**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: How Does the Rook Move?
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an $$$n \times n$$$ chessboard where you and the computer take turns alternatingly to place white rooks & black rooks on the board respectively. While placing rooks, you have to ensure that no two rooks attack each other. Two rooks attack each other if they share the same row or column regardless of color.

A valid move is placing a rook on a position ($$$r$$$, $$$c$$$) such that it doesn't attack any other rook.

You start first, and when you make a valid move in your turn, placing a white rook at position ($$$r$$$, $$$c$$$), the computer will mirror you and place a black rook at position ($$$c$$$, $$$r$$$) in its turn. If $$$r = c$$$, then the computer can't mirror your move, and skips its turn.

You have already played $$$k$$$ moves with the computer (the computer tries to mirror these moves too), and you must continue playing the game until there are no valid moves remaining. How many different final configurations are possible when you continue the game after the $$$k$$$ moves? It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid. Since the answer may be large, print it modulo $$$10^9+7$$$.

Two configurations are considered different if there exists a coordinate ($$$r$$$, $$$c$$$) which has a rook in one configuration, but not in the other or the color of the rook on the coordinate is different.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 3 \cdot 10^5$$$, $$$0 \leq k \leq n$$$) — the size of the chessboard and the number of moves you have already played respectively.

Each of the next $$$k$$$ lines of the test case contains two integers $$$r_i$$$ and $$$c_i$$$, denoting the $$$i$$$-th move you made.

It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, output a single integer on a new line — the total number of possible final configurations modulo $$$10^9+7$$$.

Examples:
input:
3
4 1
1 2
8 1
7 6
1000 4
4 4
952 343
222 333
90 91
output:
3
331
671968183

Note: In the first test case, we have a $$$4 \times 4$$$ grid and you've already played $$$1$$$ move. After you and the computer play a turn, we have a white rook at ($$$1$$$, $$$2$$$), and a black rook at ($$$2$$$, $$$1$$$). There are three possible configurations from this state —

1. You place a white rook at ($$$3$$$, $$$4$$$) and the computer places a black rook at ($$$4$$$, $$$3$$$) as a response.
2. You place a white rook at ($$$4$$$, $$$3$$$) and the computer places a black rook at ($$$3$$$, $$$4$$$) as a response.
3. You place a white rook at ($$$3$$$, $$$3$$$) and then at ($$$4$$$, $$$4$$$), or the other way around. They both result in the same configuration.

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

public class Main {
    public static int INF = 0x3f3f3f3f, mod = 1000000007, mod9 = 998244353;
    public static void main(String args[]){
        try {
            PrintWriter o = new PrintWriter(System.out);
            boolean multiTest = true;
            // init
            if(multiTest) {
                int t = nextInt(), loop = 0;
                while (loop < t) {loop++;solve(o);}
            } else solve(o);
            o.close();
        } catch (Exception e) {e.printStackTrace();}
    }
    static void solve(PrintWriter o) {
        try {
            int n = nextInt(), k = nextInt();
            int sub = 0;
            for(int i=0;i<k;i++) {
                int r = nextInt();
                int c = nextInt();
                sub += 2-(r==c?1:0);
            }
            int m = n-sub;
            long[] dp = new long[m+1];
            dp[0] = 1;
            dp[1] = 1;
            for(int i=2;i<=m;i++) dp[i] = (dp[i-1]+2l*(i-1)*dp[i-2]%mod)%mod;
            o.println(dp[m]);
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
    public static int upper_bound(int[] a, int val){
        int l = 0, r = a.length;
        while(l < r){
            int mid = l + (r - l) / 2;
            if(a[mid] <= val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int lower_bound(int[] a, int val){
        int l = 0, r = a.length;
        while(l < r){
            int mid = l + (r - l) / 2;
            if(a[mid] < val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a%b);
    }
    public static long[] extgcd(long a, long b) {
        if(b == 0) return new long[]{1, 0};
        long[] it = extgcd(b, a%b);
        long x = it[1], y = it[0];
        y -= a/b*x;
        return new long[]{x, y};
    }
    public static long lcm(long a, long b){
        return a / gcd(a,b)*b;
    }
    public static long qpow(long a, long n, int md){
        a %= md;
        long ret = 1l;
        while(n > 0){
            if((n & 1) == 1) ret = ret * a % md;
            n >>= 1;
            a = a * a % md;
        }
        return ret;
    }
    public static class FenWick {
        int n;
        long[] a;
        long[] tree;
        public FenWick(int n){
            this.n = n;
            a = new long[n+1];
            tree = new long[n+1];
        }
        private void add(int x, long val){
            while(x <= n){
                tree[x] += val;
                x += x&-x;
            }
        }
        private void addMx(int x, long val) {
            a[x] += val;
            tree[x] = a[x];
            while(x <= n) {
                for(int i=1;i<(x&-x);i<<=1) tree[x] = Math.max(tree[x], tree[x-i]);
                x += x&-x;
            }
        }
        private long query(int x){
            long ret = 0l;
            while(x > 0){
                ret += tree[x];
                x -= x&-x;
            }
            return ret;
        }
        private long queryMx(int l, int r) {
            long res = 0l;
            while(l <= r) {
                if(r-(r&-r) >= l) {
                    res = Math.max(res, tree[r]);
                    r -= r&-r;
                }
                else {
                    res = Math.max(res, a[r]);
                    r--;
                }
            }
            return res;
        }
    }
    static class DSU {
        int[] parent;
        public DSU(int n) {
            parent = new int[n];
            for(int i=0;i<n;i++) parent[i] = i;
        }
        public int find(int x) {
            if(parent[x] != x) parent[x] = find(parent[x]);
            return parent[x];
        }
        public void union(int x, int y) {
            int root_x = find(x);
            int root_y = find(y);
            if(root_x == root_y) return;
            parent[root_x] = root_y;
        }
    }
    private static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    private static StringTokenizer tokenizer = new StringTokenizer("");
    private static String next() throws IOException{
        while(!tokenizer.hasMoreTokens()){tokenizer = new StringTokenizer(reader.readLine());}
        return tokenizer.nextToken();
    }
    public static int nextInt() throws IOException {return Integer.parseInt(next());}
    public static Long nextLong() throws IOException {return Long.parseLong(next());}
    public static double nextDouble() throws IOException {return Double.parseDouble(next());}
    public static char nextChar() throws IOException {return next().toCharArray()[0];}
    public static String nextString() throws IOException {return next();}
    public static String nextLine() throws IOException {return reader.readLine();}
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: How Does the Rook Move?
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an $$$n \times n$$$ chessboard where you and the computer take turns alternatingly to place white rooks & black rooks on the board respectively. While placing rooks, you have to ensure that no two rooks attack each other. Two rooks attack each other if they share the same row or column regardless of color.

A valid move is placing a rook on a position ($$$r$$$, $$$c$$$) such that it doesn't attack any other rook.

You start first, and when you make a valid move in your turn, placing a white rook at position ($$$r$$$, $$$c$$$), the computer will mirror you and place a black rook at position ($$$c$$$, $$$r$$$) in its turn. If $$$r = c$$$, then the computer can't mirror your move, and skips its turn.

You have already played $$$k$$$ moves with the computer (the computer tries to mirror these moves too), and you must continue playing the game until there are no valid moves remaining. How many different final configurations are possible when you continue the game after the $$$k$$$ moves? It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid. Since the answer may be large, print it modulo $$$10^9+7$$$.

Two configurations are considered different if there exists a coordinate ($$$r$$$, $$$c$$$) which has a rook in one configuration, but not in the other or the color of the rook on the coordinate is different.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 3 \cdot 10^5$$$, $$$0 \leq k \leq n$$$) — the size of the chessboard and the number of moves you have already played respectively.

Each of the next $$$k$$$ lines of the test case contains two integers $$$r_i$$$ and $$$c_i$$$, denoting the $$$i$$$-th move you made.

It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, output a single integer on a new line — the total number of possible final configurations modulo $$$10^9+7$$$.

Examples:
input:
3
4 1
1 2
8 1
7 6
1000 4
4 4
952 343
222 333
90 91
output:
3
331
671968183

Note: In the first test case, we have a $$$4 \times 4$$$ grid and you've already played $$$1$$$ move. After you and the computer play a turn, we have a white rook at ($$$1$$$, $$$2$$$), and a black rook at ($$$2$$$, $$$1$$$). There are three possible configurations from this state —

1. You place a white rook at ($$$3$$$, $$$4$$$) and the computer places a black rook at ($$$4$$$, $$$3$$$) as a response.
2. You place a white rook at ($$$4$$$, $$$3$$$) and the computer places a black rook at ($$$3$$$, $$$4$$$) as a response.
3. You place a white rook at ($$$3$$$, $$$3$$$) and then at ($$$4$$$, $$$4$$$), or the other way around. They both result in the same configuration.

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

void solve(){
    int n,k;
    cin>>n>>k;
    while(k--){
        int a,b;
        cin>>a>>b;
        if(a==b){
            n--;
        }
        else{
            n-=2;
        }
    }
    long long int a=7;
    long long int b=3;
    long long int current=25;
    long long int count=4;
    if(n==1){
        cout<<1;
        return;
    }
    else if(n==0){
        cout<<1;
        return;
    }
    else if(n==2){
        cout<<3;
        return;
    }
    else if(n==3){
        cout<<4;
        return;
    }

    int mod=int(powl(10,9))+7;
    while(count<=n){

        current = (((count*2-2)*b )+a) % mod;
        b=a;
        a=current;
        count+=1;

    }
    cout<<a;

}
int main() {
    int t;
    cin>>t;
    while(t--){
        solve();
        cout<<"\n";
    }
	// your code goes here

}

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

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

Think step by step.

