**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 math

def gcd(a, b):
    return a if b == 0 else gcd(b, a % b)

def ktra(a, b):
    return (b * gcd(a, b) % (a + b)) == 0

t = int(input())
for _ in range(t):
    n, m = map(int, input().split())
    res = 0
    limit_n = int(math.sqrt(n))
    limit_m = int(math.sqrt(m))
    for i in range(1, int(math.sqrt(n)) + 1):
        for j in range(1, int(math.sqrt(m)) + 1):
            if gcd(i, j) == 1:
                res += int(min(n // ((i + j) * i), m // ((i + j) * j)))
    print(res)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases or specific scenarios that highlight the issues in the code. The buggy code has a limitation in its approach to calculating the valid pairs, particularly in the way it uses the square root limits and the gcd condition.

### Step-by-Step Analysis:
1. **Square Root Limits**: The code uses `int(math.sqrt(n))` and `int(math.sqrt(m))` to limit the range of `i` and `j`. This can miss valid pairs that are outside these limits.
2. **GCD Condition**: The code only considers pairs where `gcd(i, j) == 1`, which might not cover all valid pairs.

### Challenging Test Case:
To expose the bug, we should create a test case where valid pairs exist outside the square root limits or where `gcd(i, j) != 1` but still satisfy the conditions.

### Test Input:
We will use a small-scale test case with `n` and `m` values that are slightly larger than the square root limits and include pairs that are not coprime but still satisfy the conditions.

```plaintext
1
5 5
```

### Explanation:
- **Test Case**: `n = 5`, `m = 5`
- **Valid Pairs**:
  - (1, 1): \( 1 \cdot \gcd(1, 1) = 1 \) is a multiple of \( 1 + 1 = 2 \) (True)
  - (2, 2): \( 2 \cdot \gcd(2, 2) = 4 \) is a multiple of \( 2 + 2 = 4 \) (True)
  - (3, 3): \( 3 \cdot \gcd(3, 3) = 9 \) is a multiple of \( 3 + 3 = 6 \) (False)
  - (4, 4): \( 4 \cdot \gcd(4, 4) = 16 \) is a multiple of \( 4 + 4 = 8 \) (True)
  - (5, 5): \( 5 \cdot \gcd(5, 5) = 25 \) is a multiple of \( 5 + 5 = 10 \) (True)
  - (2, 4): \( 4 \cdot \gcd(2, 4) = 8 \) is a multiple of \( 2 + 4 = 6 \) (False)
  - (4, 2): \( 2 \cdot \gcd(4, 2) = 4 \) is a multiple of \( 4 + 2 = 6 \) (False)
  - (1, 4): \( 4 \cdot \gcd(1, 4) = 4 \) is a multiple of \( 1 + 4 = 5 \) (True)
  - (4, 1): \( 1 \cdot \gcd(4, 1) = 1 \) is a multiple of \( 4 + 1 = 5 \) (True)

### Expected Output:
The correct output for the test case `5 5` should be `6` (pairs: (1, 1), (2, 2), (4, 4), (5, 5), (1, 4), (4, 1)).

### Final 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 (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 <algorithm>
#include <bitset>
#include <cstdint>
#include <cstring>
#include <iostream>
#include <limits.h>
#include <math.h>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>

//#define int long long
#define loop(X, N) for(int X = 0; X < (N); X++)
#define all(V) V.begin(), V.end()
#define rall(V) V.rbegin(), V.rend()

using namespace std;

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<bool> vb;
typedef vector<vector<bool>> vvb;

typedef uint64_t u64;
typedef int64_t i64;

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

	return gcd(b, a % b);
}

inline std::vector<int> listDivisors(const std::vector<int>& primes, int n)
{
	std::vector<int> divisors = { 1 };

	for (int i = 0; i < primes.size(); i++)
	{
		if (primes[i] > n)
			break;

		if ((n % primes[i]) == 0)
		{
			int count = 0;
			while (n % primes[i] == 0)
			{
				n /= primes[i];
				count++;
			}

			int divisorSize = divisors.size();
			int primePower = primes[i];
			for (int j = 0; j < count; j++)
			{
				for (int k = 0; k < divisorSize; k++)
				{
					divisors.push_back(divisors[k] * primePower);
				}

				primePower *= primes[i];
			}
		}
	}

	if (n > 1) //implies n is prime, but not in primes
	{
		int divisorSize = divisors.size();
		for (int k = 0; k < divisorSize; k++)
		{
			divisors.push_back(divisors[k] * n);
		}
	}

	return divisors;
}

inline void generatePrimes(std::vector<int>& primes, int maxCount)
{
	primes.push_back(2);
	for (int i = 3; i < maxCount; i += 2)
	{
		bool success = true;
		int j = 0;
		while (primes[j] <= sqrt(i))
		{
			if ((i % primes[j]) == 0)
			{
				success = false;
				break;
			}
			j++;
		}

		if (success)
			primes.push_back(i);
	}
}

signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);

	int t;
	cin >> t;

	vi primes;
	generatePrimes(primes, 2e3);

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

		//a + b | b * gcd(a, b)
		// - (p, q)
		// - p + q | g * q <=> p + q | g * p <=> p + q | g
		// - q <= n / g, p <= m / g

		int result = 0;
		for (int g = min(n, m); g >= 1; g--)
		{
			int maxP = m / g;
			int maxQ = n / g;

			vi divisors = listDivisors(primes, g);
			for (int d : divisors)
			{
				for (int q = max(1, d - maxP); q <= min(maxQ, d - 1); q++)
				{
					//p + q = d
					//gcd(p, q) = 1

					if (gcd(d, q) != 1)
						continue;

					result++;
				}
			}
		}

		cout << result << endl;
	}

	return 0;
}

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

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

Think step by step.

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

1. **Small Values for \( n \) and \( m \)**: Test with small values to ensure the basic logic is correct.
2. **Edge Cases**: Test with \( n \) or \( m \) equal to 1.
3. **Special Values**: Test with values where the gcd is likely to be 1 or a small prime number.
4. **Symmetry**: Test with \( n = m \) to check for symmetry in the pairs.
5. **Large Values**: Test with values close to the upper limit to check for performance and correctness.

Given the constraints, we will generate a small-scale test input with a mix of these cases. Here is a test input that should be challenging:

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

This test input includes:
- The smallest possible values (1, 1) to check the edge case.
- Small symmetric values (2, 2), (3, 3), (4, 4), and (5, 5) to check for symmetry and small gcd values.

This should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (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 <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
#include <algorithm>
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")

using namespace __gnu_pbds;
using namespace std;

#define mp make_pair
#define pb push_back
#define all(a) (a).begin(), (a).end()
#define sz(a) (int)(a.size())
#define endl '\n'
#define F first
#define S second
#define rs resize

using ll = long long;
using ld = long double;
using str = string;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using ppi = pair<int, pair<int, int>>;
using ppl = pair<ll, pair<ll, ll>>;
using vi = vector<int>;
using vl = vector<ll>;
using vpi = vector<pair<int, int>>;
using vvi = vector<vi>;
using vvpi = vector<vpi>;
using vpl = vector<pl>;
using vvl = vector<vl>;
using vvpl = vector<vpl>;
using vppl = vector<ppl>;
using vppi = vector<ppi>;
using ull=unsigned long long;

typedef tree<
        int,
        null_type,
        less<int>,
        rb_tree_tag,
        tree_order_statistics_node_update>
        ordered_set;

int MOD =  1e9+7;
mt19937 rnd(time(0));

const int N=3e3+1;

int mul(int a, int b) {
    return (1LL * a * b) % MOD;
}

int add(int a, int b) {
    int s = (a+b);
    if (s>=MOD) s-=MOD;
    return s;
}

int sub(int a, int b) {
    int s = (a+MOD-b);
    if (s>=MOD) s-=MOD;
    return s;
}

int po(int a, ll deg)
{
    if (deg==0) return 1;
    if (deg%2==1) return mul(a, po(a, deg-1));
    int t = po(a, deg/2);
    return mul(t, t);
}

int inv(int n)
{
    return po(n, MOD-2);
}

/*
const int LIM = 200005;

vector<int> facs(LIM), invfacs(LIM);

void init()
{
    facs[0] = 1;
    for (int i = 1; i<LIM; i++) facs[i] = mul(facs[i-1], i);
    invfacs[LIM-1] = inv(facs[LIM-1]);
    for (int i = LIM-2; i>=0; i--) invfacs[i] = mul(invfacs[i+1], i+1);
}

int C(int n, int k)
{
    if (n<k) return 0;
    if (n<0 || k<0) return 0;
    return mul(facs[n], mul(invfacs[k], invfacs[n-k]));
}
*/

void solve()
{
    ll n, m;
    cin>>n>>m;

    ll cnt=0;
    for(ll i=2;i*i<=n+m;i++){
        for(ll j=i;j<=n;j+=i){
            for(ll z=1;z<=i-1;z++){
                if(__gcd(z, i-z)!=1) continue;
                ll a=z*j;
                ll b=(i-z)*j;
                if(a>n) break;
                if(b>m) continue;
                ll x1=a+b;
                ll y1=b*__gcd(a, b);
                if(y1%x1==0){
                    cnt++;
                }
            }
        }
    }
    cout<<cnt<<endl;
}

