**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.Arrays;
import java.util.Iterator;
import java.util.Map;
import java.util.BitSet;
import java.util.Comparator;
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 Logger lg = new Logger();
        int N;
        int M;
        BitSet[] A;

        public void fillInputParams(FastReader in) {
            N = in.nextInt();
            M = in.nextInt();
            A = new BitSet[N];
            for (int i = 0; i < N; ++i) {
                A[i] = new BitSet();
                for (int j = in.nextInt(); j > 0; --j) {
                    A[i].set(in.nextInt() - 1);
                }
            }
        }

        Object solveOptimised(FastReader in, StringBuilder sb) {
            MyArrayListInt[] commonInterests = new MyArrayListInt[M];
            for (int i = 0; i < commonInterests.length; ++i) commonInterests[i] = new MyArrayListInt(1);
            for (int p = 0; p < N; ++p) {
                for (int l = A[p].nextSetBit(0); l >= 0; l = A[p].nextSetBit(l + 1)) {
                    commonInterests[l].add(p);
                }
            }

            for (MyArrayListInt likes : commonInterests) {
                likes.sort(0, likes.length(), Comparator.comparingInt(p -> A[p].cardinality()));

                for (int idx = 1; idx < likes.length(); ++idx) {
                    int pLeft = likes.get(idx - 1), pCurr = likes.get(idx);
                    BitSet left = A[pLeft], curr = A[pCurr];

                    int oldCardi = left.cardinality();
                    left.and(curr);
                    if (left.cardinality() == oldCardi) continue;
                    return yes(pLeft, pCurr);
                }
            }

            return "NO";
        }

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

        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 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 MyArrayListInt implements Iterable<Integer> {
        private int len;
        private int[] A;
        int itrInd;

        public MyArrayListInt(int initialLength) {
            A = new int[Math.max(1, initialLength)];
            len = 0;
        }

        public MyArrayListInt(MyArrayListInt src) {
            A = new int[src.A.length];
            System.arraycopy(src.A, 0, A, 0, src.len);
            len = src.len;
        }

        public void add(int ele) {
            if (len == A.length) {
                int[] newAr = new int[A.length << 1];
                System.arraycopy(A, 0, newAr, 0, len);
                A = newAr;
            }
            A[len++] = ele;
        }

        public int get(int ind) {
            if (ind >= len) throw new ArrayIndexOutOfBoundsException(ind);
            else if (ind < 0) ind += size();
            return A[ind];
        }

        public int size() {
            return len;
        }

        public int length() {
            return len;
        }

        public void sort(int fromIndex, int toIndex, Comparator<Integer> com) {
            Integer[] temp = new Integer[toIndex - fromIndex];
            for (int i = fromIndex, k = 0; i < toIndex; ++i) temp[k++] = A[i];
            Arrays.sort(temp, com);
            for (int i = fromIndex, k = 0; i < toIndex; ++i) A[i] = temp[k++];
        }

        public Iterator<Integer> iterator() {
            itrInd = 0;
            Iterator<Integer> iterator = new Iterator<Integer>() {

                public boolean hasNext() {
                    return itrInd < size();
                }

                public Integer next() {
                    return get(itrInd++);
                }
            };
            return iterator;
        }

        public MyArrayListInt clone() {
            return new MyArrayListInt(this);
        }

        public String toString() {
            StringBuilder sb = new StringBuilder();
            for (int i = 0; i < len; i++) sb.append(A[i]).append(", ");
            return "[" + sb.toString().trim() + "]";
        }

    }

    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;
        }

    }
}

**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 consider the logic of the algorithm and identify edge cases or scenarios where the code might fail. The key points to focus on are:

1. **Shared Activities**: Ensure that the users share at least one activity.
2. **Unique Activities**: Ensure that both users have at least one activity that the other user does not like.
3. **BitSet Operations**: The code uses `BitSet` to manage the activities. We need to ensure that the `BitSet` operations are correctly handled, especially when checking for shared and unique activities.

### Step-by-Step Reasoning:

1. **Shared Activities**:
   - We need at least two users who share an activity.
   - Example: User 1 likes activities {1, 2, 3} and User 2 likes activities {1, 4, 5}. They share activity 1.

2. **Unique Activities**:
   - Both users must have at least one unique activity.
   - Example: User 1 has unique activity 2 and User 2 has unique activity 4.

3. **BitSet Operations**:
   - The code uses `BitSet` to check for shared and unique activities.
   - We need to ensure that the `BitSet` operations are correctly handled.

### Test Input Generation:

Let's create a small-scale test input that satisfies the above conditions:

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

### Explanation:

- **User 1** likes activities {1, 2, 3}.
- **User 2** likes activities {1, 4, 5}.
- **User 3** likes activities {2, 5}.

### Analysis:

- **User 1 and User 2**:
  - Shared activity: 1
  - Unique activities:
    - User 1: 2, 3
    - User 2: 4, 5
  - This pair should be a good match.

- **User 1 and User 3**:
  - Shared activity: 2
  - Unique activities:
    - User 1: 1, 3
    - User 3: 5
  - This pair should also be a good match.

- **User 2 and User 3**:
  - Shared activity: 5
  - Unique activities:
    - User 2: 1, 4
    - User 3: 2
  - This pair should also be a good match.

This test input should expose any issues in the code where the `BitSet` operations are not correctly handling the shared and unique activities. The expected output should be:

```plaintext
YES
1 2
```

or any other valid pair of users that form a good match.

### Final Test Input:

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


**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.Arrays;
import java.util.Iterator;
import java.util.Map;
import java.util.BitSet;
import java.util.Comparator;
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 Logger lg = new Logger();
        int N;
        int M;
        BitSet[] A;

        public void fillInputParams(FastReader in) {
            N = in.nextInt();
            M = in.nextInt();
            A = new BitSet[N];
            for (int i = 0; i < N; ++i) {
                A[i] = new BitSet();
                for (int j = in.nextInt(); j > 0; --j) {
                    A[i].set(in.nextInt() - 1);
                }
            }
        }

        Object solveOptimised(FastReader in, StringBuilder sb) {
            MyArrayListInt[] commonInterests = new MyArrayListInt[M];
            for (int i = 0; i < commonInterests.length; ++i) commonInterests[i] = new MyArrayListInt(1);
            for (int p = 0; p < N; ++p) {
                for (int l = A[p].nextSetBit(0); l >= 0; l = A[p].nextSetBit(l + 1)) {
                    commonInterests[l].add(p);
                }
            }
            for (MyArrayListInt likes : commonInterests) {
                likes.sort(0, likes.length(), Comparator.comparingInt(p -> A[p].cardinality()));

                for (int idx = 1; idx < likes.length(); ++idx) {
                    int pLeft = likes.get(idx - 1), pCurr = likes.get(idx);
                    BitSet left = A[pLeft], curr = A[pCurr];
                    if (isSubset(left, curr)) continue;
                    return yes(pLeft, pCurr);
                }
            }

            return "NO";
        }

        private boolean isSubset(BitSet small, BitSet big) {
            for (int idx = small.nextSetBit(0); idx >= 0; idx = small.nextSetBit(idx + 1)) {
                if (!big.get(idx)) return false;
            }
            return true;
        }

