's create a small-scale test input that maximizes the failure probability in the faulty solution:

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

### Explanation:
- **Test Case 1**:
  - **Number of Programmers (n)**: 2
  - **Number of Testers (m)**: 1
  - **Total Candidates (n + m + 1)**: 4
  - **Programming Skills (a_i)**: 5, 4, 3, 2
  - **Testing Skills (b_i)**: 1, 2, 4, 5

### Expected Output:
- If the 1st candidate is excluded:
  - 2nd candidate as programmer (4)
  - 3rd candidate as programmer (3)
  - 4th candidate as tester (5)
  - Total skill: 4 + 3 + 5 = 12

- If the 2nd candidate is excluded:
  - 1st candidate as programmer (5)
  - 3rd candidate as programmer (3)
  - 4th candidate as tester (5)
  - Total skill: 5 + 3 + 5 = 13

- If the 3rd candidate is excluded:
  - 1st candidate as programmer (5)
  - 2nd candidate as programmer (4)
  - 4th candidate as tester (5)
  - Total skill: 5 + 4 + 5 = 14

- If the 4th candidate is excluded:
  - 1st candidate as programmer (5)
  - 2nd candidate as programmer (4)
  - 3rd candidate as tester (4)
  - Total skill: 5 + 4 + 4 = 13

### Output:
```plaintext
12 13 14 13
```

This test input should expose any issues in the buggy code related to handling the exclusion of candidates and the correct distribution of roles.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Job Interview
time_limit_ms: 2000
memory_limit_mb: 256
Description: Monocarp is opening his own IT company. He wants to hire $n$ programmers and $m$ testers.

There are $n+m+1$ candidates, numbered from $1$ to $n+m+1$ in chronological order of their arriving time. The $i$-th candidate has programming skill $a_i$ and testing skill $b_i$ (a person's programming skill is different from their testing skill). The skill of the team is the sum of the programming skills of all candidates hired as programmers, and the sum of the testing skills of all candidates hired as testers.

When a candidate arrives to interview, Monocarp tries to assign them to the most suitable position for them (if their programming skill is higher, then he hires them as a programmer, otherwise as a tester). If all slots for that position are filled, Monocarp assigns them to the other position.

Your task is, for each candidate, calculate the skill of the team if everyone except them comes to interview. Note that it means that exactly $n+m$ candidates will arrive, so all $n+m$ positions in the company will be filled.

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

Each test case consists of three lines:

  * the first line contains two integers $n$ and $m$ ($0 \le n, m \le 2 \cdot 10^5$; $2 \le n + m + 1 \le 2 \cdot 10^5$) — the number of programmers and the number of testers Monocarp wants to hire, respectively;    * the second line contains $n + m + 1$ integers $a_1, a_2, \dots, a_{n+m+1}$ ($1 \le a_i \le 10^9$), where $a_i$ is the programming skill of the $i$-th candidate;    * the third line contains $n + m + 1$ integers $b_1, b_2, \dots, b_{n+m+1}$ ($1 \le b_i \le 10^9$; $b_i \ne a_i$), where $b_i$ is the testing skill of the $i$-th candidate.

Additional constraint on the input: the sum of $(n + m + 1)$ over all test cases doesn't exceed $2 \cdot 10^5$.

Output: For each test case, print $n + m + 1$ integers, where the $i$-th integer should be equal to the skill of the team if everyone except the $i$-th candidate comes to interview.

Examples:
input:
4
1 0
2 1
1 2
0 2
4 5 5
5 4 1
1 2
2 1 5 4
5 2 3 1
3 1
4 3 3 4 1
5 5 4 5 2
output:
1 2
5 6 9
8 11 11 12
13 13 13 12 15

Note: Let's consider the third test case of the example:

  * if the $1$-st candidate does not arrive, the $2$-nd candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer, the $4$-th candidate gets hired as a tester. The total skill of the team will be $2 + 5 + 1 = 8$;    * if the $2$-nd candidate does not arrive, the $1$-st candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer, the $4$-th candidate gets hired as a tester. The total skill of the team will be $5 + 5 + 1 = 11$;    * if the $3$-rd candidate does not arrive, the $1$-st candidate gets hired as a tester, the $2$-nd candidate gets hired as a tester, the $4$-th candidate gets hired as a programmer. The total skill of the team will be $5 + 2 + 4 = 11$;    * if the $4$-th candidate does not arrive, the $1$-st candidate gets hired as a tester, the $2$-nd candidate gets hired as a tester, the $3$-rd candidate gets hired as a programmer. The total skill of the team will be $5 + 2 + 5 = 12$.

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

public class JobInterview {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int testCases = Integer.parseInt(sc.nextLine());
        for (int a = 0; a < testCases; a++) {
            String[] nm = sc.nextLine().split(" ");
            String[] pSkill = sc.nextLine().split(" ");
            int len = pSkill.length;
            String[] tSkill = sc.nextLine().split(" ");
            int[] result = new int[len];
            String[][] assocPositions = new String[len][2];
            for (int b = 0; b < len; b++) {
                assocPositions[b] = Arrays.copyOf(nm, nm.length);
            }

            for (int i = 1; i < len; i++) {
                for (int j = 0; j < i; j++) {
                    int pPosition = 0;
                    int tPosition = 0;
                    if (Integer.parseInt(pSkill[i]) > Integer.parseInt(tSkill[i])) {
                        pPosition = Integer.parseInt(assocPositions[j][0]);
                        tPosition = Integer.parseInt(assocPositions[j][1]);
                        if (pPosition > 0) {
                            result[j] += Integer.parseInt(pSkill[i]);
                            assocPositions[j][0] = String.valueOf(Integer.parseInt(assocPositions[j][0]) - 1);
                        } else {
                            result[j] += Integer.parseInt(tSkill[i]);
                            assocPositions[j][1] = String.valueOf(Integer.parseInt(assocPositions[j][1]) - 1);
                        }
                    } else {
                        pPosition = Integer.parseInt(assocPositions[j][0]);
                        tPosition = Integer.parseInt(assocPositions[j][1]);
                        if (tPosition > 0) {
                            result[j] += Integer.parseInt(tSkill[i]);
                            assocPositions[j][1] = String.valueOf(Integer.parseInt(assocPositions[j][1]) - 1);
                        } else {
                            result[j] += Integer.parseInt(pSkill[i]);
                            assocPositions[j][0] = String.valueOf(Integer.parseInt(assocPositions[j][0]) - 1);
                        }
                    }

                    if (Integer.parseInt(pSkill[j]) > Integer.parseInt(tSkill[j])) {
                        pPosition = Integer.parseInt(assocPositions[i][0]);
                        tPosition = Integer.parseInt(assocPositions[i][1]);
                        if (pPosition > 0) {
                            result[i] += Integer.parseInt(pSkill[j]);
                            assocPositions[i][0] = String.valueOf(Integer.parseInt(assocPositions[i][0]) - 1);
                        } else {
                            result[i] += Integer.parseInt(tSkill[j]);
                            assocPositions[i][1] = String.valueOf(Integer.parseInt(assocPositions[i][1]) - 1);
                        }
                    } else {
                        pPosition = Integer.parseInt(assocPositions[i][0]);
                        tPosition = Integer.parseInt(assocPositions[i][1]);
                        if (tPosition > 0) {
                            result[i] += Integer.parseInt(tSkill[j]);
                            assocPositions[i][1] = String.valueOf(Integer.parseInt(assocPositions[i][1]) - 1);
                        } else {
                            result[i] += Integer.parseInt(pSkill[j]);
                            assocPositions[i][0] = String.valueOf(Integer.parseInt(assocPositions[i][0]) - 1);
                        }
                    }
                }
            }
            for(int each : result) {
                System.out.print(each + " ");
            }
            System.out.println();
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
2 1
3 2 1
1 3 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Disks
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given $$$n$$$ disks in the plane. The center of each disk has integer coordinates, and the radius of each disk is a positive integer. No two disks overlap in a region of positive area, but it is possible for disks to be tangent to each other.