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    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 provided 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. **Understand the Bug**: The code iterates through possible values of \(i\), \(j\), and \(z\) to find pairs \((a, b)\) that satisfy the given conditions. The bug might be related to how it handles the gcd and the conditions for \(a\) and \(b\).

2. **Edge Cases**:
   - **Small Values**: Test with small values of \(n\) and \(m\) to ensure the basic logic is correct.
   - **Large Values**: Test with large values of \(n\) and \(m\) to check for performance issues and potential overflow.
   - **Boundary Conditions**: Test with values that are just above or below the limits.
   - **Special Values**: Test with values that might cause the gcd function to behave unexpectedly.

3. **Specific Scenarios**:
   - **Equal Values**: \(n = m\) to see how the code handles symmetry.
   - **One Value is 1**: \(n = 1\) or \(m = 1\) to check how the code handles the simplest cases.
   - **Prime Values**: \(n\) or \(m\) being prime numbers to test the gcd function.

Based on these considerations, let's generate a small-scale test input that includes these edge cases:

```plaintext
6
1 1
2 2
3 3
10 10
100 100
1000 1000
```

This test input includes:
- The smallest possible values (1, 1).
- Small values (2, 2).
- Medium values (3, 3).
- Larger values (10, 10).
- Larger values (100, 100).
- Even larger values (1000, 1000).

These cases will help in identifying any issues in the logic or performance of the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (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 <numeric>
#include <bitset>

const int SUP = 2'000'001;

int prime_divisors[SUP][7] = {0};
int prime_divisors_powers[SUP][7] = {0};
int prime_divisors_size[SUP] = {0};
int rests[SUP] = {0};

int div_x[1000];
int div_y[1000];

template < class V, class D, class P >
int fill_div(V& v, int m, D& d, P& p) {
	v[0] = 1;
	int sz = 1;
	while ( m-- ) {
		for ( int z = d[m], i = 1; i <= p[m]; i++, z *= d[m] )
				for ( int j = 0; j < sz; j++ )
					v[i*sz + j] = v[j] * z;
			sz *= p[m] + 1;
		}
	return sz;
}

int gcd(int a, int b) {
	if ( a < b )
		std::swap(a, b);
	while ( b )
		std::swap(a %= b, b);
	return a;
}

void proc() {

	int n, m;
	std::cin >> n >> m;

	int r = 0;
	for ( int b = 2; b <= m; b++ ) {
		int sz_x = fill_div(div_x, prime_divisors_size[b], prime_divisors[b], prime_divisors_powers[b]);
		std::sort(&div_x[0], &div_x[sz_x]);
		for ( int i = 0; i < sz_x; i++ ) {
			int d = b / div_x[i];
			for ( int j = i+1; j < sz_x && 1ll * (div_x[j] - div_x[i]) * d <= n; j++ )
				r += 1 == gcd(div_x[i], div_x[j]);
		}
	}

	std::cout << r << std::endl;

}

int main() {

	std::iota(&rests[0], &rests[SUP], 0);
	std::fill(&prime_divisors_size[0], &prime_divisors_size[SUP], 1);

	for ( int p = 2; p*p < SUP; p++ )
		if ( 1 < rests[p] )
			for ( int v = p; v < SUP; v += p ) {
				int k = prime_divisors_size[v];
				prime_divisors[v][k-1] = p;
				prime_divisors_powers[v][k-1] = 0;
				while ( rests[v] % p == 0 ) {
					rests[v] = rests[v] / p;
					++prime_divisors_powers[v][k-1];
				}
				if ( 1 != rests[v] ) {
					k = ++prime_divisors_size[v];
					prime_divisors[v][k-1] = rests[v];
					prime_divisors_powers[v][k-1] = 1;
				}
			}

	std::cin.tie(0);
	std::cin.sync_with_stdio(false);
	std::cout.sync_with_stdio(false);

	int t = 1;
	std::cin >> t;

	while ( t-- )
		proc();

	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 consider edge cases and scenarios that are likely to cause the code to fail. Here are the steps to think through this:

1. **Small Test Cases with Specific Conditions**:
   - Test cases where \( n \) and \( m \) are small but have specific properties that might expose the bug.
   - For example, \( n \) and \( m \) could be prime numbers, or they could have specific divisors that the code might handle incorrectly.

2. **Edge Cases**:
   - Test cases where \( n \) or \( m \) are 1, as these are boundary conditions.
   - Test cases where \( n \) and \( m \) are equal, as this might simplify the problem and expose issues in the code's logic.

3. **Symmetric Cases**:
   - Test cases where \( n \) and \( m \) are symmetric, meaning they have similar divisors or properties.

4. **Large Test Cases with Specific Properties**:
   - Test cases where \( n \) and \( m \) are large but have specific properties that might cause the code to fail, such as having many divisors or being composite numbers with specific factors.

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

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

This input includes:
- A test case where \( n \) and \( m \) are both 1.
- A test case where \( n \) and \( m \) are small but have different values.
- A test case where \( n \) and \( m \) are equal and have small values.

These test cases are small and manageable but should be sufficient to expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 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 int long long
using namespace std;
const int INF = 2e18;
const int N = 4e6+10;
const int MOD = 1e9+7;
int t,n,k,m;
vector<int> pri;
bool not_prime[N];
int phi[N];
void pre(int n) {
    phi[1] = 1;
    for (int i = 2; i <= n; i++) {
        if (!not_prime[i]) {
            pri.push_back(i);
            phi[i] = i - 1;
        }
        for (int pri_j : pri) {
            if (i * pri_j > n) break;
            not_prime[i * pri_j] = true;
            if (i % pri_j == 0) {
                phi[i * pri_j] = phi[i] * pri_j;
                break;
            }
            phi[i * pri_j] = phi[i] * phi[pri_j];
        }
    }
}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    pre(4000000);
    cin>>t;
    while(t--){
        cin>>n>>m;
        int ans = 0;
        for(int d=1;d<=n+m;d++){
            for(int x=d;x<=n+m;x+=d){
                if(x<2){
                    continue;
                }
                int mn = max((int)1,(x-n+d-1)/d);
                int mx = min(m/d,x/d-1);
                int a = x/d;
                if(d%a==0){
                    if(mn==1&&mx==x/d-1&&mx>=mn){
                        ans+= phi[a];
                    }else{
                        for(int i=mn;i<=mx;i++){
                            if(__gcd(i,a)==1){
                                ans++;
                            }
                        }
                    }
                }
            }
        }
        cout<<ans<<"\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:
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 think through this:

1. **Small Test Cases**:
   - Test with the smallest possible values for \( n \) and \( m \) to ensure the code handles basic cases correctly.
   - Example: \( n = 1 \), \( m = 1 \).

2. **Edge Cases**:
   - Test with values where \( n \) and \( m \) are equal.
   - Example: \( n = 10 \), \( m = 10 \).

3. **Prime Numbers**:
   - Test with \( n \) and \( m \) being prime numbers to check how the code handles primes.
   - Example: \( n = 11 \), \( m = 13 \).

4. **Large Values**:
   - Test with large values of \( n \) and \( m \) to see if the code can handle large inputs efficiently.
   - Example: \( n = 2000000 \), \( m = 2000000 \).

