**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 sys

input = sys.stdin.readline
output = sys.stdout

def MII():
    return list(map(int, input().rstrip().split()))

def MI():
    return int(input())

from bisect import bisect_left
from bisect import bisect_right

modv = dict()
for _ in range(MI()):
    n, m = MII()
    arr1 = MII()
    arr2 = MII()
    ans = [0] * (m+1)
    a1I = list(zip(arr1, range(n)))
    a2I = list(zip(arr2, range(n)))
    a1I.sort()
    a2I.sort()
    mod = 998244353

    def mod_inverse(x, p):
        def extended_gcd(a, b):
            if b == 0:
                return a, 1, 0
            g, y, x = extended_gcd(b, a % b)
            return g, x, y - (a // b) * x
        if x in modv:
            return modv[x]
        g, a, _ = extended_gcd(x, p)
        if g != 1:
            raise ValueError("Inverse does not exist")
        modv[x] = a%p
        return modv[x]

    def calc():
        ret = 1
        for i, (v, _) in enumerate(a1I):
            ret *= min(v, a2I[i][0])
            ret %= mod
        return ret

    def op(v, index, listI, listC, pos, last):
        l = bisect_left(listI, v, key = lambda x: x[0])
        r = bisect_right(listI, v, key = lambda x: x[0])
        if r-l==1:
            last *= mod_inverse(min(listI[l][0], listC[l][0]), mod)
            listI[l] = (v+1, listI[l][1])
            last *= min(listI[l][0], listC[l][0])
            last %= mod
            return last
        else:
            k = pos[index]
            _, rindex = listI[r-1]
            if k!=r-1:
                last *= mod_inverse(min(listI[k][0], listC[k][0]), mod)
                last *= mod_inverse(min(listI[r-1][0], listC[r-1][0]), mod)
                listI[k] = (v+1, index)
                listI[k], listI[r-1] = listI[r-1], listI[k]
                pos[rindex] = k
                pos[index] = r-1
                last *= min(listI[k][0], listC[k][0])
                last *= min(listI[r-1][0], listC[r-1][0])
                last %= mod
            else:
                last *= mod_inverse(min(listI[k][0], listC[k][0]), mod)
                listI[k] = (v+1, index)
                last *= min(listI[k][0], listC[k][0])
                last %= mod
        return last

    posA = dict()
    posB = dict()
    for i, (x, y) in enumerate(a1I):
        posA[y] = i
    for i, (x,y) in enumerate(a2I):
        posB[y] = i
    ans[0] = calc()
    last = ans[0]
    for i in range(1, m+1):
        o, x = MII()
        if o==1:
            last = op(arr1[x-1], x-1, a1I, a2I, posA, last)
            arr1[x-1]+=1
        else:
            last = op(arr2[x-1], x-1, a2I, a1I, posB, last)
            arr2[x-1]+=1
        ans[i] = last
    output.write(" ".join([str(x) for x in ans])+"\n")

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

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

Think step by step.

Response:
```plaintext
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
```


**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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class Main {
	static kattio sc = new kattio();
	static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

	public static void main(String[] args) {
		//stack=new Stack<Integer>();
		//set=new HashSet<List<Integer>>();
		//f(0, new int[3]);System.out.println(set.size());
		//p1=false;

		  int t=sc.nextint(); while (t-->0) {

		  sovle(); }

		  out.close();

	}
	static void swapp(int x,int y,int []arr) {
		int c=arr[x];arr[x]=arr[y];arr[y]=c;
		x+=y;
	}
	static long[] ram,v;
	static int []size,arr;
	static TreeSet<Integer>treeSet;
	static int gcd(int a,int b) {
		return b==0?a:gcd(b, a%b);
	}
	static Map<Integer, Integer>vpo;
	static void sovle2() {
		long n=sc.nextlong();
		long k=sc.nextlong();
		long ans=0;
		if (n%2==0) {
			long ban=(n+1)/2;
			if (n%2==0) {
				ban=n+1;
				n/=2;

			}
			//System.out.println(n+" "+ban);
			while (n>=k) {
				if (n%2==1) {
					ans+=ban;
				}
				n/=2;ban*=2;
			}
		}else {
			long ban=(n+1)/2;

			while (n>=k) {
				if (n%2==1) {
					ans+=ban;
				}
				n/=2;ban*=2;
			}
		}

		out.println(ans);
	}

	static void sovle() {
		int n=sc.nextint();
		int q=sc.nextint();
		int [][]arr=new int[n][2];
		int [][]brr=new int[n][2];
		for (int i = 0; i < brr.length; i++) {
			arr[i][0]=sc.nextint();
			arr[i][1]=i;
		}
		for (int i = 0; i < brr.length; i++) {
			brr[i][0]=sc.nextint();
			brr[i][1]=i;
		}
		Arrays.sort(arr,(a,b)->a[0]-b[0]);
		Arrays.sort(brr,(a,b)->a[0]-b[0]);
		long ans=1;
		int []wa=new int[n];
		int []wb=new int[n];
		for (int i = 0; i < brr.length; i++) {
			ans*=(long)Math.min(arr[i][0], brr[i][0]);ans%=mod;
			wa[arr[i][1]]=i;
			wb[brr[i][1]]=i;
		}
		out.print(ans+" ");
		for (int i = 0; i < q; i++) {
			int op=sc.nextint();int x=sc.nextint()-1;
			if (op==1) {
				x=wa[x];
				if (x==n-1||arr[x][0]+1<=arr[x+1][0]) {
					arr[x][0]++;
					if (arr[x][0]<=brr[x][0]) {
						ans*=pow(arr[x][0]-1, mod-2);ans%=mod;ans*=arr[x][0];ans%=mod;
					}
				}else {

					int j=er(x, arr);ans=del(ans, arr, brr, x);ans=del(ans, arr, brr, j);arr[x][0]++;
					swap(x, j, wa, arr);
					ans=add(ans, arr, brr, x);ans=add(ans, arr, brr, j);
					//System.out.println(x+" "+j);
				}
			}else {
				x=wb[x];
				if (x==n-1||brr[x][0]+1<=brr[x+1][0]) {
					brr[x][0]++;
					if (brr[x][0]<=arr[x][0]) {
						ans*=pow(brr[x][0]-1, mod-2);;ans%=mod;ans*=brr[x][0];ans%=mod;
					}
				}else {
					int j=er(x, brr);ans=del(ans, arr, brr, x);ans=del(ans, arr, brr, j);brr[x][0]++;
					//System.out.println(ans+" "+j+" "+x);
					swap(x, j, wb, brr);
					ans=add(ans, arr, brr, x);ans=add(ans, arr, brr, j);
				}
			}
			for (int j = 0; j < wb.length; j++) {
				//System.out.println(Arrays.toString(arr[j])+" "+wa[j]+" "+Arrays.toString(brr[j])+" "+wb[j]);
			}
			out.print(ans+" ");
		}
		out.println();
	}
	static int er(int i,int [][]arr) {
		int l=i;int r=arr.length-1;int mid;
		while (l<r) {
			mid=(l+r+1)/2;
			if (arr[mid][0]<arr[i][0]+1) {
				l=mid;
			}else {
				r=mid-1;
			}
		}
		return l;
	}
	static long del(long ans,int[][]arr,int [][]brr,int i) {
		//System.out.println(Math.min(arr[i][0], brr[i][0])+" "+ans+" "+(ans*pow(Math.min(arr[i][0], brr[i][0]), mod-2)%mod));
		ans=(long) (ans*pow(Math.min(arr[i][0], brr[i][0]), mod-2));ans%=mod;

		return ans%mod;
	}
	static long  add(long ans,int[][]arr,int [][]brr,int i) {
		ans*=Math.min(arr[i][0], brr[i][0]);ans%=mod;
		return ans;
	}
	static void swap(int x,int j,int []wa,int [][]arr) {
		int v=arr[x][0];int wv=arr[x][1];arr[x][0]=arr[j][0];arr[x][1]=arr[j][1];arr[j][0]=v;arr[j][1]=wv;
		wa[arr[x][1]]=x;wa[arr[j][1]]=j;
	}

	static long  pow(long x,long a) {
		if (a==0) {
			return 1;
		}else {
			long c=pow(x, a/2);
			if (a%2==0) {
				return c*c%mod;
			}else {
				return c*c%mod*x%mod;
			}
		}
	}
	static long mod = (long) (998244353);

	static void sovle1() {
		int n=sc.nextint();
		int m=sc.nextint();

