**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Quintomania
time_limit_ms: 1000
memory_limit_mb: 256
Description: Boris Notkin composes melodies. He represents them as a sequence of notes, where each note is encoded as an integer from $0$ to $127$ inclusive. The interval between two notes $a$ and $b$ is equal to $|a - b|$ semitones.

Boris considers a melody perfect if the interval between each two adjacent notes is either $5$ semitones or $7$ semitones.

After composing his latest melodies, he enthusiastically shows you his collection of works. Help Boris Notkin understand whether his melodies are perfect.

Input: The first line contains an integer $t$ ($1 \leq t \leq 1000$) — the number of melodies.

Each melody is described by two lines.

The first line contains an integer $n$ ($2 \leq n \leq 50$) — the number of notes in the melody.

The second line contains $n$ integers $a_{1}, a_{2}, \dots, a_{n}$ ($0 \leq a_{i} \leq 127$) — the notes of the melody.

Output: For each melody, output "YES", if it is perfect; otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
8
2
114 109
2
17 10
3
76 83 88
8
38 45 38 80 85 92 99 106
5
63 58 65 58 65
8
117 124 48 53 48 43 54 49
5
95 102 107 114 121
10
72 77 82 75 70 75 68 75 68 75
output:
YES
YES
YES
NO
YES
NO
YES
YES

Note:

Buggy Code:
#include <iostream>
using namespace std;
int main() {
	int t,b;
	b = 0;
	cin >> t;
	for(int j=1;j<=t;j++) {
		int n;
		cin >> n;
		int r;
		cin >> r;
		for (int i = 2; i <= n; i++) {
			int p;
			cin >> p;
			if (r - p == 5 || r - p == 7 || r - p == -5 || r - p == -7) { b=1; r = p; }
			else { b = 0; continue; }
		}
		if (b == 1)cout << "YES" <<'\n';
		else cout << "NO" << '\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
2
120 115
3
120 115 120
4
120 115 122 117
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Quintomania
time_limit_ms: 1000
memory_limit_mb: 256
Description: Boris Notkin composes melodies. He represents them as a sequence of notes, where each note is encoded as an integer from $0$ to $127$ inclusive. The interval between two notes $a$ and $b$ is equal to $|a - b|$ semitones.

Boris considers a melody perfect if the interval between each two adjacent notes is either $5$ semitones or $7$ semitones.

After composing his latest melodies, he enthusiastically shows you his collection of works. Help Boris Notkin understand whether his melodies are perfect.

Input: The first line contains an integer $t$ ($1 \leq t \leq 1000$) — the number of melodies.

Each melody is described by two lines.

The first line contains an integer $n$ ($2 \leq n \leq 50$) — the number of notes in the melody.

The second line contains $n$ integers $a_{1}, a_{2}, \dots, a_{n}$ ($0 \leq a_{i} \leq 127$) — the notes of the melody.

Output: For each melody, output "YES", if it is perfect; otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
8
2
114 109
2
17 10
3
76 83 88
8
38 45 38 80 85 92 99 106
5
63 58 65 58 65
8
117 124 48 53 48 43 54 49
5
95 102 107 114 121
10
72 77 82 75 70 75 68 75 68 75
output:
YES
YES
YES
NO
YES
NO
YES
YES

Note:

Buggy Code:
#include <bits/stdc++.h>
#define ll long long
#define pb push_back
#define vi vector<int>
#define vvi vector<vi>
#define umi unordered_map<int,int>
#define usi unordered_set<int>
#define all(x) x.begin(),x.end()
#define fr(i, n) for (long long i = 0; i < (n); i++)
using namespace std;
const int mod =1e9+7;
 void solve(){
 int n;
 cin>>n;
 vector<int>v(n);
 for(int i=0;i<n;i++){
    cin>>v[i];
 }
 int cnt=0;
 for(int i=0;i<n;i++){
    if(abs(v[i]-v[i+1])==5||abs(v[i]-v[i+1])==7){
        cnt++;
    }
 }
 if(cnt==n-1){
    cout<<"YES"<<endl;
 }
 else{
    cout<<"NO"<<endl;
 }
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int tt=1;
cin >> tt;
while(tt--){
 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 expose the bug in the provided code, we need to focus on edge cases where the code might fail. The main issue in the buggy code is that it does not handle the case where the melody has only two notes correctly. Specifically, it tries to access `v[i+1]` when `i` is the last index, which leads to an out-of-bounds error.

Let's create a test input that will highlight this issue:

1. **Single Note Melody**: This is not possible since the problem specifies that the melody must have at least 2 notes.
2. **Two Note Melody**: This is the smallest valid input and should be handled correctly.
3. **Melody with a Single Interval Error**: This should return "NO".
4. **Melody with All Correct Intervals**: This should return "YES".

Here is a small-scale test input that covers these cases:

```plaintext
3
2
10 15
2
10 17
3
10 15 20
```

Explanation:
1. The first melody has two notes (10 and 15) with an interval of 5 semitones. This should return "YES".
2. The second melody has two notes (10 and 17) with an interval of 7 semitones. This should return "YES".
3. The third melody has three notes (10, 15, 20) with intervals of 5 and 5 semitones. This should return "YES".

This test input will help expose the bug in the buggy code by ensuring that the code handles the smallest valid input correctly and does not go out of bounds.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Quintomania
time_limit_ms: 1000
memory_limit_mb: 256
Description: Boris Notkin composes melodies. He represents them as a sequence of notes, where each note is encoded as an integer from $0$ to $127$ inclusive. The interval between two notes $a$ and $b$ is equal to $|a - b|$ semitones.

Boris considers a melody perfect if the interval between each two adjacent notes is either $5$ semitones or $7$ semitones.

After composing his latest melodies, he enthusiastically shows you his collection of works. Help Boris Notkin understand whether his melodies are perfect.

Input: The first line contains an integer $t$ ($1 \leq t \leq 1000$) — the number of melodies.

Each melody is described by two lines.

The first line contains an integer $n$ ($2 \leq n \leq 50$) — the number of notes in the melody.

The second line contains $n$ integers $a_{1}, a_{2}, \dots, a_{n}$ ($0 \leq a_{i} \leq 127$) — the notes of the melody.

Output: For each melody, output "YES", if it is perfect; otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
8
2
114 109
2
17 10
3
76 83 88
8
38 45 38 80 85 92 99 106
5
63 58 65 58 65
8
117 124 48 53 48 43 54 49
5
95 102 107 114 121
10
72 77 82 75 70 75 68 75 68 75
output:
YES
YES
YES
NO
YES
NO
YES
YES

Note:

Buggy Code:
for _ in range(int(input())):
    n = int(input())
    a = list(map(int,input().split()))
    ans = "YES"
    for i in range(n-1):
        if not abs(a[i+1] - a[i]) <= 7 and abs(a[i+1] - a[i]) >= 5:
            ans = "NO"
    print(ans)

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

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

Think step by step.

Response:
```plaintext
3
2
10 15
3
10 17 22
4
10 15 20 25
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Quintomania
time_limit_ms: 1000
memory_limit_mb: 256
Description: Boris Notkin composes melodies. He represents them as a sequence of notes, where each note is encoded as an integer from $0$ to $127$ inclusive. The interval between two notes $a$ and $b$ is equal to $|a - b|$ semitones.

Boris considers a melody perfect if the interval between each two adjacent notes is either $5$ semitones or $7$ semitones.

After composing his latest melodies, he enthusiastically shows you his collection of works. Help Boris Notkin understand whether his melodies are perfect.

Input: The first line contains an integer $t$ ($1 \leq t \leq 1000$) — the number of melodies.

Each melody is described by two lines.

The first line contains an integer $n$ ($2 \leq n \leq 50$) — the number of notes in the melody.

The second line contains $n$ integers $a_{1}, a_{2}, \dots, a_{n}$ ($0 \leq a_{i} \leq 127$) — the notes of the melody.

Output: For each melody, output "YES", if it is perfect; otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
8
2
114 109
2
17 10
3
76 83 88
8
38 45 38 80 85 92 99 106
5
63 58 65 58 65
8
117 124 48 53 48 43 54 49
5
95 102 107 114 121
10
72 77 82 75 70 75 68 75 68 75
output:
YES
YES
YES
NO
YES
NO
YES
YES

Note:

Buggy Code:
t = int(input())

for T in range(t):
    n = int(input())
    arr = [int(x) for x in input().split(" ")]

    perfect = True
    for i in range(len(arr)-2):
        if abs(arr[i] - arr[i+1]) != 5 and abs(arr[i] - arr[i+1]) != 7:
            perfect = False
            break

    print("YES" if perfect else "NO")

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

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

Think step by step.

Response:
```plaintext
3
3
0 5 10
3
0 5 12
2
0 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Quintomania
time_limit_ms: 1000
memory_limit_mb: 256
Description: Boris Notkin composes melodies. He represents them as a sequence of notes, where each note is encoded as an integer from $0$ to $127$ inclusive. The interval between two notes $a$ and $b$ is equal to $|a - b|$ semitones.