5. **Special Cases**:
   - Test with values where \( n \) and \( m \) are multiples of each other.
   - Example: \( n = 10 \), \( m = 20 \).

6. **Multiple Test Cases**:
   - Include multiple test cases to ensure the code can handle a variety of scenarios.
   - Example: \( t = 5 \).

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

```plaintext
5
1 1
10 10
11 13
10 20
2000000 2000000
```

This test input includes:
- A small test case to check basic functionality.
- A test case with equal values for \( n \) and \( m \).
- A test case with prime values for \( n \) and \( m \).
- A test case with \( n \) and \( m \) being multiples of each other.
- A test case with large values to check performance and correctness.


**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 <algorithm>
#include <bitset>
#include <cstdint>
#include <cstring>
#include <iostream>
#include <limits.h>
#include <math.h>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>

//#define int long long
#define loop(X, N) for(int X = 0; X < (N); X++)
#define all(V) V.begin(), V.end()
#define rall(V) V.rbegin(), V.rend()

using namespace std;

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<bool> vb;
typedef vector<vector<bool>> vvb;

typedef uint64_t u64;
typedef int64_t i64;

inline int gcd(int a, int b)
{
	while (b != 0)
	{
		swap(a, b);
		b = b % a;
	}

	return a;
}

inline std::vector<int> listDivisors(const std::vector<int>& primes, int n)
{
	std::vector<int> divisors = { 1 };

	for (int prime : primes)
	{
		if (prime > n)
			break;

		if ((n % prime) == 0)
		{
			int count = 0;
			while (n % prime == 0)
			{
				n /= prime;
				count++;
			}

			int divisorSize = divisors.size();
			int primePower = prime;
			for (int j = 0; j < count; j++)
			{
				for (int k = 0; k < divisorSize; k++)
				{
					divisors.push_back(divisors[k] * primePower);
				}

				primePower *= prime;
			}
		}
	}

	if (n > 1) //implies n is prime, but not in primes
	{
		int divisorSize = divisors.size();
		for (int k = 0; k < divisorSize; k++)
		{
			divisors.push_back(divisors[k] * n);
		}
	}

	return divisors;
}

inline void generatePrimes(std::vector<int>& primes, int maxCount)
{
	primes.push_back(2);
	for (int i = 3; i < maxCount; i += 2)
	{
		bool success = true;
		int j = 0;
		while (primes[j] <= sqrt(i))
		{
			if ((i % primes[j]) == 0)
			{
				success = false;
				break;
			}
			j++;
		}

		if (success)
			primes.push_back(i);
	}
}

signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);

	int t;
	cin >> t;

	vi primes;
	generatePrimes(primes, 2e3);

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

		//a + b | b * gcd(a, b)
		// - (p, q)
		// - p + q | g * q <=> p + q | g * p <=> p + q | g
		// - q <= n / g, p <= m / g

		int result = 0;
		for (int g = min(n, m); g >= 1; g--)
		{
			int maxP = m / g;
			int maxQ = n / g;

			//d > maxP + maxQ

			vi divisors = listDivisors(primes, g);
			for (int d : divisors)
			{
				int l = max(1, d - maxP);
				int r = min(maxQ, d - 1);
				if (l > r)
					continue;

				/*int extra = r - l + 1;
				for (int prime : primes)
				{
					if (prime > r)
						break;

					extra = extra - r / prime + (l - 1) / prime;
				}

				result += extra;*/

				for (int q = l; q <= r; q++)
				{
					//p + q = d
					//gcd(p, q) = 1

					if (gcd(d, q) != 1)
						continue;

					result++;
				}
			}
		}

		//cout << "Result: ";
		cout << result << 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 code, we need to focus on edge cases and scenarios that might cause the algorithm to fail. Here are some potential edge cases to consider:

1. **Small values of \( n \) and \( m \)**: These can help verify the basic logic.
2. **Large values of \( n \) and \( m \)**: These can help check the performance and correctness for larger inputs.
3. **Values where \( n \) and \( m \) are close to each other**: This can test the logic where the gcd and divisors play a significant role.
4. **Values where \( n \) and \( m \) are significantly different**: This can test the logic for unbalanced inputs.
5. **Values where \( n \) or \( m \) is 1**: This can test the edge case where one of the dimensions is minimal.

Given the constraints and the need for a small-scale test input, let's focus on a combination of small and medium values that are likely to expose the bug.

Here is the test input:

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

This input includes:
- A minimal case (1, 1)
- A small case where \( n \) and \( m \) are different (2, 3)
- A small case where \( n \) and \( m \) are close (3, 5)
- A medium case where \( n \) and \( m \) are equal (10, 10)

These cases should help verify the basic logic and edge conditions of the algorithm.


**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 math
import sys
from collections import defaultdict

INF = float('inf')
MOD = 1000000007
mod = 998244353
gcd = lambda a, b: gcd(b, a % b) if b else a

def Solve():
    n, m = map(int, input().split())
    sum = 0
    for a in range(1, int(math.sqrt(n)) + 1):
        for b in range(1, min(n // a, m) + 1):
            if (a + b) * a <= n and (a + b) * b <= m and gcd(a, b) == 1:
                sum += min(n // ((a + b) * a), m // ((a + b) * b))
    print(sum)

if __name__ == "__main__":
    t = int(input())
    for _ 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 given buggy code, we need to focus on edge cases and scenarios that are likely to cause the code to fail. The code provided has a specific logic that iterates over pairs \((a, b)\) and checks certain conditions. Here are the key points to consider:

1. **Edge Cases**: Small values of \(n\) and \(m\) where the conditions are either trivially true or false.
2. **Large Values**: Values of \(n\) and \(m\) that are close to the upper limit to test performance and correctness.
3. **Special Values**: Values where \(n\) and \(m\) have a specific relationship, such as \(n = m\), \(n\) is a multiple of \(m\), or vice versa.
4. **Mixed Values**: A combination of different values to ensure the code handles various scenarios correctly.

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

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

### Explanation:
- **Test Case 1 (1 1)**: This is the smallest possible input. No pairs satisfy the conditions.
- **Test Case 2 (2 2)**: This is a small input where the conditions are more likely to be satisfied. The valid pairs are \((2, 2)\).
- **Test Case 3 (3 3)**: Another small input to test a slightly larger range. The valid pairs are \((3, 3)\).
- **Test Case 4 (10 10)**: A medium-sized input to test a broader range. The valid pairs are \((2, 2)\), \((3, 6)\), \((4, 4)\), \((6, 3)\), \((6, 6)\), \((8, 8)\).
- **Test Case 5 (100 100)**: A larger input to test the performance and correctness with a larger range. This will ensure the code can handle more complex scenarios.

This test input should be sufficient to expose any issues in the buggy code, especially in handling edge cases and larger 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:
from sys import stdin, stdout
from math import sqrt

input = stdin.readline
print = lambda s: stdout.write(s+'\n')

N = 1414
good = [[1]*N for _ in range(N)]
for i in range(2, N):
    for p in range(i, N, i):
        for q in range(i, N, i):
            good[p][q] = 0

for _ in range(int(input())):

    n, m = map(int, input().split())
    sqm, ans = int(sqrt(m-1))+1, 0
    for p in range(1, int(sqrt(n-1))+1):
        for q in range(1, sqm):
            if good[p][q]:
                ans += min(n//p, m//q)//(p+q)

    print(str(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 challenging test input that exposes the bug in the provided buggy code, we need to consider the following:

1. **Edge Cases**: Test cases with small values of \( n \) and \( m \) where the conditions are likely to be met or not met.
2. **Specific Patterns**: Values of \( n \) and \( m \) that are likely to cause the code to fail due to its incorrect handling of the conditions.
3. **Boundary Conditions**: Values of \( n \) and \( m \) that are at the boundaries of the input constraints.

Given the constraints \( 1 \leq n, m \leq 2 \cdot 10^6 \), we can focus on values that might expose the bug in the `good` array and the loop logic.

### Step-by-Step Analysis:
1. **Small Values**: Test with small values of \( n \) and \( m \) to ensure the code handles basic cases correctly.
2. **Specific Pairs**: Test with pairs \((a, b)\) that should satisfy the conditions but might be incorrectly handled by the code.
3. **Large Values**: Test with values close to the upper limit to check the performance and correctness.

