**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Refined Product Optimality
time_limit_ms: 3000
memory_limit_mb: 512
Description: Although Iris occasionally sets a problem where the solution is possibly wrong, she still insists on creating problems with her imagination; after all, everyone has always been on the road with their stubbornness... And like ever before, Iris has set a problem to which she gave a wrong solution, but Chris is always supposed to save it! You are going to play the role of Chris now:

- Chris is given two arrays $$$a$$$ and $$$b$$$, both consisting of $$$n$$$ integers.
- Iris is interested in the largest possible value of $$$P = \prod\limits_{i=1}^n \min(a_i, b_i)$$$ after an arbitrary rearrangement of $$$b$$$. Note that she only wants to know the maximum value of $$$P$$$, and no actual rearrangement is performed on $$$b$$$.
- There will be $$$q$$$ modifications. Each modification can be denoted by two integers $$$o$$$ and $$$x$$$ ($$$o$$$ is either $$$1$$$ or $$$2$$$, $$$1 \leq x \leq n$$$). If $$$o = 1$$$, then Iris will increase $$$a_x$$$ by $$$1$$$; otherwise, she will increase $$$b_x$$$ by $$$1$$$.
- Iris asks Chris the maximum value of $$$P$$$ for $$$q + 1$$$ times: once before any modification, then after every modification.
- Since $$$P$$$ might be huge, Chris only needs to calculate it modulo $$$998\,244\,353$$$.

Chris soon worked out this problem, but he was so tired that he fell asleep. Besides saying thanks to Chris, now it is your turn to write a program to calculate the answers for given input data.

Note: since the input and output are large, you may need to optimize them for this problem.

For example, in C++, it is enough to use the following lines at the start of the main() function:

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2\cdot 10^5$$$, $$$1 \leq q \leq 2\cdot 10^5$$$) — the length of the array and the number of operations.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 5\cdot 10^8$$$) — the array $$$a$$$.

The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \leq b_i \leq 5\cdot 10^8$$$) — the array $$$b$$$.

Then $$$q$$$ lines follow, each line contains two integers $$$o$$$ and $$$x$$$ ($$$o \in \{1, 2\}$$$, $$$1 \leq x \leq n$$$), representing an operation.

It's guaranteed that the sum of $$$n$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$4\cdot 10^5$$$, respectively.

Output: For each test case, output $$$q + 1$$$ integers in a line, representing the answers that Chris will calculate, modulo $$$998\,244\,353$$$.

Examples:
input:
4
3 4
1 1 2
3 2 1
1 3
2 3
1 1
2 1
6 8
1 4 2 7 3 5
7 6 5 6 3 3
2 5
1 6
1 5
1 5
1 5
2 3
2 3
1 6
13 8
7 7 6 6 5 5 5 2 2 3 4 5 1
1 4 1 9 6 6 9 1 5 1 3 8 4
2 2
2 11
2 4
2 4
1 7
1 1
2 12
1 5
5 3
10000000 20000000 30000000 40000000 50000000
10000000 20000000 30000000 40000000 50000000
1 1
2 2
2 1
output:
2 3 3 6 6
840 840 1008 1344 1680 2016 2016 2016 2352
2116800 2646000 3528000 3528000 3528000 4233600 4838400 4838400 4838400
205272023 205272023 205272023 264129429

Note: In the first test case:

- Before the modifications, Chris can rearrange $$$b$$$ to $$$[1, 2, 3]$$$ so that $$$P = \prod\limits_{i=1}^n \min(a_i, b_i) = 1 \cdot 1 \cdot 2 = 2$$$. We can prove that this is the maximum possible value. For example, if Chris rearranges $$$b = [2, 3, 1]$$$, $$$P$$$ will be equal $$$1 \cdot 1 \cdot 1 = 1 < 2$$$, which is not optimal.
- After the first modification, Chris can rearrange $$$b$$$ to $$$[1, 2, 3]$$$ so that $$$P = 1 \cdot 1 \cdot 3 = 3$$$, which is maximized.
- After the second modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 3]$$$ so that $$$P = 1 \cdot 1 \cdot 3 = 3$$$, which is maximized.
- After the third modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 3]$$$ so that $$$P = 6$$$, which is maximized.
- After the fourth modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 4]$$$ so that $$$P = 6$$$, which is maximized.

Buggy Code:
import java.io.*;
import java.util.*;

public class Main {
    static FastReader fr = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

    public static void Solve() throws IOException {

        int n = fr.nextInt();
        int m = fr.nextInt();

        int [][] a = new int [n][2];
        int [][] b = new int [n][2];

        for(int i=0; i<n; i++) {
            a[i][0] = fr.nextInt();
            a[i][1] = i;
        }
        for(int i=0; i<n; i++) {
            b[i][0] = fr.nextInt();
            b[i][1] = i;
        }

        long mul = 1l;
        Arrays.sort(a, Comparator.comparingInt(x -> x[0])); Arrays.sort(b, Comparator.comparingInt(x -> x[0]));
//        for(int [] x : a) System.out.println(Arrays.toString(x));
//        System.out.printf("\n-------------------------------------------\n");
//        for(int [] x : b) System.out.println(Arrays.toString(x));
        int [] map1 = new int [n];
        int [] map2 = new int [n];
        for(int i=0; i<n; i++) {
            mul = ModMul(mul, min(a[i][0], b[i][0]));
            map1[a[i][1]]=i;
            map2[b[i][1]]=i;
        }

        HashMap<Integer, Integer> last1 = new HashMap<>();
        HashMap<Integer, Integer> last2 = new HashMap<>();

        for(int i=n-1; i>=0; i--) {
            int curr = a[i][0];
            int ptr=i;
            while (ptr >= 0) {
                if(a[ptr][0] == a[i][0]) ptr--;
                else break;
            }
            last1.put(curr, i);
            i = ptr+1;
        }

        for(int i=n-1; i>=0; i--) {
            int curr = b[i][0];
            int ptr=i;
            while (ptr >= 0) {
                if(b[ptr][0] == b[i][0]) ptr--;
                else break;
            }
            last2.put(curr, i);
            i = ptr+1;
        }

        StringBuilder ans = new StringBuilder();
        ans.append(mul + " ");

        while (m-- > 0) {
            int o = fr.nextInt();
            int idx = fr.nextInt()-1;

            if(o==1) {
                int s= map1[idx];
                int ptr = map1[idx];
                ptr = last1.get(a[ptr][0])+1;
                int prev = Math.min(a[ptr-1][0], b[ptr-1][0]);
                int now = a[s][0]+1;
                a[s][1] = a[ptr-1][1];
                map1[a[ptr-1][1]] = s;
                a[ptr-1][1] = idx;
                map1[idx] = ptr-1;
                a[ptr-1][0] = now;
                if(ptr-1 == s) {
                    last1.remove(now-1);
                    last1.put(now, last1.getOrDefault(now, ptr-1));
                } else {
                    last1.put(now-1, ptr-2);
                    last1.put(now, last1.getOrDefault(now, ptr-1));
                }
                now = Math.min(now, b[ptr-1][0]);
                if(now != prev) {
                    mul = ModMul(mul, BinaryExpo(prev, MOD-2l));
                    mul = ModMul(mul, now);
                }
            } else {
                int s= map2[idx];
                int ptr = map2[idx];
                ptr = last2.get(b[ptr][0])+1;
                int prev = Math.min(a[ptr-1][0], b[ptr-1][0]);
                int now = b[s][0]+1;
                b[s][1] = b[ptr-1][1];
                map2[b[ptr-1][1]] = s;
                b[ptr-1][1] = idx;
                map2[idx] = ptr-1;
                b[ptr-1][0] = now;
                if(ptr-1 == s) {
                    last2.remove(now-1);
                    last2.put(now, last2.getOrDefault(now, ptr-1));
                } else {
                    last2.put(now-1, ptr-2);
                    last2.put(now, last2.getOrDefault(now, ptr-1));
                }
                now = Math.min(now, a[ptr-1][0]);
                if(now != prev) {
                    mul = ModMul(mul, BinaryExpo(prev, MOD-2l));
                    mul = ModMul(mul, now);
                }
            }
            ans.append(mul + " ");
        }
        out.println(ans);
    }
    //    -----------------------------x ^ y binary exponentiation-----------------------------------------
    static long BinaryExpo(long x, long y) {
        if (y == 0) return 1;
        long temp = BinaryExpo(x, y / 2);
        temp = ModMul(temp, temp);
        if (y % 2 != 0) return ModMul(temp, x);
        return temp;
    }

    //    -------------------------------------------------------------------------------------------------------
    public static void main(String args[]) throws IOException {
//        Factorial((int)1e6);
        int t = fr.nextInt();
//        int t = 1;
        //        long SqxtasrtTime = System.currentTimeMillis();
        while (t-- > 0) {
            Solve();
        }
        out.close();
    }

    //    --------------------------------FOR TAKING FASTER INPUTS----------------------------------------
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        public FastReader() {br = new BufferedReader(new InputStreamReader(System.in));}
        String next() {
            while (st == null || !st.hasMoreElements()) {
                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());}
        String nextLine() {
            String str = "";
            try {
                if (st.hasMoreTokens()) {str = st.nextToken("\n");}
                else {str = br.readLine();}
            } catch (IOException e) {e.printStackTrace();}
            return str;
        }
    }

    //    -------------------------------(X * Y) % MOD-------------------------------------------------------
    public static long ModMul(long a, long b){
        return ((a % MOD) * (b % MOD)) % MOD;
    }

    public static long ModMul(long a, long b, long MOD){
        return ((a % MOD) * (b % MOD)) % MOD;
    }

    static int MOD = 998244353;
    static int MOD2 = (int) 1e9 + 9;
    static int maxI = Integer.MAX_VALUE;
    static int minI = Integer.MIN_VALUE;
    static long maxL = Long.MAX_VALUE;
    static long minL = Long.MIN_VALUE;

    public static int max(int a, int b) {return Math.max(a, b);}
    public static int min(int a, int b) {return Math.min(a, b);}
    public static long max(long a, long b) {return Math.max(a, b);}
    public static long min(long a, long b) {return Math.min(a, b);}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. The main areas to consider are:

1. **Initialization and Initial Values**: Ensure that the initial values of arrays `a` and `b` are such that the sorting and mapping operations in the code can be tested effectively.
2. **Modifications**: Include a mix of operations on both arrays `a` and `b` to ensure that the code handles both types of updates correctly.
3. **Edge Cases**: Include cases where the values in the arrays are close to each other or where the modifications lead to significant changes in the product.

Here is a small-scale test input that covers these aspects:

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

### Explanation:
- **Initial Arrays**:
  - `a = [1, 2, 3, 4, 5]`
  - `b = [5, 4, 3, 2, 1]`

- **Operations**:
  - **1 1**: Increase `a[0]` by 1 (a[0] becomes 2)
  - **2 1**: Increase `b[0]` by 1 (b[0] becomes 6)
  - **1 2**: Increase `a[1]` by 1 (a[1] becomes 3)
  - **2 2**: Increase `b[1]` by 1 (b[1] becomes 5)
  - **1 3**: Increase `a[2]` by 1 (a[2] becomes 4)

### Expected Output:
- **Initial P**: `1 * 2 * 3 * 4 * 5 = 120`
- **After 1 1**: `2 * 2 * 3 * 4 * 5 = 240`
- **After 2 1**: `2 * 2 * 3 * 4 * 6 = 288`
- **After 1 2**: `2 * 3 * 3 * 4 * 6 = 432`
- **After 2 2**: `2 * 3 * 3 * 4 * 5 = 360`
- **After 1 3**: `2 * 3 * 4 * 4 * 5 = 480`

This test input is small enough to be manageable but complex enough to test the code's handling of various operations and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Refined Product Optimality
time_limit_ms: 3000
memory_limit_mb: 512
Description: Although Iris occasionally sets a problem where the solution is possibly wrong, she still insists on creating problems with her imagination; after all, everyone has always been on the road with their stubbornness... And like ever before, Iris has set a problem to which she gave a wrong solution, but Chris is always supposed to save it! You are going to play the role of Chris now:

- Chris is given two arrays $$$a$$$ and $$$b$$$, both consisting of $$$n$$$ integers.
- Iris is interested in the largest possible value of $$$P = \prod\limits_{i=1}^n \min(a_i, b_i)$$$ after an arbitrary rearrangement of $$$b$$$. Note that she only wants to know the maximum value of $$$P$$$, and no actual rearrangement is performed on $$$b$$$.
- There will be $$$q$$$ modifications. Each modification can be denoted by two integers $$$o$$$ and $$$x$$$ ($$$o$$$ is either $$$1$$$ or $$$2$$$, $$$1 \leq x \leq n$$$). If $$$o = 1$$$, then Iris will increase $$$a_x$$$ by $$$1$$$; otherwise, she will increase $$$b_x$$$ by $$$1$$$.
- Iris asks Chris the maximum value of $$$P$$$ for $$$q + 1$$$ times: once before any modification, then after every modification.
- Since $$$P$$$ might be huge, Chris only needs to calculate it modulo $$$998\,244\,353$$$.

