**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Destruction
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given a tree$$$^{\text{∗}}$$$ with $$$n$$$ vertices. You can choose two vertices $$$a$$$ and $$$b$$$ once and remove all vertices on the path from $$$a$$$ to $$$b$$$, including the vertices themselves. If you choose $$$a=b$$$, only one vertex will be removed.

Your task is to find the maximum number of connected components$$$^{\text{†}}$$$ that can be formed after removing the path from the tree.

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

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

The next $$$n-1$$$ lines contain two integers $$$u$$$ and $$$v$$$ ($$$1 \le u, v \le n$$$, $$$u \ne v$$$) — the vertices connected by an edge. It is guaranteed that the edges form a tree.

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

Output: For each test case, output one integer — the maximum number of connected components that can be achieved using the described operation.

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

Note:

Buggy Code:
import java.util.*;
import java.io.*;
public class Main
{
    // static int M=(int)(1e9+7);
    static int M=998244353;
    static FastReader in  = new FastReader();
    static FastWriter out = new FastWriter();

    static double PI=3.1415926535897932384626433832795;
    @SuppressWarnings("unchecked")

    // हर हर महादेव
    // हरे कृष्णा

    //10^6 space of int is allowed not of long
    // Yesterday was history
    // Tomorrow will be a mystery
    // but today is a gift
    // That is why it is called the present

    /* stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
	* DON'T GET STUCK ON ONE APPROACH
    */

    //Credit Striver
    // Remember you were also a novice when you started,
    // hence never be rude to anyone who wants to learn something
    // Never open a ranklist untill and unless you are done with solving problems, wastes 3/4 minuts
    // Donot treat CP as a placement thing, love it and ejoy it, you will suceed for sure.

    public static void main(String[] args)throws Exception
    {
        // in = new FastReader("input11.txt");
            // out = new FastWriter("mootube.out");
            int testcases=1;

            testcases=in.nextInt();

            // if(testcases==10000)
            // {

            //     for(int i1=1;i1<=testcases;i1++)
            //     {
            //         int n=in.nextInt();
            //         graph=new ArrayList<>();
            //         for(int i=0;i<n;++i)graph.add(new ArrayList<>());
            //         for(int i=0;i<n-1;i++)
            //         {
            //             int a=in.nextInt();int b=in.nextInt();
            //             a--;b--;
            //             graph.get(a).add(b);
            //             graph.get(b).add(a);
            //         }
            //         if(i1==50)
            //         {
            //             out.println(n);
            //             out.println(graph);
            //             out.close();
            //             return;
            //         }
            //     }
            // }
            // testcases=sc.nextInt();//For interactive problems;
            outer:
            for(int tt=1;tt<=testcases;tt++)
            {
                solve();
                // YES(solve());
                // Yes(solve());
                // out.println("Case #"+tt+": "+solve());
            }
            out.close();

    }

    static ArrayList<ArrayList<Integer>> graph;
    static ArrayList<ArrayList<Integer>> graph1;
    static ArrayList<ArrayList<Edge>> g1;
    static ArrayList<ArrayList<Edge>> g2;

    static boolean visited[];
    static int color[];
    static ArrayList<ArrayList<Pair>> graphh=new ArrayList<>();
    static Scanner sc=new Scanner(System.in);
    static int parent[];

    static long dp[];
    static long ans=0;
    @SuppressWarnings("unchecked")
    public static void solve()throws Exception
    {
        int n=in.nextInt();
        graph=new ArrayList<>();
        for(int i=0;i<n;++i)graph.add(new ArrayList<>());
        for(int i=0;i<n-1;i++)
        {
            int a=in.nextInt();int b=in.nextInt();
            a--;b--;
            graph.get(a).add(b);
            graph.get(b).add(a);
        }
        dp=new long[n];
        ans=0;
        dfx(0,-1);
        // printArray(dp);

        for(int i=0;i<n;i++)

        {
            ans=max(ans,dp[i]);
            ans=max(ans,graph.get(i).size());

        }
        out.println(ans);
    }

    public static void dfx(int u,int p) throws Exception
    {
        TreeMap<Long,Integer> t=new TreeMap<>();
        for(int child:graph.get(u))
        {
            if(child!=p)
            {
                dfx(child,u);
                dp[u]=max(dp[u],dp[child]+graph.get(u).size()-2);
                addMap(t,dp[child]+graph.get(u).size()-2);
                ans=max(ans,dp[child]+graph.get(u).size()-1);
                if(p==-1)
                {
                    // System.err.println(graph.get(u).size());
                    dp[u]=max(dp[u],dp[child]+graph.get(u).size()-1);
                }
            }
        }
        // if(u!=)
        // if(p==-1)
        // addMap(t,graph.get(u).size());
        // if(u==1)out.println(t);
        dp[u]=max(graph.get(u).size()-1,dp[u]);
        // if(u==0)return;
        for(int child:graph.get(u))
        {
            if(child!=p)
            {

                removeMap(t,dp[child]+graph.get(u).size()-2);
                if(t.size()==0)
                {
                    addMap(t,dp[child]+graph.get(u).size()-2);
                    continue;
                }

                Long x=t.lastKey();
                if(x!=null)
                {
                    // x--;
                    ans=max(ans,graph.get(child).size()-1+x);
                    if(u==1 && child==2)
                    {
                        // out.println(ans+" "+x);
                    }
                }
                addMap(t,dp[child]+graph.get(u).size()-2);
            }
        }

        return;
    }

    public static long abs(long a)
    {
        return Math.abs(a);
    }
    public static int abs(int a)
    {
        return Math.abs(a);
    }
    public static double abs(double a)
    {
        return Math.abs(a);
    }
    public static int first(Integer ar[],int k)
    {
        int n=ar.length;
        if(k>ar[n-1])return n;
        int ans=0;
        int min=0;
        int max=n-1;
        while(max>=min)
        {
            int mid=min+(max-min)/2;
            if(ar[mid]>=k)
            {
                ans=mid;
                max=mid-1;
            }
            else min=mid+1;
        }
        return ans;
    }

    public static int last(Integer ar[],int k)
    {
        if(k<ar[0])
        {
            return 0;
        }
        int n=ar.length;
        if(k>=ar[n-1])return n;
        int ans=n-1;
        int min=0;
        int max=n-1;
        while(max>=min)
        {
            int mid=min+(max-min)/2;

            if(ar[mid]>k)
            {
                ans=mid;
                max=mid-1;
            }
            else min=mid+1;
        }

        return ans;
    }

    public static long gcd(long a,long b)//Note a>b
    {
        if(b==0)return a;
        return gcd(b,a%b);
    }

    public static long lcm(long a,long b)
    {
        return a*b/gcd(a,b);
    }

    public static long pow(long a, long n, long M)
    {
        if (n == 0) return 1;
        long ans = 1;
        a %= M;
        while (n != 0) {
            if ((n & 1) == 1) {
                ans = (ans % M * a % M) % M;
            }
            a = (a % M * a % M) % M;
            n /= 2;
        }
        return ans;
    }

    public static int max(int ... a)
    {
        int max = a[0];
        for (int i : a)
        {
            if (i > max)
            {
                max = i;
            }
        }
        return max;
    }

    public static long max(long ... a)
    {
        long max = a[0];
        for (long i : a) {
            if (i > max) {
                max = i;
            }
        }
        return max;
    }

    public static double max(double ... a)
    {
        double max = a[0];
        for (double i : a) {
            if (i > max) {
                max = i;
            }
        }
        return max;
    }

    // Minimum functions
    public static int min(int ... a)
    {
        int min = a[0];
        for (int i : a) {
            if (i < min) {
                min = i;
            }
        }
        return min;
    }

    public static long min(long ... a)
    {
        long min = a[0];
        for (long i : a) {
            if (i < min) {
                min = i;
            }
        }
        return min;
    }

    public static double min(double ... a)
    {
        double min = a[0];
        for (double i : a) {
            if (i < min) {
                min = i;
            }
        }
        return min;
    }

    public static boolean isSorted(long ar[])
    {
        int n=ar.length;
        for(int i=0;i<n-1;i++)
        {
            if(ar[i]>ar[i+1])return false;
        }
        return true;
    }

    public static int maxIntArray(ArrayList<Integer> ar)
    {
        int max=ar.get(0);
        for(int i=0;i<ar.size();i++)max=Math.max(max,ar.get(i));
        return max;
    }

    private static int countDigits(long l)
    {
        if (l >= 1000000000000000000L) return 19;
        if (l >= 100000000000000000L) return 18;
        if (l >= 10000000000000000L) return 17;
        if (l >= 1000000000000000L) return 16;
        if (l >= 100000000000000L) return 15;
        if (l >= 10000000000000L) return 14;
        if (l >= 1000000000000L) return 13;
        if (l >= 100000000000L) return 12;
        if (l >= 10000000000L) return 11;
        if (l >= 1000000000L) return 10;
        if (l >= 100000000L) return 9;
        if (l >= 10000000L) return 8;
        if (l >= 1000000L) return 7;
        if (l >= 100000L) return 6;
        if (l >= 10000L) return 5;
        if (l >= 1000L) return 4;
        if (l >= 100L) return 3;
        if (l >= 10L) return 2;
        return 1;
    }
    public static int minIntArray(ArrayList<Integer> ar)
    {

        int n=ar.size();
        int max=ar.get(0);
        for(int i=0;i<n;i++)max=Math.min(max,ar.get(i));
        return max;
    }
    public static int maxIntArray(int ar[])
    {
        int max=ar[0];
        for(int i=0;i<ar.length;i++)max=Math.max(max,ar[i]);
        return max;
    }

    public static int minIntArray(int ar[])
    {

        int max=ar[0];
        for(int i=0;i<ar.length;i++)max=Math.min(max,ar[i]);
        return max;
    }

    public static void endl()throws Exception
    {
        out.println();
    }
    public static boolean isSorted(int ar[],int n)
    {
        if(n==1)return true;
        int diff=ar[1]-ar[0];
        for(int i=1;i<n;i++)
        {
            diff=Math.min(diff,ar[i]-ar[i-1]);
        }
        if(diff<0)return false;
        return true;
    }
    public static boolean isSorted(Integer ar[],int n)

    {
        if(n==1)return true;
        int diff=ar[1]-ar[0];
        for(int i=1;i<n;i++)
        {
            diff=Math.min(diff,ar[i]-ar[i-1]);
        }
        if(diff<0)return false;
        return true;
    }

    // public static void addMap(TreeMap<Integer,Integer> map,Integer element)
    // {
    //     if(map.containsKey(element))
    //     map.put(element,map.get(element)+1);
    //     else map.put(element,1);
    // }

    public static <K> void removeMap(Map<K, Integer> map, K element)
    {
        if (!map.containsKey(element)) return;
        int currentCount = map.get(element);
        if (currentCount == 1) {
            map.remove(element);
        } else {
            map.put(element, currentCount - 1);
        }
    }

    public static <K> void addMap(Map<K, Integer> map, K element)
    {
        if (map.containsKey(element)) {
            map.put(element, map.get(element) + 1);
        } else {
            map.put(element, 1);
        }
    }

