**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dyn-scripted Robot (Easy Version)
time_limit_ms: 3000
memory_limit_mb: 256
Description: This is the easy version of the problem. The only difference is that in this version $k \le n$. You can make hacks only if both versions of the problem are solved.      Given a $w \times h$ rectangle on the $Oxy$ plane, with points $(0, 0)$ at the bottom-left and $(w, h)$ at the top-right of the rectangle.

You also have a robot initially at point $(0, 0)$ and a script $s$ of $n$ characters. Each character is either L, R, U, or D, which tells the robot to move left, right, up, or down respectively.

The robot can only move inside the rectangle; otherwise, it will change the script $s$ as follows:

  * If it tries to move outside a vertical border, it changes all L characters to R's (and vice versa, all R's to L's).    * If it tries to move outside a horizontal border, it changes all U characters to D's (and vice versa, all D's to U's).

Then, it will execute the changed script starting from the character which it couldn't execute.

![](CDN_BASE_URL/ff5ae9758c965c2d8398c936e9581dab) An example of the robot's movement process, $s = \texttt{"ULULURD"}$

The script $s$ will be executed for $k$ times continuously. All changes to the string $s$ will be retained even when it is repeated. During this process, how many times will the robot move to the point $(0, 0)$ in total? Note that the initial position does NOT count.

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

The first line of each test case contains four integers $n$, $k$, $w$, and $h$ ($1 \le n, w, h \le 10^6$; $1 \le k \le n$).

The second line contains a single string $s$ of size $n$ ($s_i \in \\{\texttt{L}, \texttt{R}, \texttt{U}, \texttt{D}\\}$) — the script to be executed.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print a single integer — the number of times the robot reaches $(0, 0)$ when executing script $s$ for $k$ times continuously.

Examples:
input:
5
2 2 2 2
UR
4 2 1 1
LLDD
6 3 3 1
RLRRRL
5 5 3 3
RUURD
7 5 3 4
RRDLUUU
output:
0
4
3
0
1

Note: In the first test case, the robot only moves up and right. In the end, it occupies the position $(2, 2)$ but never visits $(0, 0)$. So the answer is $0$.

In the second test case, each time executing the script the robot visits the origin twice. And since $k=2$, it visits the origin $2 \cdot 2 = 4$ times overall.

![](CDN_BASE_URL/ea799abf84f8a212d5a0d782be4d5e9c)

In the third test case, the visualization is shown as below:

![](CDN_BASE_URL/c6a5be14982b3d57a5103783598d4683)

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
template<class T,T Q=2>constexpr T inf=numeric_limits<T>::max()/Q-1;
template<class T,class S=T>inline bool umin(T&m,S&&x){return x<m?m=x,1:0;}
template<class T,class S=T>inline bool umax(T&m,S&&x){return m<x?m=x,1:0;}
auto&operator<<(ostream&o,auto&&t)requires requires{tuple_cat(t);}{int f=0,u=&o!=&cerr;o<<"<"+u;apply([&](auto&...x){(((f++?o<<", "+u:o)<<x),...);},t);return o<<">"+u;}
auto&operator<<(ostream&o,ranges::range auto&&r)requires(!requires{o<<r;}){int f=0,u=&o!=&cerr;o<<"["+u;for(auto&&x:r)(f++?o<<", "+u:o)<<x;return o<<"]"+u;}
#ifdef BIZON
	#define rr(v...) [](auto&&...x){ cerr << boolalpha << "\e[1;38;5;68m" << #v << " "; int _=0; ((cerr<<"\e[0;38;5;61m"<<",="[!_++]<<"\e[0m "<<x),...)<<endl; }(v);
#else
	#define rr(...) ;
	#define endl '\n'
#endif
#define bb(c) begin(c), end(c)
#define ff(T, name, args...) function<T(args)> name = [&](args)->T
#define ee(e) [&](auto&&it) { return e; }
#define jj(v...) v; [](auto&...x){(cin>>...>>x);}(v);
#define ii(v...) int jj(v)
using ll = decltype(1LL);

void run_case(const size_t ____case) { // rr(____case)
	ii(n, k, w, h)
	string jj(s)

	const ll rw = w * 2;
	const ll rh = h * 2;

	ll x = 0, y = 0;
	vector<pair<ll,ll>> pts;
	for(char c : s) {
		if(c == 'U') ++y;
		if(c == 'D') --y;
		if(c == 'L') --x;
		if(c == 'R') ++x;
		pts.emplace_back(x%rw, y%rh);
	}

	map<pair<ll,ll>, ll> cnt;
	for(auto it : pts) cnt[it]++;

	ll ans = 0;
	for(int it=0; it<k; ++it) {
		auto [x, y] = pts.back();
		x *= it;
		y *= it;
		for(auto [cx, cy] : pts) {
			if((x+cx)%rw == 0 && (y+cy)%rh == 0) ++ans;
		}
	}

	cout << ans << endl;

}

int main() {
	if(auto f="in.txt"; fopen(f,"r") && freopen(f,"r",stdin));
	cin.tie(0)->sync_with_stdio(0);

	size_t tn = 1; cin>>tn;
	for(size_t t=1; t<=tn; ++t) run_case(t);

	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that can maximize the failure probability. Here are the steps to consider:

1. **Edge Cases for Rectangle Dimensions**:
   - When the rectangle dimensions \( w \) and \( h \) are very small (e.g., \( w = 1 \) or \( h = 1 \)).
   - When \( w \) or \( h \) are even or odd, as this can affect the periodicity of the robot's movement.

2. **Edge Cases for Script Length and Content**:
   - When the script \( s \) is very short (e.g., \( n = 1 \) or \( n = 2 \)).
   - When the script \( s \) contains only one type of move (e.g., all 'U', all 'D', all 'L', all 'R').
   - When the script \( s \) is designed to make the robot move back and forth between two points, including the origin.

3. **Edge Cases for Repeating the Script**:
   - When \( k \) is very small (e.g., \( k = 1 \) or \( k = 2 \)).
   - When \( k \) is large enough to cause significant changes in the script due to boundary crossings.

Given these considerations, a challenging test input could be:

- A small rectangle with \( w = 1 \) and \( h = 1 \).
- A short script that makes the robot move in such a way that it frequently crosses the boundaries.
- A small \( k \) value to ensure the script is repeated a few times.

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

```plaintext
1
4 2 1 1
LLDD
```

Explanation:
- \( t = 1 \): One test case.
- \( n = 4 \), \( k = 2 \), \( w = 1 \), \( h = 1 \): The script is "LLDD" and is executed 2 times in a 1x1 rectangle.
- The script "LLDD" will cause the robot to move left, left, down, down. Since the robot is in a 1x1 rectangle, every move will be invalid and will change the script. This will lead to frequent changes and a high probability of exposing any bugs in the code.

This test input is designed to be small and valid while maximizing the chances of exposing issues 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: Dyn-scripted Robot (Easy Version)
time_limit_ms: 3000
memory_limit_mb: 256
Description: This is the easy version of the problem. The only difference is that in this version $k \le n$. You can make hacks only if both versions of the problem are solved.      Given a $w \times h$ rectangle on the $Oxy$ plane, with points $(0, 0)$ at the bottom-left and $(w, h)$ at the top-right of the rectangle.

You also have a robot initially at point $(0, 0)$ and a script $s$ of $n$ characters. Each character is either L, R, U, or D, which tells the robot to move left, right, up, or down respectively.

The robot can only move inside the rectangle; otherwise, it will change the script $s$ as follows:

  * If it tries to move outside a vertical border, it changes all L characters to R's (and vice versa, all R's to L's).    * If it tries to move outside a horizontal border, it changes all U characters to D's (and vice versa, all D's to U's).

Then, it will execute the changed script starting from the character which it couldn't execute.

![](CDN_BASE_URL/ff5ae9758c965c2d8398c936e9581dab) An example of the robot's movement process, $s = \texttt{"ULULURD"}$

The script $s$ will be executed for $k$ times continuously. All changes to the string $s$ will be retained even when it is repeated. During this process, how many times will the robot move to the point $(0, 0)$ in total? Note that the initial position does NOT count.

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

The first line of each test case contains four integers $n$, $k$, $w$, and $h$ ($1 \le n, w, h \le 10^6$; $1 \le k \le n$).

The second line contains a single string $s$ of size $n$ ($s_i \in \\{\texttt{L}, \texttt{R}, \texttt{U}, \texttt{D}\\}$) — the script to be executed.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print a single integer — the number of times the robot reaches $(0, 0)$ when executing script $s$ for $k$ times continuously.

Examples:
input:
5
2 2 2 2
UR
4 2 1 1
LLDD
6 3 3 1
RLRRRL
5 5 3 3
RUURD
7 5 3 4
RRDLUUU
output:
0
4
3
0
1

Note: In the first test case, the robot only moves up and right. In the end, it occupies the position $(2, 2)$ but never visits $(0, 0)$. So the answer is $0$.

In the second test case, each time executing the script the robot visits the origin twice. And since $k=2$, it visits the origin $2 \cdot 2 = 4$ times overall.