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

1. **Edge Cases**: Test with the smallest and largest possible values of `n` and `k`.
2. **Special Moves**: Test with moves where `r == c` and where `r != c`.
3. **Initial Configuration**: Ensure the initial configuration of the board is valid and can lead to a variety of final configurations.

### Step-by-Step Analysis

1. **Small Board with No Moves**:
   - This tests the base case where `n` is small and `k` is 0.
   - Expected output: 1 (only one configuration possible with an empty board).

2. **Small Board with One Move**:
   - This tests the case where `n` is small and `k` is 1.
   - Expected output: 3 (as shown in the example).

3. **Small Board with Multiple Moves**:
   - This tests the case where `n` is small and `k` is greater than 1.
   - Expected output: Depends on the moves made, but should be a small number.

4. **Large Board with No Moves**:
   - This tests the case where `n` is large and `k` is 0.
   - Expected output: A large number, but should be computable within the time limit.

5. **Large Board with One Move**:
   - This tests the case where `n` is large and `k` is 1.
   - Expected output: A large number, but should be computable within the time limit.

6. **Large Board with Multiple Moves**:
   - This tests the case where `n` is large and `k` is greater than 1.
   - Expected output: A large number, but should be computable within the time limit.

### Challenging Test Input

Let's choose a test input that is small enough to be manageable but large enough to expose potential issues in the code. We will use a small board with multiple moves to ensure the code handles the logic correctly.

```plaintext
3
4 1
1 2
4 2
1 2
2 3
1000 1
1 2
```

### Explanation

1. **Test Case 1**: A 4x4 board with 1 move.
   - Move: (1, 2)
   - Expected output: 3 (as shown in the example).

2. **Test Case 2**: A 4x4 board with 2 moves.
   - Moves: (1, 2) and (2, 3)
   - Expected output: 1 (only one configuration possible after these moves).

3. **Test Case 3**: A 1000x1000 board with 1 move.
   - Move: (1, 2)
   - Expected output: A large number, but should be computable within the time limit.

This test input should help expose any bugs in the code, particularly in handling multiple moves and large boards.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: How Does the Rook Move?
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an $$$n \times n$$$ chessboard where you and the computer take turns alternatingly to place white rooks & black rooks on the board respectively. While placing rooks, you have to ensure that no two rooks attack each other. Two rooks attack each other if they share the same row or column regardless of color.

A valid move is placing a rook on a position ($$$r$$$, $$$c$$$) such that it doesn't attack any other rook.

You start first, and when you make a valid move in your turn, placing a white rook at position ($$$r$$$, $$$c$$$), the computer will mirror you and place a black rook at position ($$$c$$$, $$$r$$$) in its turn. If $$$r = c$$$, then the computer can't mirror your move, and skips its turn.

You have already played $$$k$$$ moves with the computer (the computer tries to mirror these moves too), and you must continue playing the game until there are no valid moves remaining. How many different final configurations are possible when you continue the game after the $$$k$$$ moves? It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid. Since the answer may be large, print it modulo $$$10^9+7$$$.

Two configurations are considered different if there exists a coordinate ($$$r$$$, $$$c$$$) which has a rook in one configuration, but not in the other or the color of the rook on the coordinate is different.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 3 \cdot 10^5$$$, $$$0 \leq k \leq n$$$) — the size of the chessboard and the number of moves you have already played respectively.

Each of the next $$$k$$$ lines of the test case contains two integers $$$r_i$$$ and $$$c_i$$$, denoting the $$$i$$$-th move you made.

It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, output a single integer on a new line — the total number of possible final configurations modulo $$$10^9+7$$$.

Examples:
input:
3
4 1
1 2
8 1
7 6
1000 4
4 4
952 343
222 333
90 91
output:
3
331
671968183

Note: In the first test case, we have a $$$4 \times 4$$$ grid and you've already played $$$1$$$ move. After you and the computer play a turn, we have a white rook at ($$$1$$$, $$$2$$$), and a black rook at ($$$2$$$, $$$1$$$). There are three possible configurations from this state —

1. You place a white rook at ($$$3$$$, $$$4$$$) and the computer places a black rook at ($$$4$$$, $$$3$$$) as a response.
2. You place a white rook at ($$$4$$$, $$$3$$$) and the computer places a black rook at ($$$3$$$, $$$4$$$) as a response.
3. You place a white rook at ($$$3$$$, $$$3$$$) and then at ($$$4$$$, $$$4$$$), or the other way around. They both result in the same configuration.

Buggy Code:
#include<iostream>
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
ll mod = 1000000007;
void solve(){
    ll n,k,a,b;
    cin>>n>>k;
    ll c = n;
    for(int i=0; i<k; i++){
        cin>>a>>b;
        c-=2;
        if(a==b) c++;
    }
    vector<ll> dp(c+1,0);
    dp[1] = 1;
    dp[2] = 3;
    for(ll i=3; i<=c; i++){
        dp[i] = (dp[i-1] + ((i-1)*(2*(dp[i-2]%mod))%mod)%mod)%mod;
    }
    cout<<dp[c]<<"\n";
}
int main(){
    int t=1;
    cin>>t;
    while(t--)solve();
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
4 2
1 2
2 1
8 2
7 6
6 7
1000 2
1 1
999 999
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: How Does the Rook Move?
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an $$$n \times n$$$ chessboard where you and the computer take turns alternatingly to place white rooks & black rooks on the board respectively. While placing rooks, you have to ensure that no two rooks attack each other. Two rooks attack each other if they share the same row or column regardless of color.

A valid move is placing a rook on a position ($$$r$$$, $$$c$$$) such that it doesn't attack any other rook.

You start first, and when you make a valid move in your turn, placing a white rook at position ($$$r$$$, $$$c$$$), the computer will mirror you and place a black rook at position ($$$c$$$, $$$r$$$) in its turn. If $$$r = c$$$, then the computer can't mirror your move, and skips its turn.

You have already played $$$k$$$ moves with the computer (the computer tries to mirror these moves too), and you must continue playing the game until there are no valid moves remaining. How many different final configurations are possible when you continue the game after the $$$k$$$ moves? It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid. Since the answer may be large, print it modulo $$$10^9+7$$$.

Two configurations are considered different if there exists a coordinate ($$$r$$$, $$$c$$$) which has a rook in one configuration, but not in the other or the color of the rook on the coordinate is different.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 3 \cdot 10^5$$$, $$$0 \leq k \leq n$$$) — the size of the chessboard and the number of moves you have already played respectively.

Each of the next $$$k$$$ lines of the test case contains two integers $$$r_i$$$ and $$$c_i$$$, denoting the $$$i$$$-th move you made.

It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, output a single integer on a new line — the total number of possible final configurations modulo $$$10^9+7$$$.

Examples:
input:
3
4 1
1 2
8 1
7 6
1000 4
4 4
952 343
222 333
90 91
output:
3
331
671968183

Note: In the first test case, we have a $$$4 \times 4$$$ grid and you've already played $$$1$$$ move. After you and the computer play a turn, we have a white rook at ($$$1$$$, $$$2$$$), and a black rook at ($$$2$$$, $$$1$$$). There are three possible configurations from this state —

1. You place a white rook at ($$$3$$$, $$$4$$$) and the computer places a black rook at ($$$4$$$, $$$3$$$) as a response.
2. You place a white rook at ($$$4$$$, $$$3$$$) and the computer places a black rook at ($$$3$$$, $$$4$$$) as a response.
3. You place a white rook at ($$$3$$$, $$$3$$$) and then at ($$$4$$$, $$$4$$$), or the other way around. They both result in the same configuration.

Buggy Code:
mod=10**9+7
def solve(n):
    dp = [0]*(n+1)
    if(n>=1):
        dp[0] = 1
    if(n>=2):
        dp[1] = 1
    for i in range(2, n+1):
        dp[i] = ((dp[i-1]%mod) + ((((i-1)%mod)*2)%mod)*(dp[i-2])%mod) % mod
    return dp[n]%mod

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

    s=set()

    for i in range(k):
        r,c=map(int,input().split())
        s.add(r)
        s.add(c)
    ans=solve(n-len(s))
    print(ans%mod)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: How Does the Rook Move?
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an $$$n \times n$$$ chessboard where you and the computer take turns alternatingly to place white rooks & black rooks on the board respectively. While placing rooks, you have to ensure that no two rooks attack each other. Two rooks attack each other if they share the same row or column regardless of color.

A valid move is placing a rook on a position ($$$r$$$, $$$c$$$) such that it doesn't attack any other rook.

You start first, and when you make a valid move in your turn, placing a white rook at position ($$$r$$$, $$$c$$$), the computer will mirror you and place a black rook at position ($$$c$$$, $$$r$$$) in its turn. If $$$r = c$$$, then the computer can't mirror your move, and skips its turn.

You have already played $$$k$$$ moves with the computer (the computer tries to mirror these moves too), and you must continue playing the game until there are no valid moves remaining. How many different final configurations are possible when you continue the game after the $$$k$$$ moves? It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid. Since the answer may be large, print it modulo $$$10^9+7$$$.

Two configurations are considered different if there exists a coordinate ($$$r$$$, $$$c$$$) which has a rook in one configuration, but not in the other or the color of the rook on the coordinate is different.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 3 \cdot 10^5$$$, $$$0 \leq k \leq n$$$) — the size of the chessboard and the number of moves you have already played respectively.

Each of the next $$$k$$$ lines of the test case contains two integers $$$r_i$$$ and $$$c_i$$$, denoting the $$$i$$$-th move you made.

