**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scooter
time_limit_ms: 2000
memory_limit_mb: 512
Description: The Czech Technical University campus consists of $$$n$$$ buildings, indexed from $$$1$$$ to $$$n$$$. In each building, there can be a math class scheduled, or a computer science class, or neither (but not both). Additionally, in each building, there is at most one professor, and each professor is either an expert in mathematics or in computer science.

As an intern at University Express Inc., your job is to quickly transport the professors to their classes. For this, you have been granted a brand new two-person scooter, able to accommodate yourself, plus at most one passenger.

Initially, you are the only person on the scooter. When you arrive at a building, you may drop off or pick up professors to/from that building. However, in order to improve the efficiency of your task, you are allowed to drive to each of the $$$n$$$ buildings at most once, in the order of your choice (you can also decide where to start the itinerary).

After the end of your itinerary, in each building where a math class is scheduled, there must be a professor expert in math, and in each building where a computer science class is scheduled, there must be a professor expert in computer science.

Devise an itinerary that makes it possible to teach all classes.

Input: The first line contains an integer $$$n$$$ ($$$1\le n \le 2000$$$) — the number of buildings in the campus.

The second line contains a string of $$$c$$$ of length $$$n$$$ consisting of the characters $$$\texttt{-}$$$, $$$\texttt{C}$$$, $$$\texttt{M}$$$ — the $$$i$$$-th character denotes the subject of the class scheduled in the $$$i$$$-th building. $$$\texttt{C}$$$ stands for computer science, $$$\texttt{M}$$$ stands for mathematics, while $$$\texttt{-}$$$ means that there is no class scheduled in the $$$i$$$-th building.

The third line contains a string $$$p$$$ of length $$$n$$$ consisting of the characters $$$\texttt{-}$$$, $$$\texttt{C}$$$, $$$\texttt{M}$$$ — the $$$i$$$-th character denotes the expertise of the professor in the $$$i$$$-th building (if there is a professor). $$$\texttt{C}$$$ stands for computer science, $$$\texttt{M}$$$ stands for mathematics, while $$$\texttt{-}$$$ means that there is no professor in the $$$i$$$-th building.

It is guaranteed that, for all tests given to your program, there exists a valid itinerary.

Output: In the first line print an integer $$$l$$$ — the number of operations in your chosen itinerary.

The $$$i$$$-th ($$$1 \leq i \leq l$$$) of the next $$$l$$$ lines must contain one of three commands:

1. $$$\texttt{DRIVE } x$$$ — go to the building with the number $$$x$$$ ($$$1 \leq x \leq n$$$);
2. $$$\texttt{PICKUP}$$$ — pick up the professor which was initially at the current building;
3. $$$\texttt{DROPOFF}$$$ — drop off the passenger professor at the current building.

In order for the itinerary to be valid, the following conditions must hold:

1. No two $$$\texttt{DRIVE}$$$ instructions should go to the same building;
2. At most one $$$\texttt{DROPOFF}$$$ and one $$$\texttt{PICKUP}$$$ instruction in this order should be issued at each specific building;
3. For all $$$\texttt{PICKUP}$$$ instructions, there must be a professor initially at the building, as well as no one already riding along on the scooter;
4. For all $$$\texttt{DROPOFF}$$$ instructions, there must be a professor riding along at the time of the command;
5. After the itinerary, in each building, if there is a class in that building, there must be a professor expert in the class' subject (either initially, or because they were dropped off there).

Note that, in particular, you cannot pick up a professor that you just dropped off for an itinerary to be valid.

Examples:
input:
3
CM-
-CM
output:
7
DRIVE 3
PICKUP
DRIVE 2
DROPOFF
PICKUP
DRIVE 1
DROPOFF

input:
1
C
C
output:
0

input:
2
-M
MC
output:
4
DRIVE 1
PICKUP
DRIVE 2
DROPOFF

Note: In the first sample, You start by driving to building number $$$3$$$. You then pick up the mathematics professor. After dropping him off at building number $$$2$$$, where a mathematics class is being held, you pick up the computer science professor from there, and drop her off at building number $$$1$$$, finishing your itinerary.

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

using namespace std;

using ll = long long;
#define int ll
using pii = pair<int, int>;

const int MOD=1e9+7;
const int INF=1e9;
const int dx[4]={-1, 1, 0, 0}, dy[4]={0, 0, -1, 1}; // N S E V
const int MAXN=2e5+5;

char a[MAXN];
char b[MAXN];
int t, n, m, k;
vector<int> ps[7];
vector<pii> outp;
// 1 DRIVE
// 2 PICKUP
// 3 DROPOFF

string conv[4]={"", "DRIVE", "PICKUP", "DROPOFF"};
int c0, m0, c1, m1;

void solve()
{
    cin>>n;
    for(int i=1; i<=n; i++)
    {
        cin>>a[i];
        if(a[i] == 'C')
            c0++;
        else if(a[i] == 'M')
            m0++;
    }
    for(int i=1; i<=n; i++)
    {
        cin>>b[i];
        int tip;
        if(a[i] == '-' && b[i] == 'C')
            tip=1;
        if(a[i] == '-' && b[i] == 'M')
            tip=2;
        if(a[i] == 'C' && b[i] == 'M')
            tip=3;
        if(a[i] == 'M' && b[i] == 'C')
            tip=4;
        if(a[i] == 'C' && b[i] == '-')
            tip=5;
        if(a[i] == 'M' && b[i] == '-')
            tip=6;
        ps[tip].push_back(i);
        if(b[i] == 'C')
            c1++;
        else if(b[i] == 'M')
            m1++;
    }
    // cout<<c0<<' '<<c1<<' '<<m0<<' '<<m1<<'\n';
    while(c1 > c0 && !ps[4].empty())
    {
        ps[6].push_back(ps[4].back());
        ps[4].pop_back();
        c1--;
    }
    while(m1 > m0 && !ps[3].empty())
    {
        ps[5].push_back(ps[3].back());
        ps[3].pop_back();
        m1--;
    }
    while(ps[3].size() > ps[4].size())
    {
        outp.push_back({1, ps[1].back()});
        outp.push_back({2, ps[1].back()});
        ps[1].pop_back();
        outp.push_back({1, ps[3].back()});
        outp.push_back({3, ps[3].back()});
        outp.push_back({2, ps[3].back()});
        ps[3].pop_back();
        outp.push_back({1, ps[6].back()});
        outp.push_back({3, ps[6].back()});
        ps[6].pop_back();
    }
    while(ps[4].size() > ps[3].size())
    {
        outp.push_back({1, ps[2].back()});
        outp.push_back({2, ps[2].back()});
        ps[2].pop_back();
        outp.push_back({1, ps[4].back()});
        outp.push_back({3, ps[4].back()});
        outp.push_back({2, ps[4].back()});
        ps[4].pop_back();
        outp.push_back({1, ps[5].back()});
        outp.push_back({3, ps[5].back()});
        ps[5].pop_back();
    }
    while(ps[5].size() > 0)
    {
        outp.push_back({1, ps[1].back()});
        outp.push_back({2, ps[1].back()});
        ps[1].pop_back();
        outp.push_back({1, ps[5].back()});
        outp.push_back({3, ps[5].back()});
        ps[5].pop_back();
    }
    while(ps[6].size() > 0)
    {
        outp.push_back({1, ps[2].back()});
        outp.push_back({2, ps[2].back()});
        ps[2].pop_back();
        outp.push_back({1, ps[6].back()});
        outp.push_back({3, ps[6].back()});
        ps[6].pop_back();
    }
    if(outp.size() >= 3)
    {
        pii last=outp.back();
        outp.pop_back();
        pii last0=outp.back();
        outp.pop_back();
        int a=3, b=4;
        if(last.second == 5) // 3 4 3 4...
        {
            // nothing
        }
        else // 4 3 4 3...
        {
            a=4;
            b=3;
        }
        while(ps[a].size() > 0)
        {
            outp.push_back({1, ps[a].back()});
            outp.push_back({3, ps[a].back()});
            outp.push_back({2, ps[a].back()});
            ps[a].pop_back();
            outp.push_back({1, ps[b].back()});
            outp.push_back({3, ps[b].back()});
            outp.push_back({2, ps[b].back()});
            ps[b].pop_back();
        }
        outp.push_back(last0);
        outp.push_back(last);
    }
    cout<<outp.size()<<'\n';
    for(pii p:outp)
    {
        cout<<conv[p.first]<<' ';
        if(p.first == 1)
            cout<<p.second;
        cout<<'\n';
    }
}

int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases that might cause the algorithm to fail. The code seems to handle most general cases, but it might have issues with specific edge conditions. Let's consider the following scenarios:

1. **No Professors at All**: This should be trivially handled, but it's a good edge case to check.
2. **All Professors and Classes Match Perfectly**: This should also be trivial, but it's another edge case.
3. **Mismatched Professors and Classes**: This is a more complex case where the number of professors and classes do not match perfectly.
4. **Single Building with Both Class and Professor**: This is a simple but important edge case.
5. **Multiple Buildings with Complex Interdependencies**: This is a more complex scenario to test the logic of the algorithm.

Let's start with a small-scale test input that includes a mix of these edge cases to see if the code handles them correctly.

```plaintext
4
M-C
M-C
```

### Explanation:
- **Building 1**: Math class, Math professor.
- **Building 2**: No class, No professor.
- **Building 3**: No class, Computer Science professor.
- **Building 4**: Computer Science class, Computer Science professor.

This test case includes:
- A building with a math class and a math professor.
- A building with no class and no professor.
- A building with a computer science professor but no class.
- A building with a computer science class and a computer science professor.

This should be a good starting point to see if the code can handle these simple but varied conditions. If the code fails, it will help identify the specific part of the logic that is not robust.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scooter
time_limit_ms: 2000
memory_limit_mb: 512
Description: The Czech Technical University campus consists of $$$n$$$ buildings, indexed from $$$1$$$ to $$$n$$$. In each building, there can be a math class scheduled, or a computer science class, or neither (but not both). Additionally, in each building, there is at most one professor, and each professor is either an expert in mathematics or in computer science.

As an intern at University Express Inc., your job is to quickly transport the professors to their classes. For this, you have been granted a brand new two-person scooter, able to accommodate yourself, plus at most one passenger.

Initially, you are the only person on the scooter. When you arrive at a building, you may drop off or pick up professors to/from that building. However, in order to improve the efficiency of your task, you are allowed to drive to each of the $$$n$$$ buildings at most once, in the order of your choice (you can also decide where to start the itinerary).

After the end of your itinerary, in each building where a math class is scheduled, there must be a professor expert in math, and in each building where a computer science class is scheduled, there must be a professor expert in computer science.

Devise an itinerary that makes it possible to teach all classes.

Input: The first line contains an integer $$$n$$$ ($$$1\le n \le 2000$$$) — the number of buildings in the campus.