	}

	static long []c;
	static int lowbit(int x) {
		return x&-x;
	}
	static void add(int x,long v) {
		while (x<c.length) {
			//System.out.println(x);
			c[x]+=v;
			x+=lowbit(x);
		}
	}
	static long sum(int x) {
		long ans=0;
		//System.out.println(x+"dasdsafadf");
		while (x>0) {
			//System.out.println(x);
			ans+=c[x];
			x-=lowbit(x);
		}
		return ans;
	}
	static Map<Integer, List<Integer>>map;

}

class kattio extends PrintWriter {

	BufferedReader r;
	StringTokenizer st;

	public kattio() {
		this(System.in, System.out);
	}

	public kattio(InputStream i, OutputStream o) {
		super(o);
		r = new BufferedReader(new InputStreamReader(i));
	}

	public kattio(String in, String out) throws IOException {
		super(out);
		r = new BufferedReader(new FileReader(in));
	}

	public String next() {
		try {
			while (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(r.readLine());
			}
			return st.nextToken();
		} catch (Exception e) {
			// TODO: handle exception
			return null;
		}
	}

	public int nextint() {
		return Integer.parseInt(next());
	}

	public long nextlong() {
		return Long.parseLong(next());
	}

	public double nextdouble() {
		return Double.parseDouble(next());
	}
}

**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 cause the faulty solution to fail. Here are the steps to consider:

1. **Initial Setup:**
   - Choose a small but non-trivial size for the arrays `a` and `b`.
   - Ensure that the initial values of `a` and `b` are such that the product `P` is not immediately maximized and can be influenced by the modifications.

2. **Modifications:**
   - Design modifications that can cause the algorithm to make incorrect decisions.
   - Include a mix of operations on both `a` and `b` to test the handling of different scenarios.

3. **Edge Cases:**
   - Include cases where the values in `a` and `b` are very close to each other.
   - Include cases where the values in `a` and `b` are significantly different.
   - Include cases where the values in `a` and `b` are equal.

4. **Output:**
   - Ensure the test input is valid and follows the format specified in the problem description.

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

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

**Explanation:**
- **Initial Arrays:**
  - `a = [1, 2, 3, 4, 5]`
  - `b = [5, 4, 3, 2, 1]`
  - The initial product `P` before any modifications is `1 * 2 * 3 * 2 * 1 = 12`.

- **Modifications:**
  - **1 1:** Increase `a[1]` by 1, so `a` becomes `[2, 2, 3, 4, 5]`.
  - **2 2:** Increase `b[2]` by 1, so `b` becomes `[5, 5, 3, 2, 1]`.
  - **1 3:** Increase `a[3]` by 1, so `a` becomes `[2, 2, 4, 4, 5]`.

- **Expected Output:**
  - The product `P` after each modification should be calculated and printed.

This test input is designed to challenge the algorithm's ability to correctly handle the rearrangement and product calculation, especially when the values in `a` and `b` are close to each other and the modifications affect the optimal product.


**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.PrintWriter;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Scanner;

public class Main {

    static final long MOD = 998244353;

    public static void main(String[] args) {

        var sc = new Scanner(System.in);
        var pw = new PrintWriter(System.out);

        int T = Integer.parseInt(sc.next());
        for(int t = 0; t < T; t++){
            int n = Integer.parseInt(sc.next());
            int Q = Integer.parseInt(sc.next());
            var a = new Pair[n];
            for(int i = 0; i < n; i++){
                a[i] = new Pair(i, Long.parseLong(sc.next()));
            }
            var b = new Pair[n];
            for(int i = 0; i < n; i++){
                b[i] = new Pair(i, Long.parseLong(sc.next()));
            }

            Arrays.sort(a);
            Arrays.sort(b);

            var posA = new int[n];
            var posB = new int[n];
            for(int i = 0; i < n; i++){
                posA[a[i].id] = i;
                posB[b[i].id] = i;
            }

            var mapA = new HashMap<Long, Integer>();
            var leftA = new int[n+Q];
            var rightA = new int[n+Q];
            int indexA = 0;
            int l = 0;
            int r = 0;
            while(l < n){
                while(r < n-1 && a[l].value == a[r+1].value){
                    r++;
                }
                mapA.put(a[l].value, indexA);
                leftA[indexA] = l;
                rightA[indexA] = r;
                indexA++;
                l = r+1;
                r = l;
            }

            var mapB = new HashMap<Long, Integer>();
            var leftB = new int[n+Q];
            var rightB = new int[n+Q];
            int indexB = 0;
            l = 0;
            r = 0;
            while(l < n){
                while(r < n-1 && b[l].value == b[r+1].value){
                    r++;
                }
                mapB.put(b[l].value, indexB);
                leftB[indexB] = l;
                rightB[indexB] = r;
                indexB++;
                l = r+1;
                r = l;
            }

            long p = 1;
            for(int i = 0; i < n; i++){
                p *= Math.min(a[i].value, b[i].value);
                p %= MOD;
            }
            pw.print(p);

            for(int q = 0; q < Q; q++){
                int o = Integer.parseInt(sc.next());
                int x = Integer.parseInt(sc.next()) - 1;

                switch(o){
                    case 1 -> {
                        int pos1 = posA[x];
                        long value = a[pos1].value;
                        int index = mapA.get(value);
                        int pos2 = rightA[index];

                        if(pos1 != pos2){
                            int tmp = a[pos1].id;
                            a[pos1].id = a[pos2].id;
                            a[pos2].id = tmp;
                            posA[x] = pos2;
                            posA[a[pos1].id] = pos1;
                        }

                        p *= modinv(Math.min(value, b[pos2].value));
                        p %= MOD;

                        if(leftA[index] == rightA[index]){
                            mapA.remove(value);
                        }else{
                            rightA[index] = pos2-1;
                        }

                        a[pos2].value++;
                        value = a[pos2].value;

                        if(mapA.containsKey(value)){
                            index = mapA.get(value);
                            leftA[index] = pos2;
                        }else{
                            mapA.put(value, indexA);
                            leftA[indexA] = pos2;
                            rightA[indexA] = pos2;
                            indexA++;
                        }

                        p *= Math.min(value, b[pos2].value);
                        p %= MOD;
                    }
                    case 2 -> {
                        int pos1 = posB[x];
                        long value = b[pos1].value;
                        int index = mapB.get(value);
                        int pos2 = rightB[index];

                        if(pos1 != pos2){
                            int tmp = b[pos1].id;
                            b[pos1].id = b[pos2].id;
                            b[pos2].id = tmp;
                            posB[x] = pos2;
                            posB[b[pos1].id] = pos1;
                        }

                        p *= modinv(Math.min(value, a[pos2].value));
                        p %= MOD;

                        if(leftB[index] == rightB[index]){
                            mapB.remove(value);
                        }else{
                            rightB[index] = pos2-1;
                        }

                        b[pos2].value++;
                        value = b[pos2].value;

                        if(mapB.containsKey(value)){
                            index = mapB.get(value);
                            leftB[index] = pos2;
                        }else{
                            mapB.put(value, indexB);
                            leftB[indexB] = pos2;
                            rightB[indexB] = pos2;
                            indexB++;
                        }

                        p *= Math.min(value, a[pos2].value);
                        p %= MOD;
                    }
                }

                pw.print(" " + p);
            }
            pw.println();
        }
        pw.flush();
    }

    static class Pair implements Comparable<Pair> {
        int id;
        long value;
        public Pair(int id, long value) {
            this.id = id;
            this.value = value;
        }
        @Override
        public int compareTo(Pair p) {
            return Long.compare(this.value, p.value);
        }
    }

    static long modinv(long a){
        long b = MOD;
        long u = 1;
        long v = 0;
        long t, tmp;
        while(b != 0){
            t = a/b;
            a -= t*b;
            tmp = a;
            a = b;
            b = tmp;
            u -= t*v;
            tmp = u;
            u = v;
            v = tmp;
        }
        u %= MOD;
        if(u < 0) u += MOD;
        return u;
    }
}

**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 cause the code to fail. Here are some steps to consider:

1. **Edge Cases with Small Arrays**: Small arrays can help in verifying the correctness of the initial setup and the first few operations.
2. **Equal Values in Arrays**: Arrays with equal values can challenge the sorting and mapping logic.
3. **Large Values in Arrays**: Large values can test the modulo operation and the handling of large numbers.
4. **Consecutive Operations on the Same Element**: Repeated operations on the same element can test the update logic.
5. **Operations on Different Elements**: Mixed operations on different elements can test the overall logic flow.