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

        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 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 MyArrayListInt implements Iterable<Integer> {
        private int len;
        private int[] A;
        int itrInd;

        public MyArrayListInt(int initialLength) {
            A = new int[Math.max(1, initialLength)];
            len = 0;
        }

        public MyArrayListInt(MyArrayListInt src) {
            A = new int[src.A.length];
            System.arraycopy(src.A, 0, A, 0, src.len);
            len = src.len;
        }

        public void add(int ele) {
            if (len == A.length) {
                int[] newAr = new int[A.length << 1];
                System.arraycopy(A, 0, newAr, 0, len);
                A = newAr;
            }
            A[len++] = ele;
        }

        public int get(int ind) {
            if (ind >= len) throw new ArrayIndexOutOfBoundsException(ind);
            else if (ind < 0) ind += size();
            return A[ind];
        }

        public int size() {
            return len;
        }

        public int length() {
            return len;
        }

        public void sort(int fromIndex, int toIndex, Comparator<Integer> com) {
            Integer[] temp = new Integer[toIndex - fromIndex];
            for (int i = fromIndex, k = 0; i < toIndex; ++i) temp[k++] = A[i];
            Arrays.sort(temp, com);
            for (int i = fromIndex, k = 0; i < toIndex; ++i) A[i] = temp[k++];
        }

        public Iterator<Integer> iterator() {
            itrInd = 0;
            Iterator<Integer> iterator = new Iterator<Integer>() {

                public boolean hasNext() {
                    return itrInd < size();
                }

                public Integer next() {
                    return get(itrInd++);
                }
            };
            return iterator;
        }

        public MyArrayListInt clone() {
            return new MyArrayListInt(this);
        }

        public String toString() {
            StringBuilder sb = new StringBuilder();
            for (int i = 0; i < len; i++) sb.append(A[i]).append(", ");
            return "[" + sb.toString().trim() + "]";
        }

    }

    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;
        }

    }
}

**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 the conditions under which the algorithm might fail. The key points to focus on are:

1. **Shared Activities**: Ensure that the users share at least one activity.
2. **Unique Activities**: Ensure that each user likes at least one activity that the other user does not like.
3. **Edge Cases**: Consider scenarios with minimal or maximal values for \( n \) and \( m \).

Given the constraints \( 2 \leq n \leq 200,000 \) and \( 1 \leq m \leq 10^6 \), we can create a small-scale test input that is complete and valid. Let's create a test case with \( n = 4 \) and \( m = 5 \):

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

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

### Analysis:
- **User 1 and User 2**:
  - Shared activity: 2
  - Unique activities:
    - User 1: 1
    - User 2: 3
  - This is a good match.

- **User 2 and User 3**:
  - Shared activity: 3
  - Unique activities:
    - User 2: 2
    - User 3: 4
  - This is a good match.

- **User 3 and User 4**:
  - Shared activity: 4
  - Unique activities:
    - User 3: 3
    - User 4: 5
  - This is a good match.

- **User 1 and User 3**:
  - No shared activities.
  - This is not a good match.

- **User 1 and User 4**:
  - No shared activities.
  - This is not a good match.

- **User 2 and User 4**:
  - No shared activities.
  - This is not a good match.

### Expected Output:
The expected output for this test case should be:
```plaintext
YES
1 2
```

This test case ensures that the algorithm must correctly identify a good match, and it is small enough to be easily verified.


**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 inf64 INT64_MAX/2
#define inf32 INT32_MAX/2
#define ll long long
#define int ll
#define pii pair<int,int>
#define endl '\n'
#define vv vector
#define cy cout<<"Yes"<<endl
#define cn cout<<"No"<<endl
#define its(a)  a.begin() + 1,a.end()
#define minV *min_element
#define maxV *max_element

int _;
ll lcm(ll a, ll b) { return a / __gcd(a, b) * b; }
//数学公式
ll mpow(ll x, ll y, ll mod);

void solve() {

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

	vv<vv<int>>all(n + 1);
	vv<int>id(n + 1);//id数组的id_i表示的是第i大的数组是下标为id_i的
	for (int i = 1; i <= n; i++) {
		int sz; cin >> sz;
		while (sz--) {
			int a; cin >> a;
			all[i].push_back(a);
		}
		id[i] = i;
	}
	auto cmp = [&](int a, int b)->int {
		return all[a].size() > all[b].size();
	};
	sort(its(id), cmp);
	vv<int>f(m + 1);
	vv<int>rs;
	for (int i = 1; i <= n; i++) {
		int n_id = id[i];
		int pos = f[all[n_id][0]];
		for (auto a : all[n_id]) {
			if (f[a] != pos) {
				cout << "Yes" << endl;
				int mx = max(pos, f[a]);
				cout << id[mx] << ' ' << n_id << endl;
				return;
			}
			else {
				f[a] = i;
			}
		}
	}

	cout << "NO" << endl;
}
signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);

	_ = 1;
	//cin >> _;

	while (_--) {
		solve();
	}

	return 0;
}

ll mpow(ll x, ll y, ll mod) {
	x %= mod;
	ll s = 1;
	while (y) {
		if (y & 1) {
			s *= x;
			s %= mod;
		}
		x *= x;
		x %= mod;
		y >>= 1;
	}
	return s;
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can highlight the issues in the logic. The primary issue in the buggy code is the way it handles the matching condition. Specifically, the code seems to be overly optimistic in finding a match and may incorrectly identify pairs that do not satisfy the conditions.

### Step-by-Step Reasoning:

1. **Understanding the Bug**:
   - The code uses a frequency array `f` to track the position of the first occurrence of each activity.
   - It checks if two users share an activity and if they have different positions in the `f` array, it assumes a match.
   - However, it does not properly check if both users have at least one activity that the other does not like.

2. **Creating a Test Case**:
   - We need to create a test case where users share activities but do not satisfy the second condition (both users must like at least one activity that the other does not like).

3. **Test Case**:
   - Let's consider a simple case with 3 users and 2 activities.
   - User 1 likes activity 1.
   - User 2 likes activity 1.
   - User 3 likes activity 2.

   In this case:
   - Users 1 and 2 share activity 1, but neither of them likes an activity that the other does not like.
   - Users 1 and 3 do not share any activities.
   - Users 2 and 3 do not share any activities.

   Therefore, there should be no good match, and the output should be `NO`.

