**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Startup
time_limit_ms: 2000
memory_limit_mb: 256
Description: Arseniy came up with another business plan — to sell soda from a vending machine! For this, he purchased a machine with $n$ shelves, as well as $k$ bottles, where the $i$-th bottle is characterized by the brand index $b_i$ and the cost $c_i$.

You can place any number of bottles on each shelf, but all bottles on the same shelf must be of the same brand.

Arseniy knows that all the bottles he puts on the shelves of the machine will be sold. Therefore, he asked you to calculate the maximum amount he can earn.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 2 \cdot 10^5$), where $n$ is the number of shelves in the machine, and $k$ is the number of bottles available to Arseniy.

The next $k$ lines contain two integers $b_i$ and $c_i$ ($1 \le b_i \le k, 1 \le c_i \le 1000$) — the brand and cost of the $i$-th bottle.

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

Output: For each test case, output one integer — the maximum amount that Arseniy can earn.

Examples:
input:
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
output:
28
15
12
1000

Note: In the first test case, Arseniy has $3$ shelves in the vending machine. He can place, for example, two bottles of the brand $2$ on the first shelf and a bottle of the brand $1$ on the second shelf. Then the total cost of the bottles would be $6 + 7 + 15 = 28$.

In the second test case, he has only one shelf. It is not difficult to show that the optimal option is to place a bottle of the brand $1$ on it. Then the total cost will be $15$.

In the third test case, he has as many as $6$ shelves, so he can place all available bottles with a total cost of $7 + 5 = 12$.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <string>
#include <queue>
#include <set>
#include <numeric>
#include <map>
#include <functional>
using namespace std;

#define kevo ios_base::sync_with_stdio(false); cin.tie(0);
const int MOD = 1e9+7;

/***************************************Sparse Table****************************************/
struct SparseTable{
    int N;
    vector<vector<int>> lookup;
    SparseTable(int n) {
        N = n;
        int P = floor(log2(N)) + 1;
        lookup.resize(P, vector<int>(N,0));
    }
    int operation(int x, int y) {
        return gcd(x,y);
    }
    void build(vector<int>& vec) {
        for (int i = 0; i < N; i++) {
            lookup[0][i] = vec[i];
        }
        for (int i = 1; (1 << i) <= N; i++) {
            for (int j = 0; (j + (1 << i) - 1) < N; j++) {
                lookup[i][j] = operation(lookup[i-1][j], lookup[i-1][j + (1 << (i-1))]);
            }
        }
    }
    int query(int l, int r) {
        int cur = r-l+1;
        int k = floor(log2(cur));
        return operation(lookup[k][l], lookup[k][r-(1<<k)+1]);
    }
};
/*******************************************************************************************/

/***************************************Segment Tree****************************************/
struct SegmentTree{
    int N;
    vector<int> tree;
    SegmentTree(int n){
        N = 1;
        while(N < n) N<<=1;
        tree.resize(N<<1);
    }
};
/*******************************************************************************************/

/*************************************start † here******************************************/

vector<bool> sieve(1001, true);
void sieveCalc(vector<bool>& sieve) {
    sieve[0] = sieve[1] = false;
    for (int i = 2; i * i < sieve.size()+1; i++) {
        if (sieve[i]) {
            for (int j = i * i; j < sieve.size()+1; j += i) {
                sieve[j] = false;
            }
        }
    }
}

void solve() {
    int n, k;cin>>n>>k;
    vector<int> num(1001, 0);
    for (int i = 0; i < k; i++)
    {
        int x,y;cin>>x>>y;
        num[x] += y;
    }
    sort(num.begin(), num.end(), greater<int>());
    int l = 0, ans = 0;
    while(n-- && num[l] != 0) {
        ans += num[l++];
    }
    cout<<ans<<endl;
}

int main() {
    kevo
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif

    sieveCalc(sieve);
    int t = 1;cin>>t;
    while (t--) {
        solve();
    }
    return 0;
}
/***************************************End † here******************************************/

**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 craft a test case that highlights the issue. The code is intended to maximize the earnings by placing bottles on shelves, with the constraint that all bottles on the same shelf must be of the same brand.

The provided code sorts the total cost of bottles by brand and then sums the top `n` costs. However, this approach might fail in scenarios where the number of shelves is greater than the number of unique brands, or when the cost distribution is very uneven.

Let's consider a test case where the number of shelves is greater than the number of unique brands, and the costs are distributed in a way that the current implementation might not handle correctly.

### Test Case
- **Number of test cases (t):** 1
- **Number of shelves (n):** 5
- **Number of bottles (k):** 3
- **Bottles:**
  - Brand 1, Cost 10
  - Brand 2, Cost 20
  - Brand 3, Cost 30

### Expected Output
The expected output should be the sum of the top 3 costs, which is `10 + 20 + 30 = 60`.

However, the provided code might only sum the top 3 costs without considering that there are 5 shelves available. This means it will not place any bottles on the remaining 2 shelves, leading to an incorrect result.

### Test Input
```plaintext
1
5 3
1 10
2 20
3 30
```

This test case will expose the bug in the code by ensuring that the number of shelves is greater than the number of unique brands, forcing the code to handle the extra shelves correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Startup
time_limit_ms: 2000
memory_limit_mb: 256
Description: Arseniy came up with another business plan — to sell soda from a vending machine! For this, he purchased a machine with $n$ shelves, as well as $k$ bottles, where the $i$-th bottle is characterized by the brand index $b_i$ and the cost $c_i$.

You can place any number of bottles on each shelf, but all bottles on the same shelf must be of the same brand.

Arseniy knows that all the bottles he puts on the shelves of the machine will be sold. Therefore, he asked you to calculate the maximum amount he can earn.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 2 \cdot 10^5$), where $n$ is the number of shelves in the machine, and $k$ is the number of bottles available to Arseniy.

The next $k$ lines contain two integers $b_i$ and $c_i$ ($1 \le b_i \le k, 1 \le c_i \le 1000$) — the brand and cost of the $i$-th bottle.

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

Output: For each test case, output one integer — the maximum amount that Arseniy can earn.

Examples:
input:
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
output:
28
15
12
1000

Note: In the first test case, Arseniy has $3$ shelves in the vending machine. He can place, for example, two bottles of the brand $2$ on the first shelf and a bottle of the brand $1$ on the second shelf. Then the total cost of the bottles would be $6 + 7 + 15 = 28$.

In the second test case, he has only one shelf. It is not difficult to show that the optimal option is to place a bottle of the brand $1$ on it. Then the total cost will be $15$.

In the third test case, he has as many as $6$ shelves, so he can place all available bottles with a total cost of $7 + 5 = 12$.

Buggy Code:
#include <bits/stdc++.h>
#define se second
#define fi first
#define ll long long
#define fo(var, n) for(ll var=0;var<n;var++)
#define fore(var, n) for(ll var=n-1;var>=0;var--)
using namespace std;
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

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

    int tt=1;
    cin>>tt;

    while(tt--)
    {
        int n,k;cin>>n>>k;
        int a[1000] ={0};
        fo(i,1000) a[i]=0;
        fo(i,k)
        {
            int kk,j; cin>>kk>>j;
            kk--;
            a[kk]+=j;
        }
        sort(a,a+1000);
        int ans=0;
        n=min(1000,n);
        while(n>0)
        {
            ans+=a[1000-n];
            n--;
        }
        cout<<ans<<"\n";

    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Startup
time_limit_ms: 2000
memory_limit_mb: 256
Description: Arseniy came up with another business plan — to sell soda from a vending machine! For this, he purchased a machine with $n$ shelves, as well as $k$ bottles, where the $i$-th bottle is characterized by the brand index $b_i$ and the cost $c_i$.

You can place any number of bottles on each shelf, but all bottles on the same shelf must be of the same brand.

Arseniy knows that all the bottles he puts on the shelves of the machine will be sold. Therefore, he asked you to calculate the maximum amount he can earn.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 2 \cdot 10^5$), where $n$ is the number of shelves in the machine, and $k$ is the number of bottles available to Arseniy.

The next $k$ lines contain two integers $b_i$ and $c_i$ ($1 \le b_i \le k, 1 \le c_i \le 1000$) — the brand and cost of the $i$-th bottle.

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

Output: For each test case, output one integer — the maximum amount that Arseniy can earn.

Examples:
input:
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
output:
28
15
12
1000

Note: In the first test case, Arseniy has $3$ shelves in the vending machine. He can place, for example, two bottles of the brand $2$ on the first shelf and a bottle of the brand $1$ on the second shelf. Then the total cost of the bottles would be $6 + 7 + 15 = 28$.

In the second test case, he has only one shelf. It is not difficult to show that the optimal option is to place a bottle of the brand $1$ on it. Then the total cost will be $15$.

In the third test case, he has as many as $6$ shelves, so he can place all available bottles with a total cost of $7 + 5 = 12$.