It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, output a single integer on a new line — the total number of possible final configurations modulo $$$10^9+7$$$.

Examples:
input:
3
4 1
1 2
8 1
7 6
1000 4
4 4
952 343
222 333
90 91
output:
3
331
671968183

Note: In the first test case, we have a $$$4 \times 4$$$ grid and you've already played $$$1$$$ move. After you and the computer play a turn, we have a white rook at ($$$1$$$, $$$2$$$), and a black rook at ($$$2$$$, $$$1$$$). There are three possible configurations from this state —

1. You place a white rook at ($$$3$$$, $$$4$$$) and the computer places a black rook at ($$$4$$$, $$$3$$$) as a response.
2. You place a white rook at ($$$4$$$, $$$3$$$) and the computer places a black rook at ($$$3$$$, $$$4$$$) as a response.
3. You place a white rook at ($$$3$$$, $$$3$$$) and then at ($$$4$$$, $$$4$$$), or the other way around. They both result in the same configuration.

Buggy Code:
import sys
input = sys.stdin.readline

for _ in range(int(input())):
    n, k = map(int, input().split())
    if n <= 1:
        print(n)
    if n == 2:
        if k > 0:
            print(1)
        else:
            print(3)
    for _ in range(k):
        row, col = map(int, input().split())
        if row == col:
            n -= 1
        else:
            n -= 2
    a = 1
    b = 3
    count = 4
    for i in range(3, n + 1):
        temp = (b + (count * a)) % 1000000007
        a = b
        b = temp
        count += 2
    print(b)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: How Does the Rook Move?
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an $$$n \times n$$$ chessboard where you and the computer take turns alternatingly to place white rooks & black rooks on the board respectively. While placing rooks, you have to ensure that no two rooks attack each other. Two rooks attack each other if they share the same row or column regardless of color.

A valid move is placing a rook on a position ($$$r$$$, $$$c$$$) such that it doesn't attack any other rook.

You start first, and when you make a valid move in your turn, placing a white rook at position ($$$r$$$, $$$c$$$), the computer will mirror you and place a black rook at position ($$$c$$$, $$$r$$$) in its turn. If $$$r = c$$$, then the computer can't mirror your move, and skips its turn.

You have already played $$$k$$$ moves with the computer (the computer tries to mirror these moves too), and you must continue playing the game until there are no valid moves remaining. How many different final configurations are possible when you continue the game after the $$$k$$$ moves? It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid. Since the answer may be large, print it modulo $$$10^9+7$$$.

Two configurations are considered different if there exists a coordinate ($$$r$$$, $$$c$$$) which has a rook in one configuration, but not in the other or the color of the rook on the coordinate is different.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 3 \cdot 10^5$$$, $$$0 \leq k \leq n$$$) — the size of the chessboard and the number of moves you have already played respectively.

Each of the next $$$k$$$ lines of the test case contains two integers $$$r_i$$$ and $$$c_i$$$, denoting the $$$i$$$-th move you made.

It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, output a single integer on a new line — the total number of possible final configurations modulo $$$10^9+7$$$.

Examples:
input:
3
4 1
1 2
8 1
7 6
1000 4
4 4
952 343
222 333
90 91
output:
3
331
671968183

Note: In the first test case, we have a $$$4 \times 4$$$ grid and you've already played $$$1$$$ move. After you and the computer play a turn, we have a white rook at ($$$1$$$, $$$2$$$), and a black rook at ($$$2$$$, $$$1$$$). There are three possible configurations from this state —

1. You place a white rook at ($$$3$$$, $$$4$$$) and the computer places a black rook at ($$$4$$$, $$$3$$$) as a response.
2. You place a white rook at ($$$4$$$, $$$3$$$) and the computer places a black rook at ($$$3$$$, $$$4$$$) as a response.
3. You place a white rook at ($$$3$$$, $$$3$$$) and then at ($$$4$$$, $$$4$$$), or the other way around. They both result in the same configuration.

Buggy Code:
import java.util.*;

public class Main {
    static int mod = (int)1e9 + 7;
    public static void main(String[] args) {
        Scanner io = new Scanner(System.in);

        long[] a = new long[300005];
        a[1] = 1;
        a[2] = 3;
        for(int i = 3; i < a.length; i++) {
            a[i] = (((a[i - 2] * 2) % mod * (i - 1)) % mod + a[i - 1]) % mod;
        }
        int t = io.nextInt();
        while(t -- > 0) {
            int n = io.nextInt();
            int k = io.nextInt();
            while(k -- > 0) {
                int aa = io.nextInt();
                int bb = io.nextInt();
                if(aa != bb)    n -= 2;
                else    n--;
            }
            System.out.println(a[n]);
        }
    }

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: How Does the Rook Move?
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an $$$n \times n$$$ chessboard where you and the computer take turns alternatingly to place white rooks & black rooks on the board respectively. While placing rooks, you have to ensure that no two rooks attack each other. Two rooks attack each other if they share the same row or column regardless of color.

A valid move is placing a rook on a position ($$$r$$$, $$$c$$$) such that it doesn't attack any other rook.

You start first, and when you make a valid move in your turn, placing a white rook at position ($$$r$$$, $$$c$$$), the computer will mirror you and place a black rook at position ($$$c$$$, $$$r$$$) in its turn. If $$$r = c$$$, then the computer can't mirror your move, and skips its turn.

You have already played $$$k$$$ moves with the computer (the computer tries to mirror these moves too), and you must continue playing the game until there are no valid moves remaining. How many different final configurations are possible when you continue the game after the $$$k$$$ moves? It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid. Since the answer may be large, print it modulo $$$10^9+7$$$.

Two configurations are considered different if there exists a coordinate ($$$r$$$, $$$c$$$) which has a rook in one configuration, but not in the other or the color of the rook on the coordinate is different.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 3 \cdot 10^5$$$, $$$0 \leq k \leq n$$$) — the size of the chessboard and the number of moves you have already played respectively.

Each of the next $$$k$$$ lines of the test case contains two integers $$$r_i$$$ and $$$c_i$$$, denoting the $$$i$$$-th move you made.

It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, output a single integer on a new line — the total number of possible final configurations modulo $$$10^9+7$$$.

Examples:
input:
3
4 1
1 2
8 1
7 6
1000 4
4 4
952 343
222 333
90 91
output:
3
331
671968183

Note: In the first test case, we have a $$$4 \times 4$$$ grid and you've already played $$$1$$$ move. After you and the computer play a turn, we have a white rook at ($$$1$$$, $$$2$$$), and a black rook at ($$$2$$$, $$$1$$$). There are three possible configurations from this state —

1. You place a white rook at ($$$3$$$, $$$4$$$) and the computer places a black rook at ($$$4$$$, $$$3$$$) as a response.
2. You place a white rook at ($$$4$$$, $$$3$$$) and the computer places a black rook at ($$$3$$$, $$$4$$$) as a response.
3. You place a white rook at ($$$3$$$, $$$3$$$) and then at ($$$4$$$, $$$4$$$), or the other way around. They both result in the same configuration.

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

public class C_How_Does_the_Rook_Move {
    static long mod = (long) 1e9 + 7;

    // My code is my Identity
    public static void main(String[] args) {
        FastScanner sc = new FastScanner(System.in);
        PrintWriter pr = new PrintWriter(System.out);
        long t = sc.nextLong();
        long dp[] = new long[(int) 1e6];
        Arrays.fill(dp, -1);
        dp[0] = 0;
        dp[1] = 1;
        dp[2] = 3;
        first: while (t-- > 0) {
            // Vector <Pair> = new Vector<>();
            // Pair[] pairs=new Pair[n];
            int n = sc.nextInt();
            int k = sc.nextInt();
            for (int i = 0; i < k; i++) {
                int l = sc.nextInt();
                int r = sc.nextInt();
                if (l == r)
                    n--;
                else
                    n -= 2;
            }
            long ans = recur(dp, n);
            System.out.println(ans);
        }
        pr.close();
    }

    static long recur(long dp[], int n) {
        if (dp[n] != -1)
            return dp[n];
        long ans = (recur(dp,n-1)%mod +(2*(n-1)*(recur(dp, n - 2)) % mod)) % mod;
        return dp[n] = ans;
    }

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

    static long andOperator(long a, long b) {
        int shiftcount = 0;
        while (a != b && a > 0) {
            shiftcount++;
            a = a >> 1;
            b = b >> 1;
        }
        return (a << shiftcount);
    }

}

class ArrayOfPairsSorter {
    static void sort(Pair[] arr) {
        Comparator<Pair> comparator = new Comparator<>() {
            @Override
            public int compare(Pair p1, Pair p2) {
                return p2.y
                        - p1.y; // To compare the first element
                                // just
                                // change the variable from p1.y
                                // - p2.y to p1.x-p2.x.
            }
        };
        Arrays.sort(arr, comparator);
    }
}

class Pair {
    int x;
    int y;

    // Constructor
    public Pair(int x, int y) {
        this.x = x;
        this.y = y;
    }

    public int getX() {
        return x;
    }

    public int getY() {
        return y;
    }

    // Overriding toString method
    // for beautiful printing of pairs
    @Override
    public String toString() {
        return "(" + x + "," + y + ")";
    }
}

class FastScanner {
    private BufferedReader reader = null;
    private StringTokenizer tokenizer = null;

    public FastScanner(InputStream in) {
        reader = new BufferedReader(new InputStreamReader(in));
        tokenizer = null;
    }

    public String next() {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }

    public String nextLine() {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                return reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }

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

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

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

    public int[] nextIntArray(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++)
            a[i] = nextInt();
        return a;
    }

    public long[] nextLongArray(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++)
            a[i] = nextLong();
        return a;
    }
}

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

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

Think step by step.