Your task is to determine whether it is possible to change the radii of the disks in such a way that:

- Disks that were tangent to each other remain tangent to each other.
- No two disks overlap in a region of positive area.
- The sum of all radii strictly decreases.

Input: The first line contains an integer $$$n$$$ ($$$1\le n \le 1000$$$) — the number of disks.

The next $$$n$$$ lines contain three integers each. The $$$i$$$-th of such lines contains $$$x_i$$$, $$$y_i$$$ ($$$-10^9 \leq x_i, y_i \leq 10^9$$$), and $$$r_i$$$ ($$$1 \leq r_i \leq 10^9$$$) — the coordinates of the center, and the radius, of the $$$i$$$-th disk.

Output: Print $$$\texttt{YES}$$$ if it is possible to change the radii in the desired manner. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
5
0 2 1
0 0 1
4 -3 4
11 0 3
11 5 2
output:
YES

input:
4
2 2 2
7 2 3
7 7 2
2 7 3
output:
NO

Note: In the first sample, one can decrease the radii of the first and third disk by $$$0.5$$$, and increase the radius of the second disk by $$$0.5$$$. This way, the sum of all radii decreases by $$$0.5$$$. The situation before and after changing the radii is depicted below.

First sample (left) and a valid way to change the radii of the disks (right).

In the second sample, depicted below, there is no way to change the radii of the disks in the desired manner.

Second sample.

Buggy Code:
#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
#include <string>
#include <queue>
#include <numeric>
#include <map>
#include <stack>
#include <math.h>
#include <unordered_map>
#include <deque>
#include <stdlib.h>
#include <string.h>

using namespace std;

typedef long long ll;

const int MAX_INT = 1e9 + 10;
const ll MAX_LL = 1e18 + 10;
const ll MOD = 998244353;

const int N = 1e3 + 10;

int n;
vector<double> x(N);
vector<double> y(N);
vector<double> r(N);

int red = 0; // 0
int blue = 0; // 1
bool is_bipar;

vector<vector<int>> adj(N);
vector<bool> visited(N, false);
vector<int> colored(N, -1);

void dfs(int node, int color) {
    if (visited[node] == true) {
        if (colored[node] != color) {
            is_bipar = false;
        }
        return;
    }

    colored[node] = color;
    visited[node] = true;

    if (color == 0)
        red ++;
    else
        blue ++;

    int next_color = color ^ 1;

    for (auto next_node: adj[node]) {
        dfs(next_node, next_color);
    }
}

// condition is bipartite graph with at least 2 node and not the same number of red and blue

void solve() {

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < i; j++) {
            if (i == j)
                continue;

            double d = sqrt(x[i] * x[i] + y[i] * y[i]);

            if (abs(d - r[i] - r[j]) < 1e-6)
            {
                adj[i].push_back(j);
                adj[j].push_back(i);
            }
        }
    }

    for (int i = 0; i < n; i++) {
        red = 0; blue = 0, is_bipar = true;
        if (!visited[i]) {
            dfs(i, 0);

            if (is_bipar && red != blue) {
                printf("YES\n");
                return;
            }
        }
    }
    printf("NO\n");
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int test;
    //cin >> test;
    test = 1;
    while (test--) {
        cin >> n;
        for (int i = 0; i < n; i++) {
            cin >> x[i] >> y[i] >> r[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 generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that can maximize the failure probability. The primary issues in the provided code are:

1. The distance calculation between disk centers is incorrect. It should be the distance between the centers of two disks, not the distance from the origin.
2. The code assumes that the graph is bipartite and checks if the number of red and blue nodes are not equal. However, this is not a sufficient condition to determine if the radii can be changed to meet the problem's requirements.

### Step-by-Step Analysis:

1. **Distance Calculation Error**:
   - The distance between the centers of two disks should be calculated using the formula:
     \[
     d = \sqrt{(x_i - x_j)^2 + (y_i - y_j)^2}
     \]
   - The current code uses:
     \[
     d = \sqrt{x_i^2 + y_i^2}
     \]
   - This will lead to incorrect adjacency list construction.

2. **Bipartite Graph Check**:
   - The code checks if the graph is bipartite and if the number of red and blue nodes are not equal.
   - This is not a sufficient condition to determine if the radii can be changed to meet the problem's requirements.

### Challenging Test Input:

To expose these issues, we can create a test case where the distance calculation error is significant and the bipartite graph check is insufficient.

### Test Input:
```plaintext
3
0 0 1
2 0 1
4 0 1
```

### Explanation:
- **Disks**:
  - Disk 1: Center (0, 0), Radius 1
  - Disk 2: Center (2, 0), Radius 1
  - Disk 3: Center (4, 0), Radius 1

- **Distance Calculation**:
  - Distance between Disk 1 and Disk 2: \(\sqrt{(0-2)^2 + (0-0)^2} = 2\)
  - Distance between Disk 2 and Disk 3: \(\sqrt{(2-4)^2 + (0-0)^2} = 2\)
  - Distance between Disk 1 and Disk 3: \(\sqrt{(0-4)^2 + (0-0)^2} = 4\)

- **Adjacency List**:
  - Disk 1 is tangent to Disk 2.
  - Disk 2 is tangent to Disk 3.
  - Disk 1 is not tangent to Disk 3.

- **Bipartite Graph**:
  - The graph is a simple path (1-2-3), which is bipartite.
  - The number of red and blue nodes will be equal (1 red, 1 blue, 1 red).

- **Expected Output**:
  - The correct answer should be **NO** because it is not possible to change the radii in such a way that the sum of all radii strictly decreases while maintaining tangency and non-overlapping conditions.

### Conclusion:
This test input should expose the incorrect distance calculation and the insufficient bipartite graph check in the provided code. The code will incorrectly output **YES** due to the flawed adjacency list construction and bipartite graph check.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Disks
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given $$$n$$$ disks in the plane. The center of each disk has integer coordinates, and the radius of each disk is a positive integer. No two disks overlap in a region of positive area, but it is possible for disks to be tangent to each other.

Your task is to determine whether it is possible to change the radii of the disks in such a way that:

- Disks that were tangent to each other remain tangent to each other.
- No two disks overlap in a region of positive area.
- The sum of all radii strictly decreases.

Input: The first line contains an integer $$$n$$$ ($$$1\le n \le 1000$$$) — the number of disks.

The next $$$n$$$ lines contain three integers each. The $$$i$$$-th of such lines contains $$$x_i$$$, $$$y_i$$$ ($$$-10^9 \leq x_i, y_i \leq 10^9$$$), and $$$r_i$$$ ($$$1 \leq r_i \leq 10^9$$$) — the coordinates of the center, and the radius, of the $$$i$$$-th disk.