Buggy Code:
import java.lang.reflect.Array;
import java.util.Arrays;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        for (int i = 0; i < t; i++) {
            int n = scanner.nextInt();
            int k = scanner.nextInt();
            int[] arr1 = new int[k];
            int[] arr2 = new int[k];
            for (int j = 0; j < k; j++) {
                arr1[j] = scanner.nextInt();
                arr2[j] = scanner.nextInt();
            }
            System.out.println(mine(n,k,arr1,arr2));
        }
    }
    static int sum(int[] arr){
        int sum =0;
        for (int i = 0; i < arr.length; i++) {
            sum = sum + arr[i];
        }
        return sum;
    }
    static int max(int[] arr){
        int max= arr[0];
        for (int i = 0; i < arr.length; i++) {
            if(arr[i]>max){
                max=arr[i];
            }
        }
        return max;
    }
    static int nonZero(int[] arr){
        int count =0;
        for (int i = 0; i < arr.length; i++) {
            if(arr[i]!=0){
                count++;
            }
        }
        return count;
    }
    static int mine(int n, int k, int[] arr1,int[] arr2){
        if(n>=k){
            return sum(arr2);
        }
        int[] sums = new int[k];
        int x=0;
        int ans =0;
        for (int i = 1; i <= k; i++) {
            ans = 0;
            for (int j = 0; j < k; j++) {
                if(arr1[j]==i){
                    ans=ans+arr2[j];
                }
            }
            sums[x]=ans;
            x++;
        }
        if(nonZero(sums)<=n){
            return sum(sums);
        }else{
            Arrays.sort(sums);
            int z =0;
            for (int i = 0; i < n; i++) {
                z=z+sums[sums.length-1-i];
            }
            return z;
        }
    }

    }

**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 that might cause the code to fail. Here are a few scenarios to consider:

1. **All bottles of the same brand**: This could cause issues if the code does not handle the case where all bottles are of the same brand and the number of shelves is less than the number of bottles.
2. **Mixed brands with varying costs**: This could expose issues in the logic that calculates the maximum earnings.
3. **Single bottle with high cost**: This could test the logic for handling a single bottle.
4. **Many bottles with low cost**: This could test the logic for handling a large number of bottles with low costs.

Let's create a test input that combines these scenarios to maximize the failure probability:

```plaintext
4
3 3
1 10
1 10
1 10
1 3
1 5
2 10
3 15
6 3
1 1
2 2
3 3
190000 1
1 1000
```

This test input includes:
1. A case with all bottles of the same brand and more bottles than shelves.
2. A case with mixed brands and varying costs.
3. A case with more shelves than bottles.
4. A case with a single bottle and a high cost.

This should help expose any issues 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: Startup
time_limit_ms: 2000
memory_limit_mb: 256
Description: Arseniy came up with another business plan — to sell soda from a vending machine! For this, he purchased a machine with $n$ shelves, as well as $k$ bottles, where the $i$-th bottle is characterized by the brand index $b_i$ and the cost $c_i$.

You can place any number of bottles on each shelf, but all bottles on the same shelf must be of the same brand.

Arseniy knows that all the bottles he puts on the shelves of the machine will be sold. Therefore, he asked you to calculate the maximum amount he can earn.

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

The first line of each test case contains two integers $n$ and $k$ ($1 \le n, k \le 2 \cdot 10^5$), where $n$ is the number of shelves in the machine, and $k$ is the number of bottles available to Arseniy.

The next $k$ lines contain two integers $b_i$ and $c_i$ ($1 \le b_i \le k, 1 \le c_i \le 1000$) — the brand and cost of the $i$-th bottle.

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

Output: For each test case, output one integer — the maximum amount that Arseniy can earn.

Examples:
input:
4
3 3
2 6
2 7
1 15
1 3
2 6
2 7
1 15
6 2
1 7
2 5
190000 1
1 1000
output:
28
15
12
1000

Note: In the first test case, Arseniy has $3$ shelves in the vending machine. He can place, for example, two bottles of the brand $2$ on the first shelf and a bottle of the brand $1$ on the second shelf. Then the total cost of the bottles would be $6 + 7 + 15 = 28$.

In the second test case, he has only one shelf. It is not difficult to show that the optimal option is to place a bottle of the brand $1$ on it. Then the total cost will be $15$.

In the third test case, he has as many as $6$ shelves, so he can place all available bottles with a total cost of $7 + 5 = 12$.

Buggy Code:
import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;
import java.util.Arrays;
//problem number 2036B Startup

public class Startup
{
    public static void main(String []args)
    {
        Scanner sc = new Scanner(System.in);
        int cycle = sc.nextInt();
        Integer xCount[] = new Integer[cycle];

        for(int i=0; i<cycle; i++){
            //n is number of shelves, k is number of bottoles
            int n = sc.nextInt();
            int k = sc.nextInt();
            Integer[] brand = new Integer[k];
            Integer[] cost = new Integer[k];
            for(int j=0; j<k; j++){
                brand[j]= sc.nextInt();
                cost[j] = sc.nextInt();
            }

            Integer earnings = 0;

            //if number of bottles is less than number of shelves - total earning
            if(k<=n){
                for(int j=0; j<k; j++){
                    earnings = earnings + cost[j];
                }
            }

            //if number of bottles is more than number of shelves - total earning
            if(k>n){
                ArrayList<Integer> uniqueBrands = new ArrayList<Integer>();
                for(int j=0; j<k; j++){
                    if(search(brand[j], uniqueBrands)){
                        uniqueBrands.add(brand[j]);
                    }
                }

                //if unique brands are less than numebers of shelves
                int size = uniqueBrands.size();
                if(size <= n){
                    for(int j=0; j<k; j++){
                        earnings = earnings + cost[j];
                    }
                }

                //if unique brands are more than numebers of shelves
                if(size>n) {
                    Integer[] brandCost = new Integer[size];
                    for (int j = 0; j < size; j++) {
                        brandCost[j] = 0;  // Initialize each element to 0
                    }

                    for(int j=0; j<k; j++){
                        for(int l=0; l<size; l++){
                            if(brand[j] == uniqueBrands.get(l)){
                                brandCost[l] = brandCost[l] + cost[j];
                            }
                        }
                    }

                    Arrays.sort(brandCost, Collections.reverseOrder());
                    for(int j=0; j<n; j++){
                        earnings = earnings + brandCost[j];
                    }
                }
            }

            xCount[i] = earnings;
        }

        for(int i=0; i<cycle; i++){
            System.out.println(xCount[i]);
        }

        sc.close();
    }

    public static boolean search(Integer n, ArrayList<Integer> numbers){
        for(int i=0; i<numbers.size(); i++){
            if(numbers.get(i) == n){
                return false;
            }
        }
        return true;
    }
};

**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 4
1 5
2 10
2 15
3 20
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dating
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are the developer of a dating app which ignores gender completely. The app has $$$n$$$ users, indexed from $$$1$$$ to $$$n$$$. Each user's profile features a list of the activities they enjoy doing. There are $$$m$$$ possible activities, indexed from $$$1$$$ to $$$m$$$.

A match between two users is good if they share at least one activity and, at the same time, both of them like at least one activity that the other user does not like.

Find a good match if it exists.

Input: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 200\,000$$$, $$$1 \leq m \leq 10^6$$$) — the number of users and the number of activities.

Each of the following $$$n$$$ lines contains a number $$$k_i$$$ ($$$0 \leq k_i \leq m$$$) — the number of activities that user $$$i$$$ likes — followed by $$$k_i$$$ distinct integers from $$$1$$$ to $$$m$$$ — the activities user $$$i$$$ likes.

It is guaranteed that $$$k_1+k_2+\cdots+k_n$$$ does not exceed $$$10^6$$$.

Output: Print $$$\texttt{YES}$$$ if a good match exists. Otherwise, print $$$\texttt{NO}$$$.

If a good match exists, on the next line print two integers — the indexes of two users that make a match.

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

input:
3 3
1 1
1 2
3 2 3 1
output:
NO

Note: In the first sample, users $$$1$$$ and $$$3$$$ form a match, because they share activity $$$1$$$, and, furthermore, user $$$3$$$ likes activity $$$5$$$ (which user $$$1$$$ does not like) and user $$$1$$$ likes activity $$$4$$$ (which user $$$3$$$ does not like). Note that users $$$1$$$ and $$$2$$$, as well as users $$$2$$$ and $$$3$$$, do not form a match, as there is no activity that users $$$1$$$ or $$$3$$$ like, and user $$$2$$$ doesn't like.

Buggy Code:
import sys
arr=[[int(x) for x in line.split()] for line in sys.stdin.read().strip().split('\n')][1:]
r=0
for i in range(len(arr)-1):
    for j in range(1,len(arr)):
        l1=set(arr[i])-set(arr[j])
        l2=set(arr[j])-set(arr[i])
        k=set(arr[i])&set(arr[j])
        if len(k)>=2 and len(l1)>=1 and len(l2)>=1:
            print('YES')
            print(j+1,i+1)
    r+=1
if r==len(arr)-2:
    print('NO')

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's consider the following points:

1. **Understanding the Bug**: The buggy code checks for a match if there are at least two common activities and each user has at least one activity that the other user does not like. However, the problem statement only requires at least one common activity, not two.