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

1. **Initial Moves on the Diagonal**: Moves where `r = c` should reduce the board size by 1, not 2.
2. **Initial Moves Off the Diagonal**: Moves where `r ≠ c` should reduce the board size by 2.
3. **Small Board Size**: Smaller board sizes can help in verifying the correctness of the dynamic programming (DP) array.
4. **Multiple Test Cases**: Including multiple test cases to ensure the code handles different scenarios.

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

- A test case with a small board size and a move on the diagonal.
- A test case with a small board size and a move off the diagonal.
- A test case with a small board size and multiple moves.

Here is the test input:

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

### Explanation:
1. **Test Case 1**:
   - Board size: 4
   - Moves: (1, 1)
   - Since `r = c`, the board size should be reduced by 1, resulting in a 3x3 board.
   - Expected configurations: 3 (as the DP array should correctly handle the reduced board size).

2. **Test Case 2**:
   - Board size: 4
   - Moves: (1, 2)
   - Since `r ≠ c`, the board size should be reduced by 2, resulting in a 2x2 board.
   - Expected configurations: 3 (as the DP array should correctly handle the reduced board size).

3. **Test Case 3**:
   - Board size: 4
   - Moves: (1, 2), (2, 1)
   - Both moves are off the diagonal, so the board size should be reduced by 2 for each move, resulting in a 0x0 board.
   - Expected configurations: 1 (as the board is fully occupied).

This test input should help in identifying the bug in the DP array initialization and the logic for reducing the board size based on the moves.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: How Does the Rook Move?
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an $$$n \times n$$$ chessboard where you and the computer take turns alternatingly to place white rooks & black rooks on the board respectively. While placing rooks, you have to ensure that no two rooks attack each other. Two rooks attack each other if they share the same row or column regardless of color.

A valid move is placing a rook on a position ($$$r$$$, $$$c$$$) such that it doesn't attack any other rook.

You start first, and when you make a valid move in your turn, placing a white rook at position ($$$r$$$, $$$c$$$), the computer will mirror you and place a black rook at position ($$$c$$$, $$$r$$$) in its turn. If $$$r = c$$$, then the computer can't mirror your move, and skips its turn.

You have already played $$$k$$$ moves with the computer (the computer tries to mirror these moves too), and you must continue playing the game until there are no valid moves remaining. How many different final configurations are possible when you continue the game after the $$$k$$$ moves? It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid. Since the answer may be large, print it modulo $$$10^9+7$$$.

Two configurations are considered different if there exists a coordinate ($$$r$$$, $$$c$$$) which has a rook in one configuration, but not in the other or the color of the rook on the coordinate is different.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 3 \cdot 10^5$$$, $$$0 \leq k \leq n$$$) — the size of the chessboard and the number of moves you have already played respectively.

Each of the next $$$k$$$ lines of the test case contains two integers $$$r_i$$$ and $$$c_i$$$, denoting the $$$i$$$-th move you made.

It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, output a single integer on a new line — the total number of possible final configurations modulo $$$10^9+7$$$.

Examples:
input:
3
4 1
1 2
8 1
7 6
1000 4
4 4
952 343
222 333
90 91
output:
3
331
671968183

Note: In the first test case, we have a $$$4 \times 4$$$ grid and you've already played $$$1$$$ move. After you and the computer play a turn, we have a white rook at ($$$1$$$, $$$2$$$), and a black rook at ($$$2$$$, $$$1$$$). There are three possible configurations from this state —

1. You place a white rook at ($$$3$$$, $$$4$$$) and the computer places a black rook at ($$$4$$$, $$$3$$$) as a response.
2. You place a white rook at ($$$4$$$, $$$3$$$) and the computer places a black rook at ($$$3$$$, $$$4$$$) as a response.
3. You place a white rook at ($$$3$$$, $$$3$$$) and then at ($$$4$$$, $$$4$$$), or the other way around. They both result in the same configuration.

Buggy Code:
#ifndef ONLINE_JUDGE
#include "debug.cpp"
#else
#define debug(...)
#define debugArr(...)
#endif
#include<bits/stdc++.h>
#define ll long long
#define ld long double
#define nl '\n'
#define no "NO\n"
#define yes "YES\n"
#define forin(s, n, k) for (ll i = s; i < n; i += k)
#define forjn(s, n, k) for (ll j = s; j < n; j += k)
#define forde(s, n, k) for (ll i = s; i > n; i -= k)
#define all(v) v.begin(), v.end()
#define allr(v) v.rbegin(), v.rend()
#define pb push_back
#define allin(s, n, vec) forin(s, n, 1) cin >> vec[i]
#define vll vector<ll>
#define vi vector<int>
#define vc vector<char>
#define vs vector<string>
ll min(ll a, ll b) { return (a < b) ? a : b; }
ll max(ll a, ll b) { return (a > b) ? a : b; }
ll GCD(ll a, ll b) { return b == 0 ? a : GCD(b, a % b); }
ll LCM(ll a, ll b) { return (a * b) / GCD(a, b); }
ll MOD=1e9+7;

using namespace std;

ll help(ll n, vll &dp)
{
    if(n<=1)
        return 1;
    else if(dp[n]!=0)
        return dp[n];
    else
        return dp[n] = help(n-1,dp)%MOD + ((help(n-2,dp)%MOD*(n-1)%MOD)*2%MOD)%MOD;
}

void solve()
{
    ll n,k;  cin >> n >>k;
    ll cnt=n;
    while(k--)
    {
        ll a,b;    cin >> a >> b;
        if(a==b) cnt--;
        else    cnt-=2;
    }
    vll dp(cnt+1,0);
    ll out=help(cnt,dp);
    cout << out << nl;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll tc;
    cin >> tc;
    while (tc--)
        solve();
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases or scenarios that can highlight issues in the logic. The key points to consider are:

1. **Initial Moves and Board Size**: The code processes the initial moves and updates the count of available positions. We need to ensure that the initial moves are valid and do not violate the rules of the game.
2. **Mirroring Moves**: The computer mirrors the player's moves, so we need to ensure that the initial moves are such that the computer can mirror them.
3. **Final Configurations**: The code calculates the number of possible final configurations. We need to ensure that the initial moves and the board size are such that the final configurations can be challenging to compute.

Given these considerations, let's create a test case that maximizes the complexity and exposes potential issues:

1. **Small Board Size**: Start with a small board size to ensure the test case is manageable.
2. **Initial Moves**: Place initial moves that are valid and ensure the computer can mirror them.
3. **Edge Cases**: Include moves where the player and the computer place rooks on the same row or column to test the logic.

Here is a test input that meets these criteria:

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

### Explanation:
- **Board Size**: 5x5.
- **Initial Moves**:
  - The first move places a white rook at (1, 2) and the computer mirrors it by placing a black rook at (2, 1).
  - The second move places a white rook at (2, 1) and the computer mirrors it by placing a black rook at (1, 2).

This test case is designed to:
- Ensure the initial moves are valid and the computer can mirror them.
- Test the logic for counting available positions and calculating the number of possible final configurations.
- Check how the code handles the mirroring of moves and the reduction of available positions.

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: How Does the Rook Move?
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an $$$n \times n$$$ chessboard where you and the computer take turns alternatingly to place white rooks & black rooks on the board respectively. While placing rooks, you have to ensure that no two rooks attack each other. Two rooks attack each other if they share the same row or column regardless of color.

A valid move is placing a rook on a position ($$$r$$$, $$$c$$$) such that it doesn't attack any other rook.

You start first, and when you make a valid move in your turn, placing a white rook at position ($$$r$$$, $$$c$$$), the computer will mirror you and place a black rook at position ($$$c$$$, $$$r$$$) in its turn. If $$$r = c$$$, then the computer can't mirror your move, and skips its turn.

You have already played $$$k$$$ moves with the computer (the computer tries to mirror these moves too), and you must continue playing the game until there are no valid moves remaining. How many different final configurations are possible when you continue the game after the $$$k$$$ moves? It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid. Since the answer may be large, print it modulo $$$10^9+7$$$.

Two configurations are considered different if there exists a coordinate ($$$r$$$, $$$c$$$) which has a rook in one configuration, but not in the other or the color of the rook on the coordinate is different.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 3 \cdot 10^5$$$, $$$0 \leq k \leq n$$$) — the size of the chessboard and the number of moves you have already played respectively.

Each of the next $$$k$$$ lines of the test case contains two integers $$$r_i$$$ and $$$c_i$$$, denoting the $$$i$$$-th move you made.

It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, output a single integer on a new line — the total number of possible final configurations modulo $$$10^9+7$$$.

Examples:
input:
3
4 1
1 2
8 1
7 6
1000 4
4 4
952 343
222 333
90 91
output:
3
331
671968183

Note: In the first test case, we have a $$$4 \times 4$$$ grid and you've already played $$$1$$$ move. After you and the computer play a turn, we have a white rook at ($$$1$$$, $$$2$$$), and a black rook at ($$$2$$$, $$$1$$$). There are three possible configurations from this state —

1. You place a white rook at ($$$3$$$, $$$4$$$) and the computer places a black rook at ($$$4$$$, $$$3$$$) as a response.
2. You place a white rook at ($$$4$$$, $$$3$$$) and the computer places a black rook at ($$$3$$$, $$$4$$$) as a response.
3. You place a white rook at ($$$3$$$, $$$3$$$) and then at ($$$4$$$, $$$4$$$), or the other way around. They both result in the same configuration.