Given these considerations, let's create a test input that includes these scenarios:

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

### Explanation:
- **Test Case 1**:
  - **Array Length (n)**: 3
  - **Number of Operations (q)**: 4
  - **Initial Arrays**:
    - `a = [1, 1, 1]`
    - `b = [3, 2, 1]`
  - **Operations**:
    - `1 3`: Increase `a[3-1]` by 1 (i.e., `a[2]` becomes 2)
    - `2 3`: Increase `b[3-1]` by 1 (i.e., `b[2]` becomes 2)
    - `1 1`: Increase `a[1-1]` by 1 (i.e., `a[0]` becomes 2)
    - `2 1`: Increase `b[1-1]` by 1 (i.e., `b[0]` becomes 4)

This test input is designed to:
- Test the initial setup with equal values in `a`.
- Test the sorting and mapping logic with different values in `b`.
- Test the update logic with consecutive operations on the same element.
- Test the overall logic flow with mixed operations on different elements.

This should help in exposing any bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Painting the Ribbon
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob have bought a ribbon consisting of $$$n$$$ parts. Now they want to paint it.

First, Alice will paint every part of the ribbon into one of $$$m$$$ colors. For each part, she can choose its color arbitrarily.

Then, Bob will choose at most $$$k$$$ parts of the ribbon and repaint them into the same color (he chooses the affected parts and the color arbitrarily).

Bob would like all parts to have the same color. However, Alice thinks that this is too dull, so she wants to paint the ribbon in such a way that Bob cannot make all parts have the same color.

Is it possible to paint the ribbon in such a way?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of one line containing three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$1 \le m, k \le n \le 50$$$) — the number of parts, the number of colors and the number of parts Bob can repaint, respectively.

Output: For each test case, print YES if Alice can paint the ribbon so that Bob cannot make all parts have the same color. Otherwise, print NO.

You can print every letter in any register. For example, Yes, yes, yEs will all be recognized as positive answer.

Examples:
input:
5
1 1 1
5 1 1
5 2 1
5 2 2
5 5 3
output:
NO
NO
YES
NO
YES

Note: In the first test case, a ribbon consists of $$$1$$$ part. So all its parts will always have the same color.

In the second test case, there is only $$$1$$$ color.

In the third test case, Alice can paint the ribbon as follows: $$$[1, 2, 1, 2, 1]$$$. It's impossible to change the color of at most $$$1$$$ part so that all parts have the same color.

In the fourth test case, no matter how Alice paints the ribbon, Bob will always be able to repaint $$$2$$$ parts so that all parts have the same color.

In the fifth test case, Alice can paint the ribbon as follows: $$$[1, 2, 3, 4, 5]$$$. It's impossible to change the color of at most $$$3$$$ parts so that all parts have the same color.

Buggy Code:
#include <iostream>
using namespace std;

int main() {
    int t;
    cin >> t;

    while (t--) {
        int n, m, k;
        cin >> n >> m >> k;

        if (m == 1 || m == k)
            cout << "NO" << endl;
        else
            cout << "YES" << endl;
    }

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on the logic of the algorithm. The buggy code incorrectly assumes that if \( m == k \), then Bob can always repaint the ribbon to a single color. However, this is not always true, especially when \( n \) is large enough and \( k \) is not sufficient to repaint all parts to a single color.

Let's break down the conditions:
1. If \( m = 1 \), Alice cannot use more than one color, so Bob can always make all parts the same color.
2. If \( m \le k \), Bob can always repaint up to \( k \) parts to any color, so he can make all parts the same color.
3. If \( m > k \) and \( n > k \), Alice can paint the ribbon in such a way that Bob cannot make all parts the same color.

The buggy code fails to consider the case where \( m > k \) and \( n > k \). Specifically, it incorrectly outputs "NO" when \( m = k \) and \( n > k \).

Let's create a test case where \( m = k \) and \( n > k \):

```plaintext
1
4 2 2
```

In this test case:
- \( n = 4 \): The ribbon has 4 parts.
- \( m = 2 \): There are 2 colors available.
- \( k = 2 \): Bob can repaint up to 2 parts.

Alice can paint the ribbon as follows: \([1, 2, 1, 2]\). It is impossible for Bob to repaint 2 parts to make all 4 parts the same color. Therefore, the correct output should be "YES".

This test case will expose the bug in the buggy code, which currently outputs "NO" for this input.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Painting the Ribbon
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob have bought a ribbon consisting of $$$n$$$ parts. Now they want to paint it.

First, Alice will paint every part of the ribbon into one of $$$m$$$ colors. For each part, she can choose its color arbitrarily.

Then, Bob will choose at most $$$k$$$ parts of the ribbon and repaint them into the same color (he chooses the affected parts and the color arbitrarily).

Bob would like all parts to have the same color. However, Alice thinks that this is too dull, so she wants to paint the ribbon in such a way that Bob cannot make all parts have the same color.

Is it possible to paint the ribbon in such a way?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of one line containing three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$1 \le m, k \le n \le 50$$$) — the number of parts, the number of colors and the number of parts Bob can repaint, respectively.

Output: For each test case, print YES if Alice can paint the ribbon so that Bob cannot make all parts have the same color. Otherwise, print NO.

You can print every letter in any register. For example, Yes, yes, yEs will all be recognized as positive answer.

Examples:
input:
5
1 1 1
5 1 1
5 2 1
5 2 2
5 5 3
output:
NO
NO
YES
NO
YES

Note: In the first test case, a ribbon consists of $$$1$$$ part. So all its parts will always have the same color.

In the second test case, there is only $$$1$$$ color.

In the third test case, Alice can paint the ribbon as follows: $$$[1, 2, 1, 2, 1]$$$. It's impossible to change the color of at most $$$1$$$ part so that all parts have the same color.

In the fourth test case, no matter how Alice paints the ribbon, Bob will always be able to repaint $$$2$$$ parts so that all parts have the same color.

In the fifth test case, Alice can paint the ribbon as follows: $$$[1, 2, 3, 4, 5]$$$. It's impossible to change the color of at most $$$3$$$ parts so that all parts have the same color.

Buggy Code:
#include <stdio.h>

int
main ()
{
  int t, n, m, k, i;
  scanf ("%d", &t);
  for (i = 0; i < t; i++)
	{
	  scanf ("%d%d%d", &n, &m, &k);
	  if (m > k)
		{
		  printf ("Yes\n");
		}
	  else if (m == 4 && k == 6)
		{
		  printf ("Yes\n");
		}
	  else
		{
		  printf ("No\n");
		}
	}
  return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Painting the Ribbon
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob have bought a ribbon consisting of $$$n$$$ parts. Now they want to paint it.

First, Alice will paint every part of the ribbon into one of $$$m$$$ colors. For each part, she can choose its color arbitrarily.

Then, Bob will choose at most $$$k$$$ parts of the ribbon and repaint them into the same color (he chooses the affected parts and the color arbitrarily).

Bob would like all parts to have the same color. However, Alice thinks that this is too dull, so she wants to paint the ribbon in such a way that Bob cannot make all parts have the same color.

Is it possible to paint the ribbon in such a way?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of one line containing three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$1 \le m, k \le n \le 50$$$) — the number of parts, the number of colors and the number of parts Bob can repaint, respectively.

Output: For each test case, print YES if Alice can paint the ribbon so that Bob cannot make all parts have the same color. Otherwise, print NO.

You can print every letter in any register. For example, Yes, yes, yEs will all be recognized as positive answer.

Examples:
input:
5
1 1 1
5 1 1
5 2 1
5 2 2
5 5 3
output:
NO
NO
YES
NO
YES

Note: In the first test case, a ribbon consists of $$$1$$$ part. So all its parts will always have the same color.

In the second test case, there is only $$$1$$$ color.

In the third test case, Alice can paint the ribbon as follows: $$$[1, 2, 1, 2, 1]$$$. It's impossible to change the color of at most $$$1$$$ part so that all parts have the same color.

In the fourth test case, no matter how Alice paints the ribbon, Bob will always be able to repaint $$$2$$$ parts so that all parts have the same color.

In the fifth test case, Alice can paint the ribbon as follows: $$$[1, 2, 3, 4, 5]$$$. It's impossible to change the color of at most $$$3$$$ parts so that all parts have the same color.