The second line contains a string of $$$c$$$ of length $$$n$$$ consisting of the characters $$$\texttt{-}$$$, $$$\texttt{C}$$$, $$$\texttt{M}$$$ — the $$$i$$$-th character denotes the subject of the class scheduled in the $$$i$$$-th building. $$$\texttt{C}$$$ stands for computer science, $$$\texttt{M}$$$ stands for mathematics, while $$$\texttt{-}$$$ means that there is no class scheduled in the $$$i$$$-th building.

The third line contains a string $$$p$$$ of length $$$n$$$ consisting of the characters $$$\texttt{-}$$$, $$$\texttt{C}$$$, $$$\texttt{M}$$$ — the $$$i$$$-th character denotes the expertise of the professor in the $$$i$$$-th building (if there is a professor). $$$\texttt{C}$$$ stands for computer science, $$$\texttt{M}$$$ stands for mathematics, while $$$\texttt{-}$$$ means that there is no professor in the $$$i$$$-th building.

It is guaranteed that, for all tests given to your program, there exists a valid itinerary.

Output: In the first line print an integer $$$l$$$ — the number of operations in your chosen itinerary.

The $$$i$$$-th ($$$1 \leq i \leq l$$$) of the next $$$l$$$ lines must contain one of three commands:

1. $$$\texttt{DRIVE } x$$$ — go to the building with the number $$$x$$$ ($$$1 \leq x \leq n$$$);
2. $$$\texttt{PICKUP}$$$ — pick up the professor which was initially at the current building;
3. $$$\texttt{DROPOFF}$$$ — drop off the passenger professor at the current building.

In order for the itinerary to be valid, the following conditions must hold:

1. No two $$$\texttt{DRIVE}$$$ instructions should go to the same building;
2. At most one $$$\texttt{DROPOFF}$$$ and one $$$\texttt{PICKUP}$$$ instruction in this order should be issued at each specific building;
3. For all $$$\texttt{PICKUP}$$$ instructions, there must be a professor initially at the building, as well as no one already riding along on the scooter;
4. For all $$$\texttt{DROPOFF}$$$ instructions, there must be a professor riding along at the time of the command;
5. After the itinerary, in each building, if there is a class in that building, there must be a professor expert in the class' subject (either initially, or because they were dropped off there).

Note that, in particular, you cannot pick up a professor that you just dropped off for an itinerary to be valid.

Examples:
input:
3
CM-
-CM
output:
7
DRIVE 3
PICKUP
DRIVE 2
DROPOFF
PICKUP
DRIVE 1
DROPOFF

input:
1
C
C
output:
0

input:
2
-M
MC
output:
4
DRIVE 1
PICKUP
DRIVE 2
DROPOFF

Note: In the first sample, You start by driving to building number $$$3$$$. You then pick up the mathematics professor. After dropping him off at building number $$$2$$$, where a mathematics class is being held, you pick up the computer science professor from there, and drop her off at building number $$$1$$$, finishing your itinerary.

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

#define endl '\n'
#define IOS ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define ll long long
#define debug(x) cout << #x << ' ' << x << endl;

const int inf = 1e9 + 10;
const ll INF = 1e18 + 10;

string tostring(int x){
	string ss = "";
	while(x){
		ss = (char)('0' + x % 10) + ss;
		x /= 10;
	}
	return ss;
}

void test(){
	int n;cin >> n;
	vector<int> c(n + 1,0);
	vector<int> p(n + 1,0);
	string s;cin >> s;
	for(int i = 1;i <= n;i++){
		if(s[i - 1] == 'C') p[i] = 1;
		else if(s[i - 1] == 'M') p[i] = 2;
	}
	vector<int> c_;//have c not need
	vector<int> m_;//have m not need
	vector<int> _c;//no need c
	vector<int> _m;//no need m
	vector<int> cm;//have c need m
	vector<int> mc;//have m need c
	vector<int> __;//no
	cin >> s;
	for(int i = 1;i <= n;i++){
		if(s[i - 1] == 'C') c[i] = 1;
		else if(s[i - 1] == 'M') c[i] = 2;
		if(c[i] == 0 && p[i] == 0) __.push_back(i);
		else if(c[i] == 1 && p[i] == 0) c_.push_back(i);//c
		else if(c[i] == 2 && p[i] == 0) m_.push_back(i);//m
		else if(c[i] == 0 && p[i] == 1) _c.push_back(i);//get c
		else if(c[i] == 2 && p[i] == 1)	mc.push_back(i);//c to m
		else if(c[i] == 0 && p[i] == 2) _m.push_back(i);//get m
		else if(c[i] == 1 && p[i] == 2) cm.push_back(i);//m to c
	}
	int l[6] = {0},r[6] = {0};
	r[2] = _c.size();
	r[3] = _m.size();
	r[4] = mc.size();
	r[5] = cm.size();
	r[0] = min((int)c_.size(),r[2] - r[5] + r[4]);
	r[1] = min((int)m_.size(),r[3] - r[4] + r[5]);
	if(r[0] == 0 && r[1] == 0){
		if(!c_.empty()) r[0] = 1;
		else r[1] = 1;
	}
	queue<string> qu;
	for(int i = 0;i < r[0];i++){
		s = "";
		s += "DRIVE ";
		s += tostring(c_[i]);
		s += "\n";
		qu.push(s);
		s = "PICKUP\n";
		qu.push(s);
		bool f = 0;
		while(1){
			if(f == 1 && l[5] < r[5]){
				s = "";
				s += "DRIVE ";
				s += tostring(cm[l[5]]);
				s += "\n";
				qu.push(s);
				s = "DROPOFF\n";
				qu.push(s);
				s = "PICKUP\n";
				qu.push(s);
				l[5]++;
				f = !f;
			}else if(f == 0 && l[4] < r[4]){
				s = "";
				s += "DRIVE ";
				s += tostring(mc[l[4]]);
				s += "\n";
				qu.push(s);
				s = "DROPOFF\n";
				qu.push(s);
				s = "PICKUP\n";
				qu.push(s);
				l[4]++;
				f = !f;
			}else
				break;
		}
		if(f && _m.size() > l[3]){
			s = "";
			s += "DRIVE ";
			s += tostring(_m[l[3]++]);
			s += "\n";
			qu.push(s);
			s = "DROPOFF" + '\n';
			qu.push(s);
		}
		else if(_c.size() > l[2]){
			s = "";
			s += "DRIVE ";
			s += tostring(_c[l[2]++]);
			s += "\n";
			qu.push(s);
			s = "DROPOFF\n";
			qu.push(s);
		}
	}

	for(int i = 0;i < r[1];i++){
		s = "";
		s += "DRIVE ";
		s += tostring(m_[i]);
		s += "\n";
		qu.push(s);
		s = "PICKUP\n";
		qu.push(s);
		bool f = 1;
		while(1){
			if(f == 1 && l[5] < r[5]){
				s = "";
				s += "DRIVE ";
				s += tostring(cm[l[5]]);
				s += "\n";
				qu.push(s);
				s = "DROPOFF\n";
				qu.push(s);
				s = "PICKUP\n";
				qu.push(s);
				l[5]++;
				f = !f;
			}else if(f == 0 && l[4] < r[4]){
				s = "";
				s += "DRIVE ";
				s += tostring(mc[l[4]]);
				s += "\n";
				qu.push(s);
				s = "DROPOFF\n";
				qu.push(s);
				s = "PICKUP\n";
				qu.push(s);
				l[4]++;
				f = !f;
			}else
				break;
		}
		if(f && _m.size() > l[3]){
			s = "";
			s += "DRIVE ";
			s += tostring(_m[l[3]++]);
			s += "\n";
			qu.push(s);
			s = "DROPOFF" + '\n';
			qu.push(s);
		}
		else if(_c.size() > l[2]){
			s = "";
			s += "DRIVE ";
			s += tostring(_c[l[2]++]);
			s += "\n";
			qu.push(s);
			s = "DROPOFF\n";
			qu.push(s);
		}
	}
	cout << qu.size() << '\n';
	while(!qu.empty()){
		cout << qu.front();
		qu.pop();
	}
}

int main(){
	IOS
	int t = 1;//cin >> t;
	while(t--){
		test();
	}
}

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

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

Think step by step.