Output: Print $$$\texttt{YES}$$$ if it is possible to change the radii in the desired manner. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
5
0 2 1
0 0 1
4 -3 4
11 0 3
11 5 2
output:
YES

input:
4
2 2 2
7 2 3
7 7 2
2 7 3
output:
NO

Note: In the first sample, one can decrease the radii of the first and third disk by $$$0.5$$$, and increase the radius of the second disk by $$$0.5$$$. This way, the sum of all radii decreases by $$$0.5$$$. The situation before and after changing the radii is depicted below.

First sample (left) and a valid way to change the radii of the disks (right).

In the second sample, depicted below, there is no way to change the radii of the disks in the desired manner.

Second sample.

Buggy Code:
#include<iostream>
#include<vector>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
typedef long long ll;
const int inf = 1e9 + 1;
const int MAXN = 1e3 + 1;
int n = 0;
vector<vector<ll>> radii(MAXN, vector<ll>(3));

// 检查图是否是二分图
bool isBipartited(vector<int>& color, vector<vector<int>>& graph, int node) {
    queue<int> q;
    color[node] = 0; // 起始节点染成颜色 0
    q.push(node);
    while (!q.empty()) {
        int now = q.front();
        q.pop();
        for (int neighbor : graph[now]) {
            if (color[neighbor] == -1) { // 如果邻居未染色
                color[neighbor] = 1 - color[now]; // 染成相反颜色
                q.push(neighbor);
            } else if (color[neighbor] == color[now]) { // 如果邻居颜色相同
                return false; // 不是二分图
            }
        }
    }
    return true; // 是二分图
}

void solve() {
    cin >> n;
    vector<vector<int>> graph(n); // 图的邻接表
    vector<int> color(n, -1); // 节点颜色数组，初始化为 -1

    // 读取圆的属性
    for (int i = 0; i < n; i++) {
        cin >> radii[i][0] >> radii[i][1] >> radii[i][2];
    }

    // 构建图
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            ll dx = radii[i][0] - radii[j][0];
            ll dy = radii[i][1] - radii[j][1];
            ll d_sq = dx * dx + dy * dy;
            ll r_sum = radii[i][2] + radii[j][2];
            if (d_sq == r_sum * r_sum) { // 如果相切
                graph[i].push_back(j);
                graph[j].push_back(i);
            }
        }
    }

    // 检查每个连通分量是否是二分图
    bool isBipartite = true;
    for (int i = 0; i < n; i++) {
        if (color[i] == -1) { // 如果节点未染色
            if (!isBipartited(color, graph, i)) { // 检查连通分量是否是二分图
                isBipartite = false;
                break;
            }
        }
    }

    // 输出结果
    if (isBipartite) {
        cout << "YES" << endl;
    } else {
        cout << "NO" << endl;
    }
}

int main() {
    solve();
    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases where the algorithm might fail. The code checks if the graph formed by the tangent disks is a bipartite graph. If it is, it assumes that it is possible to change the radii in the desired manner. However, this assumption is not always correct.

### Key Points to Consider:
1. **Bipartite Graph Condition**: The code correctly identifies if the graph is bipartite, but it does not consider the sum of radii.
2. **Sum of Radii**: The problem requires that the sum of all radii strictly decreases. The code does not check this condition.

### Test Case:
We need a test case where the graph is bipartite, but it is impossible to decrease the sum of the radii while maintaining the tangency conditions.

### Example Test Case:
Consider a simple case with 3 disks where the graph is bipartite, but the radii cannot be adjusted to decrease the total sum.

```plaintext
3
0 0 1
2 0 1
1 1 2
```

### Explanation:
- **Disk 1**: Center at (0, 0) with radius 1.
- **Disk 2**: Center at (2, 0) with radius 1.
- **Disk 3**: Center at (1, 1) with radius 2.

### Tangency Conditions:
- Disk 1 and Disk 2 are tangent.
- Disk 1 and Disk 3 are tangent.
- Disk 2 and Disk 3 are tangent.

### Graph:
- The graph formed by these disks is a triangle, which is bipartite.

### Radii Adjustment:
- If we try to decrease the radii of Disk 1 and Disk 2, the sum of the radii will not decrease because Disk 3's radius must increase to maintain the tangency.
- Similarly, if we try to decrease the radius of Disk 3, the sum of the radii will not decrease because Disk 1 and Disk 2 must increase to maintain the tangency.

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

This test case should expose the bug in the code because the graph is bipartite, but it is impossible to decrease the sum of the radii while maintaining the tangency conditions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Disks
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given $$$n$$$ disks in the plane. The center of each disk has integer coordinates, and the radius of each disk is a positive integer. No two disks overlap in a region of positive area, but it is possible for disks to be tangent to each other.

Your task is to determine whether it is possible to change the radii of the disks in such a way that:

- Disks that were tangent to each other remain tangent to each other.
- No two disks overlap in a region of positive area.
- The sum of all radii strictly decreases.

Input: The first line contains an integer $$$n$$$ ($$$1\le n \le 1000$$$) — the number of disks.

The next $$$n$$$ lines contain three integers each. The $$$i$$$-th of such lines contains $$$x_i$$$, $$$y_i$$$ ($$$-10^9 \leq x_i, y_i \leq 10^9$$$), and $$$r_i$$$ ($$$1 \leq r_i \leq 10^9$$$) — the coordinates of the center, and the radius, of the $$$i$$$-th disk.

Output: Print $$$\texttt{YES}$$$ if it is possible to change the radii in the desired manner. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
5
0 2 1
0 0 1
4 -3 4
11 0 3
11 5 2
output:
YES

input:
4
2 2 2
7 2 3
7 7 2
2 7 3
output:
NO

Note: In the first sample, one can decrease the radii of the first and third disk by $$$0.5$$$, and increase the radius of the second disk by $$$0.5$$$. This way, the sum of all radii decreases by $$$0.5$$$. The situation before and after changing the radii is depicted below.

First sample (left) and a valid way to change the radii of the disks (right).

In the second sample, depicted below, there is no way to change the radii of the disks in the desired manner.

Second sample.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=1e5+10;
LL n;
LL a[N],b[N],r[N];
vector<LL>v[N];
LL d[N],vis[N]={-1};
bool ct=false;
LL t[3]={0};
void bfs(LL x,LL c)
{
	vis[x]=c;
	queue<pair<LL,LL> > q;
	q.push({x,c});
	t[c]++;
	while(!q.empty())
	{
		pair<LL,LL> u=q.front();
		q.pop();
		//if(vis[u.first]!=-1)continue;
		vis[u.first]=u.second;

		for(int i=0;i<v[u.first].size();i++)
		{
			if(vis[v[u.first][i]]==-1)
			{
				q.push({v[u.first][i],1-u.second});
				t[1-u.second]++;
			}
			else if(vis[v[u.first][i]]!=1-u.second&&vis[v[u.first][i]]!=-1)
			{
				ct=true;
				return;
			}
		}
	}
}
int main()
{
	memset(vis,-1,sizeof(vis));
	bool ct1=false;
	ios::sync_with_stdio(false);cin.tie(nullptr);
    cin>>n;
    for(int i=1;i<=n;i++)
    {
    	cin>>a[i]>>b[i]>>r[i];
    }
    for(int i=1;i<=n;i++)
    {
    	for(int j=i+1;j<=n;j++)
    	{
    		if((abs(a[i]-a[j])*abs(a[i]-a[j])+abs(b[i]-b[j])*abs(b[i]-b[j]))<=(r[i]+r[j])*(r[i]+r[j]))
    		{
    			v[i].push_back(j);
    			v[j].push_back(i);
    			d[i]++,d[j]++;
    		}
    	}
    }
    for(int i=1;i<=n;i++)
    {
    	if(vis[i]==-1)
    	{
    		t[1]=t[0]=0;
    		ct=false;
  			//vis[i]=1;
    		bfs(i,0);
    		//cout<<t[0]<<" "<<t[1]<<endl;
    		if(ct==false&&t[0]!=t[1])
    		{
    			ct1=true;
    			break;
    		}
    	}
    }
    if(ct1==true)
    {
    	cout<<"YES"<<endl;
    }
    else cout<<"NO"<<endl;
    //for(int i=1;i<=n;i++)cout<<vis[i]<<" ";
    return 0;
}

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

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

Think step by step.

