**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sofia and the Lost Operations
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sofia had an array of $n$ integers $a_1, a_2, \ldots, a_n$. One day she got bored with it, so she decided to sequentially apply $m$ modification operations to it.

Each modification operation is described by a pair of numbers $\langle c_j, d_j \rangle$ and means that the element of the array with index $c_j$ should be assigned the value $d_j$, i.e., perform the assignment $a_{c_j} = d_j$. After applying all modification operations sequentially, Sofia discarded the resulting array.

Recently, you found an array of $n$ integers $b_1, b_2, \ldots, b_n$. You are interested in whether this array is Sofia's array. You know the values of the original array, as well as the values $d_1, d_2, \ldots, d_m$. The values $c_1, c_2, \ldots, c_m$ turned out to be lost.

Is there a sequence $c_1, c_2, \ldots, c_m$ such that the sequential application of modification operations $\langle c_1, d_1, \rangle, \langle c_2, d_2, \rangle, \ldots, \langle c_m, d_m \rangle$ to the array $a_1, a_2, \ldots, a_n$ transforms it into the array $b_1, b_2, \ldots, b_n$?

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

Then follow the descriptions of the test cases.

The first line of each test case contains an integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the size of the array.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the original array.

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \le b_i \le 10^9$) — the elements of the found array.

The fourth line contains an integer $m$ ($1 \le m \le 2 \cdot 10^5$) — the number of modification operations.

The fifth line contains $m$ integers $d_1, d_2, \ldots, d_m$ ($1 \le d_j \le 10^9$) — the preserved value for each modification operation.

It is guaranteed that the sum of the values of $n$ for all test cases does not exceed $2 \cdot 10^5$, similarly the sum of the values of $m$ for all test cases does not exceed $2 \cdot 10^5$.

Output: Output $t$ lines, each of which is the answer to the corresponding test case. As an answer, output "YES" if there exists a suitable sequence $c_1, c_2, \ldots, c_m$, and "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).

Examples:
input:
7
3
1 2 1
1 3 2
4
1 3 1 2
4
1 2 3 5
2 1 3 5
2
2 3
5
7 6 1 10 10
3 6 1 11 11
3
4 3 11
4
3 1 7 8
2 2 7 10
5
10 3 2 2 1
5
5 7 1 7 9
4 10 1 2 9
8
1 1 9 8 7 2 10 4
4
1000000000 203 203 203
203 1000000000 203 1000000000
2
203 1000000000
1
1
1
5
1 3 4 5 1
output:
YES
NO
NO
NO
YES
NO
YES

Note:

Buggy Code:
import java.io.*;
import java.util.*;
public class Main {
    public static void main(String[] args) {
        int T=Int();
        while(T-->0){
            int n=Int();
            int[] a=new int[n];
            int[] b=new int[n];
            for(int i=0;i<n;i++) a[i]=Int();
            Map<Integer,Integer> map2=new HashMap<>();
            for(int i=0;i<n;i++) {
                b[i] = Int();
                map2.put(b[i],map2.getOrDefault(b[i],0)+1);
            }
            int m=Int();
            int[] c=new int[m];
            Map<Integer,Integer> map=new HashMap<>();
            for(int i=0;i<m;i++) {
                c[i] = Int();
                map.put(c[i],map.getOrDefault(c[i],0)+1);
            }
            boolean ans=true;
            boolean[] tag=new boolean[n];
            for(int i=0;i<n;i++){
                if(map.containsKey(b[i])&&map.get(b[i])>=map2.get(b[i])){
                    ans=true;
                }else{
                    ans=false;
                    break;
                }
            }

            if(ans){
                pw.println("YES");
            }else{
                pw.println("NO");
            }

        }
        pw.flush();
    }

    private static BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
    private static PrintWriter pw=new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    private static StreamTokenizer st=new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    public static  int Int(){
        try{
            st.nextToken();
        }catch(IOException e){
            e.printStackTrace();
        }
        return (int) st.nval;
    }
    public static  double Dou(){
        try{
            st.nextToken();
        }catch(IOException e){
            e.printStackTrace();
        }
        return st.nval;
    }
    public static  long Lon(){
        try{
            st.nextToken();
        }catch(IOException e){
            e.printStackTrace();
        }
        return (long) st.nval;
    }
    public static String Str(){
        String p="";
        try{
            p=bf.readLine();
        }catch (IOException e){
            e.printStackTrace();
        }
        return 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 generate a challenging test input that exposes the bug in the given code, we need to consider the following points:

1. **Understanding the Bug**: The buggy code incorrectly checks if the values in the `b` array can be achieved by the operations in the `d` array. It only checks if the frequency of each value in `b` is less than or equal to the frequency of the corresponding value in `d`, without considering the order of operations.

2. **Edge Cases**:
   - **Reusing Values**: The code does not handle cases where values in `d` can be reused to achieve the final array `b`.
   - **Order of Operations**: The code does not consider the order in which operations are applied.

3. **Test Input**: We need to create a test input where the values in `b` can be achieved from `a` using the values in `d`, but the order of operations matters. This will expose the bug in the code.

Let's create a small-scale test input:

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

### Explanation:
- **Test Case**: 1
- **Array Size**: 3
- **Original Array (a)**: 1 2 3
- **Target Array (b)**: 3 2 1
- **Number of Operations (m)**: 3
- **Values for Operations (d)**: 3 2 1

### Why This Test Input?
- **Original Array**: [1, 2, 3]
- **Target Array**: [3, 2, 1]
- **Operations**: [3, 2, 1]

The target array can be achieved by applying the operations in the following order:
1. Set `a[0]` to 3 (using `d[0]`).
2. Set `a[2]` to 1 (using `d[2]`).
3. Set `a[1]` to 2 (using `d[1]`).

However, the buggy code will incorrectly check the frequencies and might not consider the order of operations, leading to an incorrect "NO" output.

This test input should expose the bug in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sofia and the Lost Operations
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sofia had an array of $n$ integers $a_1, a_2, \ldots, a_n$. One day she got bored with it, so she decided to sequentially apply $m$ modification operations to it.

Each modification operation is described by a pair of numbers $\langle c_j, d_j \rangle$ and means that the element of the array with index $c_j$ should be assigned the value $d_j$, i.e., perform the assignment $a_{c_j} = d_j$. After applying all modification operations sequentially, Sofia discarded the resulting array.

Recently, you found an array of $n$ integers $b_1, b_2, \ldots, b_n$. You are interested in whether this array is Sofia's array. You know the values of the original array, as well as the values $d_1, d_2, \ldots, d_m$. The values $c_1, c_2, \ldots, c_m$ turned out to be lost.

Is there a sequence $c_1, c_2, \ldots, c_m$ such that the sequential application of modification operations $\langle c_1, d_1, \rangle, \langle c_2, d_2, \rangle, \ldots, \langle c_m, d_m \rangle$ to the array $a_1, a_2, \ldots, a_n$ transforms it into the array $b_1, b_2, \ldots, b_n$?

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

Then follow the descriptions of the test cases.

The first line of each test case contains an integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the size of the array.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the original array.

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \le b_i \le 10^9$) — the elements of the found array.

The fourth line contains an integer $m$ ($1 \le m \le 2 \cdot 10^5$) — the number of modification operations.

The fifth line contains $m$ integers $d_1, d_2, \ldots, d_m$ ($1 \le d_j \le 10^9$) — the preserved value for each modification operation.

It is guaranteed that the sum of the values of $n$ for all test cases does not exceed $2 \cdot 10^5$, similarly the sum of the values of $m$ for all test cases does not exceed $2 \cdot 10^5$.

Output: Output $t$ lines, each of which is the answer to the corresponding test case. As an answer, output "YES" if there exists a suitable sequence $c_1, c_2, \ldots, c_m$, and "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).

Examples:
input:
7
3
1 2 1
1 3 2
4
1 3 1 2
4
1 2 3 5
2 1 3 5
2
2 3
5
7 6 1 10 10
3 6 1 11 11
3
4 3 11
4
3 1 7 8
2 2 7 10
5
10 3 2 2 1
5
5 7 1 7 9
4 10 1 2 9
8
1 1 9 8 7 2 10 4
4
1000000000 203 203 203
203 1000000000 203 1000000000
2
203 1000000000
1
1
1
5
1 3 4 5 1
output:
YES
NO
NO
NO
YES
NO
YES

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define mod 1000000007
#define pb push_back
#define F first
#define S second
#define all(v) v.begin(), v.end()
#define YES cout << "YES\n"
#define NO cout << "NO\n"
#define pp pair<ll, bool>
#define vi vector<ll>
#define vpi vector<pair<ll, ll>>
#define kash ios_base::sync_with_stdio(false)
#define f(i, a, b) for (ll i = a; i < b; i++)
ll func(ll x){
	if(x>0){
		return abs(x);

	}
	return 0;
}
void solve()
{
ll n;
cin>>n;
//original
ll x;
vi a(n);
f(i,0,n){
	cin>>x;
	a[i]=x;
}
//found
vi b(n);
f(i,0,n){
	cin>>x;
	b[i]=x;
}

// sort(b.begin(),b.end());
ll m;
cin>>m;
vi d(m);
f(i,0,m){
	cin>>d[i];
}
vi d2(m);
f(i,0,m){
	d2[i]=d[i];
}
sort(d.begin(),d.end());
vi ans(m,0);
int f=0;
ll i=0, j=0,k=-1;
ll end=-1;
while(i<n && d.size()>=0 && k<m){
	if(a[i]==b[i]){
		i++;
		continue;
	}
	else{
		if(lower_bound(d.begin(),d.end(),b[i])!=d.end() && d[lower_bound(d.begin(),d.end(),b[i])-d.begin()]==b[i]){
			// if(find(d.begin(),d.end(),b[i])-d.begin()>k){
				// k=find(d.begin(),d.end(),b[i])-d.begin();
				d.erase(lower_bound(d.begin(),d.end(),b[i]));
			//j++;
			i++;

		}
		else{
			f=-1;
			break;

		}

	}
}
// 	auto it=find(b.begin(),b.end(),d[j]);
// 	if(it!=b.end()){
// 		auto i=it-b.begin();
// 		if(b[i]==a[i]){
// 			//b.erase(find(b.begin(),b.end(),d[j]));
// 			j++;
// 			continue;
// 		}
// 		else{
// 			j++;
// 		}

// 	}
// 	else{
// 		f=-1;
// 		break;
// 	}
// }

set<ll> s;
if(f==0 && i==n){
	for(ll i=d2.size()-1;i>=0;i--){
		ll x=d2[i];
		if(find(b.begin(),b.end(),d2[i])!=b.end()  ){
			cout<<"YES"<<endl;
			return;

		}
		else{
			cout<<"NO"<<endl;
			return;
		}

	}
}

else{
	cout<<"NO"<<endl;
 }

}

int main()
{
    kash;
    ll t;
    cin >> t;
    // t = 1;
    while (t--)
    {
        solve();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sofia and the Lost Operations
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sofia had an array of $n$ integers $a_1, a_2, \ldots, a_n$. One day she got bored with it, so she decided to sequentially apply $m$ modification operations to it.

