**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $a+b$ is a multiple of $b \cdot \gcd(a,b)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
1
3
4
14
153
1643498

Note: In the first test case, only $(1,1)$ satisfies the conditions.

In the fourth test case, $(1,1),(2,1),(2,2),(3,1),(4,1),(5,1),(6,1),(6,2),(6,3),(7,1),(8,1),(9,1),(10,1),(10,2)$ satisfy the conditions.

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

public class Ex3 {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int t = scanner.nextInt();
        for (int q = 0; q < t; q++) {
            int n = scanner.nextInt();
            int m = scanner.nextInt();
            int count = n;
            for (int b = 2; b <= m; b++) {
                int d = (((n + b + 1) / b) / b);
                count += d;
            }
            pw.println(count);
        }
        pw.flush();
        pw.close();
        scanner.close();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios where the code might fail. The bug in the code is in the calculation of the number of valid pairs. Specifically, the line:

```java
int d = (((n + b + 1) / b) / b);
```

This formula is incorrect and does not correctly count the number of valid pairs. Let's consider the following steps to create a test case:

1. **Small Test Cases**: Start with small values of `n` and `m` to ensure the input is manageable and can be easily verified.
2. **Edge Cases**: Test with `n = 1` and `m = 1` to check the simplest case.
3. **Boundary Values**: Test with values where `n` and `m` are close to the upper limit of 2,000,000.
4. **Special Cases**: Test with values where `n` and `m` are equal, and where `n` is much larger than `m` or vice versa.

Given these considerations, a challenging test input that exposes the bug could be:

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

This input includes:
- The simplest case (1, 1)
- A small case where `n` and `m` are different (2, 3)
- A slightly larger case where `n` and `m` are different (3, 5)
- A case where `n` and `m` are equal and larger (10, 10)

This set of test cases should help expose the bug in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $a+b$ is a multiple of $b \cdot \gcd(a,b)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
1
3
4
14
153
1643498

Note: In the first test case, only $(1,1)$ satisfies the conditions.

In the fourth test case, $(1,1),(2,1),(2,2),(3,1),(4,1),(5,1),(6,1),(6,2),(6,3),(7,1),(8,1),(9,1),(10,1),(10,2)$ satisfy the conditions.

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

#define x first
#define y second
typedef long long ll;
typedef int64_t lll;
typedef pair<int,int> pii;
int n,m,T;
string s;
//bool check(int mid){
//	if(mid*(mid-1)>n) return true;
//	else return false;
//}
int main(){
	cin>>T;
	while(T--){
		int ans=0;
		cin>>n>>m;
//		for(int i=1;i<=n;i++){
//			for(int j=1;j<=m;j++){
//				int t1=i+j;
//				int t2=j*__gcd(i,j);
//				if(t1%t2==0&&j==10){
//
//					cout<<i<<' '<<j<<endl;
//				}
//			}
//		}
		int k;
//		int l=1,r=m;
//		while(l<r){
//			int mid=(l+r)>>1;
//			if(check(mid)) r=mid;
//			else l=mid+1;
//		}
//		k=l-1;
		//cout<<k<<endl;
		for(int i=1;i<=m;i++){
			if(i*(i-1)>n){
				k=i-1;
				//cout<<i;
				break;
			}
		}
		int sum=0;
		//cout<<ans<<endl;
		for(int i=2;i<=k;i++){
			int fr=i*(i-1);
			//cout<<fr<<endl;
			int ans=1;
			while(fr<=n){
				fr+=i*i;
				//cout<<fr<<endl;
				ans++;
			}
			//cout<<"======"<<endl;
			//cout<<ans<<endl;
			sum+=ans-1;
		}
		cout<<sum+n<<endl;
	}
}
/*
1
100 1233
/*
UDDUD
UDUU
UUD
DU
U

*/

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

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

Think step by step.

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

1. **Understand the Bug**:
   - The buggy code seems to have issues with the loop that calculates the valid pairs. Specifically, it does not correctly handle the condition where \(a + b\) is a multiple of \(b \cdot \gcd(a, b)\).

2. **Identify Edge Cases**:
   - Small values of \(n\) and \(m\) to test basic functionality.
   - Large values of \(n\) and \(m\) to test performance and correctness.
   - Cases where \(n\) and \(m\) are equal.
   - Cases where \(n\) is much larger than \(m\) or vice versa.

3. **Generate Test Input**:
   - We will create a small-scale test input with a mix of these edge cases to ensure the code is thoroughly tested.

Given the constraints and the need for a small-scale test input, let's create a test input with a few test cases:

```plaintext
4
1 1
2 3
3 5
10 8
```

This test input includes:
- A single pair (1, 1) to test the simplest case.
- A pair (2, 3) to test a small range with more pairs.
- A pair (3, 5) to test another small range with different values.
- A pair (10, 8) to test a slightly larger range.

These test cases will help identify if the buggy code correctly handles the conditions and calculations for valid pairs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $a+b$ is a multiple of $b \cdot \gcd(a,b)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
1
3
4
14
153
1643498

Note: In the first test case, only $(1,1)$ satisfies the conditions.

In the fourth test case, $(1,1),(2,1),(2,2),(3,1),(4,1),(5,1),(6,1),(6,2),(6,3),(7,1),(8,1),(9,1),(10,1),(10,2)$ satisfy the conditions.

Buggy Code:
#pragma GCC optimize ("Ofast,unroll-loops")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef vector<int > vi;
typedef vector<ll > vl;
#define pb push_back
#define f(i, a, b) for(ll i = (ll)a; i < (ll)b; i++)
#define fr(i, b, a) for(ll i = (ll)a - 1; i >= (ll)b; i--)
#define ff first
#define ss second
#define fastio ios_base::sync_with_stdio(0); cin.tie(0)
#define all(v) (v).begin(), (v).end()
#define sz() size()
#define trace(...) fff(#__VA_ARGS__, __VA_ARGS__)
template<typename t> void fff(const char* x, t&& val1) { cout<<x<< " : "<<val1<<"\n";}
template<typename t1, typename... t2> void fff(const char* x, t1&& val1, t2&&... val2){
    const char* xd=strchr(x+1, ',');
    cout.write(x, xd-x)<<" : "<<val1<<" | ";
    fff(xd+1, val2...);
}
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
const ll mod = 1e9 + 7, inf = (1 << 30) - 1 + (1 << 30);
ll add(ll a, ll b){
    if(a + b >= mod){
        return a + b - mod;
    }
    return a + b;
}
ll rem(ll a, ll b){
    if( a <= b){
        return a - b + mod;
    }
    return a - b;
}
ll mul(ll a, ll b){
    return (ll) (a * b) % mod;
}
const ll N = 2e6 + 5;
int main(){
    fastio;
    int t; cin >> t;
    while(t--){
        ll n, m; cin >> n >> m;
        ll ans = n;
        f(i, 2, m){
            if((n/i) %i == i - 1) ans ++;
			ans +=((n/i))/i;
        }
        cout << ans <<'\n';
    }
    return 0;
}