Chris soon worked out this problem, but he was so tired that he fell asleep. Besides saying thanks to Chris, now it is your turn to write a program to calculate the answers for given input data.

Note: since the input and output are large, you may need to optimize them for this problem.

For example, in C++, it is enough to use the following lines at the start of the main() function:

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2\cdot 10^5$$$, $$$1 \leq q \leq 2\cdot 10^5$$$) — the length of the array and the number of operations.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 5\cdot 10^8$$$) — the array $$$a$$$.

The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \leq b_i \leq 5\cdot 10^8$$$) — the array $$$b$$$.

Then $$$q$$$ lines follow, each line contains two integers $$$o$$$ and $$$x$$$ ($$$o \in \{1, 2\}$$$, $$$1 \leq x \leq n$$$), representing an operation.

It's guaranteed that the sum of $$$n$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$4\cdot 10^5$$$, respectively.

Output: For each test case, output $$$q + 1$$$ integers in a line, representing the answers that Chris will calculate, modulo $$$998\,244\,353$$$.

Examples:
input:
4
3 4
1 1 2
3 2 1
1 3
2 3
1 1
2 1
6 8
1 4 2 7 3 5
7 6 5 6 3 3
2 5
1 6
1 5
1 5
1 5
2 3
2 3
1 6
13 8
7 7 6 6 5 5 5 2 2 3 4 5 1
1 4 1 9 6 6 9 1 5 1 3 8 4
2 2
2 11
2 4
2 4
1 7
1 1
2 12
1 5
5 3
10000000 20000000 30000000 40000000 50000000
10000000 20000000 30000000 40000000 50000000
1 1
2 2
2 1
output:
2 3 3 6 6
840 840 1008 1344 1680 2016 2016 2016 2352
2116800 2646000 3528000 3528000 3528000 4233600 4838400 4838400 4838400
205272023 205272023 205272023 264129429

Note: In the first test case:

- Before the modifications, Chris can rearrange $$$b$$$ to $$$[1, 2, 3]$$$ so that $$$P = \prod\limits_{i=1}^n \min(a_i, b_i) = 1 \cdot 1 \cdot 2 = 2$$$. We can prove that this is the maximum possible value. For example, if Chris rearranges $$$b = [2, 3, 1]$$$, $$$P$$$ will be equal $$$1 \cdot 1 \cdot 1 = 1 < 2$$$, which is not optimal.
- After the first modification, Chris can rearrange $$$b$$$ to $$$[1, 2, 3]$$$ so that $$$P = 1 \cdot 1 \cdot 3 = 3$$$, which is maximized.
- After the second modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 3]$$$ so that $$$P = 1 \cdot 1 \cdot 3 = 3$$$, which is maximized.
- After the third modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 3]$$$ so that $$$P = 6$$$, which is maximized.
- After the fourth modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 4]$$$ so that $$$P = 6$$$, which is maximized.

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

using ll = long long;
using vi = vector<ll>;
using vl = vector<ll>;
using pii = pair<ll, ll>;
using pll = pair<ll, ll>;

#define endl '\n'
#define pb push_back
#define mp make_pair
#define f first
#define s second
#define all(x) x.begin(), x.end()
#define mod 998244353
const int N = 2e5 + 2;
#define int ll
int a[N];
int b[N];
int n,q,l,t,x,y,res,test;
int tree[4*N];
map <int,int> ma;
map <int,int> mb;

int mrg(int x, int y){
    return 1LL*(1LL*(x%mod)*(y%mod))%mod;
}

void build(int id=0,int ns = 0, int ne = n-1){
    if(ns==ne){
        tree[id] = min(a[ns],b[ns]);
        return ;
    }
    int l = 2*id+1;
    int r = l+1;
    int md = ns+(ne-ns)/2;
    build(l, ns, md);
    build(r, md+1, ne);
    tree[id] = mrg(tree[l],tree[r]);
}

int query(int qs, int qe, int id=0, int ns=0, int ne=n-1){
    if(ns>qe || qs>ne){
        return 1LL;
    }
    if(qs<=ns && qe>=ne){
        return (tree[id]%mod);
    }
    int l = 2*id+1;
    int r = l+1;
    int md = ns+(ne-ns)/2;
    return mrg(query(qs, qe, l, ns, md), query(qs, qe,r, md+1,ne));
}

void upd(int pos , int val , int id=0, int ns=0,int ne=n-1){
    if(ns>pos || pos>ne){
        return;
    }
    if(ns==ne){
        tree[id]=(val%mod);
        return ;
    }
    int l = 2*id+1;
    int r = l+1;
    int md = ns+(ne-ns)/2;
    upd(pos, val,l, ns, md);
    upd(pos, val, r, md+1, ne);
    tree[id] = mrg(tree[l],tree[r]);
}

void solve(){
    cin >> n >> q;
    for(int i=0;i<n;i++){
        cin >> a[i];
    }
    vector <pii> v(n);
    for (int i=0;i<n;i++){
        v[i].f=a[i];
        v[i].s=i;
    }
    sort(v.begin(),v.end());
    for (int i=0;i<n;i++){
        ma[v[i].s]=i;
    }
    for (int i=0;i<n;i++){
        a[i]=v[i].f;
    }

    for(int i=0;i<n;i++){
        cin >> b[i];
    }
    vector <pii> u(n);
    for (int i=0;i<n;i++){
        u[i].f=b[i];
        u[i].s=i;
    }
    sort(u.begin(),u.end());
    for (int i=0;i<n;i++){
        mb[u[i].s]=i;
    }
    for (int i=0;i<n;i++){
        b[i]=u[i].f;
    }
    map<int ,int>ia ;
    map<int ,int>ib ;
    for (auto i:ma){
        ia[i.s]=i.f;
    }
    for (auto i:mb){
        ib[i.s]=i.f;
    }

    build();
    res=1;
    for(int i=0;i<n;i++)
    {
        res=(1LL * res * min(a[i],b[i])%mod)%mod;
    }
    cout << res << " ";
    while(q--){
        int id=0;
        cin >> t >> x ;
        x--;
        test=0;
        switch (t){
        case 1:
            if(ma[x]==n-1){
                a[n-1]++;
                //cout<<"###"<<q<<"####";
                if(b[n-1]>=a[n-1] ){

                    upd(n-1,min(a[n-1],b[n-1]));
                    break;
                }
                test=1;
                break;
            }
                id=upper_bound(a,a+n,a[ma[x]])-a;
                id--;
                if(id==ma[x]){
                    //cout << "###"<<q<<"####";
                    a[ma[x]]++;
                    if(b[ma[x]]>=a[ma[x]]){

                    upd(ma[x],min(a[ma[x]],b[ma[x]]));
                    break;
                    }
                    test=1;
                    break;
                }
                 y=ia[id];
                ia[id]=x;
                ia[ma[x]]=y;
                ma[y]=ma[x];
                ma[x]=id;
                a[id]++;
                if (b[id]>=a[id] ){
                    upd(id,min(a[id],b[id]));
                    break;
                }
                test=1;
                break;

        case 2 :
            if(mb[x]==n-1){
                b[n-1]++;
                if(a[n-1]>=b[n-1] ){

                upd(n-1,min(b[n-1],a[n-1]));
                break;
                }
                test =1;
                break;
            }
                id=upper_bound(b,b+n,b[mb[x]])-b;
                id--;
                if(id==mb[x]){
                    //cout << "###"<<q<<"####";
                    b[mb[x]]++;
                    if(a[mb[x]]>=b[mb[x]]){

                    upd(mb[x],min(b[mb[x]],a[mb[x]]));
                    break;
                    }
                    test=1;

                    break;
                }
                //cout << "###"<<q<<"####";
                y=ib[id];
                ib[id]=x;
                ib[mb[x]]=y;
                mb[y]=mb[x];
                mb[x]=id;
                b[id]++;
                if(a[id]>=b[id] ){

                upd(id,min(b[id],a[id]));
                break;
                }
                test=1;
                break;
        }
        if(test){
            cout << res << " ";
        }else{res=query(0,n-1);cout <<res  << " ";}

    }
    cout << endl;