Each modification operation is described by a pair of numbers $\langle c_j, d_j \rangle$ and means that the element of the array with index $c_j$ should be assigned the value $d_j$, i.e., perform the assignment $a_{c_j} = d_j$. After applying all modification operations sequentially, Sofia discarded the resulting array.

Recently, you found an array of $n$ integers $b_1, b_2, \ldots, b_n$. You are interested in whether this array is Sofia's array. You know the values of the original array, as well as the values $d_1, d_2, \ldots, d_m$. The values $c_1, c_2, \ldots, c_m$ turned out to be lost.

Is there a sequence $c_1, c_2, \ldots, c_m$ such that the sequential application of modification operations $\langle c_1, d_1, \rangle, \langle c_2, d_2, \rangle, \ldots, \langle c_m, d_m \rangle$ to the array $a_1, a_2, \ldots, a_n$ transforms it into the array $b_1, b_2, \ldots, b_n$?

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

Then follow the descriptions of the test cases.

The first line of each test case contains an integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the size of the array.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the original array.

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \le b_i \le 10^9$) — the elements of the found array.

The fourth line contains an integer $m$ ($1 \le m \le 2 \cdot 10^5$) — the number of modification operations.

The fifth line contains $m$ integers $d_1, d_2, \ldots, d_m$ ($1 \le d_j \le 10^9$) — the preserved value for each modification operation.

It is guaranteed that the sum of the values of $n$ for all test cases does not exceed $2 \cdot 10^5$, similarly the sum of the values of $m$ for all test cases does not exceed $2 \cdot 10^5$.

Output: Output $t$ lines, each of which is the answer to the corresponding test case. As an answer, output "YES" if there exists a suitable sequence $c_1, c_2, \ldots, c_m$, and "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).

Examples:
input:
7
3
1 2 1
1 3 2
4
1 3 1 2
4
1 2 3 5
2 1 3 5
2
2 3
5
7 6 1 10 10
3 6 1 11 11
3
4 3 11
4
3 1 7 8
2 2 7 10
5
10 3 2 2 1
5
5 7 1 7 9
4 10 1 2 9
8
1 1 9 8 7 2 10 4
4
1000000000 203 203 203
203 1000000000 203 1000000000
2
203 1000000000
1
1
1
5
1 3 4 5 1
output:
YES
NO
NO
NO
YES
NO
YES

Note:

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

int t,n,m,a[200010],b[200010],d[200010];

int main(){
	std::ios::sync_with_stdio(false);
	std::cin.tie(0);
	cin>>t;
	while(t--){
		cin>>n;
		for(int i=0;i<n;++i)cin>>a[i];
		for(int i=0;i<n;++i)cin>>b[i];
		cin>>m;
		for(int i=0;i<m;++i)cin>>d[i];
		int q=d[m-1],sum=0;
		for(int i=0;i<n;++i){
			if(b[i]==q)q=-1;
			if(a[i]==b[i]){
				a[i]=-1;b[i]=-1;
				sum++;
			}
		}
		sort(b,b+n);
		for(int i=sum;i<n;++i){
			if(b[i]!=-1){
			for(int j=0;j<m;++j){
				if(b[i]==d[j]){
					b[i]=-1;
					d[j]=-1;
					break;
				}
			}
			}
		}
		int k=0;
		for(int i=0;i<n;++i)if(b[i]!=-1)k=1;
		if(k==0&&q==-1)cout<<"YES"<<endl;
		else cout<<"NO"<<endl;
	}
	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sofia and the Lost Operations
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sofia had an array of $n$ integers $a_1, a_2, \ldots, a_n$. One day she got bored with it, so she decided to sequentially apply $m$ modification operations to it.

Each modification operation is described by a pair of numbers $\langle c_j, d_j \rangle$ and means that the element of the array with index $c_j$ should be assigned the value $d_j$, i.e., perform the assignment $a_{c_j} = d_j$. After applying all modification operations sequentially, Sofia discarded the resulting array.

Recently, you found an array of $n$ integers $b_1, b_2, \ldots, b_n$. You are interested in whether this array is Sofia's array. You know the values of the original array, as well as the values $d_1, d_2, \ldots, d_m$. The values $c_1, c_2, \ldots, c_m$ turned out to be lost.

Is there a sequence $c_1, c_2, \ldots, c_m$ such that the sequential application of modification operations $\langle c_1, d_1, \rangle, \langle c_2, d_2, \rangle, \ldots, \langle c_m, d_m \rangle$ to the array $a_1, a_2, \ldots, a_n$ transforms it into the array $b_1, b_2, \ldots, b_n$?

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

Then follow the descriptions of the test cases.

The first line of each test case contains an integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the size of the array.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the original array.

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \le b_i \le 10^9$) — the elements of the found array.

The fourth line contains an integer $m$ ($1 \le m \le 2 \cdot 10^5$) — the number of modification operations.

The fifth line contains $m$ integers $d_1, d_2, \ldots, d_m$ ($1 \le d_j \le 10^9$) — the preserved value for each modification operation.

It is guaranteed that the sum of the values of $n$ for all test cases does not exceed $2 \cdot 10^5$, similarly the sum of the values of $m$ for all test cases does not exceed $2 \cdot 10^5$.

Output: Output $t$ lines, each of which is the answer to the corresponding test case. As an answer, output "YES" if there exists a suitable sequence $c_1, c_2, \ldots, c_m$, and "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).

Examples:
input:
7
3
1 2 1
1 3 2
4
1 3 1 2
4
1 2 3 5
2 1 3 5
2
2 3
5
7 6 1 10 10
3 6 1 11 11
3
4 3 11
4
3 1 7 8
2 2 7 10
5
10 3 2 2 1
5
5 7 1 7 9
4 10 1 2 9
8
1 1 9 8 7 2 10 4
4
1000000000 203 203 203
203 1000000000 203 1000000000
2
203 1000000000
1
1
1
5
1 3 4 5 1
output:
YES
NO
NO
NO
YES
NO
YES

Note:

Buggy Code:
for t in range(int(input())):
    n = int(input())
    a = [int(i) for i in input().split()]
    b = [int(i) for i in input().split()]
    m = int(input())
    d = [int(i) for i in input().split()]
    c = 0
    l = {}
    for i in range(n):
        if a[i] != b[i]:
            if b[i] not in l:
                l[b[i]] = 0
            l[b[i]] += 1
    for i in range(len(d)):
        if d[i] in b:
            if d[i] in l:
                l[d[i]] -= 1
                if l[d[i]] == 0:
                    l.pop(d[i])
            d[i] = 0
            while c:
                d[i - c] = 0
                c -= 1
            c = 0
        else:
            c += 1
    if d.count(0) == len(d) and not l:
        print("YES")
    else:
        print("NO")

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

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

Think step by step.