![](CDN_BASE_URL/ea799abf84f8a212d5a0d782be4d5e9c)

In the third test case, the visualization is shown as below:

![](CDN_BASE_URL/c6a5be14982b3d57a5103783598d4683)

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
string su(string s){
    string sU = s;
    for(int h = 0;h<s.length();h++){
            if(sU[h] == 'D'){
                sU[h] = 'U';
            }
            else if(sU[h] == 'U'){
                sU[h] = 'D';
            }
        }
        return sU;
}
string sr(string s){
    string sR = s;
     for(int h = 0;h<s.length();h++){
            if(sR[h] == 'L'){
                sR[h] = 'R';
            }
            else if(sR[h] == 'R'){
                sR[h] = 'L';
            }
        }
        return sR;
}
int main(){
    int t;
    cin >> t;
    for(int i = 0;i<t;i++){
        int n,k,w,v;
        cin >> n >> k >> w >> v;
        string s;
        cin >> s;

        int x = 0;
        int y = 0;
        int ans = 0;

        for(int l = 0;l<k;l++){
        for(int j = 0;j<n;j++){
            if(s[j] == 'U'){
                y++;
            }
            if(s[j] == 'D'){
                y--;
            }
            if(s[j] == 'L'){
                x--;
            }
            if(s[j] == 'R'){
                x++;
            }
            if(x==0 && y ==0){
                ans++;
            }
            if(x < 0 || x > w){
                s = sr(s);
                if(x<0) x++;
                else x--;
                j--;
            }
            if( y < 0 || y > v){
                if(y<0) y++;
                else y--;
               s = su(s);
               j--;
            }
        }
            }
        cout << ans << 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
3 3 1 1
RDU
```

### Explanation:
- **Test Case:**
  - **n = 3**: The script length is 3.
  - **k = 3**: The script is executed 3 times.
  - **w = 1**: The width of the rectangle is 1.
  - **h = 1**: The height of the rectangle is 1.
  - **s = "RDU"**: The script is "RDU".

- **Expected Behavior:**
  - The robot starts at (0, 0).
  - The first move is 'R' (right), which moves the robot to (1, 0).
  - The second move is 'D' (down), but since the robot is already at the bottom border, it will not move and the script will be changed to "LUD".
  - The third move is 'U' (up), which moves the robot to (1, 1).

  - The script is now "LUD" and the robot is at (1, 1).

  - The second execution of the script:
    - The first move is 'L' (left), which moves the robot to (0, 1).
    - The second move is 'U' (up), which moves the robot to (0, 0), and the robot reaches the origin.
    - The third move is 'D' (down), which moves the robot to (0, 1).

  - The script is still "LUD" and the robot is at (0, 1).

  - The third execution of the script:
    - The first move is 'L' (left), but the robot is already at the left border, so it will not move and the script will be changed to "RUD".
    - The second move is 'U' (up), which moves the robot to (0, 0), and the robot reaches the origin.
    - The third move is 'D' (down), which moves the robot to (0, 1).

- **Expected Output:**
  - The robot reaches the origin (0, 0) **2 times** during the 3 executions of the script.

This test case is designed to expose the bug in the code where the script is not correctly changed when the robot tries to move outside the borders, and the robot's position is not correctly updated.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dyn-scripted Robot (Easy Version)
time_limit_ms: 3000
memory_limit_mb: 256
Description: This is the easy version of the problem. The only difference is that in this version $k \le n$. You can make hacks only if both versions of the problem are solved.      Given a $w \times h$ rectangle on the $Oxy$ plane, with points $(0, 0)$ at the bottom-left and $(w, h)$ at the top-right of the rectangle.

You also have a robot initially at point $(0, 0)$ and a script $s$ of $n$ characters. Each character is either L, R, U, or D, which tells the robot to move left, right, up, or down respectively.

The robot can only move inside the rectangle; otherwise, it will change the script $s$ as follows:

  * If it tries to move outside a vertical border, it changes all L characters to R's (and vice versa, all R's to L's).    * If it tries to move outside a horizontal border, it changes all U characters to D's (and vice versa, all D's to U's).

Then, it will execute the changed script starting from the character which it couldn't execute.

![](CDN_BASE_URL/ff5ae9758c965c2d8398c936e9581dab) An example of the robot's movement process, $s = \texttt{"ULULURD"}$

The script $s$ will be executed for $k$ times continuously. All changes to the string $s$ will be retained even when it is repeated. During this process, how many times will the robot move to the point $(0, 0)$ in total? Note that the initial position does NOT count.

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

The first line of each test case contains four integers $n$, $k$, $w$, and $h$ ($1 \le n, w, h \le 10^6$; $1 \le k \le n$).

The second line contains a single string $s$ of size $n$ ($s_i \in \\{\texttt{L}, \texttt{R}, \texttt{U}, \texttt{D}\\}$) — the script to be executed.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print a single integer — the number of times the robot reaches $(0, 0)$ when executing script $s$ for $k$ times continuously.

Examples:
input:
5
2 2 2 2
UR
4 2 1 1
LLDD
6 3 3 1
RLRRRL
5 5 3 3
RUURD
7 5 3 4
RRDLUUU
output:
0
4
3
0
1

Note: In the first test case, the robot only moves up and right. In the end, it occupies the position $(2, 2)$ but never visits $(0, 0)$. So the answer is $0$.

In the second test case, each time executing the script the robot visits the origin twice. And since $k=2$, it visits the origin $2 \cdot 2 = 4$ times overall.

![](CDN_BASE_URL/ea799abf84f8a212d5a0d782be4d5e9c)

In the third test case, the visualization is shown as below:

![](CDN_BASE_URL/c6a5be14982b3d57a5103783598d4683)

Buggy Code:
#include <bits/stdc++.h>
#define int long long
using namespace std;
int n,k,w,h;
void chang(string& str,char s)
{
	if(s=='L'||s=='R')
	{
		for(int i=0;i<n;i++)
		{
			if(str[i]=='L')
			{
				str[i]='R';
			}
			else if(str[i]=='R')
			{
				str[i]='L';
			}
		}
	}
	if(s=='D'||s=='U')
	{
		for(int i=0;i<n;i++)
		{
			if(str[i]=='U')
			{
				str[i]='D';
			}
			else if(str[i]=='D')
			{
				str[i]='U';
			}
		}
	}
}

void slove()
{

	cin>>n>>k>>w>>h;
	string str;
	cin>>str;
	int sum=0;
	int x1=0,y1=0;
	while(k--)
	{
		for(int i=0;i<n;)
	{
		if(str[i]=='L')
		{
			if(x1-1<0)
			{
				chang(str,'L');
			}
			else
			{
				x1--;
				i++;
				if(x1==0&&y1==0)
		{
			sum++;
		}
			}
		}
		else if(str[i]=='R')
		{
			if(x1+1>w)
			{
				chang(str,'R');
			}
			else
			{
				x1++;i++;
				if(x1==0&&y1==0)
		{
			sum++;
		}
			}
		}
		else if(str[i]=='U')
		{
			if(y1+1>h)
			{
				chang(str,'U');
			}
			else
			{
				y1++;i++;
				if(x1==0&&y1==0)
		{
			sum++;
		}
			}
		}
		else if(str[i]=='D')
		{
			if(y1-1<0)
			{
				chang(str,'D');
			}
			else
			{
				y1--;i++;
				if(x1==0&&y1==0)
		{
			sum++;
		}
			}
		}

	}
	}

	cout<<sum<<endl;
}
signed main()
{
 	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
 	int t;
	cin>>t;
	while(t--)
	{
		slove();
	}
	return 0;
}

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

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

Think step by step.

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

1. The script modification logic does not handle the case where the robot tries to move outside the bounds of the rectangle.
2. The script modification logic is applied incorrectly, potentially leading to incorrect script transformations.
3. The logic for counting the number of times the robot reaches the origin is flawed.

Let's create a test case that can expose these issues:

1. **Script that causes multiple boundary violations**: This will test the script modification logic.
2. **Script that causes the robot to frequently visit the origin**: This will test the counting logic.

Here's a small-scale test input that can expose the bugs:

```plaintext
3
2 2 1 1
LR
4 2 1 1
UDUD
6 3 3 1
RLRRRL
```

### Explanation:
1. **Test Case 1**:
   - **n = 2, k = 2, w = 1, h = 1, s = "LR"**
   - The robot starts at (0, 0) and tries to move left, which is out of bounds. The script should change to "RL".
   - The robot then moves right to (1, 0) and tries to move right again, which is out of bounds. The script should change back to "LR".
   - The robot then moves left to (0, 0) and tries to move left again, which is out of bounds. The script should change back to "RL".
   - The robot then moves right to (1, 0) and tries to move right again, which is out of bounds. The script should change back to "LR".
   - The robot should visit the origin 2 times in total.

2. **Test Case 2**:
   - **n = 4, k = 2, w = 1, h = 1, s = "UDUD"**
   - The robot starts at (0, 0) and tries to move up, which is out of bounds. The script should change to "DUUD".
   - The robot then moves down to (0, 0) and tries to move up again, which is out of bounds. The script should change back to "UDUD".
   - The robot then moves up to (0, 1) and tries to move up again, which is out of bounds. The script should change back to "DUUD".
   - The robot then moves down to (0, 0) and tries to move down again, which is out of bounds. The script should change back to "UDUD".
   - The robot should visit the origin 4 times in total.

3. **Test Case 3**:
   - **n = 6, k = 3, w = 3, h = 1, s = "RLRRRL"**
   - The robot starts at (0, 0) and moves right to (1, 0), then moves left to (0, 0), then moves right to (1, 0), then moves right to (2, 0), then moves left to (1, 0), and finally moves left to (0, 0).
   - The script "RLRRRL" will be repeated 3 times, and the robot should visit the origin 3 times in total.

This test input should expose the bugs in the script modification logic and the counting logic of the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dyn-scripted Robot (Easy Version)
time_limit_ms: 3000
memory_limit_mb: 256
Description: This is the easy version of the problem. The only difference is that in this version $k \le n$. You can make hacks only if both versions of the problem are solved.      Given a $w \times h$ rectangle on the $Oxy$ plane, with points $(0, 0)$ at the bottom-left and $(w, h)$ at the top-right of the rectangle.

You also have a robot initially at point $(0, 0)$ and a script $s$ of $n$ characters. Each character is either L, R, U, or D, which tells the robot to move left, right, up, or down respectively.

The robot can only move inside the rectangle; otherwise, it will change the script $s$ as follows:

  * If it tries to move outside a vertical border, it changes all L characters to R's (and vice versa, all R's to L's).    * If it tries to move outside a horizontal border, it changes all U characters to D's (and vice versa, all D's to U's).

Then, it will execute the changed script starting from the character which it couldn't execute.

![](CDN_BASE_URL/ff5ae9758c965c2d8398c936e9581dab) An example of the robot's movement process, $s = \texttt{"ULULURD"}$

The script $s$ will be executed for $k$ times continuously. All changes to the string $s$ will be retained even when it is repeated. During this process, how many times will the robot move to the point $(0, 0)$ in total? Note that the initial position does NOT count.

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

The first line of each test case contains four integers $n$, $k$, $w$, and $h$ ($1 \le n, w, h \le 10^6$; $1 \le k \le n$).

The second line contains a single string $s$ of size $n$ ($s_i \in \\{\texttt{L}, \texttt{R}, \texttt{U}, \texttt{D}\\}$) — the script to be executed.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print a single integer — the number of times the robot reaches $(0, 0)$ when executing script $s$ for $k$ times continuously.

Examples:
input:
5
2 2 2 2
UR
4 2 1 1
LLDD
6 3 3 1
RLRRRL
5 5 3 3
RUURD
7 5 3 4
RRDLUUU
output:
0
4
3
0
1

Note: In the first test case, the robot only moves up and right. In the end, it occupies the position $(2, 2)$ but never visits $(0, 0)$. So the answer is $0$.

In the second test case, each time executing the script the robot visits the origin twice. And since $k=2$, it visits the origin $2 \cdot 2 = 4$ times overall.

![](CDN_BASE_URL/ea799abf84f8a212d5a0d782be4d5e9c)

In the third test case, the visualization is shown as below:

![](CDN_BASE_URL/c6a5be14982b3d57a5103783598d4683)

Buggy Code:
t = int(input())
answers = [0] * t
for i in range(t):
    inp = input().split()
    n = int(inp[0])
    k = int(inp[1])
    w = int(inp[2])
    h = int(inp[3])
    s = input()
    s = s * k
    coords = [0,0]
    side_flips = 0
    vert_flips = 0
    for j in range(len(s)):
        if(s[j] == 'U'):
            if(vert_flips == 0):
                if(coords[1] + 1 > h):
                    vert_flips = 1
                    coords[1] = coords[1] - 1
                else:
                    coords[1] = coords[1] + 1
            else:
                if(coords[1] == 0):
                    vert_flips = 0
                    coords[1] = 1
                else:
                    coords[1] = coords[1] - 1

        elif(s[j] == 'D'):
            if(vert_flips == 0):
                if(coords[1] == 0):
                    vert_flips = 1
                    coords[1] = coords[1] + 1
                else:
                    coords[1] = coords[1] - 1
            elif(coords[1] + 1 > h):
                    vert_flips = 0
                    coords[1] = coords[1] - 1
            else:
                coords[1] = coords[1] + 1
        elif(s[j] == 'R'):
            if(side_flips == 0):
                if(coords[0] + 1 > w):
                    side_flips = 1
                    coords[0] = coords[0] - 1
                else:
                    coords[0] = coords[0] + 1
            elif(coords[0] == 0):
                    side_flips = 0
                    coords[0] = 1
            else:
                coords[0] = coords[0] - 1
        elif(s[j] == 'L'):
            if(side_flips == 0):
                if(coords[0] == 0):
                    side_flips = 1
                    coords[0] = 1
                else:
                    coords[0] = coords[0] - 1
            elif(coords[0] + 1 > w):
                    side_flips = 0
                    coords[0] = coords[0] - 1
            else:
                coords[0] = coords[0] + 1

        if(coords[0] == 0 and coords[1] == 0):
            answers[i] = int(answers[i]) + 1

b = "\n".join([str(answer) for answer in answers])
print(b)

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

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

Think step by step.

Response:
```plaintext
1
10 2 2 2
RRDDLULURD
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dyn-scripted Robot (Easy Version)
time_limit_ms: 3000
memory_limit_mb: 256
Description: This is the easy version of the problem. The only difference is that in this version $k \le n$. You can make hacks only if both versions of the problem are solved.      Given a $w \times h$ rectangle on the $Oxy$ plane, with points $(0, 0)$ at the bottom-left and $(w, h)$ at the top-right of the rectangle.

You also have a robot initially at point $(0, 0)$ and a script $s$ of $n$ characters. Each character is either L, R, U, or D, which tells the robot to move left, right, up, or down respectively.

The robot can only move inside the rectangle; otherwise, it will change the script $s$ as follows:

  * If it tries to move outside a vertical border, it changes all L characters to R's (and vice versa, all R's to L's).    * If it tries to move outside a horizontal border, it changes all U characters to D's (and vice versa, all D's to U's).

Then, it will execute the changed script starting from the character which it couldn't execute.

![](CDN_BASE_URL/ff5ae9758c965c2d8398c936e9581dab) An example of the robot's movement process, $s = \texttt{"ULULURD"}$

The script $s$ will be executed for $k$ times continuously. All changes to the string $s$ will be retained even when it is repeated. During this process, how many times will the robot move to the point $(0, 0)$ in total? Note that the initial position does NOT count.

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

The first line of each test case contains four integers $n$, $k$, $w$, and $h$ ($1 \le n, w, h \le 10^6$; $1 \le k \le n$).

The second line contains a single string $s$ of size $n$ ($s_i \in \\{\texttt{L}, \texttt{R}, \texttt{U}, \texttt{D}\\}$) — the script to be executed.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print a single integer — the number of times the robot reaches $(0, 0)$ when executing script $s$ for $k$ times continuously.

Examples:
input:
5
2 2 2 2
UR
4 2 1 1
LLDD
6 3 3 1
RLRRRL
5 5 3 3
RUURD
7 5 3 4
RRDLUUU
output:
0
4
3
0
1

Note: In the first test case, the robot only moves up and right. In the end, it occupies the position $(2, 2)$ but never visits $(0, 0)$. So the answer is $0$.

In the second test case, each time executing the script the robot visits the origin twice. And since $k=2$, it visits the origin $2 \cdot 2 = 4$ times overall.

![](CDN_BASE_URL/ea799abf84f8a212d5a0d782be4d5e9c)

In the third test case, the visualization is shown as below:

![](CDN_BASE_URL/c6a5be14982b3d57a5103783598d4683)

Buggy Code:
def checkorigin(pos,count,w,h):
    if pos[0]%(2*w)==0 and pos[1]%(2*h)==0:
        return count+1
    else:
        return count

t = int(input())
#print(ud(input()))

for i in range(t):
    listnum = list(map(int,input().split()))
    n = listnum[0]
    k = listnum[1]
    w = listnum[2]
    h = listnum[3]
    seq = input()
    times = 0
    maxpos = [w,h]
    pos=[0,0]

    for j in range(k):
        for a in range(len(seq)):
            #print(seq[a])
            if seq[a] == "R":
                pos[0]+=1
            elif seq[a] == "U":
                pos[1]+=1
            elif seq[a] == "L":
                pos[0]-=1
            else:
                pos[1]-=1
            #print(pos)
            times = checkorigin(pos,times,w,h)
        if pos[0]%(2*w)==0 and pos[1]%(2*h)==0 and j==0:
            times = k*times
            break
    print(times)

**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 1 1
RDLUR
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dyn-scripted Robot (Easy Version)
time_limit_ms: 3000
memory_limit_mb: 256
Description: This is the easy version of the problem. The only difference is that in this version $k \le n$. You can make hacks only if both versions of the problem are solved.      Given a $w \times h$ rectangle on the $Oxy$ plane, with points $(0, 0)$ at the bottom-left and $(w, h)$ at the top-right of the rectangle.

You also have a robot initially at point $(0, 0)$ and a script $s$ of $n$ characters. Each character is either L, R, U, or D, which tells the robot to move left, right, up, or down respectively.

The robot can only move inside the rectangle; otherwise, it will change the script $s$ as follows:

  * If it tries to move outside a vertical border, it changes all L characters to R's (and vice versa, all R's to L's).    * If it tries to move outside a horizontal border, it changes all U characters to D's (and vice versa, all D's to U's).

Then, it will execute the changed script starting from the character which it couldn't execute.

![](CDN_BASE_URL/ff5ae9758c965c2d8398c936e9581dab) An example of the robot's movement process, $s = \texttt{"ULULURD"}$

The script $s$ will be executed for $k$ times continuously. All changes to the string $s$ will be retained even when it is repeated. During this process, how many times will the robot move to the point $(0, 0)$ in total? Note that the initial position does NOT count.

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

The first line of each test case contains four integers $n$, $k$, $w$, and $h$ ($1 \le n, w, h \le 10^6$; $1 \le k \le n$).

The second line contains a single string $s$ of size $n$ ($s_i \in \\{\texttt{L}, \texttt{R}, \texttt{U}, \texttt{D}\\}$) — the script to be executed.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print a single integer — the number of times the robot reaches $(0, 0)$ when executing script $s$ for $k$ times continuously.

Examples:
input:
5
2 2 2 2
UR
4 2 1 1
LLDD
6 3 3 1
RLRRRL
5 5 3 3
RUURD
7 5 3 4
RRDLUUU
output:
0
4
3
0
1

Note: In the first test case, the robot only moves up and right. In the end, it occupies the position $(2, 2)$ but never visits $(0, 0)$. So the answer is $0$.

In the second test case, each time executing the script the robot visits the origin twice. And since $k=2$, it visits the origin $2 \cdot 2 = 4$ times overall.

