**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$x \oplus y$$$ is divisible$$$^{\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

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

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
7 10
2 3
6 4
1 6
4 1
output:
3
2
2
6
1

Note: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.

- $$$y = 1$$$ is valid because $$$x \oplus y = 7 \oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 7 \oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.
- $$$y = 9$$$ is valid because $$$x \oplus y = 7 \oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.

Buggy Code:
#include<iostream>
#include<algorithm>
#include<string.h>
#include<vector>
#include<numeric>
#include<map>
#include<set>
#define ll long long
#define inf 0x7FFFFFFF
using namespace std;
int cal(int x) {
	int ans=0;
	while (x) {
		x=x >> 1;
		ans++;
	}
	return ans;
}
void solve()
{
	ll x, m;
	cin >> x >> m;
	ll n = cal(x);
	int t = 1 << (n + 1) - 1;
	t = t & m;
	int mx = 0;
	for (int i = 0; i <= t; i++) {
		mx = max(mx, i ^ t);
	}
	n = m - t + mx;
	ll ans = max(m,n) / x;
	for (int i = 1; i < min(x,m+1); i++) {
		int temp = x ^ i;
		ans += (temp % i == 0 && temp % x != 0);
	}
	cout << min(ans,m) << endl;
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int T;
	cin >> T;
	while (T--) {
		solve();
	}
	return 0;
}
//111111110001
//        1000

**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 issues. Here are the steps to think through this:

1. **Understanding the Buggy Code**:
   - The function `cal(int x)` calculates the number of bits in the binary representation of `x`.
   - The variable `t` is calculated as `1 << (n + 1) - 1`, which is a power of 2 minus 1.
   - The loop `for (int i = 0; i <= t; i++)` is intended to find the maximum value of `i ^ t`.
   - The variable `n` is then calculated as `m - t + mx`.
   - The final answer is calculated using a combination of these values and a loop to check additional conditions.

2. **Identifying Potential Bugs**:
   - The logic in the loop `for (int i = 0; i <= t; i++)` might be incorrect or inefficient.
   - The condition `temp % i == 0 && temp % x != 0` might not cover all valid cases.
   - The final answer calculation might be flawed, especially when `m` is very large.

3. **Generating a Challenging Test Input**:
   - **Edge Case 1**: `x = 1` and `m` is a large value. This should return `m` because every `y` from 1 to `m` is valid.
   - **Edge Case 2**: `x` is a power of 2 and `m` is a large value. This should test the bitwise operations and the loop logic.
   - **Edge Case 3**: `x` and `m` are both large values, but `x` is not a power of 2. This should test the general logic and edge cases.

Based on these considerations, a small-scale but challenging test input could be:

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

This input includes:
- `x = 1` and `m = 10`, which should return `10` because every `y` from 1 to 10 is valid.
- `x = 2` and `m = 10`, which should return `5` because the valid `y` values are 1, 2, 3, 6, and 10.
- `x = 3` and `m = 10`, which should return `3` because the valid `y` values are 1, 3, and 9.

This input will help expose any issues in the buggy code's handling of different values of `x` and `m`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$x \oplus y$$$ is divisible$$$^{\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

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

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
7 10
2 3
6 4
1 6
4 1
output:
3
2
2
6
1

Note: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.

- $$$y = 1$$$ is valid because $$$x \oplus y = 7 \oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 7 \oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.
- $$$y = 9$$$ is valid because $$$x \oplus y = 7 \oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.

Buggy Code:
for __ in range(int(input())):

    x, m = map(int, input().split())
    count = 0
    y = 1
    while y <= min(6*x,m):
        if ((x^y)%x ==0) or ((x^y)%y == 0):
            count += 1
        y += 1

    print(count)

**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 or scenarios that maximize the failure probability. The buggy code has a loop that iterates up to `min(6*x, m)`, which is not always sufficient to cover all possible values of `y` up to `m`. This can lead to incorrect results when `m` is much larger than `6*x`.

### Step-by-Step Analysis:
1. **Loop Limitation**: The loop only goes up to `min(6*x, m)`, which means if `m` is significantly larger than `6*x`, the loop will terminate early, potentially missing valid `y` values.
2. **Edge Cases**:
   - **Small `x` and large `m`**: This is a critical scenario where the loop will terminate early, missing many valid `y` values.
   - **Special values of `y`**: Values where `x^y` is divisible by `x` or `y` might be missed if the loop limit is too small.