### Test Case Generation:
Let's generate a test case with a few small values and a few specific pairs that should be valid but might be incorrectly handled by the code.

```plaintext
5
1 1
2 3
3 5
10 8
12 18
```

### Explanation:
1. **Test Case 1: \( n = 1, m = 1 \)**
   - Expected Output: 0 (No pairs satisfy the conditions)
2. **Test Case 2: \( n = 2, m = 3 \)**
   - Expected Output: 1 (Pair (2, 2) satisfies the conditions)
3. **Test Case 3: \( n = 3, m = 5 \)**
   - Expected Output: 1 (Pair (3, 6) satisfies the conditions)
4. **Test Case 4: \( n = 10, m = 8 \)**
   - Expected Output: 6 (Pairs (2, 2), (3, 6), (4, 4), (6, 3), (6, 6), (8, 8) satisfy the conditions)
5. **Test Case 5: \( n = 12, m = 18 \)**
   - Expected Output: 10 (Pairs (2, 2), (3, 6), (4, 4), (6, 3), (6, 6), (8, 8), (9, 18), (12, 12), (12, 18), (18, 12) satisfy the conditions)

This test case is designed to cover a range of values and specific pairs that should expose any issues in the `good` array and the loop logic of 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;

void solve() {
    int n, m;
    cin >> n >> m;
    vector<pair<int, int>> ans;
    for (int j = 1; j <= min(n, m); j++) {
        for (int i = j; i <= n + m; i += j) {
            if (((long long) j * j) % i) continue;
            for (int b = max(j, (i - n + j - 1) / j * j); b <= min(i - 1, m); b += j) {
                if (((long long) b * gcd(i, b)) % i == 0) {
                    ans.push_back({i - b, b});
                }
            }
        }
    }
    sort(ans.begin(), ans.end());
    ans.erase(unique(ans.begin(), ans.end()), ans.end());
    cout << ans.size() << '\n';
}

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

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that might cause the code to fail. The code is designed to find pairs \((a, b)\) such that \(1 \le a \le n\), \(1 \le b \le m\), and \(b \cdot \gcd(a, b)\) is a multiple of \(a + b\).

### Analysis of the Buggy Code:
1. **Loop Structure**: The code uses nested loops to iterate over possible values of \(a\) and \(b\).
2. **Condition Check**: The condition \(((long long) b * gcd(i, b)) % i == 0\) is used to check if the pair \((a, b)\) is valid.
3. **Unique Pairs**: The code attempts to remove duplicate pairs using `sort` and `unique`.

### Potential Issues:
1. **Range of Loops**: The outer loop iterates from 1 to `min(n, m)`, which might miss some valid pairs if \(n\) and \(m\) are significantly different.
2. **Condition Calculation**: The condition \(((long long) b * gcd(i, b)) % i == 0\) might be incorrectly evaluated for certain values of \(a\) and \(b\).
3. **Edge Cases**: The code might not handle edge cases where \(n\) or \(m\) are small or very large correctly.

### Challenging Test Input:
To expose these potential issues, we can create a test case where:
- \(n\) and \(m\) are small but have a specific relationship that might cause the condition to fail.
- \(n\) and \(m\) are large, but the pairs are sparse, making it harder for the code to find valid pairs.

### Example Test Input:
```plaintext
1
5 10
```

### Explanation:
- **Small \(n\) and \(m\)**: This test case is small enough to be manageable but large enough to contain a few valid pairs.
- **Specific Relationship**: The values \(5\) and \(10\) are chosen to test the condition \(((long long) b * gcd(i, b)) % i == 0\) for a range of pairs.

### Expected Output:
The correct output for this test case should be the number of valid pairs \((a, b)\) that satisfy the given conditions. For \(n = 5\) and \(m = 10\), the valid pairs are:
- \((2, 2)\)
- \((3, 6)\)
- \((4, 4)\)
- \((6, 3)\)
- \((6, 6)\)
- \((8, 8)\)

Thus, the expected output is:
```plaintext
6
```

This test case should help expose any issues in the loop structure or condition evaluation 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 <vector>
#include <string>
#include <map>
#include <set>
#include <unordered_set>
#include <unordered_map>
#include <algorithm>
#include <numeric>
#include <cassert>
#include <queue>

#define LOAD(v)       \
    for (auto &x : v) \
        cin >> x;

#define INT(x) \
    int x;     \
    cin >> x;

using namespace std;

using ll = long long;
using ull = unsigned long long;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using pi = pair<int, int>;
using pll = pair<ll, ll>;
using vpi = vector<pi>;
using vpll = vector<pll>;

template <typename T>
using min_queue = priority_queue<T, vector<T>, greater<T>>;

void ans(bool a)
{
    if (a)
        cout << "Yes\n";
    else
        cout << "No\n";
}

void ans(auto a)
{
    cout << a << '\n';
}

template <typename T>
void ans(const vector<T> &a)
{
    for (const auto &x : a)
        cout << x << ' ';
    cout << '\n';
}

constexpr ll maxa = 4e6 + 5;

vi phi(maxa);
vi era(maxa);

void solve()
{
    ll n, m;
    cin >> n >> m;

    ll res = 0;
    for (ll g = 1; g <= min(m, n); ++g)
    {
        for (ll s = g + g; s <= n + m; s += g)
        {
            ll mul = lcm(g, s / g);
            if (gcd(s / g, mul / g) == 1)
            {
                ll mn = max(s - n, 1ll);
                ll mx = min(s - 1, m);
                for (ll b = mn - mn % mul; b <= mx - mx % mul; b += mul)
                {
                    ll a = s - b;
                    if (a > n || gcd(s, b) != g)
                    {
                        continue;
                    }
                    ++res;
                }
            }
        }
    }
    ans(res);
}

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

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

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

Think step by step.

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

1. **Edge Cases**: Test with the smallest and largest possible values for `n` and `m`.
2. **Special Values**: Test with values where `n` and `m` are equal, or where one is a multiple of the other.
3. **Boundary Conditions**: Test with values close to the boundaries of the input constraints.
4. **Symmetry**: Test with symmetric values where `n` and `m` are the same.
5. **Prime Numbers**: Test with prime numbers for `n` and `m` to ensure the GCD calculations are handled correctly.

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

```plaintext
5
1 1
2 3
3 5
10 10
11 11
```

### Explanation:
1. **(1, 1)**: This is the smallest possible input, testing the edge case where both `n` and `m` are 1.
2. **(2, 3)**: This is a small input where `n` and `m` are different, testing the basic functionality.
3. **(3, 5)**: Another small input where `n` and `m` are different, testing the GCD and LCM calculations.
4. **(10, 10)**: This is a symmetric input where `n` and `m` are the same, testing the symmetry in the problem.
5. **(11, 11)**: This is another symmetric input where `n` and `m` are the same, but with a prime number, testing the GCD and LCM calculations with prime numbers.

These test cases should help expose any bugs in the provided code by covering a range of scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (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;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<double,double> pdd;
//typedef complex<double> cpx;
#define fastio cin.tie(0)->sync_with_stdio(0); cout.tie(0);
#define all(x) x.begin(),x.end()
#define compress(x) x.erase(unique(all(x)),x.end())
#define ff first
#define ss second
#define INF 1e15
#define MAX 410
#define SIZE 10010
#define MOD 1000000007

void solve(){
    ll n,m; cin >> n >> m;
    ll cnt = 0;
    for(int i=1;i<=1500;i++){
        for(int j=1;j<=1500;j++){
            if(__gcd(i,j)!=1) continue;
            cnt += min(n/i/(i+j),m/j/(i+j));
        }
    }
    cout << cnt << "\n";
}

int main(){
    fastio;
    ll 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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios where the code might fail. The main issue in the buggy code is that it only iterates up to 1500 for both `i` and `j`, which can miss valid pairs when `n` or `m` is larger than 1500.

Here are the steps to generate a small-scale test input:

1. **Choose a small number of test cases**: Since we want a small-scale test, we can choose a small number of test cases, say 2 or 3.
2. **Select values for `n` and `m`**: Choose values for `n` and `m` that are greater than 1500 to ensure that the code's iteration limit is exceeded.
3. **Ensure the values are within the problem constraints**: The values should be within the range specified in the problem, i.e., \(1 \le n, m \le 2 \cdot 10^6\).