![](CDN_BASE_URL/ea799abf84f8a212d5a0d782be4d5e9c)

In the third test case, the visualization is shown as below:

![](CDN_BASE_URL/c6a5be14982b3d57a5103783598d4683)

Buggy Code:
import java.util.*;

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

        int t = sc.nextInt();

        while(t-- > 0) {
            int n = sc.nextInt();
            int k = sc.nextInt();

            int w = sc.nextInt();
            int h = sc.nextInt();

            String s = sc.next();

            System.out.println(move(n,k,w,h,s));
        }

        sc.close();
    }

    public static int move(int n, int k, int w, int h, String s) {
        boolean LRrev = false;
        boolean UDRev = false;

        int count = 0;

        int x=0,y=0;
        for(int i=0; i<n*k; i++) {
            char ch = s.charAt(i%n);

            if(ch=='U' && !UDRev) {
                if(y+1 <= h)
                    y++;
                else {
                    UDRev=true;
                    y--;
                }
            } else if(ch=='U' && UDRev) {
                if(y-1 >= 0)
                    y--;
                else {
                    UDRev=false;
                    y++;
                }
            } else if(ch=='D' && !UDRev) {
                if(y-1 >= 0)
                    y--;
                else {
                    UDRev = true;
                    y++;
                }
            } else if(ch=='D' && UDRev) {
                if(y+1 <= h)
                    y++;
                else {
                    UDRev = false;
                    y--;
                }
            } else if(ch=='L' && !LRrev) {
                if(x-1 >=0)
                    x--;
                else {
                    LRrev = true;
                    x++;
                }
            } else if(ch=='L' && LRrev) {
                if(x+1 <= w)
                    x++;
                else {
                    LRrev = false;
                    x--;
                }
            } else if(ch=='R' && !LRrev) {
                if(x+1 <= w)
                    x++;
                else {
                    LRrev = true;
                    x--;
                }
            } else if(ch=='R' && LRrev) {
                if(x-1 >=0)
                    x--;
                else {
                    LRrev = false;
                    x++;
                }
            }

            if(x==0 && y==0)
                count++;
        }

        return count;
    }
}

**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 1 1
RULDR
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Dyn-scripted Robot (Easy Version)
time_limit_ms: 3000
memory_limit_mb: 256
Description: This is the easy version of the problem. The only difference is that in this version $k \le n$. You can make hacks only if both versions of the problem are solved.      Given a $w \times h$ rectangle on the $Oxy$ plane, with points $(0, 0)$ at the bottom-left and $(w, h)$ at the top-right of the rectangle.

You also have a robot initially at point $(0, 0)$ and a script $s$ of $n$ characters. Each character is either L, R, U, or D, which tells the robot to move left, right, up, or down respectively.

The robot can only move inside the rectangle; otherwise, it will change the script $s$ as follows:

  * If it tries to move outside a vertical border, it changes all L characters to R's (and vice versa, all R's to L's).    * If it tries to move outside a horizontal border, it changes all U characters to D's (and vice versa, all D's to U's).

Then, it will execute the changed script starting from the character which it couldn't execute.

![](CDN_BASE_URL/ff5ae9758c965c2d8398c936e9581dab) An example of the robot's movement process, $s = \texttt{"ULULURD"}$

The script $s$ will be executed for $k$ times continuously. All changes to the string $s$ will be retained even when it is repeated. During this process, how many times will the robot move to the point $(0, 0)$ in total? Note that the initial position does NOT count.

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

The first line of each test case contains four integers $n$, $k$, $w$, and $h$ ($1 \le n, w, h \le 10^6$; $1 \le k \le n$).

The second line contains a single string $s$ of size $n$ ($s_i \in \\{\texttt{L}, \texttt{R}, \texttt{U}, \texttt{D}\\}$) — the script to be executed.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print a single integer — the number of times the robot reaches $(0, 0)$ when executing script $s$ for $k$ times continuously.

Examples:
input:
5
2 2 2 2
UR
4 2 1 1
LLDD
6 3 3 1
RLRRRL
5 5 3 3
RUURD
7 5 3 4
RRDLUUU
output:
0
4
3
0
1

Note: In the first test case, the robot only moves up and right. In the end, it occupies the position $(2, 2)$ but never visits $(0, 0)$. So the answer is $0$.

In the second test case, each time executing the script the robot visits the origin twice. And since $k=2$, it visits the origin $2 \cdot 2 = 4$ times overall.

![](CDN_BASE_URL/ea799abf84f8a212d5a0d782be4d5e9c)

In the third test case, the visualization is shown as below:

![](CDN_BASE_URL/c6a5be14982b3d57a5103783598d4683)

Buggy Code:
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class F {

    static Map<Character, Character> hmH = Map.of( 'R', 'L', 'L', 'R');
    static Map<Character, Character> hmV = Map.of('U', 'D', 'D', 'U');

    public static void rev(char[] path, boolean v) {
        if (v) {
            for (int i = 0; i < path.length; i++) {
                if (path[i] == 'L' || path[i] == 'R')
                    path[i] = hmH.get(path[i]);
            }
        } else {
            for (int i = 0; i < path.length; i++) {
                if (path[i] == 'U' || path[i] == 'D')
                    path[i] = hmV.get(path[i]);
            }
        }
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t != 0) {
            t--;
            int n = sc.nextInt();
            int k = sc.nextInt();
            int w = sc.nextInt();
            int h = sc.nextInt();
            sc.nextLine();
            char[] path = sc.nextLine().toCharArray();
            int ans = 0;
            int x = 0;
            int y = 0;
            while (k != 0) {
                k--;
                for (int i = 0; i < path.length; i++) {
                    if (path[i] == 'U') {
                        if (y + 1 > h) {
                            rev(path, false);
                            i--;
                            continue;
                        }
                        y++;
                    }
                    if (path[i] == 'D') {
                        if (y - 1 < 0) {
                            rev(path, false);
                            i--;
                            continue;
                        }
                        y--;
                    }
                    if (path[i] == 'R') {
                        if (x + 1 > w) {
                            rev(path, true);
                            i--;
                            continue;
                        }
                        x++;
                    }
                    if (path[i] == 'L') {
                        if (x - 1 < 0) {
                            rev(path, true);
                            i--;
                            continue;
                        }
                        x--;
                    }
                    if (x == 0 && y == 0) ans++;
                }
            }
            System.out.println(ans);
        }
    }
}

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

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

Think step by step.

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

1. **Boundary Conditions**: The robot should be pushed to the boundaries of the rectangle to test the logic that reverses the commands.
2. **Repetitive Commands**: Commands that cause the robot to repeatedly hit the same boundary can expose issues with the reversal logic.
3. **Complex Paths**: Paths that involve multiple boundary crossings can help in identifying if the reversal is applied correctly and consistently.

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

```plaintext
3
1 1 1 1
U
2 2 1 1
UD
4 2 2 2
RDLU
```

### Explanation:
1. **Test Case 1**:
   - **n = 1, k = 1, w = 1, h = 1, s = "U"**
   - The robot starts at (0, 0) and tries to move up to (0, 1), which is outside the boundary. This should trigger the reversal of 'U' to 'D'.
   - The robot then moves down to (0, 0), which is the origin. The answer should be 1.

2. **Test Case 2**:
   - **n = 2, k = 2, w = 1, h = 1, s = "UD"**
   - The robot starts at (0, 0) and moves up to (0, 1), which is outside the boundary. This should trigger the reversal of 'U' to 'D'.
   - The robot then moves down to (0, 0), which is the origin.
   - The script "UD" is repeated, and the same sequence happens again. The answer should be 2 * 2 = 4.

3. **Test Case 3**:
   - **n = 4, k = 2, w = 2, h = 2, s = "RDLU"**
   - The robot starts at (0, 0) and moves right to (1, 0), down to (1, 1), left to (0, 1), and up to (0, 0), which is the origin. The answer should be 1 for the first iteration.
   - The script "RDLU" is repeated, and the same sequence happens again. The answer should be 2 * 1 = 2.

This test input should help in identifying the bug in the reversal logic and the handling of boundary conditions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Contest Proposal
time_limit_ms: 1000
memory_limit_mb: 256
Description: A contest contains $n$ problems and the difficulty of the $i$-th problem is expected to be at most $b_i$. There are already $n$ problem proposals and the difficulty of the $i$-th problem is $a_i$. Initially, both $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ are sorted in non-decreasing order.

Some of the problems may be more difficult than expected, so the writers must propose more problems. When a new problem with difficulty $w$ is proposed, the most difficult problem will be deleted from the contest, and the problems will be sorted in a way that the difficulties are non-decreasing.

In other words, in each operation, you choose an integer $w$, insert it into the array $a$, sort array $a$ in non-decreasing order, and remove the last element from it.

Find the minimum number of new problems to make $a_i\le b_i$ for all $i$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 100$). The description of the test cases follows.

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of problems.

The second line of each test case contains an array $a$ of length $n$ ($1\le a_1\le a_2\le\cdots\le a_n\le 10^9$).

The third line of each test case contains an array $b$ of length $n$ ($1\le b_1\le b_2\le\cdots\le b_n\le 10^9$).

Output: For each test case, print an integer as your answer in a new line.

Examples:
input:
2
6
1000 1400 2000 2000 2200 2700
800 1200 1500 1800 2200 3000
6
4 5 6 7 8 9
1 2 3 4 5 6
output:
2
3

Note: In the first test case:

  * Propose a problem with difficulty $w=800$ and $a$ becomes $[800,1000,1400,2000,2000,2200]$.    * Propose a problem with difficulty $w=1800$ and $a$ becomes $[800,1000,1400,1800,2000,2000]$.

It can be proved that it's impossible to reach the goal by proposing fewer new problems.

In the second test case:

  * Propose a problem with difficulty $w=1$ and $a$ becomes $[1,4,5,6,7,8]$.    * Propose a problem with difficulty $w=2$ and $a$ becomes $[1,2,4,5,6,7]$.    * Propose a problem with difficulty $w=3$ and $a$ becomes $[1,2,3,4,5,6]$.

It can be proved that it's impossible to reach the goal by proposing fewer new problems.

Buggy Code:
#include <cmath>
#include <functional>
#include <fstream>
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <list>
#include <time.h>
#include <math.h>
#include <random>
#include <deque>
#include <queue>
#include <cassert>
#include <unordered_map>
#include <unordered_set>
#include <iomanip>
#include <bitset>
#include <sstream>
#include <chrono>
#include <cstring>

using namespace std;
using ll = long long;

void solve(){
	int n; cin >> n;
	vector<int> a(n), b(n);
	for(int i = 0; i < n; i++) cin >> a[i];
	for(int i = 0; i < n; i++) cin >> b[i];
	int ans = 0;
	for(int i = 0; i < n; i++){
		for(int j = 0; j < n; j++){
			if(a[i] <= b[j]){
				ans = max(ans, abs(j - i));
				break;
			}
		}
	}
	cout << ans << "\n";
}

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int t;
	cin >> t;
	while(t--){
		solve();
	}
}

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

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

Think step by step.

