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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
#include<bits/stdc++.h>
#define pb push_back
#define ff first
#define ss second
#define endl '\n'
#define all(a) (a).begin(), (a).end()
#define max3(a,b,c) max(max(a,b),c)
#define max4(a,b,c,d) max(max(a,b),max(c,d))
#define min3(a,b,c) min(min(a,b),c)
#define min4(a,b,c,d) min(min(a,b),min(c,d))
#define gap " "
#define vi vector<int>
#define vll vector<ll>
#define vpll vector<pair<ll,ll>>
#define rev_sort(v) sort(all(v),greater<int>())
#define ll long long
#define ll long long
#define yes() "YES\n"
#define no() "NO\n"
#define vin(a,n) for(int i = 0; i < n; ++i)cin>>a[i];
#define vin1(a,n) for(int i = 1; i <= n; ++i)cin>>a[i];
/*
    ॐ गं गणपतये नमः
	जय श्री राम
	ॐ नमः शिवाय
*/
using namespace std;
int main() {
    int t;
    cin >> t;
while(t--) {
ll n,m;
cin>>n>>m;
if(n==1 && m==1) cout<<1<<endl;
else if(n==m && n!=1 && m!=1){
    cout<<n+1<<endl;

}
else{ll ct=0;
for(int i=1;i<=m;i++){
ct+=((n+i)/(1ll*i*i));

}

cout<<(ct-1)<<endl;
}

            }}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios where the code's logic might fail. The code has a specific logic for handling cases where \( n = m \) and \( n \neq 1 \) and \( m \neq 1 \). It also has a general loop for other cases. We need to create test cases that can highlight the issues in these conditions.

### Step-by-Step Analysis:
1. **Edge Case: \( n = 1 \) and \( m = 1 \)**:
   - This is already handled correctly in the code, so no need to focus on this.

2. **Edge Case: \( n = m \) and \( n \neq 1 \)**:
   - The code outputs \( n + 1 \) for this case. This is a potential bug because it might not correctly count all valid pairs. We should test with different values of \( n \) to see if this logic holds.

3. **General Case: \( n \neq m \)**:
   - The code uses a loop to count valid pairs. We need to ensure that the loop logic is correct. We can test with values of \( n \) and \( m \) where the loop might not handle the conditions correctly.

### Challenging Test Input:
Let's create a test input with the following cases:
1. \( n = 2 \) and \( m = 2 \) (to test the \( n = m \) case).
2. \( n = 3 \) and \( m = 3 \) (to further test the \( n = m \) case).
3. \( n = 4 \) and \( m = 5 \) (to test the general case).

### Test Input:
```plaintext
3
2 2
3 3
4 5
```

### Explanation:
1. **Case 1: \( n = 2 \) and \( m = 2 \)**:
   - The code outputs \( 2 + 1 = 3 \). The correct output should be 3 because the valid pairs are \((1,1)\), \((2,1)\), and \((2,2)\).

2. **Case 2: \( n = 3 \) and \( m = 3 \)**:
   - The code outputs \( 3 + 1 = 4 \). The correct output should be 4 because the valid pairs are \((1,1)\), \((2,1)\), \((2,2)\), and \((3,1)\).

3. **Case 3: \( n = 4 \) and \( m = 5 \)**:
   - The code uses the loop to count valid pairs. We need to ensure that the loop logic correctly counts all valid pairs. The correct output should be 7 because the valid pairs are \((1,1)\), \((2,1)\), \((2,2)\), \((3,1)\), \((4,1)\), \((4,2)\), and \((4,4)\).

This test input should help expose any issues in the code's logic for handling the \( n = m \) case and the general case.


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

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

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

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

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

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

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

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

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

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

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

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

#define int long long
#define pii pair<int,int>
#define vti vector<int>

void solve(stringstream&cerr, stringstream&cout)
{
	int a, b;
	cin>> a>> b;

	int r = 0;

	for(int i=1; i<=b; i++)
	{
		for(int j=-i+i*i; j<=a; j+=i)
		{
			if(j==0)
				continue;
			cerr <<i <<' ' <<j <<endl;
			r+=((i+j)%(i*(i)) == 0);
		}
	}

	cout <<r <<' ';
}

signed main()
{
	stringstream b;
	int t;
	cin>> t;
	while(t--)
	{
		stringstream a;
		solve(a,b);
		cerr <<a.str().substr(0,5000);
	}
	cout <<b.str();
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, let's analyze the code and identify potential issues:

1. **Inner Loop Condition**: The inner loop iterates from `j = -i + i*i` to `j <= a` in steps of `i`. This means `j` starts at `i * (i - 1)` and increments by `i`. This is unusual and might be incorrect.

2. **Condition Check**: The condition `r+=((i+j)%(i*(i)) == 0);` checks if `i + j` is a multiple of `i * i`. This seems to be a bug because it should check if `i + j` is a multiple of `i * gcd(i, j)`.

3. **Edge Cases**: We should consider edge cases where `n` and `m` are small, and where `n` and `m` are large to see if the code handles them correctly.

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

- A small test case to check basic functionality.
- A test case where `n` and `m` are equal to expose any symmetry issues.
- A test case where `n` and `m` are large to test performance and correctness.

Here is the test input:

```plaintext
3
1 1
2 2
100 100
```

This input includes:
- The simplest case (`1 1`) to verify basic functionality.
- A case where `n` and `m` are equal (`2 2`) to check for symmetry.
- A larger case (`100 100`) to test performance and correctness with more pairs.

This should help expose any issues in the provided code.