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

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

- **Expected Output**:
  ```plaintext
  NO
  ```

This test case should expose the bug in the buggy code because it will incorrectly identify a match between users 1 and 2, even though they do not satisfy the second condition of the problem.


**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;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define au auto&&
#define ar auto&
#define car const ar
#define int long long
#define V(T) using v##T=vector<T>; using set##T=set<T>;
#define T(T, t...) using T=t; V(T) V(v##T)
T(z, int)   T(pzz, pair<z,z>)   T(ld, long double) T(uz, make_unsigned_t<z>)
#undef T
        z INF = 1e18;   const z O = 0;
#define fe(x...) for(au x)
            au operator>>(au i, au v) { return (i >> v.first) >> v.second;}
#define in(x...) x; [](au... a){((cin>>a), ...);}(x);
#define ind(x...) x; [](au... a){((cin>>a, --a), ...);}(x);
#define inv(v, sz) v(sz); fe(_:v){in(_)}
#define invd(v, sz) v(sz); fe(_:v){ind(_)}
   au pr = cout;   au operator-(ostream& o, au t) {return o << t << ' ';}   au operator+(ostream& o, au t) {return o << t << '\n';}   au operator*(ostream& o, au v) {fe(t:v) o-t; return o+"";}
#define DBT(...) |;
       template<typename T> z mine(T& x, T y) {return y<x?x=y,1:0;}    template<typename T> z maxe(T& x, T y) {return y>x?x=y,1:0;}
#define fn [&](au a)
#define cfn [](au a)
#define fn2 [&](au a, au b)
#define cfn2 [](au a, au b)
#define r(x...) {return x;}
#define cmpby(x...) fn2 r(fn r(x)(a) < fn r(x)(b))
#define yesno(p) {pr+((p)?"YES":"NO"); return;}
#define yes yesno(1)
#define no yesno(0)
#define dbg(x) (cerr<<__LINE__<<": \t"<<#x+1<<" = ")x;
    auto& cache(auto&& coll) { 	static z c = 0; 	static decltype(coll) copy = {}; 	return (c=!c) ? copy = coll : copy; }; auto& cache(auto& coll) { 	return coll; };
#define be(coll...) cache(coll).begin(), cache(coll).end()
   auto mpp(au coll, au f) {     return vector(be(coll | views::transform(f)));      }
#define mp(coll, f...) mpp(coll, fn r(f))
   auto zip(auto... colls) {   	assert(0);   	  	set s{colls.size()...}; 	assert(s.size() == 1); 	au g = [...i=colls.begin()] () mutable { 		if constexpr(sizeof...(i)==2) return pair{*i++...}; 		else return tuple{*i++...}; 	}; 	return mp(vz(*s.begin()), [...i=colls.begin()] () mutable { 			if constexpr(sizeof...(i)==2) return pair{*i++...}; 			else return tuple{*i++...}; 	}()); } auto zip2(auto... colls) { 	set s{colls.size()...}; 	assert(s.size() == 1); 	return mp(vz(*s.begin()), 			  [...i=colls.begin()] () mutable { if constexpr(sizeof...(i)==2) return pair{ *i++... };        else return tuple{ *i++... }; }()); }  template<size_t... Idx> auto unzip(au tuple_or_pair_coll, index_sequence<Idx...>) { 	return tuple{mp(tuple_or_pair_coll, get<Idx>(a))...}; } auto unzip(auto tuple_or_pair_coll) { 	make_index_sequence<tuple_size_v<remove_reference_t<decltype(*tuple_or_pair_coll.begin())>>> idx; 	return unzip(tuple_or_pair_coll, idx); }  template<typename T> auto incr(vector<T> v, function<bool(T,T)> cmp = less()) {     auto r = zip(v, views::iota(0U, size(v)));     sort(r.begin(), r.end(), fn2{auto [a0,a1]=a; auto [b0,b1]=b; return cmp(a0, b0) or !cmp(b0, a0) && a1<b1;});     return r; } auto incrBy(auto v, auto by) {     return incr(v, cmpby(by(a))); }  template<typename T> auto decr(vector<T> v, function<bool(T,T)> cmp = less()) {     return incr(v, cmpby(cmp(b,a))); } auto decrBy(auto v, auto by) {     return decr(v, cmpby(by(a))); }        const z DM = 1e9+7;   z modmul(z a, z b, z M = DM) {       return (__int128_t) a * b % M; }    z modpow(z a, z b, z M = DM) {     if(b < 0) b = modmul(-b, M-2, M);     if(b <= 1) return b ? a : 1;     z r = 1;     for (z x = a; b; x = modmul(x, x, M), b /= 2)         if(b & 1) r = modmul(r, x, M);     return r; }  z modinv(z a, z M = DM) {     return modpow(a, -1, M); }  template<z M = DM> struct mz {     using T = mz<M>;     z a;     mz(z a = 0) : a(a) {}     T operator+(T o) {         return (a + o.a) % M;     }     T operator-() {         return (M - a) % M;     }     T operator-(T o) {         return *this + -o;     }     T operator*(T o) {         return modmul(a, o.a, M);     }           T operator^(z b) {         return modpow(a, b, M);     }     T operator/(T o) {         return *this * (o^-1);     }
#define assOp(op) T operator op##=(T o) {return *this = *this op o;}
#define assOps assOp(+)	assOp(-) assOp(*) assOp(/)
	assOps };  template<z M = DM> struct mfr {     using T = mfr<M>;     mz<M> p = 0, q = 1;     T operator*(T o) {         return {p * o.p, q * o.q};     }     T operator-() {         return {-p, q};     }     T operator+(T o) {         return {p * o.q + q * o.p, q * o.q};     }     T operator-(T o) {         return *this + -o;     }           T operator^(z b) {         return b < 0 ? T{q, p} ^ -b : T{p ^ b, q ^ b};     }           T operator/(T o) {         return *this * (o^-1);     }  	assOps };  au wit = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};   bool prime(z n) {     if(n < 41) return count(be(wit), n);      z d = n - 1, s = 0;     while(d % 2 == 0) d /= 2, ++s;      for (z a : wit) {         z x = modpow(a, d, n), r = 0;         if(x != 1 && x != n-1)             while(++r < s && (x = modmul(x, x, n)) != n-1);         if(r >= s) return 0;       }     return 1; }      z rho(z n) {     if (n % 2 == 0) return 2;     auto f = [=](z x) { return (modmul(x, x, n) + 1) % n; };     for (z x0 = 0; x0 < n; ++x0) {         z x = x0, y = f(x), g;         while ((g = gcd(x - y, n)) == 1)             x = f(x), y = f(f(y));         if(g != n) return g;     }     throw; }    vz factors(z n) {     if(n == 1) return {};     if(prime(n)) return {n};     z r = rho(n);     vz v0 = factors(r), v1 = factors(n / r);     v0.insert(v0.end(), be(v1));     return v0; }    array<z, 3> gcd_ext(z a, z b) {     if(!a) return {b, 0, 1};     auto [d, x1, y1] = gcd_ext(b % a, a);     return {d, y1 - (b / a) * x1, x1}; }    pzz _crt(z a1, z m1, z a2, z m2) {       auto [g, x, y] = gcd_ext(m1, m2);     z l = m1 / g * m2;     z ret = (a1 + ((x * (a2-a1) / g) % (l/m1)) * m1) % l;     return {(a2-a1) % g ? -1 : (ret + l) % l, l}; }        pzz crt(vz a, vz m) {       z sol = a[0], l = m[0];     for(z i=1; sol+1 && i<a.size(); i++)         tie(sol, l) = _crt(sol, l, a[i], m[i]);     return {sol, l}; }
#pragma ide diagnostic ignored "misc-no-recursion"