}

signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    ll 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
1
3 4
1 1 2
3 2 1
1 3
2 3
1 1
2 1
```

### Explanation:
This test case is designed to expose potential issues in the buggy code by focusing on a small-scale scenario that can still highlight the problems in the algorithm. Here’s a breakdown:

1. **Initial Arrays:**
   - `a = [1, 1, 2]`
   - `b = [3, 2, 1]`

2. **Initial Product Calculation:**
   - The initial maximum value of \( P \) is \( \min(1, 3) \cdot \min(1, 2) \cdot \min(2, 1) = 1 \cdot 1 \cdot 1 = 1 \).

3. **Operations:**
   - **Operation 1 (1 3):** Increase \( a_3 \) by 1.
     - New `a = [1, 1, 3]`
     - New \( P \) is \( \min(1, 3) \cdot \min(1, 2) \cdot \min(3, 1) = 1 \cdot 1 \cdot 1 = 1 \).
   - **Operation 2 (2 3):** Increase \( b_3 \) by 1.
     - New `b = [3, 2, 2]`
     - New \( P \) is \( \min(1, 3) \cdot \min(1, 2) \cdot \min(3, 2) = 1 \cdot 1 \cdot 2 = 2 \).
   - **Operation 3 (1 1):** Increase \( a_1 \) by 1.
     - New `a = [2, 1, 3]`
     - New \( P \) is \( \min(2, 3) \cdot \min(1, 2) \cdot \min(3, 2) = 2 \cdot 1 \cdot 2 = 4 \).
   - **Operation 4 (2 1):** Increase \( b_1 \) by 1.
     - New `b = [4, 2, 2]`
     - New \( P \) is \( \min(2, 4) \cdot \min(1, 2) \cdot \min(3, 2) = 2 \cdot 1 \cdot 2 = 4 \).

### Expected Output:
```plaintext
1 1 2 4 4
```

This test case is designed to ensure that the code correctly handles the updates and recalculates the product \( P \) after each operation. The small scale allows for easy verification of the correctness of the solution.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Refined Product Optimality
time_limit_ms: 3000
memory_limit_mb: 512
Description: Although Iris occasionally sets a problem where the solution is possibly wrong, she still insists on creating problems with her imagination; after all, everyone has always been on the road with their stubbornness... And like ever before, Iris has set a problem to which she gave a wrong solution, but Chris is always supposed to save it! You are going to play the role of Chris now:

- Chris is given two arrays $$$a$$$ and $$$b$$$, both consisting of $$$n$$$ integers.
- Iris is interested in the largest possible value of $$$P = \prod\limits_{i=1}^n \min(a_i, b_i)$$$ after an arbitrary rearrangement of $$$b$$$. Note that she only wants to know the maximum value of $$$P$$$, and no actual rearrangement is performed on $$$b$$$.
- There will be $$$q$$$ modifications. Each modification can be denoted by two integers $$$o$$$ and $$$x$$$ ($$$o$$$ is either $$$1$$$ or $$$2$$$, $$$1 \leq x \leq n$$$). If $$$o = 1$$$, then Iris will increase $$$a_x$$$ by $$$1$$$; otherwise, she will increase $$$b_x$$$ by $$$1$$$.
- Iris asks Chris the maximum value of $$$P$$$ for $$$q + 1$$$ times: once before any modification, then after every modification.
- Since $$$P$$$ might be huge, Chris only needs to calculate it modulo $$$998\,244\,353$$$.

Chris soon worked out this problem, but he was so tired that he fell asleep. Besides saying thanks to Chris, now it is your turn to write a program to calculate the answers for given input data.

Note: since the input and output are large, you may need to optimize them for this problem.

For example, in C++, it is enough to use the following lines at the start of the main() function:

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2\cdot 10^5$$$, $$$1 \leq q \leq 2\cdot 10^5$$$) — the length of the array and the number of operations.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 5\cdot 10^8$$$) — the array $$$a$$$.

The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \leq b_i \leq 5\cdot 10^8$$$) — the array $$$b$$$.

Then $$$q$$$ lines follow, each line contains two integers $$$o$$$ and $$$x$$$ ($$$o \in \{1, 2\}$$$, $$$1 \leq x \leq n$$$), representing an operation.

It's guaranteed that the sum of $$$n$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$4\cdot 10^5$$$, respectively.

Output: For each test case, output $$$q + 1$$$ integers in a line, representing the answers that Chris will calculate, modulo $$$998\,244\,353$$$.

Examples:
input:
4
3 4
1 1 2
3 2 1
1 3
2 3
1 1
2 1
6 8
1 4 2 7 3 5
7 6 5 6 3 3
2 5
1 6
1 5
1 5
1 5
2 3
2 3
1 6
13 8
7 7 6 6 5 5 5 2 2 3 4 5 1
1 4 1 9 6 6 9 1 5 1 3 8 4
2 2
2 11
2 4
2 4
1 7
1 1
2 12
1 5
5 3
10000000 20000000 30000000 40000000 50000000
10000000 20000000 30000000 40000000 50000000
1 1
2 2
2 1
output:
2 3 3 6 6
840 840 1008 1344 1680 2016 2016 2016 2352
2116800 2646000 3528000 3528000 3528000 4233600 4838400 4838400 4838400
205272023 205272023 205272023 264129429

Note: In the first test case:

- Before the modifications, Chris can rearrange $$$b$$$ to $$$[1, 2, 3]$$$ so that $$$P = \prod\limits_{i=1}^n \min(a_i, b_i) = 1 \cdot 1 \cdot 2 = 2$$$. We can prove that this is the maximum possible value. For example, if Chris rearranges $$$b = [2, 3, 1]$$$, $$$P$$$ will be equal $$$1 \cdot 1 \cdot 1 = 1 < 2$$$, which is not optimal.
- After the first modification, Chris can rearrange $$$b$$$ to $$$[1, 2, 3]$$$ so that $$$P = 1 \cdot 1 \cdot 3 = 3$$$, which is maximized.
- After the second modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 3]$$$ so that $$$P = 1 \cdot 1 \cdot 3 = 3$$$, which is maximized.
- After the third modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 3]$$$ so that $$$P = 6$$$, which is maximized.
- After the fourth modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 4]$$$ so that $$$P = 6$$$, which is maximized.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
#define MOD 998244353
const int N = 500000;
int seg[4*N], seg2[4*N];
ll binpow(ll a,int b){
    ll res=1;
    while(b){
        if(b&1)
            res=(res*a%MOD);
        a=(a*a%MOD);
        b/=2;
    }
    return res;
}
ll bol(int n){
    ll ans = binpow(n, MOD-2);
    return ans;
}
int f(int i, int l, int r, int ql, int qr){
    if(l > qr || r < ql) return 0;
    if(l >= ql && r <= qr) return seg[i];
    int m = (l+r) / 2;
    return f(i*2, l, m, ql, qr) + f(i*2+1, m+1, r, ql, qr);
}
void up(int i, int l, int r, int p, int v){
    if(l > p || r < p) return;
    if(l == r){
        seg[i] += v;
        return;
    }
    int m = (l+r) / 2;
    up(i*2, l, m, p, v);
    up(i*2+1, m+1, r, p, v);
    seg[i] = seg[i*2] + seg[i*2+1];
}
int f2(int i, int l, int r, int ql, int qr){
    if(l > qr || r < ql) return 0;
    if(l >= ql && r <= qr) return seg2[i];
    int m = (l+r) / 2;
    return f2(i*2, l, m, ql, qr) + f2(i*2+1, m+1, r, ql, qr);
}
void up2(int i, int l, int r, int p, int v){
    if(l > p || r < p) return;
    if(l == r){
        seg2[i] += v;
        return;
    }
    int m = (l+r) / 2;
    up2(i*2, l, m, p, v);
    up2(i*2+1, m+1, r, p, v);
    seg2[i] = seg2[i*2] + seg2[i*2+1];
}
void solve(){
    int n, m;
    cin >> n >> m;
    int a[n], b[n], c[n], d[n], q[n][2];
    set<pair<int, int> > A, B;
    set<int> lanet;
    for(int i = 0; i < n; i++){
        cin >> a[i];
        c[i] = a[i];
        lanet.insert(a[i]);
        A.insert({a[i], i+1});
        // update(&rootA, a[i], 1);
    }
    for(int i = 0; i < n; i++){
        cin >> b[i];
        d[i] = b[i];
        lanet.insert(d[i]);
        B.insert({b[i], i+1});
        // update(&rootB, b[i], 1);
    }

    //init
    ll ans = 1, kkk = 0;
    int bk[n];
    for(auto i:A) bk[kkk++] = i.first;
        kkk =0;
    for(auto i:B) bk[kkk] = min(bk[kkk], i.first), kkk++ ;
    for(int i:bk) ans = ans * i % MOD;
    cout << ans << " ";

    for(int i = 0; i < m; i++){
        cin >> q[i][0] >> q[i][1];
        if(q[i][0] == 1){
            c[q[i][1]-1]++;
            lanet.insert(c[q[i][1]-1]);
        }
        else{
            d[q[i][1]-1]++;
            lanet.insert(d[q[i][1]-1]);
        }
    }
    //compress that btch
    map<int, int> encoder, decoder;
    int cnt = 1;
    for(int i:lanet){
        encoder[i] = cnt, decoder[cnt] = i;
        cnt++;
    }
    cnt--;

    for(int i:a){
        up(1, 1, cnt, encoder[i], 1);
    }
    for(int i:b){
        up2(1, 1, cnt, encoder[i], 1);
    }

    for(int i = 0; i < m; i++){
        int x, y;
        x = q[i][0], y = q[i][1];
        if(x == 1){
            int change = f(1, 1, cnt, 1, encoder[a[y-1]]);
            up(1, 1, cnt, encoder[a[y-1]], -1);
            up(1, 1, cnt, encoder[a[y-1]+1], 1);
            a[y-1]++;
            int l = 1, r = cnt;
            while(l != r){
                int m = (l + r) / 2;
                int x = f2(1, 1, cnt, 1, m);
                if(x < change) l = m+1;
                else r = m;
            }
            //cout << change << " " << l << endl;
            ans = (ans * bol(min(a[y-1]-1, decoder[l])) % MOD);
            ans = (ans * min(a[y-1], decoder[l]) % MOD);
        }
        else{
            int change = f2(1, 1, cnt, 1, encoder[b[y-1]]);
            up2(1, 1, cnt, encoder[b[y-1]], -1);
            up2(1, 1, cnt, encoder[b[y-1]+1], 1);
            b[y-1]++;
            int l = 1, r = cnt;
            while(l != r){
                int m = (l + r) / 2;
                int x = f(1, 1, cnt, 1, m);
                if(x < change) l = m+1;
                else r = m;
            }
            //cout << change << " " << l << endl;
            ans = (ans * bol(min(b[y-1]-1, decoder[l])) % MOD);
            ans = (ans * min(b[y-1], decoder[l]) % MOD);
        }
        cout << ans << " ";
    }
    for(int i:a) up(1, 1, cnt, encoder[i], -1);
    for(int i:b) up2(1, 1, cnt, encoder[i], -1);
    cout << endl;
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL), cout.tie(NULL);
    int t;
    cin >> t;
    while(t--){
        solve();
    }
}

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

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

Think step by step.

Response:
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. Here are the steps to create such a test input:

1. **Edge Case with Small Values**: Use small values for `n` and `q` to ensure the test case is small and manageable.
2. **Initial Arrays with Identical Elements**: Use arrays `a` and `b` with identical elements to test the edge case where the product is maximized by rearranging the same values.
3. **Modifications that Affect the Product Significantly**: Use modifications that can significantly change the product, such as increasing elements that are already large or small.

Let's create a test input based on these criteria:

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

### Explanation:
- **Test Case Count**: 1
- **Array Length and Modifications**: `n = 3`, `q = 4`
- **Initial Arrays**:
  - `a = [1, 1, 1]`
  - `b = [1, 1, 1]`
- **Modifications**:
  - Increase `a[0]` by 1
  - Increase `b[0]` by 1
  - Increase `a[1]` by 1
  - Increase `b[1]` by 1

### Expected Output:
- **Initial Product**: `1 * 1 * 1 = 1`
- **After First Modification**: `2 * 1 * 1 = 2`
- **After Second Modification**: `2 * 2 * 1 = 4`
- **After Third Modification**: `2 * 2 * 2 = 8`
- **After Fourth Modification**: `2 * 3 * 2 = 12`

This test input is designed to expose any issues in the buggy code by starting with identical elements and then applying modifications that significantly change the product. The small scale ensures that the test case is manageable and easy to verify.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Refined Product Optimality
time_limit_ms: 3000
memory_limit_mb: 512
Description: Although Iris occasionally sets a problem where the solution is possibly wrong, she still insists on creating problems with her imagination; after all, everyone has always been on the road with their stubbornness... And like ever before, Iris has set a problem to which she gave a wrong solution, but Chris is always supposed to save it! You are going to play the role of Chris now:

- Chris is given two arrays $$$a$$$ and $$$b$$$, both consisting of $$$n$$$ integers.
- Iris is interested in the largest possible value of $$$P = \prod\limits_{i=1}^n \min(a_i, b_i)$$$ after an arbitrary rearrangement of $$$b$$$. Note that she only wants to know the maximum value of $$$P$$$, and no actual rearrangement is performed on $$$b$$$.
- There will be $$$q$$$ modifications. Each modification can be denoted by two integers $$$o$$$ and $$$x$$$ ($$$o$$$ is either $$$1$$$ or $$$2$$$, $$$1 \leq x \leq n$$$). If $$$o = 1$$$, then Iris will increase $$$a_x$$$ by $$$1$$$; otherwise, she will increase $$$b_x$$$ by $$$1$$$.
- Iris asks Chris the maximum value of $$$P$$$ for $$$q + 1$$$ times: once before any modification, then after every modification.
- Since $$$P$$$ might be huge, Chris only needs to calculate it modulo $$$998\,244\,353$$$.

Chris soon worked out this problem, but he was so tired that he fell asleep. Besides saying thanks to Chris, now it is your turn to write a program to calculate the answers for given input data.

Note: since the input and output are large, you may need to optimize them for this problem.

For example, in C++, it is enough to use the following lines at the start of the main() function:

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2\cdot 10^5$$$, $$$1 \leq q \leq 2\cdot 10^5$$$) — the length of the array and the number of operations.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 5\cdot 10^8$$$) — the array $$$a$$$.

The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \leq b_i \leq 5\cdot 10^8$$$) — the array $$$b$$$.

Then $$$q$$$ lines follow, each line contains two integers $$$o$$$ and $$$x$$$ ($$$o \in \{1, 2\}$$$, $$$1 \leq x \leq n$$$), representing an operation.

It's guaranteed that the sum of $$$n$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$4\cdot 10^5$$$, respectively.

Output: For each test case, output $$$q + 1$$$ integers in a line, representing the answers that Chris will calculate, modulo $$$998\,244\,353$$$.

Examples:
input:
4
3 4
1 1 2
3 2 1
1 3
2 3
1 1
2 1
6 8
1 4 2 7 3 5
7 6 5 6 3 3
2 5
1 6
1 5
1 5
1 5
2 3
2 3
1 6
13 8
7 7 6 6 5 5 5 2 2 3 4 5 1
1 4 1 9 6 6 9 1 5 1 3 8 4
2 2
2 11
2 4
2 4
1 7
1 1
2 12
1 5
5 3
10000000 20000000 30000000 40000000 50000000
10000000 20000000 30000000 40000000 50000000
1 1
2 2
2 1
output:
2 3 3 6 6
840 840 1008 1344 1680 2016 2016 2016 2352
2116800 2646000 3528000 3528000 3528000 4233600 4838400 4838400 4838400
205272023 205272023 205272023 264129429

Note: In the first test case:

- Before the modifications, Chris can rearrange $$$b$$$ to $$$[1, 2, 3]$$$ so that $$$P = \prod\limits_{i=1}^n \min(a_i, b_i) = 1 \cdot 1 \cdot 2 = 2$$$. We can prove that this is the maximum possible value. For example, if Chris rearranges $$$b = [2, 3, 1]$$$, $$$P$$$ will be equal $$$1 \cdot 1 \cdot 1 = 1 < 2$$$, which is not optimal.
- After the first modification, Chris can rearrange $$$b$$$ to $$$[1, 2, 3]$$$ so that $$$P = 1 \cdot 1 \cdot 3 = 3$$$, which is maximized.
- After the second modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 3]$$$ so that $$$P = 1 \cdot 1 \cdot 3 = 3$$$, which is maximized.
- After the third modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 3]$$$ so that $$$P = 6$$$, which is maximized.
- After the fourth modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 4]$$$ so that $$$P = 6$$$, which is maximized.

Buggy Code:
from collections import defaultdict
def modpower(x, n, m):
    if n == 0:
        return 1 % m
    u = modpower(x, n // 2, m)
    u = (u * u) % m
    if n % 2 == 1:
        u = (u * x) % m
    return u

tt=int(input())
m=998244353

for t in range(tt):
    n,q=map(int,input().split())
    arr1=list(map(int,input().split()))
    arr2=list(map(int,input().split()))
    arr3=list(arr1)
    arr4=list(arr2)
    ans=1
    arr3.sort()
    arr4.sort()
    for i in range(n):
        ans=(ans*min(arr3[i],arr4[i]))%m
    d1=dict()
    f1=defaultdict(int)
    f2=defaultdict(int)
    d2=dict()
    i=0
    while i<n:
        j=i
        c=0
        while j<n and arr3[i]==arr3[j]:
            j+=1
            c+=1
        d1[arr3[i]]=j-1
        f1[arr3[i]]=c
        c=0
        i=j
    i=0
    while i<n:
        j=i
        c=0
        while j<n and arr4[i]==arr4[j]:
            j+=1
            c+=1
        d2[arr4[i]]=j-1
        f2[arr4[i]]=c
        c=0
        i=j
    print(ans,end=" ")

    for i in range(q):
        a,b=map(int,input().split())
        if a==1:
            ele1=arr1[b-1]
            newele=ele1+1
            ind=d1[ele1]
            ele2=arr4[ind]
            if newele>ele2:
                print(ans,end=" ")
            else:
                ans=(ans+(ans%m)*((modpower(ele1,m-2,m))))%m
                print(ans,end=" ")
            if newele not in d1:
                d1[newele]=ind
            f1[ele1]-=1
            f1[newele]+=1
            if f1[ele1]:
                d1[ele1]-=1
            else:
                d1[ele1]=-1
            arr1[b-1]+=1
            arr3[ind]+=1

        else:
            ele1=arr2[b-1]
            newele=ele1+1
            ind=d2[ele1]
            ele2=arr3[ind]
            if newele>ele2:
                print(ans,end=" ")
            else:
                ans=(ans+(ans%m)*((modpower(ele1,m-2,m))))%m
                print(ans,end=" ")
            if newele not in d2:
                d2[newele]=ind
            f2[ele1]-=1
            f2[newele]+=1
            if f2[ele1]:
                d2[ele1]-=1
            else:
                d2[ele1]=-1
            arr2[b-1]+=1
            arr4[ind]+=1
    print()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Refined Product Optimality
time_limit_ms: 3000
memory_limit_mb: 512
Description: Although Iris occasionally sets a problem where the solution is possibly wrong, she still insists on creating problems with her imagination; after all, everyone has always been on the road with their stubbornness... And like ever before, Iris has set a problem to which she gave a wrong solution, but Chris is always supposed to save it! You are going to play the role of Chris now:

- Chris is given two arrays $$$a$$$ and $$$b$$$, both consisting of $$$n$$$ integers.
- Iris is interested in the largest possible value of $$$P = \prod\limits_{i=1}^n \min(a_i, b_i)$$$ after an arbitrary rearrangement of $$$b$$$. Note that she only wants to know the maximum value of $$$P$$$, and no actual rearrangement is performed on $$$b$$$.
- There will be $$$q$$$ modifications. Each modification can be denoted by two integers $$$o$$$ and $$$x$$$ ($$$o$$$ is either $$$1$$$ or $$$2$$$, $$$1 \leq x \leq n$$$). If $$$o = 1$$$, then Iris will increase $$$a_x$$$ by $$$1$$$; otherwise, she will increase $$$b_x$$$ by $$$1$$$.
- Iris asks Chris the maximum value of $$$P$$$ for $$$q + 1$$$ times: once before any modification, then after every modification.
- Since $$$P$$$ might be huge, Chris only needs to calculate it modulo $$$998\,244\,353$$$.

Chris soon worked out this problem, but he was so tired that he fell asleep. Besides saying thanks to Chris, now it is your turn to write a program to calculate the answers for given input data.

Note: since the input and output are large, you may need to optimize them for this problem.

For example, in C++, it is enough to use the following lines at the start of the main() function:

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2\cdot 10^5$$$, $$$1 \leq q \leq 2\cdot 10^5$$$) — the length of the array and the number of operations.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 5\cdot 10^8$$$) — the array $$$a$$$.

The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \leq b_i \leq 5\cdot 10^8$$$) — the array $$$b$$$.

Then $$$q$$$ lines follow, each line contains two integers $$$o$$$ and $$$x$$$ ($$$o \in \{1, 2\}$$$, $$$1 \leq x \leq n$$$), representing an operation.

It's guaranteed that the sum of $$$n$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$4\cdot 10^5$$$, respectively.

Output: For each test case, output $$$q + 1$$$ integers in a line, representing the answers that Chris will calculate, modulo $$$998\,244\,353$$$.

Examples:
input:
4
3 4
1 1 2
3 2 1
1 3
2 3
1 1
2 1
6 8
1 4 2 7 3 5
7 6 5 6 3 3
2 5
1 6
1 5
1 5
1 5
2 3
2 3
1 6
13 8
7 7 6 6 5 5 5 2 2 3 4 5 1
1 4 1 9 6 6 9 1 5 1 3 8 4
2 2
2 11
2 4
2 4
1 7
1 1
2 12
1 5
5 3
10000000 20000000 30000000 40000000 50000000
10000000 20000000 30000000 40000000 50000000
1 1
2 2
2 1
output:
2 3 3 6 6
840 840 1008 1344 1680 2016 2016 2016 2352
2116800 2646000 3528000 3528000 3528000 4233600 4838400 4838400 4838400
205272023 205272023 205272023 264129429

Note: In the first test case:

- Before the modifications, Chris can rearrange $$$b$$$ to $$$[1, 2, 3]$$$ so that $$$P = \prod\limits_{i=1}^n \min(a_i, b_i) = 1 \cdot 1 \cdot 2 = 2$$$. We can prove that this is the maximum possible value. For example, if Chris rearranges $$$b = [2, 3, 1]$$$, $$$P$$$ will be equal $$$1 \cdot 1 \cdot 1 = 1 < 2$$$, which is not optimal.
- After the first modification, Chris can rearrange $$$b$$$ to $$$[1, 2, 3]$$$ so that $$$P = 1 \cdot 1 \cdot 3 = 3$$$, which is maximized.
- After the second modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 3]$$$ so that $$$P = 1 \cdot 1 \cdot 3 = 3$$$, which is maximized.
- After the third modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 3]$$$ so that $$$P = 6$$$, which is maximized.
- After the fourth modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 4]$$$ so that $$$P = 6$$$, which is maximized.

Buggy Code:
from bisect import bisect_left

k = 998244353

for _ in range(int(input())):
    n, q = map(int, input().split())

    a = list(map(int, input().split()))
    b = list(map(int, input().split()))

    a2 = sorted([[i, a[i]] for i in range(n)], key=lambda x : x[1])
    b2 = sorted([[i, b[i]] for i in range(n)], key=lambda x : x[1])
    ad = [0] * n
    bd = [0] * n

    for i in range(n):
        ad[a2[i][0]] = i
        bd[b2[i][0]] = i

    p = 1
    for i in range(n):
        p = p * min(a2[i][1], b2[i][1]) % k
    print(p, end=' ')

    while q:
        o, x = map(int, input().split())
        x -= 1

        if o == 1:
            a[x] += 1

            src = ad[x]
            dst = bisect_left(a2, a[x], key=lambda x : x[1]) - 1

            g = min(a2[dst][1], b2[dst][1])
            if src != dst:
                g = g * min(a2[src][1], b2[src][1]) % k

            a2[src], a2[dst] = a2[dst], a2[src]
            a2[dst][1] += 1

            ad[x] = dst
            ad[a2[src][0]] = src

            p = p * min(a2[dst][1], b2[dst][1])
            if src != dst:
                p = p * min(a2[src][1], b2[src][1])

            g2 = pow(g, k - 2, k)
            p = (p * g2) % k

        else:
            b[x] += 1

            src = bd[x]
            dst = bisect_left(b2, b[x], key=lambda x : x[1]) - 1

            g = min(a2[dst][1], b2[dst][1])
            if src != dst:
                g = g * min(a2[src][1], b2[src][1]) % k

            b2[src], b2[dst] = b2[dst], b2[src]
            b2[dst][1] += 1

            bd[x] = dst
            bd[b2[src][0]] = src

            p = p * min(a2[dst][1], b2[dst][1])
            if src != dst:
                p = p * min(a2[src][1], b2[src][1])

            g2 = pow(g, k - 2, k)
            p = (p * g2) % k

        print(p, end=' ')

        q -= 1
    print()

'''
1
5 3
10000000 20000000 30000000 40000000 50000000
10000000 20000000 30000000 40000000 50000000
1 1
2 2
2 1
'''

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Refined Product Optimality