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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
print('1')
print('3')
print('4')
print('14')
print('153')
print('1643498')

**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 buggy code, we need to focus on edge cases and scenarios that can reveal the incorrect logic in the provided output. The buggy code simply prints a fixed set of outputs for the given examples, which means it will fail if the input deviates from the specific examples provided.

Let's consider the following aspects:
1. **Single Test Case**: Since the buggy code only handles the specific examples, a single test case that is different from the given examples will expose the bug.
2. **Small Scale**: The input should be small enough to be manageable but different enough to show the bug.

Let's choose a small but different input:
- \( n = 5 \)
- \( m = 5 \)

This input is small and different from the provided examples. It should be able to expose the bug in the buggy code.

Here is the test input:
```plaintext
1
5 5
```


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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
import collections,sys,math,functools,operator,itertools,bisect,heapq,decimal,string,time,random
#sys.setrecursionlimit(10**9)
#sys.set_int_max_str_digits(0)
input = sys.stdin.readline
t= int(input())
#alist = list(map(int,input().split()))
#alist = []
#s = input()
#n,m = map(int,input().split())
#for i in range(n):
#    alist.append(list(map(int,input().split())))
# a + b
c = [[] for i in range(2*10**6+10)]
"""
for a in range(1,51):
    for b in range(1,51):
        if (a+b) % (b*math.gcd(a,b)) == 0:
            print(a,b,math.gcd(a,b))
            c[math.gcd(a,b)].append((a,b))
i*(i-1) + ki**2,i

"""
for b in range(1,2*10**6+1):
    a = b*(b-1)
    bb = b**2
    for k in range(2*10**6):
        if a >= 1:
            c[b].append(a)
        a += bb
        if a > 2000000:
            break
#print(c[1][:20])

def solve():
    n,m = map(int,input().split())
    ans = 0
    for i in range(1,m+1):
        x = bisect.bisect_right(c[i],n)
        ans += x
    print(ans)

for i in range(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 that are likely to cause the code to fail. Here are the steps to think through this:

1. **Understand the Bug**:
   - The buggy code seems to be precomputing the values of `a` for each `b` based on the condition \((a + b) \mod (b \cdot \gcd(a, b)) = 0\).
   - It uses a fixed increment \(b^2\) to generate `a` values, which might not cover all valid pairs.

2. **Identify Edge Cases**:
   - Small values of `n` and `m` where the condition is more likely to be met.
   - Large values of `n` and `m` to test the limits of the precomputation.
   - Cases where `n` and `m` are prime numbers or have specific properties that might not be covered by the fixed increment.

3. **Generate Test Input**:
   - We will create a small test case with a combination of small and large values for `n` and `m`.

Given the constraints and the need for a small-scale test input, let's create a test case that includes both small and large values to maximize the probability of exposing the bug.

```plaintext
6
1 1
2 3
3 5
10 8
100 1233
2000000 2000000
```

This test input includes:
- The smallest possible values for `n` and `m` (1, 1).
- Small values where the condition is more likely to be met (2, 3 and 3, 5).
- A medium-sized value (10, 8) to check the precomputation.
- A large value (100, 1233) to test the limits.
- A very large value (2000000, 2000000) to stress the precomputation and the bisect function.

This should help in exposing any issues with the precomputation and the logic in the `solve` function.


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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
import sys
from math import gcd
#input functions
readint = lambda: int(sys.stdin.readline())
readints = lambda: map(int,sys.stdin.readline().split())
readar = lambda: list(map(int,sys.stdin.readline().split()))
flush = lambda: sys.stdout.flush()

"""
ar = list()
br = list()
n = 10

for a in range(1,n+1):
    for b in range(1,n+1):
        c = a + b
        d = b*gcd(a,b)
        if c % d == 0: ar.append((a,b))
        if d % c == 0: br.append((a,b))
print(ar)
print(br)
"""
"""
only factors look to be involved

"""
#def gcd(a,b):
#    while b != 0:
#        (a,b) = (b,a%b)
#        return a

def f(a,b):
    c = a + b
    d = b*gcd(a,b)
    if c % d == 0: return 1
    return 0

for _ in range(readint()):
    ans = 0
    #h = 0
    n,m = readints()
    if n <= m:
        for i in range(1,n+1):
            if i <= m:
                ans += f(i,i)
                #h += 1
                for j in range(2*i,m+1,i):
                    ans += f(i,j)
                    #h += 1
                    if i <= m and j <= n:
                        ans += f(j,i)
                        #h += 1
    else:
        for i in range(1,m+1):
            if i <= n:
                ans += f(i,i)
                #h += 1
                for j in range(2*i,n+1,i):
                    ans += f(j,i)
                    #h += 1
                    if i <= n and j <= m:
                        ans += f(i,j)
                        #h += 1
    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:
To generate a test input that exposes the bug in the provided code, we need to consider the following:

1. **Edge Cases**: Test with the smallest and largest possible values of `n` and `m`.
2. **Boundary Conditions**: Test with values where `n` and `m` are close to each other.
3. **Symmetry**: Test with `n = m` to ensure the code handles symmetric cases correctly.
4. **Special Values**: Test with values that have many factors (e.g., highly composite numbers) to ensure the code handles complex gcd calculations correctly.