Response:
```plaintext
5
C-MC
-M-M
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scooter
time_limit_ms: 2000
memory_limit_mb: 512
Description: The Czech Technical University campus consists of $$$n$$$ buildings, indexed from $$$1$$$ to $$$n$$$. In each building, there can be a math class scheduled, or a computer science class, or neither (but not both). Additionally, in each building, there is at most one professor, and each professor is either an expert in mathematics or in computer science.

As an intern at University Express Inc., your job is to quickly transport the professors to their classes. For this, you have been granted a brand new two-person scooter, able to accommodate yourself, plus at most one passenger.

Initially, you are the only person on the scooter. When you arrive at a building, you may drop off or pick up professors to/from that building. However, in order to improve the efficiency of your task, you are allowed to drive to each of the $$$n$$$ buildings at most once, in the order of your choice (you can also decide where to start the itinerary).

After the end of your itinerary, in each building where a math class is scheduled, there must be a professor expert in math, and in each building where a computer science class is scheduled, there must be a professor expert in computer science.

Devise an itinerary that makes it possible to teach all classes.

Input: The first line contains an integer $$$n$$$ ($$$1\le n \le 2000$$$) — the number of buildings in the campus.

The second line contains a string of $$$c$$$ of length $$$n$$$ consisting of the characters $$$\texttt{-}$$$, $$$\texttt{C}$$$, $$$\texttt{M}$$$ — the $$$i$$$-th character denotes the subject of the class scheduled in the $$$i$$$-th building. $$$\texttt{C}$$$ stands for computer science, $$$\texttt{M}$$$ stands for mathematics, while $$$\texttt{-}$$$ means that there is no class scheduled in the $$$i$$$-th building.

The third line contains a string $$$p$$$ of length $$$n$$$ consisting of the characters $$$\texttt{-}$$$, $$$\texttt{C}$$$, $$$\texttt{M}$$$ — the $$$i$$$-th character denotes the expertise of the professor in the $$$i$$$-th building (if there is a professor). $$$\texttt{C}$$$ stands for computer science, $$$\texttt{M}$$$ stands for mathematics, while $$$\texttt{-}$$$ means that there is no professor in the $$$i$$$-th building.

It is guaranteed that, for all tests given to your program, there exists a valid itinerary.

Output: In the first line print an integer $$$l$$$ — the number of operations in your chosen itinerary.

The $$$i$$$-th ($$$1 \leq i \leq l$$$) of the next $$$l$$$ lines must contain one of three commands:

1. $$$\texttt{DRIVE } x$$$ — go to the building with the number $$$x$$$ ($$$1 \leq x \leq n$$$);
2. $$$\texttt{PICKUP}$$$ — pick up the professor which was initially at the current building;
3. $$$\texttt{DROPOFF}$$$ — drop off the passenger professor at the current building.

In order for the itinerary to be valid, the following conditions must hold:

1. No two $$$\texttt{DRIVE}$$$ instructions should go to the same building;
2. At most one $$$\texttt{DROPOFF}$$$ and one $$$\texttt{PICKUP}$$$ instruction in this order should be issued at each specific building;
3. For all $$$\texttt{PICKUP}$$$ instructions, there must be a professor initially at the building, as well as no one already riding along on the scooter;
4. For all $$$\texttt{DROPOFF}$$$ instructions, there must be a professor riding along at the time of the command;
5. After the itinerary, in each building, if there is a class in that building, there must be a professor expert in the class' subject (either initially, or because they were dropped off there).

Note that, in particular, you cannot pick up a professor that you just dropped off for an itinerary to be valid.

Examples:
input:
3
CM-
-CM
output:
7
DRIVE 3
PICKUP
DRIVE 2
DROPOFF
PICKUP
DRIVE 1
DROPOFF

input:
1
C
C
output:
0

input:
2
-M
MC
output:
4
DRIVE 1
PICKUP
DRIVE 2
DROPOFF

Note: In the first sample, You start by driving to building number $$$3$$$. You then pick up the mathematics professor. After dropping him off at building number $$$2$$$, where a mathematics class is being held, you pick up the computer science professor from there, and drop her off at building number $$$1$$$, finishing your itinerary.

Buggy Code:
#include <iostream>
#include <vector>
#include <cstdio>
#include <string>

using namespace std;

vector<string>ans;
vector<int> order;

void drive(int i) {
	char ret[20];
	sprintf(ret, "DRIVE %d", i+1);
	ans.push_back(string(ret));
}

void dropoff() {
	ans.push_back("DROPOFF");
}

void pickup() {
	ans.push_back("PICKUP");
}

int main() {
	int n;
	scanf("%d", &n);

	char input[2][2101];
	scanf("%s", input[0]);
	scanf("%s", input[1]);

	vector<int> mc, cm, xc, cx, xm, mx;
	int used_mc = 0, used_cm = 0, used_xc = 0, used_cx = 0, used_xm = 0, used_mx = 0;

	for (int i = 0; i < n; ++i) {
		if (input[0][i] == 'M' && input[1][i] == 'C') {
			mc.push_back(i);
		}
		if (input[0][i] == 'C' && input[1][i] == 'M') {
			cm.push_back(i);
		}
		if (input[0][i] == '-' && input[1][i] == 'C') {
			xc.push_back(i);
		}
		if (input[0][i] == 'C' && input[1][i] == '-') {
			cx.push_back(i);
		}
		if (input[0][i] == '-' && input[1][i] == 'M') {
			xm.push_back(i);
		}
		if (input[0][i] == 'M' && input[1][i] == '-') {
			mx.push_back(i);
		}
	}

	// printf("%zu %zu %zu %zu %zu %zu\n", cm.size(), mc.size(), xc.size(), cx.size(), xm.size(), mx.size());

	// cm mc xc cx xm mx
	// 0  1  0  1  1  0
	//

	char need = 'x';
	while(used_cx < cx.size() || used_mx < mx.size() || used_cm < cm.size() || used_mc < mc.size() || need != 'x') {
		if (need == 'x') {
			if (used_cx < cx.size()) {
				order.push_back(cx[used_cx++]);
				need = 'c';
				continue;
			}
			if (used_mx < mx.size()) {
				order.push_back(mx[used_mx++]);
				need = 'm';
				continue;
			}
			if (used_cm < cm.size() && used_xm < xm.size()) {
				order.push_back(cm[used_cm++]);
				need = 'c';
				continue;
			}
			if (used_mc < mc.size() && used_xc < xc.size()) {
				order.push_back(mc[used_mc++]);
				need = 'm';
				continue;
			}
		} else if (need == 'c') {
			if (used_mc < mc.size()) {
				order.push_back(mc[used_mc++]);
				need = 'm';
				continue;
			} else {
				order.push_back(xc[used_xc++]);
				need = 'x';
				continue;
			}
		} else if (need == 'm') {
			if (used_cm < cm.size()) {
				order.push_back(cm[used_cm++]);
				need = 'c';
				continue;
			} else {
				order.push_back(xm[used_xm++]);
				need = 'x';
				continue;
			}
		}
	}

//	printf("%zu\n", order.size());
	for (int i = order.size() - 1; i >= 0; i--) {
//		printf("%d ", order[i]);
		drive(order[i]);
		if (input[0][order[i]] != '-') {
			dropoff();
		}
		if (input[1][order[i]] != '-' && (i > 0 && input[0][order[i-1]] != '-')) {
			pickup();
		}
	}
//	printf("\n");

	if (ans.size() > 0 && ans[ans.size() - 1] == "PICKUP") {
		ans.pop_back();
	}

	printf("%d\n", int(ans.size()));
	for(const string& s : ans) {
		printf("%s\n", s.c_str());
	}

	return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to craft a test case that challenges the logic and order of operations. The key points to consider are:

1. **Initial Conditions**: Ensure that there are buildings with professors that need to be picked up and dropped off.
2. **Order of Operations**: The code should handle the correct sequence of picking up and dropping off professors.
3. **Edge Cases**: Buildings with no classes or no professors, and buildings with both classes and professors.

Given the code's logic, a critical edge case is when there are buildings with professors that need to be moved to buildings with classes, but the order of operations is not correctly managed.

### Test Case:
Let's create a small-scale test case with 4 buildings:
- Building 1: Math class, no professor
- Building 2: No class, Computer Science professor
- Building 3: Computer Science class, no professor
- Building 4: No class, Math professor

This test case will challenge the code to correctly pick up the Math professor from Building 4 and drop them at Building 1, and pick up the Computer Science professor from Building 2 and drop them at Building 3.

### Test Input:
```plaintext
4
M-C
-CM-
```

### Explanation:
- **Building 1**: Math class, no professor.
- **Building 2**: No class, Computer Science professor.
- **Building 3**: Computer Science class, no professor.
- **Building 4**: No class, Math professor.

This test case will force the code to:
1. Drive to Building 4, pick up the Math professor.
2. Drive to Building 1, drop off the Math professor.
3. Drive to Building 2, pick up the Computer Science professor.
4. Drive to Building 3, drop off the Computer Science professor.

If the code fails to manage the order of operations correctly, it will fail to drop off the professors at the correct buildings.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scooter
time_limit_ms: 2000
memory_limit_mb: 512
Description: The Czech Technical University campus consists of $$$n$$$ buildings, indexed from $$$1$$$ to $$$n$$$. In each building, there can be a math class scheduled, or a computer science class, or neither (but not both). Additionally, in each building, there is at most one professor, and each professor is either an expert in mathematics or in computer science.

As an intern at University Express Inc., your job is to quickly transport the professors to their classes. For this, you have been granted a brand new two-person scooter, able to accommodate yourself, plus at most one passenger.

Initially, you are the only person on the scooter. When you arrive at a building, you may drop off or pick up professors to/from that building. However, in order to improve the efficiency of your task, you are allowed to drive to each of the $$$n$$$ buildings at most once, in the order of your choice (you can also decide where to start the itinerary).

After the end of your itinerary, in each building where a math class is scheduled, there must be a professor expert in math, and in each building where a computer science class is scheduled, there must be a professor expert in computer science.

Devise an itinerary that makes it possible to teach all classes.

Input: The first line contains an integer $$$n$$$ ($$$1\le n \le 2000$$$) — the number of buildings in the campus.

The second line contains a string of $$$c$$$ of length $$$n$$$ consisting of the characters $$$\texttt{-}$$$, $$$\texttt{C}$$$, $$$\texttt{M}$$$ — the $$$i$$$-th character denotes the subject of the class scheduled in the $$$i$$$-th building. $$$\texttt{C}$$$ stands for computer science, $$$\texttt{M}$$$ stands for mathematics, while $$$\texttt{-}$$$ means that there is no class scheduled in the $$$i$$$-th building.

The third line contains a string $$$p$$$ of length $$$n$$$ consisting of the characters $$$\texttt{-}$$$, $$$\texttt{C}$$$, $$$\texttt{M}$$$ — the $$$i$$$-th character denotes the expertise of the professor in the $$$i$$$-th building (if there is a professor). $$$\texttt{C}$$$ stands for computer science, $$$\texttt{M}$$$ stands for mathematics, while $$$\texttt{-}$$$ means that there is no professor in the $$$i$$$-th building.

It is guaranteed that, for all tests given to your program, there exists a valid itinerary.

Output: In the first line print an integer $$$l$$$ — the number of operations in your chosen itinerary.

The $$$i$$$-th ($$$1 \leq i \leq l$$$) of the next $$$l$$$ lines must contain one of three commands:

1. $$$\texttt{DRIVE } x$$$ — go to the building with the number $$$x$$$ ($$$1 \leq x \leq n$$$);
2. $$$\texttt{PICKUP}$$$ — pick up the professor which was initially at the current building;
3. $$$\texttt{DROPOFF}$$$ — drop off the passenger professor at the current building.

In order for the itinerary to be valid, the following conditions must hold:

1. No two $$$\texttt{DRIVE}$$$ instructions should go to the same building;
2. At most one $$$\texttt{DROPOFF}$$$ and one $$$\texttt{PICKUP}$$$ instruction in this order should be issued at each specific building;
3. For all $$$\texttt{PICKUP}$$$ instructions, there must be a professor initially at the building, as well as no one already riding along on the scooter;
4. For all $$$\texttt{DROPOFF}$$$ instructions, there must be a professor riding along at the time of the command;
5. After the itinerary, in each building, if there is a class in that building, there must be a professor expert in the class' subject (either initially, or because they were dropped off there).

Note that, in particular, you cannot pick up a professor that you just dropped off for an itinerary to be valid.

Examples:
input:
3
CM-
-CM
output:
7
DRIVE 3
PICKUP
DRIVE 2
DROPOFF
PICKUP
DRIVE 1
DROPOFF

input:
1
C
C
output:
0

input:
2
-M
MC
output:
4
DRIVE 1
PICKUP
DRIVE 2
DROPOFF

Note: In the first sample, You start by driving to building number $$$3$$$. You then pick up the mathematics professor. After dropping him off at building number $$$2$$$, where a mathematics class is being held, you pick up the computer science professor from there, and drop her off at building number $$$1$$$, finishing your itinerary.

Buggy Code:
def eular(neigh,start,visited):

    seq = []
    stack = []
    stack.append(start)

    visited.add(start)

    while stack:
        index = stack[-1]
        if len(neigh[index]) > 0:
            nextindex = -1
            while neigh[index][-1] in visited:
                nextindex = neigh[index].pop()
            if nextindex >= 0 and nextindex not in visited:
                stack.append(nextindex)
                visited.add(nextindex)
            else:
                seq.append(index)
                stack.pop()
        else:
            seq.append(index)
            stack.pop()
    return seq[::-1]

def getresult(index, s1,s2, ans):

    ans.append(["DRIVE", index+1])
    if len(ans) > 1 and ans[-2]==["PICKUP"]:
        ans.append(["DROPOFF"])
    if s2[index]!='-':
        ans.append(["PICKUP"])
    return

def main():
    n = int(input())
    s1 = input()
    s2 = input()

    if s1==s2:
        print(0)
        return

    neigh = [[] for _ in range(n)]

    kind = {"-M":[], "-C":[], "MC":[], "CM":[], "C-":[], "M-":[]}

    path = []

    for i in range(n):
        if s1[i]+s2[i] in kind:
            kind[s1[i]+s2[i]].append(i)

    num1 = len(kind["-M"])
    num2 = len(kind["-C"])

    num3 = len(kind["M-"])
    num4 = len(kind["C-"])

    for _ in range(min(num1,num3)):
        path.append(kind["-M"].pop())
        path.append(kind["M-"].pop())
    for _ in range(min(num2,num4)):
        path.append(kind["-C"].pop())
        path.append(kind["C-"].pop())

    while len(kind["M-"]) > 0:
        path.append(kind["-C"].pop())
        path.append(kind["CM"].pop())
        path.append(kind["M-"].pop())

    while len(kind["C-"]) > 0:
        path.append(kind["-M"].pop())
        path.append(kind["MC"].pop())
        path.append(kind["C-"].pop())

    temp = []
    while len(kind["CM"]) > 0:
        temp.append(kind["CM"].pop())
        temp.append(kind["MC"].pop())

    if s1[path[-2]] + s2[path[-2]] == "M-":
        temp = temp[::-1]

    path = path[:-1] + temp + path[-1:]

    ans = []
    for index in path:
        getresult(index,s1,s2,ans)

    print(len(ans))
    for ele in ans:
        print(*ele)

main()

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

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

Think step by step.

Response:
```plaintext
4
M-C-
MC-C
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scooter
time_limit_ms: 2000
memory_limit_mb: 512
Description: The Czech Technical University campus consists of $$$n$$$ buildings, indexed from $$$1$$$ to $$$n$$$. In each building, there can be a math class scheduled, or a computer science class, or neither (but not both). Additionally, in each building, there is at most one professor, and each professor is either an expert in mathematics or in computer science.