2. **Edge Cases**:
   - **Single Common Activity**: Ensure there is exactly one common activity between two users.
   - **Unique Activities**: Ensure each user has activities that the other user does not like.

3. **Small Scale**: The input should be small enough to be easily manageable and fit within the constraints.

Given these points, let's create a test input where there is exactly one common activity between two users, and each user has unique activities:

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

### Explanation:
- **4 users** and **5 activities**.
- **User 1** likes activities 1 and 2.
- **User 2** likes activities 1 and 3.
- **User 3** likes activities 4 and 5.
- **User 4** likes activities 2 and 5.

### Analysis:
- **User 1** and **User 2** share activity 1.
- **User 1** likes activity 2, which **User 2** does not like.
- **User 2** likes activity 3, which **User 1** does not like.

This input should expose the bug in the buggy code, as it only requires one common activity, not two. The correct output for this input should be:

```plaintext
YES
1 2
```

However, the buggy code will fail to find this match because it incorrectly checks for at least two common activities.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dating
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are the developer of a dating app which ignores gender completely. The app has $$$n$$$ users, indexed from $$$1$$$ to $$$n$$$. Each user's profile features a list of the activities they enjoy doing. There are $$$m$$$ possible activities, indexed from $$$1$$$ to $$$m$$$.

A match between two users is good if they share at least one activity and, at the same time, both of them like at least one activity that the other user does not like.

Find a good match if it exists.

Input: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 200\,000$$$, $$$1 \leq m \leq 10^6$$$) — the number of users and the number of activities.

Each of the following $$$n$$$ lines contains a number $$$k_i$$$ ($$$0 \leq k_i \leq m$$$) — the number of activities that user $$$i$$$ likes — followed by $$$k_i$$$ distinct integers from $$$1$$$ to $$$m$$$ — the activities user $$$i$$$ likes.

It is guaranteed that $$$k_1+k_2+\cdots+k_n$$$ does not exceed $$$10^6$$$.

Output: Print $$$\texttt{YES}$$$ if a good match exists. Otherwise, print $$$\texttt{NO}$$$.

If a good match exists, on the next line print two integers — the indexes of two users that make a match.

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

input:
3 3
1 1
1 2
3 2 3 1
output:
NO

Note: In the first sample, users $$$1$$$ and $$$3$$$ form a match, because they share activity $$$1$$$, and, furthermore, user $$$3$$$ likes activity $$$5$$$ (which user $$$1$$$ does not like) and user $$$1$$$ likes activity $$$4$$$ (which user $$$3$$$ does not like). Note that users $$$1$$$ and $$$2$$$, as well as users $$$2$$$ and $$$3$$$, do not form a match, as there is no activity that users $$$1$$$ or $$$3$$$ like, and user $$$2$$$ doesn't like.

Buggy Code:
import sys
arr=[[int(x) for x in line.split()] for line in sys.stdin.read().strip().split('\n')][1:]
r=0
for i in range(len(arr)-1):
    for j in range(1,len(arr)):
        l1=set(arr[i])-set(arr[j])
        l2=set(arr[j])-set(arr[i])
        k=set(arr[i])&set(arr[j])
        if len(k)>=2 and len(l1)>=1 and len(l2)>=1:
            print('YES')
            print(max(j,i)+1,min(i,j)+1)
    r+=1
if r==len(arr)-2:
    print('NO')

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

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

Think step by step.

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

1. **Edge Cases**:
   - Users with no common activities.
   - Users with exactly one common activity.
   - Users with all activities in common.

2. **Faulty Logic**:
   - The buggy code checks if the intersection of activities has a length of at least 2, which is incorrect. It should check if the intersection has a length of at least 1.
   - The buggy code does not handle the case where the intersection is exactly 1 and both users have at least one unique activity.

3. **Small-Scale Test Input**:
   - We need a small number of users and activities to ensure the input is manageable and can be easily verified.

Let's construct a test input that exposes the bug:

- We will have 3 users.
- User 1 will have activities {1, 2}.
- User 2 will have activities {1, 3}.
- User 3 will have activities {1, 4}.

This setup ensures:
- Users 1 and 2 share activity 1, and each has a unique activity (2 for User 1, 3 for User 2).
- Users 1 and 3 share activity 1, and each has a unique activity (2 for User 1, 4 for User 3).
- Users 2 and 3 share activity 1, and each has a unique activity (3 for User 2, 4 for User 3).

This test input should expose the bug because the intersection of activities for any pair of users is exactly 1, and each pair has at least one unique activity.

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

This input should output `YES` followed by the indexes of any two users that form a match. The buggy code, however, will not output `YES` because it incorrectly checks for an intersection length of at least 2.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dating
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are the developer of a dating app which ignores gender completely. The app has $$$n$$$ users, indexed from $$$1$$$ to $$$n$$$. Each user's profile features a list of the activities they enjoy doing. There are $$$m$$$ possible activities, indexed from $$$1$$$ to $$$m$$$.

A match between two users is good if they share at least one activity and, at the same time, both of them like at least one activity that the other user does not like.

Find a good match if it exists.

Input: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 200\,000$$$, $$$1 \leq m \leq 10^6$$$) — the number of users and the number of activities.

Each of the following $$$n$$$ lines contains a number $$$k_i$$$ ($$$0 \leq k_i \leq m$$$) — the number of activities that user $$$i$$$ likes — followed by $$$k_i$$$ distinct integers from $$$1$$$ to $$$m$$$ — the activities user $$$i$$$ likes.

It is guaranteed that $$$k_1+k_2+\cdots+k_n$$$ does not exceed $$$10^6$$$.

Output: Print $$$\texttt{YES}$$$ if a good match exists. Otherwise, print $$$\texttt{NO}$$$.

If a good match exists, on the next line print two integers — the indexes of two users that make a match.

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

input:
3 3
1 1
1 2
3 2 3 1
output:
NO

Note: In the first sample, users $$$1$$$ and $$$3$$$ form a match, because they share activity $$$1$$$, and, furthermore, user $$$3$$$ likes activity $$$5$$$ (which user $$$1$$$ does not like) and user $$$1$$$ likes activity $$$4$$$ (which user $$$3$$$ does not like). Note that users $$$1$$$ and $$$2$$$, as well as users $$$2$$$ and $$$3$$$, do not form a match, as there is no activity that users $$$1$$$ or $$$3$$$ like, and user $$$2$$$ doesn't like.

Buggy Code:
n, m = map(int, input().split())
k = -1
a = []
for i in range(n):
    b = [int(x) for x in input().split()]
    a.append(b[1:])
dop = []
for i in range(len(a)):
    for j in range(i + 1, len(a)):
        dop.extend(a[i])
        dop.extend(a[j])
        c = set(dop)
        l1 = len(dop) - len(c)
        if len(c) > 0 and l1 != len(a[i]) and l1 != a[j]:
            print("YES")
            print(i + 1, j + 1)
            k = 0
            break
        dop = []
    if k == 0:
        break
if k == -1:
    print("NO")

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The key issues in the buggy code are:

1. The logic for checking if two users form a good match is flawed.
2. The code does not correctly handle cases where users have no common activities or where the sets of activities are subsets of each other.

Let's consider the following scenario:

- Two users with a small number of activities.
- One user has a subset of the activities of the other user.
- No other users are present to avoid unnecessary complexity.

This scenario should expose the bug because the code incorrectly checks for the conditions of a good match.

Here is a small-scale test input that fits this scenario:

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

Explanation:
- User 1 likes activities 1, 2, and 3.
- User 2 likes activities 1 and 2.

In this case:
- Users 1 and 2 share activities 1 and 2.
- User 1 likes activity 3, which User 2 does not like.
- User 2 does not like any activity that User 1 does not like.

This should be a good match, but the buggy code might fail to correctly identify it due to the flawed logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dating
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are the developer of a dating app which ignores gender completely. The app has $$$n$$$ users, indexed from $$$1$$$ to $$$n$$$. Each user's profile features a list of the activities they enjoy doing. There are $$$m$$$ possible activities, indexed from $$$1$$$ to $$$m$$$.

A match between two users is good if they share at least one activity and, at the same time, both of them like at least one activity that the other user does not like.

Find a good match if it exists.

Input: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 200\,000$$$, $$$1 \leq m \leq 10^6$$$) — the number of users and the number of activities.

Each of the following $$$n$$$ lines contains a number $$$k_i$$$ ($$$0 \leq k_i \leq m$$$) — the number of activities that user $$$i$$$ likes — followed by $$$k_i$$$ distinct integers from $$$1$$$ to $$$m$$$ — the activities user $$$i$$$ likes.

It is guaranteed that $$$k_1+k_2+\cdots+k_n$$$ does not exceed $$$10^6$$$.

Output: Print $$$\texttt{YES}$$$ if a good match exists. Otherwise, print $$$\texttt{NO}$$$.