Given the constraints and the need for a small-scale test, let's construct a test input with a few test cases that cover these scenarios:

1. `n = 1, m = 1` (smallest possible values)
2. `n = 2, m = 3` (small values to check basic functionality)
3. `n = 3, m = 5` (small values to check more pairs)
4. `n = 10, m = 8` (values where `n` and `m` are close)
5. `n = 12, m = 12` (symmetric case)
6. `n = 100, m = 100` (larger symmetric case)
7. `n = 120, m = 120` (highly composite number)

Here is the test input:

```plaintext
7
1 1
2 3
3 5
10 8
12 12
100 100
120 120
```

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


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

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

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

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

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

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

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

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

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

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

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

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

public class ABC {
    static Scanner sc = new Scanner(System.in);
    public static long solve() {
        long n = sc.nextLong();
        long m = sc.nextLong();
        long answer = 0;

        for(int b = 1; b <= 10000; b++) {
            // binary search for upper bound on b
            long lo = 1;
            long hi = 2000000L;
            lo--;
            while(lo < hi) {
                long mid = lo + (hi - lo + 1)/2;
                if(mid * b * b - b <= n) lo = mid;
                else {
                    hi = mid - 1;
                }
            }
            answer += lo;
        }
        answer--;
        return answer;
    }

    public static void main(String[] args) {
        int t = sc.nextInt();

        for(int i = 1; i <= t; i++) {
            System.out.println(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 the following aspects:

1. **Edge Cases**: Test with small values of \( n \) and \( m \) where the conditions are more likely to be met.
2. **Large Values**: Test with large values of \( n \) and \( m \) to see if the binary search and loop logic handle them correctly.
3. **Specific Values**: Test with specific values that might cause the binary search to fail or produce incorrect results.

Given the constraints and the need for a small-scale test input, let's consider the following scenarios:

1. **Small \( n \) and \( m \)**: These should be easy to handle and can help verify basic correctness.
2. **Medium \( n \) and \( m \)**: These can help identify issues in the loop and binary search logic.
3. **Large \( n \) and \( m \)**: These can help identify performance and correctness issues.

Here is a small-scale test input that includes these scenarios:

```plaintext
5
1 1
2 3
10 10
100 100
2000000 2000000
```

This test input includes:
- A very small test case (1, 1) to check basic functionality.
- A small test case (2, 3) to check for more pairs.
- A medium test case (10, 10) to check for a larger set of pairs.
- A larger test case (100, 100) to check for even more pairs.
- A very large test case (2000000, 2000000) to check for performance and correctness with 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 (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

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

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

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

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

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

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

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

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

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

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

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.StringTokenizer;

public class Code {

    public static void main(String[] args) {
        Code code = new Code();
        code.start();
    }

    public void start() {
        int t = 1;
        this.loadSolve();
        t = ini();
        while (t-- > 0) postSolve();
        out.close();
    }

    public void postSolve() {
        long n = inl();
        long m = inl();
        long cnt = 0;

        for (long i = 1; i <= n; i++) {
            int index = Collections.binarySearch(nodes[(int) i].list, (int) m);
            if (index >= 0) {
                cnt = cnt + index + 1;
            } else {
                cnt = cnt + ((-index) - 1);
            }
        }
        pl(cnt);
    }

    public void loadSolve() {
        long n = 2000000;
        for (int i = 1; i <= n; i++) {
            this.nodes[i] = new Node();
        }
        for (long i = 2; i <= 1414; i++) {
            long mul = i * i;
            long start = mul;
            while (start < n) {
                this.nodes[(int) (start - i)].list.add((int) i);
                start += mul;
            }
        }
    }

    public Node[] nodes = new Node[2000001];

    public class Node {
        public List<Integer> list = new ArrayList<>();

        public Node() {
            this.list.add(1);
        }
    }

    /*
    ==========================================================================================================================
    */
    PrintWriter out = new PrintWriter(System.out);
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st;

    void p(String s) {
        out.print(s);
    }

    void p(int s) {
        out.print(s);
    }

    void p(long s) {
        out.print(s);
    }

    void p(double s) {
        out.print(s);
    }

    void p(char s) {
        out.print(s);
    }

    void pl(String s) {
        out.println(s);
    }

    void pl(int s) {
        out.println(s);
    }

    void pl(long s) {
        out.println(s);
    }

    void pl(double s) {
        out.println(s);
    }

    void pl(char c) {
        out.println(c);
    }

    void pl() {
        out.println();
    }

    void ps(int s) {
        out.print(s + " ");
    }

    void ps(long s) {
        out.print(s + " ");
    }

    void ps(double s) {
        out.print(s + " ");
    }

    void ps(char c) {
        out.print(c + " ");
    }

    void ps(String s) {
        out.print(s + " ");
    }

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

    char[] inc() {
        return ins().toCharArray();
    }

    int ini() {
        return Integer.parseInt(ins());
    }

    long inl() {
        return Long.parseLong(ins());
    }

    double ind() {
        return Double.parseDouble(ins());
    }

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

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios that are likely to cause the algorithm to fail. The code seems to be using a precomputed list of values to speed up the calculation, but there are a few potential issues:

1. **Precomputation Errors**: The precomputation in `loadSolve` might not correctly handle all cases, especially for larger values of `i` and `mul`.
2. **Binary Search Issues**: The binary search in `postSolve` might not correctly handle cases where the list is empty or does not contain the value `m`.

Given these potential issues, let's consider the following test cases:

### Test Case 1: Small Values
- **n = 1, m = 1**: This is a simple case where only one pair (1,1) should satisfy the condition.
- **n = 2, m = 3**: This is a slightly more complex case with multiple pairs.

### Test Case 2: Edge Case for Precomputation
- **n = 1414, m = 1414**: This is a case where `i` and `mul` are at the upper limit of the precomputation range.

### Test Case 3: Large Values
- **n = 2000000, m = 2000000**: This is a large case to test the performance and correctness of the precomputation.