As an intern at University Express Inc., your job is to quickly transport the professors to their classes. For this, you have been granted a brand new two-person scooter, able to accommodate yourself, plus at most one passenger.

Initially, you are the only person on the scooter. When you arrive at a building, you may drop off or pick up professors to/from that building. However, in order to improve the efficiency of your task, you are allowed to drive to each of the $$$n$$$ buildings at most once, in the order of your choice (you can also decide where to start the itinerary).

After the end of your itinerary, in each building where a math class is scheduled, there must be a professor expert in math, and in each building where a computer science class is scheduled, there must be a professor expert in computer science.

Devise an itinerary that makes it possible to teach all classes.

Input: The first line contains an integer $$$n$$$ ($$$1\le n \le 2000$$$) — the number of buildings in the campus.

The second line contains a string of $$$c$$$ of length $$$n$$$ consisting of the characters $$$\texttt{-}$$$, $$$\texttt{C}$$$, $$$\texttt{M}$$$ — the $$$i$$$-th character denotes the subject of the class scheduled in the $$$i$$$-th building. $$$\texttt{C}$$$ stands for computer science, $$$\texttt{M}$$$ stands for mathematics, while $$$\texttt{-}$$$ means that there is no class scheduled in the $$$i$$$-th building.

The third line contains a string $$$p$$$ of length $$$n$$$ consisting of the characters $$$\texttt{-}$$$, $$$\texttt{C}$$$, $$$\texttt{M}$$$ — the $$$i$$$-th character denotes the expertise of the professor in the $$$i$$$-th building (if there is a professor). $$$\texttt{C}$$$ stands for computer science, $$$\texttt{M}$$$ stands for mathematics, while $$$\texttt{-}$$$ means that there is no professor in the $$$i$$$-th building.

It is guaranteed that, for all tests given to your program, there exists a valid itinerary.

Output: In the first line print an integer $$$l$$$ — the number of operations in your chosen itinerary.

The $$$i$$$-th ($$$1 \leq i \leq l$$$) of the next $$$l$$$ lines must contain one of three commands:

1. $$$\texttt{DRIVE } x$$$ — go to the building with the number $$$x$$$ ($$$1 \leq x \leq n$$$);
2. $$$\texttt{PICKUP}$$$ — pick up the professor which was initially at the current building;
3. $$$\texttt{DROPOFF}$$$ — drop off the passenger professor at the current building.

In order for the itinerary to be valid, the following conditions must hold:

1. No two $$$\texttt{DRIVE}$$$ instructions should go to the same building;
2. At most one $$$\texttt{DROPOFF}$$$ and one $$$\texttt{PICKUP}$$$ instruction in this order should be issued at each specific building;
3. For all $$$\texttt{PICKUP}$$$ instructions, there must be a professor initially at the building, as well as no one already riding along on the scooter;
4. For all $$$\texttt{DROPOFF}$$$ instructions, there must be a professor riding along at the time of the command;
5. After the itinerary, in each building, if there is a class in that building, there must be a professor expert in the class' subject (either initially, or because they were dropped off there).

Note that, in particular, you cannot pick up a professor that you just dropped off for an itinerary to be valid.

Examples:
input:
3
CM-
-CM
output:
7
DRIVE 3
PICKUP
DRIVE 2
DROPOFF
PICKUP
DRIVE 1
DROPOFF

input:
1
C
C
output:
0

input:
2
-M
MC
output:
4
DRIVE 1
PICKUP
DRIVE 2
DROPOFF

Note: In the first sample, You start by driving to building number $$$3$$$. You then pick up the mathematics professor. After dropping him off at building number $$$2$$$, where a mathematics class is being held, you pick up the computer science professor from there, and drop her off at building number $$$1$$$, finishing your itinerary.

Buggy Code:
import java.io.*;
import java.lang.reflect.Member;
import java.math.BigInteger;
import java.util.*;
import java.util.function.Function;

public class Main {

    public static void main(String[] args) throws java.lang.Exception {
        int n = in.nextInt();
        String b = in.nextLine();
        String p = in.nextLine();
        Queue<Integer> cm = new LinkedList<>(), ce = new LinkedList<>(), mc = new LinkedList<>(), me = new LinkedList<>(), ec = new LinkedList<>(), em = new LinkedList<>();
        for (int i = 0; i < n; i++) {
            if(b.charAt(i) != p.charAt(i)){
                if(b.charAt(i) == 'C'){
                    if(p.charAt(i) == 'M') cm.add(i+1);
                    else ce.add(i+1);
                } else if(b.charAt(i) == 'M'){
                    if(p.charAt(i) == 'C') mc.add(i+1);
                    else me.add(i+1);
                } else{
                    if(p.charAt(i) == 'C') ec.add(i+1);
                    else em.add(i+1);
                }
            }
        }
        List<String> res = new ArrayList<>();
        while (!ec.isEmpty()){
            drive(ec.poll(), res); pickup(res);
            if(cm.isEmpty()){
                if(!ce.isEmpty()) drive(ce.poll(), res);
                dropOff(res);
            } else {
                drive(cm.poll(), res); dropOff(res); pickup(res);
                deliverCM(mc, cm, ce, me, res);
            }
        }
        while (!em.isEmpty()){
            drive(em.poll(), res); pickup(res);
            if(mc.isEmpty()){
                if(!me.isEmpty()){
                    drive(me.poll(), res);
                }
                dropOff(res);
            } else{
                drive(mc.poll(), res); dropOff(res); pickup(res);
                deliverMC(mc, cm, ce, me, res);
            }
        }
        System.out.println(res.size());
        for(var s : res) System.out.println(s);
    }

    static void deliverCM(Queue<Integer> mc, Queue<Integer> cm, Queue<Integer> ce, Queue<Integer> me, List<String> res){
        boolean flag = false;
        while (true){
            if(!flag){
                if(mc.isEmpty()) break;
                drive(mc.poll(), res);
            } else{
                if(cm.isEmpty()) break;
                drive(cm.poll(), res);
            }
            dropOff(res);
            pickup(res);
            flag = !flag;
        }
        if(flag){
            if(!ce.isEmpty()) drive(ce.poll(), res);
        } else if(!me.isEmpty()) drive(me.poll(), res);
        dropOff(res);
    }

    static void deliverMC(Queue<Integer> mc, Queue<Integer> cm, Queue<Integer> ce, Queue<Integer> me, List<String> res){
        boolean flag = false;
        while (true){
            if(!flag){
                if(cm.isEmpty()) break;
                drive(cm.poll(), res);
            } else{
                if(mc.isEmpty()) break;
                drive(mc.poll(), res);
            }
            flag = !flag;
            dropOff(res);
            pickup(res);
        }
        if(flag){
            if(!me.isEmpty()){
                drive(me.poll(), res);
            }
        } else{
            if(!ce.isEmpty()) drive(ce.poll(), res);
        }
        dropOff(res);
    }

    static void pickup(List<String> res){
        res.add("PICKUP");
    }

    static void dropOff(List<String> res){
        res.add("DROPOFF");
    }

    static void drive(int j, List<String> res){
        res.add("DRIVE " + j);
    }

    static class Pair<T, U> {
        T f;
        U s;

        public Pair(T first, U sec) {
            this.f = first;
            this.s = sec;
        }

        @Override
        public String toString() {
            return f + " " + s;
        }

    }

    static long nCrModP(int n, int[] fac, int[] mods, int r, int p) {

        if (n < r)
            return 0;

        if (r == 0)
            return 1;

        return ((long) fac[n] * mods[r]
                % p * mods[n - r]
                % p)
                % p;
    }

    static long modInverse(int n, int p) {
        return pow(n, p - 2, p);
    }

    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());
        }

        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("").append(String.valueOf(object));
        }

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

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

    static FastReader in = new FastReader();

    static int[] ai(int n) {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) arr[i] = in.nextInt();
        return arr;
    }

    static long[] al(int n) {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++) arr[i] = in.nextLong();
        return arr;
    }

    static long gcd(long a, long b) {
        while (b != 0) {
            long temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }

    static int aMax(int[] arr) {
        int a = Integer.MIN_VALUE;
        for (int i = 0; i < arr.length; i++) a = Math.max(a, arr[i]);
        return a;
    }

    static int aMin(int[] arr) {
        int a = Integer.MAX_VALUE;
        for (int i = 0; i < arr.length; i++) a = Math.min(a, arr[i]);
        return a;
    }

    static int mod = (int) 1e9 + 7;

    static long lcm(long u, long v) {
        return (u / gcd(u, v)) * v;
    }

    static long sumOfNumbers(long start, long end) {
        if (end < start) return 0;
        else if (end == start) return end;
        return (((end * (end + 1)) / 2) - ((start * (start - 1)) / 2));
    }

    static long pow(long a, long b, int mod) {
        long res = 1;
        while (b > 0) {
            if ((b&1) == 1) {
                res = (res * a % mod) % mod;
            }
            a = (a % mod * a % mod) % mod;
            b >>= 1;
        }
        return res % mod;
    }

    static long pow(long a, long b) {
        long res = 1;
        while (b > 0) {
            if ((b&1) == 1) {
                res = (res * a % mod) % mod;
            }
            a = (a % mod * a % mod) % mod;
            b >>= 1;
        }
        return res % mod;
    }

    static class Triplet<T> {
        T x, y, z;

        public Triplet(T x, T y, T z) {
            this.x = x;
            this.y = y;
            this.z = z;
        }
    }

    static Triplet<Long> extendedEuclid(long a, long b) { // a>b
        if (b == 0) {
            return new Triplet<>(a, 0L, a);
        }