Response:
```plaintext
3
3
1 2 3
1 2 4
1
4
4
1 2 3 4
1 2 3 4
0
4
1 2 3 4
4 3 2 1
3
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sofia and the Lost Operations
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sofia had an array of $n$ integers $a_1, a_2, \ldots, a_n$. One day she got bored with it, so she decided to sequentially apply $m$ modification operations to it.

Each modification operation is described by a pair of numbers $\langle c_j, d_j \rangle$ and means that the element of the array with index $c_j$ should be assigned the value $d_j$, i.e., perform the assignment $a_{c_j} = d_j$. After applying all modification operations sequentially, Sofia discarded the resulting array.

Recently, you found an array of $n$ integers $b_1, b_2, \ldots, b_n$. You are interested in whether this array is Sofia's array. You know the values of the original array, as well as the values $d_1, d_2, \ldots, d_m$. The values $c_1, c_2, \ldots, c_m$ turned out to be lost.

Is there a sequence $c_1, c_2, \ldots, c_m$ such that the sequential application of modification operations $\langle c_1, d_1, \rangle, \langle c_2, d_2, \rangle, \ldots, \langle c_m, d_m \rangle$ to the array $a_1, a_2, \ldots, a_n$ transforms it into the array $b_1, b_2, \ldots, b_n$?

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

Then follow the descriptions of the test cases.

The first line of each test case contains an integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the size of the array.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the original array.

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \le b_i \le 10^9$) — the elements of the found array.

The fourth line contains an integer $m$ ($1 \le m \le 2 \cdot 10^5$) — the number of modification operations.

The fifth line contains $m$ integers $d_1, d_2, \ldots, d_m$ ($1 \le d_j \le 10^9$) — the preserved value for each modification operation.

It is guaranteed that the sum of the values of $n$ for all test cases does not exceed $2 \cdot 10^5$, similarly the sum of the values of $m$ for all test cases does not exceed $2 \cdot 10^5$.

Output: Output $t$ lines, each of which is the answer to the corresponding test case. As an answer, output "YES" if there exists a suitable sequence $c_1, c_2, \ldots, c_m$, and "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).

Examples:
input:
7
3
1 2 1
1 3 2
4
1 3 1 2
4
1 2 3 5
2 1 3 5
2
2 3
5
7 6 1 10 10
3 6 1 11 11
3
4 3 11
4
3 1 7 8
2 2 7 10
5
10 3 2 2 1
5
5 7 1 7 9
4 10 1 2 9
8
1 1 9 8 7 2 10 4
4
1000000000 203 203 203
203 1000000000 203 1000000000
2
203 1000000000
1
1
1
5
1 3 4 5 1
output:
YES
NO
NO
NO
YES
NO
YES

Note:

Buggy Code:
def ans():
    n = int(input())
    # l1,f,k = (map(int, input().split()))
    na = list((map(int, input().split())))
    l = list((map(int, input().split())))
    m = int(input())
    x = list((map(int, input().split())))
    p={}
    if x[-1] not in l:
        print('NO')
        return
    for j in range(0,n):
        if na[j]!=l[j]:
            if l[j] not in x:
                print("NO")
                return
            else:
                x.remove(l[j])
    print('YES')
t = int(input())
for _ in range(t):

    (ans())

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sofia and the Lost Operations
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sofia had an array of $n$ integers $a_1, a_2, \ldots, a_n$. One day she got bored with it, so she decided to sequentially apply $m$ modification operations to it.

Each modification operation is described by a pair of numbers $\langle c_j, d_j \rangle$ and means that the element of the array with index $c_j$ should be assigned the value $d_j$, i.e., perform the assignment $a_{c_j} = d_j$. After applying all modification operations sequentially, Sofia discarded the resulting array.

Recently, you found an array of $n$ integers $b_1, b_2, \ldots, b_n$. You are interested in whether this array is Sofia's array. You know the values of the original array, as well as the values $d_1, d_2, \ldots, d_m$. The values $c_1, c_2, \ldots, c_m$ turned out to be lost.

Is there a sequence $c_1, c_2, \ldots, c_m$ such that the sequential application of modification operations $\langle c_1, d_1, \rangle, \langle c_2, d_2, \rangle, \ldots, \langle c_m, d_m \rangle$ to the array $a_1, a_2, \ldots, a_n$ transforms it into the array $b_1, b_2, \ldots, b_n$?

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

Then follow the descriptions of the test cases.

The first line of each test case contains an integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the size of the array.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the original array.

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \le b_i \le 10^9$) — the elements of the found array.

The fourth line contains an integer $m$ ($1 \le m \le 2 \cdot 10^5$) — the number of modification operations.

The fifth line contains $m$ integers $d_1, d_2, \ldots, d_m$ ($1 \le d_j \le 10^9$) — the preserved value for each modification operation.

It is guaranteed that the sum of the values of $n$ for all test cases does not exceed $2 \cdot 10^5$, similarly the sum of the values of $m$ for all test cases does not exceed $2 \cdot 10^5$.

Output: Output $t$ lines, each of which is the answer to the corresponding test case. As an answer, output "YES" if there exists a suitable sequence $c_1, c_2, \ldots, c_m$, and "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).

Examples:
input:
7
3
1 2 1
1 3 2
4
1 3 1 2
4
1 2 3 5
2 1 3 5
2
2 3
5
7 6 1 10 10
3 6 1 11 11
3
4 3 11
4
3 1 7 8
2 2 7 10
5
10 3 2 2 1
5
5 7 1 7 9
4 10 1 2 9
8
1 1 9 8 7 2 10 4
4
1000000000 203 203 203
203 1000000000 203 1000000000
2
203 1000000000
1
1
1
5
1 3 4 5 1
output:
YES
NO
NO
NO
YES
NO
YES

Note:

Buggy Code:
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.*;
public class C_Sofia_and_the_Lost_Operations {

    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while (t-->0) {
            int n=sc.nextInt();
            long[] a=new long[n];
            long[] b=new long[n];
            HashMap<Long,Integer> hmb=new HashMap<>();
            for (int i = 0; i < n; i++) {
                a[i]=sc.nextLong();
            }
            for (int i = 0; i < n; i++) {
                b[i]=sc.nextLong();
                if (a[i]!=b[i]) {
                    if (!hmb.containsKey(b[i])) {
                        hmb.put(b[i],1);
                    }
                    else{
                        int k=hmb.get(b[i]);
                        hmb.put(b[i], ++k);
                    }
                }
            }
            int m=sc.nextInt();
            long[] d=new long[m];
            HashMap<Long,Integer> hm=new HashMap<>();
            for (int i = 0; i < m; i++) {
                d[i]=sc.nextLong();
                if (!hm.containsKey(d[i])) {
                    hm.put(d[i], 1);
                }
                else{
                    int k=hm.get(d[i]);
                    hm.put(d[i], ++k);
                }
            }
            boolean flag=true;
            for (long s:hmb.keySet()) {
                int freq=hmb.get(s);
                int freq2=0;
                if (hm.containsKey(s)) {
                    freq2=hm.get(s);
                }
                else{
                    flag=false;
                    break;
                }
                if (freq>freq2) {
                    flag=false;
                    break;
                }
            }
            if (flag) {
                boolean isThere=false;
                for (int i = 0; i < n; i++) {
                    if (d[m-1]==b[i]) {
                        System.out.println("YES");
                        isThere=true;
                        break;
                    }
                }
                if (!isThere) {
                    System.out.println("NO");
                }
            }
            else{
                System.out.println("NO");
            }
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sofia and the Lost Operations
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sofia had an array of $n$ integers $a_1, a_2, \ldots, a_n$. One day she got bored with it, so she decided to sequentially apply $m$ modification operations to it.

Each modification operation is described by a pair of numbers $\langle c_j, d_j \rangle$ and means that the element of the array with index $c_j$ should be assigned the value $d_j$, i.e., perform the assignment $a_{c_j} = d_j$. After applying all modification operations sequentially, Sofia discarded the resulting array.

Recently, you found an array of $n$ integers $b_1, b_2, \ldots, b_n$. You are interested in whether this array is Sofia's array. You know the values of the original array, as well as the values $d_1, d_2, \ldots, d_m$. The values $c_1, c_2, \ldots, c_m$ turned out to be lost.

Is there a sequence $c_1, c_2, \ldots, c_m$ such that the sequential application of modification operations $\langle c_1, d_1, \rangle, \langle c_2, d_2, \rangle, \ldots, \langle c_m, d_m \rangle$ to the array $a_1, a_2, \ldots, a_n$ transforms it into the array $b_1, b_2, \ldots, b_n$?

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

Then follow the descriptions of the test cases.

The first line of each test case contains an integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the size of the array.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the original array.

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \le b_i \le 10^9$) — the elements of the found array.

The fourth line contains an integer $m$ ($1 \le m \le 2 \cdot 10^5$) — the number of modification operations.

The fifth line contains $m$ integers $d_1, d_2, \ldots, d_m$ ($1 \le d_j \le 10^9$) — the preserved value for each modification operation.

It is guaranteed that the sum of the values of $n$ for all test cases does not exceed $2 \cdot 10^5$, similarly the sum of the values of $m$ for all test cases does not exceed $2 \cdot 10^5$.

Output: Output $t$ lines, each of which is the answer to the corresponding test case. As an answer, output "YES" if there exists a suitable sequence $c_1, c_2, \ldots, c_m$, and "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).

Examples:
input:
7
3
1 2 1
1 3 2
4
1 3 1 2
4
1 2 3 5
2 1 3 5
2
2 3
5
7 6 1 10 10
3 6 1 11 11
3
4 3 11
4
3 1 7 8
2 2 7 10
5
10 3 2 2 1
5
5 7 1 7 9
4 10 1 2 9
8
1 1 9 8 7 2 10 4
4
1000000000 203 203 203
203 1000000000 203 1000000000
2
203 1000000000
1
1
1
5
1 3 4 5 1
output:
YES
NO
NO
NO
YES
NO
YES

Note:

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

public class CP62 {
        public static void main(String args[])
        throws IOException{

            BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
//Scanner sc = new Scanner(System.in);

int t = Integer.parseInt(br.readLine()); //sc.nextInt();

while(t--!=0){

    int n = Integer.parseInt(br.readLine()); //sc.nextInt();

    String a[]; // = new int[n];
    a = br.readLine().split(" ");

     String b[];
    b = br.readLine().split(" ");

    String d[];
   int m = Integer.parseInt(br.readLine()); //sc.nextInt();
    d = br.readLine().split(" ");

    int trans = 0;
        ArrayList<Integer> dCopy = new ArrayList<Integer>();
        for(int i=0; i<m; i++){
            dCopy.add(Integer.valueOf(d[i]));
        }

        boolean ifYes = false;
        //for(int i = 0; i<n; i++){

           // System.out.println(d[m-1]);

           // for(String c : b) System.out.print(c + " ");
           // System.out.println();
           // System.out.println(Arrays.toString(b));
            if(Arrays.asList(b).contains(d[m-1])) ifYes = true;

            //if(d[m-1].equals(b[i]))ifYes = true;

        //}

        if(!ifYes){
            System.out.println("NO");
        }
else {
    boolean isAv= true;
        for(int i=0; i<n; i++){
            isAv = true;
            if(!a[i].equals(b[i])){
                 trans++;
            isAv =  dCopy.remove(Integer.valueOf(b[i]));

            if(!isAv)break;
        }

        }

        if(!isAv || trans>m){
            System.out.println("NO");
        }
        else System.out.println("YES");

        }
    }

}
        }

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the probability of the code failing. Here are the key points to consider:

1. **Different Initial and Final Arrays**: The initial array `a` and the final array `b` should be different to ensure that some modifications are necessary.
2. **Modification Operations**: The number of modification operations `m` should be such that it can potentially transform `a` into `b`, but the code should fail to correctly handle the operations.
3. **Edge Cases**: Consider edge cases where the arrays have repeated elements, or where the final array `b` contains elements that are not in the initial array `a` but are in the modification values `d`.

Given these considerations, let's construct a small-scale test input:

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

### Explanation:

- **Test Case 1**:
  - **Initial Array `a`**: [1, 2, 3]
  - **Final Array `b`**: [3, 2, 1]
  - **Number of Modification Operations `m`**: 3
  - **Modification Values `d`**: [3, 2, 1]

### Why This Test Input is Challenging:

1. **Different Initial and Final Arrays**: The arrays `a` and `b` are different, requiring modifications.
2. **Valid Modification Operations**: The modification values `d` are exactly the elements needed to transform `a` into `b`.
3. **Order of Operations**: The order of operations is crucial. The code should correctly handle the order of modifications to determine if the transformation is possible.

### Expected Output:
The correct output for this test input should be:
```plaintext
YES
```

However, the buggy code might fail to handle the order of operations correctly, leading to an incorrect "NO" output. This test input will help expose the bug in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cutting Game
time_limit_ms: 3000
memory_limit_mb: 256
Description: Alice and Bob were playing a game again. They have a grid of size $a \times b$ ($1 \le a, b \le 10^9$), on which there are $n$ chips, with at most one chip in each cell. The cell at the intersection of the $x$-th row and the $y$-th column has coordinates $(x, y)$.

Alice made the first move, and the players took turns. On each move, a player could cut several (but not all) rows or columns from the beginning or end of the remaining grid and earn a point for each chip that was on the cut part of the grid. Each move can be described by the character 'U', 'D', 'L', or 'R' and an integer $k$:

  * If the character is 'U', then the first $k$ remaining rows will be cut;    * If the character is 'D', then the last $k$ remaining rows will be cut;    * If the character is 'L', then the first $k$ remaining columns will be cut;    * If the character is 'R', then the last $k$ remaining columns will be cut.

Based on the initial state of the grid and the players' moves, determine the number of points earned by Alice and Bob, respectively.

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

The first line of each test case contains four integers $a$, $b$, $n$, and $m$ ($2 \le a, b \le 10^9$, $1 \le n, m \le 2 \cdot 10^5$) — the dimensions of the grid, the number of chips, and the number of moves.

Each of the next $n$ lines contain two integers $x_i$ and $y_i$ ($1 \le x_i \le a$, $1 \le y_i \le b$) — the coordinates of the chips. All pairs of coordinates are distinct.

Each of the next $m$ lines contain a character $c_j$ and an integer $k_j$ — the description of the $j$-th move. It is guaranteed that $k$ is less than the number of rows/columns in the current grid. In other words, a player cannot cut the entire remaining grid on their move.

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

Output: For each test case, output two integers — the number of points earned by Alice and Bob, respectively.

Examples:
input:
6
4 4 3 2
4 1
3 3
2 4
D 2
R 1
4 4 3 3
4 1
3 2
2 3
D 1
L 1
U 2
3 5 3 2
1 3
2 2
3 3
R 2
R 2
6 4 4 2
1 4
2 3
5 3
1 1
R 1
U 1
9 3 2 1
6 1
3 3
D 8
10 10 2 5
7 5
9 1
R 1
L 2
D 1
U 4
D 1
output:
2 1
2 0
0 3
1 1
2 0
0 1

Note: Below is the game from the first example:

![](CDN_BASE_URL/3730e4e7ac2159abc9c10c7ca7091e03)

On her turn, Alice cut $2$ rows from the bottom and scored $2$ points, then Bob cut $1$ column from the right and scored one point. Note that if Bob had cut $1$ row from the bottom, he would have also scored $1$ point.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define _                         \
    ios_base::sync_with_stdio(0); \
    cin.tie(0);
#define fi first
#define se second
typedef long long ll;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3fll;

int main()
{
    int testes; cin >> testes;
    while (testes--)
    {
        ll a,ia,b,ib,n,m;
        cin >> a >> b >> n >> m;
        int bob_points = 0;
        int alice_points = 0;
        bool player = 0; // 0 is Alcie
        bool grid[a][b];
        ia = 0;
        ib = 0;
        for (int i = 0; i < a; i++)
            for (int j = 0; j < b; j++)
                grid[i][j] = false;
        while (n--)
        {
        ll x,y;
        cin >> x >> y;
        grid[x -1][y -1] = true;
        }
        while (m--)
        {
            char c;
            ll k;
            cin >> c >> k;
        switch (c)
        {
        case 'U':
            for (int i = ia; i < k; i++)
                for(int j = ib; j < b; j++)
                    if (grid[i][j] == true)
                    {
                        if(player == 0)
                            alice_points++;
                        else
                            bob_points++;
                        grid[i][j] = false;
                    }
            ia += k;
        break;
        case 'D':
            for (int i= a-1; i > a-1-k; i--)
                for(int j = ib; j < b; j++)
                    if (grid[i][j] == true)
                    {
                        if(player == 0)
                            alice_points++;
                        else
                            bob_points++;
                        grid[i][j] = false;
                    }
            a -= k;
             break;
        case 'L':
        for (int i= ia; i < a; i++)
                for(int j = ib; j < k; j++)
                    if (grid[i][j] == true)
                    {
                        if(player == 0)
                            alice_points++;
                        else
                            bob_points++;
                        grid[i][j] = false;
                    }
        ib  += k;
            break;
        case 'R':
            for (int i= ia; i < a; i++)
                for(int j = b-1; j > b-1-k; j--)
                    if (grid[i][j] == true)
                    {
                        if(player == 0)
                            alice_points++;
                        else
                            bob_points++;
                        grid[i][j] = false;
                    }
            b -= k;
            break;
        }
        player = !player;
        }
        cout << alice_points << " " << bob_points << 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 generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on scenarios that can cause issues in the code's logic. Here are some considerations:

1. **Edge Cases for Grid Dimensions**: Test with the smallest and largest possible values for \(a\) and \(b\).
2. **Edge Cases for Number of Chips and Moves**: Test with the smallest and largest possible values for \(n\) and \(m\).
3. **Grid Overflow**: Ensure that the grid size is not exceeded when cutting rows or columns.
4. **Player Turn Alternation**: Ensure that the player turn alternation is correctly handled.

Given these considerations, let's create a small-scale test input that can expose potential issues:

1. **Test Case 1**: Small grid with a few chips and moves to ensure basic functionality.
2. **Test Case 2**: Grid with a large number of chips and moves to test performance and correctness.
3. **Test Case 3**: Edge case with the maximum grid size and a single chip to test boundary conditions.
4. **Test Case 4**: Test with a move that cuts the entire remaining grid (should be invalid but might expose issues).

Here is the test input:

```plaintext
4
3 3 2 2
1 1
3 3
U 1
R 1
10 10 10 10
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 10
U 1
D 1
L 1
R 1
U 1
D 1
L 1
R 1
U 1
D 1
L 1
R 1
1000000000 1000000000 1 1
1 1
D 1
2 2 1 1
1 1
U 2
```

Explanation:
1. **Test Case 1**: A 3x3 grid with 2 chips and 2 moves. This tests basic functionality.
2. **Test Case 2**: A 10x10 grid with 10 chips and 10 moves. This tests performance and correctness with a larger number of chips and moves.
3. **Test Case 3**: A large grid with 1 chip and 1 move. This tests boundary conditions.
4. **Test Case 4**: A 2x2 grid with 1 chip and 1 move that attempts to cut the entire grid. This tests the invalid move condition.

This test input should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cutting Game
time_limit_ms: 3000
memory_limit_mb: 256
Description: Alice and Bob were playing a game again. They have a grid of size $a \times b$ ($1 \le a, b \le 10^9$), on which there are $n$ chips, with at most one chip in each cell. The cell at the intersection of the $x$-th row and the $y$-th column has coordinates $(x, y)$.

Alice made the first move, and the players took turns. On each move, a player could cut several (but not all) rows or columns from the beginning or end of the remaining grid and earn a point for each chip that was on the cut part of the grid. Each move can be described by the character 'U', 'D', 'L', or 'R' and an integer $k$:

  * If the character is 'U', then the first $k$ remaining rows will be cut;    * If the character is 'D', then the last $k$ remaining rows will be cut;    * If the character is 'L', then the first $k$ remaining columns will be cut;    * If the character is 'R', then the last $k$ remaining columns will be cut.

Based on the initial state of the grid and the players' moves, determine the number of points earned by Alice and Bob, respectively.

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

The first line of each test case contains four integers $a$, $b$, $n$, and $m$ ($2 \le a, b \le 10^9$, $1 \le n, m \le 2 \cdot 10^5$) — the dimensions of the grid, the number of chips, and the number of moves.

Each of the next $n$ lines contain two integers $x_i$ and $y_i$ ($1 \le x_i \le a$, $1 \le y_i \le b$) — the coordinates of the chips. All pairs of coordinates are distinct.

Each of the next $m$ lines contain a character $c_j$ and an integer $k_j$ — the description of the $j$-th move. It is guaranteed that $k$ is less than the number of rows/columns in the current grid. In other words, a player cannot cut the entire remaining grid on their move.

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

Output: For each test case, output two integers — the number of points earned by Alice and Bob, respectively.

Examples:
input:
6
4 4 3 2
4 1
3 3
2 4
D 2
R 1
4 4 3 3
4 1
3 2
2 3
D 1
L 1
U 2
3 5 3 2
1 3
2 2
3 3
R 2
R 2
6 4 4 2
1 4
2 3
5 3
1 1
R 1
U 1
9 3 2 1
6 1
3 3
D 8
10 10 2 5
7 5
9 1
R 1
L 2
D 1
U 4
D 1
output:
2 1
2 0
0 3
1 1
2 0
0 1

Note: Below is the game from the first example:

![](CDN_BASE_URL/3730e4e7ac2159abc9c10c7ca7091e03)

On her turn, Alice cut $2$ rows from the bottom and scored $2$ points, then Bob cut $1$ column from the right and scored one point. Note that if Bob had cut $1$ row from the bottom, he would have also scored $1$ point.

Buggy Code:
#include <bits/stdc++.h>
#define ll long long int
#define endl '\n'
#define M 1000000007
#define eps 1e-6
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
const int N=1e6;
using namespace __gnu_pbds;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
using namespace std;

int main(){
   ios_base::sync_with_stdio(false);
   cin.tie(NULL);
   cout.tie(NULL);
  ll t;
  ll ct = 0;
  cin >> t;
  while(t--){
      ct++;
      ll a, b, n, m, k, ct1 = 0, ct2 = 0;
      cin >> a >> b >> n >> m;
      if(ct == 30){
          cout << a << " " << b << " " << n << " " << m << endl;
      }
      char c;
      vector<pair<ll, ll>> v1(n), v2(n);
      for(ll i = 0 ; i < n ; i++){
        cin >> v1[i].first >> v1[i].second;
        if(ct == 30){
        cout << v1[i].first << " " << v1[i].second << endl;
        }
        v2[i].first = v1[i].second;
        v2[i].second = v1[i].first;
      }
      sort(v1.begin(), v1.end());
      sort(v2.begin(), v2.end());
      ll r1 = 1, r2 = a, c1 = 1, c2 = b, i1 = 0, j1 = n - 1, i2 = 0, j2 = n - 1;
      for(ll i = 0 ; i < m ; i++){
        cin >> c >> k;
        if(ct == 30){
        cout << c << " " << k << endl;
        }
        if(i % 2 == 0){
          if(c == 'U'){
            while(1){
              ll tp1 = v1[i1].first, tp2 = v1[i1].second;
              if(tp1 >= r1 && tp1 <= r2 && tp1 <= r1 + k - 1 && tp2 >= c1 && tp2 <= c2){
                ct1++;
                i1++;
                if(i1 > j1){
                  break;
                }
              }
              else{
                break;
              }
            }
            r1 += k;
            // for(ll j = i1 ; j <= i2 ; j++){
            //   if(j >= r1 && j <= r2 && j <= r1 + k - 1){

            //   }
            //   else{
            //     break;
            //   }
            // }
            // auto it = upper_bound(v1.begin() + r1 - 1, v1.begin() + r1 + k - 1, {r1 + k - 1, })
          }
          else if(c == 'D'){
            while(1){
              ll tp1 = v1[j1].first, tp2 = v1[j1].second;
              if(tp1 >= r1 && tp1 <= r2 && tp1 >= r2 - k + 1 && tp2 >= c1 && tp2 <= c2){
                ct1++;
                j1--;
                if(i1 > j1){
                  break;
                }
              }
              else{
                break;
              }
            }
            r2 -= k;
        }
        else if(c == 'L'){
            while(1){
              ll tp1 = v2[i2].second, tp2 = v2[i2].first;
              if(tp1 >= r1 && tp1 <= r2 && tp2 >= c1 && tp2 <= c2 && tp2 <= c1 + k - 1){
                ct1++;
                i2++;
                if(i2 > j2){
                  break;
                }
              }
              else{
                break;
              }
            }
            c1 += k;
        }
        else if(c == 'R'){
           while(1){
              ll tp1 = v2[j2].second, tp2 = v2[j2].first;
              // cout << r1 << " " << r2 << " " << c1 << " " << c2 << endl;
              // cout << tp1 << " " << tp2 << endl;
              if(tp1 >= r1 && tp1 <= r2 && tp2 >= c1 && tp2 <= c2 && tp2 >= c2 - k + 1){
                ct1++;
                j2--;
                if(i2 > j2){
                  break;
                }
              }
              else{
                break;
              }
            }
            c2 -= k;
        }
        }
        else{
          if(c == 'U'){
            while(1){
              ll tp1 = v1[i1].first, tp2 = v1[i1].second;
              if(tp1 >= r1 && tp1 <= r2 && tp1 <= r1 + k - 1 && tp2 >= c1 && tp2 <= c2){
                ct2++;
                i1++;
                if(i1 > j1){
                  break;
                }
              }
              else{
                break;
              }
            }
            r1 += k;
            // for(ll j = i1 ; j <= i2 ; j++){
            //   if(j >= r1 && j <= r2 && j <= r1 + k - 1){

            //   }
            //   else{
            //     break;
            //   }
            // }
            // auto it = upper_bound(v1.begin() + r1 - 1, v1.begin() + r1 + k - 1, {r1 + k - 1, })
          }
          else if(c == 'D'){
            while(1){
              ll tp1 = v1[j1].first, tp2 = v1[j1].second;
              if(tp1 >= r1 && tp1 <= r2 && tp1 >= r2 - k + 1 && tp2 >= c1 && tp2 <= c2){
                ct2++;
                j1--;
                if(i1 > j1){
                  break;
                }
              }
              else{
                break;
              }
            }
            r2 -= k - 1;
        }
        else if(c == 'L'){
            while(1){
              ll tp1 = v2[i2].second, tp2 = v2[i2].first;
              if(tp1 >= r1 && tp1 <= r2 && tp2 >= c1 && tp2 <= c2 && tp2 <= c1 + k - 1){
                ct2++;
                i2++;
                if(i2 > j2){
                  break;
                }
              }
              else{
                break;
              }
            }
            c1 += k;
        }
        else if(c == 'R'){
           while(1){
              ll tp1 = v2[j2].second, tp2 = v2[j2].first;
              // cout << r1 << " " << r2 << " " << c1 << " " << c2 << endl;
              // cout << tp1 << " " << tp2 << endl;
              if(tp1 >= r1 && tp1 <= r2 && tp2 >= c1 && tp2 <= c2 && tp2 >= c2 - k + 1){
                ct2++;
                j2--;
                if(i2 > j2){
                  break;
                }
              }
              else{
                break;
              }
            }
            c2 -= k;
        }
      }
      }
      cout << ct1 << " " << ct2 << endl;
  }
   return 0;
}

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

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

Think step by step.

Response:
To generate a test input that can expose the bug in the provided buggy code, we need to consider the following:

1. **Edge Cases:**
   - Test with the minimum and maximum values of `a` and `b`.
   - Test with the minimum and maximum values of `n` and `m`.
   - Test with a grid that has chips in the corners and along the edges.
   - Test with a grid that has chips in the middle.

2. **Logical Errors:**
   - The buggy code has an off-by-one error in the `D` move when it's Bob's turn (`r2 -= k - 1` instead of `r2 -= k`).
   - The code might not handle the case where a move cuts more rows or columns than available correctly.

3. **Small-Scale Test:**
   - Use a small grid to make the test input manageable and easy to verify.

Let's create a test input that focuses on the off-by-one error in the `D` move and includes a few chips to ensure the bug is exposed.

### Test Input