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

1. **Edge Cases**: Test cases where the arrays `a` and `b` have specific values that can cause the algorithm to fail.
2. **Small Scale**: Ensure the test case is small enough to fit within the output limit and is easy to verify.
3. **Failure Scenarios**: Focus on scenarios where the current logic in the buggy code fails to produce the correct result.

### Analysis of the Buggy Code
The buggy code iterates through the arrays `a` and `b` and calculates the maximum distance `abs(j - i)` where `a[i] <= b[j]`. This logic is flawed because it does not correctly handle the insertion and sorting of new elements to ensure `a[i] <= b[i]` for all `i`.

### Challenging Test Input
Let's create a test case where the current logic fails to correctly determine the minimum number of new problems needed. Consider a case where the initial values of `a` and `b` are such that the current logic will not correctly identify the required number of insertions.

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

### Explanation
- **Initial Arrays**:
  - `a = [1, 2, 3, 4, 5]`
  - `b = [2, 3, 4, 5, 6]`

- **Expected Output**:
  - The correct answer is `1` because we need to propose one new problem with difficulty `1` to ensure `a[0] <= b[0]`.

- **Buggy Code Output**:
  - The current logic in the buggy code will incorrectly determine the number of new problems. It will not correctly handle the insertion and sorting of new elements to ensure `a[i] <= b[i]` for all `i`.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Contest Proposal
time_limit_ms: 1000
memory_limit_mb: 256
Description: A contest contains $n$ problems and the difficulty of the $i$-th problem is expected to be at most $b_i$. There are already $n$ problem proposals and the difficulty of the $i$-th problem is $a_i$. Initially, both $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ are sorted in non-decreasing order.

Some of the problems may be more difficult than expected, so the writers must propose more problems. When a new problem with difficulty $w$ is proposed, the most difficult problem will be deleted from the contest, and the problems will be sorted in a way that the difficulties are non-decreasing.

In other words, in each operation, you choose an integer $w$, insert it into the array $a$, sort array $a$ in non-decreasing order, and remove the last element from it.

Find the minimum number of new problems to make $a_i\le b_i$ for all $i$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 100$). The description of the test cases follows.

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of problems.

The second line of each test case contains an array $a$ of length $n$ ($1\le a_1\le a_2\le\cdots\le a_n\le 10^9$).

The third line of each test case contains an array $b$ of length $n$ ($1\le b_1\le b_2\le\cdots\le b_n\le 10^9$).

Output: For each test case, print an integer as your answer in a new line.

Examples:
input:
2
6
1000 1400 2000 2000 2200 2700
800 1200 1500 1800 2200 3000
6
4 5 6 7 8 9
1 2 3 4 5 6
output:
2
3

Note: In the first test case:

  * Propose a problem with difficulty $w=800$ and $a$ becomes $[800,1000,1400,2000,2000,2200]$.    * Propose a problem with difficulty $w=1800$ and $a$ becomes $[800,1000,1400,1800,2000,2000]$.

It can be proved that it's impossible to reach the goal by proposing fewer new problems.

In the second test case:

  * Propose a problem with difficulty $w=1$ and $a$ becomes $[1,4,5,6,7,8]$.    * Propose a problem with difficulty $w=2$ and $a$ becomes $[1,2,4,5,6,7]$.    * Propose a problem with difficulty $w=3$ and $a$ becomes $[1,2,3,4,5,6]$.

It can be proved that it's impossible to reach the goal by proposing fewer new problems.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main(){
    int testcase;
    cin>>testcase;
    vector<int>a,b;
    int size;
    int pairs;
    for(int i=0;i<testcase;i++){
        cin>>size;
        a.resize(size);
        b.resize(size);
        pairs=0;
        for(int j=0;j<size;j++){
            cin>>a[j];
        }
        for(int j=0;j<size;j++){
            cin>>b[j];
        }
        for(int j=0;j<size;j++){
            if(a[j]>b[j]){
                pairs++;
            }
        }
        if(pairs % 2 == 0){
            cout<<(pairs/2)<<endl;
        }
        else{
            cout<<((pairs/2)+1)<<endl;
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
5
1 2 3 4 5
1 2 3 4 6
5
1 2 3 4 5
1 2 3 5 5
6
1 2 3 4 5 6
1 2 3 4 5 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Contest Proposal
time_limit_ms: 1000
memory_limit_mb: 256
Description: A contest contains $n$ problems and the difficulty of the $i$-th problem is expected to be at most $b_i$. There are already $n$ problem proposals and the difficulty of the $i$-th problem is $a_i$. Initially, both $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ are sorted in non-decreasing order.

Some of the problems may be more difficult than expected, so the writers must propose more problems. When a new problem with difficulty $w$ is proposed, the most difficult problem will be deleted from the contest, and the problems will be sorted in a way that the difficulties are non-decreasing.

In other words, in each operation, you choose an integer $w$, insert it into the array $a$, sort array $a$ in non-decreasing order, and remove the last element from it.

Find the minimum number of new problems to make $a_i\le b_i$ for all $i$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 100$). The description of the test cases follows.

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of problems.

The second line of each test case contains an array $a$ of length $n$ ($1\le a_1\le a_2\le\cdots\le a_n\le 10^9$).

The third line of each test case contains an array $b$ of length $n$ ($1\le b_1\le b_2\le\cdots\le b_n\le 10^9$).

Output: For each test case, print an integer as your answer in a new line.

Examples:
input:
2
6
1000 1400 2000 2000 2200 2700
800 1200 1500 1800 2200 3000
6
4 5 6 7 8 9
1 2 3 4 5 6
output:
2
3

Note: In the first test case:

  * Propose a problem with difficulty $w=800$ and $a$ becomes $[800,1000,1400,2000,2000,2200]$.    * Propose a problem with difficulty $w=1800$ and $a$ becomes $[800,1000,1400,1800,2000,2000]$.

It can be proved that it's impossible to reach the goal by proposing fewer new problems.

In the second test case:

  * Propose a problem with difficulty $w=1$ and $a$ becomes $[1,4,5,6,7,8]$.    * Propose a problem with difficulty $w=2$ and $a$ becomes $[1,2,4,5,6,7]$.    * Propose a problem with difficulty $w=3$ and $a$ becomes $[1,2,3,4,5,6]$.

It can be proved that it's impossible to reach the goal by proposing fewer new problems.

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

#define ll long long
#define int long long

#define pb push_back
#define endl '\n'
#define all(x) x.begin(), x.end()

const int mod = 1e9 + 7;
const int inf = 1e18;
const int N = 2e5 + 5;

int binpow(int a, int n) {
    if (n == 0)
        return 1;
    if (n % 2 == 1)
        return binpow(a, n - 1) * a;
    else {
        int b = binpow(a, n / 2);
        return b * b;
    }
}

void zhanIsa() {
    int n;
    cin >> n;

    vector<int> a(n), b(n);

    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }

    for (int i = 0; i < n; i++) {
        cin >> b[i];
    }

    int l = 0, r = 0;

    while (l < n and r < n) {
        if (a[l] > b[r]) {
            r++;
        }
        l++;
        r++;
    }

    cout << r - l << endl;

}

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

	int tt = 1;
    cin >> tt;

	while (tt--) {
		zhanIsa();
	}
}

// being into me

**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 code and identify potential issues:

### Analysis of the Buggy Code

The buggy code attempts to find the minimum number of new problems needed to make the array `a` such that `a[i] <= b[i]` for all `i`. However, the logic in the `while` loop is flawed. Specifically, the loop increments both `l` and `r` in every iteration, which is incorrect. The correct approach should be to only increment `r` when `a[l] > b[r]` and only increment `l` when `a[l] <= b[r]`.

### Challenging Test Input

To expose this bug, we need a test case where the incorrect logic in the loop will lead to an incorrect result. A good candidate is a case where the arrays `a` and `b` are such that `a[i]` is greater than `b[i]` for multiple indices, but the faulty logic will incorrectly count the number of new problems needed.

### Test Case

Let's consider the following test case:

```plaintext
1
4
1000 2000 3000 4000
1000 2000 3000 3000
```

### Explanation

- **Input:**
  - `n = 4`
  - `a = [1000, 2000, 3000, 4000]`
  - `b = [1000, 2000, 3000, 3000]`

- **Expected Output:**
  - The correct number of new problems needed is 1. By proposing a problem with difficulty 3000, the array `a` becomes `[1000, 2000, 3000, 3000]` after removing the last element, which satisfies the condition `a[i] <= b[i]` for all `i`.

- **Buggy Code Output:**
  - The faulty logic in the loop will incorrectly count the number of new problems needed. It will likely output a different (and incorrect) number.

### Test Input

```plaintext
1
4
1000 2000 3000 4000
1000 2000 3000 3000
```

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Contest Proposal
time_limit_ms: 1000
memory_limit_mb: 256
Description: A contest contains $n$ problems and the difficulty of the $i$-th problem is expected to be at most $b_i$. There are already $n$ problem proposals and the difficulty of the $i$-th problem is $a_i$. Initially, both $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ are sorted in non-decreasing order.

Some of the problems may be more difficult than expected, so the writers must propose more problems. When a new problem with difficulty $w$ is proposed, the most difficult problem will be deleted from the contest, and the problems will be sorted in a way that the difficulties are non-decreasing.

In other words, in each operation, you choose an integer $w$, insert it into the array $a$, sort array $a$ in non-decreasing order, and remove the last element from it.

Find the minimum number of new problems to make $a_i\le b_i$ for all $i$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 100$). The description of the test cases follows.

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of problems.

The second line of each test case contains an array $a$ of length $n$ ($1\le a_1\le a_2\le\cdots\le a_n\le 10^9$).

The third line of each test case contains an array $b$ of length $n$ ($1\le b_1\le b_2\le\cdots\le b_n\le 10^9$).

Output: For each test case, print an integer as your answer in a new line.

Examples:
input:
2
6
1000 1400 2000 2000 2200 2700
800 1200 1500 1800 2200 3000
6
4 5 6 7 8 9
1 2 3 4 5 6
output:
2
3

Note: In the first test case:

  * Propose a problem with difficulty $w=800$ and $a$ becomes $[800,1000,1400,2000,2000,2200]$.    * Propose a problem with difficulty $w=1800$ and $a$ becomes $[800,1000,1400,1800,2000,2000]$.