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

template<bool persistent=0>
struct segT {
	z sz; //always initialize!

	using A = z;
	using U = z;
	A e = 0;
	U id = 0;
	A aggr(A a, A b) {return a+b;}
	A app(A a, U u, z s) {return a+u*s;}
	U comp(U later, U first) {return later+first;}

	/*using A = z;
	using U = z;
	A e = -INF;
	U id = 0;
	A aggr(A a, A b) {return max(a, b);}
	A app(A a, U u, z s) {return a+u;}
	U comp(U later, U first) {return later + first;}*/

	/*using A = z;
	using U = z;
	A e = -INF;
	U id = -1;
	A aggr(A a, A b) {return max(a, b);}
	A app(A a, U u, z s) {return u!=id ? u : a;}
	U comp(U later, U first) {return later!=id ? later : first;}*/

	/*//main1:
	using A = z;
	using U = z;
	A e = 0;
	U id = 0;
	A aggr(A a, A b) {return a+b;}
	A app(A a, U u, z s) {return a+u*s;}
	U comp(U later, U first) {return later+first;}*/

	/*//main2:
	using A = z;
	using U = pzz;
	z M = 998244353;
	A e = 0;
	U id = {1,0};
	A aggr(A a, A b) {return (a+b)%M;}
	A app(A a, U u, z s) {return (u.first*a + u.second*s)%M;}
	U comp(U later, U first) {return {(later.first*first.first)%M, (later.first*first.second + later.second)%M};}*/

	struct node {
		int32_t l, r;
		A agg; //or value (for leaves)
		U lazy;
	};
	shared_ptr<vector<node>> v{new vector<node>{{0, 0, e, id}}}; //Array wegen Speicher (int32_t < node*). Is that worth it??? // Linus: probably this is also faster because of caching
	//shared_ptr: bei vielen Updates/SegTs ist es wichtig, dass der Speicher für v wieder freigegeben wird
#define v (*v)
#define N v[n]
	z root = 0;

	z set(z n, node x) {
		return persistent || !n ? v.push_back(x), v.size()-1 : (v[n]=x, n);
	}
	z apply(z n, U u, z s) {
		return set(n, {N.l, N.r, app(N.agg, u, s), comp(u, N.lazy)});
	}
	/*old version (if persistent: 1/3 more memory. TODO: Delete when tested NE!!!: mit dem TODO unten könnte das hier einfacher sein)
	z update(z a, z bex, z l, z rex, z n, U u) {
		if(bex <= l || rex <= a) return n;
		if(a <= l && rex <= bex) return apply(n, u, rex - l);
		z m = (l + rex) / 2;
		z nl = update(a, bex, l, m, apply(N.l, N.lazy, m - l), u); //here, it's important that v[0] is an empty node
		z nr = update(a, bex, m, rex, apply(N.r, N.lazy, rex - m), u);
		return set(n, {nl, nr, aggr(v[nl].agg, v[nr].agg), id});
	}*/
	z update(z a, z bex, z l, z rex, z n, U u, U lazy) { //TODO: make update and apply return nodes (that need to be pushed into v manually, if desired)!!?
		if (bex <= l || rex <= a) return apply(n,lazy,rex-l);
		if (a <= l && rex <= bex) return apply(n, comp(u, lazy), rex-l);
		z m = (l+rex)/2;
		z nl = update(a, bex, l, m, N.l, u, comp(lazy, N.lazy)); //here, it's important that v[0] is an empty node
		z nr = update(a, bex, m, rex, N.r, u, comp(lazy, N.lazy));
		return set(n, {nl, nr, aggr(v[nl].agg, v[nr].agg), id});
	}
	A query(z a, z bex, z l, z rex, z n) {
		if(bex <= l || rex <= a) return e;
		if(a <= l && rex <= bex) return N.agg;
		//push(n, rex - l); //might need to much memory if more queries than updates allowed
		z m = (l + rex) / 2;
		A la = query(a, bex, l, m, N.l);
		A ra = query(a, bex, m, rex, N.r);
		return app(aggr(la, ra), N.lazy, min(bex, rex) - max(a, l));
	}

	z lower_bound(z a, z bex, z l, z rex, z n, au cur_agg, U lazy, au p, bool rtl) {
		if(bex <= l || rex <= a) return -2; //todo: bei Link anders -- passt das so (immer)? Chris fragen!?
		A n_agg = app(N.agg, lazy, rex - l);
		A new_agg = rtl ? aggr(n_agg, cur_agg) : aggr(cur_agg, n_agg);
		if (a <= l && rex <= bex && !p(new_agg)) return cur_agg = new_agg, -2;
		if (l + 1 == rex) return l;

		//push(n, rex - l); //might need to much memory if more queries than updates allowed
		z m = (l + rex) / 2;
		z res = lower_bound(a, bex, rtl?m:l, rtl?rex:m, rtl?N.r:N.l, cur_agg, lazy = comp(lazy, N.lazy), p, rtl);
		if (res != -2) return res;
		return lower_bound(a, bex, rtl?l:m, rtl?m:rex, rtl?N.l:N.r, cur_agg, lazy, p, rtl);
	}
#undef N
#undef v

	segT update(z l, z rex, U u) {
		assert(0 <= l && l <= rex && rex <= sz);
		segT t = *this;
		t.root = update(l, rex, 0, sz, root, u, id);
		if(!persistent) root = t.root;
		return t;
	}

	A query(z l, z rex) {
		assert(0 <= l && l <= rex && rex <= sz);
		return query(l, rex, 0, sz, root);
	}

	z lower_bound(z l, z rex, function<bool(A&)> p, bool rtl = false) {
		assert(0 <= l && l <= rex && rex <= sz);
		if(p(e)) return rtl?rex:l;
		if(l==rex) return -1;
		return lower_bound(l, rex, 0, sz, root, (A)e, id, p, rtl) + !rtl;
	}

	void print(z n=0) {
		for (int i = 0; i < (n?n:sz); ++i) {
			pr-query(i,i+1);
		}
		pr+"";
	}
};

//TODO: in fn a,b,c -> _,__,___ or _,_2,_3 or or A,B,C or ä,ö,ü or $,$2,$3 or ß,ßß,ßßß?
//        A geht schneller als _ oder $. Also nur Auswahl zw A,B,C und ß,ßß,ßßß und ä,ö,ü

//TODO: complex für 2D merken!