### Test Case 4: Empty List Scenario
- **n = 2000001, m = 1**: This case should expose any issues with handling empty lists.

### Test Case 5: Large m with Small n
- **n = 1, m = 2000000**: This case should expose any issues with handling large values of `m` when `n` is small.

### Test Input
```plaintext
5
1 1
2 3
1414 1414
2000000 2000000
2000001 1
```

This test input includes a mix of small, large, and edge cases to thoroughly test the algorithm and expose any bugs in the precomputation and binary search logic.


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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
#include <iostream>

using namespace std;

int main()
{
    int t ;
    cin >> t ;
    while (t--)
    {
        int a , b ;
        cin >> a >>b ;
        b = min (a,b);
        int ans = 0 ;
        for (int i = 2; i <= b ; i++)
        {
            int j = 2;
            while (j*i <= a)
            {
                if ((j+1) % i == 0)
                    ans++;

                j++ ;
            }
        }
        ans = ans + a + 1;
        if (a==1)
            ans-- ;
        cout << ans << endl ;
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases and scenarios where the faulty logic in the code is most likely to fail. The code has a loop that iterates from 2 to `b` and another nested loop that iterates from 2 to `a`. The condition `if ((j+1) % i == 0)` is used to count valid pairs, but this logic is incorrect.

### Key Points to Consider:
1. **Edge Cases:**
   - When `a` or `b` is 1.
   - When `a` and `b` are small values.
   - When `a` and `b` are equal.

2. **Faulty Logic:**
   - The code incorrectly counts pairs by checking `if ((j+1) % i == 0)`, which is not the correct condition for the problem.

### Test Input:
We will create a test input with multiple test cases to cover different scenarios:

1. **Case 1:** `a = 1, b = 1` (Edge case where both `a` and `b` are 1)
2. **Case 2:** `a = 2, b = 2` (Small values where `a` and `b` are equal)
3. **Case 3:** `a = 3, b = 5` (Small values where `a` and `b` are different)
4. **Case 4:** `a = 10, b = 8` ( Larger values where `a` and `b` are different)

### Test Input:
```plaintext
4
1 1
2 2
3 5
10 8
```

### Explanation:
- **Case 1:** `1 1` should output `1` (only (1,1) is valid).
- **Case 2:** `2 2` should output `3` (valid pairs are (1,1), (2,1), (2,2)).
- **Case 3:** `3 5` should output `4` (valid pairs are (1,1), (2,1), (3,1), (3,3)).
- **Case 4:** `10 8` should output `14` (valid pairs are (1,1), (2,1), (2,2), (3,1), (4,1), (5,1), (6,1), (6,2), (6,3), (7,1), (8,1), (9,1), (10,1), (10,2)).

This test input should help expose the bug 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 (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

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

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

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

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

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

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

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

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

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

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

Buggy Code:
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define INT __int128
#define ld long double
#define pii pair<int,int>
#define pll pair<ll,ll>
#define fi first
#define se second
#define pb emplace_back
#define pc __builtin_popcount
#define pcl __builtin_popcountll
#ifndef ONLINE_JUDGE
#define debug(x) cerr<<#x<<": "<<x<<endl;
#else
#define debug(x)
#endif
#define lb(x) ((x)&-(x))
using namespace std;
const ll P=1e9+7;
ll qpow(ll x,ll y=P-2){ll z=1;while(y){if(y&1)z=z*x%P;x=x*x%P,y>>=1;}return z;}
ll n,m;
ll solve(ll x,ll r){
	return r/x;
}
void work(){
	ll b;
	cin>>n>>m;ll ans=0;
	for(b=1;b<=n;b++){
		ans+=solve(b,n/b+1)-solve(b,1);
	}
	cout<<ans<<'\n';
}
int main(){
	cin.tie(nullptr)->sync_with_stdio(false);
	int _;cin>>_;while(_--) work();
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause the code to fail. The buggy code appears to have issues with the logic for counting valid pairs \((a, b)\) that satisfy the given conditions.

### Analysis of the Buggy Code:
1. **Function `solve(ll x, ll r)`**: This function returns `r / x`. It seems to be used to count the number of valid pairs for a given \(b\).
2. **Loop in `work()`**: The loop iterates over all possible values of \(b\) from 1 to \(n\). For each \(b\), it calculates the number of valid pairs using the `solve` function.

### Potential Issues:
- The `solve` function might not correctly count the valid pairs for each \(b\).
- The loop might be missing some valid pairs or counting some pairs incorrectly.

### Challenging Test Input:
To expose the bug, we should test with values of \(n\) and \(m\) that have specific properties:
- **Small values**: To ensure the test case is small and manageable.
- **Edge cases**: Values of \(n\) and \(m\) that are close to each other or have specific properties (e.g., \(n = m\), \(n\) is a multiple of \(m\), etc.).

Let's create a test input with \(n = 6\) and \(m = 6\). This is a small enough input to be manageable and has the property that \(n\) and \(m\) are equal, which might expose issues in the counting logic.

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

This test input will check if the code correctly counts the number of valid pairs for \(n = 6\) and \(m = 6\). If the code is buggy, it might not count all valid pairs correctly, and the output will differ from the expected result.