Buggy Code:
#include <iostream>

using namespace std;

int main() {
    int t;
    cin >> t;

    while (t--) {
        int n, m, k;
        cin >> n >> m >> k;

        if (n == 1) {
            cout << "NO\n";
        } else if (m == k) {
            cout << "NO\n";
        } else {
            if ( m< k) {
                cout << "NO\n";
            } else {
                cout << "YES\n";
            }
        }
    }

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios where the logic in the code might fail. The code has several conditions, and we need to test the boundaries of these conditions to ensure they are correctly handled.

### Analysis of the Buggy Code:
1. **First Condition:** `if (n == 1)`
   - This condition is correct. If there is only one part, Bob can always make it one color.

2. **Second Condition:** `if (m == k)`
   - This condition is incorrect. If the number of colors (`m`) is equal to the number of parts Bob can repaint (`k`), it does not necessarily mean Bob can make all parts the same color. This condition should be more complex.

3. **Third Condition:** `if (m < k)`
   - This condition is also incorrect. If the number of colors is less than the number of parts Bob can repaint, it does not always mean Bob can make all parts the same color. This condition should be more complex.

### Challenging Test Input:
We need to create a test case where `m` is equal to `k` or `m` is less than `k`, but Alice can still paint the ribbon in such a way that Bob cannot make all parts the same color.

### Test Case:
- **Test Case 1:** `n = 5`, `m = 2`, `k = 2`
  - Alice can paint the ribbon as follows: `[1, 2, 1, 2, 1]`.
  - Bob can only repaint 2 parts, but he cannot make all parts the same color.

### Output:
```plaintext
5 2 2
```

This test case should expose the bug in the code, as the current logic in the code will incorrectly output `NO`, while the correct output should be `YES`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Painting the Ribbon
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob have bought a ribbon consisting of $$$n$$$ parts. Now they want to paint it.

First, Alice will paint every part of the ribbon into one of $$$m$$$ colors. For each part, she can choose its color arbitrarily.

Then, Bob will choose at most $$$k$$$ parts of the ribbon and repaint them into the same color (he chooses the affected parts and the color arbitrarily).

Bob would like all parts to have the same color. However, Alice thinks that this is too dull, so she wants to paint the ribbon in such a way that Bob cannot make all parts have the same color.

Is it possible to paint the ribbon in such a way?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of one line containing three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$1 \le m, k \le n \le 50$$$) — the number of parts, the number of colors and the number of parts Bob can repaint, respectively.

Output: For each test case, print YES if Alice can paint the ribbon so that Bob cannot make all parts have the same color. Otherwise, print NO.

You can print every letter in any register. For example, Yes, yes, yEs will all be recognized as positive answer.

Examples:
input:
5
1 1 1
5 1 1
5 2 1
5 2 2
5 5 3
output:
NO
NO
YES
NO
YES

Note: In the first test case, a ribbon consists of $$$1$$$ part. So all its parts will always have the same color.

In the second test case, there is only $$$1$$$ color.

In the third test case, Alice can paint the ribbon as follows: $$$[1, 2, 1, 2, 1]$$$. It's impossible to change the color of at most $$$1$$$ part so that all parts have the same color.

In the fourth test case, no matter how Alice paints the ribbon, Bob will always be able to repaint $$$2$$$ parts so that all parts have the same color.

In the fifth test case, Alice can paint the ribbon as follows: $$$[1, 2, 3, 4, 5]$$$. It's impossible to change the color of at most $$$3$$$ parts so that all parts have the same color.

Buggy Code:
for test in range (int(input())):
    n, m, k = map(int,input().split())
    if m > k :print("YES")
    else:
        if n > k and m > 1 and n//m != k:print("YES")
        else:
            print("NO")

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

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

Think step by step.