If a good match exists, on the next line print two integers — the indexes of two users that make a match.

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

input:
3 3
1 1
1 2
3 2 3 1
output:
NO

Note: In the first sample, users $$$1$$$ and $$$3$$$ form a match, because they share activity $$$1$$$, and, furthermore, user $$$3$$$ likes activity $$$5$$$ (which user $$$1$$$ does not like) and user $$$1$$$ likes activity $$$4$$$ (which user $$$3$$$ does not like). Note that users $$$1$$$ and $$$2$$$, as well as users $$$2$$$ and $$$3$$$, do not form a match, as there is no activity that users $$$1$$$ or $$$3$$$ like, and user $$$2$$$ doesn't like.

Buggy Code:
import java.util.*;

public class DatingAppMatch {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt(); // Number of users
        int m = scanner.nextInt(); // Number of activities

        // Map to store activities and their corresponding users
        Map<Integer, List<Integer>> activityMap = new HashMap<>();

        // Populate the map
        for (int i = 1; i <= m; i++) {
            activityMap.put(i, new ArrayList<>());
        }

        for (int i = 0; i < n; i++) {
            int k = scanner.nextInt(); // Number of activities user i likes
            for (int j = 0; j < k; j++) {
                int activity = scanner.nextInt();
                activityMap.get(activity).add(i + 1); // Add user to the list of users for this activity
            }
        }

        // Check for a good match
        for (int activity : activityMap.keySet()) {
            List<Integer> likedBy = activityMap.get(activity);
            List<Integer> notLikedBy = new ArrayList<>();

            // Collect users who do not like this activity
            for (int i = 1; i <= n; i++) {
                if (!likedBy.contains(i)) {
                    notLikedBy.add(i);
                }
            }

            // Find a match if there are users who like and don't like this activity
            if (!likedBy.isEmpty() && !notLikedBy.isEmpty()) {
                System.out.println("YES");
                System.out.println(likedBy.get(0) + " " + notLikedBy.get(0));
                return;
            }
        }

        // If no match is found
        System.out.println("NO");
    }
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, we need to consider the logic and potential flaws in the algorithm. The primary issue in the code is that it only checks for a match based on a single activity and does not consider the requirement that both users must like at least one activity that the other does not like.

Here's a step-by-step approach to generate a test input:

1. **Identify the Bug**: The code checks if there is at least one user who likes an activity and at least one user who does not like it. However, it does not ensure that the two users have a shared activity and that each user has an activity the other does not like.

2. **Construct a Test Case**:
   - **Number of Users and Activities**: Choose a small number of users and activities to keep the input manageable.
   - **User Activities**:
     - User 1 should like a set of activities.
     - User 2 should like a subset of User 1's activities plus an additional activity.
     - User 3 should like a different set of activities that do not overlap with User 1 or User 2 but has one activity in common with User 1.

3. **Expected Output**: The expected output should be "NO" because, while User 1 and User 3 share an activity, User 3 does not have an activity that User 1 does not like, and vice versa.

Here is a small-scale test input that fits the criteria:

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

**Explanation**:
- User 1 likes activities 1, 2, and 3.
- User 2 likes activities 1 and 4.
- User 3 likes activities 3 and 5.

**Analysis**:
- User 1 and User 2 share activity 1, but User 2 does not have an activity that User 1 does not like.
- User 1 and User 3 share activity 3, but User 3 does not have an activity that User 1 does not like.
- User 2 and User 3 do not share any activities.

Thus, no good match exists, and the expected output should be "NO". This test input should expose the bug in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dating
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are the developer of a dating app which ignores gender completely. The app has $$$n$$$ users, indexed from $$$1$$$ to $$$n$$$. Each user's profile features a list of the activities they enjoy doing. There are $$$m$$$ possible activities, indexed from $$$1$$$ to $$$m$$$.

A match between two users is good if they share at least one activity and, at the same time, both of them like at least one activity that the other user does not like.

Find a good match if it exists.

Input: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 200\,000$$$, $$$1 \leq m \leq 10^6$$$) — the number of users and the number of activities.

Each of the following $$$n$$$ lines contains a number $$$k_i$$$ ($$$0 \leq k_i \leq m$$$) — the number of activities that user $$$i$$$ likes — followed by $$$k_i$$$ distinct integers from $$$1$$$ to $$$m$$$ — the activities user $$$i$$$ likes.

It is guaranteed that $$$k_1+k_2+\cdots+k_n$$$ does not exceed $$$10^6$$$.

Output: Print $$$\texttt{YES}$$$ if a good match exists. Otherwise, print $$$\texttt{NO}$$$.

If a good match exists, on the next line print two integers — the indexes of two users that make a match.

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

input:
3 3
1 1
1 2
3 2 3 1
output:
NO

Note: In the first sample, users $$$1$$$ and $$$3$$$ form a match, because they share activity $$$1$$$, and, furthermore, user $$$3$$$ likes activity $$$5$$$ (which user $$$1$$$ does not like) and user $$$1$$$ likes activity $$$4$$$ (which user $$$3$$$ does not like). Note that users $$$1$$$ and $$$2$$$, as well as users $$$2$$$ and $$$3$$$, do not form a match, as there is no activity that users $$$1$$$ or $$$3$$$ like, and user $$$2$$$ doesn't like.

Buggy Code:
#include <bits/stdc++.h>
#define int int64_t
using namespace std;

signed main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	//freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);

	int n, m; cin>>n>>m;

	vector<vector<int>>fred(n);
	vector<set<int>>s(n);

	for(int i = 0; i<n; ++i){
		int k; cin>>k;
		for(int j = 0; j<k; ++j){
			int a; cin>>a;
			fred[i].push_back(a);
			s[i].insert(a);
		}
		fred[i].push_back(i);
	}

	sort(fred.begin(), fred.end(), [&] (auto a, auto b){
		return (int)a.size() > (int)b.size();
	});

	int flag = 0;
	pair<int, int>ans;

	vector<int>node(m+1);
	vector<int>dist(n);

	for(int i = 0; i<n; ++i){
		if(flag) break;
		int cur_node = fred[i].back();
		int cur_par = cur_node;
		int maxi = 0;
		for(int j = 0; j<(int)fred[i].size()-1; ++j){
			if(dist[node[fred[i][j]]] > maxi){
				maxi = dist[node[fred[i][j]]];
				cur_par = node[fred[i][j]];
			}
		}
		dist[cur_node] = maxi+1;
		for(int j = 0; j<(int)fred[i].size()-1; ++j){
			node[fred[i][j]] = cur_node;
			if(!s[cur_par].count(fred[i][j])){
				flag = 1;
				ans = {cur_node+1, cur_par+1};
			}

		}
	}

	if(flag){
		cout<<"YES"<<'\n';
		cout<<ans.first<<" "<<ans.second;
	}
	else cout<<"NO";

	cout<<'\n';
}

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

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

Think step by step.

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

1. **Edge Cases**: Test with the minimum and maximum values for `n` and `m`.
2. **Special Cases**: Test with users having no activities, all users having the same activities, and users having unique activities.
3. **Complex Cases**: Test with a mix of shared and unique activities to ensure the matching logic is correct.

Given the constraints, a small-scale test input that can expose the bug is:

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

Explanation:
- **4 users** and **5 activities**.
- **User 1** likes activities 1, 2, and 3.
- **User 2** likes activities 1, 2, and 4.
- **User 3** likes activities 1, 2, and 5.
- **User 4** likes activities 1, 2, and 3.

This test case is designed to:
- Ensure that there are multiple users with overlapping activities.
- Check if the code correctly identifies that users 1 and 2, as well as users 1 and 3, form a good match.
- Verify that the code does not incorrectly identify user 1 and user 4 as a match, since they do not have unique activities that the other does not like.

This input should help expose any issues in the matching 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: Dating
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are the developer of a dating app which ignores gender completely. The app has $$$n$$$ users, indexed from $$$1$$$ to $$$n$$$. Each user's profile features a list of the activities they enjoy doing. There are $$$m$$$ possible activities, indexed from $$$1$$$ to $$$m$$$.

A match between two users is good if they share at least one activity and, at the same time, both of them like at least one activity that the other user does not like.

Find a good match if it exists.

Input: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 200\,000$$$, $$$1 \leq m \leq 10^6$$$) — the number of users and the number of activities.

Each of the following $$$n$$$ lines contains a number $$$k_i$$$ ($$$0 \leq k_i \leq m$$$) — the number of activities that user $$$i$$$ likes — followed by $$$k_i$$$ distinct integers from $$$1$$$ to $$$m$$$ — the activities user $$$i$$$ likes.

It is guaranteed that $$$k_1+k_2+\cdots+k_n$$$ does not exceed $$$10^6$$$.

Output: Print $$$\texttt{YES}$$$ if a good match exists. Otherwise, print $$$\texttt{NO}$$$.

If a good match exists, on the next line print two integers — the indexes of two users that make a match.