        Triplet<Long> smallAns = extendedEuclid(b, a % b);
        long y = smallAns.x - (a / b) * smallAns.y;
        return new Triplet<>(smallAns.y, y, smallAns.z);

    }

    static long modularMultiplicativeInverse(long a, long m) {
        long gcd = gcd(a, m);
        if (gcd != 1) {
            return -1;
        }
        long x = extendedEuclid(a, m).x;
        x = (x % m + m) % m;
        return x;
    }

    static long pow(long a, long b, long mod) {
        long res = 1;
        while (b > 0) {
            if (b % 2 == 1) {
                res = (res * a % mod) % mod;
            }
            a = (a % mod * a % mod) % mod;
            b >>= 1;
        }
        return res % mod;
    }

    static long bigPow(long base, long pow, long mod) {
        BigInteger res = new BigInteger("1");
        BigInteger bBase = BigInteger.valueOf(base);
        BigInteger mod2 = BigInteger.valueOf(mod);
        while (pow > 0) {
            if (pow % 2 == 1) {
                res = res.multiply(bBase);
                res = res.mod(mod2);
            }
            bBase = bBase.pow(2);
            bBase = bBase.mod(mod2);
            pow >>= 1;
        }
        res = res.mod(BigInteger.valueOf(mod));
        return res.longValue();
    }

    static boolean millerRobin(long n, long d, long a) {
        long x = bigPow(a, d, n);

        if (x == 1 || x == n - 1) return true;

        while (d != n - 1) {
            BigInteger bx = BigInteger.valueOf(x);
            bx = bx.multiply(bx).mod(BigInteger.valueOf(n));
            x = bx.longValue();
            d *= 2;

            if (x == 1) return false;
            if (x == n - 1) return true;
        }

        return false;
    }

    public static boolean isPrime(long n) {
        if (n <= 1 || n == 4) return false;
        if (n <= 3) return true;
        if (n % 2 == 0) return false;

        long d = n - 1;
        while ((d & 1) == 0) d /= 2;

        long[] arr = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};
        for (long j : arr) {
            if (n == j) return true;
            if (!millerRobin(n, d, j)) return false;
        }

        return true;
    }

    static long nCrModPFermat(int n, int r, int p, long[] fac) {
        if (n < r) return 0;
        if (r == 0) return 1;

        return ((fac[n] * modularMultiplicativeInverse(fac[r], p)) % p *
                (modularMultiplicativeInverse(fac[n - r], p)) % p) % p;
    }
}

**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 buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code seems to handle certain conditions incorrectly, especially when dealing with the order of operations and the availability of professors.

Let's consider the following scenarios:
1. **No professors available for some classes**: This can lead to a situation where the algorithm fails to find a valid itinerary.
2. **Multiple buildings with the same class type but no professors available**: This can test the algorithm's ability to handle resource allocation efficiently.
3. **Buildings with both classes and professors, but in a way that requires careful planning**: This can test the algorithm's ability to handle complex itineraries.

Given these considerations, let's create a small-scale test input that is complete and valid:

```plaintext
5
CMMMC
----C
```

**Explanation:**
- **5 buildings**: This is a small number to keep the test input manageable.
- **Class schedule**: "CMMMC" means:
  - Building 1: Computer Science class
  - Building 2: Mathematics class
  - Building 3: Mathematics class
  - Building 4: Mathematics class
  - Building 5: Computer Science class
- **Professor schedule**: "----C" means:
  - Building 1: No professor
  - Building 2: No professor
  - Building 3: No professor
  - Building 4: No professor
  - Building 5: Computer Science professor

This test input is designed to:
- **Challenge the algorithm** to find a valid itinerary that ensures all classes are covered.
- **Expose potential bugs** in the logic that handles the order of operations and resource allocation.

The correct solution should:
- **Pick up the Computer Science professor** from Building 5.
- **Drive to Building 1** and drop off the Computer Science professor.
- **Drive to Building 5** again to pick up the Computer Science professor.
- **Drive to Building 5** and drop off the Computer Science professor (if needed).
- **Handle the Mathematics classes** in Buildings 2, 3, and 4, which requires careful planning to ensure all classes are covered.

This test input should help in identifying and fixing any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scooter
time_limit_ms: 2000
memory_limit_mb: 512
Description: The Czech Technical University campus consists of $$$n$$$ buildings, indexed from $$$1$$$ to $$$n$$$. In each building, there can be a math class scheduled, or a computer science class, or neither (but not both). Additionally, in each building, there is at most one professor, and each professor is either an expert in mathematics or in computer science.

As an intern at University Express Inc., your job is to quickly transport the professors to their classes. For this, you have been granted a brand new two-person scooter, able to accommodate yourself, plus at most one passenger.

Initially, you are the only person on the scooter. When you arrive at a building, you may drop off or pick up professors to/from that building. However, in order to improve the efficiency of your task, you are allowed to drive to each of the $$$n$$$ buildings at most once, in the order of your choice (you can also decide where to start the itinerary).

After the end of your itinerary, in each building where a math class is scheduled, there must be a professor expert in math, and in each building where a computer science class is scheduled, there must be a professor expert in computer science.

Devise an itinerary that makes it possible to teach all classes.

Input: The first line contains an integer $$$n$$$ ($$$1\le n \le 2000$$$) — the number of buildings in the campus.

The second line contains a string of $$$c$$$ of length $$$n$$$ consisting of the characters $$$\texttt{-}$$$, $$$\texttt{C}$$$, $$$\texttt{M}$$$ — the $$$i$$$-th character denotes the subject of the class scheduled in the $$$i$$$-th building. $$$\texttt{C}$$$ stands for computer science, $$$\texttt{M}$$$ stands for mathematics, while $$$\texttt{-}$$$ means that there is no class scheduled in the $$$i$$$-th building.

The third line contains a string $$$p$$$ of length $$$n$$$ consisting of the characters $$$\texttt{-}$$$, $$$\texttt{C}$$$, $$$\texttt{M}$$$ — the $$$i$$$-th character denotes the expertise of the professor in the $$$i$$$-th building (if there is a professor). $$$\texttt{C}$$$ stands for computer science, $$$\texttt{M}$$$ stands for mathematics, while $$$\texttt{-}$$$ means that there is no professor in the $$$i$$$-th building.

It is guaranteed that, for all tests given to your program, there exists a valid itinerary.

Output: In the first line print an integer $$$l$$$ — the number of operations in your chosen itinerary.

The $$$i$$$-th ($$$1 \leq i \leq l$$$) of the next $$$l$$$ lines must contain one of three commands:

1. $$$\texttt{DRIVE } x$$$ — go to the building with the number $$$x$$$ ($$$1 \leq x \leq n$$$);
2. $$$\texttt{PICKUP}$$$ — pick up the professor which was initially at the current building;
3. $$$\texttt{DROPOFF}$$$ — drop off the passenger professor at the current building.

In order for the itinerary to be valid, the following conditions must hold:

1. No two $$$\texttt{DRIVE}$$$ instructions should go to the same building;
2. At most one $$$\texttt{DROPOFF}$$$ and one $$$\texttt{PICKUP}$$$ instruction in this order should be issued at each specific building;
3. For all $$$\texttt{PICKUP}$$$ instructions, there must be a professor initially at the building, as well as no one already riding along on the scooter;
4. For all $$$\texttt{DROPOFF}$$$ instructions, there must be a professor riding along at the time of the command;
5. After the itinerary, in each building, if there is a class in that building, there must be a professor expert in the class' subject (either initially, or because they were dropped off there).

Note that, in particular, you cannot pick up a professor that you just dropped off for an itinerary to be valid.

Examples:
input:
3
CM-
-CM
output:
7
DRIVE 3
PICKUP
DRIVE 2
DROPOFF
PICKUP
DRIVE 1
DROPOFF

input:
1
C
C
output:
0

input:
2
-M
MC
output:
4
DRIVE 1
PICKUP
DRIVE 2
DROPOFF

Note: In the first sample, You start by driving to building number $$$3$$$. You then pick up the mathematics professor. After dropping him off at building number $$$2$$$, where a mathematics class is being held, you pick up the computer science professor from there, and drop her off at building number $$$1$$$, finishing your itinerary.

Buggy Code:
#include <algorithm>
#include <assert.h>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cstring>
#include <vector>
#include <set>
#include <unordered_set>
#include <utility>
#include <limits.h>
#include <climits>
#include <numeric>
#include <map>
#include <unordered_map>
#include <cfloat>
#define ll long long
#define int long long
#define DEBUG false
using namespace std;

void debug(string s) {
    if (DEBUG) cout << s << endl;
}

void debug(int s) {
    if (DEBUG) cout << s << endl;
}

void debug(vector<vector<int> > & v) {
    if (!DEBUG) return;
    for (int i = 0; i != v.size(); ++i) {
        for (int j = 0; j != v[i].size(); ++j) {
            cout << v[i][j] << " ";
        }
        cout << endl;
    }
}

//flat debug of vector of vectors
void fdebug(vector<vector<int> > & v) {
    if (!DEBUG) return;
    for (int i = 0; i != v.size(); ++i) {
        for (int j = 0; j != v[i].size(); ++j) {
            cout << v[i][j];
            if (j != v[i].size() - 1) {
                cout << " ";
            }
        }
        cout << "; ";
    }
    cout << endl;
}

void debug(vector<int> & v) {
    if (!DEBUG) return;
    for (int i = 0; i != v.size(); ++i) {
        cout << v[i] << " ";
    }
    cout << endl;
}

class Solution {
    int n;
    string has;
    string need;
public:
    void init() {
        cin >> n;
        cin >> need;
        cin >> has;
    }

    int tr(char ch) {
        if (ch == '-') {
            return 0;
        } else if (ch == 'M') {
            return 1;
        }
        // ch == 'C'
        return 2;
    }

    ll solve() {
        init();
        vector<int> cnt_need(3, 0);
        vector<int> cnt_has(3, 0);
        for (int i = 0; i != n; ++i) {
            if (need[i] == has[i]) continue;
            ++cnt_need[tr(need[i])];
            ++cnt_has[tr(has[i])];
        }
        vector<bool> keep(n, true);
        for (int i = 0; i != n; ++i) {
            if (cnt_has[1] > cnt_need[1]) {
                if ((has[i] == 'M') and (need[i] == '-')) {
                    --cnt_has[1];
                    keep[i] = false;
                }
            }
            if (cnt_has[2] > cnt_need[2]) {
                if ((has[i] == 'C') and (need[i] == '-')) {
                    --cnt_has[2];
                    keep[i] = false;
                }
            }
            if (has[i] == need[i]) keep[i] = false;
        }
        if (count(keep.begin(), keep.end(), true) == 0) {
            cout << "0\n";
            return 0;
        }
        if ((cnt_need[1] > cnt_has[1]) or (cnt_need[2] > cnt_has[2]) or (cnt_has[0] == 0)) {
            cout << "0\n";
            return 0;
        }

        vector<vector<int>> m(3, vector<int>(3, 0));
        vector<vector<vector<int>>> ind(3, vector<vector<int>>(3));
        int cnt(0);
        for (int i = 0; i != n; ++i) {
            if (!keep[i]) continue;
            ++m[tr(need[i])][tr(has[i])];
            ind[tr(need[i])][tr(has[i])].push_back(i);
            ++cnt;
        }
        debug(m);

        vector<string> res;
        vector<int> vis;
        int cur = 0;
        while (cnt != 0) {
            for (int i = 0; i != 3; ++i) {
                if (m[cur][i] > 0) {
                    --cnt;
                    --m[cur][i];
                    res.push_back("DRIVE");
                    int next = ind[cur][i][m[cur][i]];
                    vis.push_back(next);
                    if (cur != 0) {
                        res.push_back("DROPOFF");
                        vis.push_back(-1);
                    }
                    if (next != 0) {
                        res.push_back("PICKUP");
                        vis.push_back(-1);
                    }
                    cur = i;
                }
            }
        }
        cout << res.size() << "\n";
        for (int i = 0; i != res.size(); ++i) {
            cout << res[i] << " ";
            if (vis[i] != -1) cout << vis[i] + 1;
            cout << "\n";
        }

        return 0;
    }

};

signed main()
{
    std::ios_base::sync_with_stdio (false);
    std::cin.tie(NULL);
    Solution sol;
    sol.solve();