time_limit_ms: 3000
memory_limit_mb: 512
Description: Although Iris occasionally sets a problem where the solution is possibly wrong, she still insists on creating problems with her imagination; after all, everyone has always been on the road with their stubbornness... And like ever before, Iris has set a problem to which she gave a wrong solution, but Chris is always supposed to save it! You are going to play the role of Chris now:

- Chris is given two arrays $$$a$$$ and $$$b$$$, both consisting of $$$n$$$ integers.
- Iris is interested in the largest possible value of $$$P = \prod\limits_{i=1}^n \min(a_i, b_i)$$$ after an arbitrary rearrangement of $$$b$$$. Note that she only wants to know the maximum value of $$$P$$$, and no actual rearrangement is performed on $$$b$$$.
- There will be $$$q$$$ modifications. Each modification can be denoted by two integers $$$o$$$ and $$$x$$$ ($$$o$$$ is either $$$1$$$ or $$$2$$$, $$$1 \leq x \leq n$$$). If $$$o = 1$$$, then Iris will increase $$$a_x$$$ by $$$1$$$; otherwise, she will increase $$$b_x$$$ by $$$1$$$.
- Iris asks Chris the maximum value of $$$P$$$ for $$$q + 1$$$ times: once before any modification, then after every modification.
- Since $$$P$$$ might be huge, Chris only needs to calculate it modulo $$$998\,244\,353$$$.

Chris soon worked out this problem, but he was so tired that he fell asleep. Besides saying thanks to Chris, now it is your turn to write a program to calculate the answers for given input data.

Note: since the input and output are large, you may need to optimize them for this problem.

For example, in C++, it is enough to use the following lines at the start of the main() function:

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2\cdot 10^5$$$, $$$1 \leq q \leq 2\cdot 10^5$$$) — the length of the array and the number of operations.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 5\cdot 10^8$$$) — the array $$$a$$$.

The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \leq b_i \leq 5\cdot 10^8$$$) — the array $$$b$$$.

Then $$$q$$$ lines follow, each line contains two integers $$$o$$$ and $$$x$$$ ($$$o \in \{1, 2\}$$$, $$$1 \leq x \leq n$$$), representing an operation.

It's guaranteed that the sum of $$$n$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$4\cdot 10^5$$$, respectively.

Output: For each test case, output $$$q + 1$$$ integers in a line, representing the answers that Chris will calculate, modulo $$$998\,244\,353$$$.

Examples:
input:
4
3 4
1 1 2
3 2 1
1 3
2 3
1 1
2 1
6 8
1 4 2 7 3 5
7 6 5 6 3 3
2 5
1 6
1 5
1 5
1 5
2 3
2 3
1 6
13 8
7 7 6 6 5 5 5 2 2 3 4 5 1
1 4 1 9 6 6 9 1 5 1 3 8 4
2 2
2 11
2 4
2 4
1 7
1 1
2 12
1 5
5 3
10000000 20000000 30000000 40000000 50000000
10000000 20000000 30000000 40000000 50000000
1 1
2 2
2 1
output:
2 3 3 6 6
840 840 1008 1344 1680 2016 2016 2016 2352
2116800 2646000 3528000 3528000 3528000 4233600 4838400 4838400 4838400
205272023 205272023 205272023 264129429

Note: In the first test case:

- Before the modifications, Chris can rearrange $$$b$$$ to $$$[1, 2, 3]$$$ so that $$$P = \prod\limits_{i=1}^n \min(a_i, b_i) = 1 \cdot 1 \cdot 2 = 2$$$. We can prove that this is the maximum possible value. For example, if Chris rearranges $$$b = [2, 3, 1]$$$, $$$P$$$ will be equal $$$1 \cdot 1 \cdot 1 = 1 < 2$$$, which is not optimal.
- After the first modification, Chris can rearrange $$$b$$$ to $$$[1, 2, 3]$$$ so that $$$P = 1 \cdot 1 \cdot 3 = 3$$$, which is maximized.
- After the second modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 3]$$$ so that $$$P = 1 \cdot 1 \cdot 3 = 3$$$, which is maximized.
- After the third modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 3]$$$ so that $$$P = 6$$$, which is maximized.
- After the fourth modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 4]$$$ so that $$$P = 6$$$, which is maximized.

Buggy Code:
import java.util.*;

public class D_Refined_Product_Optimality {

    static final int MODULUS = 998244353;

    static long modInverse(long x, long modulus) {
        long result = 1;
        long exponent = modulus - 2;

        while (exponent > 0) {
            if ((exponent & 1) == 1) {
                result = result * x % modulus;
            }
            x = x * x % modulus;
            exponent >>= 1;
        }

        return result;
    }

    static void addMod(long[] result, long value) {
        result[0] = ((result[0] % MODULUS) + (value % MODULUS)) % MODULUS;
    }

    static void subtractMod(long[] result, long value) {
        result[0] = ((result[0] % MODULUS) - (value % MODULUS) + MODULUS) % MODULUS;
    }

    static void multiplyMod(long[] result, long value) {
        result[0] = ((result[0] % MODULUS) * (value % MODULUS)) % MODULUS;
    }

    static int findUpperBound(long[] array, long value) {
        int left = 0;
        int right = array.length;

        while (left < right) {
            int mid = left + (right - left) / 2;
            if (array[mid] > value) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }

        return left;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        int testCases = scanner.nextInt();

        while (testCases-- > 0) {
            int n = scanner.nextInt();
            int q = scanner.nextInt();

            long[] arrA = new long[n];
            long[] arrB = new long[n];

            for (int i = 0; i < n; i++) {
                arrA[i] = scanner.nextLong();
            }

            for (int i = 0; i < n; i++) {
                arrB[i] = scanner.nextLong();
            }

            long[] sortedA = Arrays.copyOf(arrA, arrA.length);
            long[] sortedB = Arrays.copyOf(arrB, arrB.length);

            Arrays.sort(sortedA);
            Arrays.sort(sortedB);

            long product = 1;
            for (int i = 0; i < n; i++) {
                product = (product * Math.min(sortedA[i], sortedB[i])) % MODULUS;
            }

            System.out.print(product + " ");

            for (int i = 0; i < q; i++) {
                int operation = scanner.nextInt();
                int index = scanner.nextInt() - 1;
                int position;

                if (operation == 1) {

                    position = findUpperBound(sortedA, arrA[index]) - 1;
                    arrA[index]++;

                    if (sortedA[position] < sortedB[position]) {
                        product = product * (sortedA[position] + 1) % MODULUS;
                        product = product * modInverse(sortedA[position], MODULUS) % MODULUS;
                    }
                    sortedA[position]++;
                    System.out.print(product + " ");
                } else {

                    position = findUpperBound(sortedB, arrB[index]) - 1;
                    arrB[index]++;

                    if (sortedB[position] < sortedA[position]) {
                        product = product * (sortedB[position] + 1) % MODULUS;
                        product = product * modInverse(sortedB[position], MODULUS) % MODULUS;
                    }
                    sortedB[position]++;
                    System.out.print(product + " ");
                }
            }

            System.out.println();
        }

        scanner.close();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on scenarios that can cause issues in the logic. The primary issues in the code are related to how the product is updated and the handling of the sorted arrays when elements are incremented.

### Key Points to Consider:
1. **Initial Product Calculation:**
   - The initial product is calculated using the sorted arrays.
   - Ensure that the initial arrays have elements that will cause the product to be non-trivial.

2. **Increment Operations:**
   - The code updates the product by multiplying and dividing by the updated and original values.
   - Ensure that the increments cause the product to change significantly, especially when elements are equal or close to each other.

3. **Edge Cases:**
   - Consider cases where the arrays have repeated elements.
   - Consider cases where the arrays have elements that are close to each other.
   - Consider cases where the arrays have elements that are far apart.

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

### Explanation:
- **Initial Arrays:**
  - `a = [1, 2, 3, 4, 5]`
  - `b = [5, 4, 3, 2, 1]`
  - The initial product is calculated as:
    - `P = min(1, 5) * min(2, 4) * min(3, 3) * min(4, 2) * min(5, 1) = 1 * 2 * 3 * 2 * 1 = 12`

- **Operations:**
  - **Operation 1: Increase `a[5]` by 1**
    - `a = [1, 2, 3, 4, 6]`
    - `sortedA = [1, 2, 3, 4, 6]`
    - `sortedB = [1, 2, 3, 4, 5]`
    - `P = 12 * 6 / 5 = 12 * 6 * 629145601 = 7551683212 (mod 998244353) = 12 * 6 = 72 / 5 = 14.4 (mod 998244353) = 14.4 = 14.4 * 629145601 = 9012124812 (mod 998244353) = 72`
  - **Operation 2: Increase `b[5]` by 1**
    - `b = [5, 4, 3, 2, 2]`
    - `sortedA = [1, 2, 3, 4, 6]`
    - `sortedB = [2, 2, 3, 4, 5]`
    - `P = 72 * 2 / 1 = 144`
  - **Operation 3: Increase `a[1]` by 1**
    - `a = [2, 2, 3, 4, 6]`
    - `sortedA = [2, 2, 3, 4, 6]`
    - `sortedB = [2, 2, 3, 4, 5]`
    - `P = 144 * 2 / 1 = 288`
  - **Operation 4: Increase `b[1]` by 1**
    - `b = [6, 4, 3, 2, 2]`
    - `sortedA = [2, 2, 3, 4, 6]`
    - `sortedB = [2, 2, 3, 4, 6]`
    - `P = 288 * 6 / 2 = 864`
  - **Operation 5: Increase `a[3]` by 1**
    - `a = [2, 2, 4, 4, 6]`
    - `sortedA = [2, 2, 4, 4, 6]`
    - `sortedB = [2, 2, 3, 4, 6]`
    - `P = 864 * 4 / 3 = 1152`

This test input is designed to challenge the code by causing significant changes in the product and testing the handling of repeated and close elements.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Refined Product Optimality
time_limit_ms: 3000
memory_limit_mb: 512
Description: Although Iris occasionally sets a problem where the solution is possibly wrong, she still insists on creating problems with her imagination; after all, everyone has always been on the road with their stubbornness... And like ever before, Iris has set a problem to which she gave a wrong solution, but Chris is always supposed to save it! You are going to play the role of Chris now:

- Chris is given two arrays $$$a$$$ and $$$b$$$, both consisting of $$$n$$$ integers.
- Iris is interested in the largest possible value of $$$P = \prod\limits_{i=1}^n \min(a_i, b_i)$$$ after an arbitrary rearrangement of $$$b$$$. Note that she only wants to know the maximum value of $$$P$$$, and no actual rearrangement is performed on $$$b$$$.
- There will be $$$q$$$ modifications. Each modification can be denoted by two integers $$$o$$$ and $$$x$$$ ($$$o$$$ is either $$$1$$$ or $$$2$$$, $$$1 \leq x \leq n$$$). If $$$o = 1$$$, then Iris will increase $$$a_x$$$ by $$$1$$$; otherwise, she will increase $$$b_x$$$ by $$$1$$$.
- Iris asks Chris the maximum value of $$$P$$$ for $$$q + 1$$$ times: once before any modification, then after every modification.
- Since $$$P$$$ might be huge, Chris only needs to calculate it modulo $$$998\,244\,353$$$.

Chris soon worked out this problem, but he was so tired that he fell asleep. Besides saying thanks to Chris, now it is your turn to write a program to calculate the answers for given input data.

Note: since the input and output are large, you may need to optimize them for this problem.

For example, in C++, it is enough to use the following lines at the start of the main() function:

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2\cdot 10^5$$$, $$$1 \leq q \leq 2\cdot 10^5$$$) — the length of the array and the number of operations.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 5\cdot 10^8$$$) — the array $$$a$$$.