It can be proved that it's impossible to reach the goal by proposing fewer new problems.

In the second test case:

  * Propose a problem with difficulty $w=1$ and $a$ becomes $[1,4,5,6,7,8]$.    * Propose a problem with difficulty $w=2$ and $a$ becomes $[1,2,4,5,6,7]$.    * Propose a problem with difficulty $w=3$ and $a$ becomes $[1,2,3,4,5,6]$.

It can be proved that it's impossible to reach the goal by proposing fewer new problems.

Buggy Code:
t=int(input())
for i in range(t):
    n=int(input())
    a=list(map(int,input().split()))
    b=list(map(int,input().split()))
    c=0
    i=0
    j=0
    while j<n-1:
        if a[i]>b[j]:
            c+=1
            j+=1
        else:
            i+=1
            j+=1
    print(c)

**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 where the code might fail. The buggy code has a logical error in the loop that checks if `a[i] > b[j]`. It increments `j` even when `a[i]` is greater than `b[j]`, which can lead to incorrect results.

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

1. **Create a scenario where `a[i]` is greater than `b[j]` for multiple consecutive elements**:
   - This will ensure that the loop incorrectly increments `j` multiple times without properly handling the insertion of new problems.

2. **Ensure that the arrays `a` and `b` are sorted**:
   - This is a requirement of the problem and ensures the input is valid.

3. **Use a small scale to keep the input manageable**:
   - We will use a small number of test cases and small values for `n`.

Given these considerations, here is a challenging test input:

```plaintext
2
4
100 200 300 400
100 200 300 400
6
1 2 3 4 5 1000
1 2 3 4 5 6
```

### Explanation:
1. **Test Case 1**:
   - `n = 4`
   - `a = [100, 200, 300, 400]`
   - `b = [100, 200, 300, 400]`
   - This is a simple case where no new problems are needed, and the expected output is `0`.

2. **Test Case 2**:
   - `n = 6`
   - `a = [1, 2, 3, 4, 5, 1000]`
   - `b = [1, 2, 3, 4, 5, 6]`
   - In this case, the last element of `a` (1000) is much greater than the last element of `b` (6). The correct solution would require inserting 5 new problems with difficulties 1, 2, 3, 4, and 5 to make `a` match `b`. The expected output is `5`.

This input will expose the bug in the buggy code because it will incorrectly handle the large discrepancy between the last elements of `a` and `b`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Contest Proposal
time_limit_ms: 1000
memory_limit_mb: 256
Description: A contest contains $n$ problems and the difficulty of the $i$-th problem is expected to be at most $b_i$. There are already $n$ problem proposals and the difficulty of the $i$-th problem is $a_i$. Initially, both $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ are sorted in non-decreasing order.

Some of the problems may be more difficult than expected, so the writers must propose more problems. When a new problem with difficulty $w$ is proposed, the most difficult problem will be deleted from the contest, and the problems will be sorted in a way that the difficulties are non-decreasing.

In other words, in each operation, you choose an integer $w$, insert it into the array $a$, sort array $a$ in non-decreasing order, and remove the last element from it.

Find the minimum number of new problems to make $a_i\le b_i$ for all $i$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 100$). The description of the test cases follows.

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of problems.

The second line of each test case contains an array $a$ of length $n$ ($1\le a_1\le a_2\le\cdots\le a_n\le 10^9$).

The third line of each test case contains an array $b$ of length $n$ ($1\le b_1\le b_2\le\cdots\le b_n\le 10^9$).

Output: For each test case, print an integer as your answer in a new line.

Examples:
input:
2
6
1000 1400 2000 2000 2200 2700
800 1200 1500 1800 2200 3000
6
4 5 6 7 8 9
1 2 3 4 5 6
output:
2
3

Note: In the first test case:

  * Propose a problem with difficulty $w=800$ and $a$ becomes $[800,1000,1400,2000,2000,2200]$.    * Propose a problem with difficulty $w=1800$ and $a$ becomes $[800,1000,1400,1800,2000,2000]$.

It can be proved that it's impossible to reach the goal by proposing fewer new problems.

In the second test case:

  * Propose a problem with difficulty $w=1$ and $a$ becomes $[1,4,5,6,7,8]$.    * Propose a problem with difficulty $w=2$ and $a$ becomes $[1,2,4,5,6,7]$.    * Propose a problem with difficulty $w=3$ and $a$ becomes $[1,2,3,4,5,6]$.

It can be proved that it's impossible to reach the goal by proposing fewer new problems.

Buggy Code:
for _ in range(int(input())):
  n = int(input())
  A, B = (list(map(int, input().split())) for _ in range(2))
  ia = ib = n-1
  while ia > 0:
    if A[ia] <= B[ib]:
      ib -= 1
    ia -= 1
  print(ib-ia)

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Contest Proposal
time_limit_ms: 1000
memory_limit_mb: 256
Description: A contest contains $n$ problems and the difficulty of the $i$-th problem is expected to be at most $b_i$. There are already $n$ problem proposals and the difficulty of the $i$-th problem is $a_i$. Initially, both $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ are sorted in non-decreasing order.

Some of the problems may be more difficult than expected, so the writers must propose more problems. When a new problem with difficulty $w$ is proposed, the most difficult problem will be deleted from the contest, and the problems will be sorted in a way that the difficulties are non-decreasing.

In other words, in each operation, you choose an integer $w$, insert it into the array $a$, sort array $a$ in non-decreasing order, and remove the last element from it.

Find the minimum number of new problems to make $a_i\le b_i$ for all $i$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 100$). The description of the test cases follows.

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of problems.

The second line of each test case contains an array $a$ of length $n$ ($1\le a_1\le a_2\le\cdots\le a_n\le 10^9$).

The third line of each test case contains an array $b$ of length $n$ ($1\le b_1\le b_2\le\cdots\le b_n\le 10^9$).

Output: For each test case, print an integer as your answer in a new line.

Examples:
input:
2
6
1000 1400 2000 2000 2200 2700
800 1200 1500 1800 2200 3000
6
4 5 6 7 8 9
1 2 3 4 5 6
output:
2
3

Note: In the first test case:

  * Propose a problem with difficulty $w=800$ and $a$ becomes $[800,1000,1400,2000,2000,2200]$.    * Propose a problem with difficulty $w=1800$ and $a$ becomes $[800,1000,1400,1800,2000,2000]$.

It can be proved that it's impossible to reach the goal by proposing fewer new problems.

In the second test case:

  * Propose a problem with difficulty $w=1$ and $a$ becomes $[1,4,5,6,7,8]$.    * Propose a problem with difficulty $w=2$ and $a$ becomes $[1,2,4,5,6,7]$.    * Propose a problem with difficulty $w=3$ and $a$ becomes $[1,2,3,4,5,6]$.

It can be proved that it's impossible to reach the goal by proposing fewer new problems.

Buggy Code:
test=int(input())
while(test!=0):
    cases=int(input())
    a1=input()
    b1=input()
    a_string_list=a1.split()
    b_string_list=b1.split()
    a=list(map(int,a_string_list))
    b=list(map(int,b_string_list))
    x=0
    y=0
    while(True):
        if(x==len(b)):
            break
        while(a[y]>b[x]):
            x=x+1
        x=x+1
        y=y+1
    print(len(a)-y)

    test=test-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 consider the following steps:

1. **Understand the Bug**: The bug in the code seems to be in the logic where it tries to count the number of new problems needed. Specifically, the inner while loop is incorrectly incrementing the indices without properly handling the insertion and sorting of new problems.

2. **Edge Cases**: We need to create a test case where the array `a` has elements that are significantly higher than the corresponding elements in `b`, requiring multiple new problems to be inserted. This will help in exposing the bug.

3. **Small-Scale Test Case**: Since the problem specifies that the test case should be small-scale, we will create a test case with a small number of elements.

4. **Test Case Construction**:
   - Choose a small number of test cases, say `t = 1`.
   - Choose a small number of problems, say `n = 5`.
   - Ensure that the array `a` has elements that are significantly higher than the corresponding elements in `b`.

Given these considerations, here is a challenging test input:

```plaintext
1
5
1000 2000 3000 4000 5000
100 200 300 400 500
```

This test case will require a significant number of new problems to be inserted to make `a_i <= b_i` for all `i`. The expected output for this test case is `5`, as each element in `a` is much larger than the corresponding element in `b`, and we need to insert 5 new problems with difficulties 100, 200, 300, 400, and 500 to make the arrays match.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Contest Proposal
time_limit_ms: 1000
memory_limit_mb: 256
Description: A contest contains $n$ problems and the difficulty of the $i$-th problem is expected to be at most $b_i$. There are already $n$ problem proposals and the difficulty of the $i$-th problem is $a_i$. Initially, both $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ are sorted in non-decreasing order.

Some of the problems may be more difficult than expected, so the writers must propose more problems. When a new problem with difficulty $w$ is proposed, the most difficult problem will be deleted from the contest, and the problems will be sorted in a way that the difficulties are non-decreasing.

In other words, in each operation, you choose an integer $w$, insert it into the array $a$, sort array $a$ in non-decreasing order, and remove the last element from it.

Find the minimum number of new problems to make $a_i\le b_i$ for all $i$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 100$). The description of the test cases follows.

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of problems.

The second line of each test case contains an array $a$ of length $n$ ($1\le a_1\le a_2\le\cdots\le a_n\le 10^9$).

The third line of each test case contains an array $b$ of length $n$ ($1\le b_1\le b_2\le\cdots\le b_n\le 10^9$).

Output: For each test case, print an integer as your answer in a new line.

Examples:
input:
2
6
1000 1400 2000 2000 2200 2700
800 1200 1500 1800 2200 3000
6
4 5 6 7 8 9
1 2 3 4 5 6
output:
2
3

Note: In the first test case:

  * Propose a problem with difficulty $w=800$ and $a$ becomes $[800,1000,1400,2000,2000,2200]$.    * Propose a problem with difficulty $w=1800$ and $a$ becomes $[800,1000,1400,1800,2000,2000]$.