    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases that might cause the algorithm to fail. The code has several potential issues, such as:

1. **Incorrect handling of the `keep` array**: The `keep` array is used to determine which buildings to visit, but the logic for setting `keep[i]` to `false` might be flawed.
2. **Incorrect handling of `cnt_need` and `cnt_has`**: The code might not correctly balance the number of professors needed with the number of professors available.
3. **Incorrect handling of `DRIVE`, `PICKUP`, and `DROPOFF` commands**: The logic for generating these commands might not be robust enough to handle all edge cases.

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

### Test Input
```plaintext
4
M-CM
-CMM
```

### Explanation
- **Building 1**: Needs a math professor, has no professor.
- **Building 2**: Needs no class, has a computer science professor.
- **Building 3**: Needs a math professor, has a math professor.
- **Building 4**: Needs a computer science professor, has a math professor.

### Expected Behavior
- **Building 1**: The math professor from Building 4 should be picked up and dropped off here.
- **Building 2**: The computer science professor should be picked up.
- **Building 3**: The math professor should be dropped off here (if not already there).
- **Building 4**: The computer science professor should be dropped off here.

### Why This Input is Challenging
- **Building 1 and Building 4**: This requires the scooter to pick up a professor from one building and drop them off in another, which tests the `PICKUP` and `DROPOFF` logic.
- **Building 2 and Building 3**: This tests the handling of buildings with no class and buildings with the correct professor already present.
- **Multiple Professors**: This tests the handling of multiple professors and the correct balancing of `cnt_need` and `cnt_has`.

This input should expose any issues in the logic for handling the `keep` array, the `DRIVE`, `PICKUP`, and `DROPOFF` commands, and the balancing of professors.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scooter
time_limit_ms: 2000
memory_limit_mb: 512
Description: The Czech Technical University campus consists of $$$n$$$ buildings, indexed from $$$1$$$ to $$$n$$$. In each building, there can be a math class scheduled, or a computer science class, or neither (but not both). Additionally, in each building, there is at most one professor, and each professor is either an expert in mathematics or in computer science.

As an intern at University Express Inc., your job is to quickly transport the professors to their classes. For this, you have been granted a brand new two-person scooter, able to accommodate yourself, plus at most one passenger.

Initially, you are the only person on the scooter. When you arrive at a building, you may drop off or pick up professors to/from that building. However, in order to improve the efficiency of your task, you are allowed to drive to each of the $$$n$$$ buildings at most once, in the order of your choice (you can also decide where to start the itinerary).

After the end of your itinerary, in each building where a math class is scheduled, there must be a professor expert in math, and in each building where a computer science class is scheduled, there must be a professor expert in computer science.

Devise an itinerary that makes it possible to teach all classes.

Input: The first line contains an integer $$$n$$$ ($$$1\le n \le 2000$$$) — the number of buildings in the campus.

The second line contains a string of $$$c$$$ of length $$$n$$$ consisting of the characters $$$\texttt{-}$$$, $$$\texttt{C}$$$, $$$\texttt{M}$$$ — the $$$i$$$-th character denotes the subject of the class scheduled in the $$$i$$$-th building. $$$\texttt{C}$$$ stands for computer science, $$$\texttt{M}$$$ stands for mathematics, while $$$\texttt{-}$$$ means that there is no class scheduled in the $$$i$$$-th building.

The third line contains a string $$$p$$$ of length $$$n$$$ consisting of the characters $$$\texttt{-}$$$, $$$\texttt{C}$$$, $$$\texttt{M}$$$ — the $$$i$$$-th character denotes the expertise of the professor in the $$$i$$$-th building (if there is a professor). $$$\texttt{C}$$$ stands for computer science, $$$\texttt{M}$$$ stands for mathematics, while $$$\texttt{-}$$$ means that there is no professor in the $$$i$$$-th building.

It is guaranteed that, for all tests given to your program, there exists a valid itinerary.

Output: In the first line print an integer $$$l$$$ — the number of operations in your chosen itinerary.

The $$$i$$$-th ($$$1 \leq i \leq l$$$) of the next $$$l$$$ lines must contain one of three commands:

1. $$$\texttt{DRIVE } x$$$ — go to the building with the number $$$x$$$ ($$$1 \leq x \leq n$$$);
2. $$$\texttt{PICKUP}$$$ — pick up the professor which was initially at the current building;
3. $$$\texttt{DROPOFF}$$$ — drop off the passenger professor at the current building.

In order for the itinerary to be valid, the following conditions must hold:

1. No two $$$\texttt{DRIVE}$$$ instructions should go to the same building;
2. At most one $$$\texttt{DROPOFF}$$$ and one $$$\texttt{PICKUP}$$$ instruction in this order should be issued at each specific building;
3. For all $$$\texttt{PICKUP}$$$ instructions, there must be a professor initially at the building, as well as no one already riding along on the scooter;
4. For all $$$\texttt{DROPOFF}$$$ instructions, there must be a professor riding along at the time of the command;
5. After the itinerary, in each building, if there is a class in that building, there must be a professor expert in the class' subject (either initially, or because they were dropped off there).

Note that, in particular, you cannot pick up a professor that you just dropped off for an itinerary to be valid.

Examples:
input:
3
CM-
-CM
output:
7
DRIVE 3
PICKUP
DRIVE 2
DROPOFF
PICKUP
DRIVE 1
DROPOFF

input:
1
C
C
output:
0

input:
2
-M
MC
output:
4
DRIVE 1
PICKUP
DRIVE 2
DROPOFF

Note: In the first sample, You start by driving to building number $$$3$$$. You then pick up the mathematics professor. After dropping him off at building number $$$2$$$, where a mathematics class is being held, you pick up the computer science professor from there, and drop her off at building number $$$1$$$, finishing your itinerary.

Buggy Code:
def eular(neigh,start):

    seq = []
    stack = []
    stack.append(start)
    while stack:
        index = stack[-1]
        if len(neigh[index]) > 0:
            nextindex = neigh[index].pop()
            stack.append(nextindex)
        else:
            seq.append(index)
            stack.pop()
    return seq[::-1]

def getresult(index, s1,s2, ans):

    ans.append(["DRIVE", index+1])
    if s1[index]!='-':
        ans.append(["DROPOFF"])
    if s2[index]!='-':
        ans.append(["PICKUP"])
    return

def main():
    n = int(input())
    s1 = input()
    s2 = input()

    if s1==s2:
        print(0)
        return

    neigh = [[] for _ in range(n)]

    kind = {"-M":[], "-C":[], "MC":[], "CM":[], "C-":[], "M-":[]}

    count = 0
    for i in range(n):
        c = s1[i]+s2[i]
        if c in kind:
            kind[c].append(i)
            count += 1

    indegree = [0]*n

    for key1 in kind:
        for key2 in kind:
            if key1[1]==key2[0] and key1[1]!='-':
                for index1 in kind[key1]:
                    for index2 in kind[key2]:
                        neigh[index1].append(index2)
                        indegree[index2] += 1

    path = []
    visited = [False]*n

    for i in range(n):
        if s1[i]==s2[i]:  visited[i] = True

    while len(path) < count:

        for i in range(n):
            if visited[i]:  continue
            if indegree[i] > 0:  continue

            start = i
            break

        res = eular(neigh, start)
        path.extend(res)

    ans = []
    for index in path:
        getresult(index,s1,s2,ans)

    print(len(ans))
    for ele in ans:
        print(*ele)

main()

**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
M-C
C-M
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scooter
time_limit_ms: 2000
memory_limit_mb: 512
Description: The Czech Technical University campus consists of $$$n$$$ buildings, indexed from $$$1$$$ to $$$n$$$. In each building, there can be a math class scheduled, or a computer science class, or neither (but not both). Additionally, in each building, there is at most one professor, and each professor is either an expert in mathematics or in computer science.

As an intern at University Express Inc., your job is to quickly transport the professors to their classes. For this, you have been granted a brand new two-person scooter, able to accommodate yourself, plus at most one passenger.

Initially, you are the only person on the scooter. When you arrive at a building, you may drop off or pick up professors to/from that building. However, in order to improve the efficiency of your task, you are allowed to drive to each of the $$$n$$$ buildings at most once, in the order of your choice (you can also decide where to start the itinerary).

After the end of your itinerary, in each building where a math class is scheduled, there must be a professor expert in math, and in each building where a computer science class is scheduled, there must be a professor expert in computer science.

Devise an itinerary that makes it possible to teach all classes.

Input: The first line contains an integer $$$n$$$ ($$$1\le n \le 2000$$$) — the number of buildings in the campus.

The second line contains a string of $$$c$$$ of length $$$n$$$ consisting of the characters $$$\texttt{-}$$$, $$$\texttt{C}$$$, $$$\texttt{M}$$$ — the $$$i$$$-th character denotes the subject of the class scheduled in the $$$i$$$-th building. $$$\texttt{C}$$$ stands for computer science, $$$\texttt{M}$$$ stands for mathematics, while $$$\texttt{-}$$$ means that there is no class scheduled in the $$$i$$$-th building.

The third line contains a string $$$p$$$ of length $$$n$$$ consisting of the characters $$$\texttt{-}$$$, $$$\texttt{C}$$$, $$$\texttt{M}$$$ — the $$$i$$$-th character denotes the expertise of the professor in the $$$i$$$-th building (if there is a professor). $$$\texttt{C}$$$ stands for computer science, $$$\texttt{M}$$$ stands for mathematics, while $$$\texttt{-}$$$ means that there is no professor in the $$$i$$$-th building.

It is guaranteed that, for all tests given to your program, there exists a valid itinerary.