// n<=1e18 hat O(n^1/3) Teiler  (https://codeforces.com/blog/entry/14463)

struct iterSegT1 {
	using A = z;
	A e = 0;
	A agg(A a, A b) {
		return a+b;
	}

	/*using A = z;
	A e = 0;
	A agg(A a, A b) {return a+b;}*/

	/*struct A {
		z save,add,addsafe;
	};
    A e = {0,0,0};
    A agg(A a, A b) {z m=min(a.add,b.save); return {a.save+b.save-m,a.add-m+b.add, a.addsafe + b.addsafe + m};}*/
	z n;
	vector<A> v;

	iterSegT1(z n): n(n), v(2 * n, e) {}

	// point set
	iterSegT1(vector<A> s): n(s.size()), v(n, e) {
		v.insert(v.end(), s.begin(), s.end());
		for (z i = n; i-- > 0;)
			v[i] = agg(v[2 * i], v[2 * i + 1]);
	}
	void set(z i, A a) {
		v[i += n] = a;
		while (i /= 2) v[i] = agg(v[2 * i], v[2 * i + 1]);
	}
	A get(z i) {return v[i+n];}
	A query(z l, z rex) {
		A la = e, ra = e;
		for (l += n, rex += n; l < rex; l /= 2, rex /= 2) {
			if (l&1) la = agg(la, v[l++]);
			if (rex&1) ra = agg(v[--rex], ra);
		}
		return agg(la, ra);
	}

/*
	z lower_bound(z thresh) {
		z l = 1; // start at root

		while (l < n) {
			if (v[l*2] >= thresh) {
				// links
				l *= 2;
			} else {
				thresh -= v[l*2];
				l = l*2 + 1;
			}
		}
		return l-n+1;
	}*/

	// range update
	void update(z l, z rex, A a) {
		for (l += n, rex += n; l < rex; l /= 2, rex /= 2) {
			if (l&1) {v[l] = agg(v[l], a), l++;}
			if (rex&1) {--rex, v[rex] = agg(v[rex], a);}
		}}
	A query(z i) {
		A r = v[i += n];
		while(i /= 2) r = agg(r, v[i]);
		return r;
	}
};

/*
template<typename T>
struct coco {
	map<T,z> m;
	vector<T> r;

	coco(vector<T> a) {

	}

	template<typename U>
	coco(vector<U> a, function<T&(U&)> coord) {

	}
 	auto operator+(auto& v2) {
 		fe(x:v2) x=m[x]; //eher return mp(v2, m[a]) für vec<not z>z& v2
 		return *this;
 	}
};
template<typename... T, typename... X>
void coco(tuple<pair<T,function<z&(X&)>>...> rest) {

}*/

z lis(au a) {
	vz lmo2ln(a.size(), INF); //lenMinusOne2lastNum
	fe(x:a) mine(*lower_bound(be(lmo2ln), x), x); ///strictly increasing. Otherwise just change to upper_bound
	return a.size() - count(be(lmo2ln), INF);
}

template<typename T>
using Tree = tree<T, null_type, less<T>, rb_tree_tag,
		tree_order_statistics_node_update>;

set<z> divs(z n) {
	set<z> d;
	for (int i = 1; i * i <= n; ++i)
		if(n%i==0) d.insert({i, n / i});
	return d;
}

vector<mz<>> fac={1};
void init() {
	for (int i = 1; i <= 3e5; ++i) fac.push_back(fac.back() * i);

}

vector<int> Centroid(const vector<vector<int>> &g) {
	int n = g.size();
	vector<int> centroid;
	vector<int> sz(n);
	function<void (int, int)> dfs = [&](int u, int prev) {
		sz[u] = 1;
		bool is_centroid = true;
		for (auto v : g[u]) if (v != prev) {
				dfs(v, u);
				sz[u] += sz[v];
				if (sz[v] > n / 2) is_centroid = false;
			}
		if (n - sz[u] > n / 2) is_centroid = false;
		if (is_centroid) centroid.push_back(u);
	};
	dfs(0, -1);
	return centroid;
}

void solve(z T) {
	z in(n, m)
	vz a2groupRepr(m+1);
	vz i2Sz(m+1);
	vector<pair<vz,z>> D;
	for (int i = 1; i <= n; ++i) {
		z in(k)
		vz invd(a,k)
		D.emplace_back(a,i);
		i2Sz[i] = a.size();
	}
	std::sort(D.begin(), D.end(), cmpby(-a.first.size()));
	fe([a,i]:D) {
		auto c = cmpby(i2Sz[a]);
		set<z, decltype(c)> groups(c);
		fe(a:a) {
			au repr = a2groupRepr[a];
			groups.insert(repr);
			repr=i;
		}
		if(groups.size()>1) {
			groups.erase(0);
			pr+"YES"-*groups.begin()+i;
			return;
		}
	}
	no

	vvz A;
	set<vz> s;
	vvz l(m);
	for (int i = 1; i <= n; ++i) {
		z in(k)
		vz invd(aa,k)
		A.push_back(aa);
		std::sort(aa.begin(), aa.end());
		if(aa.size()>1 && !s.contains(aa)) {
			s.insert(aa);
			fe(a:aa) l[a].push_back(i);
		}
	}
	fe(l:l) {
		if(l.size()>2) {
			std::sort(l.begin(), l.end(), cmpby(A[a].size()));
			vz cur_ixs=l;
			pr+"YES"-l[0]+l[1];
			return;
		}
	}
	no

	/*z in(n)
	vector<pair<pzz,z>> inv(s,n)
	vvz g(n);
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < n; ++j) {
			auto [xy1,r1]=s[i];
			auto [x1,y1]=xy1;
			auto [xy2,r2]=s[j];
			auto [x2,y2]=xy2;
			if(i!=j && (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)==(r1+r2)*(r1+r2)) g[i].push_back(j);
		}
	}
	vz vis(n,0);
	vz col(n,1);
	z d;
	au x=[&](z u, z p, au x) -> bool {
		vis[u]=1;
		col[u] = -col[p];
		d += col[u];
		fe(v:g[u]) if(v!=p) {
			if(vis[v]) {
				if(col[v]==col[u]) return 0;
			}
			else if(!x(v,u,x)) return 0;
		}
		return 1;
	};
	z r=0;
	for (int u = 0; u < n; ++u) {
		d=0;
		if(!vis[u]) r |= (x(u, u, x) && d);
	}
	yesno(r)
	//bis -03.55*/

	/*string in(s,b)
	fe([s,b]:zip2(s,b)) {
		pr-s+b;
	}*/

	/*vz v(9);
	auto l = v | views::transform([i = 0](z s)mutable { return i++; });
	*l.begin();
	pr+(*--l.end());*/
}

signed main() {
#ifdef ONLINE_JUDGE
	//cin.tie(0)->sync_with_stdio(0);
#endif
	pr.precision(16);
	fixed(pr);
	init();
	z t = 1;
	//in(t)
	for (z i = 1; i <= t; ++i) {
#ifndef ONLINE_JUDGE
		//pr-"__________________________"+i;
#endif
		solve(i);
	}
}

**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 craft a test input that can highlight the issue. The code seems to be focusing on finding a good match between users based on their activities. A good match is defined as two users who share at least one activity and each has at least one activity that the other does not like.