/*
#################################################################################################
#                                        You told me                                            #
#           At your absolute best, you still won't be good enough for the wrong person          #
#                   At your worst, you'll still be worth it to the right person                 #
#                           It was good while it lasted, good bye                               #
#    I believe I really loved you... to that point that I always wanted to hear your voice      #
#    But before my hand could reach you... you seem to be slowly disappearing from my sight     #
#################################################################################################
*/

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on edge cases where the logic might fail. The code seems to have an issue with the way it calculates the number of valid pairs, particularly in the loop where it iterates over `i` from 2 to `m`.

### Step-by-Step Analysis:

1. **Understanding the Bug:**
   - The code calculates `ans` by starting with `n` and then adding to it based on the condition `(n/i) % i == i - 1`.
   - The condition `(n/i) % i == i - 1` is incorrect. For a given `i`, `n/i` should be an integer, and the condition should check if `n % i == i - 1` to ensure that `a + b` is a multiple of `b * gcd(a, b)`.

2. **Edge Cases to Consider:**
   - **Small values of `n` and `m`:** Test cases where `n` and `m` are small can help us understand if the logic is correct.
   - **Large values of `n` and `m`:** Test cases where `n` and `m` are large can help us see if the code handles large inputs correctly.
   - **Special values of `n` and `m`:** Test cases where `n` or `m` are powers of small numbers (like 2, 3, etc.) can expose issues in the loop.

3. **Constructing the Test Input:**
   - We will create a test input with a few test cases that include small values, large values, and special values to maximize the chance of exposing the bug.

### Test Input:
```plaintext
4
1 1
2 3
10 10
1000000 1000000
```

### Explanation:
- **Test Case 1: (1, 1)**
  - Only one pair (1, 1) satisfies the condition.
- **Test Case 2: (2, 3)**
  - The pairs (1, 1), (2, 1), (2, 2), and (2, 3) should be checked.
- **Test Case 3: (10, 10)**
  - This test case will check the logic for a larger but manageable range.
- **Test Case 4: (1000000, 1000000)**
  - This test case will check the performance and correctness for very large inputs.

This test input should help expose the bug in the given code by providing a variety of scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $a+b$ is a multiple of $b \cdot \gcd(a,b)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
1
3
4
14
153
1643498

Note: In the first test case, only $(1,1)$ satisfies the conditions.

In the fourth test case, $(1,1),(2,1),(2,2),(3,1),(4,1),(5,1),(6,1),(6,2),(6,3),(7,1),(8,1),(9,1),(10,1),(10,2)$ satisfy the conditions.

Buggy Code:
# (1(1n-1), 1)
# (2(2n-1), 2)
# (3(3n-1), 3)
# (4(4n-1), 4)

for _ in range(int(input())):
    n, m = map(int, input().split())
    result = n
    for i in range(2, m):
        result += (n+i)//i**2
    print(result)