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

input:
3 3
1 1
1 2
3 2 3 1
output:
NO

Note: In the first sample, users $$$1$$$ and $$$3$$$ form a match, because they share activity $$$1$$$, and, furthermore, user $$$3$$$ likes activity $$$5$$$ (which user $$$1$$$ does not like) and user $$$1$$$ likes activity $$$4$$$ (which user $$$3$$$ does not like). Note that users $$$1$$$ and $$$2$$$, as well as users $$$2$$$ and $$$3$$$, do not form a match, as there is no activity that users $$$1$$$ or $$$3$$$ like, and user $$$2$$$ doesn't like.

Buggy Code:
#include <cstdio>
#include <cmath>
#include <algorithm>
#define lld long long
const int N =1e4+9;
using namespace std;
struct node{
    int ac[1005];
    int tag=1;
    int len;
    int tong[1006];
};
node man[N];
int n,m;
bool cmp(int a, int b) {
    return a < b;
}
int minn(int a,int b){
    return a<b?a:b;
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;++i){
        scanf("%d",&man[i].len);
        if(man[i].len==m) man[i].tag=0;
        for(int j=1;j<=man[i].len;++j){
            scanf("%d",&man[i].ac[j]);
            man[i].tong[man[i].ac[j]]=1;
        }
        sort(man[i].ac+1, man[i].ac + man[i].len+1, cmp);
   }
   int cnt=0;
   int tag1=0,tag2=0;
   int tag3=0;
   for(int i=1;i<=n;++i){
    if(man[i].tag==0) continue;
        for(int j=i+1;j<=n;++j){
            tag1=0,tag2=0;
            int pos1,pos2;
            int leng;
            if(man[j].tag==0) continue;

            if(man[j].len>man[i].len) {
                leng=man[j].len;
                pos1=i;
                pos2=j;
            }
            if(man[i].len>man[j].len) {
                leng=man[i].len;
                pos1=j;
                pos2=i;
            }
            else leng=man[j].len,pos1=i,pos2=j;
            for(int k=1;k<=leng;++k){
              //printf("%d %d\n",j,man[j].ac[k]);
                 //printf("%d\n",man[j].ac[k],man[i]);
                 if(man[pos1].tong[man[pos2].ac[k]]) tag1=1;
                 else tag2=1;
            }
            if((tag1 && tag2 ) && (tag3==0)) tag3=1;
            if(tag3==1) printf("YES\n");
            tag3=100;
            if(tag1 && tag2) printf("%d %d\n",i,j),cnt++;
        }
   }
   if(cnt==0) printf("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:
To create a challenging test input that exposes the bug in the provided buggy code, let's consider the following steps:

1. **Understand the Bug**: The main issue in the code is the way it checks for a good match. It seems to rely on the length of the activity lists and the presence of activities in one user's list that are not in the other's list. The code might fail when the lengths of the activity lists are the same or when the activities are distributed in a way that makes the matching logic incorrect.

2. **Edge Cases**:
   - **Equal Length Lists**: If two users have the same number of activities, the code might not correctly identify a good match.
   - **Shared and Unique Activities**: Ensure that there are shared activities and unique activities for each user.

3. **Small-Scale Test Input**:
   - We will create a test case with a small number of users and activities to ensure it is manageable and valid.
   - We will ensure that the test case includes users with equal lengths of activity lists and shared activities to challenge the code's logic.

Here is the test input:

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

### Explanation:
- **Users and Activities**:
  - User 1: Activities 1, 2, 3
  - User 2: Activities 2, 3, 4
  - User 3: Activities 1, 4, 5
  - User 4: Activities 3, 5, 6

- **Expected Output**:
  - The code should output `YES` and a pair of users that form a good match. For example, User 1 and User 3 form a good match:
    - Shared activity: 1
    - Unique to User 1: 2, 3
    - Unique to User 3: 4, 5

This test input should expose the bug in the code by challenging the logic for matching users with equal lengths of activity lists and ensuring that the shared and unique activities are correctly identified.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dating
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are the developer of a dating app which ignores gender completely. The app has $$$n$$$ users, indexed from $$$1$$$ to $$$n$$$. Each user's profile features a list of the activities they enjoy doing. There are $$$m$$$ possible activities, indexed from $$$1$$$ to $$$m$$$.

A match between two users is good if they share at least one activity and, at the same time, both of them like at least one activity that the other user does not like.

Find a good match if it exists.

Input: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 200\,000$$$, $$$1 \leq m \leq 10^6$$$) — the number of users and the number of activities.

Each of the following $$$n$$$ lines contains a number $$$k_i$$$ ($$$0 \leq k_i \leq m$$$) — the number of activities that user $$$i$$$ likes — followed by $$$k_i$$$ distinct integers from $$$1$$$ to $$$m$$$ — the activities user $$$i$$$ likes.

It is guaranteed that $$$k_1+k_2+\cdots+k_n$$$ does not exceed $$$10^6$$$.

Output: Print $$$\texttt{YES}$$$ if a good match exists. Otherwise, print $$$\texttt{NO}$$$.

If a good match exists, on the next line print two integers — the indexes of two users that make a match.

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

input:
3 3
1 1
1 2
3 2 3 1
output:
NO

Note: In the first sample, users $$$1$$$ and $$$3$$$ form a match, because they share activity $$$1$$$, and, furthermore, user $$$3$$$ likes activity $$$5$$$ (which user $$$1$$$ does not like) and user $$$1$$$ likes activity $$$4$$$ (which user $$$3$$$ does not like). Note that users $$$1$$$ and $$$2$$$, as well as users $$$2$$$ and $$$3$$$, do not form a match, as there is no activity that users $$$1$$$ or $$$3$$$ like, and user $$$2$$$ doesn't like.

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

using namespace std;

using ll = long long;

const int INF = 1e9;
const ll MOD = 1e9 + 7;

void solve() {
    int n, m;
    cin >> n >> m;

    vector<vector<int>> a(n);
    vector<int> k(n + 1);
    for (int i = 0; i < n; ++i) {
        cin >> k[i];
        a[i].push_back(i);
        for (int j = 1; j <= k[i]; ++j) {
            int x;
            cin >> x;
            --x;
            a[i].push_back(x);
        }
    }

    sort(a.begin(), a.end(),
         [](const auto &a, const auto &b) {
            return a.size() < b.size();
         });

    vector<int> color(m, 0);
    for (int i = 0; i < n; ++i) {
        map<int, int> mp;
        for (int j = 1; j < a[i].size(); ++j) {
            int x = a[i][j];
            if (color[x] != -1 && ++mp[color[x]] == k[color[x]]) {
                mp.erase(color[x]);
            }

            color[x] = a[i][0];
        }

        if (!mp.empty()) {
            cout << "YES" << endl;
            cout << a[i][0] + 1 << " " << mp.begin()->first + 1 << endl;
            return;
        }
    }

    cout << "NO" << endl;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t = 1;
//    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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are the steps to think through this:

1. **Identify the Bug**: The bug in the code seems to be in the logic used to determine a good match. Specifically, the code is not correctly checking if two users share at least one activity and both like at least one activity that the other does not like.

2. **Edge Cases**:
   - **Users with no activities**: This should result in no matches.
   - **Users with the same activities**: This should result in no matches.
   - **Users with partially overlapping activities**: This should result in a match if both users have unique activities.
   - **Users with a single activity**: This can be tricky if the single activity is shared.

3. **Constructing the Test Case**:
   - We will create a small-scale test case with a few users to ensure it is easy to verify.
   - We will ensure that the test case includes users with partially overlapping activities and unique activities to test the match condition.

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

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

### Explanation:
- **User 1** likes activities 1 and 2.
- **User 2** likes activities 1 and 3.
- **User 3** likes activities 2 and 4.
- **User 4** likes activities 3 and 4.

### Expected Output:
- **User 1 and User 2** should form a good match because they share activity 1, and User 1 likes activity 2 (which User 2 does not like), and User 2 likes activity 3 (which User 1 does not like).
- **User 3 and User 4** should also form a good match because they share activity 4, and User 3 likes activity 2 (which User 4 does not like), and User 4 likes activity 3 (which User 3 does not like).

The buggy code might fail to correctly identify these matches due to the flawed logic in the `color` and `mp` usage. This test case should help expose the bug.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dating
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are the developer of a dating app which ignores gender completely. The app has $$$n$$$ users, indexed from $$$1$$$ to $$$n$$$. Each user's profile features a list of the activities they enjoy doing. There are $$$m$$$ possible activities, indexed from $$$1$$$ to $$$m$$$.

A match between two users is good if they share at least one activity and, at the same time, both of them like at least one activity that the other user does not like.

Find a good match if it exists.

Input: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 200\,000$$$, $$$1 \leq m \leq 10^6$$$) — the number of users and the number of activities.

Each of the following $$$n$$$ lines contains a number $$$k_i$$$ ($$$0 \leq k_i \leq m$$$) — the number of activities that user $$$i$$$ likes — followed by $$$k_i$$$ distinct integers from $$$1$$$ to $$$m$$$ — the activities user $$$i$$$ likes.