Buggy Code:
#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
#include<unordered_map>
using namespace std;
#define for1 for(int i = 1;i <= n;i++)
#define for0 for(int i = 0;i < n;i++)
#define forn1 for(int j = 1;j <= n;j++)
#define forn0 for(int j = 0;j < n;j++)
#define form1 for (int j = 1; j <= m; j++)
#define form0 for (int j = 0; j < m; j++)
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define arrn int* arr = new int[n+1];arr[0] = 0
#define carr cin >> arr[i]
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#define mod  1000000007
double gcd(int a, int b)
{
	if (b == 0)return a;
	return gcd(b, a % b);
}
const int MOD = 1e9 + 7;
#define int long long
signed main() {
    int t;
    cin >> t;
    while (t--) {
        int n, k;
        cin >> n >> k;
        int l = n;
        for (int i = 1; i <= k; i++)
        {
            int x, y;
            cin >> x >> y;
            if (x != y)l -= 2;
            else l -= 1;
        }
        long long arr[10000] = { 0 };
        arr[0] = 1;
        arr[1] = 1, arr[2] = 3;
        for (int i = 3; i <= l+1; i++)
        {
            arr[i] = (((2*i - 2) * (arr[i - 2]))%MOD + arr[i - 1]%MOD )% MOD;
        }
        cout << arr[l] << endl;
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that maximize the failure probability. The code has a few potential issues:

1. **Handling of initial moves**: The code reduces the available positions based on the initial moves, but it might not correctly handle the case where the number of moves is very small or very large.
2. **Dynamic programming array size**: The array `arr` is initialized with a fixed size of 10000, which might not be sufficient for larger values of `l`.
3. **Modulo operation**: The code uses modulo operations, but it might not handle large intermediate values correctly.

Let's create a small-scale test input that can expose these issues:

1. **Small board with a few initial moves**: This can test the handling of the initial moves and the reduction of available positions.
2. **Large board with a few initial moves**: This can test the handling of large values of `l` and the dynamic programming array size.

Here is a small-scale test input:

```plaintext
3
4 1
1 2
8 1
7 6
100 1
1 1
```

Explanation:
- **Test Case 1**: A 4x4 board with 1 initial move at (1, 2). This should test the handling of a small board with a few initial moves.
- **Test Case 2**: An 8x8 board with 1 initial move at (7, 6). This should test the handling of a larger board with a few initial moves.
- **Test Case 3**: A 100x100 board with 1 initial move at (1, 1). This should test the handling of a very large board with a few initial moves and the dynamic programming array size.

This input should help expose any issues in the code's handling of initial moves and the dynamic programming array size.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: How Does the Rook Move?
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an $$$n \times n$$$ chessboard where you and the computer take turns alternatingly to place white rooks & black rooks on the board respectively. While placing rooks, you have to ensure that no two rooks attack each other. Two rooks attack each other if they share the same row or column regardless of color.

A valid move is placing a rook on a position ($$$r$$$, $$$c$$$) such that it doesn't attack any other rook.

You start first, and when you make a valid move in your turn, placing a white rook at position ($$$r$$$, $$$c$$$), the computer will mirror you and place a black rook at position ($$$c$$$, $$$r$$$) in its turn. If $$$r = c$$$, then the computer can't mirror your move, and skips its turn.

You have already played $$$k$$$ moves with the computer (the computer tries to mirror these moves too), and you must continue playing the game until there are no valid moves remaining. How many different final configurations are possible when you continue the game after the $$$k$$$ moves? It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid. Since the answer may be large, print it modulo $$$10^9+7$$$.

Two configurations are considered different if there exists a coordinate ($$$r$$$, $$$c$$$) which has a rook in one configuration, but not in the other or the color of the rook on the coordinate is different.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 3 \cdot 10^5$$$, $$$0 \leq k \leq n$$$) — the size of the chessboard and the number of moves you have already played respectively.

Each of the next $$$k$$$ lines of the test case contains two integers $$$r_i$$$ and $$$c_i$$$, denoting the $$$i$$$-th move you made.

It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, output a single integer on a new line — the total number of possible final configurations modulo $$$10^9+7$$$.

Examples:
input:
3
4 1
1 2
8 1
7 6
1000 4
4 4
952 343
222 333
90 91
output:
3
331
671968183

Note: In the first test case, we have a $$$4 \times 4$$$ grid and you've already played $$$1$$$ move. After you and the computer play a turn, we have a white rook at ($$$1$$$, $$$2$$$), and a black rook at ($$$2$$$, $$$1$$$). There are three possible configurations from this state —

1. You place a white rook at ($$$3$$$, $$$4$$$) and the computer places a black rook at ($$$4$$$, $$$3$$$) as a response.
2. You place a white rook at ($$$4$$$, $$$3$$$) and the computer places a black rook at ($$$3$$$, $$$4$$$) as a response.
3. You place a white rook at ($$$3$$$, $$$3$$$) and then at ($$$4$$$, $$$4$$$), or the other way around. They both result in the same configuration.

Buggy Code:
t=int(input())
for i in range(t):
    n, k=list(map(int, input().split()))
    moves=[]
    for j in range(k):
        moves.append((list(map(int, input().split()))))
    dim=n
    for j in moves:
        if j[0]==j[1]:
            dim-=1
        else:
            dim-=2
    # print('dim', dim)
    non_diag=dim//2
    diag=dim-2*non_diag
    sum=0
    while non_diag>=0:
        product=1
        for j in range(dim-diag+1, dim+1):
            product*=j%1000000007
        for j in range(1, diag+1):
            product//=j%1000000007
        for j in range(non_diag+1, 2*non_diag+1):
            product*=j%1000000007
        sum+=product%1000000007
        sum=sum%1000000007 # (dim!/diag! (dim-diag)!) * 2*non_diag!/non_diag!
        #print(f'diag {diag}, non_diag {non_diag} sum {sum%1000000007}')
        diag+=2
        non_diag-=1
    print(sum%1000000007)

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

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

Think step by step.