It can be proved that it's impossible to reach the goal by proposing fewer new problems.

In the second test case:

  * Propose a problem with difficulty $w=1$ and $a$ becomes $[1,4,5,6,7,8]$.    * Propose a problem with difficulty $w=2$ and $a$ becomes $[1,2,4,5,6,7]$.    * Propose a problem with difficulty $w=3$ and $a$ becomes $[1,2,3,4,5,6]$.

It can be proved that it's impossible to reach the goal by proposing fewer new problems.

Buggy Code:
import java.util.*;
public class Main {
   	public static void main(String[] args) {
	    Scanner in = new Scanner(System.in);
	    byte t,n,s;
	    t=in.nextByte();
	    int [][] a;

	    while(t-->0){
            n=in.nextByte();
            s=0;
            a=new int[2][n];
            for(int i=0;i<2;i++)
            for (int j=0;j<n;j++){
                a[i][j]=in.nextInt();
            }
            for(int i=0,j=0;i<n-s-1;)
                if (a[0][i]<=a[1][j]){
                    i++;
                    j++;
                }
                else{
                    s++;
                    j++;
                }
            System.out.println(s);

	    }

 	}
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Contest Proposal
time_limit_ms: 1000
memory_limit_mb: 256
Description: A contest contains $n$ problems and the difficulty of the $i$-th problem is expected to be at most $b_i$. There are already $n$ problem proposals and the difficulty of the $i$-th problem is $a_i$. Initially, both $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ are sorted in non-decreasing order.

Some of the problems may be more difficult than expected, so the writers must propose more problems. When a new problem with difficulty $w$ is proposed, the most difficult problem will be deleted from the contest, and the problems will be sorted in a way that the difficulties are non-decreasing.

In other words, in each operation, you choose an integer $w$, insert it into the array $a$, sort array $a$ in non-decreasing order, and remove the last element from it.

Find the minimum number of new problems to make $a_i\le b_i$ for all $i$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 100$). The description of the test cases follows.

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of problems.

The second line of each test case contains an array $a$ of length $n$ ($1\le a_1\le a_2\le\cdots\le a_n\le 10^9$).

The third line of each test case contains an array $b$ of length $n$ ($1\le b_1\le b_2\le\cdots\le b_n\le 10^9$).

Output: For each test case, print an integer as your answer in a new line.

Examples:
input:
2
6
1000 1400 2000 2000 2200 2700
800 1200 1500 1800 2200 3000
6
4 5 6 7 8 9
1 2 3 4 5 6
output:
2
3

Note: In the first test case:

  * Propose a problem with difficulty $w=800$ and $a$ becomes $[800,1000,1400,2000,2000,2200]$.    * Propose a problem with difficulty $w=1800$ and $a$ becomes $[800,1000,1400,1800,2000,2000]$.

It can be proved that it's impossible to reach the goal by proposing fewer new problems.

In the second test case:

  * Propose a problem with difficulty $w=1$ and $a$ becomes $[1,4,5,6,7,8]$.    * Propose a problem with difficulty $w=2$ and $a$ becomes $[1,2,4,5,6,7]$.    * Propose a problem with difficulty $w=3$ and $a$ becomes $[1,2,3,4,5,6]$.

It can be proved that it's impossible to reach the goal by proposing fewer new problems.

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

public class A_Contest_Proposal {
    public static void main(String[] args) {
        try {
            FastReader in=new FastReader();
            FastWriter out=new FastWriter();
            int testcases=in.nextInt();
            while(testcases-- >0){
                int n=in.nextInt();
                int a[]=new int[n];
                int b[]=new int[n];
                for(int i=0;i<n;i++){
                    a[i]=in.nextInt();
                }
                for(int i=0;i<n;i++){
                    b[i]=in.nextInt();
                }
                Arrays.sort(a);
                Arrays.sort(b);
                int x=0;
                boolean flag=false;
                for(int i=0;i<n;i++){
                    flag=false;
                    for(int j=i+1;j<n;j++){
                        if(a[i]<=b[j]){
                            x=Math.max(j-i,x);
                            flag=true;
                            break;
                        }
                    }
                    if(!flag){
                        x=Math.max(n-i,x);
                    }
                }
                out.println(x);

            }

            out.close();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
    static class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br=new BufferedReader(new InputStreamReader(System.in));
        }
        String next(){
            while(st==null || !st.hasMoreTokens()){
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt(){
            return Integer.parseInt(next());
        }
        long nextLong(){
            return Long.parseLong(next());
        }
        double nextDouble(){
            return Double.parseDouble(next());
        }
        char nextChar() {
            return next().charAt(0);
        }
        String nextLine(){
            String str="";
            try {
                str=br.readLine().trim();
            } catch (Exception e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    static class FastWriter {
        private final BufferedWriter bw;

        public FastWriter() {
            this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
        }

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

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

        public void close() throws IOException {
            bw.close();
        }
    }

//    static return_type function_name(input parameters) { ... }

    static int add(int a, int b) {
        return a+b;
    }

}

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

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

Think step by step.

Response:
```plaintext
3
4
1 2 3 4
1 2 3 4
5
1 2 3 4 5
2 3 4 5 6
6
1 2 3 4 5 6
2 3 4 5 7 8
```

### Explanation:
1. **Test Case 1:**
   - `n = 4`
   - `a = [1, 2, 3, 4]`
   - `b = [1, 2, 3, 4]`
   - The arrays are already in the required order, so no new problems are needed. The expected output is `0`.

2. **Test Case 2:**
   - `n = 5`
   - `a = [1, 2, 3, 4, 5]`
   - `b = [2, 3, 4, 5, 6]`
   - The first element of `a` is less than the first element of `b`, but the rest are in order. One new problem with difficulty `2` is needed to make `a` meet the requirements. The expected output is `1`.

3. **Test Case 3:**
   - `n = 6`
   - `a = [1, 2, 3, 4, 5, 6]`
   - `b = [2, 3, 4, 5, 7, 8]`
   - The first element of `a` is less than the first element of `b`, and the last two elements of `a` are less than the last two elements of `b`. Two new problems with difficulties `2` and `7` are needed to make `a` meet the requirements. The expected output is `2`.

These test cases are designed to cover different scenarios, including the case where no new problems are needed, the case where a single new problem is needed, and the case where multiple new problems are needed. This will help expose any 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: Contest Proposal
time_limit_ms: 1000
memory_limit_mb: 256
Description: A contest contains $n$ problems and the difficulty of the $i$-th problem is expected to be at most $b_i$. There are already $n$ problem proposals and the difficulty of the $i$-th problem is $a_i$. Initially, both $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ are sorted in non-decreasing order.

Some of the problems may be more difficult than expected, so the writers must propose more problems. When a new problem with difficulty $w$ is proposed, the most difficult problem will be deleted from the contest, and the problems will be sorted in a way that the difficulties are non-decreasing.

In other words, in each operation, you choose an integer $w$, insert it into the array $a$, sort array $a$ in non-decreasing order, and remove the last element from it.

Find the minimum number of new problems to make $a_i\le b_i$ for all $i$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 100$). The description of the test cases follows.

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of problems.

The second line of each test case contains an array $a$ of length $n$ ($1\le a_1\le a_2\le\cdots\le a_n\le 10^9$).

The third line of each test case contains an array $b$ of length $n$ ($1\le b_1\le b_2\le\cdots\le b_n\le 10^9$).

Output: For each test case, print an integer as your answer in a new line.

Examples:
input:
2
6
1000 1400 2000 2000 2200 2700
800 1200 1500 1800 2200 3000
6
4 5 6 7 8 9
1 2 3 4 5 6
output:
2
3

Note: In the first test case:

  * Propose a problem with difficulty $w=800$ and $a$ becomes $[800,1000,1400,2000,2000,2200]$.    * Propose a problem with difficulty $w=1800$ and $a$ becomes $[800,1000,1400,1800,2000,2000]$.

It can be proved that it's impossible to reach the goal by proposing fewer new problems.

In the second test case:

  * Propose a problem with difficulty $w=1$ and $a$ becomes $[1,4,5,6,7,8]$.    * Propose a problem with difficulty $w=2$ and $a$ becomes $[1,2,4,5,6,7]$.    * Propose a problem with difficulty $w=3$ and $a$ becomes $[1,2,3,4,5,6]$.

It can be proved that it's impossible to reach the goal by proposing fewer new problems.

Buggy Code:
#include<bits/stdc++.h>
#define _for(i,x,y) for(int i=x;i<=y;++i)
#define _forp(i,x,y,z) for(int i=x;i<=y;i+=z)
#define _rep(i,x,y) for(int i=x;i<y;++i)
using namespace std;
int T,n,a[105],b[105];
int main(){
//	freopen("test.in","r",stdin);
//	freopen("test.out","w",stdout);
    ios::sync_with_stdio(false);
    cin.tie(0);
	cin>>T;
	while(T--){
		cin>>n;
		_for(i,1,n){
			cin>>a[i];
		}
		int cnt=0;
		_for(i,1,n) cin>>b[i],cnt+=(a[i]>b[i]);
		cout<<(int)(ceil(cnt*1.0/2))<<'\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 provided code, we need to focus on scenarios where the code's logic fails. The code incorrectly assumes that the number of new problems needed is `ceil(cnt / 2)`, where `cnt` is the number of problems where `a[i] > b[i]`. This assumption is flawed because it does not consider the actual difficulties and the need to maintain the non-decreasing order of the array.

### Key Considerations:
1. **Edge Cases**: Scenarios where the arrays `a` and `b` are close to each other but still require adjustments.
2. **Large Discrepancies**: Scenarios where there are significant differences between `a` and `b` that need multiple adjustments.
3. **Non-trivial Sorting**: Scenarios where the insertion and sorting of new problems are necessary to meet the conditions.