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

1. **Tangency and Overlap**: Ensure that some disks are tangent to each other and that no two disks overlap in a region of positive area.
2. **Sum of Radii**: Ensure that the sum of the radii can or cannot be strictly decreased.
3. **Bipartite Graph**: The code uses a bipartite graph to check if the disks can be divided into two groups where each group has a different radius adjustment. We need to create a scenario where this bipartite check fails or passes incorrectly.

Let's create a small-scale test input that can expose the bug:

### Test Input
```plaintext
4
0 0 1
1 0 1
2 0 1
3 0 1
```

### Explanation
- **Disks**: We have four disks arranged in a line on the x-axis.
- **Coordinates and Radii**:
  - Disk 1: Center (0, 0), Radius 1
  - Disk 2: Center (1, 0), Radius 1
  - Disk 3: Center (2, 0), Radius 1
  - Disk 4: Center (3, 0), Radius 1

### Analysis
- **Tangency**: Each disk is tangent to its adjacent disk.
- **Sum of Radii**: The sum of the radii is 4.
- **Bipartite Check**: The code should check if the disks can be divided into two groups such that the sum of the radii can be strictly decreased. In this case, the disks can be divided into two groups (e.g., Group 1: {1, 3}, Group 2: {2, 4}) and the radii can be adjusted to decrease the total sum.

### Expected Output
- **Correct Output**: `YES`
- **Buggy Code Output**: The buggy code might incorrectly output `NO` because it does not handle the bipartite check correctly.

This test input should expose the bug in the bipartite check and the incorrect handling of the radius adjustments.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Disks
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given $$$n$$$ disks in the plane. The center of each disk has integer coordinates, and the radius of each disk is a positive integer. No two disks overlap in a region of positive area, but it is possible for disks to be tangent to each other.

Your task is to determine whether it is possible to change the radii of the disks in such a way that:

- Disks that were tangent to each other remain tangent to each other.
- No two disks overlap in a region of positive area.
- The sum of all radii strictly decreases.

Input: The first line contains an integer $$$n$$$ ($$$1\le n \le 1000$$$) — the number of disks.

The next $$$n$$$ lines contain three integers each. The $$$i$$$-th of such lines contains $$$x_i$$$, $$$y_i$$$ ($$$-10^9 \leq x_i, y_i \leq 10^9$$$), and $$$r_i$$$ ($$$1 \leq r_i \leq 10^9$$$) — the coordinates of the center, and the radius, of the $$$i$$$-th disk.

Output: Print $$$\texttt{YES}$$$ if it is possible to change the radii in the desired manner. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
5
0 2 1
0 0 1
4 -3 4
11 0 3
11 5 2
output:
YES

input:
4
2 2 2
7 2 3
7 7 2
2 7 3
output:
NO

Note: In the first sample, one can decrease the radii of the first and third disk by $$$0.5$$$, and increase the radius of the second disk by $$$0.5$$$. This way, the sum of all radii decreases by $$$0.5$$$. The situation before and after changing the radii is depicted below.

First sample (left) and a valid way to change the radii of the disks (right).

In the second sample, depicted below, there is no way to change the radii of the disks in the desired manner.

Second sample.

Buggy Code:
import sys

def main():
    n = int(input())
    disks = []

    for _ in range(n):
        x, y, r = map(int, input().split())
        disks.append((x, y, r))

    for i in range(n):
        for j in range(i+1, n):
            dist_centers_squared = (disks[i][0] - disks[j][0])**2 + (disks[i][1] - disks[j][1])**2
            sum_radii_squared = (disks[i][2] + disks[j][2])**2

            if dist_centers_squared == sum_radii_squared:
                continue
            elif dist_centers_squared < sum_radii_squared:
                print("NO")
                return

    print("YES")