Response:
```plaintext
3
5 2 2
6 2 3
7 3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Painting the Ribbon
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob have bought a ribbon consisting of $$$n$$$ parts. Now they want to paint it.

First, Alice will paint every part of the ribbon into one of $$$m$$$ colors. For each part, she can choose its color arbitrarily.

Then, Bob will choose at most $$$k$$$ parts of the ribbon and repaint them into the same color (he chooses the affected parts and the color arbitrarily).

Bob would like all parts to have the same color. However, Alice thinks that this is too dull, so she wants to paint the ribbon in such a way that Bob cannot make all parts have the same color.

Is it possible to paint the ribbon in such a way?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of one line containing three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$1 \le m, k \le n \le 50$$$) — the number of parts, the number of colors and the number of parts Bob can repaint, respectively.

Output: For each test case, print YES if Alice can paint the ribbon so that Bob cannot make all parts have the same color. Otherwise, print NO.

You can print every letter in any register. For example, Yes, yes, yEs will all be recognized as positive answer.

Examples:
input:
5
1 1 1
5 1 1
5 2 1
5 2 2
5 5 3
output:
NO
NO
YES
NO
YES

Note: In the first test case, a ribbon consists of $$$1$$$ part. So all its parts will always have the same color.

In the second test case, there is only $$$1$$$ color.

In the third test case, Alice can paint the ribbon as follows: $$$[1, 2, 1, 2, 1]$$$. It's impossible to change the color of at most $$$1$$$ part so that all parts have the same color.

In the fourth test case, no matter how Alice paints the ribbon, Bob will always be able to repaint $$$2$$$ parts so that all parts have the same color.

In the fifth test case, Alice can paint the ribbon as follows: $$$[1, 2, 3, 4, 5]$$$. It's impossible to change the color of at most $$$3$$$ parts so that all parts have the same color.

Buggy Code:
cases = int(input())

while cases > 0 :
  alice = []
  n, m, k = map(int, input().split())
  if  len(set(range(1, m + 1))) > 1:
    for i in list(range(1, m + 1)) * n :
      if len(alice) < n:
        alice.append(i)
      else:
        break
    if len(set(alice)) == 2 and k > 1:
      alico = list(set(alice))
      for ii in alico :
        if alice.count(ii) == k:
          print("NO")
          break
    else:
      print("YES")
  else:
    print("NO")
  cases -= 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:
To generate a challenging test input that exposes the bug in the provided buggy code, let's analyze the logic and identify the edge cases where the code might fail.

### Analysis of the Buggy Code
1. **Initialization and Input Reading:**
   - The code reads the number of test cases and then processes each test case.
   - It reads `n`, `m`, and `k` for each test case.

2. **Color Assignment:**
   - It creates a list `alice` and attempts to fill it with colors from 1 to `m`, repeating the sequence until it has `n` elements.
   - This part of the code is correct in generating a list of colors.

3. **Condition Check:**
   - If `m` is greater than 1, it checks if the set of colors in `alice` has exactly 2 unique colors and `k` is greater than 1.
   - If this condition is true, it checks if any color appears exactly `k` times and prints "NO" if it does.
   - Otherwise, it prints "YES".
   - If `m` is 1, it prints "NO".

### Identifying the Bug
The main issue with the code is in the condition check:
- It only checks if the set of colors has exactly 2 unique colors and `k` is greater than 1.
- It does not correctly handle cases where the number of unique colors is more than 2 or where the distribution of colors is such that Bob can still make all parts have the same color.

### Challenging Test Input
To expose the bug, we need a test case where:
- The number of unique colors is more than 2.
- The distribution of colors is such that Bob can still make all parts have the same color, but the code incorrectly prints "YES".

### Test Input
Let's create a test case with `n = 5`, `m = 3`, and `k = 2`.

- Alice can paint the ribbon as `[1, 2, 3, 2, 1]`.
- Bob can repaint the parts at indices 2 and 4 to color 1, making the ribbon `[1, 1, 1, 1, 1]`.

### Expected Output
The correct output for this test case should be "NO" because Bob can make all parts have the same color. However, the buggy code will likely print "YES" because it does not correctly handle the distribution of colors.

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

This test input should expose the bug in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Painting the Ribbon
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob have bought a ribbon consisting of $$$n$$$ parts. Now they want to paint it.

First, Alice will paint every part of the ribbon into one of $$$m$$$ colors. For each part, she can choose its color arbitrarily.

Then, Bob will choose at most $$$k$$$ parts of the ribbon and repaint them into the same color (he chooses the affected parts and the color arbitrarily).

Bob would like all parts to have the same color. However, Alice thinks that this is too dull, so she wants to paint the ribbon in such a way that Bob cannot make all parts have the same color.

Is it possible to paint the ribbon in such a way?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of one line containing three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$1 \le m, k \le n \le 50$$$) — the number of parts, the number of colors and the number of parts Bob can repaint, respectively.

Output: For each test case, print YES if Alice can paint the ribbon so that Bob cannot make all parts have the same color. Otherwise, print NO.

You can print every letter in any register. For example, Yes, yes, yEs will all be recognized as positive answer.

Examples:
input:
5
1 1 1
5 1 1
5 2 1
5 2 2
5 5 3
output:
NO
NO
YES
NO
YES

Note: In the first test case, a ribbon consists of $$$1$$$ part. So all its parts will always have the same color.

In the second test case, there is only $$$1$$$ color.

In the third test case, Alice can paint the ribbon as follows: $$$[1, 2, 1, 2, 1]$$$. It's impossible to change the color of at most $$$1$$$ part so that all parts have the same color.

In the fourth test case, no matter how Alice paints the ribbon, Bob will always be able to repaint $$$2$$$ parts so that all parts have the same color.

In the fifth test case, Alice can paint the ribbon as follows: $$$[1, 2, 3, 4, 5]$$$. It's impossible to change the color of at most $$$3$$$ parts so that all parts have the same color.

Buggy Code:
t = int(input())

for _ in range(t):
    n, m, k = map(int, input().split())

    if m == 1 or k >= m:
        print("NO")
    elif (n % k == 0 and n // k >= 2) or (n % k != 0 and m >= k+1):
        print("YES")
    else:
        print("NO")

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Painting the Ribbon
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob have bought a ribbon consisting of $$$n$$$ parts. Now they want to paint it.

First, Alice will paint every part of the ribbon into one of $$$m$$$ colors. For each part, she can choose its color arbitrarily.

Then, Bob will choose at most $$$k$$$ parts of the ribbon and repaint them into the same color (he chooses the affected parts and the color arbitrarily).

Bob would like all parts to have the same color. However, Alice thinks that this is too dull, so she wants to paint the ribbon in such a way that Bob cannot make all parts have the same color.

Is it possible to paint the ribbon in such a way?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of one line containing three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$1 \le m, k \le n \le 50$$$) — the number of parts, the number of colors and the number of parts Bob can repaint, respectively.

Output: For each test case, print YES if Alice can paint the ribbon so that Bob cannot make all parts have the same color. Otherwise, print NO.

You can print every letter in any register. For example, Yes, yes, yEs will all be recognized as positive answer.

Examples:
input:
5
1 1 1
5 1 1
5 2 1
5 2 2
5 5 3
output:
NO
NO
YES
NO
YES

Note: In the first test case, a ribbon consists of $$$1$$$ part. So all its parts will always have the same color.

In the second test case, there is only $$$1$$$ color.

In the third test case, Alice can paint the ribbon as follows: $$$[1, 2, 1, 2, 1]$$$. It's impossible to change the color of at most $$$1$$$ part so that all parts have the same color.

In the fourth test case, no matter how Alice paints the ribbon, Bob will always be able to repaint $$$2$$$ parts so that all parts have the same color.

In the fifth test case, Alice can paint the ribbon as follows: $$$[1, 2, 3, 4, 5]$$$. It's impossible to change the color of at most $$$3$$$ parts so that all parts have the same color.

Buggy Code:
import java.util.Scanner;

public class A
{
 static Scanner sc = new Scanner(System.in);

     public static void main(String[] args){
         int T = sc.nextInt();

         while(T-->0){

            int part = sc.nextInt();

            int color = sc.nextInt();

            int bob = sc.nextInt();

            if(part == color && color == bob && bob == part){
                    System.out.println("NO");
            }else if(color%2==0 && bob%2==0 || color == bob){
                System.out.println("NO");
            }
            else{
                System.out.println("YES");
            }

         }

     }

}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases where the conditions in the if-else statements might not handle the logic correctly. The primary issues in the code are:

1. The condition `if(part == color && color == bob && bob == part)` is too restrictive and doesn't cover all scenarios where Bob can always make all parts the same color.
2. The condition `else if(color%2==0 && bob%2==0 || color == bob)` is also too restrictive and might incorrectly handle some cases.

To generate a test input that exposes these bugs, we can focus on the following scenarios:

1. **Small values where `n`, `m`, and `k` are equal**: This should always result in "NO" because Bob can always repaint all parts to the same color.
2. **Small values where `m` and `k` are even and `n` is larger**: This should result in "NO" if Bob can always make all parts the same color, but the code might incorrectly handle this.
3. **Small values where `m` and `k` are odd and `n` is larger**: This should result in "YES" if Alice can paint the ribbon such that Bob cannot make all parts the same color, but the code might incorrectly handle this.

Let's construct a small-scale test input that covers these scenarios:

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

Explanation:
1. **1 1 1**: Bob can always repaint the single part to the same color, so the output should be "NO".
2. **3 2 2**: Alice can paint the ribbon as `[1, 2, 1]`, and Bob cannot make all parts the same color, so the output should be "YES".
3. **4 2 2**: Bob can always repaint two parts to the same color, so the output should be "NO".
4. **5 3 2**: Alice can paint the ribbon as `[1, 2, 3, 1, 2]`, and Bob cannot make all parts the same color, so the output should be "YES".
5. **5 3 3**: Bob can always repaint three parts to the same color, so the output should be "NO".

This test input should help expose the bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Painting the Ribbon
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob have bought a ribbon consisting of $$$n$$$ parts. Now they want to paint it.

First, Alice will paint every part of the ribbon into one of $$$m$$$ colors. For each part, she can choose its color arbitrarily.

Then, Bob will choose at most $$$k$$$ parts of the ribbon and repaint them into the same color (he chooses the affected parts and the color arbitrarily).

Bob would like all parts to have the same color. However, Alice thinks that this is too dull, so she wants to paint the ribbon in such a way that Bob cannot make all parts have the same color.

Is it possible to paint the ribbon in such a way?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of one line containing three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$1 \le m, k \le n \le 50$$$) — the number of parts, the number of colors and the number of parts Bob can repaint, respectively.

Output: For each test case, print YES if Alice can paint the ribbon so that Bob cannot make all parts have the same color. Otherwise, print NO.

You can print every letter in any register. For example, Yes, yes, yEs will all be recognized as positive answer.

Examples:
input:
5
1 1 1
5 1 1
5 2 1
5 2 2
5 5 3
output:
NO
NO
YES
NO
YES

Note: In the first test case, a ribbon consists of $$$1$$$ part. So all its parts will always have the same color.

In the second test case, there is only $$$1$$$ color.

In the third test case, Alice can paint the ribbon as follows: $$$[1, 2, 1, 2, 1]$$$. It's impossible to change the color of at most $$$1$$$ part so that all parts have the same color.

In the fourth test case, no matter how Alice paints the ribbon, Bob will always be able to repaint $$$2$$$ parts so that all parts have the same color.

In the fifth test case, Alice can paint the ribbon as follows: $$$[1, 2, 3, 4, 5]$$$. It's impossible to change the color of at most $$$3$$$ parts so that all parts have the same color.

Buggy Code:
/* package whatever; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Main
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while(t--!=0){
			int n=sc.nextInt();
			int m=sc.nextInt();
			int k=sc.nextInt();
			if(k%m==0){
				System.out.println("NO");
			}
			else{
				System.out.println("YES");
			}
		}
	}
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Painting the Ribbon
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob have bought a ribbon consisting of $$$n$$$ parts. Now they want to paint it.

First, Alice will paint every part of the ribbon into one of $$$m$$$ colors. For each part, she can choose its color arbitrarily.

Then, Bob will choose at most $$$k$$$ parts of the ribbon and repaint them into the same color (he chooses the affected parts and the color arbitrarily).

Bob would like all parts to have the same color. However, Alice thinks that this is too dull, so she wants to paint the ribbon in such a way that Bob cannot make all parts have the same color.

Is it possible to paint the ribbon in such a way?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of one line containing three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$1 \le m, k \le n \le 50$$$) — the number of parts, the number of colors and the number of parts Bob can repaint, respectively.

Output: For each test case, print YES if Alice can paint the ribbon so that Bob cannot make all parts have the same color. Otherwise, print NO.

You can print every letter in any register. For example, Yes, yes, yEs will all be recognized as positive answer.

Examples:
input:
5
1 1 1
5 1 1
5 2 1
5 2 2
5 5 3
output:
NO
NO
YES
NO
YES

Note: In the first test case, a ribbon consists of $$$1$$$ part. So all its parts will always have the same color.

In the second test case, there is only $$$1$$$ color.

In the third test case, Alice can paint the ribbon as follows: $$$[1, 2, 1, 2, 1]$$$. It's impossible to change the color of at most $$$1$$$ part so that all parts have the same color.

In the fourth test case, no matter how Alice paints the ribbon, Bob will always be able to repaint $$$2$$$ parts so that all parts have the same color.

In the fifth test case, Alice can paint the ribbon as follows: $$$[1, 2, 3, 4, 5]$$$. It's impossible to change the color of at most $$$3$$$ parts so that all parts have the same color.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main ()
{
    int n,m,k,t;
    cin>>t;
    while(t--)
    {
        cin>>n>>m>>k;
        if(n<=k||m==k)
            cout<<"NO\n";
        else cout<<"YES\n";
    }

    return 0;
}

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

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

Think step by step.

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

### Explanation:
1. **Test Case 1: `1 1 1`**
   - **Input:** A single part, a single color, and Bob can repaint 1 part.
   - **Expected Output:** `NO` because Bob can always repaint the single part to any color.
   - **Buggy Code Output:** `NO` (Correct)

2. **Test Case 2: `5 1 1`**
   - **Input:** 5 parts, a single color, and Bob can repaint 1 part.
   - **Expected Output:** `NO` because all parts are already the same color.
   - **Buggy Code Output:** `NO` (Correct)

3. **Test Case 3: `5 2 1`**
   - **Input:** 5 parts, 2 colors, and Bob can repaint 1 part.
   - **Expected Output:** `YES` because Alice can paint the ribbon such that Bob cannot make all parts the same color (e.g., `[1, 2, 1, 2, 1]`).
   - **Buggy Code Output:** `YES` (Correct)

4. **Test Case 4: `5 2 2`**
   - **Input:** 5 parts, 2 colors, and Bob can repaint 2 parts.
   - **Expected Output:** `NO` because Bob can always repaint 2 parts to make all parts the same color (e.g., `[1, 2, 1, 2, 1]` can be changed to `[1, 1, 1, 1, 1]` by repainting the 2nd and 4th parts).
   - **Buggy Code Output:** `YES` (Incorrect)

5. **Test Case 5: `5 5 3`**
   - **Input:** 5 parts, 5 colors, and Bob can repaint 3 parts.
   - **Expected Output:** `YES` because Alice can paint the ribbon such that Bob cannot make all parts the same color (e.g., `[1, 2, 3, 4, 5]`).
   - **Buggy Code Output:** `YES` (Correct)

### Key Points:
- The buggy code incorrectly assumes that `m == k` is a condition for Bob to always win, which is not always true.
- The test case `5 2 2` is particularly challenging because it exposes the bug in the logic where the code incorrectly outputs `YES` when the correct answer should be `NO`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Painting the Ribbon
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob have bought a ribbon consisting of $$$n$$$ parts. Now they want to paint it.

First, Alice will paint every part of the ribbon into one of $$$m$$$ colors. For each part, she can choose its color arbitrarily.

Then, Bob will choose at most $$$k$$$ parts of the ribbon and repaint them into the same color (he chooses the affected parts and the color arbitrarily).

Bob would like all parts to have the same color. However, Alice thinks that this is too dull, so she wants to paint the ribbon in such a way that Bob cannot make all parts have the same color.

Is it possible to paint the ribbon in such a way?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of one line containing three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$1 \le m, k \le n \le 50$$$) — the number of parts, the number of colors and the number of parts Bob can repaint, respectively.

Output: For each test case, print YES if Alice can paint the ribbon so that Bob cannot make all parts have the same color. Otherwise, print NO.

You can print every letter in any register. For example, Yes, yes, yEs will all be recognized as positive answer.

Examples:
input:
5
1 1 1
5 1 1
5 2 1
5 2 2
5 5 3
output:
NO
NO
YES
NO
YES

Note: In the first test case, a ribbon consists of $$$1$$$ part. So all its parts will always have the same color.

In the second test case, there is only $$$1$$$ color.

In the third test case, Alice can paint the ribbon as follows: $$$[1, 2, 1, 2, 1]$$$. It's impossible to change the color of at most $$$1$$$ part so that all parts have the same color.

In the fourth test case, no matter how Alice paints the ribbon, Bob will always be able to repaint $$$2$$$ parts so that all parts have the same color.

In the fifth test case, Alice can paint the ribbon as follows: $$$[1, 2, 3, 4, 5]$$$. It's impossible to change the color of at most $$$3$$$ parts so that all parts have the same color.

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

vector<int> sieve(int n) {
    vector<int> primes;
    vector<bool> is_prime(n+1, true);
    for (int p=2; p*p<=n; ++p) if (is_prime[p]) for (int i=p*p; i<=n; i+=p) is_prime[i] = false;
    for (int p=2; p<=n; ++p) if (is_prime[p]) primes.push_back(p);
    return primes;
}

int main(){
    // vector<int> Prime = sieve(n);
    // cout << Prime.size() << endl;
    // for (auto it:Prime) cout << it << " ";

    long long int t;
    cin >> t;
    while (t--){
        int n,m,k;
        cin >> n >> m>>k;
        if (n==1) cout << "NO" << endl;
        else if (k>=n) cout << "NO" << endl;
        else if (m > k){
            cout  << "YES" << endl;
        }
        else cout << "NO" << endl;
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5 2 2
```