Let's create a test input with the following test cases:
- One test case where both `n` and `m` are just above 1500.
- One test case where `n` is 1500 and `m` is significantly larger.
- One test case where `m` is 1500 and `n` is significantly larger.

Here is the test input:

```plaintext
3
1501 1501
1500 20000
20000 1500
```

This test input will help expose the bug in the code by ensuring that the iteration limit of 1500 is exceeded, thus missing 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:
#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#include<bits/stdc++.h>
using namespace std;
int t,n,m;
inline void solute()
{
	int ans=0;
	cin>>n>>m;
	for(int i=1;i<=min(n,1700);i++)
	{
		for(int j=1;j<=min(m,1700);j++)
		{
			if(__gcd(i,j)==1)
			{
				ans+=min(n/i,m/j)/(i+j);
			}

		}
	}
	cout<<ans<<"\n";
}
int main()
{
	ios::sync_with_stdio(false);
	cin>>t;
	while(t--)
	{
		solute();
	}
	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 buggy code, we need to focus on edge cases where the code might fail. The main issue in the buggy code is that it only iterates up to `min(n, 1700)` and `min(m, 1700)`, which means it might miss valid pairs for larger values of `n` and `m`.

Let's consider the following scenarios:

1. **Small values of `n` and `m`**: The code might work correctly for small values.
2. **Large values of `n` and `m`**: The code might miss valid pairs because it only iterates up to 1700.
3. **Boundary values**: Values around 1700 might be particularly interesting to test.

Given the constraints, a small-scale test input that can expose the bug would be:

```plaintext
1
1700 1700
```

This input will ensure that the code iterates up to its limit, but it might still miss valid pairs for larger values. However, since the code is limited to 1700, we can also test with a slightly larger value to ensure it fails:

```plaintext
1
1701 1701
```

This input will ensure that the code does not consider pairs where `i` or `j` is greater than 1700, which might lead to incorrect results.

Let's use the second input to expose the bug:

```plaintext
1
1701 1701
```


**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;
int t;
long long n,m,ans;
long long gcd(long long x,long long y){
	if(y==0){
		return x;
	}
	return gcd(y,x%y);
}
int main(){
	cin>>t;
	while(t--){
		ans=0;
		scanf("%lld%lld",&n,&m);
		for(long long i=1;i<=n;i++){
			for(long long j=1;j<=min(m,m/i-i);j++){
				if(gcd(i,j)==1&&(i+j)*i<=n&&(i+j)*j<=m){
					ans+=min(n/i/(i+j),m/j/(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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios where the code's logic might fail. Let's analyze the code and identify potential issues:

1. **GCD Calculation**: The GCD calculation seems correct.
2. **Loop Conditions**: The loop conditions and the inner condition checks are where the bugs might be. Specifically, the inner loop runs from `1` to `min(m, m/i - i)`, which might not cover all valid pairs.
3. **Condition Check**: The condition `gcd(i, j) == 1 && (i + j) * i <= n && (i + j) * j <= m` is used to filter valid pairs. However, this might miss some valid pairs.

Given these points, let's construct a test case that might expose the issue:

- **Small `n` and `m`**: Start with small values to ensure the test case is manageable.
- **Edge Cases**: Consider values where the conditions might be tight, such as when `n` and `m` are close to each other or when `i` and `j` are close to the boundaries.

Let's try the following test case:

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

This test case includes a variety of small and large values, which should help in identifying any issues in the loop conditions and the condition checks.

However, to make it more challenging and focused, let's add a specific test case that might expose the issue in the loop conditions:

```plaintext
1
10 10
```

This test case has both `n` and `m` equal to 10, which should cover a range of values and might expose any issues in the loop conditions.

Here is the final test input:

```plaintext
1
10 10
```


**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:
/*
* Powered by ANDRIY POPYK
* Last update: 4/24/2024
*/

#include <bits/stdc++.h>

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

using namespace std;

#define int long long
#define ll long long
#define ld long double
#define pi pair<int, int>
#define ull unsigned long long

// vector
#define sz(a) (int)((a).size())
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define maxel(x) (*max_element(all(x)))
#define minel(x) (*min_element(all(x)))
#define maxpos(x) (max_element(all(x)) - x.begin())
#define minpos(x) (min_element(all(x)) - x.begin())
#define srt(x) (sort(all(x)))
#define rev(x) (reverse(all(x)))
// bitwise operations
#define cnt_bit(n) __builtin_popcountll(n)
#define low_bit(n) ((n) & (-(n)))
#define bit(n, i) (((n) >> (i)) & 1)
#define set_bit(n, i) ((n) | (1LL << (i)))
#define reset_bit(n, i) ((n) & ~(1LL << (i)))
#define flip_bit(n, i) ((n) ^ (1LL << (i)))
// math
#define sqr(n) ((n) * (n))
#define divup(a, b) (((a) + (b) - 1) / (b))
// ostream
#define Fixed(a) cout << fixed << setprecision(12) << a;

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

template<int mod> struct modint {
	int v;
	modint(int _v = 0) { v = _v % mod; if (v < 0) v += mod; }
	modint& operator+=(const modint& b) { v += b.v; if (v >= mod) v -= mod; return *this; }
	modint& operator-=(const modint& b) { v -= b.v; if (v < 0) v += mod; return *this; }
	modint& operator*=(const modint& b) { v = v * b.v % mod; return *this; }
	modint& operator/=(const modint& b) { return (*this) *= pow(b, mod - 2); }
	friend modint pow(modint a, int p) {
		modint res = 1; while (p) { if (p & 1) res *= a; a *= a; p >>= 1; }
		return res;
	}
	friend modint operator+(modint a, const modint& b) { return a += b; }
	friend modint operator-(modint a, const modint& b) { return a -= b; }
	friend modint operator*(modint a, const modint& b) { return a *= b; }
	friend modint operator/(modint a, const modint& b) { return a /= b; }
	friend istream& operator>>(istream& is, modint& a) { int x; is >> x; a = modint(x); return is; }
	friend ostream& operator<<(ostream& os, modint& a) { return os << a.v; }
};

template <typename T>
using min_queue = priority_queue <T, vector <T>, greater <T>>;
template <typename T>
using max_queue = priority_queue <T, vector <T>, less <T>>;

const int mod = 1e9 + 7; // 998244353 1e9 + 7
const ll inf = (int)(1e18) + 7;
const int inf_s = 1e9 + 7;
const ld eps = 1e-8;
using mint = modint<mod>;

template<class T> using V = vector <T>;
using vi = V<int>;
using vd = V<ld>;
using vb = V<bool>;
using vm = V<mint>;
using vpi = V<pi>;
using vvm = V<vm>;
using vvi = V<vi>;
using vvb = V<vb>;

const int B = 32;
const int maxq = 2e5 + 7;
const int logn = 20;
const int N = 2000 + 3;
const int maxn = 2e5 + 7;
//////////////////////////solve//////////////////////////

int gcd(int a, int b) {
	while (b) {
		a %= b;
		swap(a, b);
	}
	return a;
}

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

	int ans = 0;
	for (int p = 1; p <= min(N, n); ++p) {
		for (int q = 1; q <= min(N, m); ++q) {
			if (gcd(p, q) != 1) continue;
			ans += min(n / p, m / q) / (p + q);
		}
	}

	cout << ans << '\n';
}

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

#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
#endif

	int t = 1;
	cin >> t;

	while (t--)
		solve();

	cerr << '\n' << "Time: " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\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. The buggy code has a limitation in its approach, particularly in the way it handles the gcd and the loop bounds. Let's break it down:

1. **Loop Bounds**: The code uses a fixed upper limit `N` for the loops, which is set to 2000. This means it only checks pairs `(p, q)` up to 2000, which can miss valid pairs if `n` or `m` are larger than 2000.

2. **GCD Calculation**: The code correctly calculates the gcd, but it only checks pairs where the gcd is 1. This might miss valid pairs where the gcd is greater than 1.

3. **Optimization**: The code attempts to optimize by using `min(n / p, m / q) / (p + q)`, but this might not correctly account for all valid pairs, especially when `p` and `q` are large.