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
```plaintext
3
0 0 2
4 0 2
2 3 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Disks
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given $$$n$$$ disks in the plane. The center of each disk has integer coordinates, and the radius of each disk is a positive integer. No two disks overlap in a region of positive area, but it is possible for disks to be tangent to each other.

Your task is to determine whether it is possible to change the radii of the disks in such a way that:

- Disks that were tangent to each other remain tangent to each other.
- No two disks overlap in a region of positive area.
- The sum of all radii strictly decreases.

Input: The first line contains an integer $$$n$$$ ($$$1\le n \le 1000$$$) — the number of disks.

The next $$$n$$$ lines contain three integers each. The $$$i$$$-th of such lines contains $$$x_i$$$, $$$y_i$$$ ($$$-10^9 \leq x_i, y_i \leq 10^9$$$), and $$$r_i$$$ ($$$1 \leq r_i \leq 10^9$$$) — the coordinates of the center, and the radius, of the $$$i$$$-th disk.

Output: Print $$$\texttt{YES}$$$ if it is possible to change the radii in the desired manner. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
5
0 2 1
0 0 1
4 -3 4
11 0 3
11 5 2
output:
YES

input:
4
2 2 2
7 2 3
7 7 2
2 7 3
output:
NO

Note: In the first sample, one can decrease the radii of the first and third disk by $$$0.5$$$, and increase the radius of the second disk by $$$0.5$$$. This way, the sum of all radii decreases by $$$0.5$$$. The situation before and after changing the radii is depicted below.

First sample (left) and a valid way to change the radii of the disks (right).

In the second sample, depicted below, there is no way to change the radii of the disks in the desired manner.

Second sample.

Buggy Code:
def can_change_radii(n, disks):
    for i in range(n):
        for j in range(i + 1, n):
            xi, yi, ri = disks[i]
            xj, yj, rj = disks[j]
            distance_squared = (xi - xj) ** 2 + (yi - yj) ** 2
            if distance_squared == (ri + rj) ** 2:
                continue
            elif distance_squared < (ri + rj) ** 2:
                return "NO"
    disks.sort(key=lambda x: x[2])
    prev_sum_radii = sum(r for _, _, r in disks)
    for i in range(1, n):
        curr_sum_radii = sum(r for _, _, r in disks[i:])
        if curr_sum_radii >= prev_sum_radii:
            return "NO"
        prev_sum_radii = curr_sum_radii

    return "YES"

n = int(input())
disks = []
for _ in range(n):
    xi, yi, ri = map(int, input().split())
    disks.append((xi, yi, ri))
print(can_change_radii(n, disks))

**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
0 0 1
0 2 1
2 0 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Disks
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given $$$n$$$ disks in the plane. The center of each disk has integer coordinates, and the radius of each disk is a positive integer. No two disks overlap in a region of positive area, but it is possible for disks to be tangent to each other.

Your task is to determine whether it is possible to change the radii of the disks in such a way that:

- Disks that were tangent to each other remain tangent to each other.
- No two disks overlap in a region of positive area.
- The sum of all radii strictly decreases.

Input: The first line contains an integer $$$n$$$ ($$$1\le n \le 1000$$$) — the number of disks.

The next $$$n$$$ lines contain three integers each. The $$$i$$$-th of such lines contains $$$x_i$$$, $$$y_i$$$ ($$$-10^9 \leq x_i, y_i \leq 10^9$$$), and $$$r_i$$$ ($$$1 \leq r_i \leq 10^9$$$) — the coordinates of the center, and the radius, of the $$$i$$$-th disk.

Output: Print $$$\texttt{YES}$$$ if it is possible to change the radii in the desired manner. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
5
0 2 1
0 0 1
4 -3 4
11 0 3
11 5 2
output:
YES

input:
4
2 2 2
7 2 3
7 7 2
2 7 3
output:
NO

Note: In the first sample, one can decrease the radii of the first and third disk by $$$0.5$$$, and increase the radius of the second disk by $$$0.5$$$. This way, the sum of all radii decreases by $$$0.5$$$. The situation before and after changing the radii is depicted below.

First sample (left) and a valid way to change the radii of the disks (right).

In the second sample, depicted below, there is no way to change the radii of the disks in the desired manner.

Second sample.

Buggy Code:
from math import sqrt, floor

def is_tangent(disk1, disk2):
  """Checks if two disks are tangent."""
  distance = sqrt((disk1[0] - disk2[0])**2 + (disk1[1] - disk2[1])**2)
  return distance == (disk1[2] + disk2[2])

def can_decrease_radius(disk, disks, min_separation):
  """Checks if a disk's radius can be decreased without overlap."""
  for other_disk in disks:
    if other_disk is not disk:
      center_distance = sqrt((disk[0] - other_disk[0])**2 + (disk[1] - other_disk[1])**2)
      if center_distance < (disk[2] + other_disk[2] - min_separation):
        return False
  return True

def solve(disks):
  """Solves the problem of adjusting disk radii."""
  total_decrease = 0.0
  for disk in disks:
    min_separation = 0.0
    for other_disk in disks:
      if is_tangent(disk, other_disk):
        min_separation = max(min_separation, disk[2] + other_disk[2])
    new_radius = max(disk[2] - min_separation, 0.0)
    decrease = disk[2] - new_radius
    if can_decrease_radius(disk, disks, min_separation):
      total_decrease += decrease
      disk[2] = new_radius
  return total_decrease > 0

def main():
  """Reads input and prints the solution."""
  n = int(input())
  disks = []
  for _ in range(n):
    x, y, r = map(int, input().split())
    disks.append([x, y, r])
  if solve(disks):
    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 generate a challenging test input that exposes the bug in the provided code, we need to consider scenarios where the algorithm might fail. The main issues in the code are:

1. The `is_tangent` function checks for exact tangency using floating-point arithmetic, which can lead to precision issues.
2. The `can_decrease_radius` function does not properly account for the minimum separation required to maintain tangency.
3. The `solve` function does not handle the overall reduction in radius correctly, especially when multiple disks are involved.

A challenging test input should include:
- Disks that are almost tangent but not exactly tangent.
- Disks that are tangent to multiple other disks.
- Disks with very small radii that might cause precision issues.

Here is a small-scale test input that exposes these issues:

```plaintext
3
0 0 2
4 0 2
2 0 1
```

### Explanation:
- The first disk is centered at (0, 0) with radius 2.
- The second disk is centered at (4, 0) with radius 2.
- The third disk is centered at (2, 0) with radius 1.

### Analysis:
- The first and second disks are tangent to each other.
- The third disk is tangent to both the first and second disks.

This configuration is challenging because:
- The third disk is a common tangent to both the first and second disks.
- The radius of the third disk is smaller, which can cause precision issues when checking for tangency.
- The algorithm needs to correctly handle the tangency and ensure that the sum of the radii decreases without causing overlap.

The expected output for this test input is `NO` because it is not possible to decrease the radii while maintaining tangency and ensuring the sum of radii decreases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Disks
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given $$$n$$$ disks in the plane. The center of each disk has integer coordinates, and the radius of each disk is a positive integer. No two disks overlap in a region of positive area, but it is possible for disks to be tangent to each other.

Your task is to determine whether it is possible to change the radii of the disks in such a way that:

- Disks that were tangent to each other remain tangent to each other.
- No two disks overlap in a region of positive area.
- The sum of all radii strictly decreases.

Input: The first line contains an integer $$$n$$$ ($$$1\le n \le 1000$$$) — the number of disks.

The next $$$n$$$ lines contain three integers each. The $$$i$$$-th of such lines contains $$$x_i$$$, $$$y_i$$$ ($$$-10^9 \leq x_i, y_i \leq 10^9$$$), and $$$r_i$$$ ($$$1 \leq r_i \leq 10^9$$$) — the coordinates of the center, and the radius, of the $$$i$$$-th disk.

Output: Print $$$\texttt{YES}$$$ if it is possible to change the radii in the desired manner. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
5
0 2 1
0 0 1
4 -3 4
11 0 3
11 5 2
output:
YES

input:
4
2 2 2
7 2 3
7 7 2
2 7 3
output:
NO

Note: In the first sample, one can decrease the radii of the first and third disk by $$$0.5$$$, and increase the radius of the second disk by $$$0.5$$$. This way, the sum of all radii decreases by $$$0.5$$$. The situation before and after changing the radii is depicted below.

First sample (left) and a valid way to change the radii of the disks (right).

In the second sample, depicted below, there is no way to change the radii of the disks in the desired manner.

Second sample.

Buggy Code:
import java.util.Scanner;

public class DiskRadii {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int[][] disks = new int[n][3];

        for (int i = 0; i < n; i++) {
            disks[i][0] = scanner.nextInt(); // x-coordinate
            disks[i][1] = scanner.nextInt(); // y-coordinate
            disks[i][2] = scanner.nextInt(); // radius
        }

        boolean possible = isPossible(disks);
        System.out.println(possible ? "YES" : "NO");
    }

    public static boolean isPossible(int[][] disks) {
        double sumBefore = calculateSumOfRadii(disks);
        double sumAfter = calculateSumOfRadii(disks);

        for (int i = 0; i < disks.length; i++) {
            double decrease = disks[i][2] - 0.5;
            if (decrease < 0) return false; // radius cannot be negative
            sumAfter -= decrease;
        }

        return sumAfter < sumBefore;
    }

    public static double calculateSumOfRadii(int[][] disks) {
        double sum = 0;
        for (int i = 0; i < disks.length; i++) {
            sum += disks[i][2];
        }
        return sum;
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to create a test case that highlights the issue with the `isPossible` method. The bug lies in the assumption that decreasing the radius of each disk by 0.5 will always result in a valid configuration where the disks remain tangent if they were tangent before. This assumption is incorrect because it does not consider the tangency conditions between disks.

### Step-by-Step Reasoning:
1. **Tangency Condition**: Two disks are tangent if the distance between their centers is equal to the sum of their radii.
2. **Decreasing Radii**: If we decrease the radii of all disks by 0.5, the distance between the centers of any two disks will no longer be equal to the sum of their new radii, thus breaking the tangency condition.
3. **Valid Configuration**: We need to ensure that the test case includes disks that are tangent to each other, and when the radii are decreased by 0.5, they are no longer tangent.

### Test Case:
Let's consider a simple case with three disks that are tangent to each other. If we decrease the radii of these disks by 0.5, they will no longer be tangent.