This test case is designed to challenge the buggy code. Here's the reasoning:

1. **Test Case Parameters**:
   - \( n = 5 \): The ribbon has 5 parts.
   - \( m = 2 \): There are 2 colors available.
   - \( k = 2 \): Bob can repaint up to 2 parts.

2. **Expected Behavior**:
   - Alice can paint the ribbon in such a way that Bob cannot make all parts have the same color.
   - For example, Alice can paint the ribbon as \([1, 2, 1, 2, 1]\). Bob can only repaint 2 parts, and he cannot make all 5 parts the same color.

3. **Buggy Code Analysis**:
   - The buggy code has a logical flaw in the condition `else if (m > k)`. It incorrectly assumes that if the number of colors \( m \) is greater than the number of parts Bob can repaint \( k \), then Alice can always win.
   - However, this is not always true. For example, if \( n = 5 \), \( m = 2 \), and \( k = 2 \), Alice can still win, but the buggy code would incorrectly output "NO".

By providing this test case, we can expose the bug in the code and demonstrate that the condition `else if (m > k)` is insufficient to determine the outcome correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Painting the Ribbon
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob have bought a ribbon consisting of $$$n$$$ parts. Now they want to paint it.

First, Alice will paint every part of the ribbon into one of $$$m$$$ colors. For each part, she can choose its color arbitrarily.

Then, Bob will choose at most $$$k$$$ parts of the ribbon and repaint them into the same color (he chooses the affected parts and the color arbitrarily).

Bob would like all parts to have the same color. However, Alice thinks that this is too dull, so she wants to paint the ribbon in such a way that Bob cannot make all parts have the same color.

Is it possible to paint the ribbon in such a way?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of one line containing three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$1 \le m, k \le n \le 50$$$) — the number of parts, the number of colors and the number of parts Bob can repaint, respectively.

Output: For each test case, print YES if Alice can paint the ribbon so that Bob cannot make all parts have the same color. Otherwise, print NO.

You can print every letter in any register. For example, Yes, yes, yEs will all be recognized as positive answer.

Examples:
input:
5
1 1 1
5 1 1
5 2 1
5 2 2
5 5 3
output:
NO
NO
YES
NO
YES

Note: In the first test case, a ribbon consists of $$$1$$$ part. So all its parts will always have the same color.

In the second test case, there is only $$$1$$$ color.

In the third test case, Alice can paint the ribbon as follows: $$$[1, 2, 1, 2, 1]$$$. It's impossible to change the color of at most $$$1$$$ part so that all parts have the same color.

In the fourth test case, no matter how Alice paints the ribbon, Bob will always be able to repaint $$$2$$$ parts so that all parts have the same color.

In the fifth test case, Alice can paint the ribbon as follows: $$$[1, 2, 3, 4, 5]$$$. It's impossible to change the color of at most $$$3$$$ parts so that all parts have the same color.

Buggy Code:
T = int(input())

while T:
    T -= 1
    x = [int(e) for e in input().split(' ')]
    n, m, k = x[0], x[1], x[2]

    if n == 1:
        print('NO')
    elif m == 1:
        print('NO')
    elif k >= min(n, m):
        print('NO')
    else:
        print('YES')

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's analyze the conditions under which the code might fail:

1. **n == 1**: The code correctly prints 'NO' for this case.
2. **m == 1**: The code correctly prints 'NO' for this case.
3. **k >= min(n, m)**: The code prints 'NO' for this case, but this condition is not always correct. Specifically, if `k` is greater than or equal to `n`, Bob can always repaint all parts to the same color, but if `k` is greater than or equal to `m` but less than `n`, Alice can still paint the ribbon in such a way that Bob cannot make all parts the same color.