**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
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
```

### Explanation:
1. **Test Case 1 (1 1):**
   - The only pair is (1, 1), and it satisfies the condition.
   - Expected output: 1

2. **Test Case 2 (2 3):**
   - Valid pairs are (1, 1), (2, 1), and (2, 2).
   - Expected output: 3

3. **Test Case 3 (3 5):**
   - Valid pairs are (1, 1), (2, 1), (2, 2), and (3, 1).
   - Expected output: 4

4. **Test Case 4 (10 8):**
   - Valid pairs include (1, 1), (2, 1), (2, 2), (3, 1), (4, 1), (5, 1), (6, 1), (6, 2), (6, 3), (7, 1), (8, 1), (9, 1), (10, 1), and (10, 2).
   - Expected output: 14

5. **Test Case 5 (100 1233):**
   - This test case is larger and should expose any inefficiencies in the algorithm.
   - Expected output: 153

6. **Test Case 6 (1000000 1145141):**
   - This test case is very large and should expose any significant bugs or performance issues.
   - Expected output: 1643498

These test cases are designed to cover a range of inputs, from small to very large, and to ensure that the algorithm handles edge cases correctly. The buggy code provided has a flaw in the way it calculates the number of valid pairs, and these test cases should help in identifying and fixing that flaw.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $a+b$ is a multiple of $b \cdot \gcd(a,b)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
1
3
4
14
153
1643498

Note: In the first test case, only $(1,1)$ satisfies the conditions.

In the fourth test case, $(1,1),(2,1),(2,2),(3,1),(4,1),(5,1),(6,1),(6,2),(6,3),(7,1),(8,1),(9,1),(10,1),(10,2)$ satisfy the conditions.

Buggy Code:
import sys

def readline() -> str:
    return sys.stdin.readline().strip()

def inputArray() -> list:
    return list(map(int, sys.stdin.readline().strip().split()))

def tc(solve):
    for _ in range(int(readline())):
        solve()
    return

@tc
def solve():
    n,m = inputArray()
    def gcd(x, y):
        while y:
            x, y = y, x % y
        return x

    def count_ordered_pairs(n, m):
        count = n

        for b in range(2,m):
            x=b-1
            if x*b>n:
                break
            while x*b<=n:
                count+=1
                x=x+b

        return count

    print(count_ordered_pairs(n, m))

**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 or scenarios where the faulty logic in the code is likely to fail. The buggy code has a few issues:

1. **Incorrect Loop Bound**: The loop for `b` starts from 2 and goes up to `m-1`, which might miss some valid pairs.
2. **Incorrect Increment**: The increment `x = x + b` might not correctly count all valid pairs.

Given these issues, let's consider a small-scale test input that can highlight these problems:

### Test Input
```plaintext
3
2 2
3 3
4 4
```

### Explanation
1. **Test Case 1: (2, 2)**
   - Valid pairs: (1,1), (2,2)
   - Expected output: 2
   - Buggy code output: 2 (correct in this case, but let's check the next cases)

2. **Test Case 2: (3, 3)**
   - Valid pairs: (1,1), (2,2), (3,3), (1,2), (2,1)
   - Expected output: 5
   - Buggy code output: 3 (incorrect, missing (1,2) and (2,1))

3. **Test Case 3: (4, 4)**
   - Valid pairs: (1,1), (2,2), (3,3), (4,4), (1,2), (2,1), (1,3), (3,1), (2,4), (4,2)
   - Expected output: 10
   - Buggy code output: 4 (incorrect, missing (1,2), (2,1), (1,3), (3,1), (2,4), (4,2))

### Test Input
```plaintext
3
2 2
3 3
4 4
```

This test input is small and complete, and it should effectively expose the issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $a+b$ is a multiple of $b \cdot \gcd(a,b)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
1
3
4
14
153
1643498

Note: In the first test case, only $(1,1)$ satisfies the conditions.

In the fourth test case, $(1,1),(2,1),(2,2),(3,1),(4,1),(5,1),(6,1),(6,2),(6,3),(7,1),(8,1),(9,1),(10,1),(10,2)$ satisfy the conditions.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define ll long long

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int t;
	cin>>t;
	while(t--)
	{
		int n,m;
		cin>>n>>m;
		ll ans=n;
//		ll cnt=0;
//		for(int i=1;i<=n;i++)
//		{
//			for(int j=2;j<=m;j++)
//			{
//				ll a1=i+j;
//				ll a2=j*__gcd(i,j);
//				if(a1%a2==0) {cout<<"("<<i<<","<<j<<") "; cnt++;}
//			}
//		}
//cout<<cnt<<endl;

		for(int j=2;j<=10*sqrt(m);j++)
		{

			for(int i=j*j-j;i<=n;i+=j*j)
			{
			  ans++;
			}
		}
		cout<<ans<<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 provided buggy code, we need to focus on scenarios where the code's logic might fail. The buggy code seems to have an issue with how it calculates the number of valid pairs, especially in the nested loops where it increments `ans`.

### Analysis:
1. **Edge Cases:**
   - Small values of `n` and `m` to check basic functionality.
   - Large values of `n` and `m` to check performance and correctness.
   - Values where `n` and `m` are equal.
   - Values where `n` and `m` are prime numbers.
   - Values where `n` and `m` are powers of small integers.

2. **Specific Scenario:**
   - The code uses a loop that increments `ans` based on a condition involving `j` and `i`. This might fail when `j` is not a small integer, especially when `j` is close to the upper limit of the loop.

### Chosen Test Input:
- We will use a small-scale test input that includes a case where `n` and `m` are equal and relatively small, and another case where `n` and `m` are prime numbers. This will help us verify if the code handles these specific scenarios correctly.

### Test Input:
```plaintext
3
1 1
3 3
5 7
```

### Explanation:
1. **Test Case 1: (1, 1)**
   - Only one pair `(1, 1)` is valid.
2. **Test Case 2: (3, 3)**
   - Valid pairs: `(1, 1)`, `(2, 1)`, `(3, 1)`, `(2, 2)`.
   - Expected output: 4.
3. **Test Case 3: (5, 7)**
   - Valid pairs: `(1, 1)`, `(2, 1)`, `(3, 1)`, `(4, 1)`, `(5, 1)`, `(2, 2)`, `(4, 2)`, `(3, 3)`, `(5, 2)`, `(4, 4)`.
   - Expected output: 10.