   // public static void addMap(TreeMap<Integer,Integer> map,Integer element)
    // {
    //     if(map.containsKey(element))
    //     map.put(element,map.get(element)+1);
    //     else map.put(element,1);
    // }
    public static void removeMap(TreeMap<Integer,Integer> map,int element)
    {
        if(map.get(element)==1)map.remove(element);
        else map.put(element,map.get(element)-1);
    }
    public static void removeMap(TreeMap<Long,Integer> map,long element)
    {
        if(!map.containsKey(element))return;
        if(map.get(element)==1)map.remove(element);
        else map.put(element,map.get(element)-1);
    }
    public static void removeMap(HashMap<Long,Integer> map,long element)
    {
        if(!map.containsKey(element))return;
        if(map.get(element)==1)map.remove(element);
        else map.put(element,map.get(element)-1);
    }
    public static void removeMap(HashMap<Integer,Integer> map,int element)
    {
        if(map.get(element)==1)map.remove(element);
        else map.put(element,map.get(element)-1);
    }

    public static void removeMap(HashMap<String,Integer> map,String element)
    {
        if(map.get(element)==1)map.remove(element);
        else map.put(element,map.get(element)-1);
    }
    // public static void addMap(TreeMap<Object,Integer> map,Integer element)
    // {
    //     if(map.containsKey(element))
    //     map.put(element,map.get(element)+1);
    //     else map.put(element,1);
    // }

    public static void addMap(TreeMap<Integer,Integer> map,Integer element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1);
    }

    public static void addMap(TreeMap<String,Integer> map,String element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1);
    }

    public static void addMap(TreeMap<Long,Integer> map,long element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1);
    }
    public static void addMap(HashMap<Integer,Integer> map,int element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1);
    }
    public static void addMap(HashMap<Long,Long> map,long element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1L);
    }
    public static void addMap(HashMap<Object,Integer> map,Integer element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1);
    }
    public static void addMap(HashMap<Long,Integer> map,Long element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1);
    }
    public static void addMap(HashMap<String,Integer> map,String element)
    {
        if(map.containsKey(element))
        map.put(element,map.get(element)+1);
        else map.put(element,1);
    }

    public static int[] nextIntArray(int n)throws Exception
    {
        int ar[]=new int[n];
        for(int i=0;i<n;i++)ar[i]=in.nextInt();
        return ar;
    }
    public static long[] nextlongArray(int n)throws Exception//For long
    {
        long ar[]=new long[n];
        for(int i=0;i<n;i++)ar[i]=in.nextLong();
        return ar;
    }
    public static Integer[] nextIntegerArray(int n)throws Exception
    {
        Integer ar[]=new Integer[n];
        for(int i=0;i<n;i++)ar[i]=in.nextInt();
        return ar;
    }
    public static Long[] nextLongArray(int n)//For Long array
    {
        Long ar[]=new Long[n];
        for(int i=0;i<n;i++)ar[i]=in.nextLong();
        return ar;
    }
    public static String[] nextStringArray(int n)
    {
        String ar[]=new String[n];
        for(int i=0;i<n;i++)ar[i]=in.next();
        return ar;
    }
    public static void Yes(boolean flag)throws Exception
    {
        if(flag)out.println("Yes");
        else  out.println("No");
    }
    public static void YES(boolean flag)throws Exception
    {
        if(flag)out.println("YES");
        else  out.println("NO");
    }
    public static void YES()throws Exception
    {
        out.println("YES");
    }
    public static void NO()throws Exception
    {
        out.println("NO");
    }
    public static void Yes()throws Exception
    {
        out.println("Yes");
    }
    public static void No()throws Exception
    {
        out.println("No");
    }

    public static void printArrayListPair(ArrayList<Pair> al)throws Exception
    {
        for(int i=0;i<al.size();i++)
        {
            out.println(al.get(i).a+" "+al.get(i).b);
        }
    }

    public static void printArray(boolean ar[])throws Exception{
        for(boolean x:ar)out.print(x+" ");
        out.println();
    }
    public static void printArray(char ar[])throws Exception{
        for(char x:ar)out.print(x+" ");
        out.println();
    }

    public static void printArray(int ar[])throws Exception
    {
        for(int x:ar)out.print(x+" ");
        out.println();
    }
    public static void printArray(Object ar[])throws Exception
    {
        for(Object x:ar)out.print(x+" ");
        out.println();
    }
    public static void printArray(long ar[])throws Exception
    {
        for(long x:ar)out.print(x+" ");
        out.println();
    }
    static class FastReader
    {
        BufferedReader br;
        StringTokenizer st;
        public FastReader(String filename) throws FileNotFoundException
        {
            br = new BufferedReader(new FileReader(filename));
        }
        public FastReader()
        {
            br=new BufferedReader(new InputStreamReader(System.in));
        }
        String next()
        {
            while(st==null || !st.hasMoreTokens())
            {
                try
                {
                    st=new StringTokenizer(br.readLine());
                }
                catch (IOException e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt()
        {
            return Integer.parseInt(next());
        }
        long nextLong()
        {
            return Long.parseLong(next());
        }
        double nextDouble()
        {
            return Double.parseDouble(next());
        }
        int[] nextIntArray(int n)
        {
            int ar[]=new int[n];
            for(int i=0;i<n;i++)ar[i]=Integer.parseInt(next());
            return ar;
        }
        long[] nextlongArray(int n)
        {
            long ar[]=new long[n];
            for(int i=0;i<n;i++)ar[i]=Long.parseLong(next());
            return ar;
        }
        String nextLine()
        {
            String str="";
            try
            {
                str=br.readLine().trim();
            } catch (Exception e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
    static class FastWriter
    {
        private final BufferedWriter bw;
        public FastWriter(String filename) throws IOException
        {
            this.bw = new BufferedWriter(new FileWriter(filename));
        }
    public FastWriter()
        {
            this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
        }

    public void print(Object object) throws IOException
        {
            bw.append("" + object);
        }
    public void println(Object object) throws IOException
        {
            print(object);
            bw.append("\n");
        }
    public void println() throws IOException
        {
            bw.append("\n");
        }
    public void close() throws IOException
        {
            bw.close();
        }
    }
}

class Edge
{
    int a,b;long weight;
    Edge(int a,int b,long weight)
    {
        this.a=a;
        this.b=b;
        this.weight=weight;
    }
    Edge(int a,long weight)
    {
        this.a=a;
        this.weight=weight;
    }
}

class compareEdge implements Comparator<Edge>
{
    public int compare(Edge p1,Edge p2)
    {
        return Long.compare(p1.weight, p2.weight);
    }
}

class Pair
{
   int a,b,c;
// long a;
// long b;
    int d;
    boolean flag;
    // Pair(int a,int b,boolean flag)
    // {
    //     this.a=a;
    //     this.b=b;
    //     this.flag=flag;
    // }
    // Pair(long a,long b)
    // {
    //     this.a=a;
    //     this.b=b;
    // }
    // Pair(long a,long b,long c)
    // {
    //     this.a=a;
    //     this.b=b;
    //     this.c=c;
    // }
    // Pair(int a,int b)
    // {
    //     this.a=a;
    //     this.b=b;
    // }
    Pair(int a,int b,int c)
    {
        this.a=a;
        this.b=b;
        this.c=c;
    }
    // Pair (int a,int b,int c)
    // {
    //     this.a=a;
    //     this.b=b;
    //     this.c=c;
    // }

    // Pair (int a,int b,int c,int d)
    // {
    //     this.a=a;
    //     this.b=b;
    //     this.c=c;
    //     this.d=d;
    // }
    @Override
    public String toString() {
        // return a + " " + b + " " + c;
        return a+" "+b;
    }

    public boolean equals(Object obj) {
        if (this == obj) return true;
        if (obj == null || getClass() != obj.getClass()) return false;
        Pair pair = (Pair) obj;
        // return a == pair.a && b == pair.b && c == pair.c && d == pair.d && flag == pair.flag;

        return a==pair.a && b==pair.b;
    }

    // @Override
    // public int hashCode() {
    //     int result = a;
    //     result = 31 * result + b;
    //     result = 31 * result + c;
    //     result = 31 * result + d;
    //     result = 31 * result + (flag ? 1 : 0);
    //     return result;
    // }
}

class compare1 implements Comparator<Pair>
{
    public int compare(Pair p1,Pair p2)
    {
        // if(p1.a==p2.a && p2.b==p1.b)return 0;
        // return p2.a-p1.a;//decreasing
        // return p1.c-p2.c;
        // if(p1.a!=p2.a)
        // return p1.a-p2.a;//increasing
        // if(p1.a==p2.b)return p1.b-p2.b;

        return p1.a-p2.a;
        // return p1.b-p2.b;
        // return Long.compare(p1.a,p2.a);

        // if(p1.a!=p2.a)
        // {
        //     return p1.a-p2.a;
        // }
        // else
        // return p1.b-p2.b;

        // return Long.compare((p2.a+1)/2*p2.b ,(p1.a+1)/2*p1.b);

        // return p2.a-p1.b;
        // return p1.a-p2.a;
        // return p2.c-p1.c;
        // else
        // return p1.b-p2.b;
        // if(p1.a==p2.a)
        // {
        //     return p2.b-p1.b;
        // }
        // else return +p1.a-p2.a;
        // if(p1.a!=p2.a)
        //     return p1.a-p2.a;
        // else
        //     return p2.b-p1.b;

        // if(p1.a!=p2.a)
        // {
        //     return p2.a-p1.a;
        // }
        // else
        // {
            // return p1.b-p2.b;
        // }
    }

    public boolean equals(Pair p1,Pair p2)
    {
        return p1.a==p2.a && p2.b==p1.b;
    }
}

class MO

{
    int a,b;
    int c;
    static int size;

    MO(int a,int b,int c)
    {
        this.a=a;
        this.b=b;
        this.c=c;
    }
}
class compareMO implements Comparator<MO>
{
    public int compare(MO a,MO b)
    {
        int size=MO.size;
        if(a.a/size!=b.a/size)
        {
            return a.a/size-b.a/size;
        }
        return a.b-b.b;
    }
}

class SegmentTree {
    int[] tree;     // Segment tree array
    int[] arr;      // Original array
    int n;          // Size of the original array

    // Constructor to build the segment tree
    public SegmentTree(int[] input) {
        n = input.length;
        arr = input;
        // Allocate memory for segment tree
        // The maximum size of segment tree array is 2*n - 1 for n leaf nodes
        tree = new int[4 * n];
        build(0, 0, n - 1);
    }

    // Build the segment tree
    private void build(int node, int start, int end) {
        if (start == end) {
            // Leaf node: store the value of the array element
            tree[node] = arr[start];
        } else {
            int mid = (start + end) / 2;
            // Recursively build the left and right subtrees
            build(2 * node + 1, start, mid);
            build(2 * node + 2, mid + 1, end);
            // Internal node: store the minimum of the two children
            tree[node] = Math.min(tree[2 * node + 1], tree[2 * node + 2]);
        }
    }