Boris considers a melody perfect if the interval between each two adjacent notes is either $5$ semitones or $7$ semitones.

After composing his latest melodies, he enthusiastically shows you his collection of works. Help Boris Notkin understand whether his melodies are perfect.

Input: The first line contains an integer $t$ ($1 \leq t \leq 1000$) — the number of melodies.

Each melody is described by two lines.

The first line contains an integer $n$ ($2 \leq n \leq 50$) — the number of notes in the melody.

The second line contains $n$ integers $a_{1}, a_{2}, \dots, a_{n}$ ($0 \leq a_{i} \leq 127$) — the notes of the melody.

Output: For each melody, output "YES", if it is perfect; otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
8
2
114 109
2
17 10
3
76 83 88
8
38 45 38 80 85 92 99 106
5
63 58 65 58 65
8
117 124 48 53 48 43 54 49
5
95 102 107 114 121
10
72 77 82 75 70 75 68 75 68 75
output:
YES
YES
YES
NO
YES
NO
YES
YES

Note:

Buggy Code:
import java.util.*;
public class solution
{
    public static void main(String[] args)
    {
        Scanner s=new Scanner(System.in);
        int t=s.nextInt();
        while(t-->0)
        {
            int n=s.nextInt();
            int[] a=new int[n];
            int flag=0;
            for(int i=0;i<n;i++)
            {
                a[i]=s.nextInt();
            }
            for(int i=0;i<n-1;i++)
            {
                flag=0;

                    int k=Math.abs(a[i]-a[i+1]);
                    if(k>=5 && k<=7)
                    {
                        flag=1;
                    }
                    else
                    {
                        flag=0;
                        break;
                    }

            }
            if(flag==1)
            System.out.println("Yes");
            else
            System.out.println("No");
        }
        s.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:
```plaintext
3
2
0 5
2
0 7
2
0 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Quintomania
time_limit_ms: 1000
memory_limit_mb: 256
Description: Boris Notkin composes melodies. He represents them as a sequence of notes, where each note is encoded as an integer from $0$ to $127$ inclusive. The interval between two notes $a$ and $b$ is equal to $|a - b|$ semitones.

Boris considers a melody perfect if the interval between each two adjacent notes is either $5$ semitones or $7$ semitones.

After composing his latest melodies, he enthusiastically shows you his collection of works. Help Boris Notkin understand whether his melodies are perfect.

Input: The first line contains an integer $t$ ($1 \leq t \leq 1000$) — the number of melodies.

Each melody is described by two lines.

The first line contains an integer $n$ ($2 \leq n \leq 50$) — the number of notes in the melody.

The second line contains $n$ integers $a_{1}, a_{2}, \dots, a_{n}$ ($0 \leq a_{i} \leq 127$) — the notes of the melody.

Output: For each melody, output "YES", if it is perfect; otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
8
2
114 109
2
17 10
3
76 83 88
8
38 45 38 80 85 92 99 106
5
63 58 65 58 65
8
117 124 48 53 48 43 54 49
5
95 102 107 114 121
10
72 77 82 75 70 75 68 75 68 75
output:
YES
YES
YES
NO
YES
NO
YES
YES

Note:

Buggy Code:
import java.util.Scanner;
public class adhd
{
    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        for(int i=0;i<t;i++)
        {
            int n=sc.nextInt();
               int perfect[]=new int[n];int a=1;
               for(int k=0;k<n;k++)
               {
                   perfect[k]=sc.nextInt();
               }
               for(int j=0;j<n-1;j++)
               {
                   int s= perfect[j+1]-perfect[j];

                   if( (s==7)||( s==-7)||(s==5)||(s==-5)|| (j==7) ||(j==-7)||(j==5)||(j==-5))
                   {
                       a=a+1;
                       if(a==n)
                       System.out.println("YES");
                   }
                   else
                   System.out.println("NO");
               }
        }

    }
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided buggy code, we need to focus on the logic errors in the code. The main issues in the buggy code are:

1. The variable `a` is used to count the number of valid intervals, but it is initialized to 1 and incremented in the loop. This means `a` will always be at least 1, and the check `if(a==n)` will always be true for the first interval.
2. The condition `if( (s==7)||( s==-7)||(s==5)||(s==-5)|| (j==7) ||(j==-7)||(j==5)||(j==-5))` is incorrect because `j` is the index of the loop, not the interval value. This condition will always be true for specific values of `j`, leading to incorrect outputs.

To expose these issues, we can create a test case with a small number of notes where the intervals are not 5 or 7 semitones, but the index values in the loop might trigger the incorrect condition.

Here is a small-scale test input that should expose the bug:

```plaintext
3
3
10 15 20
2
10 12
2
10 17
```

Explanation:
1. The first melody (10, 15, 20) has intervals of 5 semitones, so it should be "YES".
2. The second melody (10, 12) has an interval of 2 semitones, which is not 5 or 7 semitones, so it should be "NO".
3. The third melody (10, 17) has an interval of 7 semitones, so it should be "YES".

This test input should help identify 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 (Hard 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:

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

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:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $$$(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$$$ satisfy the conditions.

Buggy Code:
#include <vector>
#include <algorithm>
#include <iostream>
#include <map>

using namespace std;

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int t;
	cin >> t;

	for (int tt = 0; tt < t; tt++) {
		int n, m;
		cin >> n >> m;
		long long ans = 0;
		vector<int>p(n + m + 1);
		vector<vector<int>>dell(n + m + 1);
		for (int i = 2; i <= n + m; i++) {
			if (p[i] == 0) {
				for (int j = 1; j * i <= n + m; j++) {
					dell[j * i].push_back(i);
					p[j * i] = 1;
				}
			}
		}
		for (int i = 0; i <= n + m; i++) {
			p[i] = 0;
		}
		for (int x = 2; x <= n + m; x++) {
			int b = 1;
			int z = x;
			int i = 0;
			while (i < dell[x].size()) {
				if (z % dell[x][i] == 0) {
					z = z / dell[x][i];
					p[dell[x][i]]++;
				}
				else {
					i++;
				}
			}
			for (int i = 0; i < dell[x].size(); i++) {
				if (b % dell[x][i]) {
					for (int j = 0; j < (p[dell[x][i]] + 1) / 2; j++) {
						b = b * dell[x][i];
					}
				}
				p[dell[x][i]]=0;
			}
			if (b <= m && b < x) {
				int r = min(x - 1, m) / b;
				int l = (x - n + b - 1) / b;
				l = max(1, l);
				if (l <= r) {
					ans += r - l + 1;
				}
			}
		}
		cout << ans << "\n";
	}
}

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

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

Think step by step.

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

1. **Understand the Bug**: The buggy code seems to have issues with the prime factorization and the way it calculates the valid pairs. It might not correctly handle cases where the gcd and the multiples are involved.

2. **Edge Cases**:
   - **Small Values**: Test with small values of `n` and `m` to ensure the code handles basic cases correctly.
   - **Large Values**: Test with large values of `n` and `m` to see if the code can handle them within the time limit.
   - **Special Values**: Test with values where `n` and `m` are equal, or where one is a multiple of the other.

3. **Specific Scenarios**:
   - **Prime Numbers**: Test with `n` and `m` being prime numbers.
   - **Composite Numbers**: Test with `n` and `m` being composite numbers.
   - **Boundary Conditions**: Test with `n` and `m` at the boundary of the input range.

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

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

This test input includes:
- A case with `n` and `m` both being 1.
- A case with `n` and `m` being small composite numbers.
- A case with `n` and `m` being small prime numbers.
- A case with `n` and `m` being equal and relatively small.

This should help in identifying 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 (Hard 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:

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

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:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $$$(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$$$ satisfy the conditions.

Buggy Code:
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<cmath>
#include<random>
#include<bitset>
#include<string>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<deque>
#include<chrono>
using namespace std;
typedef long long int lld;
typedef long double lf;
#define INF 1000000000000000000LL
#define MOD 1000000007LL
#define ff first
#define ss second

lld gcd(lld x, lld y){
	if(x==0) return y;
	return gcd(y%x, x);
}

void solve(){
	lld n, m; cin >> n >> m;
    if(m>n) swap(n, m);
    vector<vector<lld>> divs(m+1);
    for(lld i=2;i<=m;i++){
        for(lld j=i;j<=m;j+=i){
            divs[j].push_back(i);
        }
    }
	lld ans = 0;
	for(lld k=1;k<=m;k++){
        for(lld dv: divs[k]){
            if(dv>m/k+n/k) break;
            for(lld b=1;b<=min(dv-1, m/k);b++){
                if(dv-b<=n/k&&gcd(b,dv)==1) ans++;

            }
        }
    }
    cout << ans << "\n";
}

int main(){
	mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	lld ttt; cin>>ttt;
	while(ttt--) 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
1
2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Hard 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:

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

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:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $$$(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$$$ satisfy the conditions.

Buggy Code:
///////////////////////////////    _LeMur_
#define _CRT_SECURE_NO_WARNINGS
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cassert>
#include <chrono>
#include <random>
#include <bitset>
#include <cstdio>
#include <vector>
#include <string>
#include <stack>
#include <tuple>
#include <queue>
#include <ctime>
#include <cmath>
#include <list>
#include <map>
#include <set>

using namespace std;

const int N = 2000005;
const int inf = 1000 * 1000 * 1000;
const int mod = 998244353;
mt19937 myrand(chrono::steady_clock::now().time_since_epoch().count());

int t;
int n, m;

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

int lp[N];

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);

  for (int i = 2; i < N; i++) {
    if (lp[i] == 0) {
      for (int j = i; j < N; j += i) {
        lp[j] = i;
      }
    }
  }

  cin >> t;

  while (t--) {
    cin >> n >> m;

    if (n < m) swap(n, m);

    int answ = 0;
    for (int sum = 1; sum <= m; sum++) {
      for (int g = sum; g <= m; g += sum) {
        // a >= 1, b >= 1
        // a + b = sum
        // a * g <= n
        // b * g <= m

        for (int a = 1; a * 1ll * g <= m; a++) {
          if (a >= sum) break;

          if (gcd(a, sum - a) == 1 && sum - a <= n / g) {
            ++answ;
          }
        }
      }
    }

    cout << answ << endl;
  }
  return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Hard 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:

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

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:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $$$(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$$$ satisfy the conditions.

Buggy Code:
import java.util.*;

public class Main {
    private static int[] s = new int[2000001];
    private static ArrayList<Pair<Integer, Integer>> o = new ArrayList<>();
    private static ArrayList<Integer> d = new ArrayList<>();

    private static void dfs(int p, int v) {
        if (p == o.size()) {
            d.add(v);
            return;
        }
        for (int i = 0; i <= o.get(p).getSecond(); ++i) {
            dfs(p + 1, v);
            v *= o.get(p).getFirst();
        }
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        Arrays.fill(s, -1);
        for (int i = 2; i <= 2000000; ++i) {
            if (s[i] == -1) {
                for (int j = i; j <= 2000000; j += i) {
                    s[j] = i;
                }
            }
        }
        int t = scanner.nextInt();
        while (t-- > 0) {
            int n = scanner.nextInt();
            int m = scanner.nextInt();
            int ans = 0;
            for (int g = 2; g <= Math.min(n, m); ++g) {
                o.clear();
                d.clear();
                for (int x = g; s[x] > 0; x /= s[x]) {
                    if (o.isEmpty() || !o.get(o.size() - 1).getFirst().equals(s[x])) {
                        o.add(new Pair<>(s[x], 1));
                    } else {
                        Pair<Integer, Integer> last = o.get(o.size() - 1);
                        o.set(o.size() - 1, new Pair<>(last.getFirst(), last.getSecond() + 1));
                    }
                }
                dfs(0, 1);
                for (int v : d) {
                    for (int a = Math.max(g, (g * v - m + g - 1) / g * g); a <= n; a += g) {
                        int b = g * v - a;
                        if (b <= 0) {
                            break;
                        }
                        if (gcd(a, b) != g) {
                            continue;
                        }
                        if (b <= m) {
                            ++ans;
                        }
                    }
                }
            }
            System.out.println(ans);
        }
        scanner.close();
    }

    private static int gcd(int a, int b) {
        while (b != 0) {
            int t = b;
            b = a % b;
            a = t;
        }
        return a;
    }

    static class Pair<T1, T2> {
        private T1 first;
        private T2 second;

        public Pair(T1 first, T2 second) {
            this.first = first;
            this.second = second;
        }

        public T1 getFirst() {
            return first;
        }

        public T2 getSecond() {
            return second;
        }
    }
}

**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)**
   - This is the smallest possible input. No pairs satisfy the conditions because the only possible pair (1, 1) does not meet the condition \( b \cdot \gcd(a, b) \) being a multiple of \( a + b \).

2. **Test Case 2: (2, 3)**
   - This is a small input that includes a valid pair (2, 2). The pair (2, 2) satisfies the condition because \( 2 \cdot \gcd(2, 2) = 4 \) is a multiple of \( 2 + 2 = 4 \).

3. **Test Case 3: (3, 5)**
   - This input includes a valid pair (3, 6). The pair (3, 6) satisfies the condition because \( 6 \cdot \gcd(3, 6) = 18 \) is a multiple of \( 3 + 6 = 9 \).

4. **Test Case 4: (10, 8)**
   - This input includes multiple valid pairs: (2, 2), (3, 6), (4, 4), (6, 3), (6, 6), and (8, 8). Each of these pairs meets the condition.

5. **Test Case 5: (100, 1233)**
   - This is a larger input that includes a significant number of valid pairs. The number of valid pairs is 423, which is a more complex scenario to handle correctly.

6. **Test Case 6: (1000000, 1145141)**
   - This is a very large input that includes a large number of valid pairs. The number of valid pairs is 5933961, which is a challenging scenario for the algorithm to handle efficiently.

These test cases cover a range of scenarios, from the smallest possible input to very large inputs, and include both edge cases and more complex scenarios. They are designed to expose any bugs in the implementation, especially in the handling of large inputs and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Hard 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:

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

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:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $$$(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$$$ satisfy the conditions.

Buggy Code:
import java.util.*;

public class Main {
    private static final int MAX = 2000000;
    private static int[] smallestPrimeFactor = new int[MAX + 1];
    private static List<Map.Entry<Integer, Integer>> primeFactors = new ArrayList<>();
    private static List<Integer> divisors = new ArrayList<>();

    private static void findDivisors(int index, int currentValue) {
        if (index == primeFactors.size()) {
            divisors.add(currentValue);
            return;
        }
        int base = primeFactors.get(index).getKey();
        int power = primeFactors.get(index).getValue();
        int value = currentValue;
        for (int i = 0; i <= power; i++) {
            findDivisors(index + 1, value);
            value *= base;
        }
    }

    public static void main(String[] args) {
        try (Scanner scanner = new Scanner(System.in)) {
            Arrays.fill(smallestPrimeFactor, -1);
            for (int i = 2; i <= MAX; ++i) {
                if (smallestPrimeFactor[i] == -1) {
                    for (int j = i; j <= MAX; j += i) {
                        smallestPrimeFactor[j] = i;
                    }
                }
            }
            int cases = scanner.nextInt();
            while (cases-- > 0) {
                int n = scanner.nextInt();
                int m = scanner.nextInt();
                int countPairs = 0;
                for (int gcdValue = 2; gcdValue <= Math.min(n, m); ++gcdValue) {
                    primeFactors.clear();
                    divisors.clear();
                    for (int x = gcdValue; x > 1; x /= smallestPrimeFactor[x]) {
                        if (primeFactors.isEmpty() || !primeFactors.get(primeFactors.size() - 1).getKey().equals(smallestPrimeFactor[x])) {
                            primeFactors.add(new AbstractMap.SimpleEntry<>(smallestPrimeFactor[x], 1));
                        } else {
                            int lastIndex = primeFactors.size() - 1;
                            Map.Entry<Integer, Integer> last = primeFactors.get(lastIndex);
                            primeFactors.set(lastIndex, new AbstractMap.SimpleEntry<>(last.getKey(), last.getValue() + 1));
                        }
                    }
                    findDivisors(0, 1);
                    for (int divisor : divisors) {
                        for (int a = Math.max(gcdValue, (gcdValue * divisor - m + gcdValue - 1) / gcdValue * gcdValue); a <= n; a += gcdValue) {
                            int b = gcdValue * divisor - a;
                            if (b <= 0) break;
                            if (gcd(a, b) != gcdValue) continue;
                            if (b <= m) countPairs++;
                        }
                    }
                }
                System.out.println(countPairs);
            }
        }
    }

    private static int gcd(int a, int b) {
        while (b != 0) {
            int temp = b;
            b = a % b;
            a = temp;
        }
        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 given 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:**
   - Test with the smallest possible values for `n` and `m` (1, 1).
   - Test with `n` and `m` being equal and small (e.g., 2, 2).
   - Test with one of `n` or `m` being 1 and the other being a larger value (e.g., 1, 10).

2. **Special Values:**
   - Test with values where `gcd(a, b)` is a significant factor (e.g., 10, 10).
   - Test with values where `a` and `b` have a specific relationship (e.g., 3, 6).

3. **Large Values:**
   - Test with values that are near the upper limit (e.g., 2000000, 2000000).

4. **Mixed Values:**
   - Test with a mix of small and large values to ensure the code handles different ranges correctly (e.g., 100, 2000000).

Based on these considerations, a small-scale test input that is complete and valid is:

```plaintext
6
1 1
2 2
1 10
10 10
3 6
100 2000000
```

This input includes a variety of scenarios to test the code's ability to handle edge cases, special values, and mixed values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Hard 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:

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

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:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $$$(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$$$ satisfy the conditions.

Buggy Code:
#ifdef ONLINE_JUDGE
#include <bits/stdc++.h>
#else
#include "./stdc++.h"
#endif
#define fastio                                                                 \
  ios_base::sync_with_stdio(false);                                            \
  cin.tie(nullptr)
#define i64 long long
#define ld long double
#define F first
#define S second
#define MOD 1000000007
#define pairii pair<int, int>
#define pb push_back
#ifdef ONLINE_JUDGE
#define endl '\n'
#endif
#define matrix(type) vector<vector<type>>
#define vi vector<int>
#define vi64 vector<i64>
#define uset unordered_set
#define umap unordered_map
#define mpr make_pair
#define ALL(x) x.begin(), x.end()
#define RALL(data) data.rbegin(), data.rend()
#define TYPEMAX(type) std::numeric_limits<type>::max()
#define TYPEMIN(type) std::numeric_limits<type>::min()
#define null nullptr
#define fr(i, a, b) for (int i = a; i < b; i++)
using namespace std;

#ifdef ONLINE_JUDGE
class NullBuffer : public std::streambuf {
public:
  int overflow(int c) { return c; }
};
ostream logger(new NullBuffer());
#else
ostream &logger = cout;
#endif

#define LIM 1500
bitset<LIM> flag;
vi primes;
void setup() {
  for (int i = 2; i < LIM; i++) {
    if (flag[i]) {
      continue;
    }
    primes.pb(i);
    for (int j = i * i; j < LIM; j += i) {
      flag[j] = 1;
    }
  }
}

void solveMain() {
  int n, m;
  cin >> n >> m;
  int S = max(n, m) << 1;
  i64 ans = 0;
  for (int s = 4; s <= S; s++) {
    int r = 1, t = s;
    for (int &p : primes) {
      if (p * p > t) {
        break;
      }
      int c = 0;
      while (!(t % p)) {
        if (c++ & 1) {
          r *= p;
        }
        t /= p;
      }
    }
    t = s / r;
    int y = r - 1;
    y -= max(0, r - 1 - n / t);
    y -= max(0, r - 1 - m / t);
    ans += max(0, y);
  }
  cout << ans << endl;
}

int main() {
#ifndef ONLINE_JUDGE
  clock_t tStart = clock();
#endif
  fastio;
  int t = 1;
  cin >> t;
  setup();
  while (t--) {
    solveMain();
  }
#ifndef ONLINE_JUDGE
  printf("Time taken: %.3fs\n", (double)(clock() - tStart) / CLOCKS_PER_SEC);
#endif
  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 consider edge cases and scenarios that are likely to cause issues. The buggy code has a few potential weak points:

1. **Prime Number Handling**: The code uses a precomputed list of primes up to 1500, which might not cover all necessary primes for larger values of \( n \) and \( m \).
2. **GCD Calculation**: The code does not explicitly use a GCD function, which is crucial for the problem's conditions.
3. **Range and Edge Cases**: The code might handle small values of \( n \) and \( m \) differently than larger values.

Given these points, we can create a test case that pushes the limits of the prime handling and GCD calculation. Let's choose \( n \) and \( m \) such that they are relatively prime and large enough to ensure the code's prime handling is tested.

Here's a small-scale test input that is complete and valid:

```plaintext
1
1500 1501
```

### Explanation:
- **Test Case Count**: 1
- **Values of \( n \) and \( m \)**: 1500 and 1501
  - 1500 is just within the range of the precomputed primes.
  - 1501 is just outside the range of the precomputed primes, which might cause issues in the prime handling logic.

This test case should help expose any issues with the prime handling and GCD calculation in 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 (Hard 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:

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

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:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $$$(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$$$ satisfy the conditions.

Buggy Code:
// -------------------- Includes -------------------- //
#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <stdio.h>
#include <cstdlib>
#include <stdlib.h>
#include <array>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <math.h>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <bitset>
#include <list>
#include <iterator>
#include <numeric>
#include <complex>
#include <tuple>
#include <utility>
#include <cassert>
#include <assert.h>
#include <climits>
#include <limits.h>
#include <ctime>
#include <time.h>
#include <random>
#include <chrono>
#include <fstream>
using namespace std;

// -------------------- Typedefs -------------------- //

typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ld;

// -------------------- Defines -------------------- //

#define pr pair
#define mpr make_pair
#define ff first
#define ss second

#define mset multiset
#define mmap multimap
#define uset unordered_set
#define umap unordered_map
#define umset unordered_multiset
#define ummap unordered_multimap
#define pqueue priority_queue

#define sz(x) (int((x).size()))
#define len(x) (int((x).length()))
#define all(x) (x).begin(), (x).end()
#define clr(x) (x).clear()

#define ft front
#define bk back
#define pf push_front
#define pb push_back
#define popf pop_front
#define popb pop_back

#define lb lower_bound
#define ub upper_bound
#define bs binary_search

// -------------------- Constants -------------------- //

const int MAX = int(1e9 + 5);
const ll MAXL = ll(1e18 + 5);
const ll MOD = ll(1e9 + 7);
const ll MOD2 = ll(998244353);

// -------------------- Functions -------------------- //

void fastio()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	return;
}

void precision(int x)
{
	cout << fixed << setprecision(x);
	return;
}

ll gcd0(ll a, ll b)
{
	while (b) {
		a %= b;
		swap(a, b);
	}
	return a;
}

ll lcm0(ll a, ll b)
{
	return a / gcd0(a, b) * b;
}

bool is_prime(ll a)
{
	if (a == 1) return false;
	for (ll i = 2; i * i <= a; i++) if (a % i == 0) return false;
	return true;
}

bool is_square(ll a)
{
	ll b = ll(sqrtl(ld(a)));
	return (b * b == a);
}

bool is_cube(ll a)
{
	ll b = ll(cbrtl(ld(a)));
	return (b * b * b == a);
}

int digit_sum(ll a)
{
	int sum = 0;
	while (a) {
		sum += int(a % 10);
		a /= 10;
	}
	return sum;
}

ll binpow(ll a, int b)
{
	ll ans = 1;
	while (b) {
		if (b & 1) ans *= a;
		b >>= 1;
		a *= a;
	}
	return ans;
}

ll binpow_mod(ll a, ll b, ll mod)
{
	ll ans = 1;
	while (b) {
		if (b & 1) ans = (ans * a) % mod;
		b >>= 1;
		a = (a * a) % mod;
	}
	return ans;
}

ll factorial(int a)
{
	ll ans = 1;
	for (int i = 2; i <= a; i++) ans *= ll(i);
	return ans;
}

ll factorial_mod(int a, ll mod)
{
	ll ans = 1;
	for (int i = 2; i <= a; i++) ans = (ans * ll(i)) % mod;
	return ans;
}

// -------------------- Solution -------------------- //

/*const int N = 200005;
ll a[N], b[N];
int n;*/

const int MM = 2000000;

vector<int> v[100005];

vector<int> v1, v2;

void solve()
{
	int i, j;

	int n, m;
	ll ans = 0;

	cin >> n >> m;

	int lm = 67000;

	int g, a0, b0, s;

	ll a, b;

	for (g = 1; g <= min(min(n, m), lm); g++) {
		for (i = 0; i < sz(v[g]); i++) {
			s = v[g][i];

			for (b0 = 1; b0 < s; b0++) {
				if (ll(b0) * ll(g) > ll(m)) break;

				a0 = s - b0;

				if (ll(a0) * ll(g) > ll(n)) continue;

				if (gcd0(a0, b0) != 1) continue;

				a = ll(g) * ll(a0);
				b = ll(g) * ll(b0);

				if (b * ll(g) % (a + b) == 0) ans++;
			}
		}
	}

	for (g = lm + 1; g <= min(n, m); g++) {
		for (a = g; a <= ll(n); a += g) {
			for (b = g; b <= ll(m); b += g) {
				if (gcd0(a / g, b / g) != 1) continue;

				if (b * ll(g) % (a + b) == 0) ans++;
			}
		}
	}

	cout << ans << "\n";

	return;
}

void precalc()
{
	int i, j;

	int e = 100000;

	for (i = 1; i <= e; i++) {
		for (j = i; j <= e; j += i) {
			v[j].pb(i);
		}
	}

	return;
}

int main()
{
	fastio();

	precalc();

	int tests = 1, tc;
	cin >> tests;
	for (tc = 1; tc <= tests; 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 and scenarios that are likely to cause issues. The code has a limit on the range of `g` (the greatest common divisor) to `lm = 67000`, and it uses a precomputed list of divisors for values up to `100000`. Therefore, we should consider test cases that:

1. **Test the boundary conditions**: Values of `n` and `m` that are just above or below the limit of `lm`.
2. **Test large values**: Values of `n` and `m` that are close to the upper limit of `2 * 10^6`.
3. **Test small values**: Values of `n` and `m` that are small to ensure the code handles them correctly.
4. **Test specific pairs**: Pairs that are likely to be on the boundary of the precomputed list.

Let's create a test input with multiple test cases to cover these scenarios:

```plaintext
6
67001 67001
2000000 2000000
1 1
2 3
3 5
10 8
```

### Explanation:
1. **67001 67001**: This tests the boundary condition where `g` is just above the limit of `67000`.
2. **2000000 2000000**: This tests the upper limit of the input range.
3. **1 1**: This tests the smallest possible values.
4. **2 3**: This is a small test case that should have a known result.
5. **3 5**: Another small test case with a known result.
6. **10 8**: A slightly larger test case with a known result.

This test input should help expose any issues in the code, especially those related to the boundary conditions and the handling of large values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Hard 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:

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

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:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $$$(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$$$ satisfy the conditions.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <map>
#include <utility>
#include <unordered_set>
#include <unordered_map>
#include <cstdlib>
#include <iomanip>
#include <cassert>
#include <deque>
#include <cmath>
#include <random>
#include <queue>
#include <iterator>
#include <bitset>
#include <array>
#include <cstring>
#include <memory>
#include <chrono>
#include <functional>
#include <optional>
#include <list>
#include <complex>

#pragma GCC target("avx2")
#pragma GCC optimize("O3")

using namespace std;
//#define FAST_ALLOCATOR_MEMORY 200'000'000
#ifdef FAST_ALLOCATOR_MEMORY
size_t allocator_pos = 0;
char allocator_memory[(int)FAST_ALLOCATOR_MEMORY];
inline void * operator new ( size_t n ) {
    char *res = allocator_memory + allocator_pos;
    allocator_pos += n;
    //assert(allocator_pos <= (int)FAST_ALLOCATOR_MEMORY);
    return (void *)res;
}
inline void operator delete ( void * ) noexcept { }
//inline void * operator new [] ( size_t ) { assert(0); }
//inline void operator delete [] ( void * ) { assert(0); }
#endif

using ui64 = uint64_t;
using i64 = int64_t;
using ui32 = uint32_t;
using i32 = int32_t;
using ui16 = uint16_t;
using i16 = int16_t;
using ui8 = uint8_t;
using i8 = int8_t;
using ui128 = __uint128_t;
using i128 = __int128_t;

#define ld long double
#define pb push_back
#define all(x) x.begin(), x.end()
#define FAST ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define vi vector<int>
#define vvi vector<vi>
#define pii pair<int, int>

const int inf = 1e9 + 7;
const int64_t linf = 1e18 + 1;

template<typename T>
using V = vector<T>;

template<typename T = int>
T nxt() {
    T x;
    cin >> x;
    return x;
}

template<typename T = int>
void read(vector<T>& x) {
    for (T& i: x)
        cin >> i;
}

template<typename T>
istream& operator>>(istream& in, V<T>& v) {
    for (auto& x: v) {
        in >> x;
    }
    return in;
}

template<typename T>
struct IsVector {
    static constexpr bool value = false;
};

template<typename T>
struct IsVector<V<T>> {
    static constexpr bool value = true;
};

template<typename T>
ostream& operator<<(ostream& out, V<T>& v) {
    bool is_first = true;
    for (auto& x: v) {
        if (!is_first) {
            out << (IsVector<T>::value ? "\n" : " ");
        }
        is_first = false;
        out << x;
    }
    //out << endl;
    return out;
}

int bit(int x, int i) {
    return ((x>>i)&1);
}

int popcnt(int x) {
    return __builtin_popcount(x);
}

// todo: нормальная система assert и дебаг вывода

const int maxn = 2e6 + 6;
const int max_values = 29326330 + 10;

int values[max_values];
int last[maxn];
int pr[max_values];

void Precalc() {
    int cnt = 1;
    memset(pr, 0, sizeof(pr));
    memset(last, 0, sizeof(last));
    for (int i = 1; i < maxn; ++i) {
        for (int j = i; j < maxn; j += i) {
            pr[cnt] = last[j];
            last[j] = cnt;
            values[cnt] = i;
            ++cnt;
        }
    }
}

void Solve() {
    int n, m;
    cin >> n >> m;
    int ans = 0;
    for (i64 q = 1; q <= m; ++q) {
        for (i64 b = q, b1 = 1; b <= m; b += q, ++b1) {
            int c = gcd(b1, q);
            int x = q / c;
            for (int j = last[x]; j; j = pr[j]) {
                int ab = values[j];
                i64 a1 = ab - b1;
                if (a1 <= 0) continue;
                i64 a = a1 * q;
                if (a <= n && gcd(a1, b1) == 1) {
                    ++ans;
                }
            }
        }
    }
    cout << ans;
}

void Clear() {

}

signed main() {
    FAST;
    Precalc();
    int t = nxt();
    while (t--) {
        Clear();
        Solve();
        /// be careful!!!
        cout << "\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
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Hard 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:

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

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:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $$$(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$$$ satisfy the conditions.

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

using namespace std;

#pragma GCC target("avx2")
#pragma GCC optimize("Ofast,unroll-loops")

#define ll long long
#define pb push_back
#define pii pair<int,int>
#define all(v) v.begin(),v.end()
#define F first
#define S second
#define ub upper_bound
// #define int ll
#define sz(s) (int)s.size()

const int inf=1e9;
const int MAX=5005;
const int N=2e6+5;

int m,n;
vector<pii> vec[MAX];

void calc(){
  for(int i=1;i<=2000;i++){
    for(int j=1;j<=2000;j++){
      if(gcd(i,j)==1){
        vec[j+i].pb({i,j});
      }
    }
  }
  for(int i=1;i<=4000;i++){
    sort(all(vec[i]));
  }
}

void solve(){
//   return;
  cin>>n>>m;
  int ans=0;
  for(int g=1;g<=min(n,m);g++){
    for(int s=g;s<=n+m;s+=g){
      int sum=s/g;
      sum/=gcd(sum,g);
      if(sum!=1||s/g>2000)continue;
      int pos=ub(all(vec[s/g]),make_pair(min(s/g-1,m/g),inf))-vec[s/g].begin()-1;
      for(int b=pos;b>=0;b--){
        if(vec[s/g][b].F*g<=m&&vec[s/g][b].S*g<=n)ans++;
        else break;
      }
    }
  }
  cout<<ans<<"\n";
}

signed main(){
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t=1;
  cin>>t;
  calc();
  while(t--)solve();
}

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

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

Think step by step.

Response:
To 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 gcd and the conditions in the problem statement might not be handled correctly. Here are the steps to think through this:

1. **Understand the Bug**: The buggy code has a limitation where it only considers pairs \((i, j)\) with \( \gcd(i, j) = 1 \) and only up to 2000. This means it might miss valid pairs where the gcd is greater than 1 or where the values are larger than 2000.

2. **Edge Cases**:
   - **Small Values**: Test with small values of \(n\) and \(m\) to ensure the basic logic works.
   - **Large Values**: Test with values close to the upper limit to see if the code can handle large inputs.
   - **Specific Values**: Test with values where the gcd is greater than 1 to see if the code correctly handles such cases.

3. **Test Input**:
   - **Test Case 1**: Small values to ensure basic functionality.
   - **Test Case 2**: Values where the gcd is greater than 1.
   - **Test Case 3**: Large values to stress the code.

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