```plaintext
1
5 5 5 3
1 1
2 2
3 3
4 4
5 5
D 3
R 2
U 2
```

### Explanation

- **Grid Dimensions:** 5x5
- **Number of Chips:** 5 (placed at (1,1), (2,2), (3,3), (4,4), (5,5))
- **Number of Moves:** 3
- **Moves:**
  - `D 3`: Alice cuts the last 3 rows (rows 3, 4, and 5). This should remove the chips at (3,3), (4,4), and (5,5). However, the buggy code will incorrectly leave row 5 due to the off-by-one error.
  - `R 2`: Bob cuts the last 2 columns (columns 4 and 5). This should remove the chip at (2,2) but the grid is already in an incorrect state.
  - `U 2`: Alice cuts the first 2 rows (rows 1 and 2). This should remove the chip at (1,1) and (2,2).

### Expected Output

- **Correct Output:** Alice: 3, Bob: 1
- **Buggy Output:** Alice: 2, Bob: 1 (due to the off-by-one error in the `D` move)

This test input should expose the off-by-one error in the `D` move when it's Bob's turn.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cutting Game
time_limit_ms: 3000
memory_limit_mb: 256
Description: Alice and Bob were playing a game again. They have a grid of size $a \times b$ ($1 \le a, b \le 10^9$), on which there are $n$ chips, with at most one chip in each cell. The cell at the intersection of the $x$-th row and the $y$-th column has coordinates $(x, y)$.

Alice made the first move, and the players took turns. On each move, a player could cut several (but not all) rows or columns from the beginning or end of the remaining grid and earn a point for each chip that was on the cut part of the grid. Each move can be described by the character 'U', 'D', 'L', or 'R' and an integer $k$:

  * If the character is 'U', then the first $k$ remaining rows will be cut;    * If the character is 'D', then the last $k$ remaining rows will be cut;    * If the character is 'L', then the first $k$ remaining columns will be cut;    * If the character is 'R', then the last $k$ remaining columns will be cut.

Based on the initial state of the grid and the players' moves, determine the number of points earned by Alice and Bob, respectively.

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

The first line of each test case contains four integers $a$, $b$, $n$, and $m$ ($2 \le a, b \le 10^9$, $1 \le n, m \le 2 \cdot 10^5$) — the dimensions of the grid, the number of chips, and the number of moves.

Each of the next $n$ lines contain two integers $x_i$ and $y_i$ ($1 \le x_i \le a$, $1 \le y_i \le b$) — the coordinates of the chips. All pairs of coordinates are distinct.

Each of the next $m$ lines contain a character $c_j$ and an integer $k_j$ — the description of the $j$-th move. It is guaranteed that $k$ is less than the number of rows/columns in the current grid. In other words, a player cannot cut the entire remaining grid on their move.

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

Output: For each test case, output two integers — the number of points earned by Alice and Bob, respectively.

Examples:
input:
6
4 4 3 2
4 1
3 3
2 4
D 2
R 1
4 4 3 3
4 1
3 2
2 3
D 1
L 1
U 2
3 5 3 2
1 3
2 2
3 3
R 2
R 2
6 4 4 2
1 4
2 3
5 3
1 1
R 1
U 1
9 3 2 1
6 1
3 3
D 8
10 10 2 5
7 5
9 1
R 1
L 2
D 1
U 4
D 1
output:
2 1
2 0
0 3
1 1
2 0
0 1

Note: Below is the game from the first example:

![](CDN_BASE_URL/3730e4e7ac2159abc9c10c7ca7091e03)

On her turn, Alice cut $2$ rows from the bottom and scored $2$ points, then Bob cut $1$ column from the right and scored one point. Note that if Bob had cut $1$ row from the bottom, he would have also scored $1$ point.

Buggy Code:
#include <bits/stdc++.h>
#define fastread() ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL)
#define nl "\n"
#define ll long long
#define all(x) x.begin(), x.end()
#define int long long
using namespace std;
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
//template <typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const int N=2e5+7,M=1e9+7;
const double pi=acos(-1.0);
pair<int,int>p[N];
int mn1[4*N],mn2[4*N],mx1[4*N],mx2[4*N];
void build(int node,int st,int ed)
{
    if (st==ed)
    {
        mn1[node]=mx1[node]=p[st].first;
        mn2[node]=mx2[node]=p[st].second;
        return;
    }
    int lft=node*2,rgt=(node*2)+1,mid=(st+ed)/2;
    build(lft,st,mid);
    build(rgt,mid+1,ed);
    mn1[node]=min(mn1[lft],mn1[rgt]);
    mn2[node]=min(mn2[lft],mn2[rgt]);
    mx1[node]=max(mx1[lft],mx1[rgt]);
    mx2[node]=max(mx2[lft],mx2[rgt]);
}
void update(int node,int st,int ed,int u,int d,int l,int r)
{
    if (st==ed)
    {
        mn1[node]=mn2[node]=1e17;
        mx1[node]=mx2[node]=-1e17;
        return;
    }
    int lft=node*2,rgt=(node*2)+1,mid=(st+ed)/2;
    if (mn1[lft]<=d && mx1[lft]>=u && mn2[lft]<=r && mx2[lft]>=l)
    {
        update(lft,st,mid,u,d,l,r);
    }
    if (mn1[rgt]<=d && mx1[rgt]>=u && mn2[rgt]<=r && mx2[rgt]>=l)
    {
        update(rgt,mid+1,ed,u,d,l,r);
    }
    mn1[node]=min(mn1[lft],mn1[rgt]);
    mn2[node]=min(mn2[lft],mn2[rgt]);
    mx1[node]=max(mx1[lft],mx1[rgt]);
    mx2[node]=max(mx2[lft],mx2[rgt]);
}
int query(int node,int st,int ed,int u,int d,int l,int r)
{
    if (st==ed) return 1;
    int lft=node*2,rgt=(node*2)+1,mid=(st+ed)/2;
    int ans1=0,ans2=0;
    if (mn1[lft]<=d && mx1[lft]>=u && mn2[lft]<=r && mx2[lft]>=l)
    {
        ans1=query(lft,st,mid,u,d,l,r);
    }
    if (mn1[rgt]<=d && mx1[rgt]>=u && mn2[rgt]<=r && mx2[rgt]>=l)
    {
        ans2=query(rgt,mid+1,ed,u,d,l,r);
    }
    return ans1+ans2;
}
signed main()
{
    fastread();
    int t=1;
    cin>>t;
    for (int tc=1;tc<=t;tc++)
    {
        int a,b,n,m;
        cin>>a>>b>>n>>m;
        for (int i=0;i<n;i++)
        {
            cin>>p[i].first>>p[i].second;
        }
        build(1,0,n-1);
        int u=1,d=a,l=1,r=b;
        int alice=0,bob=0;
        for (int i=1;i<=m;i++)
        {
            char c;
            int k,points=0;
            cin>>c>>k;
            if (c=='U')
            {
                points=query(1,0,n-1,u,u+k-1,l,r);
                update(1,0,n-1,u,u+k-1,l,r);
                u+=k;
            }
            if (c=='D')
            {
                points=query(1,0,n-1,d-k+1,d,l,r);
                update(1,0,n-1,d-k+1,d,l,r);
                d-=k;
            }
            if (c=='L')
            {
                points=query(1,0,n-1,u,d,l,l+k-1);
                update(1,0,n-1,u,d,l,l+k-1);
                l+=k;
            }
            if (c=='R')
            {
                points=query(1,0,n-1,u,d,r-k+1,r);
                update(1,0,n-1,u,d,r-k+1,r);
                r-=k;
            }
            if (i%2) alice+=points;
            else bob+=points;
        }
        cout<<alice<<" "<<bob<<nl;
    }
}

**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 given buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. Here are the steps to think through this:

1. **Edge Cases for Grid Dimensions and Moves**:
   - Test with the minimum and maximum possible values for grid dimensions (`a` and `b`).
   - Test with the minimum and maximum possible values for the number of chips (`n`) and moves (`m`).

2. **Edge Cases for Chip Placement**:
   - Place chips in the corners of the grid.
   - Place chips in the middle of the grid.
   - Place chips in a pattern that can be easily cut in multiple ways.

3. **Edge Cases for Moves**:
   - Test with moves that cut the entire remaining grid (though this is guaranteed not to happen, we can test close to it).
   - Test with moves that cut the grid in such a way that the remaining grid is very small or has no chips.

4. **Edge Cases for Player Moves**:
   - Test with an even number of moves to ensure both players get a chance.
   - Test with an odd number of moves to ensure the last move is counted correctly.

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