### Step-by-Step Analysis:
1. **Identify the Logic Flaw:**
   - The code sorts the users by the number of activities they like and then attempts to group users based on shared activities.
   - The issue might arise when the code incorrectly groups users or fails to find a match when one exists.

2. **Craft a Test Case:**
   - We need a small-scale test input where the users have activities that can form a good match, but the code might fail to identify it.
   - Consider a scenario where users have activities that overlap in such a way that the sorting and grouping logic might miss the match.

### Test Input:
Let's create a test input with 4 users and 5 activities. The users will have the following activities:
- User 1: Activities 1, 2
- User 2: Activities 1, 3
- User 3: Activities 2, 4
- User 4: Activities 3, 5

### Explanation:
- Users 1 and 2 share activity 1, and User 1 likes activity 2 which User 2 does not, and User 2 likes activity 3 which User 1 does not. This forms a good match.
- Users 3 and 4 share activity 3, and User 3 likes activity 4 which User 4 does not, and User 4 likes activity 5 which User 3 does not. This also forms a good match.

### Test Input:
```plaintext
4 5
2 1 2
2 1 3
2 2 4
2 3 5
```

This input should expose the bug if the code fails to identify the good matches between Users 1 and 2, and Users 3 and 4.


**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:
#!/usr/bin/env python3

n, m = map(int, input().split())
users = [(frozenset(), i) for i in range(n + 1)]
for i in range(1, n + 1):
    users[i] = (frozenset(map(int, input().split()[1:])), i)
activities = [[] for _ in range(m + 1)]
for user in users:
    for a in user[0]:
        activities[a].append(user)
for a in activities:
    for k, (u1, i) in enumerate(a):
        for u2, j in a[:k]:
            # print(a, u1, u2)
            if (u1 - u2) and (u2 - u1):
                print("YES")
                print(i, j)
                exit()
print("NO")

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

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

Think step by step.

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


**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())
usr = [list(map(int, input().split())) for _ in range(n)]
usr = [sub[1:] for sub in usr if sub[0] != 1]

gd_mtc = False
mtc_usr = None

if len(usr) < 2:
    gd_mtc = False
else:
    usr_set = [set(sub) for sub in usr]
    for i in range(len(usr_set)):
        for j in range(i + 1, len(usr_set)):
            cmn_act = usr_set[i].intersection(usr_set[j])
            if cmn_act:
                u1_ld = usr_set[i] - usr_set[j]
                u2_ld = usr_set[j] - usr_set[i]
                if u1_ld and u2_ld:
                    gd_mtc = True
                    mtc_usr = (i + 1, j + 1)
                    break
        if gd_mtc:
            break

if gd_mtc:
    print("YES")
    if mtc_usr:
        print(*mtc_usr)
else:
    print("NO")

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

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

Think step by step.

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

**Explanation:**
This test case is designed to expose the bug in the buggy code. Here's how it works:

1. **Input Description:**
   - There are 2 users (n = 2) and 2 activities (m = 2).
   - User 1 likes activity 1.
   - User 2 likes activity 2.

2. **Expected Output:**
   - The output should be `NO` because there is no common activity between the two users, and hence, they cannot form a good match.

3. **Bug Exposure:**
   - The buggy code filters out users who like exactly one activity: `usr = [sub[1:] for sub in usr if sub[0] != 1]`.
   - This means that both users in this test case will be filtered out, and the `usr` list will be empty.
   - As a result, the code will incorrectly output `NO` because it will not find any users to compare.

This test case is small and valid, making it easy to verify the behavior of the buggy code. It specifically targets the part of the code that filters out users with exactly one activity, which is a critical flaw in the algorithm.


**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()));

                int lastPerson = peopleWithSameHobby.get(0);
                ArrayList<Integer> lastPersonHobbies = likes[lastPerson];
                for (int i = 1; i < peopleWithSameHobby.size(); ++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 (lastPersonHobbies.size() < 50) {
                        for (int currHobby : lastPersonHobbies) {
                            if (Collections.binarySearch(currentPersonHobbies, currHobby) < 0) {
                                return yes(lastPerson, currentPerson);
                            }
                        }
                    } else {
                        for (int attempt = Math.min(lastPersonHobbies.size() / 10, 50); attempt > 0; --attempt) {
                            int randomIdx = hp.getRandomInRange(0, lastPersonHobbies.size() - 1);
                            int randomHobby = lastPersonHobbies.get(randomIdx);
                            if (Collections.binarySearch(currentPersonHobbies, randomHobby) < 0) {
                                return yes(lastPerson, currentPerson);
                            }
                        }
                    }

                    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 challenging test input that exposes the bug in the buggy code, we need to focus on edge cases that can reveal issues in the algorithm's logic. The key points to consider are:

1. **Users with a single activity**: The code skips users with only one activity, which might lead to missing valid matches.
2. **Users with the same set of activities**: The code uses a hash function to compare users with the same number of activities, which might lead to false negatives if the hash function collides.
3. **Users with a large number of activities**: The code uses a random sampling approach to check for unique activities, which might miss valid matches if the random samples are not representative.

Given these points, let's create a small-scale test input that maximizes the failure probability:

1. **Users with a single activity**: Ensure there are users with only one activity to test the skipping logic.
2. **Users with the same set of activities**: Ensure there are users with the same set of activities to test the hash function.
3. **Users with a large number of activities**: Ensure there are users with a large number of activities to test the random sampling logic.

Here's a test input that covers these scenarios:

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

Explanation:
- User 1: Likes activity 1 (single activity).
- User 2: Likes activities 1 to 10 (large set of activities).
- User 3: Likes activities 1 to 10 (same set as User 2).
- User 4: Likes activities 1 to 5 (small set of activities).

This test input should expose the following issues:
- The code skips User 1, which might be incorrect if there is a valid match.
- The code uses a hash function to compare User 2 and User 3, which might lead to a false negative if the hash function collides.
- The code uses random sampling to check for unique activities between User 2 and User 4, which might miss valid matches if the random samples are not representative.