```plaintext
3
1 1
2 4
2001 2001
```

**Explanation**:
- **Test Case 1**: \(n = 1\), \(m = 1\). This is a basic case where no pairs satisfy the condition.
- **Test Case 2**: \(n = 2\), \(m = 4\). This case includes pairs where the gcd is greater than 1, such as \((2, 2)\) and \((2, 4)\).
- **Test Case 3**: \(n = 2001\), \(m = 2001\). This case is designed to test the code's handling of values larger than 2000, which the code might not handle correctly due to its precomputed vectors.

This test input should help in identifying 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 (Hard 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:

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

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:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $$$(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$$$ satisfy the conditions.

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

void run_case(const size_t ____case) { // rr(____case)
    ii(n, m)

    ll ans = 0;
    /*for(int a=1; a<=n; ++a)
    for(int b=1; b<=m; ++b) {
        int g = gcd(a, b);
        int s = a + b;
        if(g % (s / g) == 0) {
            ++ans;
            assert(a/g <= sqrt(n));
            assert(b/g <= sqrt(m));
            int w = b / (s / g);
            assert(b % g == 0);
            assert(b % w == 0);
            assert(w < g);
            assert(a == b / w * (g - w));
            assert(g % (b / w) == 0);
            assert(1LL * g * g % s == 0);

            int pa = a / g, pb = b / g;
            assert(gcd(pa, pb) == 1);
            assert(pa + pb == s / g);
        }
    }
    */

    ans = 0;
    for(int pa=1; pa*pa<=n; ++pa)
    for(int pb=1; pb*pb<=m; ++pb) if(gcd(pa, pb) == 1) {
        int d = pa + pb;
        for(int g=d; g<=max(n, m); g+=d) {
            ll a = pa * g, b = pb * g;
            if(a > n) continue ;
            if(b > m) continue ;
            if(1LL * b * g % (a + b) == 0) {
                ++ans;
            }
        }
    }
    cout << ans << endl;

}

void test() {
    int n = 2e6;
    ll cnt = 0;
    for(int d=1; d<=n; ++d)
    for(int g=d; g<=n; g+=d) if(1LL*g*d <= n*2) {
        cnt += g*d;
    }
    rr(cnt)
}

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

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

    //test();

    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 the code to fail. The code is designed to count the number of ordered pairs \((a, b)\) that satisfy specific conditions. Here are the steps to identify such a test input:

1. **Edge Cases:**
   - Small values of \(n\) and \(m\).
   - Large values of \(n\) and \(m\) within the limits.
   - \(n\) and \(m\) where one is significantly larger than the other.
   - \(n\) and \(m\) where both are equal.

2. **Critical Conditions:**
   - Pairs where \(a\) and \(b\) are both small.
   - Pairs where \(a\) and \(b\) are both large.
   - Pairs where \(a\) and \(b\) have a large greatest common divisor (GCD).
   - Pairs where \(a\) and \(b\) have a small GCD.

3. **Test Input:**
   - We will choose values that are likely to expose the bug in the code. For example, values where the GCD of \(a\) and \(b\) is large and \(a + b\) is a multiple of \(b \cdot \gcd(a, b)\).

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

```plaintext
5
1 1
2 2
5 5
10 10
100 100
```

This input includes a variety of small values for \(n\) and \(m\), including cases where \(n\) and \(m\) are equal, to ensure that the code is thoroughly tested for potential bugs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Hard 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:

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

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:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $$$(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$$$ satisfy the conditions.

Buggy Code:
#include "bits/stdc++.h"
#ifdef mlocal
#include "./Competitive-Code-Library/snippets/prettyprint.h"
#endif
using namespace std;
#define for_(i, s, e) for (int i = s; i < (int) e; i++)
#define for__(i, s, e) for (ll i = s; i < e; i++)
typedef long long ll;
typedef vector<int> vi;
typedef array<int, 2> ii;
#define endl '\n'

void solve() {
    int n, m; cin >> n >> m;
    int ans = 0;

    for_(g, 1, m + 1) {
        for (int s = g; s <= n + m; s += g) {
            int bfac = s / g;
            ll mul_of = ((ll) bfac) * (g / __gcd(g, bfac));
            if (mul_of > min(m, s)) continue;

            for (int b = mul_of; b <= m; b += mul_of) {
                if (b >= s) break;

                int a = s - b;
                if (a <= n and __gcd(a, b) == g) {
                    ans++;
                }
            }
        }
    }

    cout << ans << endl;
}

int main() {
#ifdef mlocal
    freopen("test.in", "r", stdin);
#endif

    ios_base::sync_with_stdio(false);
    cin.tie(0);

    int t; cin >> t;
    while (t--) {
        solve();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Hard 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:

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

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:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $$$(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$$$ satisfy the conditions.

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

using ll = long long;
const int N = 4e6 + 1;
int m;
int minfac[N];
int prime[N];
/*
    a + b = k;
    k | b * (b, k) b要有k的“一半”， 素因数分解，2 * pbi >= pki. b <= m , b >= k - n, b <= k - 1;
*/
void solve(){
    int n, m, ans = 0;
    std::cin >> n >> m;
    for(int k = 2; k <= n + m; k++) {
        int t = k, x = 1;
        while(t > 1) {
            int w = minfac[t], cnt = 1;
            while(t % w == 0) {
                if(cnt) x *= w;
                t /= w;
                cnt ^= 1;
            }
        }
		int u = std::min(k - 1, m), d = std::max(k - n, 0);
		if(u > d) ans += u / x - (d - 1) / x;
    }
    std::cout << ans << '\n';
}
int main() {
    for(int i = 2; i < N; i++) {
        if(minfac[i] == 0) {
            minfac[i] = i;
            prime[++m] = i;
        }
        for(int j = 1; j <= m; j++) {
            if(i * prime[j] >= N) break;
            minfac[i * prime[j]] = prime[j];
            if(i % prime[j] == 0) break;
        }
    }
    int t = 1;
    std::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 expose the bug in the given code, we need to focus on edge cases and scenarios where the algorithm might fail. The code involves calculating the number of valid pairs \((a, b)\) that satisfy certain conditions. The primary conditions are:

1. \(1 \le a \le n\)
2. \(1 \le b \le m\)
3. \(b \cdot \gcd(a, b)\) is a multiple of \(a + b\)

The buggy code might fail in scenarios where the calculation of \(x\) (the product of prime factors of \(k\) with odd exponents) is incorrect or where the range for \(b\) is not handled properly.

Let's consider the following edge cases:

1. **Small values of \(n\) and \(m\)**: These can help verify basic functionality.
2. **Values of \(n\) and \(m\) where \(a + b\) is a prime number**: This can test the gcd condition more rigorously.
3. **Values of \(n\) and \(m\) where \(a + b\) is a composite number with specific prime factors**: This can test the gcd condition and the calculation of \(x\).

Given the constraints, a small-scale test input that can expose potential issues is:

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

This input includes:
- A simple case with \(n = 1\) and \(m = 1\) where no pairs satisfy the conditions.
- A case with \(n = 2\) and \(m = 3\) where one pair \((2, 2)\) satisfies the conditions.
- A case with \(n = 3\) and \(m = 5\) where one pair \((3, 3)\) satisfies the conditions.
- A more complex case with \(n = 10\) and \(m = 8\) where multiple pairs satisfy the conditions.

These cases should help identify any issues in the algorithm's handling of different ranges and conditions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Hard 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:

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

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:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $$$(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$$$ satisfy the conditions.

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

using namespace std;
using ll = int;
using ull = unsigned long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;

#define fir first
#define sec second
#define mpr make_pair

#define vv vector
#define eb emplace_back

#define rep(i, l, r) for (ll i = l; i <= r; ++i)
#define per(i, r, l) for (ll i = r; i >= l; --i)

template<typename T> int Max(T& x, T y) { return y > x ? (x = y, 2) : x == y; }
template<typename T> int Min(T& x, T y) { return y < x ? (x = y, 2) : x == y; }

template<typename T> void in(T& x) {
	x = 0; char ch = getchar(); bool f = false;
	while (!isdigit(ch)) f = f || ch == '-', ch = getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	if (f) x = -x;
}

template<typename T> void out(T x, char ch = '\n') {
	static int tp, sk[70];
	if (x < 0) putchar('-'), x = -x;
	do { sk[++tp] = x % 10, x /= 10; } while (x);
	while (tp) putchar('0' + sk[tp--]);
	putchar(ch);
}

ll n,m;
vv<ll> d[2000010];
ll mindiv[2000010];
int cntp,p[2000010];
vv<ll> b[2000010], ib[2000010];
ll cnt[2000010];

void solve() {
	in(n),in(m);
	ll ans=0;
	rep(g,1,m){
		rep(k2,1,m/g){
			// ++ans;
			// ll v=g/__gcd(g,k2);
			ll t2=k2;
			// ll v=g;
			// while(t2!=1){
				// v/=t2,t2=__gcd(v,k2);
			// }
			ll v=g;
			while(t2!=1){
				ll k=mindiv[t2];
				while(t2%k==0)t2/=k;
				while(v%k==0)v/=k;
			}
				// rep(j,1,v)if(v%j==0){
				// 	// ll t1=t*j;
				// 	ll t1=j;
				// 	if(t1>=k2+1&&t1<=k2+n/g){
				// 		// if(__gcd(k2,t1-k2)!=1)
				// 		// cout << g << " " << k2 << " " << t1-k2 << "\n";
				// 		// ++ans;
				// 	}
				// }
						ll L=k2+1,R=k2+n/g;
						Max(L,1),Min(R,v);
						if(L<=R){
							if(R-L<=20){rep(j,L,R)if(v%j==0)++ans;}
							else b[L].eb(v),ib[R+1].eb(v);
						}
			// auto it1=lower_bound(d[v].begin(),d[v].end(),L);
			// auto it2=upper_bound(d[v].begin(),d[v].end(),R);
			// --it2;
			// for(auto j:d[v])if(j>=L&&j<=R)++ans;
			// ans+=max(0ll,it2-it1);
		}
	}
	rep(i,1,m){
		for(auto j:b[i])++cnt[j];
		for(auto j:ib[i])--cnt[j];
		for(int j=i;j<=m;j+=i)ans+=cnt[j];
	}
	// rep(i,1,m+1)b[i].clear(),ib[i].clear();
	rep(i,1,m+1)vv<int>().swap(b[i]),vv<int>().swap(ib[i]);
	out(ans);
}

int main() {
	int T;in(T);
	const int M=2000000;
	// rep(i,1,2000000)for(int j=i;j<=2000000;j+=i)d[j].eb(i);
	rep(i,2,2000000){
		if(mindiv[i]==0)mindiv[i]=i,p[++cntp]=i;
		rep(j,1,cntp){
			if(i*p[j]>M)break;
			mindiv[i*p[j]]=p[j];
			if(i%p[j]==0)break;
		}
	}
	// rep(i,1,cntp)out(p[i]);
	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 expose the bug in the given code, we need to focus on edge cases and scenarios that might cause the code to fail. Here are some considerations:

1. **Small Values of \( n \) and \( m \)**: Small values can help us quickly identify if the basic logic is correct.
2. **Equal Values of \( n \) and \( m \)**: This can simplify the problem and help us verify if the code handles symmetry correctly.
3. **Large Values of \( n \) and \( m \)**: Large values can stress-test the code and reveal performance issues or logical errors.
4. **Special Values**: Values that might cause the gcd or other arithmetic operations to behave unexpectedly.

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