### Test Input:
Let's choose a small `x` and a large `m` to ensure the loop limit is significantly smaller than `m`. For example, `x = 1` and `m = 10^18`:

```plaintext
1
1 1000000000000000000
```

### Explanation:
- **Test Case**: `1 1000000000000000000`
- **Expected Behavior**:
  - For `x = 1`, every `y` from `1` to `10^18` should be valid because `1^y` is always `y-1`, and `y-1` is divisible by `y-1`.
  - Therefore, the expected output is `10^18`.

### Why This Test Input is Challenging:
- The loop in the buggy code will only go up to `6*1 = 6`, which is much smaller than `10^18`.
- This means the loop will terminate early, and the code will miss most of the valid `y` values, leading to an incorrect result.

### Output:
```plaintext
1
1 1000000000000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$x \oplus y$$$ is divisible$$$^{\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

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

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
7 10
2 3
6 4
1 6
4 1
output:
3
2
2
6
1

Note: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.

- $$$y = 1$$$ is valid because $$$x \oplus y = 7 \oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 7 \oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.
- $$$y = 9$$$ is valid because $$$x \oplus y = 7 \oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.

Buggy Code:
from math import *
from collections import *
from bisect import bisect_left, bisect_right, bisect
from itertools import *

R=lambda:map(int,input().split())
S=lambda:map(str,input().split())

MOD = int(1e9 + 7)
# MOD = 998244353
# MOD = 100000000669

t, = R()
# t = 1

def answer():
    x, m = R()

    if x == 1:
        return m

    cnt = 0
    for i in range(4 * x + 1):
        y = i ^ x

        # print(" .. ", i, y)

        # print(y>= 1)
        # print(y <= m)

        if y >= 1 and y <= m and (i % x == 0 or i % y == 0):
            # print(" -- ", i, y)
            cnt += 1

    return cnt

for i in range(t):
    print(answer())

**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 where the algorithm might fail. The code has a loop that iterates up to `4 * x + 1`, which is a potential source of error, especially for large values of `x` and `m`.

### Step-by-Step Analysis:

1. **Edge Case: \( x = 1 \)**:
   - The code correctly handles the case where \( x = 1 \) by returning \( m \). This is a straightforward case and is unlikely to expose a bug.

2. **Edge Case: \( x \) is a large number**:
   - The loop in the code iterates up to `4 * x + 1`. For large values of `x`, this loop can be very inefficient and might miss valid `y` values if `m` is much larger than `4 * x`.

3. **Edge Case: \( m \) is a large number**:
   - If `m` is much larger than `4 * x`, the loop might not cover all valid `y` values, leading to an incorrect count.

4. **Edge Case: \( x \) and \( m \) are powers of 2**:
   - Powers of 2 can have specific properties in bitwise operations, which might cause the loop to miss valid `y` values.

5. **Edge Case: \( x \) is a prime number**:
   - Prime numbers can have unique properties in divisibility, which might cause the loop to miss valid `y` values.

### Chosen Test Input:
Let's choose a test input where \( x \) is a large number and \( m \) is much larger than \( 4 * x \). This will maximize the chances of the loop missing valid `y` values.