Let's consider the edge cases:

- **n > m**: If `n` is greater than `m`, and `k` is less than `m`, Alice can paint the ribbon with all different colors, and Bob cannot repaint all parts to the same color.
- **k == m - 1**: If `k` is one less than `m`, Alice can still paint the ribbon in such a way that Bob cannot make all parts the same color.

A challenging test input that maximizes the failure probability in the faulty solution is:

```plaintext
4
6 5 4
6 5 5
6 6 5
6 6 6
```

Explanation:
- For the first test case (6 5 4): Alice can paint the ribbon with 5 different colors, and Bob cannot repaint 4 parts to make all parts the same color.
- For the second test case (6 5 5): Alice can paint the ribbon with 5 different colors, and Bob cannot repaint 5 parts to make all parts the same color.
- For the third test case (6 6 5): Alice can paint the ribbon with 6 different colors, and Bob cannot repaint 5 parts to make all parts the same color.
- For the fourth test case (6 6 6): Bob can repaint all 6 parts to the same color, so the output should be 'NO'.

This input will help expose the bug in the condition `k >= min(n, m)`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Painting the Ribbon
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob have bought a ribbon consisting of $$$n$$$ parts. Now they want to paint it.

First, Alice will paint every part of the ribbon into one of $$$m$$$ colors. For each part, she can choose its color arbitrarily.

Then, Bob will choose at most $$$k$$$ parts of the ribbon and repaint them into the same color (he chooses the affected parts and the color arbitrarily).

Bob would like all parts to have the same color. However, Alice thinks that this is too dull, so she wants to paint the ribbon in such a way that Bob cannot make all parts have the same color.

Is it possible to paint the ribbon in such a way?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of one line containing three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$1 \le m, k \le n \le 50$$$) — the number of parts, the number of colors and the number of parts Bob can repaint, respectively.

Output: For each test case, print YES if Alice can paint the ribbon so that Bob cannot make all parts have the same color. Otherwise, print NO.

You can print every letter in any register. For example, Yes, yes, yEs will all be recognized as positive answer.

Examples:
input:
5
1 1 1
5 1 1
5 2 1
5 2 2
5 5 3
output:
NO
NO
YES
NO
YES

Note: In the first test case, a ribbon consists of $$$1$$$ part. So all its parts will always have the same color.

In the second test case, there is only $$$1$$$ color.

In the third test case, Alice can paint the ribbon as follows: $$$[1, 2, 1, 2, 1]$$$. It's impossible to change the color of at most $$$1$$$ part so that all parts have the same color.

In the fourth test case, no matter how Alice paints the ribbon, Bob will always be able to repaint $$$2$$$ parts so that all parts have the same color.

In the fifth test case, Alice can paint the ribbon as follows: $$$[1, 2, 3, 4, 5]$$$. It's impossible to change the color of at most $$$3$$$ parts so that all parts have the same color.

Buggy Code:
res=[]
for k in range(int(input())):
    l=list(map(int,input().split()))
    if(l[1]==1 or l[1]==l[2]):
        res.append("NO")
    else:
        res.append("YES")
for i in res:
    print(i)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code has a logical error in its condition for determining whether Alice can paint the ribbon so that Bob cannot make all parts have the same color.

The key issues in the buggy code are:
1. It checks if `l[1] == 1` or `l[1] == l[2]` to determine if the answer is "NO". This is incorrect because it doesn't consider the relationship between `n`, `m`, and `k` properly.
2. It should check if `m > 1` and `k < n` to determine if Alice can paint the ribbon in such a way that Bob cannot make all parts have the same color.

Let's generate a small-scale test input that includes the following scenarios:
1. `m = 1` (only one color available)
2. `k = n` (Bob can repaint all parts)
3. `m > 1` and `k < n` (Alice can paint the ribbon to prevent Bob from making all parts the same color)
4. `m > 1` and `k = n` (Bob can always make all parts the same color)

Here is the test input:

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

Explanation:
1. `1 1 1`: Only one part and one color. Bob can always make all parts the same color. (Expected output: NO)
2. `5 1 5`: Five parts and one color. Bob can always make all parts the same color. (Expected output: NO)
3. `5 2 2`: Five parts, two colors, and Bob can repaint two parts. Bob can always make all parts the same color. (Expected output: NO)
4. `5 2 1`: Five parts, two colors, and Bob can repaint one part. Alice can paint the ribbon so that Bob cannot make all parts the same color. (Expected output: YES)


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Painting the Ribbon
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob have bought a ribbon consisting of $$$n$$$ parts. Now they want to paint it.

First, Alice will paint every part of the ribbon into one of $$$m$$$ colors. For each part, she can choose its color arbitrarily.

Then, Bob will choose at most $$$k$$$ parts of the ribbon and repaint them into the same color (he chooses the affected parts and the color arbitrarily).

Bob would like all parts to have the same color. However, Alice thinks that this is too dull, so she wants to paint the ribbon in such a way that Bob cannot make all parts have the same color.

Is it possible to paint the ribbon in such a way?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of one line containing three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$1 \le m, k \le n \le 50$$$) — the number of parts, the number of colors and the number of parts Bob can repaint, respectively.

Output: For each test case, print YES if Alice can paint the ribbon so that Bob cannot make all parts have the same color. Otherwise, print NO.

You can print every letter in any register. For example, Yes, yes, yEs will all be recognized as positive answer.

Examples:
input:
5
1 1 1
5 1 1
5 2 1
5 2 2
5 5 3
output:
NO
NO
YES
NO
YES

Note: In the first test case, a ribbon consists of $$$1$$$ part. So all its parts will always have the same color.

In the second test case, there is only $$$1$$$ color.

In the third test case, Alice can paint the ribbon as follows: $$$[1, 2, 1, 2, 1]$$$. It's impossible to change the color of at most $$$1$$$ part so that all parts have the same color.

In the fourth test case, no matter how Alice paints the ribbon, Bob will always be able to repaint $$$2$$$ parts so that all parts have the same color.

In the fifth test case, Alice can paint the ribbon as follows: $$$[1, 2, 3, 4, 5]$$$. It's impossible to change the color of at most $$$3$$$ parts so that all parts have the same color.

Buggy Code:
import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int t = sc.nextInt();
        for (int i = 1; i <=t; i++) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            int k = sc.nextInt();

            if (m == k) {
                System.out.println("NO");
            } else {
                System.out.println("YES");
            }
        }
        sc.close();

    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios where the code's logic fails. The buggy code incorrectly checks if `m == k` and prints "NO" if true, otherwise it prints "YES". This logic is flawed because it doesn't consider the actual constraints and the relationship between `n`, `m`, and `k`.

### Analysis of the Buggy Code
- The buggy code checks if the number of colors `m` is equal to the number of parts Bob can repaint `k`. This is incorrect because the ability to make all parts the same color depends on the total number of parts `n`, the number of colors `m`, and the number of parts Bob can repaint `k`.

### Correct Logic
- Alice can prevent Bob from making all parts the same color if the number of parts `n` is greater than the number of parts Bob can repaint `k` plus one. This is because Bob can only change up to `k` parts, and if `n > k + 1`, there will always be at least one part that Bob cannot change to match the others.

### Challenging Test Input
To expose the bug, we need a test case where `n > k + 1` and `m` is not necessarily equal to `k`. This will ensure that the buggy code's logic fails.

```plaintext
3
10 5 4
10 5 5
10 5 6
```

### Explanation
- **Test Case 1: 10 5 4**
  - `n = 10`, `m = 5`, `k = 4`
  - Since `n > k + 1` (10 > 4 + 1), Alice can paint the ribbon in such a way that Bob cannot make all parts the same color.
  - Expected output: `YES`
  - Buggy code output: `YES` (correct, but for the wrong reason)

- **Test Case 2: 10 5 5**
  - `n = 10`, `m = 5`, `k = 5`
  - Since `n > k + 1` (10 > 5 + 1) is false, Bob can always make all parts the same color.
  - Expected output: `NO`
  - Buggy code output: `NO` (correct, but for the wrong reason)

- **Test Case 3: 10 5 6**
  - `n = 10`, `m = 5`, `k = 6`
  - Since `n > k + 1` (10 > 6 + 1) is false, Bob can always make all parts the same color.
  - Expected output: `NO`
  - Buggy code output: `YES` (incorrect)

This test input will expose the bug in the buggy code, as it will produce incorrect output for the third test case.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Painting the Ribbon
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob have bought a ribbon consisting of $$$n$$$ parts. Now they want to paint it.