### Test Input:
Let's create a test case where the initial arrays `a` and `b` are such that the number of new problems required is more than `ceil(cnt / 2)`.

```plaintext
1
5
1 2 3 4 5
1 2 3 10 10
```

### Explanation:
- **Initial Arrays**:
  - `a = [1, 2, 3, 4, 5]`
  - `b = [1, 2, 3, 10, 10]`
- **Initial Comparison**:
  - `a[4] > b[4]` (4 > 10) -> 1 discrepancy
  - `a[5] > b[5]` (5 > 10) -> 1 discrepancy
- **Total Discrepancies**: `cnt = 2`
- **Incorrect Output**:
  - The code will output `ceil(2 / 2) = 1`
- **Correct Output**:
  - To make `a` meet the conditions of `b`, we need to propose at least 2 new problems with difficulties 10 and 10.
  - After proposing the first problem with difficulty 10, `a` becomes `[1, 2, 3, 4, 10]` and `b` remains `[1, 2, 3, 10, 10]`.
  - After proposing the second problem with difficulty 10, `a` becomes `[1, 2, 3, 10, 10]` and `b` remains `[1, 2, 3, 10, 10]`.

This test case will expose the bug in the code because the code incorrectly assumes that one new problem is sufficient, while in reality, two new problems are needed.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Contest Proposal
time_limit_ms: 1000
memory_limit_mb: 256
Description: A contest contains $n$ problems and the difficulty of the $i$-th problem is expected to be at most $b_i$. There are already $n$ problem proposals and the difficulty of the $i$-th problem is $a_i$. Initially, both $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ are sorted in non-decreasing order.

Some of the problems may be more difficult than expected, so the writers must propose more problems. When a new problem with difficulty $w$ is proposed, the most difficult problem will be deleted from the contest, and the problems will be sorted in a way that the difficulties are non-decreasing.

In other words, in each operation, you choose an integer $w$, insert it into the array $a$, sort array $a$ in non-decreasing order, and remove the last element from it.

Find the minimum number of new problems to make $a_i\le b_i$ for all $i$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 100$). The description of the test cases follows.

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of problems.

The second line of each test case contains an array $a$ of length $n$ ($1\le a_1\le a_2\le\cdots\le a_n\le 10^9$).

The third line of each test case contains an array $b$ of length $n$ ($1\le b_1\le b_2\le\cdots\le b_n\le 10^9$).

Output: For each test case, print an integer as your answer in a new line.

Examples:
input:
2
6
1000 1400 2000 2000 2200 2700
800 1200 1500 1800 2200 3000
6
4 5 6 7 8 9
1 2 3 4 5 6
output:
2
3

Note: In the first test case:

  * Propose a problem with difficulty $w=800$ and $a$ becomes $[800,1000,1400,2000,2000,2200]$.    * Propose a problem with difficulty $w=1800$ and $a$ becomes $[800,1000,1400,1800,2000,2000]$.

It can be proved that it's impossible to reach the goal by proposing fewer new problems.

In the second test case:

  * Propose a problem with difficulty $w=1$ and $a$ becomes $[1,4,5,6,7,8]$.    * Propose a problem with difficulty $w=2$ and $a$ becomes $[1,2,4,5,6,7]$.    * Propose a problem with difficulty $w=3$ and $a$ becomes $[1,2,3,4,5,6]$.

It can be proved that it's impossible to reach the goal by proposing fewer new problems.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
int t, n;
int solve() {
    cin >> n;
    int A[n];
    int B[n];
    for (int i = 0; i < n; i++) cin >> A[i];
    for (int i = 0; i < n; i++) cin >> B[i];
    int displacement = 0;
    for (int i = 0; i < n; i++) {
        if (i + displacement < n) {
            if (A[i] > B[i+displacement]) displacement++;
        }
    }
    return displacement;
}
int main() {
    cin >> t;
    for (int i = 0; i < t; i++) cout << solve() << '\n';
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and identify potential issues:

### Analysis of the Buggy Code
1. **Displacement Calculation:**
   - The `displacement` variable is intended to track how many new problems need to be added.
   - The condition `if (A[i] > B[i+displacement])` is used to determine if the current problem `A[i]` is too difficult compared to the expected difficulty `B[i+displacement]`.

2. **Potential Issues:**
   - The code does not handle cases where `A[i]` is significantly greater than `B[i+displacement]` and requires multiple new problems to be added.
   - The code does not account for the sorting and removal of the most difficult problem after each insertion.

### Challenging Test Case
To expose the bug, we need a test case where the initial problem difficulties in `A` are significantly higher than the expected difficulties in `B`, and the code's logic fails to correctly calculate the number of new problems needed.

### Example Test Case
Consider a test case where the initial difficulties in `A` are much higher than the expected difficulties in `B`, and the difference is not linear. This will force the code to incorrectly calculate the number of new problems needed.

```plaintext
1
4
10000 20000 30000 40000
1 2 3 4
```

### Explanation
- **Initial Array `A`:** [10000, 20000, 30000, 40000]
- **Expected Array `B`:** [1, 2, 3, 4]

In this case:
- The first problem `A[0] = 10000` is much greater than `B[0] = 1`.
- The second problem `A[1] = 20000` is much greater than `B[1] = 2`.
- The third problem `A[2] = 30000` is much greater than `B[2] = 3`.
- The fourth problem `A[3] = 40000` is much greater than `B[3] = 4`.

The correct solution would require a significant number of new problems to be added and the most difficult problem to be removed after each insertion. The buggy code will likely fail to correctly calculate this.

### Output
```plaintext
1
4
10000 20000 30000 40000
1 2 3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Contest Proposal
time_limit_ms: 1000
memory_limit_mb: 256
Description: A contest contains $n$ problems and the difficulty of the $i$-th problem is expected to be at most $b_i$. There are already $n$ problem proposals and the difficulty of the $i$-th problem is $a_i$. Initially, both $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ are sorted in non-decreasing order.

Some of the problems may be more difficult than expected, so the writers must propose more problems. When a new problem with difficulty $w$ is proposed, the most difficult problem will be deleted from the contest, and the problems will be sorted in a way that the difficulties are non-decreasing.

In other words, in each operation, you choose an integer $w$, insert it into the array $a$, sort array $a$ in non-decreasing order, and remove the last element from it.

Find the minimum number of new problems to make $a_i\le b_i$ for all $i$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 100$). The description of the test cases follows.

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of problems.

The second line of each test case contains an array $a$ of length $n$ ($1\le a_1\le a_2\le\cdots\le a_n\le 10^9$).

The third line of each test case contains an array $b$ of length $n$ ($1\le b_1\le b_2\le\cdots\le b_n\le 10^9$).

Output: For each test case, print an integer as your answer in a new line.

Examples:
input:
2
6
1000 1400 2000 2000 2200 2700
800 1200 1500 1800 2200 3000
6
4 5 6 7 8 9
1 2 3 4 5 6
output:
2
3

Note: In the first test case:

  * Propose a problem with difficulty $w=800$ and $a$ becomes $[800,1000,1400,2000,2000,2200]$.    * Propose a problem with difficulty $w=1800$ and $a$ becomes $[800,1000,1400,1800,2000,2000]$.

It can be proved that it's impossible to reach the goal by proposing fewer new problems.

In the second test case:

  * Propose a problem with difficulty $w=1$ and $a$ becomes $[1,4,5,6,7,8]$.    * Propose a problem with difficulty $w=2$ and $a$ becomes $[1,2,4,5,6,7]$.    * Propose a problem with difficulty $w=3$ and $a$ becomes $[1,2,3,4,5,6]$.

It can be proved that it's impossible to reach the goal by proposing fewer new problems.

Buggy Code:
t = int(input())

while(t):
    n = int(input())
    a = list(map(int,input().split()))
    b = list(map(int,input().split()))

    a.sort()
    b.sort()

    check = 0
    for ai,bi in zip(a,b):
        if bi<ai:
            check+=1

    if(check==0):
        print(0)

    i,j = 0,0
    count = 0
    while i<len(a) and j<len(b):
        if a[i] > b[j]:
            count+=1
            j+=1
        else:
            i+=1
            j+=1

    print(count)

    t-=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 given buggy code, we need to consider the following aspects:

1. **Edge Cases**: Test cases where the arrays `a` and `b` are already in a valid state or are very close to being valid.
2. **Critical Differences**: Test cases where the differences between `a` and `b` are just enough to require a specific number of operations.
3. **Large Differences**: Test cases where the differences between `a` and `b` are large, requiring multiple operations to fix.

Let's create a test input that covers these aspects:

### Test Case 1: Already Valid
- This case should have `a` and `b` already in a valid state, so no new problems are needed.
- This will help verify that the code correctly handles the case where no operations are required.

### Test Case 2: Small Differences
- This case should have `a` and `b` with small differences, requiring a minimal number of operations.
- This will help verify that the code correctly counts the number of operations needed for small differences.

### Test Case 3: Large Differences
- This case should have `a` and `b` with large differences, requiring a significant number of operations.
- This will help verify that the code correctly counts the number of operations needed for large differences.

### Test Case 4: Mixed Differences
- This case should have a mix of small and large differences, requiring a combination of operations.
- This will help verify that the code correctly handles a mixed scenario.

### Test Case 5: Edge Case with Equal Lengths
- This case should have `a` and `b` with equal lengths and some elements that are just slightly off.
- This will help verify that the code correctly handles edge cases where the lengths are the same but some elements need adjustment.

### Test Input
```plaintext
5
5
1 2 3 4 5
1 2 3 4 5
6
1 2 3 4 5 6
1 2 3 4 5 7
6
10 20 30 40 50 60
1 2 3 4 5 6
6
1 2 3 4 5 6
10 20 30 40 50 60
6
1 2 3 4 5 6
1 2 3 4 6 7
```

### Explanation
- **Test Case 1**: `a` and `b` are already valid, so the output should be `0`.
- **Test Case 2**: `a` and `b` have a small difference at the last element, so the output should be `1`.
- **Test Case 3**: `a` and `b` have large differences, so the output should be `6`.
- **Test Case 4**: `a` and `b` are completely opposite, so the output should be `6`.
- **Test Case 5**: `a` and `b` have a small difference at the last element, so the output should be `1`.

This test input will help expose any issues in the buggy code by covering a range of scenarios.