```plaintext
3
0 0 2
4 0 2
2 3 1
```

### Explanation:
- **Disk 1**: Center at (0, 0) with radius 2.
- **Disk 2**: Center at (4, 0) with radius 2.
- **Disk 3**: Center at (2, 3) with radius 1.

**Tangency Conditions:**
- Disk 1 and Disk 2 are tangent because the distance between their centers is 4, which is equal to the sum of their radii (2 + 2).
- Disk 2 and Disk 3 are tangent because the distance between their centers is \(\sqrt{(4-2)^2 + (0-3)^2} = \sqrt{4 + 9} = \sqrt{13}\), which is approximately 3.6, and the sum of their radii is 3 (2 + 1).
- Disk 1 and Disk 3 are not tangent, but this does not affect the test case.

**After Decreasing Radii by 0.5:**
- Disk 1: Radius 1.5
- Disk 2: Radius 1.5
- Disk 3: Radius 0.5

**New Tangency Conditions:**
- Distance between Disk 1 and Disk 2 is 4, but the sum of their new radii is 3 (1.5 + 1.5), so they are no longer tangent.
- Distance between Disk 2 and Disk 3 is \(\sqrt{13}\), but the sum of their new radii is 2 (1.5 + 0.5), so they are no longer tangent.

This test case will expose the bug in the `isPossible` method because it will incorrectly return `YES` despite the fact that the disks will no longer be tangent after the radii are decreased.

### Output:
```plaintext
3
0 0 2
4 0 2
2 3 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Disks
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given $$$n$$$ disks in the plane. The center of each disk has integer coordinates, and the radius of each disk is a positive integer. No two disks overlap in a region of positive area, but it is possible for disks to be tangent to each other.

Your task is to determine whether it is possible to change the radii of the disks in such a way that:

- Disks that were tangent to each other remain tangent to each other.
- No two disks overlap in a region of positive area.
- The sum of all radii strictly decreases.

Input: The first line contains an integer $$$n$$$ ($$$1\le n \le 1000$$$) — the number of disks.

The next $$$n$$$ lines contain three integers each. The $$$i$$$-th of such lines contains $$$x_i$$$, $$$y_i$$$ ($$$-10^9 \leq x_i, y_i \leq 10^9$$$), and $$$r_i$$$ ($$$1 \leq r_i \leq 10^9$$$) — the coordinates of the center, and the radius, of the $$$i$$$-th disk.

Output: Print $$$\texttt{YES}$$$ if it is possible to change the radii in the desired manner. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
5
0 2 1
0 0 1
4 -3 4
11 0 3
11 5 2
output:
YES

input:
4
2 2 2
7 2 3
7 7 2
2 7 3
output:
NO

Note: In the first sample, one can decrease the radii of the first and third disk by $$$0.5$$$, and increase the radius of the second disk by $$$0.5$$$. This way, the sum of all radii decreases by $$$0.5$$$. The situation before and after changing the radii is depicted below.

First sample (left) and a valid way to change the radii of the disks (right).

In the second sample, depicted below, there is no way to change the radii of the disks in the desired manner.

Second sample.

Buggy Code:
import java.util.Scanner;

public class Main{
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int[][] disks = new int[n][3];
        for (int i = 0; i < n; i++) {
            disks[i][0] = scanner.nextInt(); // x-coordinate
            disks[i][1] = scanner.nextInt(); // y-coordinate
            disks[i][2] = scanner.nextInt(); // radius
        }
        scanner.close();

        if (canDecreaseRadii(disks)) {
            System.out.println("YES");
        } else {
            System.out.println("NO");
        }
    }

    private static boolean canDecreaseRadii(int[][] disks) {
        // Logic to determine if it's possible to decrease the sum of radii
        // while maintaining tangency without overlap goes here.
        // This is a placeholder for the actual implementation.
        return true; // Placeholder return value.
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to consider edge cases that can cause the algorithm to fail. Specifically, we should focus on scenarios where the disks are arranged in such a way that it is impossible to decrease the sum of the radii while maintaining tangency and avoiding overlap.

Here are some key points to consider:
1. **Tangency and Overlap**: Ensure that the disks are arranged in a way that makes it impossible to decrease the radii without breaking tangency or causing overlap.
2. **Small Scale**: The input should be small enough to be manageable and fit within the output constraints.

Let's create a test input with a small number of disks that are arranged in a specific way to make it impossible to decrease the sum of the radii while maintaining the conditions.

### Test Input
```plaintext
4
0 0 2
4 0 2
0 4 2
4 4 2
```

### Explanation
- We have 4 disks, each with a radius of 2.
- The centers of the disks are arranged at the corners of a 4x4 square:
  - Disk 1: (0, 0)
  - Disk 2: (4, 0)
  - Disk 3: (0, 4)
  - Disk 4: (4, 4)

### Analysis
- Each disk is tangent to two other disks.
- If we decrease the radius of any disk, it will no longer be tangent to the adjacent disks.
- If we increase the radius of any disk, it will overlap with the adjacent disks.
- Therefore, it is impossible to change the radii in such a way that the sum of the radii decreases while maintaining tangency and avoiding overlap.

This test input should effectively expose the bug in the buggy code, as the placeholder logic in `canDecreaseRadii` will return `true` by default, which is incorrect for this input.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Disks
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given $$$n$$$ disks in the plane. The center of each disk has integer coordinates, and the radius of each disk is a positive integer. No two disks overlap in a region of positive area, but it is possible for disks to be tangent to each other.

Your task is to determine whether it is possible to change the radii of the disks in such a way that:

- Disks that were tangent to each other remain tangent to each other.
- No two disks overlap in a region of positive area.
- The sum of all radii strictly decreases.

Input: The first line contains an integer $$$n$$$ ($$$1\le n \le 1000$$$) — the number of disks.

The next $$$n$$$ lines contain three integers each. The $$$i$$$-th of such lines contains $$$x_i$$$, $$$y_i$$$ ($$$-10^9 \leq x_i, y_i \leq 10^9$$$), and $$$r_i$$$ ($$$1 \leq r_i \leq 10^9$$$) — the coordinates of the center, and the radius, of the $$$i$$$-th disk.

Output: Print $$$\texttt{YES}$$$ if it is possible to change the radii in the desired manner. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
5
0 2 1
0 0 1
4 -3 4
11 0 3
11 5 2
output:
YES

input:
4
2 2 2
7 2 3
7 7 2
2 7 3
output:
NO

Note: In the first sample, one can decrease the radii of the first and third disk by $$$0.5$$$, and increase the radius of the second disk by $$$0.5$$$. This way, the sum of all radii decreases by $$$0.5$$$. The situation before and after changing the radii is depicted below.

First sample (left) and a valid way to change the radii of the disks (right).

In the second sample, depicted below, there is no way to change the radii of the disks in the desired manner.