This test input is small enough to be manageable and includes a mix of simple and slightly more complex cases to challenge the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $a+b$ is a multiple of $b \cdot \gcd(a,b)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
1
3
4
14
153
1643498

Note: In the first test case, only $(1,1)$ satisfies the conditions.

In the fourth test case, $(1,1),(2,1),(2,2),(3,1),(4,1),(5,1),(6,1),(6,2),(6,3),(7,1),(8,1),(9,1),(10,1),(10,2)$ satisfy the conditions.

Buggy Code:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <set>
#include <unordered_map>
#include <queue>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <bitset>
#include <array>
#include <climits>
#include <iomanip>
#include <unordered_set>
#include <ranges>
//#include <x86intrin.h>

using namespace std;

// #pragma GCC optimize("O3,unroll-loops")
// #pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")

#define fr first
#define sc second
#define getunique(v)                              \
  {                                               \
    sort(v.begin(), v.end());                     \
    v.erase(unique(v.begin(), v.end()), v.end()); \
  }
#define all(x) x.begin(), x.end()
#define f0r(a, b) for (long long a = 0; a < (b); ++a)

using ll = long long;
// using ll = int;
//#pragma warning("short")
//
using vl = vector<ll>;
using pl = pair<ll, ll>;

typedef long double ld;
typedef unsigned long long ull;
typedef complex<double> base;

/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;

template <typename num_t>
using ordered_set = tree<num_t, null_type, less<num_t>, rb_tree_tag, tree_order_statistics_node_update>;
*/

#ifdef debug_local
#include "debug.h"
#define pr(...) debug(#__VA_ARGS__, __VA_ARGS__)
#define prs(...) debug_nameless(__VA_ARGS__)
const bool local_ = true;
#else
#define pr(...)  /* Nothing */
#define prs(...) /* Nothing */
const bool local_ = false;
#endif

// mt19937_64 rng(std::chrono::steady_clock::now().time_since_epoch().count());
//  mt19937_64 rng(61378913);
/* usage - just do rng() */

// #include <atcoder/all>
//  using namespace atcoder;

// const ld pi = 3.14159265358979323846;
// const ll mod = 1000000007;
//   const ll mod = 998244353;
// const ll mod = (ll)1e9 + 7;

ll n, m, q, k, l, r, x, y, z;
// string s;
string t;

const bool input = local_ ? 1 : 0;

template <class T>
bool chmin(T &a, const T &b)
{
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool chmax(T &a, const T &b)
{
  return a < b ? a = b, 1 : 0;
}

ll cdiv(ll a, ll b)
{
  return a / b + ((a ^ b) > 0 && a % b);
} // divide a by b rounded up

namespace number_theory {
	ll gcd(ll x, ll y) {
	  if (x == 0) return y;
	  if (y == 0) return x;
	  return gcd(y, x % y);
	}
	bool isprime(ll n) {
	  if (n <= 1) return false;
	  if (n <= 3) return true;

	  if (n % 2 == 0 || n % 3 == 0) return false;

	  for (ll i = 5; i * i <= n; i += 6)
		if (n % i == 0 || n % (i+2) == 0)
		  return false;

	  return true;
	}

	bool prime[15000105];
	void sieve(int n) {
	  for (ll i = 0; i <= n; i++) prime[i] = 1;
	  for (ll p = 2; p * p <= n; p++) {
		if (prime[p] == true) {
		  for (ll i = p * p; i <= n; i += p)
			prime[i] = false;
		}
	  }
	  prime[1] = prime[0] = 0;
	}

	vector<ll> primelist;
	bool __primes_generated__ = 0;

	void genprimes(int n) {
	  __primes_generated__ = 1;
	  sieve(n + 1);
	  for (ll i = 2; i <= n; i++) if (prime[i]) primelist.push_back(i);
	}

	vector<ll> factors(ll n) {
	  if (!__primes_generated__) {
		cerr << "Call genprimes you dope" << endl;
		exit(1);
	  }
	  vector<ll> facs;

	  for (ll i = 0; primelist[i] * primelist[i] <= n && i < primelist.size(); i++) {
		if (n % primelist[i] == 0) {
		  while (n % primelist[i] == 0) {
			n /= primelist[i];
			facs.push_back(primelist[i]);
		  }
		}
	  }
	  if (n > 1) {
		facs.push_back(n);
	  }
	  sort(facs.begin(), facs.end());
	  return facs;
	}

	vector<ll> getdivs(ll n) {
    vector<ll> divs;
    for (ll i = 1; i * i <= n; i++) {
      if (n % i == 0) {
        divs.push_back(i);
        divs.push_back(n / i);
      }
    }

    getunique(divs);
    return divs;
  }
}

using namespace number_theory;

void solve(int tc = 0)
{
  cin >> n >> m;
  ll ans = n; // bound is n
  ll val = 1;
  for (ll i = 2; i <= (ll)1e8; i++)
  {
	val = (i-1)*(i);
	ll delta = i*i;
	pr(val, delta);
	if (n < val) {break;}
	//pr((n-val)/delta);
	ans += min(m, (n-val)/delta+ 1);
  }
  pr(ans);
  cout << ans << '\n';
}

int main()
{
#ifdef debug_local
  auto begin = std::chrono::high_resolution_clock::now();
#endif

  ios_base::sync_with_stdio(false);
  cin.tie(NULL);

  //freopen("magic.in", "r", stdin);
  //freopen("magic.out", "w", stdout);

  cout << setprecision(16) << fixed;
  cerr << setprecision(4) << fixed;

  int tc = 1;
  // if (local_)
  cin >> tc;
  for (int t = 0; t < tc; t++)
  {
    pr(t);
    prs(string(50, '-'));
    solve(t);
    prs(string(50, '-') + "\n");
  }

#ifdef debug_local
  auto end = std::chrono::high_resolution_clock::now();
  cerr << setprecision(4) << fixed;
  cerr << "Execution time: " << std::chrono::duration_cast<std::chrono::duration<double>>(end - begin).count() << " seconds" << endl;
#endif
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. The main issue in the code is the loop that iterates up to \(10^8\), which is not necessary and can be optimized. Additionally, the logic for calculating the number of valid pairs might have issues with large values of \(n\) and \(m\).

Let's consider the following scenarios:
1. **Small values of \(n\) and \(m\)**: These can help verify the basic logic.
2. **Large values of \(n\) and \(m\)**: These can help identify performance and correctness issues.
3. **Edge cases where \(n\) and \(m\) are equal**: This can help verify the symmetry in the problem.
4. **Cases where \(n\) or \(m\) is 1**: These can help verify the boundary conditions.

Given the constraints, a small-scale test input that is complete and valid can be:

```plaintext
3
1 1
2 3
10 10
```

This test input includes:
- The smallest possible values for \(n\) and \(m\) (1, 1).
- A small but non-trivial case (2, 3).
- A case where \(n\) and \(m\) are equal and relatively small (10, 10).

These cases should help identify any issues in the logic and performance of the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $a+b$ is a multiple of $b \cdot \gcd(a,b)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
1
3
4
14
153
1643498

Note: In the first test case, only $(1,1)$ satisfies the conditions.

In the fourth test case, $(1,1),(2,1),(2,2),(3,1),(4,1),(5,1),(6,1),(6,2),(6,3),(7,1),(8,1),(9,1),(10,1),(10,2)$ satisfy the conditions.

Buggy Code:
import math

l = []
for i in range(1, 1000000):
    if i * i >= 10000000:
        break
    l.append(i ** 2)

for _ in range(int(input())):
    a, b = map(int, input().split())
    ans = 0
    for i in l:

        ans += (a + math.floor(math.sqrt(i))) // i

    print(ans-1)

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on the logic that calculates the number of valid pairs \((a, b)\). The code seems to be using a precomputed list of squares and then iterating over these squares to count the valid pairs. However, the logic used to count the pairs is incorrect.

### Analysis of the Buggy Code
1. **Precomputed List of Squares**: The code generates a list of squares up to a certain limit.
2. **Counting Valid Pairs**: The code iterates over these squares and uses a formula to count the valid pairs. However, the formula \((a + \text{math.floor}(\text{math.sqrt}(i))) // i\) is not correct for this problem.

### Correct Approach
The correct approach to count the valid pairs \((a, b)\) is to check the condition \(a + b\) being a multiple of \(b \cdot \gcd(a, b)\). This can be done by iterating over all possible values of \(a\) and \(b\) and checking the condition directly.

### Challenging Test Input
To expose the bug, we need a test input that will show the incorrect counting of pairs. A good test input is one where the correct number of valid pairs is known and the buggy code will likely fail to count them correctly.

### Example Test Input
Let's consider a small test case where \(n = 5\) and \(m = 5\). We can manually verify the valid pairs for this test case:

- Valid pairs: \((1,1), (2,1), (2,2), (3,1), (4,1), (4,2), (5,1), (5,2), (5,3)\)
- Total valid pairs: 9

### Test Input
```plaintext
1
5 5
```

This test input is small and complete, and it will expose the bug in the given code because the code will likely count the pairs incorrectly due to the faulty logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $a+b$ is a multiple of $b \cdot \gcd(a,b)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
1
3
4
14
153
1643498

Note: In the first test case, only $(1,1)$ satisfies the conditions.

In the fourth test case, $(1,1),(2,1),(2,2),(3,1),(4,1),(5,1),(6,1),(6,2),(6,3),(7,1),(8,1),(9,1),(10,1),(10,2)$ satisfy the conditions.

Buggy Code:
import math
def c(n, m):
    counter = 0
    for b in range(1, math.ceil(math.sqrt(n))+1,1):
        #for a in range(b, n+1 ,b):
        #    if (a + b) % (b ** 2) == 0:
        #        counter += 1
                #print(a,b)
        counter += math.floor((n+b)/(b**2))
        #print(b,counter)
    return counter
t = int(input(""))
for _ in range(t):
    a,b = map(int,input("").split(" "))
    print(c(a,b)-1)

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

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

Think step by step.

Response:
```plaintext
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
```

### Explanation:
1. **Edge Case: (1, 1)**
   - This is the simplest case where both \( n \) and \( m \) are 1. The only valid pair is \((1, 1)\).

2. **Small Values: (2, 3)**
   - This tests the algorithm with small values of \( n \) and \( m \). The valid pairs are \((1, 1)\), \((2, 1)\), and \((2, 2)\).

3. **Intermediate Values: (3, 5)**
   - This tests the algorithm with slightly larger values. The valid pairs are \((1, 1)\), \((2, 1)\), \((2, 2)\), and \((3, 1)\).

4. **Moderate Values: (10, 8)**
   - This tests the algorithm with moderate values. The valid pairs are \((1, 1)\), \((2, 1)\), \((2, 2)\), \((3, 1)\), \((4, 1)\), \((5, 1)\), \((6, 1)\), \((6, 2)\), \((6, 3)\), \((7, 1)\), \((8, 1)\), \((9, 1)\), \((10, 1)\), and \((10, 2)\).

5. **Large Values: (100, 1233)**
   - This tests the algorithm with larger values to ensure it can handle more significant inputs without performance issues.

6. **Very Large Values: (1000000, 1145141)**
   - This tests the algorithm with very large values to ensure it can handle the upper limits of the problem constraints.

### Why These Inputs?
- **Edge Case**: The first test case ensures that the algorithm works for the smallest possible input.
- **Small and Intermediate Values**: These cases help in identifying any issues with the basic logic and edge conditions.
- **Moderate Values**: These cases test the algorithm with a reasonable number of pairs, ensuring it can handle a moderate amount of computation.
- **Large and Very Large Values**: These cases test the performance and correctness of the algorithm with large inputs, ensuring it can handle the upper limits of the problem constraints.

By using these test cases, you can effectively expose any bugs in the buggy code and ensure that the algorithm works correctly across a range of inputs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $a+b$ is a multiple of $b \cdot \gcd(a,b)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
1
3
4
14
153
1643498

Note: In the first test case, only $(1,1)$ satisfies the conditions.

In the fourth test case, $(1,1),(2,1),(2,2),(3,1),(4,1),(5,1),(6,1),(6,2),(6,3),(7,1),(8,1),(9,1),(10,1),(10,2)$ satisfy the conditions.

Buggy Code:
import java.util.Arrays;
import java.util.Scanner;

public class CodeForces {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int N = 2 * (int) 1e6 + 1;
        int[] arr = new int[N];
        Arrays.fill(arr, 1);
        for (int i = 2; i < N; i++) {
            for (int j = i; j < N; j += i) {
                int temp = i * gcd(j, i);
                if (temp <= 0) continue;
                if ((i + j) % temp == 0) arr[j]++;
            }
        }
        for (int i = 2; i < N; i++) {
            arr[i] += arr[i - 1];
        }
        int t = scan.nextInt();
        while (t --> 0) {
            int n = scan.nextInt();
            int m = scan.nextInt();
            if (n <= m) {
                System.out.println(arr[n]);
                continue;
            }
            int ans = arr[m];
            for (int i = 1; i <= m; i++) {
                int j = (m / i) * i + i;
                if (j > n) break;
                for (; j <= n; j += i) {
                    if ((i + j) % (i * gcd(j, i)) == 0) ans++;
                }
            }
            System.out.println(ans);
        }
    }

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

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

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

Think step by step.

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

1. **Edge Cases for Small Values**: Test with small values of \( n \) and \( m \) to ensure basic functionality.
2. **Edge Cases for Large Values**: Test with large values of \( n \) and \( m \) to check performance and correctness.
3. **Boundary Values**: Test with boundary values of \( n \) and \( m \) (e.g., \( n = 1 \), \( m = 1 \), \( n = 2 \cdot 10^6 \), \( m = 2 \cdot 10^6 \)).
4. **Mixed Values**: Test with mixed values where \( n \) and \( m \) are not equal to ensure the code handles different ranges correctly.

Given these considerations, let's create a small-scale test input that includes a mix of these cases:

```plaintext
6
1 1
2 3
3 5
10 8
2000000 1
1 2000000
```

### Explanation:
1. **Case 1: (1, 1)**
   - Only one pair: (1, 1).
   - This tests the smallest possible input.

2. **Case 2: (2, 3)**
   - Pairs: (1, 1), (2, 1), (2, 2), (2, 3).
   - This tests a small range with multiple pairs.

3. **Case 3: (3, 5)**
   - Pairs: (1, 1), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5).
   - This tests a slightly larger range with more pairs.

4. **Case 4: (10, 8)**
   - Pairs: (1, 1), (2, 1), (2, 2), (3, 1), (4, 1), (5, 1), (6, 1), (6, 2), (6, 3), (7, 1), (8, 1), (9, 1), (10, 1), (10, 2).
   - This tests a medium range with a significant number of pairs.

5. **Case 5: (2000000, 1)**
   - Only one pair: (1, 1).
   - This tests the boundary condition for a very large \( n \) and a very small \( m \).

6. **Case 6: (1, 2000000)**
   - Only one pair: (1, 1).
   - This tests the boundary condition for a very small \( n \) and a very large \( m \).

This test input covers a range of scenarios and should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $a+b$ is a multiple of $b \cdot \gcd(a,b)$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
1
3
4
14
153
1643498

Note: In the first test case, only $(1,1)$ satisfies the conditions.

In the fourth test case, $(1,1),(2,1),(2,2),(3,1),(4,1),(5,1),(6,1),(6,2),(6,3),(7,1),(8,1),(9,1),(10,1),(10,2)$ satisfy the conditions.

Buggy Code:
#include<bits/stdc++.h>
#define endl '\n'
#define ll long long
#define pb push_back
#define lbt(x) (x&(-x))
#define mii map<int,int>
#define pii pair<int,int>
#define go(i,a,b) for(int i=a;i<b;i++)
#define WJXNB ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
const int maxn=200010;
const int mod=998244353;
const int inf=0x3f3f3f3f;
ll n,m,ans=0;
void solve(){
    ans=0;
    cin>>n>>m;
    go(i,1,sqrt(m)+1000){
        if(i==1)ans+=n;
        else{
            ans+=((n+i)/(i*i));
        }

    }
    cout<<ans<<endl;
    return;
}

int main(){
    WJXNB
    int t;
    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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios where the code might fail. The buggy code seems to have an issue with the way it calculates the number of valid pairs, especially when dealing with specific values of \( n \) and \( m \).

### Analysis of the Buggy Code
1. **Initialization and Input Handling:**
   - The code reads multiple test cases and processes each one.
   - For each test case, it initializes `ans` to 0 and reads `n` and `m`.

2. **Loop for Calculating Valid Pairs:**
   - The loop iterates from 1 to `sqrt(m) + 1000`.
   - For \( i = 1 \), it adds `n` to `ans`.
   - For \( i > 1 \), it adds `((n + i) / (i * i))` to `ans`.

### Potential Issues
- The loop iterates up to `sqrt(m) + 1000`, which might be too large or too small depending on the values of `n` and `m`.
- The condition \( a + b \) being a multiple of \( b \cdot \gcd(a, b) \) is not correctly handled in the loop.

### Challenging Test Case
To expose the bug, we can use a test case where the loop's range or the calculation inside the loop might not cover all valid pairs. A good candidate is a small but specific value of \( n \) and \( m \) that should have a known number of valid pairs.

### Test Case
Let's use \( n = 6 \) and \( m = 6 \). The valid pairs for this input are:
- (1,1), (2,1), (2,2), (3,1), (4,1), (5,1), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6).

There are 12 valid pairs in total.

### Test Input
```plaintext
1
6 6
```

This test input should expose the bug in the buggy code by showing that the calculated number of valid pairs is incorrect.


**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<cmath>
#define int long long
using namespace std;
const int mod = 1e9+7,maxn = 3e5+8;
int dp[maxn];
signed main(){
	int t;
	cin>>t;
	//dp[i] = dp[i-1]+(i-1)*dp[i-2]
	dp[1] = 1;
	dp[2] = 3;
	for(int i = 3;i <= maxn;i ++){
		dp[i] = (dp[i-1]+2*(i-1)*dp[i-2])%mod;
	}
	while(t--){
		int n,k;
		cin>>n>>k;
		int m = n;
		for(int i = 0;i < k;i ++){
			int x,y;
			cin>>x>>y;
			if(x == y)
				m --;
			else
				m -= 2;
		}
		cout<<dp[m]<<"\n";
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
4 2
1 2
2 1
8 4
1 2
2 1
3 4
4 3
1000 5
4 4
952 343
222 333
90 91
1 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 <iostream>
#include <cstdlib>
#include <map>
#include <vector>
#include <algorithm>
#include <climits>
#include <cmath>
#include <string>
#include <string.h>
#include <set>
#include <bitset>
#include <unordered_set>

#define ll long long int
#define ull unsigned long long int
using namespace std;

int lowerbound(vector<int>& arr,int value)
{
  int i=0,j=arr.size()-1;
  while(i<=j)
  {
    int mid=i+(j-i)/2;
    if(arr[mid]>value)j=mid-1;
    else
    i=mid+1;
  }
  return i;
}

ll nCr(ll n, ll r) {

    if (r > n) return 0;

    if (r == 0 || n == r) return 1;

    long double res = 0;

    for (int i = 0; i < r; i++) {

        res += log(n-i) - log(i+1);
    }

    return (ll)round(exp(res));
}

bool isPalindrome(int n) {
    string str = to_string(n);
    int len = str.length();
    for(int i = 0; i < len/2; i++) {
        if(str[i] != str[len-1-i])
            return false;
    }
    return true;
}

int binarySearch(vector<int> arr, int low,
                 int high, int key)
{
    if (high < low)
        return -1;

    int mid = low + (high - low)/2; /*low + (high - low)/2;*/
    if (key == arr[mid])
        return mid;

    if (key > arr[mid])
        return binarySearch(arr, (mid + 1), high, key);

    // else
    return binarySearch(arr, low, (mid - 1), key);
}

bool isPallindrome(string s)
{
    string str = s;
    int len = str.length();
    for(int i = 0; i < len/2; i++) {
        if(str[i] != str[len-1-i])
            return false;
    }
    return true;
}

int length(int n){
  return to_string(n).length();
}

bool isPrime(int n)
{
    if (n <= 1)
        return false;

    for (int i = 2; i <= sqrt(n); i++)
        if (n % i == 0)
            return false;

    return true;
}

bool cmp(pair<int, int>& a,
        pair<int, int>& b)
{
    return a.first + a.second < b.first + b.second;
}

ll setBitNumber(ll n)
{
    if (n == 0)
        return 0;

    ll msb = 0;
    n = n / 2;
    while (n != 0) {
        n = n / 2;
        msb++;
    }
    if(msb==0)return 0;
    ll val = 1;
    while(--msb!=0)
    {
      val*=2;
    }
    return val;
}

ll selfPower(ll x, ll y, ll p)
{
    ll res = 1;
    x = x % p;
    if (x == 0) return 0;
    while (y > 0)
    {
        if (y & 1)
            res = (res*x) % p;
        y = y>>1; // y = y/2
        x = (x*x) % p;
    }
    return res;
}

int maxSubstr(map<char,int> arr)
{
    int max=INT_MIN;
    for(auto i:arr)
    {
        if(i.second>max)
        max=i.second;
    }
    return max;
}

ll numZeros(int n)
{
    ll zeros=0;
    while(n%10==0)
    {
        zeros++;
        n/=10;
    }
    return zeros;
}

ll numDigits(int n)
{
    ll count=0;
    while(n>0)
    {
        count++;
        n/=10;
    }
    return count;
}
int sumOfDigitsFrom1ToNUtil(int n, int a[])
{
    if (n < 10)
        return (n * (n + 1) / 2);

    int d = (int)(log10(n));
    int p = (int)(ceil(pow(10, d)));
    int msd = n / p;

    return (msd * a[d] + (msd * (msd - 1) / 2) * p +
            msd * (1 + n % p) +
            sumOfDigitsFrom1ToNUtil(n % p, a));
}

int sumOfDigitsFrom1ToN(int n)
{
    int d = (int)(log10(n));
    int a[d + 1];
    a[0] = 0; a[1] = 45;

    for(int i = 2; i <= d; i++)
        a[i] = a[i - 1] * 10 + 45 *
               (int)(ceil(pow(10, i - 1)));

    return sumOfDigitsFrom1ToNUtil(n, a);
}

ll sumDigits(int n)
{
    ll sum=0;
    while(n>0)
    {
        sum+=n%10;
        n/=10;
    }
    return sum;
}

long long ms1(vector<long long> &a , int n)
{
    long long m1 = INT_MIN, m2 = 0,start = 0, end = 0, s = 0;

    for (int i = 0; i < n; i++) {
        m2 += a[i];