    // Query for the minimum in range [L, R]
    public int query(int L, int R) {
        return query(0, 0, n - 1, L, R);
    }

    private int query(int node, int start, int end, int L, int R) {
        if (R < start || L > end) {
            // Range [L, R] is completely outside the range [start, end]
            return Integer.MAX_VALUE;
        }
        if (L <= start && end <= R) {
            // Range [L, R] completely covers the range [start, end]
            return tree[node];
        }
        // Partial overlap
        int mid = (start + end) / 2;
        int leftMin = query(2 * node + 1, start, mid, L, R);
        int rightMin = query(2 * node + 2, mid + 1, end, L, R);
        return Math.min(leftMin, rightMin);
    }

    // Update an element in the array
    public void update(int index, int value) {
        update(0, 0, n - 1, index, value);
    }

    private void update(int node, int start, int end, int index, int value) {
        if (start == end) {
            // Leaf node: update the value of the array element
            arr[index] = value;
            tree[node] = value;
        } else {
            int mid = (start + end) / 2;
            if (index <= mid) {
                // Update the left child
                update(2 * node + 1, start, mid, index, value);
            } else {
                // Update the right child
                update(2 * node + 2, mid + 1, end, index, value);
            }
            // Update the current node after updating the children
            tree[node] = Math.min(tree[2 * node + 1], tree[2 * node + 2]);
        }
    }
}

class GCDSegmentTree {
    private int[] tree;
    private int[] arr;

    public GCDSegmentTree(int[] inputArr) {
        int n = inputArr.length;
        arr = inputArr;
        int treeSize = 4 * n;
        tree = new int[treeSize];
        buildTree(0, 0, n - 1);
    }

    // Build segment tree
    private void buildTree(int node, int start, int end) {
        if (start == end) {
            tree[node] = arr[start];  // Leaf node
        } else {
            int mid = (start + end) / 2;
            buildTree(2 * node + 1, start, mid);   // Left child
            buildTree(2 * node + 2, mid + 1, end); // Right child
            tree[node] = gcd(tree[2 * node + 1], tree[2 * node + 2]); // Combine GCD
        }
    }

    // GCD Query
    private int query(int node, int start, int end, int L, int R) {
        if (R < start || end < L) {
            return 0;  // Out of range
        }
        if (L <= start && end <= R) {
            return tree[node];  // Current segment fully in range
        }
        int mid = (start + end) / 2;
        int leftGCD = query(2 * node + 1, start, mid, L, R);
        int rightGCD = query(2 * node + 2, mid + 1, end, L, R);
        return gcd(leftGCD, rightGCD);  // Combine GCD from left and right children
    }

    public int getGCD(int L, int R) {
        return query(0, 0, arr.length - 1, L, R);
    }

    // Utility GCD function
    private int gcd(int a, int b) {
        if (b == 0) return a;
        return gcd(b, a % b);
    }

}

class DSU {
    private int[] parent;
    private int[] rank;
    private int[] size; // Array to track the size of each set
    private int groups; // Counter for the number of groups

    // Constructor to initialize the DSU
    public DSU(int n) {
        parent = new int[n];
        rank = new int[n];
        size = new int[n];
        groups = n; // Initially, every element is its own group

        // Initialize each node as its own parent (self-loop)
        for (int i = 0; i < n; i++) {
            parent[i] = i;
            rank[i] = 0; // Initial rank is 0
            size[i] = 1; // Each set initially contains only one element
        }
    }

    // Find operation with path compression
    public int find(int x) {
        if (parent[x] != x) {
            parent[x] = find(parent[x]); // Path compression
        }
        return parent[x];
    }

    // Union operation with union by rank
    public void union(int x, int y) {
        int rootX = find(x);
        int rootY = find(y);

        if (rootX != rootY) {
            if (rank[rootX] > rank[rootY]) {
                parent[rootY] = rootX;
                size[rootX] += size[rootY]; // Update size of rootX
            } else if (rank[rootX] < rank[rootY]) {
                parent[rootX] = rootY;
                size[rootY] += size[rootX]; // Update size of rootY
            } else {
                parent[rootY] = rootX;
                size[rootX] += size[rootY]; // Update size of rootX
                rank[rootX]++;
            }
            groups--; // Decrease the number of groups as two groups are merged
        }
    }

    // Check if two elements are in the same set
    public boolean isConnected(int x, int y) {
        return find(x) == find(y);
    }

    // Get the size of the set containing the element x
    public int size(int x) {
        return size[find(x)];
    }

    // Get the number of groups
    public int numberOfGroups() {
        return groups;
    }

}

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

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

Think step by step.

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

1. **Edge Cases:**
   - Trees with only two vertices.
   - Trees with a linear structure (a single path).
   - Trees with a star structure (one central vertex connected to all others).

2. **Complex Structures:**
   - Trees with multiple branches and deep structures.
   - Trees with multiple leaves and internal nodes.

3. **Special Cases:**
   - Trees where the optimal solution involves removing a path that includes the root node.
   - Trees where the optimal solution involves removing a path that includes a leaf node.

Given these considerations, let's create a small-scale test input that includes a mix of these cases to maximize the failure probability in faulty solutions.

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

This test input includes:
- A tree with only two vertices.
- A linear tree with five vertices.
- A linear tree with four vertices.
- A star tree with five vertices.
- A tree with multiple branches and leaves.
- A tree with a more complex structure.

Each of these cases can help identify potential issues in the algorithm, such as incorrect handling of paths, edge cases, and complex tree structures.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Destruction
time_limit_ms: 2000
memory_limit_mb: 256
Description: Given a tree$$$^{\text{∗}}$$$ with $$$n$$$ vertices. You can choose two vertices $$$a$$$ and $$$b$$$ once and remove all vertices on the path from $$$a$$$ to $$$b$$$, including the vertices themselves. If you choose $$$a=b$$$, only one vertex will be removed.

Your task is to find the maximum number of connected components$$$^{\text{†}}$$$ that can be formed after removing the path from the tree.

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

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

The next $$$n-1$$$ lines contain two integers $$$u$$$ and $$$v$$$ ($$$1 \le u, v \le n$$$, $$$u \ne v$$$) — the vertices connected by an edge. It is guaranteed that the edges form a tree.

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

Output: For each test case, output one integer — the maximum number of connected components that can be achieved using the described operation.

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

Note:

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class G_Tree_Destruction {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        int t = Integer.parseInt(br.readLine().trim());
        long[][] arr;
        for (int i = 0; i < t; i++) {
            String[] inputs = br.readLine().split(" ");
            long n = Long.parseLong(inputs[0]);
            arr = new long[(int) n - 1][2];
            for (int j = 0; j < n - 1; j++) {
                inputs = br.readLine().split(" ");
                arr[j][0] = Long.parseLong(inputs[0]);
                arr[j][1] = Long.parseLong(inputs[1]);
            }
            System.out.println(helper(arr, n));
        }
    }

    private static long helper(long[][] arr, long n) {
        ArrayList<Long>[] g = new ArrayList[(int) n];
        for(long[] e : arr) {
            long a = e[0] - 1;
            long b = e[1] - 1;
            if(g[(int) a] == null) {
                g[(int) a] = new ArrayList<>();
            }
            ArrayList<Long> li = g[(int) a];
            li.add(b);
            if(g[(int) b] == null) {
                g[(int) b] = new ArrayList<>();
            }
            li = g[(int) b];
            li.add(a);
        }
        long[] dp1 = new long[(int) n];
        long[] dp2 = new long[(int) n];
        dp1[0] = helper1(0, dp1, dp2, g, new boolean[(int) n]);
        long ans = 0;
        for(int i=0;i<n;i++) {
            ans = Math.max(ans, dp1[i]);
            ans = Math.max(ans, dp2[i]);
        }
        return ans;
    }

    private static long helper1(long c, long[] dp1, long[] dp2, ArrayList<Long>[] g, boolean[] v) {
        v[(int) c] = true;
        long max1 = -1;
        long max2 = -1;
        long temp = 0;
        for(long n : g[(int) c]) {
            if(!v[(int) n]) {
                temp = helper1(n, dp1, dp2, g, v);
                if(temp > max1) {
                    max2 = max1;
                    max1 = temp;
                } else {
                    max2 = temp;
                }
            }
        }
        long degree = g[(int) c].size();
        dp1[(int) c] = Math.max(max1 - 2, 0) + degree;
        if(max2 > -1) {
            dp2[(int) c] = max1 + max2 + degree - 4;
        } else {
            dp2[(int) c] = 0;
        }
        return dp1[(int) c];
    }

    // private static long helper2(long c, long[] dp1, ArrayList<Long>[] g, boolean[] v, ArrayDeque<Long> q) {
    //     long max1 = -1;
    //     long max2 = -1;
    //     if(g[(int) c].size() < 2) {
    //         return 0;
    //     }
    //     for(long n : g[(int) c]) {
    //         if(!v[(int) n] && dp1[(int) n] > max2) {
    //             q.offer(n);
    //             v[(int) n] = true;
    //             if(dp1[(int) n] > max1) {
    //                 max2 = max1;
    //                 max1 = dp1[(int) n];
    //             } else {
    //                 max2 = dp1[(int) n];
    //             }
    //         }
    //     }
    //     // System.out.println("for " + c + " we get " + (max1 + max2 + g[(int) c].size() - 4));
    //     return max1 + max2 + g[(int) c].size() - 4;
    // }
}

// import java.util.*;

// class G_Tree_Destruction {
//     static class Pair {
//         int x, y;
//         Pair(int x, int y) {
//             this.x = x;
//             this.y = y;
//         }
//     }

//     static void dfs(int v, int p, List<List<Integer>> sl, List<Pair> dp) {
//         dp.get(v).x = sl.get(v).size();
//         int m1 = -1, m2 = -1;

//         for (int u : sl.get(v)) {
//             if (u == p) continue;

//             dfs(u, v, sl, dp);
//             dp.get(v).x = Math.max(dp.get(v).x, dp.get(u).x + sl.get(v).size() - 2);
//             m2 = Math.max(m2, dp.get(u).x);

//             if (m1 < m2) {
//                 int temp = m1;
//                 m1 = m2;
//                 m2 = temp;
//             }
//         }

//         dp.get(v).y = dp.get(v).x;
//         if (m2 != -1) {
//             dp.get(v).y = m1 + m2 + sl.get(v).size() - 4;
//         }
//     }

//     static void solve(Scanner sc) {
//         int n = sc.nextInt();
//         List<List<Integer>> sl = new ArrayList<>();
//         for (int i = 0; i < n; i++) {
//             sl.add(new ArrayList<>());
//         }

//         for (int i = 1; i < n; i++) {
//             int u = sc.nextInt() - 1;
//             int v = sc.nextInt() - 1;
//             sl.get(u).add(v);
//             sl.get(v).add(u);
//         }

//         List<Pair> dp = new ArrayList<>();
//         for (int i = 0; i < n; i++) {
//             dp.add(new Pair(0, 0));
//         }

//         dfs(0, 0, sl, dp);

//         int ans = 0;
//         for (int i = 0; i < n; i++) {
//             ans = Math.max(ans, Math.max(dp.get(i).x, dp.get(i).y));
//         }

//         System.out.println(ans);
//     }

//     public static void main(String[] args) {
//         Scanner sc = new Scanner(System.in);
//         boolean multi = true;
//         int t = multi ? sc.nextInt() : 1;

//         for (int i = 1; i <= t; i++) {
//             solve(sc);
//             // System.out.println();
//         }

//         sc.close();
//     }
// }

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

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

Think step by step.