Second sample.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
struct node{
	int id,x,y,r;
}nodes[1010];
int vis[1010],k[1010],g[1010];
int find(int x){
	if(k[x]==x)return x;
		k[x]=find(k[x]);
	return k[x];
}
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0),cout.tie(0);
	int n,x,r,y;
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>nodes[i].x>>nodes[i].y>>nodes[i].r;
		k[i]=i;
	}
	for(int i=1;i<=n;i++){
		for(int j=i-1;j>=1;j--){
			if((nodes[i].x-nodes[j].x)*(nodes[i].x-nodes[j].x)+(nodes[i].y-nodes[j].y)*(nodes[i].y-nodes[j].y)==(nodes[i].r+nodes[j].r)*(nodes[i].r+nodes[j].r)){
				k[find(j)]=k[i];
			}
		}
	}
	for(int i=1;i<=n;i++){
		g[k[i]]++;
	}
	for(int i=1;i<=n;i++)
	{
		if(g[i]%2!=0)
		{
			printf("YES");
			return 0;
		}
	}
	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
3
0 0 1
2 0 1
1 1 1
```

**Explanation:**
- This test case involves three disks that are tangent to each other in a way that forms a triangle.
- Disk 1 is centered at (0, 0) with radius 1.
- Disk 2 is centered at (2, 0) with radius 1.
- Disk 3 is centered at (1, 1) with radius 1.

**Analysis:**
- The distances between the centers of the disks are:
  - Distance between Disk 1 and Disk 2: 2, which is equal to the sum of their radii (1 + 1).
  - Distance between Disk 1 and Disk 3: \(\sqrt{2}\), which is not equal to the sum of their radii (1 + 1).
  - Distance between Disk 2 and Disk 3: \(\sqrt{2}\), which is not equal to the sum of their radii (1 + 1).

- Since the distance between Disk 1 and Disk 2 is exactly equal to the sum of their radii, they are tangent.
- The other two pairs of disks are not tangent, but they are close enough to form a triangle.

**Buggy Code Issue:**
- The buggy code only checks for tangency and does not properly handle the scenario where the sum of the radii can be decreased while maintaining the tangency.
- The code might incorrectly conclude that it is possible to decrease the sum of the radii, even though it is not possible in this case.

**Expected Output:**
- The correct output for this test case should be `NO` because it is not possible to decrease the sum of the radii while maintaining the tangency between the disks.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Disks
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given $$$n$$$ disks in the plane. The center of each disk has integer coordinates, and the radius of each disk is a positive integer. No two disks overlap in a region of positive area, but it is possible for disks to be tangent to each other.

Your task is to determine whether it is possible to change the radii of the disks in such a way that:

- Disks that were tangent to each other remain tangent to each other.
- No two disks overlap in a region of positive area.
- The sum of all radii strictly decreases.

Input: The first line contains an integer $$$n$$$ ($$$1\le n \le 1000$$$) — the number of disks.

The next $$$n$$$ lines contain three integers each. The $$$i$$$-th of such lines contains $$$x_i$$$, $$$y_i$$$ ($$$-10^9 \leq x_i, y_i \leq 10^9$$$), and $$$r_i$$$ ($$$1 \leq r_i \leq 10^9$$$) — the coordinates of the center, and the radius, of the $$$i$$$-th disk.

Output: Print $$$\texttt{YES}$$$ if it is possible to change the radii in the desired manner. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
5
0 2 1
0 0 1
4 -3 4
11 0 3
11 5 2
output:
YES

input:
4
2 2 2
7 2 3
7 7 2
2 7 3
output:
NO

Note: In the first sample, one can decrease the radii of the first and third disk by $$$0.5$$$, and increase the radius of the second disk by $$$0.5$$$. This way, the sum of all radii decreases by $$$0.5$$$. The situation before and after changing the radii is depicted below.

First sample (left) and a valid way to change the radii of the disks (right).

In the second sample, depicted below, there is no way to change the radii of the disks in the desired manner.

Second sample.

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

# 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

from heapq import nsmallest, nlargest, heappushpop, heapify, heappop, heappush

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

MOD = 10 ** 9 + 7

# t, = R()

def answer():
    n, = R()

    a = []
    for i in range(n):
        a.append(list(R()))

    g = {}

    for i in range(n):
        for j in range(i):
            dx = abs(a[i][0] - a[j][0])
            dy = abs(a[i][1] - a[j][1])
            r = a[i][2] + a[j][2]

            if r * r == (dx * dx + dy * dy):
                if i not in g:
                    g[i] = []

                if j not in g:
                    g[j] = []

                g[i].append(j)
                g[j].append(i)

    for k in g.keys():
        if len(g[k]) > 2:
            return "YES"

    for i in range(n):
        if i not in g:
            return "YES"

    visited = [False] * n

    @bootstrap
    def dfs(v, p):
        cycles = False

        visited[v] = True

        cnt = 1

        for u in g[v]:
            if u != p:
                if not visited[u]:
                    ci, cyi = (yield dfs(u, v))
                    cnt += ci
                    cycles |= cyi
                else:
                    cycles = True

        yield cnt, cycles

    for i in range(n):
        if not visited[i]:
            ci, cycles = dfs(i, -1)

            if cycles:
                continue

            if ci > 2:
                return "YES"

    return "NO"

print(answer())
# 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:
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. The main issues in the code seem to be related to the graph construction and the detection of cycles or the number of connections.

### Step-by-Step Analysis:

1. **Graph Construction**: The code constructs a graph where each node represents a disk, and an edge between two nodes indicates that the corresponding disks are tangent. The condition for tangency is:
   \[
   (x_i - x_j)^2 + (y_i - y_j)^2 = (r_i + r_j)^2
   \]
   This is a strict equality check, which can be problematic if there are floating-point precision issues.

2. **Cycle Detection**: The code uses a depth-first search (DFS) to detect cycles. If a cycle is found, it returns "YES". If no cycle is found, it checks if any node has more than 2 connections and returns "YES" in that case.

3. **Special Cases**:
   - **Disks with no connections**: If a disk is not connected to any other disk, the code should return "YES".
   - **Disks forming a chain**: If disks form a chain where each disk is tangent to exactly two other disks, the code should return "NO" because no cycle is detected and no node has more than 2 connections.
   - **Disks forming a cycle**: If disks form a cycle, the code should return "YES".

### Challenging Test Input:

Let's create a test input that includes a chain of disks where each disk is tangent to exactly two other disks. This should trigger the "NO" condition in the code.

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

### Explanation:

- **Disks**: The disks are placed along the x-axis at coordinates (0,0), (2,0), (4,0), (6,0), and (8,0) with radii 1.
- **Tangency**: Each disk is tangent to its immediate neighbors, forming a chain.
- **Graph**: The graph will have 5 nodes, each connected to exactly two other nodes.
- **Cycle Detection**: The DFS will not detect any cycle, and no node will have more than 2 connections.
- **Expected Output**: The correct output should be "NO" because it is not possible to change the radii in the desired manner for this configuration.

This test input should expose the bug in the code by correctly returning "NO" for a chain of tangent disks.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Disks
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given $$$n$$$ disks in the plane. The center of each disk has integer coordinates, and the radius of each disk is a positive integer. No two disks overlap in a region of positive area, but it is possible for disks to be tangent to each other.

Your task is to determine whether it is possible to change the radii of the disks in such a way that:

- Disks that were tangent to each other remain tangent to each other.
- No two disks overlap in a region of positive area.
- The sum of all radii strictly decreases.

Input: The first line contains an integer $$$n$$$ ($$$1\le n \le 1000$$$) — the number of disks.

The next $$$n$$$ lines contain three integers each. The $$$i$$$-th of such lines contains $$$x_i$$$, $$$y_i$$$ ($$$-10^9 \leq x_i, y_i \leq 10^9$$$), and $$$r_i$$$ ($$$1 \leq r_i \leq 10^9$$$) — the coordinates of the center, and the radius, of the $$$i$$$-th disk.

Output: Print $$$\texttt{YES}$$$ if it is possible to change the radii in the desired manner. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
5
0 2 1
0 0 1
4 -3 4
11 0 3
11 5 2
output:
YES

input:
4
2 2 2
7 2 3
7 7 2
2 7 3
output:
NO

Note: In the first sample, one can decrease the radii of the first and third disk by $$$0.5$$$, and increase the radius of the second disk by $$$0.5$$$. This way, the sum of all radii decreases by $$$0.5$$$. The situation before and after changing the radii is depicted below.

First sample (left) and a valid way to change the radii of the disks (right).

In the second sample, depicted below, there is no way to change the radii of the disks in the desired manner.

Second sample.

Buggy Code:
n=int(input())
x=[]
y=[]
r=[]
for i in range(n):
    a,b,c=map(int,input().split())
    x.append(a)
    y.append(b)
    r.append(c)
count=0
for i in range(n):
    for j in range(n):
        if j!=i:
            if (x[i]-x[j])**2+(y[i]-y[j])**2==(r[i]+r[j])**2:
                count+=1
if count==n*2:
    print("NO")
else:
    print("YES")

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to create a test case that specifically targets the logic where the code incorrectly determines whether the disks can be adjusted to meet the conditions. The buggy code counts the number of tangencies and incorrectly assumes that if the total number of tangencies is not equal to `n * 2`, then it is possible to adjust the radii. This logic is flawed because it does not account for the actual feasibility of changing the radii while maintaining tangencies.

### Step-by-Step Analysis:
1. **Identify the Flaw**: The code assumes that if the number of tangencies is not exactly `n * 2`, it is possible to adjust the radii. This is incorrect because it does not consider the actual geometric constraints of the problem.

2. **Create a Test Case**:
   - We need a configuration of disks where the number of tangencies is not `n * 2`, but it is still impossible to adjust the radii to meet the conditions.
   - A good candidate is a configuration where the disks are arranged in such a way that changing the radii to maintain tangencies and reduce the total sum of radii is impossible.

### Test Case:
Consider a configuration of 4 disks arranged in a square, where each disk is tangent to its adjacent disks. This configuration has exactly 4 tangencies, which is not equal to `n * 2` (which would be 8 in this case). However, it is impossible to adjust the radii to maintain tangencies and reduce the total sum of radii.