It is guaranteed that $$$k_1+k_2+\cdots+k_n$$$ does not exceed $$$10^6$$$.

Output: Print $$$\texttt{YES}$$$ if a good match exists. Otherwise, print $$$\texttt{NO}$$$.

If a good match exists, on the next line print two integers — the indexes of two users that make a match.

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

input:
3 3
1 1
1 2
3 2 3 1
output:
NO

Note: In the first sample, users $$$1$$$ and $$$3$$$ form a match, because they share activity $$$1$$$, and, furthermore, user $$$3$$$ likes activity $$$5$$$ (which user $$$1$$$ does not like) and user $$$1$$$ likes activity $$$4$$$ (which user $$$3$$$ does not like). Note that users $$$1$$$ and $$$2$$$, as well as users $$$2$$$ and $$$3$$$, do not form a match, as there is no activity that users $$$1$$$ or $$$3$$$ like, and user $$$2$$$ doesn't like.

Buggy Code:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.PrintStream;
import java.util.stream.IntStream;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Random;
import java.util.ArrayList;
import java.util.Map;
import java.util.Comparator;
import java.util.Collections;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastReader in = new FastReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        FDating solver = new FDating();
        solver.solve(1, in, out);
        out.close();
    }

    static class FDating {
        final static Helper hp = new Helper();
        final static Logger lg = new Logger();
        int N;
        int M;
        ArrayList<Integer>[] likes;
        long[] localHash;

        private ArrayList<Integer> toSortedArrayList(int[] ar) {
            ArrayList<Integer> ret = new ArrayList<>();
            Arrays.stream(ar).sorted().forEach(ret::add);
            return ret;
        }

        public void fillInputParams(FastReader in) {
            N = in.nextInt();
            M = in.nextInt();
            likes = new ArrayList[N];
            IntStream.range(0, N).forEach(idx -> likes[idx] = toSortedArrayList(in.getIntArray(in.nextInt())));
        }

        Object solveOptimised(FastReader in, StringBuilder sb) {
            HashMap<Integer, ArrayList<Integer>> activityMap = new HashMap<>();
            for (int person = 0; person < N; ++person) {
                ArrayList<Integer> hobbies = likes[person];
                if (hobbies.size() == 1) continue;   // get a life.

                for (int hobby : hobbies) {
                    activityMap.putIfAbsent(hobby, new ArrayList<>());
                    activityMap.get(hobby).add(person);
                }
            }

            localHash = new long[N];
            for (ArrayList<Integer> peopleWithSameHobby : activityMap.values()) {
                if (peopleWithSameHobby.size() <= 1) continue;
                Collections.sort(peopleWithSameHobby, Comparator.comparing((Integer person) -> likes[person].size()).reversed());

                int lastPerson = peopleWithSameHobby.get(peopleWithSameHobby.size() - 1);
                ArrayList<Integer> lastPersonHobbies = likes[lastPerson];
                for (int i = peopleWithSameHobby.size() - 2; i >= 0; --i) {
                    int currentPerson = peopleWithSameHobby.get(i);
                    ArrayList<Integer> currentPersonHobbies = likes[currentPerson];

                    if (lastPersonHobbies.size() == currentPersonHobbies.size()) {
                        if (hash(lastPerson) != hash(currentPerson)) {
                            return yes(lastPerson, currentPerson);
                        }
                    } else if (currentPersonHobbies.size() < 50) {
                        for (int currHobby : currentPersonHobbies) {
                            if (Collections.binarySearch(lastPersonHobbies, currHobby) < 0) {
                                return yes(lastPerson, currentPerson);
                            }
                        }
                    } else {
                        return 7 / 0;
                    }

                    lastPerson = currentPerson;
                    lastPersonHobbies = currentPersonHobbies;
                }
            }
            return "NO";
        }

        private long hash(int person) {
            if (localHash[person] == 0) {
                ArrayList<Integer> toHash = likes[person];
                for (int itr : toHash) {
                    localHash[person] += hp.pow(itr, 128, hp.MOD);
                }
            }
            return localHash[person];
        }

        private String yes(int a, int b) {
            return "YES\n" + (1 + a) + " " + (1 + b);
        }

        Object solveBrute(FastReader in, StringBuilder sb) {
            if (!Logger.isDevEnv) return null;
            return null;
        }

        public void solve(int testNumber, FastReader in, PrintWriter out) {
            // out.print("Case #" + testNumber + ": ");

            // new Thread(null, new Runnable() {
            //     @Override
            //     public void run() { /* do stuff */ }
            // }, "Solver", 1l << 30).start();

            fillInputParams(in);
            Object outOptimised = solveOptimised(in, new StringBuilder());
            Object outBrute = solveBrute(in, new StringBuilder());
            if (outBrute == null) {
                if (lg.noOutputPhase) return;
                out.println(outOptimised);
            } else if (outBrute.toString().equals(outOptimised.toString())) {
                System.err.println(testNumber + ". OK Checked");
            } else {
                // print input params
                System.err.println("Actual = " + outOptimised);
                System.err.println("Expected = " + outBrute);
                System.err.println();
                throw new ArithmeticException();
            }
        }

    }

    static class FastReader {
        static final int BUFSIZE = 1 << 20;
        static byte[] buf;
        static int index;
        static int total;
        static InputStream in;

        public FastReader(InputStream is) {
            try {
                in = is;
                buf = new byte[BUFSIZE];
            } catch (Exception e) {
            }
        }

        private int scan() {
            try {
                if (index >= total) {
                    index = 0;
                    total = in.read(buf);
                    if (total <= 0)
                        return -1;
                }
                return buf[index++];
            } catch (Exception | Error e) {
                System.err.println(e.getMessage());
                return 13 / 0;
            }
        }

        public int nextInt() {
            int c, val = 0;
            for (c = scan(); c <= 32; c = scan()) ;
            boolean neg = c == '-';
            if (c == '-' || c == '+')
                c = scan();
            for (; c >= '0' && c <= '9'; c = scan())
                val = (val << 3) + (val << 1) + (c & 15);
            return neg ? -val : val;
        }

        public int[] getIntArray(int size) {
            int[] ar = new int[size];
            for (int i = 0; i < size; ++i) ar[i] = nextInt();
            return ar;
        }

    }

    static class Logger {
        public static final boolean isDevEnv = resolveIfDevEnv();
        public boolean noOutputPhase;
        public PrintStream stream;

        public Logger(PrintStream ps) {
            stream = ps;
        }

        public Logger() {
            this(System.err);
            stream = System.err;
        }

        public Logger(boolean noOutput) {
            this();
            noOutputPhase = isDevEnv && noOutput;
        }

        private static boolean resolveIfDevEnv() {
            try {
                return System.getenv().get("USERDOMAIN") != null
                        && System.getenv().get("USERDOMAIN").equals("LAPTOP-DSSUKMC1");
            } catch (Exception | Error e) {
                return false;
            }
        }

    }

    static class Helper {
        public final long MOD;
        public final int MAXN;
        final Random rnd;

        public Helper() {
            MOD = 1000_000_007;
            MAXN = 1000_006;
            rnd = new Random();
        }

        public Helper(long mod, int maxn) {
            MOD = mod;
            MAXN = maxn;
            rnd = new Random();
        }

        public long pow(long base, long exp, long MOD) {
            base %= MOD;
            long ret = 1;
            while (exp > 0) {
                if ((exp & 1) == 1) ret = ret * base % MOD;
                base = base * base % MOD;
                exp >>= 1;
            }
            return ret;
        }

        public int getRandomInRange(int l, int r) {
            return rnd.nextInt(r - l + 1) + l;
        }

    }
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided code, we need to focus on the logic that checks for a good match between users. The key points to consider are:

1. **Shared Activities**: Users must share at least one activity.
2. **Unique Activities**: Each user must have at least one activity that the other user does not like.

The code has a specific check that skips users with only one hobby, which can be a potential bug. Additionally, the code uses a hash function to compare users' hobbies, which might not be reliable for large inputs.

### Step-by-Step Approach:

1. **Users with Only One Hobby**: The code skips users with only one hobby. We can create a test case where all users have only one hobby, and none of them should form a good match.
2. **Users with Shared and Unique Hobbies**: Create a test case where users have multiple hobbies, and some of them share activities but also have unique activities.

### Test Case 1: All Users with Only One Hobby
This test case should result in "NO" because no user has more than one hobby.

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

### Test Case 2: Users with Shared and Unique Hobbies
This test case should result in "YES" because there are users who share activities but also have unique activities.

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

### Combined Test Case
To maximize the failure probability, we can combine both scenarios in a single test case. This will ensure that the code handles both edge cases correctly.

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

### Explanation:
- **User 1**: Has only one hobby (1).
- **User 2**: Has hobbies (1, 2, 4).
- **User 3**: Has hobbies (1, 2, 5).
- **User 4**: Has hobbies (1, 3).
- **User 5**: Has hobbies (1, 4).