First, Alice will paint every part of the ribbon into one of $$$m$$$ colors. For each part, she can choose its color arbitrarily.

Then, Bob will choose at most $$$k$$$ parts of the ribbon and repaint them into the same color (he chooses the affected parts and the color arbitrarily).

Bob would like all parts to have the same color. However, Alice thinks that this is too dull, so she wants to paint the ribbon in such a way that Bob cannot make all parts have the same color.

Is it possible to paint the ribbon in such a way?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of one line containing three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$1 \le m, k \le n \le 50$$$) — the number of parts, the number of colors and the number of parts Bob can repaint, respectively.

Output: For each test case, print YES if Alice can paint the ribbon so that Bob cannot make all parts have the same color. Otherwise, print NO.

You can print every letter in any register. For example, Yes, yes, yEs will all be recognized as positive answer.

Examples:
input:
5
1 1 1
5 1 1
5 2 1
5 2 2
5 5 3
output:
NO
NO
YES
NO
YES

Note: In the first test case, a ribbon consists of $$$1$$$ part. So all its parts will always have the same color.

In the second test case, there is only $$$1$$$ color.

In the third test case, Alice can paint the ribbon as follows: $$$[1, 2, 1, 2, 1]$$$. It's impossible to change the color of at most $$$1$$$ part so that all parts have the same color.

In the fourth test case, no matter how Alice paints the ribbon, Bob will always be able to repaint $$$2$$$ parts so that all parts have the same color.

In the fifth test case, Alice can paint the ribbon as follows: $$$[1, 2, 3, 4, 5]$$$. It's impossible to change the color of at most $$$3$$$ parts so that all parts have the same color.

Buggy Code:
import java.util.*;
import java.io.*;
 public class Competetive  {
    static FastReader in=new FastReader();
    static final Random random=new Random();
    static long mod=1000000007L;
    static HashMap<String,Integer>map=new HashMap<>();

    public static void main(String args[]) throws IOException {
        int t=in.nextInt();
        PrintWriter out = new PrintWriter(System.out);
        loop:
        while(t-->0)
        {
            solve(out);
        }

    }

    private static void solve(PrintWriter out) {

        int a = in.nextInt() ;
        int b = in.nextInt() ;
        int c = in.nextInt() ;

        if(a==1 || b==1){
            out.println("NO");

        } else{
            int temp = a/b + a%b ;

            if(c>=(a-temp)){
                out.println("NO");

            } else{
                out.println("YES");
            }
        }        out.flush();
    }

    static int max(int a, int b)
    {
        if(a<b)
            return b;
        return a;
    }

    static void ruffleSort(int[] a) {
        int n=a.length;
        for (int i=0; i<n; i++) {
            int oi=random.nextInt(n), temp=a[oi];
            a[oi]=a[i]; a[i]=temp;
        }
        Arrays.sort(a);
    }

    static < E > void print(E res)
    {
        System.out.println(res);
    }

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

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

    static int abs(int a)
    {
        if(a<0)
            return -1*a;
        return a;
    }

    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
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }

        int [] readintarray(int n) {
            int res [] = new int [n];
            for(int i = 0; i<n; i++)res[i] = nextInt();
            return res;
        }
        long [] readlongarray(int n) {
            long res [] = new long [n];
            for(int i = 0; i<n; i++)res[i] = nextLong();
            return res;
        }
    }

}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on the logic where the algorithm determines whether Alice can paint the ribbon in such a way that Bob cannot make all parts have the same color. The key part of the code to scrutinize is:

```java
if(a==1 || b==1){
    out.println("NO");
} else{
    int temp = a/b + a%b ;
    if(c>=(a-temp)){
        out.println("NO");
    } else{
        out.println("YES");
    }
}
```

The logic here is flawed because it doesn't correctly handle the cases where the number of parts `a` and the number of colors `b` are not trivial. Specifically, the expression `a/b + a%b` is used to calculate the number of parts that can be uniquely colored, but this is not always correct.

### Step-by-Step Reasoning:

1. **Edge Case 1: Single Part (a = 1)**
   - This is correctly handled by the code, as it prints "NO" for `a == 1`.

2. **Edge Case 2: Single Color (b = 1)**
   - This is also correctly handled by the code, as it prints "NO" for `b == 1`.

3. **General Case:**
   - The expression `a/b + a%b` is supposed to represent the number of parts that can be uniquely colored. However, this is not always correct. For example, if `a = 5`, `b = 2`, and `c = 2`, the expression `a/b + a%b` equals `3` (since `5/2 = 2` and `5%2 = 1`). This means the code incorrectly assumes that Bob can change `2` parts to make all parts the same color, which is not true.

### Test Input to Expose the Bug:

We need a test input where `a` and `b` are such that the expression `a/b + a%b` does not correctly represent the number of uniquely colored parts, and `c` is just enough to make the code fail.

```plaintext
1
5 2 2
```

### Explanation:

- **a = 5**: There are 5 parts.
- **b = 2**: There are 2 colors.
- **c = 2**: Bob can repaint 2 parts.

For this input:
- The expression `a/b + a%b` equals `3` (since `5/2 = 2` and `5%2 = 1`).
- The code will incorrectly assume that Bob can change 2 parts to make all parts the same color, and it will print "NO".
- However, Alice can paint the ribbon as `[1, 2, 1, 2, 1]`, and Bob cannot change 2 parts to make all parts the same color, so the correct answer should be "YES".

This test input should expose the bug in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Painting the Ribbon
time_limit_ms: 2000
memory_limit_mb: 512
Description: Alice and Bob have bought a ribbon consisting of $$$n$$$ parts. Now they want to paint it.

First, Alice will paint every part of the ribbon into one of $$$m$$$ colors. For each part, she can choose its color arbitrarily.

Then, Bob will choose at most $$$k$$$ parts of the ribbon and repaint them into the same color (he chooses the affected parts and the color arbitrarily).

Bob would like all parts to have the same color. However, Alice thinks that this is too dull, so she wants to paint the ribbon in such a way that Bob cannot make all parts have the same color.

Is it possible to paint the ribbon in such a way?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases.

Each test case consists of one line containing three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$1 \le m, k \le n \le 50$$$) — the number of parts, the number of colors and the number of parts Bob can repaint, respectively.

Output: For each test case, print YES if Alice can paint the ribbon so that Bob cannot make all parts have the same color. Otherwise, print NO.

You can print every letter in any register. For example, Yes, yes, yEs will all be recognized as positive answer.

Examples:
input:
5
1 1 1
5 1 1
5 2 1
5 2 2
5 5 3
output:
NO
NO
YES
NO
YES

Note: In the first test case, a ribbon consists of $$$1$$$ part. So all its parts will always have the same color.

In the second test case, there is only $$$1$$$ color.

In the third test case, Alice can paint the ribbon as follows: $$$[1, 2, 1, 2, 1]$$$. It's impossible to change the color of at most $$$1$$$ part so that all parts have the same color.

In the fourth test case, no matter how Alice paints the ribbon, Bob will always be able to repaint $$$2$$$ parts so that all parts have the same color.

In the fifth test case, Alice can paint the ribbon as follows: $$$[1, 2, 3, 4, 5]$$$. It's impossible to change the color of at most $$$3$$$ parts so that all parts have the same color.

Buggy Code:
#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N = 3e5+5;

void solve(){
	int n,k,m;
    cin >> n >> k >> m;
    if((n-n/k<=m && k<m) || k==1 || k==m) cout << "No\n";
    else cout << "Yes\n";
}

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    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:
```plaintext
1
5 3 2
```

**Explanation:**
- **n = 5**: The ribbon has 5 parts.
- **m = 3**: There are 3 colors available.
- **k = 2**: Bob can repaint at most 2 parts.

**Why this test case is challenging:**
- Alice needs to paint the ribbon in such a way that Bob cannot make all parts the same color by repainting at most 2 parts.
- If Alice paints the ribbon as `[1, 2, 1, 2, 3]`, Bob cannot make all parts the same color by repainting at most 2 parts.
- However, the buggy code might incorrectly output "No" because it incorrectly checks the condition `(n - n/k <= m && k < m) || k == 1 || k == m`, which does not correctly handle this scenario.