Response:
```plaintext
1
5
1 2
2 3
3 4
4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Game
time_limit_ms: 2000
memory_limit_mb: 512
Description: In the most popular card game in Berland, a deck of $n \times m$ cards is used. Each card has two parameters: suit and rank. Suits in the game are numbered from $1$ to $n$, and ranks are numbered from $1$ to $m$. There is exactly one card in the deck for each combination of suit and rank.

A card with suit $a$ and rank $b$ can beat a card with suit $c$ and rank $d$ in one of two cases:

  * $a = 1$, $c \ne 1$ (a card of suit $1$ can beat a card of any other suit);    * $a = c$, $b > d$ (a card can beat any other card of the same suit but of a lower rank).

Two players play the game. Before the game starts, they receive exactly half of the deck each. The first player wins if for every card of the second player, he can choose his card that can beat it, and there is no card that is chosen twice (i. e. there exists a matching of the first player's cards with the second player's cards such that in each pair the first player's card beats the second player's card). Otherwise, the second player wins.

Your task is to calculate the number of ways to distribute the cards so that the first player wins. Two ways are considered different if there exists a card such that in one way it belongs to the first player and in the other way it belongs to the second player. The number of ways can be very large, so print it modulo $998244353$.

Input: The only line contains two integers $n$ and $m$ ($1 \le n, m \le 500$).

Additional constraint on the input: $m$ is even.

Output: Print a single integer — the number of ways to distribute the cards so that the first player wins, taken modulo $998244353$.

Examples:
input:
1 4
output:
2

input:
2 2
output:
2

input:
3 6
output:
1690

input:
5 4
output:
568

input:
500 500
output:
84693741

Note:

Buggy Code:
#include <bits/stdc++.h>
#define all(v) v.begin(),v.end()
#define F first
#define S second
using namespace std;
typedef int ll;
typedef pair<ll,ll>pll;
const ll oo = 509;
const ll mod=998244353;
ll add(ll x,ll y)
{
    if(x+y<mod)
        return x+y;
    return x+y-mod;
}
ll sub(ll x,ll y)
{
    if(x>=y)
        return x-=y;
    return x-y+mod;
}
ll mul(ll x,ll y)
{
    return 1ll*x*y%mod;
}
ll n,m,dp[oo][oo][oo],a[oo],b[oo],op[oo][oo];
ll best(ll rem,ll i)
{
    if(i>n)
        return rem==0;
    ll &r = op[rem][i];
    if(r!=-1)
        return r;
    r=0;
    for(ll j=0; j<=rem; j+=2)
        r=add(r, mul(best(rem-j,i+1), a[m-j]));
    return r;
}
void solve()
{
    cin>>n>>m;
    dp[0][0][0]=1;
    for(ll i=1; i<=m; i++)
        for(ll j=0; j<=m; j++)
        {
            for(ll t=0; t<=m; t++)
            {
                dp[t][i][j] = dp[t][i-1][j+1];
                if(j&&t)
                    dp[t][i][j] = add(dp[t][i][j], dp[t-1][i-1][j-1]);
            }
        }
    for(ll i=0; i<=m; i+=2)
        a[i] = dp[m/2][i][0];
    for(ll i=0; i<=m; i++)
        for(ll t=0; t<=m; t++)
            b[i] = add(b[i],dp[t][m][i]);
    memset(op,-1,sizeof op);
    ll ans=0;
    for(ll i=0; i<=m; i++)
        ans = add(ans, mul(best(i,2),b[i]));
    cout<<ans;
}
int main()
{
    ios_base::sync_with_stdio(0),cin.tie(0);
    ll tt=1;
//    cin>>tt;
    while(tt--)
        solve();
    return 0;
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and identify potential edge cases or scenarios that could cause issues.

### Analysis of the Buggy Code

1. **Initialization and Base Cases:**
   - The `dp` array is initialized, and the base case `dp[0][0][0] = 1` is set.
   - The `a` array is used to store intermediate results for even values of `i`.

2. **Dynamic Programming:**
   - The `dp` array is updated in a nested loop, where the states are:
     - `t`: number of cards of suit 1 that the first player has.
     - `i`: current rank.
     - `j`: number of cards of the current rank that the first player has.
   - The transitions involve adding cards of the current rank to the first player or not.

3. **Best Function:**
   - The `best` function recursively calculates the number of ways to distribute the remaining cards such that the first player wins.
   - It uses memoization to store results in the `op` array.

4. **Final Calculation:**
   - The final answer is calculated by summing the results of the `best` function for different values of `i`.

### Potential Edge Cases

1. **Small Values of `n` and `m`:**
   - Small values can help in understanding the basic logic and transitions.
   - Example: `n = 2`, `m = 2`.

2. **Large Values of `n` and `m`:**
   - Large values can test the efficiency and correctness of the dynamic programming approach.
   - Example: `n = 500`, `m = 500`.

3. **Values with Specific Constraints:**
   - Values where `m` is even and the distribution of cards is critical.
   - Example: `n = 3`, `m = 6`.

### Chosen Test Input

To maximize the failure probability, we will use a small-scale test input that is complete and valid. The chosen test input is:

```plaintext
3 6
```

This input is chosen because:
- `n = 3` and `m = 6` (with `m` being even) provides a manageable number of cards to test the logic.
- The distribution of cards is non-trivial, making it a good test for the dynamic programming and recursive functions.

### Output

```plaintext
3 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Game
time_limit_ms: 2000
memory_limit_mb: 512
Description: In the most popular card game in Berland, a deck of $n \times m$ cards is used. Each card has two parameters: suit and rank. Suits in the game are numbered from $1$ to $n$, and ranks are numbered from $1$ to $m$. There is exactly one card in the deck for each combination of suit and rank.

A card with suit $a$ and rank $b$ can beat a card with suit $c$ and rank $d$ in one of two cases:

  * $a = 1$, $c \ne 1$ (a card of suit $1$ can beat a card of any other suit);    * $a = c$, $b > d$ (a card can beat any other card of the same suit but of a lower rank).

Two players play the game. Before the game starts, they receive exactly half of the deck each. The first player wins if for every card of the second player, he can choose his card that can beat it, and there is no card that is chosen twice (i. e. there exists a matching of the first player's cards with the second player's cards such that in each pair the first player's card beats the second player's card). Otherwise, the second player wins.

Your task is to calculate the number of ways to distribute the cards so that the first player wins. Two ways are considered different if there exists a card such that in one way it belongs to the first player and in the other way it belongs to the second player. The number of ways can be very large, so print it modulo $998244353$.

Input: The only line contains two integers $n$ and $m$ ($1 \le n, m \le 500$).

Additional constraint on the input: $m$ is even.

Output: Print a single integer — the number of ways to distribute the cards so that the first player wins, taken modulo $998244353$.

Examples:
input:
1 4
output:
2

input:
2 2
output:
2

input:
3 6
output:
1690

input:
5 4
output:
568

input:
500 500
output:
84693741

Note:

Buggy Code:
#include <iostream>
#include <vector>
using namespace std;
int main()
{
    int n,m;
	cin >> n >> m;
	vector<vector<long long>> dp1(m+1, vector<long long>(m+1, 0));
	vector<vector<long long>> dp2(n+1, vector<long long>(m+1, 0));
	int MOD = 998244353;
	dp1[1][1] = 1;
	for(int i = 2;i <= m;i++)
	{
		dp1[i][0] = dp1[i - 1][1];
		for(int k = 1;k <= i;k++)
		{
			dp1[i][k] = dp1[i - 1][k + 1] + dp1[i - 1][k - 1];
		    dp1[i][k] %= MOD;
		}
	}
	if(n == 1)
	{
		cout << dp1[m][0];
		return 0;
	}
	for(int i = 2;i <= n;i++)
	{
		if(i == 2)
		{
			for(int k = m;k >= 0;k--)
			{
				dp2[i][k] = dp1[m][k];
			}
			continue;
		}
		for(int k = m;k >= 0;k--)
		{
			for(int j = k;j >= 0;j -= 2)
			{
				dp2[i][k] += dp2[i - 1][k - j] * dp1[m][j];
				dp2[i][k] %= MOD;
			}
		}
	}
	long long ans = 0;
	for(int i = 0;i <= m;i++)
		ans += dp1[m][i] * dp2[n][i],ans %= MOD;
	cout << ans;
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
2 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Game
time_limit_ms: 2000
memory_limit_mb: 512
Description: In the most popular card game in Berland, a deck of $n \times m$ cards is used. Each card has two parameters: suit and rank. Suits in the game are numbered from $1$ to $n$, and ranks are numbered from $1$ to $m$. There is exactly one card in the deck for each combination of suit and rank.

A card with suit $a$ and rank $b$ can beat a card with suit $c$ and rank $d$ in one of two cases:

  * $a = 1$, $c \ne 1$ (a card of suit $1$ can beat a card of any other suit);    * $a = c$, $b > d$ (a card can beat any other card of the same suit but of a lower rank).

Two players play the game. Before the game starts, they receive exactly half of the deck each. The first player wins if for every card of the second player, he can choose his card that can beat it, and there is no card that is chosen twice (i. e. there exists a matching of the first player's cards with the second player's cards such that in each pair the first player's card beats the second player's card). Otherwise, the second player wins.

Your task is to calculate the number of ways to distribute the cards so that the first player wins. Two ways are considered different if there exists a card such that in one way it belongs to the first player and in the other way it belongs to the second player. The number of ways can be very large, so print it modulo $998244353$.

Input: The only line contains two integers $n$ and $m$ ($1 \le n, m \le 500$).

Additional constraint on the input: $m$ is even.

Output: Print a single integer — the number of ways to distribute the cards so that the first player wins, taken modulo $998244353$.