The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \leq b_i \leq 5\cdot 10^8$$$) — the array $$$b$$$.

Then $$$q$$$ lines follow, each line contains two integers $$$o$$$ and $$$x$$$ ($$$o \in \{1, 2\}$$$, $$$1 \leq x \leq n$$$), representing an operation.

It's guaranteed that the sum of $$$n$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$4\cdot 10^5$$$, respectively.

Output: For each test case, output $$$q + 1$$$ integers in a line, representing the answers that Chris will calculate, modulo $$$998\,244\,353$$$.

Examples:
input:
4
3 4
1 1 2
3 2 1
1 3
2 3
1 1
2 1
6 8
1 4 2 7 3 5
7 6 5 6 3 3
2 5
1 6
1 5
1 5
1 5
2 3
2 3
1 6
13 8
7 7 6 6 5 5 5 2 2 3 4 5 1
1 4 1 9 6 6 9 1 5 1 3 8 4
2 2
2 11
2 4
2 4
1 7
1 1
2 12
1 5
5 3
10000000 20000000 30000000 40000000 50000000
10000000 20000000 30000000 40000000 50000000
1 1
2 2
2 1
output:
2 3 3 6 6
840 840 1008 1344 1680 2016 2016 2016 2352
2116800 2646000 3528000 3528000 3528000 4233600 4838400 4838400 4838400
205272023 205272023 205272023 264129429

Note: In the first test case:

- Before the modifications, Chris can rearrange $$$b$$$ to $$$[1, 2, 3]$$$ so that $$$P = \prod\limits_{i=1}^n \min(a_i, b_i) = 1 \cdot 1 \cdot 2 = 2$$$. We can prove that this is the maximum possible value. For example, if Chris rearranges $$$b = [2, 3, 1]$$$, $$$P$$$ will be equal $$$1 \cdot 1 \cdot 1 = 1 < 2$$$, which is not optimal.
- After the first modification, Chris can rearrange $$$b$$$ to $$$[1, 2, 3]$$$ so that $$$P = 1 \cdot 1 \cdot 3 = 3$$$, which is maximized.
- After the second modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 3]$$$ so that $$$P = 1 \cdot 1 \cdot 3 = 3$$$, which is maximized.
- After the third modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 3]$$$ so that $$$P = 6$$$, which is maximized.
- After the fourth modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 4]$$$ so that $$$P = 6$$$, which is maximized.

Buggy Code:
import java.io.*;
import java.util.*;

public class Main {
    class FenwickTree {
        int size;
        long[] tree;
        long modulus;

        FenwickTree(int size, long modulus) {
            this.size = size;
            this.modulus = modulus;
            tree = new long[size + 1];
        }

        void update(int index, long delta) {
            for (; index <= size; index += index & -index) {
                tree[index] += delta;
            }
        }

        long query(int index) {
            long sum = 0;
            for (; index > 0; index -= index & -index) {
                sum += tree[index];
            }
            return sum;
        }
    }
    static final long MODULUS = 998244353L;

    static long modularExponentiation(long base, long exponent, long modulus) {
        long result = 1 % modulus;
        base %= modulus;
        while (exponent > 0) {
            if ((exponent & 1) == 1) result = (result * base) % modulus;
            base = (base * base) % modulus;
            exponent >>= 1;
        }
        return result;
    }

    static int findKthSmallest(FenwickTree fenwick, int size, long k) {
        int position = 0;
        int bitMask = 1 << 20;
        while (bitMask > 0) {
            int next = position + bitMask;
            if (next <= fenwick.size && fenwick.tree[next] < k) {
                k -= fenwick.tree[next];
                position = next;
            }
            bitMask >>= 1;
        }
        return position + 1;
    }

    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

        int testCases = Integer.parseInt(reader.readLine().trim());

        while (testCases-- > 0) {
            String[] dimensions = reader.readLine().split("\\s+");
            int size = Integer.parseInt(dimensions[0]);
            int queries = Integer.parseInt(dimensions[1]);

            long[] arrayA = new long[size];
            long[] arrayB = new long[size];

            String[] inputA = reader.readLine().split("\\s+");
            for (int i = 0; i < size; i++) {
                arrayA[i] = Long.parseLong(inputA[i]);
            }

            String[] inputB = reader.readLine().split("\\s+");
            for (int i = 0; i < size; i++) {
                arrayB[i] = Long.parseLong(inputB[i]);
            }

            StringBuilder result = new StringBuilder();

            Runnable calculateProduct = () -> {
                long[] sortedA = arrayA.clone();
                long[] sortedB = arrayB.clone();
                Arrays.sort(sortedA);
                Arrays.sort(sortedB);

                long product = 1L;
                for (int i = 0; i < size; i++) {
                    long minimum = Math.min(sortedA[i], sortedB[i]);
                    product = (product * (minimum % MODULUS)) % MODULUS;
                }
                result.append(product).append(" ");
            };

            calculateProduct.run();

            for (int i = 0; i < queries; i++) {
                String[] query = reader.readLine().split("\\s+");
                int operation = Integer.parseInt(query[0]);
                int index = Integer.parseInt(query[1]) - 1;

                if (operation == 1) {
                    arrayA[index]++;
                } else {
                    arrayB[index]++;
                }
                calculateProduct.run();
            }

            writer.println(result.toString().trim());
        }

        writer.flush();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Refined Product Optimality
time_limit_ms: 3000
memory_limit_mb: 512
Description: Although Iris occasionally sets a problem where the solution is possibly wrong, she still insists on creating problems with her imagination; after all, everyone has always been on the road with their stubbornness... And like ever before, Iris has set a problem to which she gave a wrong solution, but Chris is always supposed to save it! You are going to play the role of Chris now:

- Chris is given two arrays $$$a$$$ and $$$b$$$, both consisting of $$$n$$$ integers.
- Iris is interested in the largest possible value of $$$P = \prod\limits_{i=1}^n \min(a_i, b_i)$$$ after an arbitrary rearrangement of $$$b$$$. Note that she only wants to know the maximum value of $$$P$$$, and no actual rearrangement is performed on $$$b$$$.
- There will be $$$q$$$ modifications. Each modification can be denoted by two integers $$$o$$$ and $$$x$$$ ($$$o$$$ is either $$$1$$$ or $$$2$$$, $$$1 \leq x \leq n$$$). If $$$o = 1$$$, then Iris will increase $$$a_x$$$ by $$$1$$$; otherwise, she will increase $$$b_x$$$ by $$$1$$$.
- Iris asks Chris the maximum value of $$$P$$$ for $$$q + 1$$$ times: once before any modification, then after every modification.
- Since $$$P$$$ might be huge, Chris only needs to calculate it modulo $$$998\,244\,353$$$.

Chris soon worked out this problem, but he was so tired that he fell asleep. Besides saying thanks to Chris, now it is your turn to write a program to calculate the answers for given input data.

Note: since the input and output are large, you may need to optimize them for this problem.

For example, in C++, it is enough to use the following lines at the start of the main() function:

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2\cdot 10^5$$$, $$$1 \leq q \leq 2\cdot 10^5$$$) — the length of the array and the number of operations.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 5\cdot 10^8$$$) — the array $$$a$$$.

The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \leq b_i \leq 5\cdot 10^8$$$) — the array $$$b$$$.

Then $$$q$$$ lines follow, each line contains two integers $$$o$$$ and $$$x$$$ ($$$o \in \{1, 2\}$$$, $$$1 \leq x \leq n$$$), representing an operation.

It's guaranteed that the sum of $$$n$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$4\cdot 10^5$$$, respectively.

Output: For each test case, output $$$q + 1$$$ integers in a line, representing the answers that Chris will calculate, modulo $$$998\,244\,353$$$.

Examples:
input:
4
3 4
1 1 2
3 2 1
1 3
2 3
1 1
2 1
6 8
1 4 2 7 3 5
7 6 5 6 3 3
2 5
1 6
1 5
1 5
1 5
2 3
2 3
1 6
13 8
7 7 6 6 5 5 5 2 2 3 4 5 1
1 4 1 9 6 6 9 1 5 1 3 8 4
2 2
2 11
2 4
2 4
1 7
1 1
2 12
1 5
5 3
10000000 20000000 30000000 40000000 50000000
10000000 20000000 30000000 40000000 50000000
1 1
2 2
2 1
output:
2 3 3 6 6
840 840 1008 1344 1680 2016 2016 2016 2352
2116800 2646000 3528000 3528000 3528000 4233600 4838400 4838400 4838400
205272023 205272023 205272023 264129429

Note: In the first test case:

- Before the modifications, Chris can rearrange $$$b$$$ to $$$[1, 2, 3]$$$ so that $$$P = \prod\limits_{i=1}^n \min(a_i, b_i) = 1 \cdot 1 \cdot 2 = 2$$$. We can prove that this is the maximum possible value. For example, if Chris rearranges $$$b = [2, 3, 1]$$$, $$$P$$$ will be equal $$$1 \cdot 1 \cdot 1 = 1 < 2$$$, which is not optimal.
- After the first modification, Chris can rearrange $$$b$$$ to $$$[1, 2, 3]$$$ so that $$$P = 1 \cdot 1 \cdot 3 = 3$$$, which is maximized.
- After the second modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 3]$$$ so that $$$P = 1 \cdot 1 \cdot 3 = 3$$$, which is maximized.
- After the third modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 3]$$$ so that $$$P = 6$$$, which is maximized.
- After the fourth modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 4]$$$ so that $$$P = 6$$$, which is maximized.

Buggy Code:
#pragma GCC optimize("Ofast","unroll-loops")
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define int unsigned long long
#define N 10000005
#define pYES cout<<"YES"<<endl;
#define pNO cout<<"NO"<<endl;

#define debug(x) cerr<<#x<<" "<<x<<endl;

template<typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
    os << "[ ";
    for (const auto& elem : v) os << elem << " ";
    os << "]";
    return os;
}

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

template<typename K, typename V>
ostream& operator<<(ostream& os, const map<K, V>& m) {
    os << "{ ";
    for (const auto& pair : m) os << "(" << pair.first << ": " << pair.second << ") ";
    os << "}";
    return os;
}

template<typename T>
ostream& operator<<(ostream& os, const unordered_set<T>& us) {
    os << "{ ";
    for (const auto& elem : us) os << elem << " ";
    os << "}";
    return os;
}

template<typename K, typename V>
ostream& operator<<(ostream& os, const unordered_map<K, V>& um) {
    os << "{ ";
    for (const auto& pair : um) os << "(" << pair.first << ": " << pair.second << ") ";
    os << "}";
    return os;
}

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

template<typename T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag,tree_order_statistics_node_update>;

int MOD=1e9+7;

struct pair_hash {
    inline std::size_t operator()(const std::pair<int,int> & v) const {
        return v.first*31+v.second;
    }
};

// ^^^^^^^^^^^^ MODULAR EXPO ^^^^^^^^^^^^^^

int modExp(int a,int b,int mod=MOD){
    int ans=1;
    while(b>0){
        if(b&1){
            ans=(ans*1LL*a)%mod;
        }
        a=(a*1LL*a)%mod;
        b=b>>1;
    }
    return ans;
}

// ^^^^^^^^^^^^ PRIME CHECK ^^^^^^^^^^^^^^

bool checkPrime(int a){
    for(int i=2;i*i<=a;i++){
        if(a%i==0) return false;
    }
    return true;
}

// ^^^^^^^^^^^^ SIEVE OF ERASTOSTHENES ^^^^^^^^^^^^^^

void sievePrime(int n){
    // prime vector stores prime check boolean
    vector<bool> prime(n+1,true);
    prime[1]=false;
    for(int i=2;i<=n;i++){
        if(prime[i] && i*i<=n){
            for(int j=i*i;j<=n;j+=i){
                prime[j]=false;
            }
        }
    }

}

// ^^^^^^^^^^^^ EUCLID GCD ^^^^^^^^^^^^^^
int gcd(int a,int b)
{
    if(a==0) return b;
    return gcd(b%a,a);
}

int lcm(int a, int b) { return a / gcd(a, b) * b; }

long long mod=1e9+7;