```plaintext
1
4 4 4 4
1 1
2 2
3 3
4 4
U 1
L 1
D 1
R 1
```

### Explanation:
- **Grid Dimensions**: 4x4
- **Number of Chips**: 4
- **Number of Moves**: 4
- **Chips Placement**:
  - (1, 1)
  - (2, 2)
  - (3, 3)
  - (4, 4)
- **Moves**:
  - `U 1`: Cut the first row, removing the chip at (1, 1).
  - `L 1`: Cut the first column, removing the chip at (2, 2).
  - `D 1`: Cut the last row, removing the chip at (4, 4).
  - `R 1`: Cut the last column, removing the chip at (3, 3).

This test case ensures that each move affects a different chip and that the grid is reduced step by step. It also tests the alternating turns between Alice and Bob. This should help expose any bugs in the code related to the correct updating and querying of the segment tree.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cutting Game
time_limit_ms: 3000
memory_limit_mb: 256
Description: Alice and Bob were playing a game again. They have a grid of size $a \times b$ ($1 \le a, b \le 10^9$), on which there are $n$ chips, with at most one chip in each cell. The cell at the intersection of the $x$-th row and the $y$-th column has coordinates $(x, y)$.

Alice made the first move, and the players took turns. On each move, a player could cut several (but not all) rows or columns from the beginning or end of the remaining grid and earn a point for each chip that was on the cut part of the grid. Each move can be described by the character 'U', 'D', 'L', or 'R' and an integer $k$:

  * If the character is 'U', then the first $k$ remaining rows will be cut;    * If the character is 'D', then the last $k$ remaining rows will be cut;    * If the character is 'L', then the first $k$ remaining columns will be cut;    * If the character is 'R', then the last $k$ remaining columns will be cut.

Based on the initial state of the grid and the players' moves, determine the number of points earned by Alice and Bob, respectively.

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

The first line of each test case contains four integers $a$, $b$, $n$, and $m$ ($2 \le a, b \le 10^9$, $1 \le n, m \le 2 \cdot 10^5$) — the dimensions of the grid, the number of chips, and the number of moves.

Each of the next $n$ lines contain two integers $x_i$ and $y_i$ ($1 \le x_i \le a$, $1 \le y_i \le b$) — the coordinates of the chips. All pairs of coordinates are distinct.

Each of the next $m$ lines contain a character $c_j$ and an integer $k_j$ — the description of the $j$-th move. It is guaranteed that $k$ is less than the number of rows/columns in the current grid. In other words, a player cannot cut the entire remaining grid on their move.

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

Output: For each test case, output two integers — the number of points earned by Alice and Bob, respectively.

Examples:
input:
6
4 4 3 2
4 1
3 3
2 4
D 2
R 1
4 4 3 3
4 1
3 2
2 3
D 1
L 1
U 2
3 5 3 2
1 3
2 2
3 3
R 2
R 2
6 4 4 2
1 4
2 3
5 3
1 1
R 1
U 1
9 3 2 1
6 1
3 3
D 8
10 10 2 5
7 5
9 1
R 1
L 2
D 1
U 4
D 1
output:
2 1
2 0
0 3
1 1
2 0
0 1

Note: Below is the game from the first example:

![](CDN_BASE_URL/3730e4e7ac2159abc9c10c7ca7091e03)

On her turn, Alice cut $2$ rows from the bottom and scored $2$ points, then Bob cut $1$ column from the right and scored one point. Note that if Bob had cut $1$ row from the bottom, he would have also scored $1$ point.

Buggy Code:
def cutting_game(t, test_cases):
    results = []

    for _ in range(t):
        a, b, n, m = test_cases[_]['grid']
        chips = test_cases[_]['chips']
        moves = test_cases[_]['moves']

        # Initialize the remaining rows and columns
        top_row = 1
        bottom_row = a
        left_col = 1
        right_col = b

        # Initialize points for Alice and Bob
        alice_points = 0
        bob_points = 0

        # Track whose turn it is
        turn = 'Alice'

        for move in moves:
            direction, k = move

            if direction == 'U':
                # Cut the first k rows
                for i in range(top_row, top_row + k):
                    for j in range(left_col, right_col + 1):
                        if (i, j) in chips:
                            if turn == 'Alice':
                                alice_points += 1
                            else:
                                bob_points += 1
                top_row += k

            elif direction == 'D':
                # Cut the last k rows
                for i in range(bottom_row - k + 1, bottom_row + 1):
                    for j in range(left_col, right_col + 1):
                        if (i, j) in chips:
                            if turn == 'Alice':
                                alice_points += 1
                            else:
                                bob_points += 1
                bottom_row -= k

            elif direction == 'L':
                # Cut the first k columns
                for i in range(top_row, bottom_row + 1):
                    for j in range(left_col, left_col + k):
                        if (i, j) in chips:
                            if turn == 'Alice':
                                alice_points += 1
                            else:
                                bob_points += 1
                left_col += k

            elif direction == 'R':
                # Cut the last k columns
                for i in range(top_row, bottom_row + 1):
                    for j in range(right_col - k + 1, right_col + 1):
                        if (i, j) in chips:
                            if turn == 'Alice':
                                alice_points += 1
                            else:
                                bob_points += 1
                right_col -= k

            # Switch turns
            turn = 'Bob' if turn == 'Alice' else 'Alice'

        results.append((alice_points, bob_points))

    return results

# Example usage
if __name__ == "__main__":
    t = 6
    test_cases = [
        {'grid': (4, 4, 3, 2), 'chips': [(4, 1), (3, 3), (2, 4)], 'moves': [('D', 2), ('R', 1)]},
        {'grid': (4, 4, 3, 3), 'chips': [(4, 1), (3, 2), (2, 3)], 'moves': [('D', 1), ('L', 1), ('U', 2)]},
        {'grid': (3, 5, 3, 2), 'chips': [(1, 3), (2, 2), (3, 3)], 'moves': [('R', 2), ('R', 2)]},
        {'grid': (6, 4, 4, 2), 'chips': [(1, 4), (2, 3), (5, 3), (1, 1)], 'moves': [('R', 1), ('U', 1)]},
        {'grid': (9, 3, 2, 1), 'chips': [(6, 1), (3, 3)], 'moves': [('D', 8)]},
        {'grid': (10, 10, 2, 5), 'chips': [(7, 5), (9, 1)], 'moves': [('R', 1), ('L', 2), ('D', 1), ('U', 4), ('D', 1)]}
    ]

    results = cutting_game(t, test_cases)
    for result in results:
        print(result[0], result[1])

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

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

Think step by step.

Response:
```plaintext
1
3 3 3 2
1 1
2 2
3 3
U 1
R 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cutting Game
time_limit_ms: 3000
memory_limit_mb: 256
Description: Alice and Bob were playing a game again. They have a grid of size $a \times b$ ($1 \le a, b \le 10^9$), on which there are $n$ chips, with at most one chip in each cell. The cell at the intersection of the $x$-th row and the $y$-th column has coordinates $(x, y)$.

Alice made the first move, and the players took turns. On each move, a player could cut several (but not all) rows or columns from the beginning or end of the remaining grid and earn a point for each chip that was on the cut part of the grid. Each move can be described by the character 'U', 'D', 'L', or 'R' and an integer $k$:

  * If the character is 'U', then the first $k$ remaining rows will be cut;    * If the character is 'D', then the last $k$ remaining rows will be cut;    * If the character is 'L', then the first $k$ remaining columns will be cut;    * If the character is 'R', then the last $k$ remaining columns will be cut.

Based on the initial state of the grid and the players' moves, determine the number of points earned by Alice and Bob, respectively.

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

The first line of each test case contains four integers $a$, $b$, $n$, and $m$ ($2 \le a, b \le 10^9$, $1 \le n, m \le 2 \cdot 10^5$) — the dimensions of the grid, the number of chips, and the number of moves.

Each of the next $n$ lines contain two integers $x_i$ and $y_i$ ($1 \le x_i \le a$, $1 \le y_i \le b$) — the coordinates of the chips. All pairs of coordinates are distinct.

Each of the next $m$ lines contain a character $c_j$ and an integer $k_j$ — the description of the $j$-th move. It is guaranteed that $k$ is less than the number of rows/columns in the current grid. In other words, a player cannot cut the entire remaining grid on their move.

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

Output: For each test case, output two integers — the number of points earned by Alice and Bob, respectively.

Examples:
input:
6
4 4 3 2
4 1
3 3
2 4
D 2
R 1
4 4 3 3
4 1
3 2
2 3
D 1
L 1
U 2
3 5 3 2
1 3
2 2
3 3
R 2
R 2
6 4 4 2
1 4
2 3
5 3
1 1
R 1
U 1
9 3 2 1
6 1
3 3
D 8
10 10 2 5
7 5
9 1
R 1
L 2
D 1
U 4
D 1
output:
2 1
2 0
0 3
1 1
2 0
0 1

Note: Below is the game from the first example:

![](CDN_BASE_URL/3730e4e7ac2159abc9c10c7ca7091e03)

On her turn, Alice cut $2$ rows from the bottom and scored $2$ points, then Bob cut $1$ column from the right and scored one point. Note that if Bob had cut $1$ row from the bottom, he would have also scored $1$ point.