Examples:
input:
1 4
output:
2

input:
2 2
output:
2

input:
3 6
output:
1690

input:
5 4
output:
568

input:
500 500
output:
84693741

Note:

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

using namespace std;

typedef long long ll;

const ll MOD = 998244353;
ll n, m;

ll ad(ll a, ll b) {
    return ((a%MOD) + (b%MOD))%MOD;
}

ll sub(ll a, ll b) {
    return (a-b+MOD)%MOD;
}

ll mult(ll a, ll b) {
    return ((a%MOD) * (b%MOD))%MOD;
}

int main() {
    cin >> n >> m;

    // first count walks from (0,0) to (a,b) above y=x
    // and  a + b = m
    ll walk[m+n+5][m+n+5];
    walk[0][0] = 1;

    for(int d=1;d<=m+n;d++) {
        // d is x+y
        for(int x=0;x<=d;x++) {
            int y = d - x;

            walk[x][y] = 0;
            if(x > y) continue;

            // (x, y) comes from (x-1,y) and (x,y-1)
            if(x-1 <= y && x-1 >= 0) {
                walk[x][y] = ad(walk[x][y], walk[x-1][y]);
            }
            if(x <= y-1 && y-1 >= 0) {
                walk[x][y] = ad(walk[x][y], walk[x][y-1]);
            }
        }
    }

    // walk[i][i] is i-th Catalan number

    //  Ok so now let  dp[a][b] mean that group 1 has a left (wlog rk 1...a of m)
    //  a = m mod 2 actually... which must be even right???
    //   And  b groups (OTHER than group 1) left
    // so    answer is    dp[m][n-1]
    // base cases   dp[0][0] = 1
    //     dp[1][0] = 0
    //     dp[2][0] = 1
    // dp[3][0] = 1
    //  dp[4][0] (two pairs)
    // dp[2i][0] = C_i   and dp[2i-1][0] = 0

    ll dp[m+n+5][m+n+5];

    for(int i=0;i<=m;i++) {
        if(i % 2 == 0) dp[i][0] = walk[i/2][m-i/2];
        else dp[i][0] = 0;
    }

    // dp[a][b] =  dp[a][b-1]    dp[a-2][b-1]   dp[a-4][b-1] ...

    for(int b=1;b<=n-1;b++) {
        for(int a=0;a<=m;a++) {
            for(int k=0;a-2*k>=0;k++) {
                dp[a][b] = ad(dp[a][b],  mult( dp[a-2*k][b-1], walk[(m/2)-k][(m/2)+k] ) );
            }
        }
    }

    cout << dp[m][n-1] << endl;
}

**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 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Game
time_limit_ms: 2000
memory_limit_mb: 512
Description: In the most popular card game in Berland, a deck of $n \times m$ cards is used. Each card has two parameters: suit and rank. Suits in the game are numbered from $1$ to $n$, and ranks are numbered from $1$ to $m$. There is exactly one card in the deck for each combination of suit and rank.

A card with suit $a$ and rank $b$ can beat a card with suit $c$ and rank $d$ in one of two cases:

  * $a = 1$, $c \ne 1$ (a card of suit $1$ can beat a card of any other suit);    * $a = c$, $b > d$ (a card can beat any other card of the same suit but of a lower rank).

Two players play the game. Before the game starts, they receive exactly half of the deck each. The first player wins if for every card of the second player, he can choose his card that can beat it, and there is no card that is chosen twice (i. e. there exists a matching of the first player's cards with the second player's cards such that in each pair the first player's card beats the second player's card). Otherwise, the second player wins.

Your task is to calculate the number of ways to distribute the cards so that the first player wins. Two ways are considered different if there exists a card such that in one way it belongs to the first player and in the other way it belongs to the second player. The number of ways can be very large, so print it modulo $998244353$.

Input: The only line contains two integers $n$ and $m$ ($1 \le n, m \le 500$).

Additional constraint on the input: $m$ is even.

Output: Print a single integer — the number of ways to distribute the cards so that the first player wins, taken modulo $998244353$.

Examples:
input:
1 4
output:
2

input:
2 2
output:
2

input:
3 6
output:
1690

input:
5 4
output:
568

input:
500 500
output:
84693741

Note:

Buggy Code:
MOD = 998244353

def mod_inverse(x):
    return pow(x, MOD - 2, MOD)

def comb(n, k):
    if k > n:
        return 0
    num = den = 1
    for i in range(k):
        num = num * (n - i) % MOD
        den = den * (i + 1) % MOD
    return num * mod_inverse(den) % MOD

def factorial(n):
    result = 1
    for i in range(2, n + 1):
        result = result * i % MOD
    return result

def count_ways(n, m):
    if n == 1:
        # Special case when n = 1
        return 2

    total_cards = n * m
    first_player_cards = total_cards // 2
    second_player_cards = total_cards // 2

    # Number of ways to choose 'k' cards of suit 1 for the first player
    ways_to_choose_suit_1 = 0
    for k in range(1, min(m, first_player_cards) + 1):
        # Choose 'k' cards of suit 1 for the first player
        ways_to_choose_suit_1 += comb(m, k)

    # Remaining cards excluding suit 1
    remaining_cards = (n - 1) * m

    # Calculate the number of valid distributions of remaining cards
    valid_distributions = 1
    for suit in range(2, n + 1):
        for rank in range(1, m + 1):
            # Cards of this rank for this suit
            cards_of_rank = m // 2
            if rank <= m // 2:
                valid_distributions *= comb(cards_of_rank, cards_of_rank)
            else:
                valid_distributions *= comb(cards_of_rank, cards_of_rank - 1)
            valid_distributions %= MOD

    # Total ways
    total_ways = ways_to_choose_suit_1 * valid_distributions % MOD

    return total_ways

# Read input
n, m = map(int, input().split())

# Output the result
print(count_ways(n, m))

**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 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Game
time_limit_ms: 2000
memory_limit_mb: 512
Description: In the most popular card game in Berland, a deck of $n \times m$ cards is used. Each card has two parameters: suit and rank. Suits in the game are numbered from $1$ to $n$, and ranks are numbered from $1$ to $m$. There is exactly one card in the deck for each combination of suit and rank.

A card with suit $a$ and rank $b$ can beat a card with suit $c$ and rank $d$ in one of two cases:

  * $a = 1$, $c \ne 1$ (a card of suit $1$ can beat a card of any other suit);    * $a = c$, $b > d$ (a card can beat any other card of the same suit but of a lower rank).

Two players play the game. Before the game starts, they receive exactly half of the deck each. The first player wins if for every card of the second player, he can choose his card that can beat it, and there is no card that is chosen twice (i. e. there exists a matching of the first player's cards with the second player's cards such that in each pair the first player's card beats the second player's card). Otherwise, the second player wins.

Your task is to calculate the number of ways to distribute the cards so that the first player wins. Two ways are considered different if there exists a card such that in one way it belongs to the first player and in the other way it belongs to the second player. The number of ways can be very large, so print it modulo $998244353$.

Input: The only line contains two integers $n$ and $m$ ($1 \le n, m \le 500$).

Additional constraint on the input: $m$ is even.

Output: Print a single integer — the number of ways to distribute the cards so that the first player wins, taken modulo $998244353$.

Examples:
input:
1 4
output:
2

input:
2 2
output:
2

input:
3 6
output:
1690

input:
5 4
output:
568

input:
500 500
output:
84693741

Note:

Buggy Code:
MOD = 998244353

# Function to compute factorial modulo MOD
def factorial_mod(n, MOD):
    fact = [1] * (n + 1)
    for i in range(2, n + 1):
        fact[i] = fact[i-1] * i % MOD
    return fact

# Function to compute inverse factorial modulo MOD
def inverse_factorial_mod(fact, MOD):
    inv_fact = [1] * (len(fact))
    inv_fact[-1] = pow(fact[-1], MOD-2, MOD)  # Fermat's Little Theorem for modular inverse
    for i in range(len(fact) - 2, -1, -1):
        inv_fact[i] = inv_fact[i+1] * (i+1) % MOD
    return inv_fact

# Function to compute binomial coefficient modulo MOD
def binomial_mod(n, k, fact, inv_fact, MOD):
    if k > n or k < 0:
        return 0
    return fact[n] * inv_fact[k] % MOD * inv_fact[n-k] % MOD

def solve(n, m):
    if m % 2 != 0:
        return 0  # Since m must be even

    # Precompute factorials and inverse factorials modulo MOD
    fact = factorial_mod(m, MOD)
    inv_fact = inverse_factorial_mod(fact, MOD)

    # The number of ways to split the ranks of a suit where player 1 always wins
    rank_ways = binomial_mod(m, m // 2, fact, inv_fact, MOD)
    rank_ways = rank_ways * fact[m // 2] % MOD * fact[m // 2] % MOD

    # Total ways = pow(rank_ways, (n-1), MOD) * 2 % MOD (since suit 1 always wins)
    total_ways = pow(rank_ways, n - 1, MOD)  # For suits other than 1
    total_ways = total_ways * 2 % MOD       # Multiply by 2 for Suit 1 rule dominance

    print(total_ways)

# Read input
n, m = map(int, input().split())

# Output the result
solve(n, m)

**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 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Game
time_limit_ms: 2000
memory_limit_mb: 512
Description: In the most popular card game in Berland, a deck of $n \times m$ cards is used. Each card has two parameters: suit and rank. Suits in the game are numbered from $1$ to $n$, and ranks are numbered from $1$ to $m$. There is exactly one card in the deck for each combination of suit and rank.

A card with suit $a$ and rank $b$ can beat a card with suit $c$ and rank $d$ in one of two cases:

  * $a = 1$, $c \ne 1$ (a card of suit $1$ can beat a card of any other suit);    * $a = c$, $b > d$ (a card can beat any other card of the same suit but of a lower rank).

Two players play the game. Before the game starts, they receive exactly half of the deck each. The first player wins if for every card of the second player, he can choose his card that can beat it, and there is no card that is chosen twice (i. e. there exists a matching of the first player's cards with the second player's cards such that in each pair the first player's card beats the second player's card). Otherwise, the second player wins.

Your task is to calculate the number of ways to distribute the cards so that the first player wins. Two ways are considered different if there exists a card such that in one way it belongs to the first player and in the other way it belongs to the second player. The number of ways can be very large, so print it modulo $998244353$.

Input: The only line contains two integers $n$ and $m$ ($1 \le n, m \le 500$).

Additional constraint on the input: $m$ is even.

Output: Print a single integer — the number of ways to distribute the cards so that the first player wins, taken modulo $998244353$.

Examples:
input:
1 4
output:
2

input:
2 2
output:
2

input:
3 6
output:
1690

input:
5 4
output:
568

input:
500 500
output:
84693741

Note:

Buggy Code:
MOD = 998244353
N = 500 + 5
factorial = [1] * N
invFactorial = [1] * N
for i in range(1, N):
    factorial[i] = factorial[i - 1] * i % MOD
    invFactorial[i] = pow(factorial[i], MOD - 2, MOD)
def comb(n, k):
    return factorial[n] * invFactorial[k] * invFactorial[n - k] % MOD
dp = [[0]*N for _ in range(N)]
dp[0][0] = 1
for i in range(1, N):
    dp[i][0] = dp[i-1][0]
    for j in range(1, i+1):
        dp[i][j] = (dp[i-1][j] + dp[i-1][j-1]) % MOD
n, m = map(int, input().split())
ans = 0
for i in range((m >> 1) + 1):
    ans = (ans + comb(n, i) * dp[n*(m>>1)-1][n*i]) % MOD
print(ans)

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

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

Think step by step.