Output: In the first line print an integer $$$l$$$ — the number of operations in your chosen itinerary.

The $$$i$$$-th ($$$1 \leq i \leq l$$$) of the next $$$l$$$ lines must contain one of three commands:

1. $$$\texttt{DRIVE } x$$$ — go to the building with the number $$$x$$$ ($$$1 \leq x \leq n$$$);
2. $$$\texttt{PICKUP}$$$ — pick up the professor which was initially at the current building;
3. $$$\texttt{DROPOFF}$$$ — drop off the passenger professor at the current building.

In order for the itinerary to be valid, the following conditions must hold:

1. No two $$$\texttt{DRIVE}$$$ instructions should go to the same building;
2. At most one $$$\texttt{DROPOFF}$$$ and one $$$\texttt{PICKUP}$$$ instruction in this order should be issued at each specific building;
3. For all $$$\texttt{PICKUP}$$$ instructions, there must be a professor initially at the building, as well as no one already riding along on the scooter;
4. For all $$$\texttt{DROPOFF}$$$ instructions, there must be a professor riding along at the time of the command;
5. After the itinerary, in each building, if there is a class in that building, there must be a professor expert in the class' subject (either initially, or because they were dropped off there).

Note that, in particular, you cannot pick up a professor that you just dropped off for an itinerary to be valid.

Examples:
input:
3
CM-
-CM
output:
7
DRIVE 3
PICKUP
DRIVE 2
DROPOFF
PICKUP
DRIVE 1
DROPOFF

input:
1
C
C
output:
0

input:
2
-M
MC
output:
4
DRIVE 1
PICKUP
DRIVE 2
DROPOFF

Note: In the first sample, You start by driving to building number $$$3$$$. You then pick up the mathematics professor. After dropping him off at building number $$$2$$$, where a mathematics class is being held, you pick up the computer science professor from there, and drop her off at building number $$$1$$$, finishing your itinerary.

Buggy Code:
def main():
    # Leer la cantidad de edificios en el campus
    n = int(input())

    # Leer el estado actual de los profesores (p) y las clases asignadas (c)
    p = input().strip()  # Profesores (C: Ciencias de la Computación, M: Matemáticas, -: vacío)
    c = input().strip()  # Clases (C: Ciencias de la Computación, M: Matemáticas, -: vacío)

    # Crear 6 grupos para clasificar los edificios según su estado
    g = [[] for _ in range(6)]

    # Recorrer cada edificio para clasificarlo según el desajuste entre profesor y clase
    for i in range(n):
        if c[i] == p[i]:
            continue  # Si el profesor ya está en su clase correcta, no hacemos nada
        if c[i] == 'C':
            # Si se necesita un profesor de ciencias de la computación en el edificio
            g[p[i] == '-' and 0 or 4].append(i + 1)
        elif c[i] == 'M':
            # Si se necesita un profesor de matemáticas en el edificio
            g[p[i] == '-' and 3 or 1].append(i + 1)
        elif c[i] == '-':
            # Si no se necesita profesor en este edificio
            g[p[i] == 'C' and 5 or 2].append(i + 1)

    # Lista para almacenar los movimientos que vamos a realizar
    ans = []

    # Función auxiliar para agregar un movimiento (put) de un profesor en un edificio
    def put(i, ty):
        assert len(g[i]) > 0
        ans.append((g[i].pop(), ty))  # Se saca el último profesor del grupo y se añade el tipo de operación

    # Empezamos a balancear el número de profesores en los edificios según el desajuste
    if len(g[1]) > len(g[4]):  # Más profesores de matemáticas necesitan ser colocados
        put(0, 1)  # Mover un profesor de un edificio vacío (g[0])
        while len(g[4]):  # Mientras haya edificios que necesiten profesores de ciencias de la computación
            put(1, 3)  # Llevar profesores de matemáticas donde hacen falta
            put(4, 3)  # Llevar profesores de ciencias de la computación donde hacen falta
        put(1, 3)  # Último ajuste para completar el ciclo
        if len(g[2]):
            put(2, 2)  # Ajustar profesores de ciencias de la computación donde no hace falta
        ans[-1] = (ans[-1][0], 2)  # Última operación es un DROPOFF

        while len(g[1]):  # Repetir el proceso para todos los edificios desajustados
            put(0, 1)
            put(1, 3)
            if len(g[2]):
                put(2, 2)
            ans[-1] = (ans[-1][0], 2)
        while len(g[5]):
            put(0, 1)
            put(5, 2)
        while len(g[2]):
            put(3, 1)
            put(2, 2)

    elif len(g[1]) < len(g[4]):  # Más profesores de ciencias de la computación necesitan ser colocados
        put(3, 1)
        while len(g[1]):
            put(4, 3)
            put(1, 3)
        put(4, 3)
        if len(g[5]):
            put(5, 2)
        ans[-1] = (ans[-1][0], 2)

        while len(g[4]):
            put(3, 1)
            put(4, 3)
            if len(g[5]):
                put(5, 2)
            ans[-1] = (ans[-1][0], 2)
        while len(g[2]):
            put(3, 1)
            put(2, 2)
        while len(g[5]):
            put(0, 1)
            put(5, 2)

    else:  # Caso donde hay el mismo número de profesores desajustados
        if len(g[1]):
            if len(g[0]):
                put(0, 1)
                while len(g[1]):
                    put(1, 3)
                    put(4, 3)
                if len(g[5]):
                    put(5, 2)
                ans[-1] = (ans[-1][0], 2)
            else:
                assert len(g[3])
                put(3, 1)
                while len(g[4]):
                    put(4, 3)
                    put(1, 3)
                if len(g[2]):
                    put(2, 2)
                ans[-1] = (ans[-1][0], 2)

        while len(g[2]):
            put(3, 1)
            put(2, 2)
        while len(g[5]):
            put(0, 1)
            put(5, 2)

    # Generar la lista de comandos según los movimientos realizados
    cmds = []
    for v, c in ans:
        cmds.append(f"DRIVE {v}")
        if c // 2:
            cmds.append("DROPOFF")
        if c % 2:
            cmds.append("PICKUP")

    # Imprimir la cantidad de movimientos y los comandos correspondientes
    print(len(cmds))
    for cmd in cmds:
        print(cmd)

# Ejecutar el programa
if __name__ == "__main__":
    main()

**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 algorithm to fail. Here are the steps to think through the problem:

1. **Identify the Bug**: The buggy code has issues with balancing the number of professors in the buildings. It may fail to correctly handle cases where the number of professors needed for each subject is not balanced.

2. **Consider Edge Cases**:
   - **No professors in buildings with classes**: Ensure there are buildings with classes but no professors.
   - **No classes in buildings with professors**: Ensure there are buildings with professors but no classes.
   - **Unbalanced number of professors**: Ensure there are more professors of one subject than the other.
   - **Single building scenarios**: Test with a single building to ensure the algorithm handles it correctly.

3. **Generate a Test Input**:
   - **Small Scale**: Since we need a small-scale test input, we can use a few buildings.
   - **Complete and Valid**: Ensure the input is valid and follows the problem constraints.

Let's create a test input that includes a mix of buildings with and without professors and classes, and an unbalanced number of professors.

```plaintext
4
CM-C
M-CM
```

### Explanation:
- **Building 1**: Class = Math, Professor = Math
- **Building 2**: Class = Computer Science, Professor = None
- **Building 3**: Class = None, Professor = Computer Science
- **Building 4**: Class = None, Professor = Math

This test input is designed to:
- Test the handling of buildings with classes but no professors.
- Test the handling of buildings with professors but no classes.
- Test the handling of unbalanced numbers of professors (more Math professors than Computer Science professors).

This should expose any issues in the buggy code related to balancing the number of professors and ensuring all classes are covered.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scooter
time_limit_ms: 2000
memory_limit_mb: 512
Description: The Czech Technical University campus consists of $$$n$$$ buildings, indexed from $$$1$$$ to $$$n$$$. In each building, there can be a math class scheduled, or a computer science class, or neither (but not both). Additionally, in each building, there is at most one professor, and each professor is either an expert in mathematics or in computer science.

As an intern at University Express Inc., your job is to quickly transport the professors to their classes. For this, you have been granted a brand new two-person scooter, able to accommodate yourself, plus at most one passenger.

Initially, you are the only person on the scooter. When you arrive at a building, you may drop off or pick up professors to/from that building. However, in order to improve the efficiency of your task, you are allowed to drive to each of the $$$n$$$ buildings at most once, in the order of your choice (you can also decide where to start the itinerary).

After the end of your itinerary, in each building where a math class is scheduled, there must be a professor expert in math, and in each building where a computer science class is scheduled, there must be a professor expert in computer science.

Devise an itinerary that makes it possible to teach all classes.

Input: The first line contains an integer $$$n$$$ ($$$1\le n \le 2000$$$) — the number of buildings in the campus.

The second line contains a string of $$$c$$$ of length $$$n$$$ consisting of the characters $$$\texttt{-}$$$, $$$\texttt{C}$$$, $$$\texttt{M}$$$ — the $$$i$$$-th character denotes the subject of the class scheduled in the $$$i$$$-th building. $$$\texttt{C}$$$ stands for computer science, $$$\texttt{M}$$$ stands for mathematics, while $$$\texttt{-}$$$ means that there is no class scheduled in the $$$i$$$-th building.

The third line contains a string $$$p$$$ of length $$$n$$$ consisting of the characters $$$\texttt{-}$$$, $$$\texttt{C}$$$, $$$\texttt{M}$$$ — the $$$i$$$-th character denotes the expertise of the professor in the $$$i$$$-th building (if there is a professor). $$$\texttt{C}$$$ stands for computer science, $$$\texttt{M}$$$ stands for mathematics, while $$$\texttt{-}$$$ means that there is no professor in the $$$i$$$-th building.

It is guaranteed that, for all tests given to your program, there exists a valid itinerary.

Output: In the first line print an integer $$$l$$$ — the number of operations in your chosen itinerary.

The $$$i$$$-th ($$$1 \leq i \leq l$$$) of the next $$$l$$$ lines must contain one of three commands:

1. $$$\texttt{DRIVE } x$$$ — go to the building with the number $$$x$$$ ($$$1 \leq x \leq n$$$);
2. $$$\texttt{PICKUP}$$$ — pick up the professor which was initially at the current building;
3. $$$\texttt{DROPOFF}$$$ — drop off the passenger professor at the current building.

In order for the itinerary to be valid, the following conditions must hold:

1. No two $$$\texttt{DRIVE}$$$ instructions should go to the same building;
2. At most one $$$\texttt{DROPOFF}$$$ and one $$$\texttt{PICKUP}$$$ instruction in this order should be issued at each specific building;
3. For all $$$\texttt{PICKUP}$$$ instructions, there must be a professor initially at the building, as well as no one already riding along on the scooter;
4. For all $$$\texttt{DROPOFF}$$$ instructions, there must be a professor riding along at the time of the command;
5. After the itinerary, in each building, if there is a class in that building, there must be a professor expert in the class' subject (either initially, or because they were dropped off there).