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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
def sol(n, m):

    if m==1:
        return n

    ans = 0

    if m==2:
        return n + int((n+2)/4)

    for b in range(2,m):
        if int((n+b)/(b*b))==0 :
            return n+ans

        ans += int((n+b)/(b*b))

    return ans+n

t = int(input())

for _ in range(t):
    inp = input().split(" ")

    inp = list(map(int, inp))

    print(sol(inp[0], inp[1]))

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's consider the following aspects:

1. **Edge Cases**:
   - When `m` is 1 or 2, the code seems to handle these cases separately and correctly.
   - For `m` greater than 2, the code uses a loop that might prematurely return a result if the condition `int((n+b)/(b*b))==0` is met. This could lead to incorrect results if the loop is terminated too early.

2. **Key Conditions**:
   - The condition `a + b` is a multiple of `b * gcd(a, b)` needs to be satisfied.
   - The loop in the buggy code might not correctly count all valid pairs.

3. **Small-Scale Test Input**:
   - We need a small-scale input that can be processed quickly but still exposes the bug.
   - A good candidate would be a test case where `m` is greater than 2 and the loop might terminate early, leading to an incorrect result.

Let's consider the following test case:

- `n = 10` and `m = 4`.

For this test case:
- The valid pairs should be: (1,1), (2,1), (2,2), (3,1), (4,1), (4,2), (5,1), (6,1), (6,2), (6,3), (7,1), (8,1), (8,2), (9,1), (10,1), (10,2).
- The expected number of valid pairs is 16.

However, the buggy code might return an incorrect result because it might prematurely terminate the loop when `b = 3` and `int((10+3)/(3*3)) == 0`.

Here is the test input:

```plaintext
1
10 4
```

This test input should expose the bug in the buggy code.


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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
#include <iostream>

int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(nullptr), std::cout.tie(nullptr);
	int _; std::cin >> _; while (_--) {
		int n, m; std::cin >> n >> m;
		long long sum = 0;
		for (int i = 1; i*i <= i+n; i++) {
			sum += (n+i)/(1LL*i*i);
		}

		std::cout << --sum << '\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 or specific scenarios that might cause the code to fail. Let's analyze the code and the problem statement:

1. **Understanding the Buggy Code:**
   - The code iterates over `i` from 1 to the square root of `i + n`.
   - It calculates `sum` using the formula `(n + i) / (1LL * i * i)`.
   - It decrements `sum` by 1 before printing it.

2. **Identifying the Bug:**
   - The loop condition `for (int i = 1; i * i <= i + n; i++)` is incorrect. It should be `for (int i = 1; i * i <= n; i++)`.
   - The formula used to calculate `sum` is also incorrect. It should be `sum += m / i` for valid pairs.

3. **Challenging Test Case:**
   - We need a test case where the loop condition and the formula calculation will differ significantly from the correct solution.
   - A small-scale test case with `n` and `m` values that expose the incorrect loop condition and formula is ideal.

Let's choose `n = 10` and `m = 10` as a test case. This will ensure that the loop condition and the formula calculation are different from the correct solution.

```plaintext
1
10 10
```

This test case will:
- Test the incorrect loop condition `i * i <= i + n`.
- Test the incorrect formula `(n + i) / (1LL * i * i)`.

The correct output for this test case should be 14, but the buggy code will produce a different result due to the identified issues.


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

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

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

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

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

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

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

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

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

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

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

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

using namespace std;

const int N = 1e6+10, M = 2 * N + 10;
const int MOD = 1e9+70;

#define PII pair<int,int>
#define LL long long

LL qmi(LL a, LL b, LL p){
    LL res = 1;
    while(b){
        if(b &1) res = res * a % p;
        a = a * a % p;
        b >>= 1;
    }
    return res;
}

int gcd(int a, int b){
    return b? gcd(b, a % b) : a;
}
LL n,m,k;
LL a[N];
void solve(){
    cin >> n >> m;
    LL ans = 0;
    for(LL i = 1; i <= n; ++i) ans += (n/i + 1) / i;
    --ans;
    cout << ans;

    cout <<"\n";
}

int main() {
    int T = 1;
    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:
To generate a challenging test input that exposes the bug in the given code, let's analyze the problem and the provided solution.

### Problem Analysis
The problem requires calculating the number of ordered pairs \((a, b)\) such that:
- \(1 \le a \le n\)
- \(1 \le b \le m\)
- \(a + b\) is a multiple of \(b \cdot \gcd(a, b)\)

### Buggy Code Analysis
The provided code has a logical error in the way it calculates the number of valid pairs. The line:
```cpp
ans += (n/i + 1) / i;
```
is incorrect. This line is trying to count the number of valid pairs, but it does not correctly account for the condition \(a + b\) being a multiple of \(b \cdot \gcd(a, b)\).

### Challenging Test Case
To expose this bug, we need a test case where the condition \(a + b\) is a multiple of \(b \cdot \gcd(a, b)\) is not satisfied by the current logic. A good approach is to use small values for \(n\) and \(m\) where the valid pairs are few and can be easily verified.

### Test Case
Let's use \(n = 4\) and \(m = 4\). We can manually verify the valid pairs:
- \((1, 1)\): \(1 + 1 = 2\) is a multiple of \(1 \cdot 1 = 1\)
- \((2, 1)\): \(2 + 1 = 3\) is a multiple of \(1 \cdot 1 = 1\)
- \((2, 2)\): \(2 + 2 = 4\) is a multiple of \(2 \cdot 2 = 4\)
- \((3, 1)\): \(3 + 1 = 4\) is a multiple of \(1 \cdot 1 = 1\)
- \((4, 1)\): \(4 + 1 = 5\) is a multiple of \(1 \cdot 1 = 1\)
- \((4, 2)\): \(4 + 2 = 6\) is a multiple of \(2 \cdot 2 = 4\) (not valid)
- \((4, 4)\): \(4 + 4 = 8\) is a multiple of \(4 \cdot 4 = 16\) (not valid)

The valid pairs are \((1, 1)\), \((2, 1)\), \((2, 2)\), \((3, 1)\), and \((4, 1)\), which gives us 5 valid pairs.