Response:
```plaintext
3 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Game
time_limit_ms: 2000
memory_limit_mb: 512
Description: In the most popular card game in Berland, a deck of $n \times m$ cards is used. Each card has two parameters: suit and rank. Suits in the game are numbered from $1$ to $n$, and ranks are numbered from $1$ to $m$. There is exactly one card in the deck for each combination of suit and rank.

A card with suit $a$ and rank $b$ can beat a card with suit $c$ and rank $d$ in one of two cases:

  * $a = 1$, $c \ne 1$ (a card of suit $1$ can beat a card of any other suit);    * $a = c$, $b > d$ (a card can beat any other card of the same suit but of a lower rank).

Two players play the game. Before the game starts, they receive exactly half of the deck each. The first player wins if for every card of the second player, he can choose his card that can beat it, and there is no card that is chosen twice (i. e. there exists a matching of the first player's cards with the second player's cards such that in each pair the first player's card beats the second player's card). Otherwise, the second player wins.

Your task is to calculate the number of ways to distribute the cards so that the first player wins. Two ways are considered different if there exists a card such that in one way it belongs to the first player and in the other way it belongs to the second player. The number of ways can be very large, so print it modulo $998244353$.

Input: The only line contains two integers $n$ and $m$ ($1 \le n, m \le 500$).

Additional constraint on the input: $m$ is even.

Output: Print a single integer — the number of ways to distribute the cards so that the first player wins, taken modulo $998244353$.

Examples:
input:
1 4
output:
2

input:
2 2
output:
2

input:
3 6
output:
1690

input:
5 4
output:
568

input:
500 500
output:
84693741

Note:

Buggy Code:
import java.util.Scanner;

public class Main {
    static final int MOD = 998244353;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        long[][] dp = new long[n + 1][m + 1];

        for (int i = 1; i <= m; i++) {
            dp[1][i] = i / 2;
        }

        for (int suit = 2; suit <= n; suit++) {
            for (int rank = 1; rank <= m; rank++) {
                dp[suit][rank] = dp[suit - 1][rank] * dp[1][rank] % MOD;
            }
        }

        long result = 1;
        for (int i = 1; i <= n; i++) {
            result = result * dp[i][m] % MOD;
        }

        System.out.println(result);
    }
}

**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 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Game
time_limit_ms: 2000
memory_limit_mb: 512
Description: In the most popular card game in Berland, a deck of $n \times m$ cards is used. Each card has two parameters: suit and rank. Suits in the game are numbered from $1$ to $n$, and ranks are numbered from $1$ to $m$. There is exactly one card in the deck for each combination of suit and rank.

A card with suit $a$ and rank $b$ can beat a card with suit $c$ and rank $d$ in one of two cases:

  * $a = 1$, $c \ne 1$ (a card of suit $1$ can beat a card of any other suit);    * $a = c$, $b > d$ (a card can beat any other card of the same suit but of a lower rank).

Two players play the game. Before the game starts, they receive exactly half of the deck each. The first player wins if for every card of the second player, he can choose his card that can beat it, and there is no card that is chosen twice (i. e. there exists a matching of the first player's cards with the second player's cards such that in each pair the first player's card beats the second player's card). Otherwise, the second player wins.

Your task is to calculate the number of ways to distribute the cards so that the first player wins. Two ways are considered different if there exists a card such that in one way it belongs to the first player and in the other way it belongs to the second player. The number of ways can be very large, so print it modulo $998244353$.

Input: The only line contains two integers $n$ and $m$ ($1 \le n, m \le 500$).

Additional constraint on the input: $m$ is even.

Output: Print a single integer — the number of ways to distribute the cards so that the first player wins, taken modulo $998244353$.

Examples:
input:
1 4
output:
2

input:
2 2
output:
2

input:
3 6
output:
1690

input:
5 4
output:
568

input:
500 500
output:
84693741

Note:

Buggy Code:
import java.util.Scanner;

public class CardGame {
    private static final int MOD = 998244353;

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int m = scanner.nextInt();
        long ways = calculateWays(n, m);
        System.out.println(ways);
        scanner.close();

    }

    private static long calculateWays(int n, int m) {
        long halfM = m / 2;
        long ways = 1;
        for (long i = 0; i < halfM; i++) {
            ways = (ways * (n + i)) % MOD;
        }
        return ways;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The buggy code has a simple loop that calculates the number of ways to distribute the cards, but it does not consider the complexity of the problem, especially when dealing with the special suit (suit 1) and the matching conditions.

### Step-by-Step Reasoning:

1. **Special Suit Consideration**: The card with suit 1 can beat any card of a different suit. This means that the first player can use cards from suit 1 to beat any card of the second player from a different suit. This should be a key factor in the distribution of cards.

2. **Even Distribution**: Since the input constraint states that \( m \) is even, we need to consider how the cards are distributed. Each player should get exactly half of the cards.

3. **Matching Condition**: The first player wins if for every card of the second player, he can choose his card that can beat it. This means we need to ensure that the first player has enough cards to beat the second player's cards.

4. **Small Scale Input**: We need a small-scale input to ensure it is complete and valid. A small input will help us quickly verify the correctness of the solution.

### Challenging Test Input:

Let's consider a small input where \( n = 2 \) and \( m = 6 \). This input is small enough to be manageable but large enough to test the logic of the code.

- **Reasoning**:
  - With \( n = 2 \) and \( m = 6 \), there are 12 cards in total (2 suits, 6 ranks).
  - Each player gets 6 cards.
  - The first player can use cards from suit 1 to beat any card of the second player from a different suit.
  - The distribution should be such that the first player has a strategic advantage.

### Test Input:
```plaintext
2 6
```

This input will test the logic of the code, especially the part where the first player can use cards from suit 1 to beat any card of the second player from a different suit. If the code is not correctly handling this logic, it will fail to produce the correct number of ways to distribute the cards.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Game
time_limit_ms: 2000
memory_limit_mb: 512
Description: In the most popular card game in Berland, a deck of $n \times m$ cards is used. Each card has two parameters: suit and rank. Suits in the game are numbered from $1$ to $n$, and ranks are numbered from $1$ to $m$. There is exactly one card in the deck for each combination of suit and rank.

A card with suit $a$ and rank $b$ can beat a card with suit $c$ and rank $d$ in one of two cases:

  * $a = 1$, $c \ne 1$ (a card of suit $1$ can beat a card of any other suit);    * $a = c$, $b > d$ (a card can beat any other card of the same suit but of a lower rank).

Two players play the game. Before the game starts, they receive exactly half of the deck each. The first player wins if for every card of the second player, he can choose his card that can beat it, and there is no card that is chosen twice (i. e. there exists a matching of the first player's cards with the second player's cards such that in each pair the first player's card beats the second player's card). Otherwise, the second player wins.

Your task is to calculate the number of ways to distribute the cards so that the first player wins. Two ways are considered different if there exists a card such that in one way it belongs to the first player and in the other way it belongs to the second player. The number of ways can be very large, so print it modulo $998244353$.

Input: The only line contains two integers $n$ and $m$ ($1 \le n, m \le 500$).

Additional constraint on the input: $m$ is even.

Output: Print a single integer — the number of ways to distribute the cards so that the first player wins, taken modulo $998244353$.

Examples:
input:
1 4
output:
2

input:
2 2
output:
2

input:
3 6
output:
1690

input:
5 4
output:
568

input:
500 500
output:
84693741

Note:

Buggy Code:
#define _CRT_SECURE_NO_WARNINGS
#define ln '\n'
#define All(x) (x).begin(), (x).end()
#define Allr(x) (x).rbegin(), (x).rend()
#define Allf(x) x.begin() + 1, x.end()

#define fi(a, b) for (auto i = (a); i <= (b); i++)
#define fj(a, b) for (auto j = (a); j <= (b); j++)
#define fo(a, b) for (auto o = (a); o <= (b); o++)
#define fq(a, b) for (auto q = (a); q <= (b); q++)
#define fe(a, b) for (auto e = (a); e <= (b); e++)
#define fw(a, b) for (auto w = (a); w <= (b); w++)

#define fdi(a, b) for (auto i = (a); i >= (b); i--)
#define fdj(a, b) for (auto j = (a); j >= (b); j--)
#define fdo(a, b) for (auto o = (a); o >= (b); o--)
#define fdq(a, b) for (auto q = (a); q >= (b); q--)
#define fde(a, b) for (auto e = (a); e >= (b); e--)
#define fdw(a, b) for (auto w = (a); w >= (b); w--)

#define FI(a, b, c) for (auto i = (a); i <= (b); i += (c))
#define FJ(a, b, c) for (auto j = (a); j <= (b); j += (c))
#define FO(a, b, c) for (auto o = (a); o <= (b); o += (c))
#define FQ(a, b, c) for (auto q = (a); q <= (b); q += (c))
#define FE(a, b, c) for (auto e = (a); e <= (b); e += (c))

#define FRI(a, b, c) for (auto i = (a); i >= (b); i -= (c))
#define FRJ(a, b, c) for (auto j = (a); j >= (b); j -= (c))
#define FRO(a, b, c) for (auto o = (a); o >= (b); o -= (c))
#define FRQ(a, b, c) for (auto q = (a); q >= (b); q -= (c))
#define FRE(a, b, c) for (auto e = (a); e >= (b); e -= (c))

#define fx(A) for (auto &x : (A))
#define fy(A) for (auto &y : (A))
#define fz(A) for (auto &z : (A))

#define pb push_back
#define ins insert
#define mp make_pair
#define len(x) (int)(x).size()
#define Len(x) (int)(x).size()
#define sz(x) (int)x.size()

#include <string>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <cmath>
#include <cstdarg>
#include <cassert>
#include <ctime>
#include <tuple>
#include <unordered_set>
#include <unordered_map>
#include <random>
#include <chrono>

using namespace std;

#ifdef _DEBUG
#define LOCAL
#endif

mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;

typedef vector<bool> vb;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<ld> vd;
typedef vector<char> vc;
typedef vector<string> vs;

typedef vector<vb> vvb;
typedef vector<vi> vvi;
typedef vector<vll> vvll;
typedef vector<vc> vvc;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

template<typename A, typename B>
ostream& operator<<(ostream& os, pair<A, B> p) {
	os << "(" << p.first << ", " << p.second << ")";
	return os;
}

template<typename T>
ostream& operator<<(ostream& os, vector<T> v) {
	fi(0, sz(v) - 1) {
		os << v[i] << " ";
	}
	return os;
}

template<typename T>
ostream& operator<<(ostream& os, set<T> t) {
	for (auto z : t) {
		os << z << " ";
	}
	return os;
}

template<typename T1, typename T2>
ostream& operator<<(ostream& os, map<T1, T2> t) {
	cerr << endl;
	for (auto z : t) {
		os << "\t" << z.first << " -> " << z.second << endl;
	}
	return os;
}

#ifdef LOCAL
#define dbg(x) {cerr << __LINE__ << "\t" << #x << ": " << (x) << endl;}
#else
#define dbg(x) {}
#endif

#ifdef LOCAL
#define ass(x) if (!(x)) { cerr << __LINE__ << "\tassertion failed: " << #x << endl, abort(); }
#else
#define ass(x) assert(x)
#endif
ll INF = 1'000'000'000'000'000'000;
int tst;
int cur_test;
const ll MOD = 998244353;
ll add(ll a, ll b) {
	return (a+b)%MOD;
}
ll mult(ll a, ll b) {
	return (a*b)%MOD;
}
ll powm(ll a, ll n) {
	if(n == 0) return 1;
	if(n == 1) return a % MOD;
	ll r = powm(a,n/2);
	r = mult(r,r);
	if(n%2==0) return r;
	return mult(r,a);
}
ll C[501][501];
ll n,m;
vector<vector<vll>>d;
void build(ll m) {
	d[m][m][1] = 1;
	fdi(m - 1, 1) {
		fj(0, m) {
			ll all = m - i + 1;
			if(j - 1 >= 0) 	d[m][i][j] = d[m][i + 1][j - 1];
			if(j >= all - j) {
				d[m][i][j] = add(d[m][i][j], d[m][i + 1][j]);
			}
		}
	}
}
ll cnk(ll n, ll k) {
	if(n == 0) return 1;
	if(k == 0) return 1;
	if(n == k) return 1;
	if(n < k) return 0;
	if(C[n][k] != 0) return C[n][k];
	C[n][k] = add(cnk(n-1,k), cnk(n-1,k-1));
	return C[n][k];
}
void solve() {
	d = vector<vector<vll>>(m+1, vector<vll>(m+1, vll( m+1)));
	fi(0, m) build(i);
	ll ans = d[m][1][m/2];
	fi(2, n) {
		ans = mult(ans, d[m][1][m/2]);
	}
	ll ans2 = 1;
	ll val = 1;
	fi(3, n) {
		val = mult(val, d[m][1][m/2]);
	}
	for(ll to_first = m / 2 + 1; to_first <= m; to_first ++) {
		if(to_first % 2 == 1) continue;
		dbg(to_first);
		ll res = d[m][1][to_first];
		ll q = m - to_first;
		ll s = 0;
		fi(2, n) {
			s = cnk(m, q);
			s = mult(s, d[m - q][1][(m-q)/2]);
			if(n - 2 >= 0) {
				s = mult(s, powm(d[m][1][m/2], n - 2));
			}
			ans = add(ans, mult(res, s));
		}
	}

	cout << ans << ln;

}

int main()
{

    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    #ifdef LOCAL
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #endif
    cin >> n >> m;

 	solve();
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider edge cases and scenarios that maximize the failure probability. Here are the steps to think through this:

1. **Understand the Problem Constraints**:
   - \( n \) and \( m \) are integers where \( 1 \le n, m \le 500 \).
   - \( m \) is always even.

2. **Identify Potential Bugs**:
   - The code uses dynamic programming and combinatorial calculations, which can be complex and prone to off-by-one errors or incorrect boundary conditions.
   - The code might have issues with large values of \( n \) and \( m \) due to the size of the DP arrays and the number of combinations.
   - The code might have issues with small values of \( n \) and \( m \) due to base cases or initialization.

3. **Generate Test Input**:
   - **Small Values**: Test with small values of \( n \) and \( m \) to ensure the base cases are handled correctly.
   - **Large Values**: Test with large values of \( n \) and \( m \) to ensure the algorithm handles large inputs without overflow or incorrect results.
   - **Edge Cases**: Test with \( n = 1 \) or \( m = 2 \) to ensure the code handles the simplest cases correctly.
   - **Balanced Values**: Test with balanced values of \( n \) and \( m \) to ensure the code handles typical cases correctly.

Given these considerations, a small-scale test input that is complete and valid and can expose potential bugs is:

```plaintext
3 6
```

This input is chosen because:
- \( n = 3 \) and \( m = 6 \) are not too large but are large enough to cover a non-trivial case.
- \( m \) is even, which is a requirement.
- The number of combinations and the DP array size are manageable but still complex enough to test the algorithm's correctness.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Game
time_limit_ms: 2000
memory_limit_mb: 512
Description: In the most popular card game in Berland, a deck of $n \times m$ cards is used. Each card has two parameters: suit and rank. Suits in the game are numbered from $1$ to $n$, and ranks are numbered from $1$ to $m$. There is exactly one card in the deck for each combination of suit and rank.

A card with suit $a$ and rank $b$ can beat a card with suit $c$ and rank $d$ in one of two cases:

  * $a = 1$, $c \ne 1$ (a card of suit $1$ can beat a card of any other suit);    * $a = c$, $b > d$ (a card can beat any other card of the same suit but of a lower rank).

Two players play the game. Before the game starts, they receive exactly half of the deck each. The first player wins if for every card of the second player, he can choose his card that can beat it, and there is no card that is chosen twice (i. e. there exists a matching of the first player's cards with the second player's cards such that in each pair the first player's card beats the second player's card). Otherwise, the second player wins.

Your task is to calculate the number of ways to distribute the cards so that the first player wins. Two ways are considered different if there exists a card such that in one way it belongs to the first player and in the other way it belongs to the second player. The number of ways can be very large, so print it modulo $998244353$.

Input: The only line contains two integers $n$ and $m$ ($1 \le n, m \le 500$).

Additional constraint on the input: $m$ is even.

Output: Print a single integer — the number of ways to distribute the cards so that the first player wins, taken modulo $998244353$.

Examples:
input:
1 4
output:
2

input:
2 2
output:
2

input:
3 6
output:
1690

input:
5 4
output:
568

input:
500 500
output:
84693741

Note:

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

#define int ll
#define endl '\n' //comment for interactive
#define fast_io ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)

#define pb push_back
#define re resize
#define ff first
#define ss second

#define all(x) (x).begin(), (x).end()
#define all1(x) (x).begin()+1, (x).end()
#define loop(i, n) for(int i = 0; i < n; i++)
#define loop1(i, n) for(int i = 1; i <= n; i++)

#define print(x) cout << #x << ": " << x << endl << flush

typedef long long ll;
typedef vector<int> vi;
typedef array<int, 2> ii;
typedef array<int, 3> ti;
typedef vector<ii> vii;
typedef vector<ti> vti;
typedef vector<vi> vvi;
typedef priority_queue<int> pq;

template<class T> bool ckmin(T&a, T b) { bool B = a > b; a = min(a, b); return B; }
template<class T> bool ckmax(T&a, T b) { bool B = a < b; a = max(a, b); return B; }

const int inf = 1e17;
const int maxf = 505;
const int mod = 998244353;

int expo(int a, int b) {
    if(a == 0) return 0;
    if(b == 0) return 1;
    int x = expo(a*a % mod,b/2);
    if(b & 1) { x *= a; x %= mod; }
    return x;
}
int inv(int a) { return expo(a, mod-2); }

vector<int> pfact(maxf,-1);
int fact(int a) {
    if(pfact[a] == -1) {
        if(a == 0 or a == 1)
            pfact[a] = 1;
        else pfact[a] = (fact(a-1) * a) % mod;
    }
    return pfact[a];
}

int nCr(int n, int r) {
    return (fact(n) * inv((fact(r) * fact(n-r) % mod))) % mod;
}

void solve() {

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

    vvi dp(m+2, vi(m+2));
    dp[0][0] = 1;
    loop1(i, m) loop1(j, i-1) loop(k, j) if(i-2 >= 0) {
        dp[i][j] += dp[i-2][k];
        dp[i][j] %= mod;
    }
    vi s(m+2);
    loop(i, m+1) loop(j, m+1)
        s[i] += dp[i][j];

    vi dp2(m+2, 0);
    vi dp3(m+2, 0);
    vi emp(m+2, 0);
    dp2[0] = 1;
    loop(idx, n-1) {
        dp3 = emp;
        loop(i, m+1) loop(j, m+1)
            if(i - j >= 0) {
                dp3[i] += (((dp2[i-j] * s[m-j]) % mod) * nCr(m, j)) % mod;
                dp3[i] %= mod;
            }
        dp2 = dp3;
    }

    int res = 0;
    loop(i, m+1) {
        res += (((s[m-i] * dp2[i]) % mod) * nCr(m, i)) % mod;
        res %= mod;
    }
    cout << res << endl;

}

signed main() {

    fast_io;

    int t = 1; //cin >> t;
    while(t--)
        solve();

    return 0;

}

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

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

Think step by step.

Response:
```plaintext
2 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Game
time_limit_ms: 2000
memory_limit_mb: 512
Description: In the most popular card game in Berland, a deck of $n \times m$ cards is used. Each card has two parameters: suit and rank. Suits in the game are numbered from $1$ to $n$, and ranks are numbered from $1$ to $m$. There is exactly one card in the deck for each combination of suit and rank.

A card with suit $a$ and rank $b$ can beat a card with suit $c$ and rank $d$ in one of two cases:

  * $a = 1$, $c \ne 1$ (a card of suit $1$ can beat a card of any other suit);    * $a = c$, $b > d$ (a card can beat any other card of the same suit but of a lower rank).

Two players play the game. Before the game starts, they receive exactly half of the deck each. The first player wins if for every card of the second player, he can choose his card that can beat it, and there is no card that is chosen twice (i. e. there exists a matching of the first player's cards with the second player's cards such that in each pair the first player's card beats the second player's card). Otherwise, the second player wins.

Your task is to calculate the number of ways to distribute the cards so that the first player wins. Two ways are considered different if there exists a card such that in one way it belongs to the first player and in the other way it belongs to the second player. The number of ways can be very large, so print it modulo $998244353$.

Input: The only line contains two integers $n$ and $m$ ($1 \le n, m \le 500$).

Additional constraint on the input: $m$ is even.

Output: Print a single integer — the number of ways to distribute the cards so that the first player wins, taken modulo $998244353$.

Examples:
input:
1 4
output:
2

input:
2 2
output:
2

input:
3 6
output:
1690

input:
5 4
output:
568

input:
500 500
output:
84693741

Note:

Buggy Code:
MOD = 998244353
N = 510
fact = [0]*N
invfact = [0]*N
dp = [[0]*N for _ in range(N)]

def power(x, y):
    res = 1
    while y > 0:
        if y % 2 == 1:
            res = res * x % MOD
        y //= 2
        x = x * x % MOD
    return res

def init():
    fact[0] = invfact[0] = 1
    for i in range(1, N):
        fact[i] = fact[i - 1] * i % MOD
        invfact[i] = power(fact[i], MOD - 2)
    for i in range(N):
        dp[i][0] = dp[i][i] = 1
        for j in range(1, i):
            dp[i][j] = (dp[i - 1][j - 1] + dp[i - 1][j]) % MOD

def solve(n, m):
    init()
    ans = 0
    m >>= 1
    for i in range(1, min(n, m) + 1):
        tmp = fact[n] * invfact[i] % MOD * invfact[n - i] % MOD
        tmp = tmp * dp[m][i] % MOD * fact[i] % MOD
        ans = (ans + tmp) % MOD
    print(ans)

n, m = map(int,input().split())
assert m % 2 == 0
solve(n, m)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that are likely to cause issues. Here are the steps to think through this:

1. **Edge Cases for \( n \) and \( m \)**:
   - \( n = 1 \): This is a special case where there is only one suit. The first player can only win if they have all the cards.
   - \( m = 2 \): This is the smallest even \( m \), which can be tricky due to the limited number of cards.
   - \( n = m \): This can be a challenging case as both \( n \) and \( m \) are equal, making the distribution more complex.
   - \( n \) and \( m \) are large: This can test the limits of the algorithm and its ability to handle large inputs efficiently.