Buggy Code:
def main():
    t = int(input())
    for _ in range(t):
        a, b, n, m = map(int, input().split())
        chip = set()

        for _ in range(n):
            x, y = map(int, input().split())
            chip.add((x, y))

        alice, bob = 0, 0

        for _ in range(m):
            direction, k = input().split()
            k = int(k)

            cut_chips = 0
            if direction == 'U':
                for x in range(1, k + 1):
                    for y in range(1, b + 1):
                        if (x, y) in chip:
                            cut_chips += 1
                a -= k
            elif direction == 'D':
                for x in range(a, a - k, -1):
                    for y in range(1, b + 1):
                        if (x, y) in chip:
                            cut_chips += 1
                a -= k
            elif direction == 'L':
                for x in range(1, a + 1):
                    for y in range(1, k + 1):
                        if (x, y) in chip:
                            cut_chips += 1
                b -= k
            elif direction == 'R':
                for x in range(1, a + 1):
                    for y in range(b, b - k, -1):
                        if (x, y) in chip:
                            cut_chips += 1
                b -= k

            if _ % 2 == 0:
                alice += cut_chips
            else:
                bob += cut_chips
        print(alice, bob)
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 given code, we need to focus on edge cases and scenarios that are likely to cause issues. The main issues in the code are related to the efficiency of checking and removing chips from the set, especially when the grid size is large. Here are the steps to create such a test input:

1. **Large Grid with Few Chips**: Create a large grid with a small number of chips to ensure that the nested loops in the code will take a significant amount of time to execute.
2. **Complex Move Sequence**: Use a sequence of moves that alternates between rows and columns to ensure that the code needs to handle both row and column cuts efficiently.
3. **Edge Cases**: Include moves that cut the grid very close to the edges, which can cause issues if the code does not handle boundary conditions correctly.

Given these considerations, here is a small-scale test input that is complete and valid:

```plaintext
1
1000000000 1000000000 2 4
1 1
1000000000 1000000000
U 500000000
R 500000000
D 499999999
L 499999999
```

### Explanation:
- **Grid Size**: \(10^9 \times 10^9\) is the maximum allowed size, which ensures that the nested loops in the code will be very slow.
- **Number of Chips**: Only 2 chips are placed at opposite corners of the grid.
- **Moves**:
  - Alice cuts the first \(5 \times 10^8\) rows.
  - Bob cuts the first \(5 \times 10^8\) columns.
  - Alice cuts the last \(4.99999999 \times 10^8\) rows.
  - Bob cuts the last \(4.99999999 \times 10^8\) columns.

This test case will expose the inefficiency of the nested loops in the code, as it will take a very long time to check and remove chips from the set. Additionally, it tests the boundary conditions by cutting very close to the edges of the grid.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cutting Game
time_limit_ms: 3000
memory_limit_mb: 256
Description: Alice and Bob were playing a game again. They have a grid of size $a \times b$ ($1 \le a, b \le 10^9$), on which there are $n$ chips, with at most one chip in each cell. The cell at the intersection of the $x$-th row and the $y$-th column has coordinates $(x, y)$.

Alice made the first move, and the players took turns. On each move, a player could cut several (but not all) rows or columns from the beginning or end of the remaining grid and earn a point for each chip that was on the cut part of the grid. Each move can be described by the character 'U', 'D', 'L', or 'R' and an integer $k$:

  * If the character is 'U', then the first $k$ remaining rows will be cut;    * If the character is 'D', then the last $k$ remaining rows will be cut;    * If the character is 'L', then the first $k$ remaining columns will be cut;    * If the character is 'R', then the last $k$ remaining columns will be cut.

Based on the initial state of the grid and the players' moves, determine the number of points earned by Alice and Bob, respectively.

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

The first line of each test case contains four integers $a$, $b$, $n$, and $m$ ($2 \le a, b \le 10^9$, $1 \le n, m \le 2 \cdot 10^5$) — the dimensions of the grid, the number of chips, and the number of moves.

Each of the next $n$ lines contain two integers $x_i$ and $y_i$ ($1 \le x_i \le a$, $1 \le y_i \le b$) — the coordinates of the chips. All pairs of coordinates are distinct.

Each of the next $m$ lines contain a character $c_j$ and an integer $k_j$ — the description of the $j$-th move. It is guaranteed that $k$ is less than the number of rows/columns in the current grid. In other words, a player cannot cut the entire remaining grid on their move.

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

Output: For each test case, output two integers — the number of points earned by Alice and Bob, respectively.

Examples:
input:
6
4 4 3 2
4 1
3 3
2 4
D 2
R 1
4 4 3 3
4 1
3 2
2 3
D 1
L 1
U 2
3 5 3 2
1 3
2 2
3 3
R 2
R 2
6 4 4 2
1 4
2 3
5 3
1 1
R 1
U 1
9 3 2 1
6 1
3 3
D 8
10 10 2 5
7 5
9 1
R 1
L 2
D 1
U 4
D 1
output:
2 1
2 0
0 3
1 1
2 0
0 1

Note: Below is the game from the first example:

![](CDN_BASE_URL/3730e4e7ac2159abc9c10c7ca7091e03)

On her turn, Alice cut $2$ rows from the bottom and scored $2$ points, then Bob cut $1$ column from the right and scored one point. Note that if Bob had cut $1$ row from the bottom, he would have also scored $1$ point.

Buggy Code:
from collections import defaultdict
import sys

for p in range(int(input())):
    a,b,n,m=map(int,input().split())
    row,col=[],[]
    for i in range(n):
        l,r=map(int,input().split())
        row.append((l,r))
    row=sorted(row,key=lambda x:x[0])
    col=sorted(row,key=lambda x:x[1])
   #  print(row)
#     print(col)
    sc,sr,ec,er=1,1,b,a
    rsi,csi,rei,cei=0,0,n-1,n-1
    a,b=0,0
    for i in range(m):
        l,r=map(str,input().split())
        r=int(r)
        if l=='U':
            while row[rsi][0]<sr+r and rsi<n:
                if row[rsi][1]>=sc and  row[rsi][1]<=ec:
                    a+=1
                rsi+=1
            sr+=r
        elif l=='D':
            while row[rei][0]>er-r and  rei>=0:
                if row[rei][1]>=sc and  row[rei][1]<=ec:
                    a+=1
                rei-=1
            er-=r
        elif l=='L':
            while col[csi][1]<sc+r and csi<n:
                if col[csi][0]>=sr and  col[csi][0]<=er:
                    b+=1
                csi+=1
            sc+=r
        elif l=='R':
            while col[cei][1]>ec-r and cei>=0:
                if col[cei][0]>=sr and  col[cei][0]<=er:
                    b+=1
                cei-=1
            ec-=r
    print(a,b)

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

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

Think step by step.