Given these points, a good test input would be one where `n` and `m` are larger than 2000, and there are valid pairs that the code might miss due to its loop bounds and gcd checks.

Let's consider the following test input:

```plaintext
1
2001 2001
```

This input ensures that the code's fixed loop bounds of 2000 will not cover all possible pairs, and there are valid pairs that the code might miss. This should expose the bug in the code.

Here is the test input:

```plaintext
1
2001 2001
```


**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 = long long;
#define pb push_back
//#define int long long

int n, m;

int find_kol(int l, int r, int x) {
    if (l <= r) {
        return r / x - (l - 1) / x;
    } else {
        return 0;
    }
}

const int N = 2e6;
int p[N + 1];
vector<int> primes, b;
int l, r;
int otv = 0;

void rec(int last, int sum, int tek) {
    if (last == (int) b.size()) {
        if (tek % 2 == 1) {
            otv -= find_kol(l, r, sum);
        } else {
            otv += find_kol(l, r, sum);
        }
        return;
    }
    rec(last + 1, sum, tek);
    sum *= b[last];
    tek++;
    rec(last + 1, sum, tek);
}

signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
#ifdef LOCAL
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    for (int i = 2; i <= N; ++i) {
        if (p[i] == 0) {
            p[i] = i;
            primes.pb(i);
        }
        for (int j : primes) {
            if ((ll) j * (ll) i > (ll) N || j > p[i]) {
                break;
            }
            p[i * j] = j;
        }
    }
    int t;
    cin >> t;
    while (t--) {
        cin >> n >> m;
        ll ans = 0;
        const int f = 1900;
        for (int i = 1; i <= min(n, f); ++i) {
            for (int j = 1; j <= min(n, f); ++j) {
                if (gcd(i, j) == 1) {
                    int sum = i + j;
                    int x = n / (sum * i), y = m / (sum * j);
                    ans += min(x, y);
                }
            }
        }
        for (int i = 1; i <= min(n, 1900); ++i) {
            int i1 = i;
            b.clear();
            while (i1 > 1) {
                if (b.empty() || b.back() != p[i1]) {
                    b.pb(p[i1]);
                }
                i1 /= p[i1];
            }
            l = 0;
            r = m + 1;
            while (l + 1 < r) {
                int mid = (l + r) / 2;
                ll f1 = (ll) (mid + i) * (ll) i;
                ll f2 = (ll) (mid + i) * (ll) mid;
                if (f1 <= n && f2 <= m) {
                    l = mid;
                } else {
                    r = mid;
                }
            }
            r = l;
            l = 1900 + 1;
            otv = 0;
            if (l <= r) {
                rec(0, 1, 0);
            }
            ans += otv;
        }
        cout << ans << "\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 <iostream>
#include <vector>
#include <algorithm>
#include <random>
#include <ctime>
#include <set>
#include <string>

using namespace std;

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

int main()
{
    int t;
    cin>>t;
    while (t--)
    {
        long long n,m;
        cin>>n>>m;
        long long s=0;

        for (long long i=1;i<sqrt(m);i++)
        {
            for (long long j=1;j<=m/i-i;j++)
            {
                if (gcd(i,j)==1)
                {
                    for (long long d=1;d<=m/(j*(i+j));d++)
                    {
                        if (i*(j+i)*d<=n)
                        {
                            s++;
                        }
                        else break;
                    }
                }
            }

        }

        cout<<s<<endl;
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that are likely to cause the code to fail. The buggy code has a nested loop structure that iterates over possible values of `i` and `j`, and then checks if the greatest common divisor (gcd) of `i` and `j` is 1. It then further iterates over possible values of `d` to count valid pairs.

Here are some key points to consider:
1. **Edge Cases**: Small values of `n` and `m` where the conditions are very strict.
2. **Large Values**: Values of `n` and `m` that are large enough to stress the loop structure.
3. **Special Values**: Values of `n` and `m` that are multiples of each other or have specific properties that might cause the gcd condition to be frequently true.

Given the constraints and the need for a small-scale test input, let's create a test case that includes a mix of these scenarios.

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

This test input includes:
- The smallest possible values for `n` and `m` (1, 1).
- Small values that are likely to have a few valid pairs (2, 3).
- Small values that are more challenging (3, 5).
- A moderate value that should have more valid pairs (10, 8).
- A larger value that is within the constraints but not too large (100, 1233).
- A very large value that is close to the upper limit of the constraints (1000000, 1145141).

These test cases should help expose any issues with the nested loop structure and the gcd condition 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: Computing Machine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sasha has two binary strings $s$ and $t$ of the same length $n$, consisting of the characters 0 and 1.

There is also a computing machine that can perform two types of operations on binary strings $a$ and $b$ of the same length $k$:

  1. If $a_{i} = a_{i + 2} =$ 0, then you can assign $b_{i + 1} :=$ 1 ($1 \le i \le k - 2$).    2. If $b_{i} = b_{i + 2} =$ 1, then you can assign $a_{i + 1} :=$ 1 ($1 \le i \le k - 2$).

Sasha became interested in the following: if we consider the string $a=s_ls_{l+1}\ldots s_r$ and the string $b=t_lt_{l+1}\ldots t_r$, what is the maximum number of 1 characters in the string $a$ that can be obtained using the computing machine. Since Sasha is very curious but lazy, it is up to you to answer this question for several pairs $(l_i, r_i)$ that interest him.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the strings $s$ and $t$.

The second line of each test case contains a binary string $s$ of length $n$, consisting of the characters 0 and 1.

The third line of each test case contains a binary string $t$ of length $n$, consisting of the characters 0 and 1.

The fourth line of each test case contains a single integer $q$ ($1 \le q \le 2 \cdot 10^5$) — the number of queries.

The $i$-th of the following lines contains two integers $l_{i}$ and $r_{i}$ ($1 \le l_{i} \le r_{i} \le n$) — the boundaries of the $i$-th pair of substrings that interest Sasha.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$ and the sum of $q$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output $q$ integers — the answers to all queries.

Examples:
input:
3
4
1111
0000
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
output:
2
3
2
3
1
4
3
1
2

Note: In the first test case:

  * In the first query, $a =$ 11, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 111, so the maximum number of 1 characters is $3$.

In the second test case:

  * In the first query, $a =$ 101 and $b =$ 110. No operations can be performed, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 1010 and $b =$ 1101. Since $a_2 = a_4 =$ 0, we can assign $b_3 :=$ 1. Now $b_1 = b_3 =$ 1, so we can assign $a_2 :=$ 1. The string $a$ becomes 1110, so the maximum number of 1 characters is $3$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
using ll=long long;
ll p=1e9+7;
ll N=200002;
ll MOD = 998244353;
ll fact(ll d){
    ll f=1;
    if(d==0)return f;
    while(d>=1){
        f=f*d%MOD;
        d--;
    }
    return f;
}
ll binpow(ll a, ll b){
ll res=1;
a=a%p;
while(b>0){
    if(b&1){
        res=res*a%p;
    }
    a=a*a%p;
    b>>=1;}
    return res;
}
void solve(){
int n;
cin>>n;
string s,t;
cin>>s>>t;
string u=s,o=s;
for(int i=0;i<n-2;i++){
    if(t[i]=='1'&&t[i+2]=='1')u[i+1]='1';
}
for(int i=0;i<n-2;i++){
    if(s[i]=='0'&&s[i+2]=='0')t[i+1]='1';
}
for(int i=0;i<n-2;i++){
    if(t[i]=='1'&&t[i+2]=='1')s[i+1]='1';
}
vector<int> a(n+1);
a[0]=0;
for(int i=0;i<n;i++){
    if(s[i]=='1')a[i+1]+=1+a[i];
    else a[i+1]+=a[i];
}
int q,d=0;
cin>>q;
while(q--){
        d++;
    int l,r;
    cin>>l>>r;
    int ans=a[r]-a[l-1];
    l--;
    r--;
    if(o[l]=='0'&&s[l]=='1')ans--;
     if(o[r]=='0'&&s[r]=='1'&&l!=r)ans--;
    if(u[l+1]=='0'&&s[l+1]=='1'&&l+1<r){
        if(t[l]=='0')ans--;
        else {
            if(l+3>r||o[l+3]!='0')ans--;
        }
    }
    if(u[r-1]=='0'&&l+1<r-1&&s[r-1]=='1'){
          if(t[r]=='0')ans--;
        else {
            if(r-3<l||o[r-3]!='0')ans--;
        }
    }

cout<<ans<<endl;
}
}
int main(){
      ios_base::sync_with_stdio(false);
      cin.tie(NULL);
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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can potentially cause the code to fail. Here are the steps to think through this:

1. **Edge Cases in Substring Lengths:**
   - Test with substrings of length 1.
   - Test with substrings of length 2.
   - Test with substrings of length 3.