### Expected Output:
```plaintext
YES
2 4
```

### Output:
```plaintext
5 5
1 1
3 1 2 4
3 1 2 5
2 1 3
2 1 4
```

This test case should expose the bug in the code by ensuring that the logic correctly handles users with only one hobby and users with multiple hobbies that share and have unique activities.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dating
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are the developer of a dating app which ignores gender completely. The app has $$$n$$$ users, indexed from $$$1$$$ to $$$n$$$. Each user's profile features a list of the activities they enjoy doing. There are $$$m$$$ possible activities, indexed from $$$1$$$ to $$$m$$$.

A match between two users is good if they share at least one activity and, at the same time, both of them like at least one activity that the other user does not like.

Find a good match if it exists.

Input: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 200\,000$$$, $$$1 \leq m \leq 10^6$$$) — the number of users and the number of activities.

Each of the following $$$n$$$ lines contains a number $$$k_i$$$ ($$$0 \leq k_i \leq m$$$) — the number of activities that user $$$i$$$ likes — followed by $$$k_i$$$ distinct integers from $$$1$$$ to $$$m$$$ — the activities user $$$i$$$ likes.

It is guaranteed that $$$k_1+k_2+\cdots+k_n$$$ does not exceed $$$10^6$$$.

Output: Print $$$\texttt{YES}$$$ if a good match exists. Otherwise, print $$$\texttt{NO}$$$.

If a good match exists, on the next line print two integers — the indexes of two users that make a match.

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

input:
3 3
1 1
1 2
3 2 3 1
output:
NO

Note: In the first sample, users $$$1$$$ and $$$3$$$ form a match, because they share activity $$$1$$$, and, furthermore, user $$$3$$$ likes activity $$$5$$$ (which user $$$1$$$ does not like) and user $$$1$$$ likes activity $$$4$$$ (which user $$$3$$$ does not like). Note that users $$$1$$$ and $$$2$$$, as well as users $$$2$$$ and $$$3$$$, do not form a match, as there is no activity that users $$$1$$$ or $$$3$$$ like, and user $$$2$$$ doesn't like.

Buggy Code:
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

typedef tree<pair<int,int>, null_type, less<pair<int,int>>, rb_tree_tag, tree_order_statistics_node_update> ordered_set; // find_by_order, order_of_key
// for changing to descending, change less<pair<int,int>> to greater<pair<int,int>>

#define int long long
#define pb push_back
#define endl "\n"
#define vi vector<int>
#define pii pair<int,int>
#define all(p) p.begin(), p.end()
#define yes() cout<<"YES"<<endl
#define no() cout<<"NO"<<endl
#define vvi vector<vector<int>>
#define INF LLONG_MAX
#define print(a) for(auto x : a) cout << x << " "; cout << endl
#define fastio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)

const int mod = 1e9 + 7;

signed main(){
    fastio();

    int seed = chrono::system_clock::now().time_since_epoch().count();

    mt19937_64 gen(seed);

    uniform_int_distribution<int> dist(numeric_limits<int>::min(), numeric_limits<int>::max());
    const int N = 1e6 + 1;
    vector<int> ran(N, 0);
    for(int i=0; i<N; i++)
        ran[i] = dist(gen);

    int n, m;
    cin >> n >> m;
    vector<int> a[n];
    vector<int> b[m+1];

    map<int,int> hash;
    vector<int> hash2(n, 0);
    const int M = 1e6 + 1;
    vector<bitset<M>> v(n);

    for(int i=0; i<n; i++){
        int x;
        cin >> x;
        int h = 0;
        vi temp;
        bitset<M> bit(0);
        while(x--){
            int p;
            cin >> p;
            temp.pb(p);
            (h += ran[p]) %= mod;
            bit[p] = 1;
        }
        if (hash.find(h) == hash.end()){
            hash2[i] = h;
            hash[h] = i;
            for(auto p : temp){
                a[i].pb(p);
                b[p].pb(i);
            }
            v[i] = bit;
        }
    }

    if (hash.size() == 1) no();
    else{
        int flag = 0;
        int one = -1, two = -1;
        for(int i=1; i<=m; i++){
            int h = -1;
            for(auto x : b[i]){
                if (h != -1){
                    if ((hash2[h]) == hash2[x]) continue;
                    int ind1 = h;
                    int ind2 = x;
                    if ((v[ind1]|v[ind2]) == v[ind1]) {
                        h = x;
                        continue;
                    }
                    if ((v[ind1]|v[ind2]) == v[ind2]){
                        continue;
                    }
                    one = h + 1;
                    two = x + 1;
                    flag = 1;
                    break;
                }

                else{
                    h = x;
                }
            }
            if (flag) break;
        }

        if (flag){
            yes();
            cout << one << " " << two << "\n";
        }

        else 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
2 1
1 1
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dating
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are the developer of a dating app which ignores gender completely. The app has $$$n$$$ users, indexed from $$$1$$$ to $$$n$$$. Each user's profile features a list of the activities they enjoy doing. There are $$$m$$$ possible activities, indexed from $$$1$$$ to $$$m$$$.

A match between two users is good if they share at least one activity and, at the same time, both of them like at least one activity that the other user does not like.

Find a good match if it exists.

Input: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 200\,000$$$, $$$1 \leq m \leq 10^6$$$) — the number of users and the number of activities.

Each of the following $$$n$$$ lines contains a number $$$k_i$$$ ($$$0 \leq k_i \leq m$$$) — the number of activities that user $$$i$$$ likes — followed by $$$k_i$$$ distinct integers from $$$1$$$ to $$$m$$$ — the activities user $$$i$$$ likes.

It is guaranteed that $$$k_1+k_2+\cdots+k_n$$$ does not exceed $$$10^6$$$.

Output: Print $$$\texttt{YES}$$$ if a good match exists. Otherwise, print $$$\texttt{NO}$$$.

If a good match exists, on the next line print two integers — the indexes of two users that make a match.

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

input:
3 3
1 1
1 2
3 2 3 1
output:
NO

Note: In the first sample, users $$$1$$$ and $$$3$$$ form a match, because they share activity $$$1$$$, and, furthermore, user $$$3$$$ likes activity $$$5$$$ (which user $$$1$$$ does not like) and user $$$1$$$ likes activity $$$4$$$ (which user $$$3$$$ does not like). Note that users $$$1$$$ and $$$2$$$, as well as users $$$2$$$ and $$$3$$$, do not form a match, as there is no activity that users $$$1$$$ or $$$3$$$ like, and user $$$2$$$ doesn't like.

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

#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define per(i, a, b) for(int i = a; i > (b); --i)
#define ar array
#define sz(x) (int) (x).size()
#define pii pair<int,int>
#define fi first
#define se second
typedef long long ll;
typedef pair<ll,ll> pll;
typedef pair<double,double> pdd;
typedef pair<double,int> pdi;
typedef vector<int> vi;

template<typename T>
void min_self(T& A, T B) {
    A = min(A,B);
}
template<typename T>
void max_self(T& A, T B) {
    A = max(A,B);
}

const int mxn=2e5,mxct=5e6;
int n,m;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

void solve() {
    cin >>n >>m;
    vector st(n, set<int>());
    map<int,vector<int>> mp;
    vector<int> vk;
    rep(i,0,n) {
        int pi; cin >>pi;
        rep(j,0,pi) {
            int ki; cin >>ki;
            st[i].insert(ki);
            mp[ki].push_back(i);
        }
    }
    for(auto [ki, vid]: mp) {
        vk.push_back(ki);
    }
    ll ct = 0, ctl=0;
    set<pii> udh;
    while(true) {
        int both = vk[rng()%sz(vk)];
        vector<int>& temp = mp[both];
        ctl++;
        if(sz(temp)>1) {
            int i = temp[rng()%sz(temp)];
            int j = i;
            while(j==i) {
                j = temp[rng()%sz(temp)];
            }
            if(i>j) swap(i,j);
            if(!udh.count({i,j})) {
                udh.insert({i,j});
                ct += min(sz(st[i]),sz(st[j]));
                if(sz(st[i])>sz(st[j])) {
                    swap(i,j);
                }
                int stj=0;
                bool validi = 0;
                for(auto ki: st[i]) {
                    if(st[j].find(ki)==st[j].end()) {
                        validi = 1;
                    } else {
                        stj++;
                    }
                }
                if(validi && sz(st[j])-stj>0) {
                    cout <<"YES\n";
                    cout <<i+1 <<" " <<j+1 <<"\n";
                    return;
                }
                if(ct>mxct) {
                    break;
                }
            }
        }
        // if(ctl==1000) {
        //     cout <<1000 <<endl;
        // } else if(ctl==10000) {
        //     cout <<10000 <<endl;
        // } else if(ctl==100000) {
        //     cout <<100000 <<endl;
        // } else if(ctl==1000000) {
        //     cout <<1000000 <<endl;
        // } else if(ctl==10000000) {
        //     cout <<10000000 <<endl;
        // }
        if(ctl>mxct) break;
    }