This test input is small and complete, and it should help identify the bugs in the provided 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:
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.Timer;
import java.util.HashMap;
import java.util.Random;
import java.util.ArrayList;
import java.util.TimerTask;
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();
        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()));

                int lastPerson = peopleWithSameHobby.get(0);
                ArrayList<Integer> lastPersonHobbies = likes[lastPerson];
                for (int i = 1; i < peopleWithSameHobby.size(); ++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 (lastPersonHobbies.size() < 50) {
                        for (int currHobby : lastPersonHobbies) {
                            if (Collections.binarySearch(currentPersonHobbies, currHobby) < 0) {
                                return yes(lastPerson, currentPerson);
                            }
                        }
                    } else {
                        for (int attempt = Math.min(lastPersonHobbies.size() / 10, 50); attempt > 0; --attempt) {
                            int randomIdx = hp.getRandomInRange(0, lastPersonHobbies.size() - 1);
                            int randomHobby = lastPersonHobbies.get(randomIdx);
                            if (Collections.binarySearch(currentPersonHobbies, randomHobby) < 0) {
                                return yes(lastPerson, currentPerson);
                            }
                        }
                    }

                    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);
        }

        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();

            TimerTask task = new TimerTask() {

                public void run() {
                    System.out.println("NO");
                    System.exit(0);
                }
            };
            Timer timer = new Timer();
            timer.schedule(task, 2950);

            fillInputParams(in);
            out.println(solveOptimised(in, new StringBuilder()));
            timer.cancel();
        }

    }

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Rating
time_limit_ms: 2000
memory_limit_mb: 512
Description: Hello, Codeforces Forcescode!

Kevin used to be a participant of Codeforces. Recently, the KDOI Team has developed a new Online Judge called Forcescode.

Kevin has participated in $n$ contests on Forcescode. In the $i$-th contest, his performance rating is $a_i$.

Now he has hacked into the backend of Forcescode and will select an interval $[l,r]$ ($1\le l\le r\le n$), then skip all of the contests in this interval. After that, his rating will be recalculated in the following way:

  * Initially, his rating is $x=0$;    * For each $1\le i\le n$, after the $i$-th contest,      * If $l\le i\le r$, this contest will be skipped, and the rating will remain unchanged;      * Otherwise, his rating will be updated according to the following rules:        * If $a_i>x$, his rating $x$ will increase by $1$;        * If $a_i=x$, his rating $x$ will remain unchanged;        * If $a_i<x$, his rating $x$ will decrease by $1$.

You have to help Kevin to find his maximum possible rating after the recalculation if he chooses the interval $[l,r]$ optimally. Note that Kevin has to skip at least one contest.

Input: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\le t\le 5\cdot 10^4$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$) — the number of contests.

The second line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$) — the performance ratings in the contests.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible rating after the recalculation if Kevin chooses the interval optimally.

Examples:
input:
5
6
1 2 3 4 5 6
7
1 2 1 1 1 3 4
1
1
9
9 9 8 2 4 4 3 5 3
10
1 2 3 4 1 3 2 1 1 10
output:
5
4
0
4
5

Note: In the first test case, Kevin must skip at least one contest. If he chooses any interval of length $1$, his rating after the recalculation will be equal to $5$.

In the second test case, Kevin's optimal choice is to select the interval $[3,5]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=1} 1 \xrightarrow{a_2=2} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{a_6=3} 3 \xrightarrow{a_7=4} 4 $$

In the third test case, Kevin must skip the only contest, so his rating will remain at the initial value of $0$.

In the fourth test case, Kevin's optimal choice is to select the interval $[7,9]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=9} 1 \xrightarrow{a_2=9} 2 \xrightarrow{a_3=8} 3 \xrightarrow{a_4=2} 2 \xrightarrow{a_5=4} 3 \xrightarrow{a_6=4} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 $$

In the fifth test case, Kevin's optimal choice is to select the interval $[5,9]$.

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

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        vector<int> a(n);
        for (int i = 0; i < n; i++) {
            cin >> a[i];
        }

        vector<int> dp(n, 0);
        vector<int> sdp(n, 0);
        int ans = 0;

        dp[0] = 1;
        for (int i = 1; i < n; i++) {
            dp[i] = 1;
            for (int j = 0; j < i; j++) {
                if (a[i] > a[j]) {
                    dp[i] = max(dp[i], dp[j] + 1);
                }
                else{
                    dp[i]=1;
                }
            }
            ans = max(ans, dp[i]);
        }

        int y = 1;
        while (y < n && dp[y] > dp[y - 1]) {
            y++;
        }

        // Initialize sdp to be same as dp initially
        for (int i = 0; i < n; i++) {
            sdp[i] = dp[i];
        }

        int maxb = y;

        for (int i = y; i < n; i++) {

          sdp[i]=dp[i];
                if (a[i] > maxb) {
                    sdp[i] = max(sdp[i], y + 1);
                } else {
                    sdp[i] = max(sdp[i], a[i]);
                }

            if (i > 0 && a[i] > sdp[i - 1]) {
                sdp[i] = max(sdp[i], sdp[i - 1] + 1);
            }
               if (i > 0 && a[i] == sdp[i - 1]) {
                sdp[i] = max(sdp[i], sdp[i - 1]);
            }

            ans = max(ans, sdp[i]);

        }

        if (ans == n) {
            ans--;
        }

        cout << ans << endl;
    }
    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to create a test case that highlights the issues in the logic. The code seems to have problems with the dynamic programming (DP) approach and the way it handles the maximum possible rating after skipping a subinterval.

### Key Points to Consider:
1. **Initial Rating Calculation**: The code initializes `dp[0] = 1` and then iterates through the array to update the DP values. However, it resets `dp[i]` to 1 if `a[i] <= a[j]` for any `j < i`. This can lead to incorrect values for `dp[i]`.
2. **Optimal Subinterval Selection**: The code tries to find the optimal subinterval to skip by using a secondary DP array `sdp`. However, the logic for updating `sdp[i]` and the conditions for updating the maximum rating are complex and can introduce errors.
3. **Edge Cases**: The code might fail when all ratings are the same, or when the ratings are in a specific pattern that the DP logic does not handle correctly.

### Challenging Test Case:
Let's create a test case where the ratings are in a pattern that should expose the issues in the DP logic. Specifically, we will use a sequence where the ratings are mostly increasing, but with some small decreases that should not affect the maximum rating if the correct subinterval is chosen.

### Test Input:
```plaintext
1
10
1 2 3 4 5 4 3 2 1 10
```

### Explanation:
- **Test Case**: The test case has 10 contests with ratings `1, 2, 3, 4, 5, 4, 3, 2, 1, 10`.
- **Expected Output**: The optimal subinterval to skip is `[6, 9]` (contests with ratings `4, 3, 2, 1`), which will leave the sequence `1, 2, 3, 4, 5, 10`. The maximum possible rating after recalculation should be `5`.