Note that, in particular, you cannot pick up a professor that you just dropped off for an itinerary to be valid.

Examples:
input:
3
CM-
-CM
output:
7
DRIVE 3
PICKUP
DRIVE 2
DROPOFF
PICKUP
DRIVE 1
DROPOFF

input:
1
C
C
output:
0

input:
2
-M
MC
output:
4
DRIVE 1
PICKUP
DRIVE 2
DROPOFF

Note: In the first sample, You start by driving to building number $$$3$$$. You then pick up the mathematics professor. After dropping him off at building number $$$2$$$, where a mathematics class is being held, you pick up the computer science professor from there, and drop her off at building number $$$1$$$, finishing your itinerary.

Buggy Code:
#include<bits/stdc++.h>
#define LL long long
#define LLL __int128
#define uint unsigned
#define ldb long double
#define uLL unsigned long long
using namespace std;
const int N=1e3+5;
int n;
char s[N],t[N];
vector<string>ans;
vector<int>vec[3][3];
inline bool dfs(int k){
    if(!vec[3-k][k].empty()){
        auto x=vec[3-k][k].back();
        vec[3-k][k].pop_back();
        if(!dfs(3-k))return vec[3-k][k].emplace_back(x),0;
        ans.emplace_back("DRIVE "+to_string(x));
        ans.emplace_back("DROPOFF");
        ans.emplace_back("PICKUP");
        return 1;
    }
    else{
        if(vec[0][k].empty())return 0;
        auto x=vec[0][k].back();
        vec[0][k].pop_back();
        ans.emplace_back("DRIVE "+to_string(x));
        ans.emplace_back("PICKUP");
        return 1;
    }
}
signed main(){
    cin.tie(0)->sync_with_stdio(0);
    cin>>n>>(s+1)>>(t+1);
    for(int i=1;i<=n;++i)
        vec[s[i]=='M'?2:s[i]=='C'][t[i]=='M'?2:t[i]=='C'].emplace_back(i);
    for(auto k:{1,2})while(!vec[k][0].empty()){
        auto x=vec[k][0].back();
        vec[k][0].pop_back();
        dfs(k);
        ans.emplace_back("DRIVE "+to_string(x));
        ans.emplace_back("DROPOFF");
    }
    for(auto k:{1,2})while(!vec[k][3-k].empty()){
        auto x=vec[k][3-k].back();
        vec[k][3-k].pop_back();
        if(!dfs(k)){
            vec[k][3-k].emplace_back(x);
            break;
        }
        ans.emplace_back("DRIVE "+to_string(x));
        ans.emplace_back("DROPOFF");
    }
    cout<<ans.size()<<'\n';
    for(auto s:ans)cout<<s<<'\n';
    return 0;

}
/*

*/

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

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

Think step by step.

Response:
```plaintext
3
M-C
-CM
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Scooter
time_limit_ms: 2000
memory_limit_mb: 512
Description: The Czech Technical University campus consists of $$$n$$$ buildings, indexed from $$$1$$$ to $$$n$$$. In each building, there can be a math class scheduled, or a computer science class, or neither (but not both). Additionally, in each building, there is at most one professor, and each professor is either an expert in mathematics or in computer science.

As an intern at University Express Inc., your job is to quickly transport the professors to their classes. For this, you have been granted a brand new two-person scooter, able to accommodate yourself, plus at most one passenger.

Initially, you are the only person on the scooter. When you arrive at a building, you may drop off or pick up professors to/from that building. However, in order to improve the efficiency of your task, you are allowed to drive to each of the $$$n$$$ buildings at most once, in the order of your choice (you can also decide where to start the itinerary).

After the end of your itinerary, in each building where a math class is scheduled, there must be a professor expert in math, and in each building where a computer science class is scheduled, there must be a professor expert in computer science.

Devise an itinerary that makes it possible to teach all classes.

Input: The first line contains an integer $$$n$$$ ($$$1\le n \le 2000$$$) — the number of buildings in the campus.

The second line contains a string of $$$c$$$ of length $$$n$$$ consisting of the characters $$$\texttt{-}$$$, $$$\texttt{C}$$$, $$$\texttt{M}$$$ — the $$$i$$$-th character denotes the subject of the class scheduled in the $$$i$$$-th building. $$$\texttt{C}$$$ stands for computer science, $$$\texttt{M}$$$ stands for mathematics, while $$$\texttt{-}$$$ means that there is no class scheduled in the $$$i$$$-th building.

The third line contains a string $$$p$$$ of length $$$n$$$ consisting of the characters $$$\texttt{-}$$$, $$$\texttt{C}$$$, $$$\texttt{M}$$$ — the $$$i$$$-th character denotes the expertise of the professor in the $$$i$$$-th building (if there is a professor). $$$\texttt{C}$$$ stands for computer science, $$$\texttt{M}$$$ stands for mathematics, while $$$\texttt{-}$$$ means that there is no professor in the $$$i$$$-th building.

It is guaranteed that, for all tests given to your program, there exists a valid itinerary.

Output: In the first line print an integer $$$l$$$ — the number of operations in your chosen itinerary.

The $$$i$$$-th ($$$1 \leq i \leq l$$$) of the next $$$l$$$ lines must contain one of three commands:

1. $$$\texttt{DRIVE } x$$$ — go to the building with the number $$$x$$$ ($$$1 \leq x \leq n$$$);
2. $$$\texttt{PICKUP}$$$ — pick up the professor which was initially at the current building;
3. $$$\texttt{DROPOFF}$$$ — drop off the passenger professor at the current building.

In order for the itinerary to be valid, the following conditions must hold:

1. No two $$$\texttt{DRIVE}$$$ instructions should go to the same building;
2. At most one $$$\texttt{DROPOFF}$$$ and one $$$\texttt{PICKUP}$$$ instruction in this order should be issued at each specific building;
3. For all $$$\texttt{PICKUP}$$$ instructions, there must be a professor initially at the building, as well as no one already riding along on the scooter;
4. For all $$$\texttt{DROPOFF}$$$ instructions, there must be a professor riding along at the time of the command;
5. After the itinerary, in each building, if there is a class in that building, there must be a professor expert in the class' subject (either initially, or because they were dropped off there).

Note that, in particular, you cannot pick up a professor that you just dropped off for an itinerary to be valid.

Examples:
input:
3
CM-
-CM
output:
7
DRIVE 3
PICKUP
DRIVE 2
DROPOFF
PICKUP
DRIVE 1
DROPOFF

input:
1
C
C
output:
0

input:
2
-M
MC
output:
4
DRIVE 1
PICKUP
DRIVE 2
DROPOFF

Note: In the first sample, You start by driving to building number $$$3$$$. You then pick up the mathematics professor. After dropping him off at building number $$$2$$$, where a mathematics class is being held, you pick up the computer science professor from there, and drop her off at building number $$$1$$$, finishing your itinerary.

Buggy Code:
// https://codeforces.com/problemset/problem/1949/G
#include <cassert>
#include <iostream>
#include <map>
#include <string>
#include <vector>
using namespace std;

int n;

bool solve(map<string, vector<int>> C, string s = "") {
  if (s.size()) C[s].push_back(-1);
  if ((C["M-"].size() + C["C-"].size() + C["-M"].size() + C["-C"].size()) % 2 != 0) return true;
  int k = (s.size() > 0) * -3;
  for (int cm = 0; cm <= n; cm++) {
    int mc = cm + C["MC"].size() - C["CM"].size();
    int m = C["-M"].size() - mc, c = C["-C"].size() - cm;
    if (m < 0 || mc < 0 || cm < 0 || c < 0 || c + mc != C["C-"].size() || m + cm != C["M-"].size()) continue;
    vector<int> r;
    auto pop = [&](string s) { int i = C[s].back(); C[s].pop_back(); return i; };
    auto spam = [&](string a, string b) {
      while (C[a].size()) {
        r.push_back(pop(a));
        r.push_back(pop(b));
        k += 6;
      }
    };
    while (mc--) {
      r.push_back(pop("-M"));
      r.push_back(pop("MC"));
      if (!cm && !m && !c) spam("MC", "CM");
      r.push_back(pop("C-"));
      k += 7;
      r.push_back(-1);
    }
    while (cm--) {
      r.push_back(pop("-C"));
      r.push_back(pop("CM"));
      if (!m && !c) spam("MC", "CM");
      r.push_back(pop("M-"));
      k += 7;
      r.push_back(-1);
    }
    while (m--) {
      r.push_back(pop("-M"));
      if (!c) spam("MC", "CM");
      r.push_back(pop("M-"));
      k += 4;
      r.push_back(-1);
    }
    while (c--) {
      r.push_back(pop("-C"));
      spam("CM", "MC");
      r.push_back(pop("C-"));
      k += 4;
      r.push_back(-1);
    }
    cout << k << endl;
    /* for (int x : r) cout << x << " "; */
    /* cout << endl; */
    for (int i = 0; i < r.size(); i++) {
      if (r[i] == -1) continue;
      cout << "DRIVE " << r[i] + 1 << endl;
      if (i && r[i - 1] != -1) cout << "DROPOFF\n";
      if (i < r.size() - 1 && r[i + 1] != -1) cout << "PICKUP\n";
    }
    return false;
  }
  return true;
}

int main() {
  cin >> n;
  string a, b; cin >> a >> b;
  map<string, vector<int>> C;
  for (int i = 0; i < n; i++) C[string() + a[i] + b[i]].push_back(i);
  solve(C) && solve(C, "M-") && solve(C, "C-");
  /* for (auto [k, v] : C) cout << k << " " << v.size() << 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:
To expose the bug in the given buggy code, we need to focus on edge cases where the algorithm might fail to correctly handle the distribution of professors and classes. The key points to consider are:

1. **Imbalance in the number of professors and classes**: Ensure that the number of professors and classes for each subject is balanced in a way that might cause the algorithm to make incorrect assumptions.
2. **Order of buildings**: The order in which buildings are visited can significantly impact the algorithm's ability to correctly drop off and pick up professors.
3. **Special cases**: Buildings with no classes or no professors, and buildings with both a class and a professor of the same subject.

Let's construct a small-scale test input that might expose these issues:

```plaintext
4
M-CM
-MMC
```

### Explanation:
- **Building 1**: No class, no professor.
- **Building 2**: Math class, no professor.
- **Building 3**: Computer science class, math professor.
- **Building 4**: Computer science class, computer science professor.

### Why This Test Case?
1. **Building 1**: This building has no class and no professor, which should not affect the algorithm but can be a sanity check.
2. **Building 2**: This building has a math class but no professor. The algorithm must pick up a math professor from another building and drop them off here.
3. **Building 3**: This building has a computer science class and a math professor. The algorithm must pick up the math professor and drop off a computer science professor.
4. **Building 4**: This building has a computer science class and a computer science professor. The algorithm should not need to do anything here, but it must ensure that the professor is not moved unnecessarily.

This test case will challenge the algorithm to correctly manage the distribution of professors and classes, especially in buildings where the subjects do not match. If the algorithm fails to correctly handle this scenario, it will likely expose the bug.