Response:
```plaintext
3
4 2
1 2
2 1
8 2
7 6
6 7
100 2
50 50
50 51
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: How Does the Rook Move?
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an $$$n \times n$$$ chessboard where you and the computer take turns alternatingly to place white rooks & black rooks on the board respectively. While placing rooks, you have to ensure that no two rooks attack each other. Two rooks attack each other if they share the same row or column regardless of color.

A valid move is placing a rook on a position ($$$r$$$, $$$c$$$) such that it doesn't attack any other rook.

You start first, and when you make a valid move in your turn, placing a white rook at position ($$$r$$$, $$$c$$$), the computer will mirror you and place a black rook at position ($$$c$$$, $$$r$$$) in its turn. If $$$r = c$$$, then the computer can't mirror your move, and skips its turn.

You have already played $$$k$$$ moves with the computer (the computer tries to mirror these moves too), and you must continue playing the game until there are no valid moves remaining. How many different final configurations are possible when you continue the game after the $$$k$$$ moves? It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid. Since the answer may be large, print it modulo $$$10^9+7$$$.

Two configurations are considered different if there exists a coordinate ($$$r$$$, $$$c$$$) which has a rook in one configuration, but not in the other or the color of the rook on the coordinate is different.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 3 \cdot 10^5$$$, $$$0 \leq k \leq n$$$) — the size of the chessboard and the number of moves you have already played respectively.

Each of the next $$$k$$$ lines of the test case contains two integers $$$r_i$$$ and $$$c_i$$$, denoting the $$$i$$$-th move you made.

It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, output a single integer on a new line — the total number of possible final configurations modulo $$$10^9+7$$$.

Examples:
input:
3
4 1
1 2
8 1
7 6
1000 4
4 4
952 343
222 333
90 91
output:
3
331
671968183

Note: In the first test case, we have a $$$4 \times 4$$$ grid and you've already played $$$1$$$ move. After you and the computer play a turn, we have a white rook at ($$$1$$$, $$$2$$$), and a black rook at ($$$2$$$, $$$1$$$). There are three possible configurations from this state —

1. You place a white rook at ($$$3$$$, $$$4$$$) and the computer places a black rook at ($$$4$$$, $$$3$$$) as a response.
2. You place a white rook at ($$$4$$$, $$$3$$$) and the computer places a black rook at ($$$3$$$, $$$4$$$) as a response.
3. You place a white rook at ($$$3$$$, $$$3$$$) and then at ($$$4$$$, $$$4$$$), or the other way around. They both result in the same configuration.

Buggy Code:
from math import comb
def recursive(num):
    if num == 0:
        return 1
    else:
        result = recursive(num - 2) * (num - 1) * 2
        return (result) % (1000000007)

def recursive2(num):
    a = num
    b = 0
    result = 0
    while num >= b:
        p = comb(num, b) % 1000000007
        result += (p * recursive(a)) % (1000000007)
        result = result % (1000000007)
        b += 2
        a -= 2
    return (result) % (1000000007)

def solve():
    n, k = map(int, input().split())
    valuesleft = n
    for x in range(k):
        r, c = input().split()
        if r == c:
            valuesleft -= 1
        else:
            valuesleft -= 2
    if valuesleft == 0:
        print(1)
    elif valuesleft == 1:
        print(1)
    else:
        if valuesleft % 2 == 1:
            result = (recursive2(valuesleft + 1)) % (1000000007)
            result -= 2*valuesleft*recursive2(valuesleft-1) % 1000000007
            result = result % 1000000007
            print(result)
        else:
            result = recursive2(valuesleft) % (1000000007)
            print(result)

for _ in range(int(input())):
    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
3
4 1
1 1
8 2
3 3
7 7
1000 500
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
10 11
11 12
12 13
13 14
14 15
15 16
16 17
17 18
18 19
19 20
20 21
21 22
22 23
23 24
24 25
25 26
26 27
27 28
28 29
29 30
30 31
31 32
32 33
33 34
34 35
35 36
36 37
37 38
38 39
39 40
40 41
41 42
42 43
43 44
44 45
45 46
46 47
47 48
48 49
49 50
50 51
51 52
52 53
53 54
54 55
55 56
56 57
57 58
58 59
59 60
60 61
61 62
62 63
63 64
64 65
65 66
66 67
67 68
68 69
69 70
70 71
71 72
72 73
73 74
74 75
75 76
76 77
77 78
78 79
79 80
80 81
81 82
82 83
83 84
84 85
85 86
86 87
87 88
88 89
89 90
90 91
91 92
92 93
93 94
94 95
95 96
96 97
97 98
98 99
99 100
100 101
101 102
102 103
103 104
104 105
105 106
106 107
107 108
108 109
109 110
110 111
111 112
112 113
113 114
114 115
115 116
116 117
117 118
118 119
119 120
120 121
121 122
122 123
123 124
124 125
125 126
126 127
127 128
128 129
129 130
130 131
131 132
132 133
133 134
134 135
135 136
136 137
137 138
138 139
139 140
140 141
141 142
142 143
143 144
144 145
145 146
146 147
147 148
148 149
149 150
150 151
151 152
152 153
153 154
154 155
155 156
156 157
157 158
158 159
159 160
160 161
161 162
162 163
163 164
164 165
165 166
166 167
167 168
168 169
169 170
170 171
171 172
172 173
173 174
174 175
175 176
176 177
177 178
178 179
179 180
180 181
181 182
182 183
183 184
184 185
185 186
186 187
187 188
188 189
189 190
190 191
191 192
192 193
193 194
194 195
195 196
196 197
197 198
198 199
199 200
200 201
201 202
202 203
203 204
204 205
205 206
206 207
207 208
208 209
209 210
210 211
211 212
212 213
213 214
214 215
215 216
216 217
217 218
218 219
219 220
220 221
221 222
222 223
223 224
224 225
225 226
226 227
227 228
228 229
229 230
230 231
231 232
232 233
233 234
234 235
235 236
236 237
237 238
238 239
239 240
240 241
241 242
242 243
243 244
244 245
245 246
246 247
247 248
248 249
249 250
250 251
251 252
252 253
253 254
254 255
255 256
256 257
257 258
258 259
259 260
260 261
261 262
262 263
263 264
264 265
265 266
266 267
267 268
268 269
269 270
270 271
271 272
272 273
273 274
274 275
275 276
276 277
277 278
278 279
279


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: How Does the Rook Move?
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an $$$n \times n$$$ chessboard where you and the computer take turns alternatingly to place white rooks & black rooks on the board respectively. While placing rooks, you have to ensure that no two rooks attack each other. Two rooks attack each other if they share the same row or column regardless of color.

A valid move is placing a rook on a position ($$$r$$$, $$$c$$$) such that it doesn't attack any other rook.

You start first, and when you make a valid move in your turn, placing a white rook at position ($$$r$$$, $$$c$$$), the computer will mirror you and place a black rook at position ($$$c$$$, $$$r$$$) in its turn. If $$$r = c$$$, then the computer can't mirror your move, and skips its turn.

You have already played $$$k$$$ moves with the computer (the computer tries to mirror these moves too), and you must continue playing the game until there are no valid moves remaining. How many different final configurations are possible when you continue the game after the $$$k$$$ moves? It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid. Since the answer may be large, print it modulo $$$10^9+7$$$.

Two configurations are considered different if there exists a coordinate ($$$r$$$, $$$c$$$) which has a rook in one configuration, but not in the other or the color of the rook on the coordinate is different.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 3 \cdot 10^5$$$, $$$0 \leq k \leq n$$$) — the size of the chessboard and the number of moves you have already played respectively.

Each of the next $$$k$$$ lines of the test case contains two integers $$$r_i$$$ and $$$c_i$$$, denoting the $$$i$$$-th move you made.

It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, output a single integer on a new line — the total number of possible final configurations modulo $$$10^9+7$$$.

Examples:
input:
3
4 1
1 2
8 1
7 6
1000 4
4 4
952 343
222 333
90 91
output:
3
331
671968183

Note: In the first test case, we have a $$$4 \times 4$$$ grid and you've already played $$$1$$$ move. After you and the computer play a turn, we have a white rook at ($$$1$$$, $$$2$$$), and a black rook at ($$$2$$$, $$$1$$$). There are three possible configurations from this state —

1. You place a white rook at ($$$3$$$, $$$4$$$) and the computer places a black rook at ($$$4$$$, $$$3$$$) as a response.
2. You place a white rook at ($$$4$$$, $$$3$$$) and the computer places a black rook at ($$$3$$$, $$$4$$$) as a response.
3. You place a white rook at ($$$3$$$, $$$3$$$) and then at ($$$4$$$, $$$4$$$), or the other way around. They both result in the same configuration.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
import static java.lang.Math.*;
import static java.util.Arrays.sort;

public class Main {
    static FastReader fs;
    static PrintWriter out;
    static int mod =  (int) 1e9+7;
    static long fact[],invfact[];
    static ArrayList<Integer> adj [];

    public static void main(String[] args) {
        fs = new FastReader();
        out = new PrintWriter(System.out);
//        int tt = 1;
        int tt  = fs.fi();

        for (int test = 1; test <= tt; test++) {
            int n =fs.fi();
            int k =fs.fi();
            HashSet<Integer> set = new HashSet<>();
            for(int i = 0; i < k; i++){
                int x  =fs.fi();
                int y = fs.fi();
                set.add(x);
                set.add(y);
            }
            int size = n-set.size();
            long dp [] = new long[size+2];
            dp[1] = 1;
            dp[0] = 1;
            for(int i = 2; i <=size; i++){
                dp[i] = (((i-1L) * 2L)%mod * (dp[i-2]))%mod + dp[i-1]%mod;
            }
            out.println(dp[size]);
        }

        out.close();

    }

    static class Pair2 {
        String s;
        long count;

        public Pair2(String s, long count) {
            this.s = s;

            this.count = count;
        }

    }

    static class MergeSortTree {
        int n;
        List<Integer>[] tree;

        public MergeSortTree(int[] a) {
            this.n = a.length;
            int size = 1;
            while (size < n) {
                size *= 2;
            }
            size *= 2;
            tree = new ArrayList[size];
            for (int i = 0; i < size; i++) {
                tree[i] = new ArrayList<>();
            }
            build(a, 0, 0, n);
            for (int i = 0; i < size; i++) {
                // if (test) System.out.format("  i:%2d %s\n", i, trace(tree[i]));
            }
        }

        void build(int[] a, int x, int l, int r) {
            if (l + 1 == r) {
                tree[x].add(a[l]);
                return;
            }
            int m = (l + r) / 2;
            build(a, 2 * x + 1, l, m);
            build(a, 2 * x + 2, m, r);
            merge(tree[2 * x + 1], tree[2 * x + 2], tree[x]);
        }

        int count(int lq, int rq, int mn, int mx, int x, int l, int r) {
            if (rq <= l || r <= lq) {
                return 0;
            }
            if (lq <= l && r <= rq) {
                int e = Collections.binarySearch(tree[x], mx);
                if (e < 0) {
                    e = -(e+1);
                }
                int b = Collections.binarySearch(tree[x], mn);
                if (b < 0) {
                    b = -(b+1);
                }
                return e - b;
            }

            int m = (l + r) / 2;
            int a = count(lq, rq, mn, mx, 2 * x + 1, l, m);
            int b = count(lq, rq, mn, mx, 2 * x + 2, m, r);
            return a + b;
        }

        int count(int lq, int rq, int mn, int mx) {
            return count(lq, rq, mn, mx, 0, 0, n);
        }

        static void merge(List<Integer> a, List<Integer> b, List<Integer> out) {
            int i = 0;
            int j = 0;
            while (i < a.size() && j < b.size()) {
                if (a.get(i) <= b.get(j)) {
                    out.add(a.get(i++));
                } else {
                    out.add(b.get(j++));
                }
            }
            while (i < a.size()) {
                out.add(a.get(i++));
            }
            while (j < b.size()) {
                out.add(b.get(j++));
            }
        }
    }

    static boolean found = false;

    static class Union {
        int[] id, sz; int comps;

        // 0 to n
        public Union(int n) {
            n--;
            id = new int[n+1]; sz = new int[n+1];
            comps = n+1;
            for (int i = 0; i < n + 1; i++) {
                id[i] = i;
            }
            Arrays.fill(sz, 1);
        }

        public int find(int a) {
            return id[a]=(id[a]==a?a:find(id[a]));
        }

        public boolean con(int a, int b) {return find(a)==find(b);}

        public int size(int a) {return sz[find(a)];}

        public int comps() {return comps;}

        public void unify(int a, int b) {
            a = find(a); b = find(b);
            if(a==b)return;
            if(sz[a] >= sz[b]) {
                id[b] = a;
                sz[a] += sz[b];
            } else {
                id[a] = b;
                sz[b] += sz[a];
            }
            comps--;
        }
    }
    public static  void cal(int n){
        fact= new long[n];
        invfact = new long[n];
        fact[0] = 1;
        invfact[0] = 1;
        for(int i =1; i < n; i++){
            fact[i] = (fact[i-1]*i)%mod;
            invfact[i] = (modInverse(fact[i]));
        }

    }

    public static  long NCR(int n, int r){
        if(r < 0 || n < 0){
            return -1;
        }
        if(n < r) return 0;
        if(n == 0 || r == n) return 1;
        return (fact[n] * (invfact[r]) % mod) * invfact[n-r]%mod;
    }

    public   static  long modInverse(long n){
        return powMod(n,mod-2);
    }
    public  static long powMod(long x, long y){
        long res = 1;
        x = x%mod;
        while(y > 0){
            if((y&1) == 1){
                res = (res*x)%mod;
            }
            y = y>>1;
            x = (x*x)%mod;
        }
        return res;
    }

    public static long [] swap(long a , long b){
        return new long[]{b,a};
    }

    static class Pair implements Comparable<Pair> {
        int start;
        int end;

        Pair(int first, int second) {
            this.start = first;
            this.end = second;
        }
        @Override
        public int compareTo(Pair a) {
            return this.start - a.end;
        }
    }

    static class Node {

        int val;
        int index;

        public Node(int val, int index) {
            this.val = val;
            this.index = index;
        }
    }

    // constants
    static final int IBIG = 1000000007;
    static final int IMAX = 2147483647;
    static final long LMAX = 9223372036854775807L;
    static Random __r = new Random();
    static int[][] direction = new int[][]{{-1,-1}, {-1,0}, {-1,1},  {0,1}, {1,1},  {1,0},  {1,-1},  {0, -1}};

    // math util
    static int minof(int a, int b, int c) {
        return min(a, min(b, c));
    }

    static int minof(int... x) {
        if (x.length == 1)
            return x[0];
        if (x.length == 2)
            return min(x[0], x[1]);
        if (x.length == 3)
            return min(x[0], min(x[1], x[2]));
        int min = x[0];
        for (int i = 1; i < x.length; ++i)
            if (x[i] < min)
                min = x[i];
        return min;
    }
    static char maxc(char a, char b){
        if(a > b){
            return a;
        }
        return b;
    }
    static long minof(long a, long b, long c) {
        return min(a, min(b, c));
    }

    static long minof(long... x) {
        if (x.length == 1)
            return x[0];
        if (x.length == 2)
            return min(x[0], x[1]);
        if (x.length == 3)
            return min(x[0], min(x[1], x[2]));
        long min = x[0];
        for (int i = 1; i < x.length; ++i)
            if (x[i] < min)
                min = x[i];
        return min;
    }

    static int maxof(int a, int b, int c) {
        return max(a, max(b, c));
    }

    static int maxof(int... x) {
        if (x.length == 1)
            return x[0];
        if (x.length == 2)
            return max(x[0], x[1]);
        if (x.length == 3)
            return max(x[0], max(x[1], x[2]));
        int max = x[0];
        for (int i = 1; i < x.length; ++i)
            if (x[i] > max)
                max = x[i];
        return max;
    }

    static long maxof(long a, long b, long c) {
        return max(a, max(b, c));
    }

    static long maxof(long... x) {
        if (x.length == 1)
            return x[0];
        if (x.length == 2)
            return max(x[0], x[1]);
        if (x.length == 3)
            return max(x[0], max(x[1], x[2]));
        long max = x[0];
        for (int i = 1; i < x.length; ++i)
            if (x[i] > max)
                max = x[i];
        return max;
    }

    static int powi(int a, int b) {
        if (a == 0)
            return 0;
        int ans = 1;
        while (b > 0) {
            if ((b & 1) > 0)
                ans *= a;
            a *= a;
            b >>= 1;
        }
        return ans;
    }

    static long powl(long a, int b) {
        if (a == 0)
            return 0;
        long ans = 1;
        while (b > 0) {
            if ((b & 1) > 0)
                ans *= a;
            a *= a;
            b >>= 1;
        }
        return ans;
    }

    static int fli(double d) {
        return (int) d;
    }

    static int cei(double d) {
        return (int) ceil(d);
    }

    static long fll(double d) {
        return (long) d;
    }

    static long cel(double d) {
        return (long) ceil(d);
    }

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

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

    static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a % b);
    }

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

    static int[] exgcd(int a, int b) {
        if (b == 0)
            return new int[] { 1, 0 };
        int[] y = exgcd(b, a % b);
        return new int[] { y[1], y[0] - y[1] * (a / b) };
    }

    static long[] exgcd(long a, long b) {
        if (b == 0)
            return new long[] { 1, 0 };
        long[] y = exgcd(b, a % b);
        return new long[] { y[1], y[0] - y[1] * (a / b) };
    }

    static int randInt(int min, int max) {
        return __r.nextInt(max - min + 1) + min;
    }

    static long mix(long x) {
        x += 0x9e3779b97f4a7c15L;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9L;
        x = (x ^ (x >> 27)) * 0x94d049bb133111ebL;
        return x ^ (x >> 31);
    }

    public static boolean[] findPrimes(int limit) {
        assert limit >= 2;

        final boolean[] nonPrimes = new boolean[limit];
        nonPrimes[0] = true;
        nonPrimes[1] = true;

        int sqrt = (int) Math.sqrt(limit);
        for (int i = 2; i <= sqrt; i++) {
            if (nonPrimes[i])
                continue;
            for (int j = i; j < limit; j += i) {
                if (!nonPrimes[j] && i != j)
                    nonPrimes[j] = true;
            }
        }

        return nonPrimes;
    }

    // array util
    static void reverse(int[] a) {
        for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
            int swap = a[i];
            a[i] = a[n - i - 1];
            a[n - i - 1] = swap;
        }
    }

    static void reverse(long[] a) {
        for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
            long swap = a[i];
            a[i] = a[n - i - 1];
            a[n - i - 1] = swap;
        }
    }

    static void reverse(double[] a) {
        for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
            double swap = a[i];
            a[i] = a[n - i - 1];
            a[n - i - 1] = swap;
        }
    }

    static void reverse(char[] a) {
        for (int i = 0, n = a.length, half = n / 2; i < half; ++i) {
            char swap = a[i];
            a[i] = a[n - i - 1];
            a[n - i - 1] = swap;
        }
    }

    static void shuffle(int[] a) {
        int n = a.length - 1;
        for (int i = 0; i < n; ++i) {
            int ind = randInt(i, n);
            int swap = a[i];
            a[i] = a[ind];
            a[ind] = swap;
        }
    }

    static void shuffle(long[] a) {
        int n = a.length - 1;
        for (int i = 0; i < n; ++i) {
            int ind = randInt(i, n);
            long swap = a[i];
            a[i] = a[ind];
            a[ind] = swap;
        }
    }

    static void shuffle(double[] a) {
        int n = a.length - 1;
        for (int i = 0; i < n; ++i) {
            int ind = randInt(i, n);
            double swap = a[i];
            a[i] = a[ind];
            a[ind] = swap;
        }
    }

    static void rsort(int[] a) {
        shuffle(a);
        sort(a);
    }
    static void msort(int a[][]){
        Arrays.sort(a, new Comparator<int[]>() {
            @Override
            public int compare(int[] o1, int[] o2) {
                return 0;
            }
        });
    }
    static PriorityQueue<int []> funcPQ(){
        return new PriorityQueue<int[]>(new Comparator<int[]>() {
            @Override
            public int compare(int[] o1, int[] o2) {
                return 0;
            }
        });
    }

    static void rsort(long[] a) {
        shuffle(a);
        sort(a);
    }

    static void rsort(double[] a) {
        shuffle(a);
        sort(a);
    }

    static int[] copy(int[] a) {
        int[] ans = new int[a.length];
        for (int i = 0; i < a.length; ++i)
            ans[i] = a[i];
        return ans;
    }

    static long[] copy(long[] a) {
        long[] ans = new long[a.length];
        for (int i = 0; i < a.length; ++i)
            ans[i] = a[i];
        return ans;
    }

    static double[] copy(double[] a) {
        double[] ans = new double[a.length];
        for (int i = 0; i < a.length; ++i)
            ans[i] = a[i];
        return ans;
    }

    static char[] copy(char[] a) {
        char[] ans = new char[a.length];
        for (int i = 0; i < a.length; ++i)
            ans[i] = a[i];
        return ans;
    }

    static class FastReader {

        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(
                    new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int fi() {
            return Integer.parseInt(next());
        }

        int[] ria(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = Integer.parseInt(next());
            return a;
        }

        long fl() {
            return Long.parseLong(next());
        }

        long[] rla(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = Long.parseLong(next());
            return a;
        }

        double fd() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider scenarios that are likely to cause issues. The code seems to be dealing with a dynamic programming (DP) approach to count the number of possible final configurations. Here are the steps to think through the problem:

1. **Identify the Bug**: The code uses a DP array `dp` to calculate the number of possible configurations. The key issue is in the DP transition:
   ```java
   dp[i] = (((i-1L) * 2L)%mod * (dp[i-2]))%mod + dp[i-1]%mod;
   ```
   This formula might be incorrect for certain edge cases, especially when the number of moves `k` is large or when the board size `n` is small.

2. **Edge Cases**:
   - **Small Board Size**: Test with a small board size to see if the DP transitions handle small values correctly.
   - **Large Number of Moves**: Test with a large number of moves to see if the DP transitions handle large values correctly.
   - **Symmetric Moves**: Test with moves that are symmetric (i.e., `r = c`), which should cause the computer to skip its turn.
   - **Non-Symmetric Moves**: Test with moves that are not symmetric to see if the DP transitions handle them correctly.

3. **Generate Test Input**:
   - **Small Board Size**: `n = 4`, `k = 2`
   - **Large Number of Moves**: `n = 100`, `k = 50`
   - **Symmetric Moves**: `n = 5`, `k = 3`
   - **Non-Symmetric Moves**: `n = 10`, `k = 4`

Let's create a small-scale test input that combines these scenarios:

```plaintext
4
4 2
1 2
3 4
100 50
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
10 11
11 12
12 13
13 14
14 15
15 16
16 17
17 18
18 19
19 20
20 21
21 22
22 23
23 24
24 25
25 26
26 27
27 28
28 29
29 30
30 31
31 32
32 33
33 34
34 35
35 36
36 37
37 38
38 39
39 40
40 41
41 42
42 43
43 44
44 45
45 46
46 47
47 48
48 49
49 50
5 3
1 1
2 2
3 3
10 4
1 2
2 3
3 4
4 5
```

This test input includes:
1. A small board with a few moves.
2. A large board with many moves.
3. A board with symmetric moves.
4. A board with non-symmetric moves.

This combination should help expose any issues in the DP transitions and other parts of the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: How Does the Rook Move?
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an $$$n \times n$$$ chessboard where you and the computer take turns alternatingly to place white rooks & black rooks on the board respectively. While placing rooks, you have to ensure that no two rooks attack each other. Two rooks attack each other if they share the same row or column regardless of color.

A valid move is placing a rook on a position ($$$r$$$, $$$c$$$) such that it doesn't attack any other rook.

You start first, and when you make a valid move in your turn, placing a white rook at position ($$$r$$$, $$$c$$$), the computer will mirror you and place a black rook at position ($$$c$$$, $$$r$$$) in its turn. If $$$r = c$$$, then the computer can't mirror your move, and skips its turn.

You have already played $$$k$$$ moves with the computer (the computer tries to mirror these moves too), and you must continue playing the game until there are no valid moves remaining. How many different final configurations are possible when you continue the game after the $$$k$$$ moves? It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid. Since the answer may be large, print it modulo $$$10^9+7$$$.

Two configurations are considered different if there exists a coordinate ($$$r$$$, $$$c$$$) which has a rook in one configuration, but not in the other or the color of the rook on the coordinate is different.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \leq n \leq 3 \cdot 10^5$$$, $$$0 \leq k \leq n$$$) — the size of the chessboard and the number of moves you have already played respectively.

Each of the next $$$k$$$ lines of the test case contains two integers $$$r_i$$$ and $$$c_i$$$, denoting the $$$i$$$-th move you made.

It is guaranteed that the $$$k$$$ moves and the implied computer moves are valid.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output: For each test case, output a single integer on a new line — the total number of possible final configurations modulo $$$10^9+7$$$.

Examples:
input:
3
4 1
1 2
8 1
7 6
1000 4
4 4
952 343
222 333
90 91
output:
3
331
671968183

Note: In the first test case, we have a $$$4 \times 4$$$ grid and you've already played $$$1$$$ move. After you and the computer play a turn, we have a white rook at ($$$1$$$, $$$2$$$), and a black rook at ($$$2$$$, $$$1$$$). There are three possible configurations from this state —

1. You place a white rook at ($$$3$$$, $$$4$$$) and the computer places a black rook at ($$$4$$$, $$$3$$$) as a response.
2. You place a white rook at ($$$4$$$, $$$3$$$) and the computer places a black rook at ($$$3$$$, $$$4$$$) as a response.
3. You place a white rook at ($$$3$$$, $$$3$$$) and then at ($$$4$$$, $$$4$$$), or the other way around. They both result in the same configuration.

Buggy Code:
import java.io.*;
import java.lang.reflect.Array;
import java.util.*;
import java.util.Formatter.BigDecimalLayoutForm;

import javax.swing.text.NavigationFilter;

import java.math.BigInteger;

public class Main{
    public static HashMap<Integer, BigInteger> h;
    public static class Pair {
        int value;
        int index;
        Pair(int x, int y){
            this.value = x;
            this.index = y;
        }
    }
    public static BigInteger find(int n){

        // System.out.println("start - func");
        if(n==0){
            return BigInteger.ONE;
        }
        if(n==1){
            return BigInteger.ONE;
        }
        // if(n==2){
        //     return new BigInteger("3");
        // }
        if(h.containsKey(n)){
            return h.get(n);
        }
        BigInteger count = BigInteger.ZERO;
        count = count.add(find(n-1));
        int u = n-1;
        count = count.add(find(n-2).multiply(BigInteger.TWO).multiply(new BigInteger("" +  u)));
        // for(int i=0;i<n;i++){
                // if(i==0){
                //     count = count.add(find(n-1));
                // }else{
                //     count = count.add(find(n-2).multiply(BigInteger.TWO));
                // }

        // }
        h.put(n, count);
        return count;
    }
    public static void sortedInsert(ArrayList<Integer> list, int element) {
        int index = Collections.binarySearch(list, element);
        if (index < 0) {
            index = -(index + 1);
        }
        list.add(index, element);
    }

    public static int countDifferentCharacters(String str1, String str2) {
        int count = 0;

        // Iterate over each character and compare
        for (int i = 0; i < str1.length(); i++) {
            if (str1.charAt(i) != str2.charAt(i)) {
                count++;
            }
        }

        return count;
    }

    public static void main(String[] args) {
       MyScanner sc = new MyScanner();
       out = new PrintWriter(new BufferedOutputStream(System.out));

       // Start writing your solution here. -------------------------------------

       /*
       int n      = sc.nextInt();        // read input as integer
       long k     = sc.nextLong();       // read input as long
       double d   = sc.nextDouble();     // read input as double
       String str = sc.next();           // read input as String
       String s   = sc.nextLine();       // read whole line as String

       int result = 3*n;
       out.println(result);                    // print via PrintWriter
       */

       /*
            String str[] = sc.nextLine().split(" ");
            BigInteger ele[] = new BigInteger[n];
            int k = 0;
            for(int i=0;i<str.length;i++){
                if(str[i]==" " || str[i]==""){
                    continue;
                }
                ele[k] = new BigInteger(str[i]);
                k++;
            }
        */
        /* for sorting array with 2D
         *
         *
         *
         *   Arrays.sort(array, Comparator.comparingInt(paid -> paid.key));
         *
         */
       int t = sc.nextInt();

       ArrayList<BigInteger> arr = new ArrayList<>();

       for(int j=0;j<t;j++){
            int n = sc.nextInt();
            int k = sc.nextInt();
            int rem = n;
            for(int i=0;i<k;i++){
                int x = sc.nextInt();
                int y = sc.nextInt();
                if(x==y){
                    rem--;
                }else{
                    rem = rem - 2;
                }
            }
            // System.out.println("start");
            h = new HashMap<>();
            arr.add(find(rem).mod(new BigInteger("1000000007")));
       }

       for(int i=0;i<arr.size();i++){
           System.out.println(arr.get(i));
       }
  }

  //-----------PrintWriter for faster output---------------------------------
  public static PrintWriter out;

  //-----------MyScanner class for faster input----------
  public static class MyScanner {
     BufferedReader br;
     StringTokenizer st;

     public MyScanner() {
        br = new BufferedReader(new InputStreamReader(System.in));
     }

     String next() {
         while (st == null || !st.hasMoreElements()) {
             try {
                 st = new StringTokenizer(br.readLine());
             } catch (IOException e) {
                 e.printStackTrace();
             }
         }
         return st.nextToken();
     }

     int nextInt() {
         return Integer.parseInt(next());
     }

     long nextLong() {
         return Long.parseLong(next());
     }

     double nextDouble() {
         return Double.parseDouble(next());
     }

     String nextLine(){
         String str = "";
     try {
        str = br.readLine();
     } catch (IOException e) {
        e.printStackTrace();
     }
     return str;
     }

  }
  //--------------------------------------------------------
}

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

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

Think step by step.

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

1. **Edge Cases for `n` and `k`:**
   - Test with the minimum and maximum values for `n` and `k`.
   - Test with `k = 0` to see how the code handles no initial moves.
   - Test with `k = n` to see how the code handles the board being completely filled.

2. **Symmetry and Mirroring:**
   - Test with moves where `r = c` to see how the code handles the computer skipping its turn.
   - Test with moves where `r ≠ c` to see how the code handles mirroring.

3. **Final Configurations:**
   - Test with configurations that have multiple valid final states to ensure the code correctly counts them.

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