```plaintext
5
1 1
2 3
3 5
10 10
100 100
```

This input includes:
- The smallest possible values (1, 1).
- A small pair (2, 3) where the gcd is non-trivial.
- A pair (3, 5) where the gcd is 1.
- A pair (10, 10) where \( n \) and \( m \) are equal.
- A pair (100, 100) where \( n \) and \( m \) are equal and larger, to test performance and logic.

This should help in identifying any issues in 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 (Hard 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:

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

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:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $$$(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$$$ satisfy the conditions.

Buggy Code:
#include<bits/stdc++.h>
#define For(i,l,r) for(int i=l,i##_e=r;i<=i##_e;++i)
#define rFor(i,r,l) for(int i=r,i##_e=l;i>=i##_e;--i)
#define y0 y_zero
#define y1 y_one
#define all(a) a.begin(),a.end()
using namespace std;
using u32=unsigned;
using i64=long long;
using ll=long long;
using u64=unsigned long long;
using ull=unsigned long long;
#if __SIZEOF_POINTER__==8
using i128=__int128;
using u128=__uint128_t;
#endif
using pii=array<int,2>;
using pll=array<ll,2>;
#if __cplusplus>=201402L
void cmin(auto &a,auto b){a=min(a,b);}
void cmax(auto &a,auto b){a=max(a,b);}
#endif
#define mtc() int T; cin>>T; while(T--) work();

#define with_buffer

#if defined(with_buffer) && defined(EOF) && __cplusplus>=201103L

class in_buffer_t {
	static constexpr int len=1<<16;
	char buf[len],*p1=buf; const char *p2=buf;
	FILE *in;
   public:
	in_buffer_t():in(nullptr){}
	in_buffer_t(FILE *in):in(in){}
	char getc() {
		if(p1!=p2 || (p2=buf+fread(p1=buf,1,len,in))!=p1) return *(p1++);
		return EOF;
	}
};
class out_buffer_t {
	static constexpr int len=1<<16;
	char buf[len],*p1=buf; const char *const p2=buf+len;
	FILE *out;
   public:
	out_buffer_t():out(nullptr){}
	out_buffer_t(FILE *out):out(out){}
	void putc(char x) {
		if(p1!=p2) { *(p1++)=x; return; }
		fwrite(p1=buf,len,1,out),*(p1++)=x;
	}
	~out_buffer_t() { fwrite(buf,p1-buf,1,out); }
};
in_buffer_t stdin_buf(stdin);
out_buffer_t stdout_buf(stdout);

#define cin my_cin
#define cout my_cout
class my_istream {
	using istream=my_istream;
	in_buffer_t *const in;
	template<class T> void read_uint(T &x)const{
		char c=0; x=0;
		for(;~c && !isdigit(c);) c=in->getc();
		if(!~c) return;
		for(;isdigit(c);c=in->getc()) x=x*10+(c&15);
	}
	template<class T> void read_sint(T &x)const{
		char m=0,c=0; x=0;
		for(;~c && !isdigit(c);) m|=(c=in->getc())==45;
		if(!~c) return;
		for(;isdigit(c);c=in->getc()) x=x*10+(c&15);
		if(m) x=-x;
	}
   public:
	my_istream():in(nullptr){}
	my_istream(in_buffer_t *const in):in(in){}
#define doit(T,F)\
	friend istream operator>>(const istream in,T &x) {\
		in.F(x); return in;\
	}
	doit(unsigned,read_uint)
	doit(int,read_sint)
	doit(unsigned long,read_uint)
	doit(long,read_sint)
	doit(unsigned long long,read_uint)
	doit(long long,read_sint)
#if __SIZEOF_POINTER__==8
	doit(__uint128_t,read_uint)
	doit(__int128,read_sint)
#endif
#undef doit
	char get()const{return in->getc();}
	friend istream operator>>(const istream in,char &x) {
		for(x=32;isspace(x);x=in.in->getc()); return in;
	}
	friend istream operator>>(const istream in,char *s) {
		char c=32;
		for(;isspace(c);c=in.in->getc());
		for(;~c && !isspace(c);c=in.in->getc()) *(s++)=c; *s=0;
		return in;
	}
	friend istream operator>>(const istream in,string &s) {
		char c=32; s.clear();
		for(;isspace(c);c=in.in->getc());
		for(;~c && !isspace(c);c=in.in->getc()) s.push_back(c);
		return in;
	}
	istream getline(char *s) {
		for(char c=in->getc();~c && c!=10 && c!=13;c=in->getc()) *(s++)=c; *s=0;
		return *this;
	}
	istream getline(string &s) {
		s.clear(); for(char c=in->getc();~c && c!=10 && c!=13;c=in->getc()) s.push_back(c);
		return *this;
	}
};
class my_ostream {
	using ostream=my_ostream;
	out_buffer_t *const out;
	template<class T> void write_uint(const T x)const{
		if(x>9) write_uint(x/10);
		out->putc((x%10)|48);
	}
	template<class T> void write_sint(T x)const{
		if(x<0) out->putc(45),x=-x;
		if(x>9) write_sint(x/10);
		out->putc((x%10)|48);
	}
   public:
	my_ostream():out(nullptr){}
	my_ostream(out_buffer_t *const out):out(out){}
#define doit(T,F)\
	friend ostream operator<<(const ostream out,const T x) {\
		out.F(x); return out;\
	}
	doit(unsigned,write_uint)
	doit(int,write_sint)
	doit(unsigned long,write_uint)
	doit(long,write_sint)
	doit(unsigned long long,write_uint)
	doit(long long,write_sint)
#if __SIZEOF_POINTER__==8
	doit(__uint128_t,write_uint)
	doit(__int128,write_sint)
#endif
#undef doit
	void put(const char c)const{out->putc(c);}
	friend ostream operator<<(const ostream out,const char x) {
		out.out->putc(x); return out;
	}
	friend ostream operator<<(const ostream out,const char *s) {
		for(;*s;) out.out->putc(*(s++)); return out;
	}
	friend ostream operator<<(const ostream out,const string s) {
		for(auto x:s) out.out->putc(x); return out;
	}
};
my_istream cin(&stdin_buf);
my_ostream cout(&stdout_buf);
#define endl '\n'
#endif
#if !defined(LOCAL) && __cplusplus>=201402L
struct empty_cerr_t {
	empty_cerr_t& operator<<(auto x) {
		return *this;
	}
}empty_cerr;
#define cerr empty_cerr
#endif
#define vect basic_string
template<int n,int B>
class fgcd_t{
	bitset<n+1> b;
	int p[n+1],t; array<int,3> a[n+1]; int c[n+1];
	int s[B+1][B+1];
   public:
	fgcd_t() {
		For(i,2,n) {
			if(!b[i]) p[++t]=i,a[i][0]=i;
			For(j,1,t) {
				int x=i*p[j];
				if(x>n) break;
				b[x]=1;
				a[x]=a[i]; c[x]=c[i];
				if(a[x][c[x]]*p[j]>B) a[x][++c[x]]=1;
				a[x][c[x]]*=p[j];
				if(i%p[j]==0) break;
			}
		}
		For(i,0,B) For(j,0,i) {
			if(!j) s[i][j]=i;
			else s[i][j]=s[j][i%j];
		}
	}
	int gcd(int x,int y) {
		if(x==1 || y==1) return 1;
		int ans=1; auto ay=a[y];
		For(j,0,c[y]) {
			if(ay[j]>B) {
				if(x%ay[j]==0) ans*=ay[j];
			} else {
				int g=s[ay[j]][x%ay[j]];
				if(g!=1) ans*=g,x/=g;
			}
		}
		return ans;
	}
}; fgcd_t<2000000,2000> g;
const int N=2e6+10;
int mu[N],f[N],cnt[N];
void init(int n) {
	mu[1]=1; For(i,2,n) mu[i]=0;
	For(i,1,n) For(j,2,n/i) mu[i*j]-=mu[i];
}
int *fr[N],len[N];
void work() {
	int n,m; cin>>n>>m;
	For(i,1,m) For(j,1,m/i) cnt[i*j]++;
	For(i,1,m) fr[i]=new int[cnt[i]],len[i]=cnt[i];
	For(i,1,m) For(j,1,m/i) fr[i*j][--cnt[i*j]]=i;
	// init(max(n,m));
	ll ans=0;
	int tot=0;
	For(x,1,m) {
		For(d,1,m/x) {
			int lim=n/d+x; if(x>lim) continue;
			while(len[d] && fr[d][len[d]-1]<=x) --len[d];
			rFor(id,len[d]-1,0) {
				int s=fr[d][id]; if(s>lim) break;
				ans+=g.gcd(x,s)==1;
			}
		}
	}
	// cerr<<tot<<endl;
	cout<<ans<<endl;
}
int main() {
#ifdef LOCAL
	freopen(".in","r",stdin);
	// freopen(".out","w",stdout);
	// freopen(".debug","w",stderr);
#endif
#ifndef with_buffer
	ios::sync_with_stdio(0); cin.tie(0);
#endif
	mtc();
}

**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 for the given algorithm problem, we need to focus on edge cases and scenarios that can expose potential bugs in the code. Here are the steps to think through this:

1. **Small and Valid Input**: Ensure the test input is small enough to be manageable and valid according to the problem constraints.
2. **Edge Cases**: Consider edge cases such as the smallest and largest possible values for \( n \) and \( m \), and specific values that might cause issues in the gcd calculations.
3. **Special Conditions**: Look for pairs \((a, b)\) that might satisfy the given conditions in a non-trivial way, such as when \( a \) and \( b \) have specific relationships.

Given these considerations, let's construct a small but challenging test input:

1. **Single Test Case with Small Values**: Start with a small \( n \) and \( m \) to ensure the input is manageable.
2. **Include a Pair that Satisfies the Condition**: Include a pair \((a, b)\) that satisfies \( b \cdot \gcd(a, b) \) being a multiple of \( a + b \).

Here is the test input:

```plaintext
1
2 3
```

This input is small and valid, and it includes a pair \((2, 2)\) that satisfies the condition \( b \cdot \gcd(a, b) \) being a multiple of \( a + b \). This should help expose any issues in the gcd calculations or the logic used to count 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 (Hard 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:

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

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:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $$$(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$$$ satisfy the conditions.

Buggy Code:
#include<bits/stdc++.h>
#define ll long long
#define pii pair<int,int>
#define fi first
#define se second
using namespace std;
const int N=2e6+10;
const ll inf=2e12;
int t,n,m;
inline int gcd(int a,int b){return b==0?a:gcd(b,a%b);}
int main()
{
    // freopen("1.in","r",stdin);
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        ll ans=0;
        for(int i=1;i<=1500;i++)
            for(int j=1;j<=1500;j++)
            if(gcd(i,j)==1)
            {
                int x=min(n/i,m/j);
                ans+=x/(i+j);
            }
        printf("%lld\n",ans);
    }
    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
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
```

### Explanation:
1. **Edge Case: (1, 1)**
   - This is the smallest possible input, and no pairs satisfy the conditions. This helps to verify that the code handles the smallest input correctly.

2. **Small Input with a Single Valid Pair: (2, 3)**
   - The pair (2, 3) is the only valid pair here. This helps to check if the code can correctly identify and count a single valid pair.

3. **Small Input with a Single Valid Pair: (3, 5)**
   - The pair (3, 5) is the only valid pair here. This is another small input to verify the code's ability to handle a single valid pair.

4. **Medium Input with Multiple Valid Pairs: (10, 8)**
   - This input has multiple valid pairs, which helps to test the code's ability to handle a moderate number of pairs and correctly sum the results.

5. **Large Input with a Large Number of Valid Pairs: (100, 1233)**
   - This input has a large number of valid pairs, which helps to test the code's performance and correctness with larger inputs.

6. **Very Large Input with a Very Large Number of Valid Pairs: (1000000, 1145141)**
   - This input is very large and has a very large number of valid pairs. It helps to test the code's performance and correctness with the maximum input size.

These test cases are designed to cover a range of input sizes and scenarios, from the smallest to the largest, and to include both single and multiple valid pairs. This should help to expose any bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Hard 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:

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

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:
0
1
1
6
423
5933961

Note: In the first test case, no pair satisfies the conditions.

In the fourth test case, $$$(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$$$ satisfy the conditions.

Buggy Code:
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define int long long
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>

using namespace std;
using namespace __gnu_pbds;

main() {
	ios_base::sync_with_stdio(0);
	int t;
	cin>>t;
	while(t--){
		int n,m;
		cin>>n>>m;
		int ans=0;
		for(int i=1;i*i<=n && i*i<=m;++i){
			for(int j=1;(i+j)*i<=n && (i+j)*j<=m;++j){
				if(__gcd(i,j)==1){
					ans+=min(n/((i+j)*i),m/((i+j)*j));
				}
			}
		}
		cout<<ans<<"\n";
	}
}

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

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

Think step by step.

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

1. **Understanding the Bug**: The buggy code iterates over pairs \((i, j)\) where \(i\) and \(j\) are coprime (i.e., \(\gcd(i, j) = 1\)). It then counts the number of valid pairs \((a, b)\) where \(a = (i + j) \cdot i\) and \(b = (i + j) \cdot j\). The bug might be in the way it handles the limits for \(a\) and \(b\) or in the way it calculates the gcd and the conditions.

2. **Edge Cases**:
   - **Small Values**: Test with small values of \(n\) and \(m\) to ensure the code handles small inputs correctly.
   - **Large Values**: Test with large values of \(n\) and \(m\) to ensure the code handles large inputs within the time and memory limits.
   - **Boundary Conditions**: Test with values where \(n\) and \(m\) are equal or close to each other, and where \(n\) and \(m\) are very different.

3. **Specific Scenarios**:
   - **Single Test Case**: Start with a single test case to simplify debugging.
   - **Multiple Test Cases**: Include multiple test cases to ensure the code handles different inputs correctly.

4. **Challenging Input**:
   - Choose values that are likely to cause the code to miss valid pairs or incorrectly count pairs.
   - For example, values where \(n\) and \(m\) are multiples of each other or where the gcd conditions are particularly complex.

Based on these considerations, here is a challenging test input:

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

This input includes:
- A trivial case (1 1) where no pairs satisfy the conditions.
- A case with small values (2 3) where there is one valid pair.
- A case with slightly larger values (3 5) where there is one valid pair.
- A case with larger values (10 8) where there are multiple valid pairs.

This input should help expose any issues in the buggy code, particularly in how it handles the conditions and the limits for \(a\) and \(b\).