```plaintext
4
0 0 1
2 0 1
2 2 1
0 2 1
```

### Explanation:
- **Disk 1**: Center at (0, 0) with radius 1
- **Disk 2**: Center at (2, 0) with radius 1
- **Disk 3**: Center at (2, 2) with radius 1
- **Disk 4**: Center at (0, 2) with radius 1

These disks form a square where each disk is tangent to its adjacent disks. The total number of tangencies is 4, which is not equal to `n * 2` (8). However, it is impossible to reduce the sum of the radii while maintaining tangencies because reducing the radius of any disk will break the tangency with its adjacent disks.

This test case should expose the bug 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: Disks
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given $$$n$$$ disks in the plane. The center of each disk has integer coordinates, and the radius of each disk is a positive integer. No two disks overlap in a region of positive area, but it is possible for disks to be tangent to each other.

Your task is to determine whether it is possible to change the radii of the disks in such a way that:

- Disks that were tangent to each other remain tangent to each other.
- No two disks overlap in a region of positive area.
- The sum of all radii strictly decreases.

Input: The first line contains an integer $$$n$$$ ($$$1\le n \le 1000$$$) — the number of disks.

The next $$$n$$$ lines contain three integers each. The $$$i$$$-th of such lines contains $$$x_i$$$, $$$y_i$$$ ($$$-10^9 \leq x_i, y_i \leq 10^9$$$), and $$$r_i$$$ ($$$1 \leq r_i \leq 10^9$$$) — the coordinates of the center, and the radius, of the $$$i$$$-th disk.

Output: Print $$$\texttt{YES}$$$ if it is possible to change the radii in the desired manner. Otherwise, print $$$\texttt{NO}$$$.

Examples:
input:
5
0 2 1
0 0 1
4 -3 4
11 0 3
11 5 2
output:
YES

input:
4
2 2 2
7 2 3
7 7 2
2 7 3
output:
NO

Note: In the first sample, one can decrease the radii of the first and third disk by $$$0.5$$$, and increase the radius of the second disk by $$$0.5$$$. This way, the sum of all radii decreases by $$$0.5$$$. The situation before and after changing the radii is depicted below.

First sample (left) and a valid way to change the radii of the disks (right).

In the second sample, depicted below, there is no way to change the radii of the disks in the desired manner.

Second sample.

Buggy Code:
#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define endl '\n'
#define int long long
#define x first
#define y second
using namespace std;
const int N=1e6+10;

int f[N];int n;
int find(int x){
	if(f[x]!=x) return f[x]=find(f[x]);
	return x;
}

void solve(){
	cin>>n;
	vector<int>x(n+1);
	vector<int>y(n+1);
	vector<int>r(n+1);
	vector<int>z(n+1,0);
	for(int i=1;i<=n;i++){
		f[i]=i;
	}
	for(int i=1;i<=n;i++){
		cin>>x[i]>>y[i]>>r[i];
	}
	if(n==1){
		cout<<"YES"<<endl;
		return ;
	}
	for(int i=1;i<=n;i++){
		for(int j=i+1;j<=n;j++){
			double sum=abs(x[i]-x[j])*abs(x[i]-x[j])+abs(y[i]-y[j])*abs(y[i]-y[j]);
			sum*=1.000;
			sum=sqrt(sum);
			if(sum<=r[i]+r[j]){
				z[i]++;
				z[j]++;
				int q1=find(i);
				int q2=find(j);
				if(q1!=q2){
					f[q1]=q2;
				}
			}
		}
	}
	int vis=0;
	int res=find(f[1]);
	for(int i=1;i<=n;i++){
//		cout<<find(f[i])<<" ";
		if(find(f[i])!=res){
			vis=1;
			break;
		}
	}
	if(vis) cout<<"YES"<<endl;
	else{
		for(int i=1;i<=n;i++){
			if(z[i]>2){
				cout<<"YES"<<endl;
				return ;
			}
		}
		cout<<"NO"<<endl;
	}
}

signed main()
{
	IOS;
	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
0 0 1
2 0 1
1 1 1
```

**Explanation:**
- This test case is designed to expose the bug in the buggy code.
- The centers of the disks are (0,0), (2,0), and (1,1), with radii 1, 1, and 1, respectively.
- The disks are arranged such that:
  - Disk 1 and Disk 2 are tangent to each other.
  - Disk 2 and Disk 3 are tangent to each other.
  - Disk 1 and Disk 3 are not tangent to each other.

**Expected Behavior:**
- The algorithm should determine if it is possible to change the radii of the disks such that:
  - Disks that were tangent remain tangent.
  - No two disks overlap.
  - The sum of all radii strictly decreases.

**Why This Test Case is Challenging:**
- The buggy code might incorrectly assume that if all disks are part of the same connected component (i.e., they are all connected through tangency), the answer should be "YES". However, this is not always true.
- In this case, the correct answer is "NO" because it is impossible to decrease the sum of the radii while maintaining the tangency conditions. If you decrease the radius of any disk, it will no longer be tangent to its neighbors, and if you increase the radius of any disk, it will overlap with its neighbors.
- The buggy code might incorrectly output "YES" because it does not properly handle the condition where the sum of the radii cannot be decreased while maintaining tangency.