### Test Input
```plaintext
1
4 4
```

This test input should expose the bug in the provided code, as the incorrect logic will likely produce a different result than the expected 5 valid pairs.


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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
#include<iostream>
#include<algorithm>
#include<string>
#include<map>
#include<math.h>
#include<queue>
#include<vector>
#include<stack>
#define N 2000010
#define int long long
using namespace std;
const int mod = 1e9 + 7;

#define endl '\n'
struct node {
    int a;
    int b;
    int c;
    int d;
};
int qmi(int a, int b, int m) {
    int res = 1;
    while (b) {
        if (b & 1) {
            res = (res * a) % m;

        }
        a = (a * a) % m;
        b = b >> 1;
    }
    return res;
}
bool cmp(struct node a, struct node b) {
    return a.c < b.c;
}

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

    for (int i = 2; i <= m; i++) {
        int x = (i * i);

        int l = 0, r = m, mid;
        while (l <= r) {
            mid = (l + r) / 2;
            if (mid * x-i <= n) {
                l = mid + 1;
            }
            else {
                r = mid - 1;
            }
        }
        ans += r ;
    }
    cout << ans << endl;
}
signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int _;
    cin >> _;
    while (_--)
    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
```

### Explanation:
1. **Test Case 1: (1, 1)**
   - This is the simplest case. Only the pair (1, 1) should be valid.
   - Expected output: 1

2. **Test Case 2: (2, 3)**
   - Pairs: (1, 1), (2, 1), (2, 2), (2, 3)
   - Expected output: 3

3. **Test Case 3: (3, 5)**
   - Pairs: (1, 1), (2, 1), (2, 2), (3, 1), (3, 3)
   - Expected output: 4

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

5. **Test Case 5: (100, 1233)**
   - This case is more complex and should test the algorithm's performance and correctness with larger values.
   - Expected output: 153

6. **Test Case 6: (1000000, 1145141)**
   - This is a very large test case to check the algorithm's performance and correctness under extreme conditions.
   - Expected output: 1643498

These test cases cover a range of scenarios, including small and large values, and should effectively expose any bugs in the buggy code.


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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
#pragma GCC optimize("03")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

/*```````````````````````````****Defined macros*****``````````````````````````*/
const int mod = 1e9 + 7;
#define ll long long
#define endl "\n"
#define ull unsigned long long
#define pb push_back
#define yes cout << "YES" << endl;
#define no cout << "NO" << endl;
#define set_bits __builtin_popcountll
#define all(x) x.begin(), x.end()
#define vll vector<ll>
#define vii vector<int>
#define tll takein<ll>
#define tii takein<int>
#define loop(i, a, b) for (int i = a; i < b; i++)

template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <class K, class V>
using ordered_map = tree<K, V, less<K>, rb_tree_tag, tree_order_statistics_node_update>;

/*```````````````````````````````//DEBUGGER``````````````````````````````````*/
#ifndef ONLINE_JUDGE
#define _debug(x)      \
    cerr << #x << " "; \
    _print(x);         \
    cerr << endl;
#else
#define _debug(x)
#endif
template <typename T>
void _print(T x) { cerr << x; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p)
{
    cerr << "{";
    _print(p.first);
    cerr << ",";
    _print(p.second);
    cerr << "}";
}
template <class T>
void _print(vector<T> v)
{
    cerr << "[ ";
    for (T i : v)
    {
        _print(i);
        cerr << " ";
    }
    cerr << "]";
}
template <class T>
void _print(set<T> v)
{
    cerr << "[ ";
    for (T i : v)
    {
        _print(i);
        cerr << " ";
    }
    cerr << "]";
}
template <class T>
void _print(multiset<T> v)
{
    cerr << "[ ";
    for (T i : v)
    {
        _print(i);
        cerr << " ";
    }
    cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v)
{
    cerr << "[ ";
    for (auto i : v)
    {
        _print(i);
        cerr << " ";
    }
    cerr << "]";
}
template <typename T>
vector<T> takein(int n)
{
    vector<T> A;
    for (int i = 0; i < n; i++)
    {
        T x;
        cin >> x;
        A.push_back(x);
    }
    return A;
}
template <typename T>
vector<T> presum(vector<T> &A)
{
    vector<T> ans(A.size(), 0);
    ans[0] = A[0];
    for (int i = 1; i < A.size(); i++)
    {
        ans[i] = ans[i - 1] + A[i];
    }
    return ans;
}
template <typename T>
void idxocc(vector<T> A, vector<vector<T>> idx)
{
    for (int i = 0; i < A.size(); i++)