2. **Special Cases**:
   - \( n \) is much larger than \( m \): This can test the distribution logic when there are many more suits than ranks.
   - \( n \) is much smaller than \( m \): This can test the distribution logic when there are many more ranks than suits.

3. **Combination of Edge Cases**:
   - \( n = 1 \) and \( m = 2 \): This is a simple and small input that should be easy to verify.
   - \( n = 2 \) and \( m = 4 \): This is a slightly more complex input but still small enough to verify manually.
   - \( n = 5 \) and \( m = 10 \): This is a medium-sized input that can test the distribution logic more thoroughly.

Given these considerations, a small-scale test input that is complete and valid and can expose potential bugs is:

```plaintext
2 4
```

This input is small enough to verify manually but complex enough to test the distribution logic and the dynamic programming approach used in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Game
time_limit_ms: 2000
memory_limit_mb: 512
Description: In the most popular card game in Berland, a deck of $n \times m$ cards is used. Each card has two parameters: suit and rank. Suits in the game are numbered from $1$ to $n$, and ranks are numbered from $1$ to $m$. There is exactly one card in the deck for each combination of suit and rank.

A card with suit $a$ and rank $b$ can beat a card with suit $c$ and rank $d$ in one of two cases:

  * $a = 1$, $c \ne 1$ (a card of suit $1$ can beat a card of any other suit);    * $a = c$, $b > d$ (a card can beat any other card of the same suit but of a lower rank).

Two players play the game. Before the game starts, they receive exactly half of the deck each. The first player wins if for every card of the second player, he can choose his card that can beat it, and there is no card that is chosen twice (i. e. there exists a matching of the first player's cards with the second player's cards such that in each pair the first player's card beats the second player's card). Otherwise, the second player wins.

Your task is to calculate the number of ways to distribute the cards so that the first player wins. Two ways are considered different if there exists a card such that in one way it belongs to the first player and in the other way it belongs to the second player. The number of ways can be very large, so print it modulo $998244353$.

Input: The only line contains two integers $n$ and $m$ ($1 \le n, m \le 500$).

Additional constraint on the input: $m$ is even.

Output: Print a single integer — the number of ways to distribute the cards so that the first player wins, taken modulo $998244353$.

Examples:
input:
1 4
output:
2

input:
2 2
output:
2

input:
3 6
output:
1690

input:
5 4
output:
568

input:
500 500
output:
84693741

Note:

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

#define ll long long int
#define ld long double
#define vi vector<ll>
#define vvi vector<vi>
#define vb vector<bool>
#define rep(i, n) for (ll i = 0; i < n; i++)
#define nl cout << endl
#define all(x) x.begin(), x.end()
#define read(a, n) rep(i, n) cin >> a[i]
#define give(a, n) rep(i, n) cout << a[i] << " "
#define pii pair<ll, ll>
#define pb push_back
#define f first
#define s second
const ll inf = 1e18;
// const ll mod = 1e9 + 7;
const ll mod = 998244353;
const ll maxn = 505 ;

vi fact(maxn,1) , invfact(maxn , 1) , cat(maxn,1) ;

ll gcd(ll a, ll b){
    if (a == 0) return b;
    return gcd(b % a, a);
}

ll pw(ll x, ll y) {
    ll ans = 1;
    x = x % mod;
    y = y%(mod-1) ;

    while (y) {
        if (y & 1) ans = (ans * x) % mod;
        y = y >> 1;
        x = (x * x) % mod;
    }
    return ans;
}

ll inv(ll x) {
    return pw(x, mod - 2);
}

void yesno(bool flag) {
    if (flag) cout << "YES\n";
    else cout << "NO\n";
}

ll ncr(ll n , ll r){
    if( n < r) return 0 ;
    return (fact[n]*((invfact[r]*invfact[n-r])%mod))%mod ;
}

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

    ll testcase = 1;
    // cin >> testcase;

    for(ll i = 1; i<maxn ; i++){
        fact[i] = (i*fact[i-1])%mod ;
        invfact[i] = inv(fact[i] ) ;
    }
    for(ll i =1  ;i<maxn ; i++){
        cat[i] = (ncr(2 * i, i) * inv(i + 1)) % mod;
    }

    for(int TEST = 1; TEST <= testcase; TEST++){
        ll n , m ; cin>>n>>m ;
        vi dp(m+1 , 1) ;

            for(ll j = 0 ; 2*j <= m ; j++){
                dp[j] = (cat[j])%mod ;
            }

        ll ans = 0 ;
        vi ddp(m + 5 , 0) ;
        vi ndp = ddp ;
        // rep(i,n+1){
        //     rep(j,m+1){
        //         cout<<dp[i][j]<<" "  ;
        //     }
        //     nl ;
        // }
        // ddp , ndp se we will  construct dp such that dp[i] denotes 2 to n agar sare suits lele to sum i kitne ways se ban sakta hai
        ddp[0] = 1 ;
        for(ll i = n ; i>1 ; i--){
            fill(all(ndp), 0);

            for (ll nused = m; nused >= 0; nused -= 2) {
                ll j = (m - nused) / 2;

                // Update ndp from ddp
                for (ll k = 0; k + nused <= m; k++) {
                    ndp[k + nused] = (ndp[k + nused] + (ddp[k] * dp[j]) % mod) % mod;
                }
            }

            swap(ndp,ddp) ;
        }

        for(ll i = 0 ; i<=m ; i+=2){
            // if I take i elements from rest all
            ll j = (m-i)/2 ;
            ans = (ans + (dp[j]*ddp[i])%mod)%mod ;
        }
        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:
```plaintext
1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Card Game
time_limit_ms: 2000
memory_limit_mb: 512
Description: In the most popular card game in Berland, a deck of $n \times m$ cards is used. Each card has two parameters: suit and rank. Suits in the game are numbered from $1$ to $n$, and ranks are numbered from $1$ to $m$. There is exactly one card in the deck for each combination of suit and rank.

A card with suit $a$ and rank $b$ can beat a card with suit $c$ and rank $d$ in one of two cases:

  * $a = 1$, $c \ne 1$ (a card of suit $1$ can beat a card of any other suit);    * $a = c$, $b > d$ (a card can beat any other card of the same suit but of a lower rank).

Two players play the game. Before the game starts, they receive exactly half of the deck each. The first player wins if for every card of the second player, he can choose his card that can beat it, and there is no card that is chosen twice (i. e. there exists a matching of the first player's cards with the second player's cards such that in each pair the first player's card beats the second player's card). Otherwise, the second player wins.

Your task is to calculate the number of ways to distribute the cards so that the first player wins. Two ways are considered different if there exists a card such that in one way it belongs to the first player and in the other way it belongs to the second player. The number of ways can be very large, so print it modulo $998244353$.

Input: The only line contains two integers $n$ and $m$ ($1 \le n, m \le 500$).

Additional constraint on the input: $m$ is even.

Output: Print a single integer — the number of ways to distribute the cards so that the first player wins, taken modulo $998244353$.

Examples:
input:
1 4
output:
2

input:
2 2
output:
2

input:
3 6
output:
1690

input:
5 4
output:
568

input:
500 500
output:
84693741

Note:

Buggy Code:
#include<bits/stdc++.h>
#define ll long long
#define db double
#define INF 1100000000
#define inf 100000000000000000ll
#define ldb long double
#define pb push_back
#define put_(x) printf("%d ",x)
#define putl_(x) printf("%lld ",x)
#define get(x) x=read()
#define putl(x) printf("%lld\n",x)
#define rep(p,n,i) for(int i=p;i<=n;i+=1)
#define fep(n,p,i) for(int i=n;i>=p;--i)
#define go(x) for(int i=lin[x],tn=ver[i];i;tn=ver[i=nex[i]])
#define pii pair<int,int>
#define mk make_pair
#define gf(x) scanf("%lf",&x)
#define pf(x) ((x)*(x))
#define ui unsigned
#define sq sqrt
#define l(w) t[w].l
#define r(w) t[w].r
#define x(w) t[w].x
#define tag(w) t[w].tag
#define sum(w) t[w].sum
#define sc(A) scanf("%d",&A)
#define scl(A) scanf("%lld",&A)
#define scs(A) scanf("%s",A);
#define put(A) printf("%d\n",A)
#define min(x,y) (x>=y?y:x)
#define max(x,y) (x>=y?x:y)
#define sub(x,y) (x-y<0?x-y+mod:x-y)
#define uint unsigned int
#define mod 998244353
#define zz p<<1
using namespace std;
const int MAXN=1010<<1;
int fac[MAXN],inv[MAXN];
int n,m;
int maxx=2000;
int ksm(int b,int p)
{
	int cnt=1;
	while(p)
	{
		if(p&1)cnt=(ll)cnt*b%mod;
		p=p>>1;b=(ll)b*b%mod;
	}
	return cnt;
}
int C(int n,int m){return (ll)fac[n]*inv[m]%mod*inv[n-m]%mod;}
int K(int n)
{
	if(n==0)return 1;
	return (ll)C(2*n,n)*inv[n+1]%mod*fac[n]%mod;
	//return (C(2*n,n)-C(2*n,n-1)+mod)%mod;
}
int f[510][510];
int g[510][510];
int main()
{
	sc(n);sc(m);
	fac[0]=1;
	rep(1,maxx,i)fac[i]=(ll)fac[i-1]*i%mod;
	inv[maxx]=ksm(fac[maxx],mod-2);
	fep(maxx-1,0,i)inv[i]=(ll)inv[i+1]*(i+1)%mod;
	f[1][0]=1;
	//cout<<K(m/2)<<endl;
	rep(2,n,i)rep(0,m/2,j)
	{
		rep(0,j,k)
		{
			f[i][j]=(f[i][j]+(ll)f[i-1][k]*C(m,2*(j-k))%mod*K((m-2*(j-k))/2))%mod;
		}
	}
	g[0][0]=1;
	rep(1,m,j)
	{
		rep(0,j,k)
		{
			if(k>=1)g[j][k]=(g[j-1][k-1]+g[j][k])%mod;
			if(k>=j-k)g[j][k]=(g[j][k]+g[j-1][k])%mod;
		}
	}
	//put(g[m][m/2]);
	int ans=0;
	rep(0,m/2,j)
	{
		ans=(ans+(ll)f[n][j]*g[m][(m-2*j)/2+2*j]%mod)%mod;
	}
	put(ans);
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3 2
```