### Why This Test Case is Challenging:
- **Increasing Sequence with Decreases**: The sequence starts increasing and then has a small decrease followed by a large increase. This should test the code's ability to handle both increasing and decreasing sequences.
- **Optimal Subinterval**: The optimal subinterval to skip is not trivial and requires the code to correctly identify and handle the small decreases in the middle of the sequence.

### Expected Behavior:
- The correct output should be `5`, but the provided code might fail to handle the small decreases correctly and might output a lower value.

This test case should help expose the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: New Rating
time_limit_ms: 2000
memory_limit_mb: 512
Description: Hello, Codeforces Forcescode!

Kevin used to be a participant of Codeforces. Recently, the KDOI Team has developed a new Online Judge called Forcescode.

Kevin has participated in $n$ contests on Forcescode. In the $i$-th contest, his performance rating is $a_i$.

Now he has hacked into the backend of Forcescode and will select an interval $[l,r]$ ($1\le l\le r\le n$), then skip all of the contests in this interval. After that, his rating will be recalculated in the following way:

  * Initially, his rating is $x=0$;    * For each $1\le i\le n$, after the $i$-th contest,      * If $l\le i\le r$, this contest will be skipped, and the rating will remain unchanged;      * Otherwise, his rating will be updated according to the following rules:        * If $a_i>x$, his rating $x$ will increase by $1$;        * If $a_i=x$, his rating $x$ will remain unchanged;        * If $a_i<x$, his rating $x$ will decrease by $1$.

You have to help Kevin to find his maximum possible rating after the recalculation if he chooses the interval $[l,r]$ optimally. Note that Kevin has to skip at least one contest.

Input: Each test contains multiple test cases. The first line of the input contains a single integer $t$ ($1\le t\le 5\cdot 10^4$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($1\le n\le 3\cdot 10^5$) — the number of contests.

The second line contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1\le a_i\le n$) — the performance ratings in the contests.

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum possible rating after the recalculation if Kevin chooses the interval optimally.

Examples:
input:
5
6
1 2 3 4 5 6
7
1 2 1 1 1 3 4
1
1
9
9 9 8 2 4 4 3 5 3
10
1 2 3 4 1 3 2 1 1 10
output:
5
4
0
4
5

Note: In the first test case, Kevin must skip at least one contest. If he chooses any interval of length $1$, his rating after the recalculation will be equal to $5$.

In the second test case, Kevin's optimal choice is to select the interval $[3,5]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=1} 1 \xrightarrow{a_2=2} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{\mathtt{skip}} 2 \xrightarrow{a_6=3} 3 \xrightarrow{a_7=4} 4 $$

In the third test case, Kevin must skip the only contest, so his rating will remain at the initial value of $0$.

In the fourth test case, Kevin's optimal choice is to select the interval $[7,9]$. During the recalculation, his rating changes as follows:

$$ 0 \xrightarrow{a_1=9} 1 \xrightarrow{a_2=9} 2 \xrightarrow{a_3=8} 3 \xrightarrow{a_4=2} 2 \xrightarrow{a_5=4} 3 \xrightarrow{a_6=4} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 \xrightarrow{\mathtt{skip}} 4 $$

In the fifth test case, Kevin's optimal choice is to select the interval $[5,9]$.

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

void solve() {
    ll n;
    cin>>n;
    vector<ll>v(n);
    for(ll i=0;i<n;i++){
        cin>>v[i];
    }

    ll a = 0;   // Flag for certain conditions
    ll b = 0;   // Another flag for different conditions
    ll h = 0;   // This tracks the "height" or level
    ll l = 0;   // Counter for decrementing condition
    ll z = 0;   // Counter for consecutive equal elements
    vector<ll> g;  // Vector to store values for certain condition
    vector<ll> f;  // Another vector for a different condition

    for(ll i = 0; i < n; i++){
        if(v[i] > h){
            h++;
            if(a == 1){
                g.push_back(l);
                a = 0;
                l=0;
            }
            if(b == 1){
                f.push_back(z);
                b=0;
                z = 0;
            }
        }
        else if(v[i] == h){
            z++;
            b = 1;   // Just set the flag, don't redefine
            if(a == 1){
                g.push_back(l);
                a = 0;
                l=0;
            }
        }
        else{
            l++;
            a = 1;   // Just set the flag, don't redefine
            if(b == 1){
                f.push_back(z);
                z = 0;
                b=0;
            }
        }
    }

    sort(g.begin(), g.end());

    // Output conditions based on g and f vectors
    if(g.size() == 0 && f.size() > 0){
        cout << h << endl;
        return;
    }

    if(g.size() == 1){
        cout << h << endl;
        return;
    }

    if(g.size() == 0 && f.size() == 0){
        cout << h - 1 << endl;
        return;
    }

    if(g.size() > 1){
        for(ll i = 1; i < g.size(); i++){   // Start from i = 1
            h -= g[i];
        }
        cout << h << endl;
    }
}

int main() {
    int t;
    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
5
6
1 2 3 4 5 6
7
1 2 1 1 1 3 4
1
1
9
9 9 8 2 4 4 3 5 3
10
1 2 3 4 1 3 2 1 1 10
```

This input is designed to cover various edge cases and typical scenarios to expose potential bugs in the given buggy code. Here's a breakdown of the test cases:

1. **Case 1: Increasing Sequence**
   - `n = 6`, `a = [1, 2, 3, 4, 5, 6]`
   - This is a simple increasing sequence where the optimal interval to skip is any single contest, resulting in a final rating of 5.

2. **Case 2: Mixed Sequence with Repeated Values**
   - `n = 7`, `a = [1, 2, 1, 1, 1, 3, 4]`
   - This sequence has repeated values and a mix of increasing and decreasing segments. The optimal interval to skip is `[3, 5]`, resulting in a final rating of 4.

3. **Case 3: Single Contest**
   - `n = 1`, `a = [1]`
   - This is a single contest, and the only option is to skip it, resulting in a final rating of 0.

4. **Case 4: Complex Sequence with Repeated Values**
   - `n = 9`, `a = [9, 9, 8, 2, 4, 4, 3, 5, 3]`
   - This sequence has repeated values and a mix of increasing and decreasing segments. The optimal interval to skip is `[7, 9]`, resulting in a final rating of 4.

5. **Case 5: Mixed Sequence with a Large Value at the End**
   - `n = 10`, `a = [1, 2, 3, 4, 1, 3, 2, 1, 1, 10]`
   - This sequence has a large value at the end, which can significantly affect the final rating. The optimal interval to skip is `[5, 9]`, resulting in a final rating of 5.

These test cases are designed to cover a range of scenarios, including simple sequences, mixed sequences, and edge cases with single contests, to ensure the buggy code is thoroughly tested.