Response:
```plaintext
1
3 3 2 4
1 1
3 3
U 1
R 1
D 1
L 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cutting Game
time_limit_ms: 3000
memory_limit_mb: 256
Description: Alice and Bob were playing a game again. They have a grid of size $a \times b$ ($1 \le a, b \le 10^9$), on which there are $n$ chips, with at most one chip in each cell. The cell at the intersection of the $x$-th row and the $y$-th column has coordinates $(x, y)$.

Alice made the first move, and the players took turns. On each move, a player could cut several (but not all) rows or columns from the beginning or end of the remaining grid and earn a point for each chip that was on the cut part of the grid. Each move can be described by the character 'U', 'D', 'L', or 'R' and an integer $k$:

  * If the character is 'U', then the first $k$ remaining rows will be cut;    * If the character is 'D', then the last $k$ remaining rows will be cut;    * If the character is 'L', then the first $k$ remaining columns will be cut;    * If the character is 'R', then the last $k$ remaining columns will be cut.

Based on the initial state of the grid and the players' moves, determine the number of points earned by Alice and Bob, respectively.

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

The first line of each test case contains four integers $a$, $b$, $n$, and $m$ ($2 \le a, b \le 10^9$, $1 \le n, m \le 2 \cdot 10^5$) — the dimensions of the grid, the number of chips, and the number of moves.

Each of the next $n$ lines contain two integers $x_i$ and $y_i$ ($1 \le x_i \le a$, $1 \le y_i \le b$) — the coordinates of the chips. All pairs of coordinates are distinct.

Each of the next $m$ lines contain a character $c_j$ and an integer $k_j$ — the description of the $j$-th move. It is guaranteed that $k$ is less than the number of rows/columns in the current grid. In other words, a player cannot cut the entire remaining grid on their move.

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

Output: For each test case, output two integers — the number of points earned by Alice and Bob, respectively.

Examples:
input:
6
4 4 3 2
4 1
3 3
2 4
D 2
R 1
4 4 3 3
4 1
3 2
2 3
D 1
L 1
U 2
3 5 3 2
1 3
2 2
3 3
R 2
R 2
6 4 4 2
1 4
2 3
5 3
1 1
R 1
U 1
9 3 2 1
6 1
3 3
D 8
10 10 2 5
7 5
9 1
R 1
L 2
D 1
U 4
D 1
output:
2 1
2 0
0 3
1 1
2 0
0 1

Note: Below is the game from the first example:

![](CDN_BASE_URL/3730e4e7ac2159abc9c10c7ca7091e03)

On her turn, Alice cut $2$ rows from the bottom and scored $2$ points, then Bob cut $1$ column from the right and scored one point. Note that if Bob had cut $1$ row from the bottom, he would have also scored $1$ point.

Buggy Code:
import java.util.*;

import static java.lang.System.in;
import static java.lang.System.out;
public class C {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        while( T > 0 ) {
            int m = sc.nextInt();
            int n = sc.nextInt();
            int chips = sc.nextInt();
            int[][] grid = new int[m + 1][n + 1];
            int moves = sc.nextInt();
            for (int chip = 1; chip <= chips; chip++) {
                int i = sc.nextInt();
                int j = sc.nextInt();
                grid[i][j] = 1;
            }
            int[][] upToDownFilling = new int[m + 1][n + 1];
            int chipsYet = 0;
            for (int i = 0; i <= m; i++) {
                for (int j = 0; j <= n; j++) {
                    if (grid[i][j] == 1) chipsYet++;
                    upToDownFilling[i][j] = chipsYet;
//                    out.print(upToDownFilling[i][j] + " ");
                }
//                out.println();
            }
            chipsYet = 0;
            int[][] leftToRightFilling = new int[m+1][n+1];
            for( int j = 0; j <= n; j++ ) {
                for( int i = 0; i <= m ; i++ ) {
                    if( grid[i][j] == 1 ) chipsYet += 1;
                    leftToRightFilling[i][j] = chipsYet;
                }
            }
            boolean aliceMove = true;
            long AScore = 0; long BScore = 0;
            int up = 1; int down = m; int left = 1; int right = n;
//            out.println("moves is " + moves);
            for( int i = 1; i <= moves; i++ ) {
                char ch = sc.next().charAt(0);
//                out.println(ch);
                if( ch == 'D') {
                    int col = sc.nextInt();
                    int upNew = down - col + 1;
                    if(aliceMove) AScore += (-upToDownFilling[upNew][left] + upToDownFilling[down][right]);
                    else BScore += ( -upToDownFilling[upNew][left] + upToDownFilling[down][right]);
                    if( grid[upNew][left] == 1 ) {
                        if(aliceMove) AScore++;
                        else BScore++;
                    }
                    down -= col;
                }
                else if( ch == 'U') {
                    int col = sc.nextInt();
                    int downNew = up + col -1;
                    if(aliceMove) AScore += (- upToDownFilling[up][left] + upToDownFilling[downNew][right]);
                    else BScore += (-upToDownFilling[up][left] + upToDownFilling[downNew][right]);
                    if( grid[up][left] == 1 ) {
                        if(aliceMove) AScore++;
                        else BScore++;
                    }
                    up += col;
                }
                else if( ch == 'L') {
                    int row = sc.nextInt();
                    int rightNew = left + row - 1;
                    if(aliceMove) AScore += (-leftToRightFilling[up][left] + leftToRightFilling[down][rightNew]);
                    else BScore += (-leftToRightFilling[up][left] + leftToRightFilling[down][rightNew]);
                    if( grid[up][left] == 1 ) {
                        if(aliceMove) AScore++;
                        else BScore++;
                    }
                    left += row;
                }
                else if( ch == 'R') {
                    int row = sc.nextInt();
                    int leftNew = right - row + 1;
//                    out.println("left " + leftNew + " right " + right + "up is " + up + " down is " + down);
                    if(aliceMove) AScore += (-leftToRightFilling[up][leftNew] + leftToRightFilling[down][right]);
                    else BScore += (-leftToRightFilling[up][leftNew] + leftToRightFilling[down][right]);
                    if( grid[up][leftNew] == 1 ) {
                        if(aliceMove) AScore++;
                        else BScore++;
                    }
                    right -= row;
                }
                aliceMove = !aliceMove;
            }
            out.println(AScore + " " + BScore );
            T--;
//            chipsYet = 0;
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
4 4 4 4
1 1
2 2
3 3
4 4
D 1
R 1
U 1
L 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cutting Game
time_limit_ms: 3000
memory_limit_mb: 256
Description: Alice and Bob were playing a game again. They have a grid of size $a \times b$ ($1 \le a, b \le 10^9$), on which there are $n$ chips, with at most one chip in each cell. The cell at the intersection of the $x$-th row and the $y$-th column has coordinates $(x, y)$.

Alice made the first move, and the players took turns. On each move, a player could cut several (but not all) rows or columns from the beginning or end of the remaining grid and earn a point for each chip that was on the cut part of the grid. Each move can be described by the character 'U', 'D', 'L', or 'R' and an integer $k$:

  * If the character is 'U', then the first $k$ remaining rows will be cut;    * If the character is 'D', then the last $k$ remaining rows will be cut;    * If the character is 'L', then the first $k$ remaining columns will be cut;    * If the character is 'R', then the last $k$ remaining columns will be cut.

Based on the initial state of the grid and the players' moves, determine the number of points earned by Alice and Bob, respectively.

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

The first line of each test case contains four integers $a$, $b$, $n$, and $m$ ($2 \le a, b \le 10^9$, $1 \le n, m \le 2 \cdot 10^5$) — the dimensions of the grid, the number of chips, and the number of moves.

Each of the next $n$ lines contain two integers $x_i$ and $y_i$ ($1 \le x_i \le a$, $1 \le y_i \le b$) — the coordinates of the chips. All pairs of coordinates are distinct.

Each of the next $m$ lines contain a character $c_j$ and an integer $k_j$ — the description of the $j$-th move. It is guaranteed that $k$ is less than the number of rows/columns in the current grid. In other words, a player cannot cut the entire remaining grid on their move.

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

Output: For each test case, output two integers — the number of points earned by Alice and Bob, respectively.

Examples:
input:
6
4 4 3 2
4 1
3 3
2 4
D 2
R 1
4 4 3 3
4 1
3 2
2 3
D 1
L 1
U 2
3 5 3 2
1 3
2 2
3 3
R 2
R 2
6 4 4 2
1 4
2 3
5 3
1 1
R 1
U 1
9 3 2 1
6 1
3 3
D 8
10 10 2 5
7 5
9 1
R 1
L 2
D 1
U 4
D 1
output:
2 1
2 0
0 3
1 1
2 0
0 1

Note: Below is the game from the first example:

![](CDN_BASE_URL/3730e4e7ac2159abc9c10c7ca7091e03)

On her turn, Alice cut $2$ rows from the bottom and scored $2$ points, then Bob cut $1$ column from the right and scored one point. Note that if Bob had cut $1$ row from the bottom, he would have also scored $1$ point.

Buggy Code:
// Source: https://usaco.guide/general/io

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

public class Main {
	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		// Up to 2 * 10^5 queries
		//  For each query look at all the rows from L1 to L1 + k - 1 that HAVE A CHIP
		// In those rows check if there are any chips within the current range of columns (L2 - R2) by binsearch
		// Remove that row/column and add to chip total for Alice and Bob alternately O(log n)
		// For the columns/rows with chips remove the respective row/column from its respective set O(log n * )
		// Then adjust bounds accordingly
		int T = sc.nextInt();
		for(int i = 1; i <= T; i++){
			int A = sc.nextInt(); // rows
			int B = sc.nextInt(); // cols
			int N = sc.nextInt(); // # of chips
			int M = sc.nextInt(); // # of moves
			int U = 1;
			int D = A;
			int L = 1;
			int R = B;
			TreeMap<Integer,Set<Integer>> rows = new TreeMap<Integer,Set<Integer>>();
			TreeMap<Integer,Set<Integer>> cols = new TreeMap<Integer,Set<Integer>>();
			for(int j = 1; j <= N; j++){
				int xi = sc.nextInt();
				int yi = sc.nextInt();
				if(!rows.containsKey(xi)) rows.put(xi, new HashSet<Integer>());
				rows.get(xi).add(yi);
				if(!cols.containsKey(yi)) cols.put(yi, new HashSet<Integer>());
				cols.get(yi).add(xi);
			}
			int alice = 0;
			int bob = 0;
			//System.out.println(rows + "\n" + cols + "\n");
			for(int j = 1; j <= M; j++){
				String dir = sc.next();
				int K = sc.nextInt();
				if(dir.equals("U")){
					Set<Integer> rowsSet = rows.keySet();
					Integer topRow = rows.firstKey();
					Integer botRow = rows.floorKey(U + K - 1);
					U += K;
					if(botRow == null) continue;
					Set<Integer> removalStack = new HashSet<Integer>();
					for(Integer relevantRow : rowsSet){
						if(relevantRow > botRow) break;
						// Look at all the cols from L to R
						Set<Integer> columns = rows.get(relevantRow);
						Set<Integer> colsStack = new HashSet<Integer>();
						for(Integer relevantCol : columns){
							// There must be a chip contained here
							if(j%2 == 1) alice++;
							else bob++;
							colsStack.add(relevantCol);
						}
						removalStack.add(relevantRow); // Remove row from map
						// Also remove row from keys of all cols that contain it
						for(Integer stuff : colsStack){
							cols.get(stuff).remove(relevantRow);
						}
						if(relevantRow == botRow) break;

					}
					for(Integer row : removalStack){
						rows.remove(row);
					}

				}
				else if(dir.equals("D")){

					Set<Integer> rowsSet = rows.descendingKeySet();
					Integer topRow = rows.ceilingKey(D - K + 1);
					D -= K;
					Integer botRow = rows.lastKey();
					if(topRow == null) continue;
					Set<Integer> removalStack = new HashSet<Integer>();
					for(Integer relevantRow : rowsSet){
						if(relevantRow < topRow) break;
						// Look at all the cols from L to R
						Set<Integer> columns = rows.get(relevantRow);
						Set<Integer> colsStack = new HashSet<Integer>();
						for(Integer relevantCol : columns){
							// There must be a chip contained here
							if(j%2 == 1) alice++;
							else bob++;
							colsStack.add(relevantCol);
						}
						removalStack.add(relevantRow); // Remove row from map
						// Also remove row from keys of all cols that contain it
						for(Integer stuff : colsStack){
							cols.get(stuff).remove(relevantRow);
						}
						if(relevantRow == topRow) break;

					}
					for(Integer row : removalStack){
						rows.remove(row);
					}

				}
				else if(dir.equals("L")){
					Set<Integer> colsSet = cols.keySet();
					Integer leftCol = cols.firstKey();
					Integer rightCol = cols.floorKey(L + K - 1);
					L += K;
					if(rightCol == null) continue;
					Set<Integer> removalStack = new HashSet<Integer>();
					for(Integer relevantCol : colsSet){
						if(relevantCol > rightCol) break;
						// Look at all the cols from L to R
						Set<Integer> rowss = cols.get(relevantCol);
						Set<Integer> rowsStack = new HashSet<Integer>();
						for(Integer relevantRow : rowss){
							// There must be a chip contained here
							if(j%2 == 1) alice++;
							else bob++;
							rowsStack.add(relevantRow);
						}
						removalStack.add(relevantCol); // Remove row from map
						// Also remove row from keys of all cols that contain it
						for(Integer stuff : rowsStack){
							rows.get(stuff).remove(relevantCol);
						}
						if(relevantCol == rightCol) break;

					}
					for(Integer row : removalStack){
						rows.remove(row);
					}

				}
				else{
					Set<Integer> colsSet = cols.descendingKeySet();
					Integer leftCol = cols.ceilingKey(R - K + 1);
					R -= K;
					Integer rightCol = cols.lastKey();
					if(leftCol == null) continue;
					Set<Integer> removalStack = new HashSet<Integer>();
					for(Integer relevantCol : colsSet){
						if(relevantCol < leftCol) break;
						//System.out.println("Looking at " + relevantCol);
						// Look at all the cols from L to R
						Set<Integer> rowss = cols.get(relevantCol);
						Set<Integer> rowsStack = new HashSet<Integer>();
						for(Integer relevantRow : rowss){
							// There must be a chip contained here
							if(j%2 == 1) alice++;
							else bob++;
							rowsStack.add(relevantRow);
						}
						removalStack.add(relevantCol); // Remove row from map
						// Also remove row from keys of all cols that contain it
						for(Integer stuff : rowsStack){
							rows.get(stuff).remove(relevantCol);
						}
						if(relevantCol == leftCol) break;

					}
					for(Integer row : removalStack){
						rows.remove(row);
					}

				}
				//System.out.println(rows);
				//System.out.println(cols + "\n");

			}
			System.out.println(alice + " " + bob);
		}

	}
}

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

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

Think step by step.

Response:
```plaintext
1
4 4 4 4
1 1
2 2
3 3
4 4
U 2
R 2
D 1
L 1
```