int nCrModp(int n, int r, int p=mod)
{
    if (r > n - r)
        r = n - r;
    int C[r + 1];
    memset(C, 0, sizeof(C));

    C[0] = 1;
    for (int i = 1; i <= n; i++) {
        for (int j = min(i, r); j > 0; j--)
            C[j] = (C[j] + C[j - 1]) % p;
    }
    return C[r];
}

// ^^^^^^^^^^^^ TO CHECK IF NUMBER IS POWER OF TWO ^^^^^^^^^^^^^^

bool isPowerOfTwo(int x)
{
    return x && (!(x & (x - 1)));
}

void dfs(vector<set<int>>& adj,int node,vector<bool>& vis){
    vis[node]=true;
    for(auto i : adj[node]){
        if(!vis[i]){
            dfs(adj,i,vis);
        }
    }
}

int getPos(vector<pair<int,int>> vec,int val,bool flag){
    int n=vec.size();
    int pos=-1;
    int l=0,r=n-1;
    while(l<=r){
        int mid=l+(r-l)/2;
        if(vec[mid].first<val){
            l=mid+1;
        }
        else if(vec[mid].first==val){
            if(flag) l=mid+1;
            else r=mid-1;
            pos=mid;
        }
        else{
            r=mid-1;
        }
    }
    return pos;
}

// Function to perform modular exponentiation
int power(int base, int exp, int mod) {
    int result = 1;
    base = base % mod;
    while (exp > 0) {
        if (exp % 2 == 1) { // If exp is odd
            result = (result * base) % mod;
        }
        exp = exp >> 1;     // Divide exp by 2
        base = (base * base) % mod;
    }
    return result;
}

// Function to find modular inverse using Fermat's Little Theorem
int modInverse(int a, int m) {
    return power(a, m - 2, m);
}

long long calcInv(long long b, long long c, long long d, long long M) {
    // cout<<"calcInv: "<<b<<" "<<c<<" "<<d<<endl;
    long long b_inv = modInverse(b, M); // Find modular inverse of b
    long long c_div_b = (c * b_inv) % M; // Compute (c / b) mod M
    long long result = (c_div_b * d) % M; // Multiply by d and take modulo M
    return result;
}

void solve(){
    int m=998244353;
    int n,q;
    cin>>n>>q;
    vector<pair<int,int>> a(n),b(n);
    map<int,int> mp1, mp2;
    for(int i=0;i<n;i++){
        cin>>a[i].first;
        a[i].second=i;
    }
    for(int i=0;i<n;i++){
        cin>>b[i].first;
        // b[i].first*=(-1);
        b[i].second=i;
    }
    sort(a.begin(),a.end());
    sort(b.begin(),b.end());
    for(int i=0;i<n;i++){
        mp1[a[i].second]=i;
    }
    for(int i=0;i<n;i++){
        mp2[b[i].second]=i;
    }
    int prod=1;
    int p=1;
    for(int i=0;i<n;i++){
        prod=(prod*min(a[i].first,b[i].first))%m;
        // prod*=(min(a[i].first,b[i].first));
    }
    // cout<<p<<" ";
    cout<<prod%m<<" ";
    for(int i=0;i<q;i++){
        int o,x;
        cin>>o>>x;
        x-=1;
        if(o==1){
            int j=mp1[x];
            // cout<<"here: "<<j<<" "<<(b[j].first)<<" "<<a[j].first<<endl;
            // if(b[j].first<=a[j].first){
            //     int pos=getPos(a,a[j].first,true);
            //     // cout<<"pos: "<<pos<<endl;
            //     int tp=j;
            //     mp1[a[j].second]=pos;
            //     mp1[a[pos].second]=tp;
            //     int temp=a[j].second;
            //     a[j].second=a[pos].second;
            //     a[pos].second=temp;
            //     a[pos].first+=1;
            // }
            // else{
            //     // prod/=a[j].first;

            //     // prod*=a[pos].first;
            // }

            int pos=getPos(a,a[j].first,true);
                if(b[pos].first>a[j].first) prod = calcInv(a[j].first,prod,a[pos].first+1,m);
                // cout<<"pos: "<<pos<<endl;
                int tp=j;
                mp1[a[j].second]=pos;
                mp1[a[pos].second]=tp;
                int temp=a[j].second;
                a[j].second=a[pos].second;
                a[pos].second=temp;
                a[pos].first+=1;
        }
        else{
            int j=mp2[x];
            // if((b[j].first)>=a[j].first){
            //     int pos=getPos(b,b[j].first,true);
            //     int tp=j;
            //     mp2[b[j].second]=pos;
            //     mp2[b[pos].second]=tp;
            //     int temp=b[j].second;
            //     b[j].second=b[pos].second;
            //     b[pos].second=temp;
            //     b[pos].first+=1;
            // }
            // else{
            //     // prod/=(b[j].first);
            //     int pos=getPos(b,b[j].first,true);
            //     if(a[pos].first>b[j].first) prod = calcInv(b[j].first,prod,b[pos].first+1,m);
            //     int tp=j;
            //     mp2[b[j].second]=pos;
            //     mp2[b[pos].second]=tp;
            //     int temp=b[j].second;
            //     b[j].second=b[pos].second;
            //     b[pos].second=temp;
            //     b[pos].first+=1;
            //     // prod*=(b[pos].first);
            // }
            int pos=getPos(b,b[j].first,true);
                if(a[pos].first>b[j].first) prod = calcInv(b[j].first,prod,b[pos].first+1,m);
                int tp=j;
                mp2[b[j].second]=pos;
                mp2[b[pos].second]=tp;
                int temp=b[j].second;
                b[j].second=b[pos].second;
                b[pos].second=temp;
                b[pos].first+=1;
        }
        // for(int j=0;j<n;j++){
        //     cout<<" { "<<a[j].first<<" "<<a[j].second<<" } ";
        // }cout<<endl;
        // for(int j=0;j<n;j++){
        //     cout<<" { "<<b[j].first<<" "<<b[j].second<<" } ";
        // }cout<<endl;
        cout<<prod%m<<" ";
    }cout<<endl;
}

signed main(){
    fastio;
    #ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif
    int t=1;
    cin>>t;
    for(int i=1;i<=t;i++){
        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 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. Here are the steps to consider:

1. **Initial Arrays with Special Values**:
   - Use arrays with values that are close to each other to test the sorting and rearrangement logic.
   - Use arrays with duplicate values to test the handling of equal elements.

2. **Operations That Cause Rearrangements**:
   - Include operations that increment elements in such a way that the order of elements in the arrays changes.
   - Test operations that increment elements to values that are equal to or slightly greater than other elements in the array.

3. **Edge Cases**:
   - Test with arrays of length 1 and 2 to ensure the code handles small inputs correctly.
   - Test with arrays where all elements are the same to ensure the code handles uniform inputs correctly.

4. **Large Values**:
   - Use large values to test the modulo operation and ensure it works correctly.

Here is a small-scale test input that follows these guidelines:

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

This test input includes:
- Initial arrays with values that are close to each other.
- Operations that increment elements in such a way that the order of elements in the arrays changes.
- A small number of elements to ensure the test input is valid and complete.

This should help expose any issues in the sorting and rearrangement logic, as well as the handling of operations and the modulo operation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Refined Product Optimality
time_limit_ms: 3000
memory_limit_mb: 512
Description: Although Iris occasionally sets a problem where the solution is possibly wrong, she still insists on creating problems with her imagination; after all, everyone has always been on the road with their stubbornness... And like ever before, Iris has set a problem to which she gave a wrong solution, but Chris is always supposed to save it! You are going to play the role of Chris now:

- Chris is given two arrays $$$a$$$ and $$$b$$$, both consisting of $$$n$$$ integers.
- Iris is interested in the largest possible value of $$$P = \prod\limits_{i=1}^n \min(a_i, b_i)$$$ after an arbitrary rearrangement of $$$b$$$. Note that she only wants to know the maximum value of $$$P$$$, and no actual rearrangement is performed on $$$b$$$.
- There will be $$$q$$$ modifications. Each modification can be denoted by two integers $$$o$$$ and $$$x$$$ ($$$o$$$ is either $$$1$$$ or $$$2$$$, $$$1 \leq x \leq n$$$). If $$$o = 1$$$, then Iris will increase $$$a_x$$$ by $$$1$$$; otherwise, she will increase $$$b_x$$$ by $$$1$$$.
- Iris asks Chris the maximum value of $$$P$$$ for $$$q + 1$$$ times: once before any modification, then after every modification.
- Since $$$P$$$ might be huge, Chris only needs to calculate it modulo $$$998\,244\,353$$$.

Chris soon worked out this problem, but he was so tired that he fell asleep. Besides saying thanks to Chris, now it is your turn to write a program to calculate the answers for given input data.

Note: since the input and output are large, you may need to optimize them for this problem.

For example, in C++, it is enough to use the following lines at the start of the main() function:

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2\cdot 10^5$$$, $$$1 \leq q \leq 2\cdot 10^5$$$) — the length of the array and the number of operations.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 5\cdot 10^8$$$) — the array $$$a$$$.

The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \leq b_i \leq 5\cdot 10^8$$$) — the array $$$b$$$.

Then $$$q$$$ lines follow, each line contains two integers $$$o$$$ and $$$x$$$ ($$$o \in \{1, 2\}$$$, $$$1 \leq x \leq n$$$), representing an operation.

It's guaranteed that the sum of $$$n$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$4\cdot 10^5$$$, respectively.

Output: For each test case, output $$$q + 1$$$ integers in a line, representing the answers that Chris will calculate, modulo $$$998\,244\,353$$$.

Examples:
input:
4
3 4
1 1 2
3 2 1
1 3
2 3
1 1
2 1
6 8
1 4 2 7 3 5
7 6 5 6 3 3
2 5
1 6
1 5
1 5
1 5
2 3
2 3
1 6
13 8
7 7 6 6 5 5 5 2 2 3 4 5 1
1 4 1 9 6 6 9 1 5 1 3 8 4
2 2
2 11
2 4
2 4
1 7
1 1
2 12
1 5
5 3
10000000 20000000 30000000 40000000 50000000
10000000 20000000 30000000 40000000 50000000
1 1
2 2
2 1
output:
2 3 3 6 6
840 840 1008 1344 1680 2016 2016 2016 2352
2116800 2646000 3528000 3528000 3528000 4233600 4838400 4838400 4838400
205272023 205272023 205272023 264129429

Note: In the first test case:

- Before the modifications, Chris can rearrange $$$b$$$ to $$$[1, 2, 3]$$$ so that $$$P = \prod\limits_{i=1}^n \min(a_i, b_i) = 1 \cdot 1 \cdot 2 = 2$$$. We can prove that this is the maximum possible value. For example, if Chris rearranges $$$b = [2, 3, 1]$$$, $$$P$$$ will be equal $$$1 \cdot 1 \cdot 1 = 1 < 2$$$, which is not optimal.
- After the first modification, Chris can rearrange $$$b$$$ to $$$[1, 2, 3]$$$ so that $$$P = 1 \cdot 1 \cdot 3 = 3$$$, which is maximized.
- After the second modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 3]$$$ so that $$$P = 1 \cdot 1 \cdot 3 = 3$$$, which is maximized.
- After the third modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 3]$$$ so that $$$P = 6$$$, which is maximized.
- After the fourth modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 4]$$$ so that $$$P = 6$$$, which is maximized.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
constexpr int p = 998244353;
ll inv(ll a)
{
    a%=p;
    ll b=p-2;
    ll res=1;
    while(b)
    {
        if(b&1)res=(res*a)%p;
        a=(a*a)%p;
        b>>=1;
    }
    return res;
}
void solve()
{
    int n,q;
    cin>>n>>q;
    vector<int> a(n+1),b(n+1);
    vector<int> a2(n+1),b2(n+1);
    for(int i=1;i<=n;i++)cin>>a[i];
    for(int i=1;i<=n;i++)cin>>b[i];
    for(int i=1;i<=n;i++)
    {
        a2[i]=a[i];
        b2[i]=b[i];
    }
    map<int,pair<int,int>> mpa,mpb; //first=l second=r
    stable_sort(a2.begin()+1,a2.end());
    stable_sort(b2.begin()+1,b2.end());
    for(int i=1;i<=n;i++)
    {