    cout <<"NO\n";
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    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
4 6
2 1 2
3 1 2 3
2 4 5
3 4 5 6
```

### Explanation:
- **Users and Activities:**
  - User 1: Activities 1, 2
  - User 2: Activities 1, 2, 3
  - User 3: Activities 4, 5
  - User 4: Activities 4, 5, 6

- **Good Matches:**
  - Users 1 and 2 share activities 1 and 2. However, both like activity 3, which the other also likes, so they do not form a good match.
  - Users 3 and 4 share activities 4 and 5. User 3 likes activity 5, which User 4 does not like, and User 4 likes activity 6, which User 3 does not like. Therefore, they form a good match.

- **Expected Output:**
  - The expected output for this test input is:
    ```plaintext
    YES
    3 4
    ```

This test case is designed to expose the bug in the algorithm by ensuring that the random selection and set operations correctly identify a good match. The small scale of the input ensures that the output is within the length limit while still being a valid and challenging test case.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dating
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are the developer of a dating app which ignores gender completely. The app has $$$n$$$ users, indexed from $$$1$$$ to $$$n$$$. Each user's profile features a list of the activities they enjoy doing. There are $$$m$$$ possible activities, indexed from $$$1$$$ to $$$m$$$.

A match between two users is good if they share at least one activity and, at the same time, both of them like at least one activity that the other user does not like.

Find a good match if it exists.

Input: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 200\,000$$$, $$$1 \leq m \leq 10^6$$$) — the number of users and the number of activities.

Each of the following $$$n$$$ lines contains a number $$$k_i$$$ ($$$0 \leq k_i \leq m$$$) — the number of activities that user $$$i$$$ likes — followed by $$$k_i$$$ distinct integers from $$$1$$$ to $$$m$$$ — the activities user $$$i$$$ likes.

It is guaranteed that $$$k_1+k_2+\cdots+k_n$$$ does not exceed $$$10^6$$$.

Output: Print $$$\texttt{YES}$$$ if a good match exists. Otherwise, print $$$\texttt{NO}$$$.

If a good match exists, on the next line print two integers — the indexes of two users that make a match.

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

input:
3 3
1 1
1 2
3 2 3 1
output:
NO

Note: In the first sample, users $$$1$$$ and $$$3$$$ form a match, because they share activity $$$1$$$, and, furthermore, user $$$3$$$ likes activity $$$5$$$ (which user $$$1$$$ does not like) and user $$$1$$$ likes activity $$$4$$$ (which user $$$3$$$ does not like). Note that users $$$1$$$ and $$$2$$$, as well as users $$$2$$$ and $$$3$$$, do not form a match, as there is no activity that users $$$1$$$ or $$$3$$$ like, and user $$$2$$$ doesn't like.

Buggy Code:
def find_good_match(n, m, user_activities):
    # Iterate through all pairs of users
    for i in range(n):
        for j in range(i + 1, n):
            # Check if they share at least one activity
            shared_activities = user_activities[i] & user_activities[j]
            if shared_activities:
                # Check if each user likes at least one activity that the other does not like
                unique_activities_i = user_activities[i] - user_activities[j]
                unique_activities_j = user_activities[j] - user_activities[i]
                if unique_activities_i and unique_activities_j:
                    return True, (i + 1, j + 1)
    return False, None

# Input reading
n, m = map(int, input().split())
user_activities = [set() for _ in range(n)]
for i in range(n):
    line = input().split()
    k_i = int(line[0])
    for j in range(1, k_i + 1):
        user_activities[i].add(int(line[j]))

# Find a good match
exists, match = find_good_match(n, m, user_activities)
if exists:
    print("YES")
    print(*match)
else:
    print("NO")

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to consider edge cases where the algorithm might fail. The primary logic of the code checks for shared activities and unique activities between pairs of users. A good test case should challenge this logic.

### Step-by-Step Reasoning:

1. **Shared Activities but No Unique Activities:**
   - If two users share all their activities, they should not form a good match because neither user has a unique activity that the other does not like.

2. **No Shared Activities:**
   - If two users have no shared activities, they should not form a good match.

3. **One User with No Activities:**
   - If a user has no activities, they should not form a good match with any other user.

4. **All Users with the Same Set of Activities:**
   - If all users have the same set of activities, no good match should exist.

5. **Users with Overlapping but Not Identical Sets of Activities:**
   - If users have overlapping but not identical sets of activities, they should form a good match.

### Test Case:
Let's create a small-scale test case that fits the above criteria and is likely to expose the bug.

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

### Explanation:
- **User 1:** Likes activities 1, 2, 3.
- **User 2:** Likes activities 1, 2, 3.
- **User 3:** Likes activity 1.

### Expected Output:
- **NO**
  - User 1 and User 2 share all their activities, so they do not form a good match.
  - User 1 and User 3 share activity 1, but User 1 does not have a unique activity that User 3 does not like.
  - User 2 and User 3 share activity 1, but User 2 does not have a unique activity that User 3 does not like.

This test case should expose the bug in the code where it might incorrectly identify a good match when there are no unique activities.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dating
time_limit_ms: 3000
memory_limit_mb: 512
Description: You are the developer of a dating app which ignores gender completely. The app has $$$n$$$ users, indexed from $$$1$$$ to $$$n$$$. Each user's profile features a list of the activities they enjoy doing. There are $$$m$$$ possible activities, indexed from $$$1$$$ to $$$m$$$.

A match between two users is good if they share at least one activity and, at the same time, both of them like at least one activity that the other user does not like.

Find a good match if it exists.

Input: The first line contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq 200\,000$$$, $$$1 \leq m \leq 10^6$$$) — the number of users and the number of activities.

Each of the following $$$n$$$ lines contains a number $$$k_i$$$ ($$$0 \leq k_i \leq m$$$) — the number of activities that user $$$i$$$ likes — followed by $$$k_i$$$ distinct integers from $$$1$$$ to $$$m$$$ — the activities user $$$i$$$ likes.

It is guaranteed that $$$k_1+k_2+\cdots+k_n$$$ does not exceed $$$10^6$$$.

Output: Print $$$\texttt{YES}$$$ if a good match exists. Otherwise, print $$$\texttt{NO}$$$.

If a good match exists, on the next line print two integers — the indexes of two users that make a match.

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

input:
3 3
1 1
1 2
3 2 3 1
output:
NO

Note: In the first sample, users $$$1$$$ and $$$3$$$ form a match, because they share activity $$$1$$$, and, furthermore, user $$$3$$$ likes activity $$$5$$$ (which user $$$1$$$ does not like) and user $$$1$$$ likes activity $$$4$$$ (which user $$$3$$$ does not like). Note that users $$$1$$$ and $$$2$$$, as well as users $$$2$$$ and $$$3$$$, do not form a match, as there is no activity that users $$$1$$$ or $$$3$$$ like, and user $$$2$$$ doesn't like.

Buggy Code:
def find_good_match(n, m, user_activities):
    # Convert each user's activities to a set for efficient operations
    activities = [set(acts) for acts in user_activities]

    # Check each pair of users
    for i in range(n):
        for j in range(i + 1, n):
            # Check if they share at least one activity
            if activities[i] & activities[j]:
                # Check if there is at least one activity that one user likes and the other does not
                if (activities[i] - activities[j]) and (activities[j] - activities[i]):
                    print("YES")
                    print(i + 1, j + 1)
                    return

    # If no good match is found
    print("NO")

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

n = int(data[0])
m = int(data[1])

index = 2
user_activities = []
for _ in range(n):
    k_i = int(data[index])
    activities = list(map(int, data[index + 1:index + 1 + k_i]))
    user_activities.append(activities)
    index += 1 + k_i

# Find and print a good match
find_good_match(n, m, user_activities)

**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 edge cases and scenarios that maximize the failure probability. Here are the steps to think through this:

1. **Identify Potential Bugs**:
   - The code checks for shared activities and unique activities between pairs of users.
   - A common bug could be in handling cases where users have very similar or identical activity sets, which might not be correctly identified as a good match.

2. **Create a Test Case**:
   - We need a small-scale test input with a specific structure to expose the bug.
   - Consider a scenario where users have activities that are very similar but not identical, and ensure that the conditions for a good match are met.

3. **Example Test Case**:
   - Let's create a test case with 3 users and 4 activities.
   - User 1: Activities 1, 2
   - User 2: Activities 2, 3
   - User 3: Activities 3, 4

   This test case should ensure that:
   - User 1 and User 2 share activity 2.
   - User 1 likes activity 1, which User 2 does not like.
   - User 2 likes activity 3, which User 1 does not like.
   - User 2 and User 3 share activity 3.
   - User 2 likes activity 2, which User 3 does not like.
   - User 3 likes activity 4, which User 2 does not like.

4. **Output the Test Case**:
   - The test input should be in the format specified in the problem description.

Here is the test input:

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

This test input should expose the bug if the code does not correctly handle the conditions for a good match.