2. **Edge Cases in String Content:**
   - Test with all 0s.
   - Test with all 1s.
   - Test with alternating 0s and 1s.

3. **Edge Cases in Queries:**
   - Test with queries that cover the entire string.
   - Test with queries that cover the start and end of the string.
   - Test with queries that cover the middle of the string.

4. **Edge Cases in Operations:**
   - Test with strings where the operations can be applied multiple times.
   - Test with strings where the operations cannot be applied.

Based on these considerations, here is a small-scale test input that should expose the bug in the buggy code:

```plaintext
3
4
0000
1111
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
```

This test input includes:
- Multiple test cases with different string lengths and content.
- Queries that cover various parts of the strings.
- Strings where the operations can be applied and where they cannot.

This 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: Computing Machine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sasha has two binary strings $s$ and $t$ of the same length $n$, consisting of the characters 0 and 1.

There is also a computing machine that can perform two types of operations on binary strings $a$ and $b$ of the same length $k$:

  1. If $a_{i} = a_{i + 2} =$ 0, then you can assign $b_{i + 1} :=$ 1 ($1 \le i \le k - 2$).    2. If $b_{i} = b_{i + 2} =$ 1, then you can assign $a_{i + 1} :=$ 1 ($1 \le i \le k - 2$).

Sasha became interested in the following: if we consider the string $a=s_ls_{l+1}\ldots s_r$ and the string $b=t_lt_{l+1}\ldots t_r$, what is the maximum number of 1 characters in the string $a$ that can be obtained using the computing machine. Since Sasha is very curious but lazy, it is up to you to answer this question for several pairs $(l_i, r_i)$ that interest him.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the strings $s$ and $t$.

The second line of each test case contains a binary string $s$ of length $n$, consisting of the characters 0 and 1.

The third line of each test case contains a binary string $t$ of length $n$, consisting of the characters 0 and 1.

The fourth line of each test case contains a single integer $q$ ($1 \le q \le 2 \cdot 10^5$) — the number of queries.

The $i$-th of the following lines contains two integers $l_{i}$ and $r_{i}$ ($1 \le l_{i} \le r_{i} \le n$) — the boundaries of the $i$-th pair of substrings that interest Sasha.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$ and the sum of $q$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output $q$ integers — the answers to all queries.

Examples:
input:
3
4
1111
0000
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
output:
2
3
2
3
1
4
3
1
2

Note: In the first test case:

  * In the first query, $a =$ 11, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 111, so the maximum number of 1 characters is $3$.

In the second test case:

  * In the first query, $a =$ 101 and $b =$ 110. No operations can be performed, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 1010 and $b =$ 1101. Since $a_2 = a_4 =$ 0, we can assign $b_3 :=$ 1. Now $b_1 = b_3 =$ 1, so we can assign $a_2 :=$ 1. The string $a$ becomes 1110, so the maximum number of 1 characters is $3$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e3 + 5;

int n, lft, rgt;
int zerone(string a, string b, int l, int r, int i) {
    lft = i - 1, rgt = i + 1;
    if (b[i - 1] == b[i + 1] and b[i + 1] == '1') return 1;
    bool lf = b[i - 1] == '1', rt = b[i + 1] == '1';
    if (l <= i - 2 and b[i - 1] == '0' and a[i - 2] == '0') lf = 1, lft = i - 2;
    if (i + 2 <= r and b[i + 1] == '0' and a[i + 2] == '0') rt = 1, rgt = i + 2;
    return lf * rt;
}
int left(string a, string b, int l, int r) {
    int rs = 0;
    for (int i = l; i <= r; ++i)
        rs += (a[i] == '0' and zerone(a, b, 1, n, i) and lft < l);

    return rs;
}
int right(string a, string b, int l, int r) {
    int rs = 0;
    for (int i = l; i <= r; ++i)
        rs += (a[i] == '0' and zerone(a, b, 1, n, i) and r < rgt);

    return rs;
}
void solve() {
    string a, b;
    cin >> n >> a >> b;

    a = '$' + a;
    b = '$' + b;
    int c_one[n + 1] = {0};
    c_one[1] = a.front() == '1';

    for (int i = 2; i < n; ++i)
        c_one[i] = c_one[i - 1] + (a[i] == '0' ? zerone(a, b, 1, n, i) : 1);
    c_one[n] = c_one[n - 1] + (a.back() == '1');

    int q, cnt = 0;
    cin >> q;
    while (q--) {
        cnt++;
        int l, r;
        cin >> l >> r;
        if (l == r) {
            cout << (a[l] == '1') << '\n';
            continue;
        }
        if (r - l + 1 <= 5) {
            if(cnt == 19706) cout << "Yes";
            int res = 0;
            for (int i = l + 1; i < r; ++i)
                res += (a[i] == '0' ? zerone(a, b, l, r, i) : 1);
            cout << ((a[l] == '1') + res + (a[r] == '1')) << '\n';
            continue;
        }
        cout << (c_one[r] - c_one[l - 1]) - (left(a, b, l, l + 5) + right(a, b, r - 5, r)) << '\n';
    }
}
/*
l   r
00000
11001
*/
int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int tc; cin >> tc; while (tc--)
        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 edge cases and scenarios that maximize the failure probability in faulty solutions. Here are the steps to think through this:

1. **Identify the Bug**: The provided code has a few potential issues:
   - The `zerone` function might not correctly handle edge cases, especially when `i - 2` or `i + 2` go out of bounds.
   - The `left` and `right` functions might not correctly handle the boundaries of the substrings.
   - The main logic in the `solve` function might not correctly handle small substrings (e.g., when `r - l + 1 <= 5`).

2. **Edge Cases**:
   - Substrings of length 1.
   - Substrings of length 2.
   - Substrings of length 3.
   - Substrings of length 4.
   - Substrings of length 5.
   - Substrings where the operations can significantly change the number of 1s.

3. **Generate Test Input**:
   - Use a small string length to ensure the test input is manageable.
   - Include a variety of substrings to test different lengths and edge cases.
   - Ensure the input is complete and valid according to the problem constraints.

Here is a small-scale test input that includes various edge cases:

```plaintext
3
5
00000
11001
2
1 1
1 5
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
4
1111
0000
2
1 2
2 4
```

This test input includes:
- A test case with a string of length 5, testing substrings of length 1 and 5.
- A test case with a string of length 6, testing substrings of lengths 2, 6, 4, 1, and 4.
- A test case with a string of length 4, testing substrings of lengths 2 and 3.

These inputs should help expose any issues in the provided code, particularly in handling small substrings and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Computing Machine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sasha has two binary strings $s$ and $t$ of the same length $n$, consisting of the characters 0 and 1.

There is also a computing machine that can perform two types of operations on binary strings $a$ and $b$ of the same length $k$:

  1. If $a_{i} = a_{i + 2} =$ 0, then you can assign $b_{i + 1} :=$ 1 ($1 \le i \le k - 2$).    2. If $b_{i} = b_{i + 2} =$ 1, then you can assign $a_{i + 1} :=$ 1 ($1 \le i \le k - 2$).

Sasha became interested in the following: if we consider the string $a=s_ls_{l+1}\ldots s_r$ and the string $b=t_lt_{l+1}\ldots t_r$, what is the maximum number of 1 characters in the string $a$ that can be obtained using the computing machine. Since Sasha is very curious but lazy, it is up to you to answer this question for several pairs $(l_i, r_i)$ that interest him.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the strings $s$ and $t$.

The second line of each test case contains a binary string $s$ of length $n$, consisting of the characters 0 and 1.

The third line of each test case contains a binary string $t$ of length $n$, consisting of the characters 0 and 1.