        mpa[a2[i]].first=2e9;
        mpa[a2[i]].second=0;
        mpb[b2[i]].first=2e9;
        mpb[b2[i]].second=0;
    }
    for(int i=1;i<=n;i++)
    {
        mpa[a2[i]].first=min(mpa[a2[i]].first,i);
        mpa[a2[i]].second=max(mpa[a2[i]].second,i);
        mpb[b2[i]].first=min(mpb[b2[i]].first,i);
        mpb[b2[i]].second=max(mpb[b2[i]].second,i);
    }
    ll ans=1;
    for(int i=1;i<=n;i++)
    {
        ans*=min(a2[i],b2[i]);
        ans%=p;
    }
    cout<<ans<<' ';
    vector<pair<int,int>> c(q);
    // for(int i=0;i<q;i++)cin>>c[i].first>>c[i].second;
    for(int i=0;i<q;i++)
    {
        cin>>c[i].first>>c[i].second;
        if(c[i].first==1)
        {
            int ch=a[c[i].second];
            auto iter1=mpa.find(ch);
            auto iter2=mpa.find(ch+1);
            if(a2[mpa[ch].second]<b2[mpa[ch].second])
            {
                ans*=inv(min(a2[mpa[ch].second],b2[mpa[ch].second]));
                ans%=p;
                ans*=min(a2[mpa[ch].second]+1,b2[mpa[ch].second]);
                ans%=p;
            }
            a[c[i].second]++;
            a2[mpa[ch].second]++;
            if(iter2!=mpa.end())
            {
                mpa[ch+1].first--;
                mpa[ch].second--;
            }
            else
            {
                mpa[ch+1].first=mpa[ch].second;
                mpa[ch+1].second=mpa[ch].second;
                mpa[ch].second--;
            }
            if(mpa[ch].first>mpa[ch].second)mpa.erase(iter1);
        }
        else
        {
            int ch=b[c[i].second];
            auto iter1=mpb.find(ch);
            auto iter2=mpb.find(ch+1);
            if(b2[mpb[ch].second]<a2[mpb[ch].second])
            {
                ans*=inv(min(a2[mpb[ch].second],b2[mpb[ch].second]));
                ans%=p;
                ans*=min(a2[mpb[ch].second],b2[mpb[ch].second]+1);
                ans%=p;
            }
            b[c[i].second]++;
            b2[mpb[ch].second]++;
            if(iter2!=mpb.end())
            {
                mpb[ch+1].first--;
                mpb[ch].second--;
            }
            else
            {
                mpb[ch+1].first=mpb[ch].second;
                mpb[ch+1].second=mpb[ch].second;
                mpb[ch].second--;
            }
            if(mpb[ch].first>mpb[ch].second)mpb.erase(iter1);
        }
        // for(int i=1;i<=n;i++)cout<<a2[i]<<" \n"[i==n];
        // for(int i=1;i<=n;i++)cout<<b2[i]<<" \n"[i==n];
        cout<<ans<<' ';
    }
    cout<<'\n';
}
int main()
{
    // ios::sync_with_stdio(false);cin.tie(nullptr);
    int t;cin>>t;while(t--)
        solve();
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can highlight potential issues in the algorithm. Here are the steps to consider:

1. **Initial Arrays**: Choose initial values for arrays `a` and `b` that are close to each other but not identical. This will make the rearrangement and modification operations more critical.
2. **Modifications**: Include a mix of operations on both arrays `a` and `b` to ensure that the algorithm handles both types of operations correctly.
3. **Edge Cases**: Include cases where the values in `a` and `b` are equal, and cases where they are very different.
4. **Boundary Conditions**: Test with the smallest and largest possible values for `n` and `q` within the constraints.

Here is a small-scale test input that meets these criteria:

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

Explanation:
- **Initial Arrays**:
  - `a = [1, 2, 3, 4, 5]`
  - `b = [5, 4, 3, 2, 1]`
- **Modifications**:
  - Increase `a[1]` by 1: `a[1]` becomes 2
  - Increase `b[1]` by 1: `b[1]` becomes 6
  - Increase `a[5]` by 1: `a[5]` becomes 6
  - Increase `b[5]` by 1: `b[5]` becomes 2
  - Increase `a[3]` by 1: `a[3]` becomes 4

This test input will:
- Test the initial product calculation.
- Test the handling of modifications on both arrays.
- Test the handling of equal and different values in the arrays.
- Test the boundary conditions of the arrays and modifications.

By running this test input, we can observe if the buggy code correctly handles the rearrangements and modifications, and if it produces the expected output.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Refined Product Optimality
time_limit_ms: 3000
memory_limit_mb: 512
Description: Although Iris occasionally sets a problem where the solution is possibly wrong, she still insists on creating problems with her imagination; after all, everyone has always been on the road with their stubbornness... And like ever before, Iris has set a problem to which she gave a wrong solution, but Chris is always supposed to save it! You are going to play the role of Chris now:

- Chris is given two arrays $$$a$$$ and $$$b$$$, both consisting of $$$n$$$ integers.
- Iris is interested in the largest possible value of $$$P = \prod\limits_{i=1}^n \min(a_i, b_i)$$$ after an arbitrary rearrangement of $$$b$$$. Note that she only wants to know the maximum value of $$$P$$$, and no actual rearrangement is performed on $$$b$$$.
- There will be $$$q$$$ modifications. Each modification can be denoted by two integers $$$o$$$ and $$$x$$$ ($$$o$$$ is either $$$1$$$ or $$$2$$$, $$$1 \leq x \leq n$$$). If $$$o = 1$$$, then Iris will increase $$$a_x$$$ by $$$1$$$; otherwise, she will increase $$$b_x$$$ by $$$1$$$.
- Iris asks Chris the maximum value of $$$P$$$ for $$$q + 1$$$ times: once before any modification, then after every modification.
- Since $$$P$$$ might be huge, Chris only needs to calculate it modulo $$$998\,244\,353$$$.

Chris soon worked out this problem, but he was so tired that he fell asleep. Besides saying thanks to Chris, now it is your turn to write a program to calculate the answers for given input data.

Note: since the input and output are large, you may need to optimize them for this problem.

For example, in C++, it is enough to use the following lines at the start of the main() function:

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

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \leq n \leq 2\cdot 10^5$$$, $$$1 \leq q \leq 2\cdot 10^5$$$) — the length of the array and the number of operations.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 5\cdot 10^8$$$) — the array $$$a$$$.

The third line of each test case contains $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \leq b_i \leq 5\cdot 10^8$$$) — the array $$$b$$$.

Then $$$q$$$ lines follow, each line contains two integers $$$o$$$ and $$$x$$$ ($$$o \in \{1, 2\}$$$, $$$1 \leq x \leq n$$$), representing an operation.

It's guaranteed that the sum of $$$n$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$4\cdot 10^5$$$, respectively.

Output: For each test case, output $$$q + 1$$$ integers in a line, representing the answers that Chris will calculate, modulo $$$998\,244\,353$$$.

Examples:
input:
4
3 4
1 1 2
3 2 1
1 3
2 3
1 1
2 1
6 8
1 4 2 7 3 5
7 6 5 6 3 3
2 5
1 6
1 5
1 5
1 5
2 3
2 3
1 6
13 8
7 7 6 6 5 5 5 2 2 3 4 5 1
1 4 1 9 6 6 9 1 5 1 3 8 4
2 2
2 11
2 4
2 4
1 7
1 1
2 12
1 5
5 3
10000000 20000000 30000000 40000000 50000000
10000000 20000000 30000000 40000000 50000000
1 1
2 2
2 1
output:
2 3 3 6 6
840 840 1008 1344 1680 2016 2016 2016 2352
2116800 2646000 3528000 3528000 3528000 4233600 4838400 4838400 4838400
205272023 205272023 205272023 264129429

Note: In the first test case:

- Before the modifications, Chris can rearrange $$$b$$$ to $$$[1, 2, 3]$$$ so that $$$P = \prod\limits_{i=1}^n \min(a_i, b_i) = 1 \cdot 1 \cdot 2 = 2$$$. We can prove that this is the maximum possible value. For example, if Chris rearranges $$$b = [2, 3, 1]$$$, $$$P$$$ will be equal $$$1 \cdot 1 \cdot 1 = 1 < 2$$$, which is not optimal.
- After the first modification, Chris can rearrange $$$b$$$ to $$$[1, 2, 3]$$$ so that $$$P = 1 \cdot 1 \cdot 3 = 3$$$, which is maximized.
- After the second modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 3]$$$ so that $$$P = 1 \cdot 1 \cdot 3 = 3$$$, which is maximized.
- After the third modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 3]$$$ so that $$$P = 6$$$, which is maximized.
- After the fourth modification, Chris can rearrange $$$b$$$ to $$$[2, 2, 4]$$$ so that $$$P = 6$$$, which is maximized.

Buggy Code:
# ◇─◇────────────────────────────────────────◇─◇
import atexit
import os
from collections import deque
from io import BytesIO
from random import seed as _seed
from sys import stdin

input = BytesIO(os.read(0,os.fstat(0).st_size)).readline
read = input
read_int = lambda: int(input())
read_ints = lambda: [int(_) for _ in input().split()]
# ◇─◇────────────────────────────────────────◇─◇
from bisect import bisect_right

MOD = 998244353

def prepare(nums):
    n = len(nums)

    srt = [(x, idx) for idx, x in enumerate(nums)]
    srt.sort()

    to_srt = []

    to_real = {}
    to_fake = {}

    for idx, (x, jdx) in enumerate(srt):
        to_srt.append(x)
        to_real[idx] = jdx
        to_fake[jdx] = idx

    return to_srt, to_real, to_fake

def solve():
    n, q = read_ints()
    a = read_ints()
    b = read_ints()

    a_srt, a_to_real, a_to_fake = prepare(a)
    b_srt, b_to_real, b_to_fake = prepare(b)

    k = 1
    for l, r in zip(a_srt, b_srt):
        k *= min(l, r)
        k %= MOD

    ans = [str(k)]

    for _ in range(q):
        o, x = read_ints()
        x -= 1

        before = -1
        after = -1

        # update a
        if o == 1:
            real_x = x
            fake_x = a_to_fake[x]
            value_x = a_srt[fake_x]

            fake_y = bisect_right(a_srt, value_x) - 1
            real_y = a_to_real[fake_y]
            value_y = a_srt[fake_y]

            # need to update
            before = min(a_srt[fake_x], b_srt[fake_x])
            before *= min(a_srt[fake_y], b_srt[fake_y]) if fake_x != fake_y else 1

            # swap
            a_to_real[fake_x] = real_y
            a_to_real[fake_y] = real_x
            a_to_fake[real_x] = fake_y
            a_to_fake[real_y] = fake_x
            a_srt[fake_x], a_srt[fake_y] = a_srt[fake_y], a_srt[fake_x]

            # don't forget increase value by one!
            a_srt[fake_y] += 1

            # need to update
            after = min(a_srt[fake_x], b_srt[fake_x])
            after *= min(a_srt[fake_y], b_srt[fake_y]) if fake_x != fake_y else 1

        # update b
        if o == 2:
            real_x = x
            fake_x = b_to_fake[x]
            value_x = b_srt[fake_x]

            fake_y = bisect_right(b_srt, value_x) - 1
            real_y = b_to_real[fake_y]
            value_y = b_srt[fake_y]

            # need to update
            before = min(a_srt[fake_x], b_srt[fake_x])
            before *= min(a_srt[fake_y], b_srt[fake_y]) if fake_x != fake_y else 1

            # swap
            b_to_real[fake_x] = real_y
            b_to_real[fake_y] = real_x
            b_to_fake[real_x] = fake_y
            b_to_fake[real_y] = fake_x
            b_srt[fake_x], b_srt[fake_y] = b_srt[fake_y], b_srt[fake_x]

            # don't forget increase value by one!
            b_srt[fake_y] += 1

            # need to update
            after = min(a_srt[fake_x], b_srt[fake_x])
            after *= min(a_srt[fake_y], b_srt[fake_y]) if fake_x != fake_y else 1

        # update answer
        before %= MOD
        after %= MOD

        k *= pow(before, -1, MOD)
        k %= MOD
        k *= after
        k %= MOD
        ans.append(str(k))

    return ' '.join(ans)

if __name__ == '__main__':
    for _ in range(read_int()):
        print(solve())

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

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

Think step by step.

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