```plaintext
4
1 0
2 1
1 1
4 2
1 2
2 1
```

### Explanation of the Test Input:
1. **Test Case 1:**
   - `n = 1`, `k = 0`
   - No initial moves. The only possible configuration is an empty board.
   - Expected output: `1`

2. **Test Case 2:**
   - `n = 2`, `k = 1`
   - One initial move at position (1, 1).
   - The computer skips its turn.
   - Expected output: `1` (only one possible configuration with a rook at (1, 1))

3. **Test Case 3:**
   - `n = 2`, `k = 1`
   - One initial move at position (1, 1).
   - The computer skips its turn.
   - This is the same as Test Case 2, but it helps to verify consistency.
   - Expected output: `1`

4. **Test Case 4:**
   - `n = 4`, `k = 2`
   - Two initial moves at positions (1, 2) and (2, 1).
   - The computer mirrors these moves, resulting in rooks at (2, 1) and (1, 2).
   - There are multiple valid final configurations:
     - Place a rook at (3, 4) and the computer places a rook at (4, 3).
     - Place a rook at (4, 3) and the computer places a rook at (3, 4).
     - Place a rook at (3, 3) and then at (4, 4), or the other way around.
   - Expected output: `3`

This test input should help expose any issues in the buggy code by covering a range of scenarios including edge cases and multiple valid configurations.