    {
        T a;
        cin >> a;
        idx[a].push_back(i);
    }
}
template <typename T>
void print_arr(vector<T> ans)
{
    for (auto i : ans)
    {
        cout << i << " ";
    }
    cout << endl;
}
ll fastExp(int N, int R)
{
    ll ans = 1;
    while (R > 0)
    {
        if (R % 2 == 1)
        {
            ans = (ans * 1LL * N);
        }
        N = (N * 1LL * N);
        R = R >> 1;
    }
    return ans;
}
/*```````````````````````````````//predefined f(x)``````````````````````````````````*/

int x_Nx = 1e5 + 1;
vector<bool> prime(x_Nx, 1);
void primality()
{
    prime[1] = prime[0] = 0;
    for (int i = 2; i * i <= x_Nx; i++)
    {
        if (prime[i] == 1)
        {
            for (int j = i * i; j <= x_Nx; j = j + i)
            {
                prime[j] = 0;
            }
        }
    }
    return;
}
// use binsea snippet for Binary_search code

// ceil(A/B)=(A+B-1)/B
// log2(A)==Highest set bit in A
/*```````````````````````````````//solution``````````````````````````````````*/

void solve()
{
    // solve the question here
    ll n, m;
    cin >> n >> m;
    ll ans = 0;
    unordered_set<ll> div;
    ll lim=ceil(sqrt(max(n,m)));
    for (ll b = 1; b * b <= lim*lim; b++)
    {
        for (ll a = b; a <= n; a += b)
        {
            ll x=a/b;
            x++;
            ll g=(a,b);
            if(x%g==0){
                ans++;
            }
        }
    }
    cout << ans << endl;
}

int main()
{
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    srand(chrono::high_resolution_clock::now().time_since_epoch().count());
    /*#ifndef ONLINE_JUDGE
    freopen("Error.txt", "w", stderr);
    #endif*/
    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
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 (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

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

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

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

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

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

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

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

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

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

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

Buggy Code:
/*
 *   author: brownFalcon
 */
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pii pair<ll, ll>
#define piii pair<ll, pii>
#define endl "\n"
#define TC ll t;cin>>t;while(t--)
#define CHECK_BIT(var,pos) (((var)>>(pos)) & 1)

#define read freopen("input.txt", "r", stdin)
#define write freopen("output.txt", "w", stdout)

#define fastio()                        \
    ios_base::sync_with_stdio(false);   \
    cin.tie(NULL);                      \
    cout.tie(0)

const ll INFI = 1e18;
const ll MOD =  1e9+7;

const int N = 1e6+1;
bool is_prime[N];

void sieve() {
  memset(is_prime, true, sizeof(is_prime));
  is_prime[0] = is_prime[1] = false;

  for(int i=2;i<=N;i++) {
    if(is_prime[i] && (ll)i*i<=N) {
      for(int j=i*i;j<=N;j+=i) {
        is_prime[j] = false;
      }
    }
  }
}

ll sqRoot(ll a) {
  ll low = 1, high = a;
  while(low<=high) {
    ll mid = (low + high)/2;
    if(mid > 1000000) {
      high = mid -1;
    }
    else if(mid * mid == a) return mid;
    else if(mid * mid > a) {
      high = mid - 1;
    }
    else {
      low = mid + 1;
    }
  }
  return -1;
}

int main() {
#ifndef ONLINE_JUDGE
	read;
	write;
#else
	fastio();
#endif

  TC {
    ll n, m;cin>>n>>m;

    ll ans = n;
    for(int i=2;i*i<=n+m;i++) {
      ans += (n+i)/((ll)(i*i));
    }
    cout<<ans<<endl;
  }
	return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The code seems to have an issue with the logic used to count the valid pairs. Specifically, the loop that iterates over potential divisors and the way it calculates the number of valid pairs might not be correct.

Let's consider the following scenarios:

1. **Small values of \( n \) and \( m \)**: These can help us quickly verify the correctness of the logic.
2. **\( n \) and \( m \) being equal**: This can help us check the symmetry in the pairs.
3. **\( n \) and \( m \) being large but within the constraints**: This can help us check the performance and correctness for larger inputs.
4. **\( n \) and \( m \) being prime numbers**: This can help us check the behavior when the GCD is 1.
5. **\( n \) and \( m \) being powers of 2**: This can help us check the behavior when the GCD is a power of 2.

Given the constraints, a small-scale test input that can expose the bug might look like this:

```plaintext
6
1 1
2 3
3 5
10 8
100 100
1000000 1000000
```

This test input includes a variety of cases:
- A single pair (1, 1)
- A pair where \( n \) and \( m \) are small and different (2, 3)
- A pair where \( n \) and \( m \) are small and different (3, 5)
- A pair where \( n \) and \( m \) are larger but within the constraints (10, 8)
- A pair where \( n \) and \( m \) are equal and larger (100, 100)
- A pair where \( n \) and \( m \) are very large (1000000, 1000000)

This should help us identify if the code correctly handles different edge cases and larger inputs.


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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
/*
                    वक्रतुण्ड महाकाय सूर्यकोटि समप्रभ।
                    निर्विघ्नं कुरु मे देव सर्वकार्येषु सर्वदा।
    */

    import java.io.BufferedReader;
    import java.time.Duration;
    import java.time.Instant;
    import java.io.IOException;
    import java.io.InputStreamReader;
    import java.io.PrintWriter;
    import java.lang.reflect.Array;
    import java.math.BigInteger;
    import java.util.*;
    public class B_1_Reverse_Card_Easy_Version{

        // Time complexity = O(sqrt(n))
        public static boolean isPrime(int n) {
            if (n <= 1) {
                return false;
            }
            for (int i = 2; i <= Math.sqrt(n); i++) {
                if (n % i == 0) {
                    return false;
                }
            }
            return true;
        }

        //GCD OF TWO NUMBERS
        public static long gcd(long a, long b){
            if(a == 0){
                return b;
            }

            return gcd(b % a, a);
        }

        //LCM OF TWO NUMBERS
        public static long LCM(int a, int b){
            return  ((long) a * b) / gcd(a, b);
        }

        //    FOR TAKING FASTER INPUTS
        static class FastReader {
            BufferedReader br;
            StringTokenizer st;

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

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

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

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

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

            String nextLine()
            {
                String str = "";
                try {
                    if(st.hasMoreTokens()){
                        str = st.nextToken("\n");
                    }
                    else{
                        str = br.readLine();
                    }
                }
                catch (IOException e) {
                    e.printStackTrace();
                }
                return str;
            }
        }

        public static long decimalToBinary(long decimal) {
            if (decimal == 0) {
                return 0;
            }
            long binary = 0;
            long place = 1;
            while (decimal > 0) {
                long remainder = decimal % 2;
                binary += remainder * place;
                place *= 10;
                decimal = decimal / 2;
            }
            return binary;
        }

        public static void ha() {
            System.out.println("YES");
        }

        public static void na() {
            System.out.println("NO");
        }

        public static void line() {
            System.out.println();
        }

        public static int mod = 1000000007;

        public static long pow(long x, long p) {
            x %= mod;
            long ans = 1;
            while (p > 0) {
                if (p % 2 == 1) ans = ans * x % mod;
                x = x * x % mod;
                p /= 2;
            }
            return ans;
        }

        public static void main(String args[]) {
            FastReader sc = new FastReader();
            PrintWriter out = new PrintWriter(System.out);
            int t = sc.nextInt();
            while (t-- != 0) {
                long n = sc.nextLong();
                long m = sc.nextLong();
                long ans = n;
                for(long i=2;i<=n;i++) {
                    ans += (n+i)/(i*i);
                }
                System.out.println(ans);
            }
            out.flush();
        }
}

**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 the core logic of the problem and identify potential edge cases or scenarios that might cause the code to fail. The code in question calculates the number of valid pairs \((a, b)\) such that \(1 \le a \le n\), \(1 \le b \le m\), and \(a + b\) is a multiple of \(b \cdot \gcd(a, b)\).

### Analysis of the Code
The provided code has a loop that iterates from \(2\) to \(n\) and calculates the number of valid pairs based on the formula:
\[ \text{ans} += \frac{n + i}{i^2} \]

This formula seems to be incorrect or incomplete because it does not correctly account for the condition \(a + b\) being a multiple of \(b \cdot \gcd(a, b)\).

### Identifying a Challenging Test Case
To expose the bug, we should test with values of \(n\) and \(m\) that have a specific structure or properties that the code might handle incorrectly. Here are a few potential edge cases to consider:

1. **Small values of \(n\) and \(m\)**: Test with small values to see if the code handles basic cases correctly.
2. **Prime values of \(n\) and \(m\)**: Test with prime numbers to see if the code handles prime cases correctly.
3. **Values where \(n\) and \(m\) are powers of 2**: Test with powers of 2 to see if the code handles these specific values correctly.
4. **Large values of \(n\) and \(m\)**: Test with large values to see if the code handles large inputs efficiently and correctly.

### Chosen Test Case
Let's choose a test case with \(n = 6\) and \(m = 6\). This is a small-scale test case that is complete and valid, and it should help us identify if the code is handling the conditions correctly.

### Expected Output
For \(n = 6\) and \(m = 6\), the valid pairs \((a, b)\) are:
- \((1, 1)\)
- \((2, 1)\)
- \((2, 2)\)
- \((3, 1)\)
- \((4, 1)\)
- \((5, 1)\)
- \((6, 1)\)
- \((6, 2)\)
- \((6, 3)\)

So, the expected output should be \(9\).

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

This test input should help us identify if the code is handling the conditions correctly and expose any 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 (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

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

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

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

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

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

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

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

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

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

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

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

public class ABC {
    static Scanner sc = new Scanner(System.in);
    public static long solve() {
        long n = sc.nextLong();
        long m = sc.nextLong();
        long answer = 0;

        for(int b = 1; b <= 1500; b++) {
            // binary search for upper bound on b
            long lo = 1;
            long hi = 20000000L;
            lo--;
            while(lo < hi) {
                long mid = lo + (hi - lo + 1)/2;
                if(mid * b * b - b <= n) lo = mid;
                else {
                    hi = mid - 1;
                }
            }
            answer += lo;
        }
        answer--;
        return answer;
    }

    public static void main(String[] args) {
        int t = sc.nextInt();

        for(int i = 1; i <= t; i++) {
            System.out.println(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 code, we need to focus on the specific logic and constraints of the problem. The buggy code has a few issues:

1. **Incorrect Binary Search Logic**: The binary search is used to find the upper bound on `a` for a given `b`, but the logic is flawed. The condition `if(mid * b * b - b <= n)` does not correctly determine the valid pairs.
2. **Fixed Loop Bound**: The loop iterates only up to `1500` for `b`, which is not sufficient for larger values of `m`.
3. **Subtraction and Increment/Decrement Issues**: The use of `lo--` and `answer--` is incorrect and can lead to off-by-one errors.

To expose these issues, we can create test cases that:
- Have large values of `n` and `m` to ensure the loop bound is insufficient.
- Have values that make the binary search logic fail.

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

```plaintext
3
1 1
2 3
1500 1500
```

### Explanation:
1. **Test Case 1: (1, 1)**
   - This is a simple case where the only valid pair is `(1, 1)`. The output should be `1`.

2. **Test Case 2: (2, 3)**
   - This case is also simple and should have the valid pairs `(1, 1)`, `(2, 1)`, and `(2, 2)`. The output should be `3`.

3. **Test Case 3: (1500, 1500)**
   - This case is more complex and should expose the issues with the loop bound and binary search logic. The number of valid pairs should be significantly larger than what the buggy code might output.

By using these test cases, we can ensure that the bugs in the provided code are exposed, as the incorrect loop bound and binary search logic will lead to incorrect results for the more complex test case.