        if (m1 < m2) {
            m1 = m2;
            start = s;
            end = i;
        }

        if (m2 < 0) {
            m2 = 0;
            s = i + 1;
        }
    }
   return m1;

}

int reverseNum(int n)
{
    int rev=0;
    while(n>0)
    {
        rev=rev*10+n%10;
        n/=10;
    }
    return rev;
}

bool cmpi(int a,int b)
{
    return numZeros(a)>=numZeros(b);
}

int minPos(vector<int> &val)
{
    int min=INT_MAX,pos=0;
    for(int i=0;i<val.size();i++)
    {
        if(val[i]<min)
        {
            min=val[i];
            pos=i;
        }
    }
    return pos;
}

void primeFactors(int n,vector<int> &arr)
{
    // Print the number of 2s that divide n
    while (n % 2 == 0)
    {
        arr.push_back(2);
        n = n/2;
    }

    // n must be odd at this point. So we can skip
    // one element (Note i = i +2)
    for (int i = 3; i <= sqrt(n); i = i + 2)
    {
        // While i divides n, print i and divide n
        while (n % i == 0)
        {
            arr.push_back(i);
            n = n/i;
        }
    }

    // This condition is to handle the case when n
    // is a prime number greater than 2
    if (n > 2)
        arr.push_back(n);
}

bool scheck(string a,string b)
{
    int c=0;
    for(int i=0;i<a.size();i++)
    {
        if(a[i]!=b[i])c++;
    }
    if(c==0 || c==1)
    return true;
    else
    return false;
}

int MEX(vector<int> &arr)
{
    int min=INT_MAX;
    set<int> mins;
    bool flag=false;
    for(int i=0;i<arr.size();i++)
    {
        mins.insert(i);
    }
    for(int i=0;i<arr.size();i++)
    {
        mins.erase(arr[i]);
        if(arr[i]<min)
        min=arr[i];
        if(arr[i]==0)
        flag=true;
    }

    if(!flag)return 0;

    int val=INT_MAX;
    for(auto i:mins)
    {
        val=i;
        break;
    }
    return (val<min-1)?(val):(min-1);
}

bool good(int n)
{
    bool flag=false;
    while(n>0)
    {
        if(n%3==2)flag=true;
        n/=3;
    }
    return flag;
}

void solve()
{
    int n,k;
    cin>>n>>k;
    int cnt=0;
    for(int i=0;i<k;i++)
    {
        int a,b;
        cin>>a>>b;
        if(a==b)
        cnt++;
        else
        cnt+=2;
    }
    ull dp[n];
    for(int i=0;i<n;i++)
    {
        dp[i]=0;
    }
    dp[0]=1;
    dp[1]=1;
    ull MOD= 1e9+7;
    for(int i=2;i<n;i++)
    {
        dp[i] = (dp[i-1] + 2ll * (i-1) * dp[i-2] % MOD) % MOD;
    }
    cout<<dp[n-cnt]<<endl;
}

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

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

1. **Understand the Bug**: The provided code seems to have issues with handling the initial moves and the dynamic programming (DP) approach for calculating the number of configurations. Specifically, the DP array is initialized and updated in a way that might not correctly handle the given moves.

2. **Edge Cases**:
   - **Small Board**: A small board (e.g., 2x2 or 3x3) can help us quickly verify the correctness of the solution.
   - **Initial Moves**: Moves that place rooks in such a way that the computer's mirroring causes a specific pattern.
   - **Diagonal Moves**: Moves where rooks are placed on the diagonal (r = c) to see how the code handles skipped computer moves.

3. **Generate Test Input**:
   - **Test Case 1**: A 2x2 board with one move.
   - **Test Case 2**: A 3x3 board with one move on the diagonal.
   - **Test Case 3**: A 4x4 board with multiple moves.

Let's generate a test input that includes these scenarios:

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

### Explanation:
1. **Test Case 1**:
   - **Board Size**: 2x2
   - **Initial Moves**: 1 move at (1, 2)
   - **Expected Output**: 1 (only one configuration possible)

2. **Test Case 2**:
   - **Board Size**: 3x3
   - **Initial Moves**: 1 move at (1, 1)
   - **Expected Output**: 2 (since the computer skips its turn, only two configurations are possible)

3. **Test Case 3**:
   - **Board Size**: 4x4
   - **Initial Moves**: 2 moves at (1, 2) and (2, 1)
   - **Expected Output**: 3 (the initial moves and their mirrors occupy the board in a specific pattern, leaving three possible configurations)

This test input should help expose any issues in the buggy code, particularly in how it handles the initial moves and the DP calculations.


**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;

long long dp[400005];

long long dpThing(long long n) {
    //cout<<n<<endl;
    if (dp[n] != 0) {return dp[n];}
    //dp[n] = 1 + 2*n*dpThing(n-2);
    //cout<<n<<" "<<dpThing(n-1)<<" "<<2*(n-1)*dpThing(n-2)<<endl;
    dp[n] = dpThing(n-1) + 2*(n-1)*dpThing(n-2);
    dp[n] %= 1000000007;
    return dp[n];
}

void solve() {
    long long n, k;
    cin>>n>>k;

    long long total = n;
    long long x, y;
    for (long long i = 0; i<k; i++) {
        cin>>x>>y;
        if (x == y) {n -= 1;}
        else {n -= 2;}

    }

    dp[1] = 1;
    dp[2] = 3;
    dp[0] = 0;

    if (dp[n]) {cout<<dp[n]<<"\n"; return;}
    if (n <= 2) {cout<<dp[n]<<"\n"; return;}

    cout<<dpThing(n) % 1000000007<<"\n";
    //cout<<dp[n]<<"\n";

    /*long long total = n;
    long long ans = 0;

    for (int i = 0; i<k; i++) {
        int x, y;
        cin>>x>>y;
        //bannedRows[x-1] = true;
        //bannedCols[y-1] = true;
        //bannedRows[y-1] = true;
        //bannedCols[x-1] = true;
        if (x == y) {n -= 1;}
        else {n -= 2;}

    }
    while (total > 0) {
        ans += n*(n-1);
        ans += (n*n-n)*(n-2);
        total--;
        ans %= 1000000007;
    }

    cout<<ans<<"\n";*/

}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    fill(dp, dp+400005, 0);
    int t;
    cin>>t;
    while (t){
        t--;
        solve();
    }

}

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

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

Think step by step.

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

1. **Handling of Initial DP Values**: The code initializes `dp[0] = 0`, `dp[1] = 1`, and `dp[2] = 3`. These initial values are used in the recursive function `dpThing`, and any incorrect initialization can lead to incorrect results.
2. **Reduction of `n`**: The code reduces `n` by 1 if the move is on the diagonal (i.e., `x == y`) and by 2 if the move is not on the diagonal (i.e., `x != y`). This reduction should be correct, but we need to ensure that it handles edge cases properly.
3. **Recursive Calculation**: The recursive calculation in `dpThing` might have issues with large values of `n` due to the modulo operation and the recursive nature of the function.

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

1. **Test Case 1**: A small board with a move on the diagonal.
2. **Test Case 2**: A small board with a move not on the diagonal.
3. **Test Case 3**: A larger board with multiple moves to test the recursive calculation and reduction of `n`.

Here is the test input:

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

Explanation:
1. **Test Case 1**: A 4x4 board with one move on the diagonal (1, 1). This should reduce `n` by 1.
2. **Test Case 2**: A 4x4 board with one move not on the diagonal (1, 2). This should reduce `n` by 2.
3. **Test Case 3**: A 6x6 board with two moves (2, 3) and (3, 2). This should reduce `n` by 4 (2 moves, each reducing `n` by 2).

This test input will help in verifying the correctness of the initial DP values, the reduction of `n`, and the recursive calculation.


**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 = map(int, input().split())
    m = n
    for j in range(k):
        r,c = map(int, input().split())
        if r!=c:
            m-=2
        else:
            m-=1
    dp = [0]*(m+1)
    dp[0]=1
    dp[1]=1
    for j in range(2,m+1):
        dp[j]=dp[j-1]%1000000007+(2*(j-1)*dp[j-2])%1000000007
        dp[j]%=1000000007
    print(dp[m])

**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
4 4
90 91
```