The fourth line of each test case contains a single integer $q$ ($1 \le q \le 2 \cdot 10^5$) — the number of queries.

The $i$-th of the following lines contains two integers $l_{i}$ and $r_{i}$ ($1 \le l_{i} \le r_{i} \le n$) — the boundaries of the $i$-th pair of substrings that interest Sasha.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$ and the sum of $q$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output $q$ integers — the answers to all queries.

Examples:
input:
3
4
1111
0000
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
output:
2
3
2
3
1
4
3
1
2

Note: In the first test case:

  * In the first query, $a =$ 11, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 111, so the maximum number of 1 characters is $3$.

In the second test case:

  * In the first query, $a =$ 101 and $b =$ 110. No operations can be performed, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 1010 and $b =$ 1101. Since $a_2 = a_4 =$ 0, we can assign $b_3 :=$ 1. Now $b_1 = b_3 =$ 1, so we can assign $a_2 :=$ 1. The string $a$ becomes 1110, so the maximum number of 1 characters is $3$.

Buggy Code:
#include <bits/stdc++.h>
#define int long long
#define mp make_pair
#define pb push_back
#define ld long double
#define pii pair<int,int>
#define sz(x) (int)x.size()
#define piii pair<pii,pii>
#define precise cout<<fixed<<setprecision(10)
#define st first
#define nd second
#define ins insert
#define vi vector<int>
#define BOOST ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
using namespace std;
const int MAX=2e5+10;
int pref[MAX];
char a[MAX];
char b[MAX];
char c[MAX];
char d[MAX];
int sum(int l,int r){
  if (l>r)return 0;
  return pref[r]-pref[l-1];
}
void solve(){
  int n;
  cin>>n;
  for (int i=1;i<=n;i++)cin>>a[i];
  for (int i=1;i<=n;i++)cin>>b[i];
  for (int i=1;i<=n;i++){
    c[i]=b[i];
    d[i]=a[i];
  }
  for (int i=1;i<=n-2;i++){
    if (a[i]=='0' && a[i+2]=='0')c[i+1]='1';
  }

  for (int i=1;i<=n-2;i++){
    if (c[i]=='1' && c[i+2]=='1')d[i+1]='1';
  }

  for (int i=1;i<=n;i++)pref[i]=pref[i-1]+(d[i]=='1');

  int q;
  cin>>q;
  for (int z=0;z<q;z++){
    int l,r;
    cin>>l>>r;
    int ans=sum(l+5,r-5);
    set<int>ids;
    for (int i=l;i<=min(r,l+4);i++)ids.ins(i);
    for (int i=r;i>=max(l,r-4);i--)ids.ins(i);
    for (int i=l;i<=min(r+1,l+5);i++)c[i]=b[i],d[i]=a[i];
    for (int i=r;i>=max(l-1,r-5);i--)c[i]=b[i],d[i]=a[i];

    for (int i=l;i<=min(r-2,l+5);i++){
      if (a[i]=='0' && a[i+2]=='0')c[i+1]='1';
    }

   // for (int i=l;i<=r;i++)cout<<c[i];
   // cout<<"\n";

    for (int i=l;i<=min(r-2,l+5);i++){
      if (c[i]=='1' && c[i+2]=='1')d[i+1]='1';
    }

    for (int i=r;i>=max(l+2,r-5);i--){
      if (a[i]=='0' && a[i+2]=='0')c[i+1]='1';
    }

    for (int i=r;i>=max(l+2,r-5);i--){
      if (c[i]=='1' && c[i+2]=='1')d[i+1]='1';
    }
    for (auto it:ids){
      //cout<<"TERAZ "<<it<<" "<<d[it]<<"\n";
      ans+=(d[it]=='1');
    }
    cout<<ans<<"\n";
  }
}
int32_t main()
{
  BOOST;
  int t;
  cin>>t;
  for (int z=0;z<t;z++)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
6
000000
111111
2
1 6
2 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Computing Machine
time_limit_ms: 2000
memory_limit_mb: 256
Description: Sasha has two binary strings $s$ and $t$ of the same length $n$, consisting of the characters 0 and 1.

There is also a computing machine that can perform two types of operations on binary strings $a$ and $b$ of the same length $k$:

  1. If $a_{i} = a_{i + 2} =$ 0, then you can assign $b_{i + 1} :=$ 1 ($1 \le i \le k - 2$).    2. If $b_{i} = b_{i + 2} =$ 1, then you can assign $a_{i + 1} :=$ 1 ($1 \le i \le k - 2$).

Sasha became interested in the following: if we consider the string $a=s_ls_{l+1}\ldots s_r$ and the string $b=t_lt_{l+1}\ldots t_r$, what is the maximum number of 1 characters in the string $a$ that can be obtained using the computing machine. Since Sasha is very curious but lazy, it is up to you to answer this question for several pairs $(l_i, r_i)$ that interest him.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 10^{4}$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the strings $s$ and $t$.

The second line of each test case contains a binary string $s$ of length $n$, consisting of the characters 0 and 1.

The third line of each test case contains a binary string $t$ of length $n$, consisting of the characters 0 and 1.

The fourth line of each test case contains a single integer $q$ ($1 \le q \le 2 \cdot 10^5$) — the number of queries.

The $i$-th of the following lines contains two integers $l_{i}$ and $r_{i}$ ($1 \le l_{i} \le r_{i} \le n$) — the boundaries of the $i$-th pair of substrings that interest Sasha.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$ and the sum of $q$ over all test cases does not exceed $2 \cdot 10^5$.

Output: For each test case, output $q$ integers — the answers to all queries.

Examples:
input:
3
4
1111
0000
2
1 2
2 4
4
1010
1101
2
1 3
1 4
6
010101
011010
5
2 3
1 6
2 5
4 4
3 6
output:
2
3
2
3
1
4
3
1
2

Note: In the first test case:

  * In the first query, $a =$ 11, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 111, so the maximum number of 1 characters is $3$.

In the second test case:

  * In the first query, $a =$ 101 and $b =$ 110. No operations can be performed, so the maximum number of 1 characters is $2$.    * In the second query, $a =$ 1010 and $b =$ 1101. Since $a_2 = a_4 =$ 0, we can assign $b_3 :=$ 1. Now $b_1 = b_3 =$ 1, so we can assign $a_2 :=$ 1. The string $a$ becomes 1110, so the maximum number of 1 characters is $3$.

Buggy Code:
import heapq
import math
'''
БИНАРНЫЙ ПОИСК
for i in range(strlen):
        left = 0
        right = alphlen
        while (left < right):
            mid = (left + right) // 2
            if alphabet[mid] < b[i]:
                left = mid + 1
            else:
                right = mid
'''

for _ in range(int(input())):
    n = int(input())
    s = [int(ch) for ch in input()]
    t = [int(ch) for ch in input()]
    s_changed = list(s)
    t_changed = list(t)
    q = int(input())
    inds = []
    for __ in range(q):
        inds.append(tuple(map(int, input().split())))
    for i in range(n-2):
        if s[i] == 0 and s[i+2] == 0:
            t_changed[i+1] = 1

    for i in range(n-2):
        if t_changed[i] == 1 and t_changed[i+2] == 1:
            s_changed[i+1] = 1

    cumsum = [0]
    cur_sum = 0
    for i in range(n):
        cur_sum += s_changed[i]
        cumsum.append(cur_sum)

    for j in range(q):
        left, right = inds[j]
        basic_counter = cumsum[right] - cumsum[left-1]
        #print(basic_counter)
        res = basic_counter - (s_changed[left-1] != s[left-1])
        if right == left:
            print(res)
        else:
            res -= (s_changed[right-1] != s[right-1])
            if right - left == 1:
                print(res)
            else:
                check = 0
                if left > 1 and s_changed[left] != s[left] and t_changed[left-1] != t[left-1]:
                    res -= 1
                    check += 1
                if right < n and s_changed[right-2] != s[right-2] and t_changed[right-1] != t[right-1]:
                    res -= 1
                    check += 1
                if left - right == 2 and check == 2:
                    res += 1
                print(max(0, res))

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

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

Think step by step.

Response:
```plaintext
3
4
1010
1101
2
1 3
2 4
4
0000
1111
2
1 4
2 4
6
101010
010101
5
1 3
2 4
3 5
4 6
1 6
```