**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:
# Rookie : Tanmay Patil

def func(i,s):
    if i == 0 :
        return 0
    if i == 2 :
        return 3
    if i == 1 :
        return 1
    if s[i] != -1 :
        return s[i]
    ans = (func(i-1,s) + (2*i-2)*func(i-2,s))%1000000007
    s[i] = ans
    return ans

def solve() :
    # Solve here
    n,k = map(int, input().split())
    moves = set()
    for i in range(k) :
        a = list(map(int,input().split()))
        moves.add(a[0])
        moves.add(a[1])
    l = len(moves)
    ans = 1
    n = n - l
    s = [-1]*(n+1)

    ans = func(n,s)
    print(ans)

testCases = int(input())
for _ in range(testCases) :
    solve()

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

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

Think step by step.

Response:
```plaintext
1
3 1
1 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
Lans = [0,1,3]
MOD = 10**9 + 7
for i in range(3,3*(10**5)+1):
    Lans.append((Lans[i-1]+(2*(i-1)*Lans[i-2])%MOD)%MOD)
def solve():
    L = list(map(int, sys.stdin.readline().split()))
    L2 = []
    count = 0
    for i in range(L[1]):
        L1 = list(map(int, sys.stdin.readline().split()))
        if L1[0]!=L1[1]:
            count += 2
        else:
            count += 1
    cursize = L[0]-count
    print(Lans[cursize])
    #L1 = list(map(int, sys.stdin.readline().split()))
    #st = sys.stdin.readline().strip()
for _ in range(int(sys.stdin.readline().strip())):
    solve()

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

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

Think step by step.

Response:
```plaintext
1
3 1
1 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.*;
import java.io.*;
public class aaaa {
    public static void main(String[] args) {
    	try {
            FastReader sc = new FastReader();
            FastWriter out = new FastWriter();
            int t=sc.nextInt();
            while(t-->0) {
            	int n=sc.nextInt();
            	int k=sc.nextInt();
            	for(int i=0;i<k;i++) {
            		int x=sc.nextInt();
            		int y=sc.nextInt();
            		if(x==y)
            			n--;
            		else
            			n-=2;
            	}
            	if(n==0) {
            		out.println(1);
            		continue;
            	}
            	long[] dp= new long[n+1];
            	dp[0]=1;
            	dp[1]=1;
            	for(int i=2;i<=n;i++) {
            		dp[i]=(dp[i-1]+((2*(i-1)%1000000007)*dp[i-2])%1000000007)%1000000007;
            	}
            	System.out.println(dp[n]);
            }
            out.close();
        } catch (Exception e) {
            e.printStackTrace();
        }

    }
    public static int gcd(int x, int y) {
        while (y != 0) {
            int temp = y;
            y = x % y;
            x = temp;
        }
        return x;
    }

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

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

    static class FastWriter {
        private final BufferedWriter bw;

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

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

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

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

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases where the behavior of the algorithm might differ from the expected behavior. The main issues in the code are related to how it handles the reduction of available positions for rooks and the dynamic programming approach to count the configurations.

Here are the key points to consider:

1. **Reduction of Available Positions**:
   - The code reduces `n` by 2 for each move where the rook is placed at a non-diagonal position.
   - The code reduces `n` by 1 for each move where the rook is placed at a diagonal position.

2. **Dynamic Programming**:
   - The DP array is used to count the number of configurations, but the transition formula might be incorrect or incomplete.

To create a challenging test case, we should focus on a scenario where the reduction of available positions is critical and the dynamic programming transitions might lead to incorrect results.

### Challenging Test Case
Let's consider a small board with a few moves where the rooks are placed in a way that maximizes the complexity of the reduction and the dynamic programming transitions.

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

### Explanation
- **Test Case**:
  - `n = 4`: A 4x4 chessboard.
  - `k = 2`: Two moves have already been played.
  - `1 2`: The first move places a white rook at (1, 2).
  - `2 1`: The second move places a white rook at (2, 1).

- **Computer's Moves**:
  - The computer will place a black rook at (2, 1) in response to the first move.
  - The computer will place a black rook at (1, 2) in response to the second move.

- **Available Positions**:
  - After the two moves, the board will have rooks at (1, 2) and (2, 1) for both white and black.
  - The remaining available positions are (3, 3), (3, 4), (4, 3), and (4, 4).

- **Expected Output**:
  - The number of possible final configurations is 3, as shown in the example.

This test case should expose the bug in the reduction of available positions and the dynamic programming transitions, as the code might not handle the diagonal and non-diagonal reductions correctly.