```plaintext
1
1000000 1000000000000
```

This test input sets \( x = 1000000 \) and \( m = 10^{12} \), which is much larger than \( 4 * x \). This will test the loop's efficiency and coverage, potentially exposing the bug in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$x \oplus y$$$ is divisible$$$^{\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

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

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
7 10
2 3
6 4
1 6
4 1
output:
3
2
2
6
1

Note: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.

- $$$y = 1$$$ is valid because $$$x \oplus y = 7 \oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 7 \oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.
- $$$y = 9$$$ is valid because $$$x \oplus y = 7 \oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.

Buggy Code:
import string
import sys
from collections import deque

input = lambda: sys.stdin.readline().rstrip()

# k, n = map(int, input().split())
for _ in range(int(input())):
    x, m = map(int, input().split())
    if m == 1:
        print(1)
        continue
    ans = 0
    for i in range(1, min(x + 1, m + 1)):
        if (x ^ i) % i == 0 or (x ^ i) % x == 0:
            ans += 1
    cut = m // x * x
    ans += m // x - (x < cut)
    if 1 <= (x ^ cut) <= m:
        ans += 1
    cut += x
    if 1 <= (x ^ cut) <= m:
        ans += 1
    if m > x:
        ans -= 1
    print(ans)

# x = 17
# print(bin(x))
# for i in range(2, 200):
#     if (x ^ i) % i == 0 or (x ^ i) % x == 0:
#         print(i, bin(i), x ^ i, bin(x ^ i))

**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
17 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$x \oplus y$$$ is divisible$$$^{\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

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

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
7 10
2 3
6 4
1 6
4 1
output:
3
2
2
6
1

Note: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.

- $$$y = 1$$$ is valid because $$$x \oplus y = 7 \oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 7 \oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.
- $$$y = 9$$$ is valid because $$$x \oplus y = 7 \oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.

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

public class Main
{
    public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    public static  BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
    public static StringTokenizer tokenizer;
    public static void read() throws IOException {
        tokenizer = new StringTokenizer(br.readLine());
    }
    public static int Int() throws IOException {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) read();
        return Integer.parseInt(tokenizer.nextToken());
    }
    public static long Long() throws IOException {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) read();
        return Long.parseLong(tokenizer.nextToken());
    }
    public static String Str() throws IOException {
        return br.readLine();
    }
    //-----------------------------------------------

    public static long quickPow(long b, long x, long mod) {
        if (x == 0) {
            return 1;
        }
        if (x == 1) {
            return b % mod;
        }
        long half = quickPow(b, x / 2, mod) % mod;
        long result = (half * half) % mod;
        result = (result * quickPow(b, x % 2, mod)) % mod;
        return result;
    }

    public static int check (long x, long m) {
        int count = 0;
        for (long y = 1; y <= m; y++) {
            long xor = x ^ y;
            if (xor != x && xor != 0 && (x % xor == 0 || y % xor == 0)) {
                count++;
            }
        }
        return count;
    }

    public static boolean f(long mult, long x, long m) {
        long max = mult * x;
        long val = 0;
        for (long i = 64; i >= 0; i--) {
            if ((max & (1L << i)) == 0 && (x & (1L << i)) > 0) {
                val |= (1L << i);
            }
            if ((max & (1L << i)) > 0 && (x & (1L << i)) == 0) {
                val |= (1L << i);
            }
        }
        return val <= m;
    }
    public static long bin(long x, long m) {
        long mult = 0;
        for (long i = 60; i >= 0; i--) {
            long maybeMult = mult + (1L << i);
            if (f(maybeMult, x, m)) {
                mult = maybeMult;
            }
        }
        return mult;
    }
    public static void main (String[] args) throws IOException {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        Random random = new Random();
        while (t > 0) {
            t--;
            long x = scanner.nextLong(), m = scanner.nextLong();
            long count = 0;
            for (long y = 1; y <= Math.min(x, m); y++) {
                long xor = x ^ y;
                if ((xor % y == 0) && (xor % x != 0) && xor <= m) {
                    count++;
                }
            }

            long mult = bin(x, m);
            count += mult;
            for (int i = 1; i <= 10; i++) {
                if (f(mult + i, x, m)) {
                    count++;
                }
            }
            System.out.println(count);
        }
    }
}

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

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

Think step by step.

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

1. **Edge Cases for `x` and `m`**:
   - Small values of `x` and `m`.
   - Large values of `x` and `m`.
   - Cases where `x` is a power of 2.
   - Cases where `m` is a power of 2.
   - Cases where `x` and `m` are equal.

2. **Special Values**:
   - `x = 1` and `m = 10` (since `x = 1` should always be valid for all `y`).
   - `x = 2` and `m = 3` (since `x = 2` has a specific pattern).
   - `x = 7` and `m = 10` (from the example, already known to be valid).

3. **Edge Cases for the Loop**:
   - Values that might cause the loop to run for a long time.
   - Values that might cause overflow in the bitwise operations.

Based on these considerations, let's create a test input that includes these edge cases:

```plaintext
4
1 10
2 3
7 10
1000000 1000000000000000000
```

This test input includes:
- A small `x` (1) and a medium `m` (10).
- A small `x` (2) and a small `m` (3).
- A small `x` (7) and a medium `m` (10).
- A large `x` (1,000,000) and a very large `m` (10^18).

These cases should help expose any bugs in the code, especially those related to edge cases and large values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$x \oplus y$$$ is divisible$$$^{\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

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

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
7 10
2 3
6 4
1 6
4 1
output:
3
2
2
6
1

Note: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.

- $$$y = 1$$$ is valid because $$$x \oplus y = 7 \oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 7 \oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.
- $$$y = 9$$$ is valid because $$$x \oplus y = 7 \oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.

Buggy Code:
import java.util.*;

public class Main {

    // Helper method to find GCD of two numbers
    public static int findGCD(int a, int b) {
        if (b == 0) {
            return a;
        }
        return findGCD(b, a % b);
    }

    // Method to find GCD of an ArrayList
    public static int findGCDOfArrayList(ArrayList<Integer> numbers) {
        if (numbers == null || numbers.isEmpty()) {
            return 0;
        }

        int gcd = numbers.get(0); // Start with the first element
        for (int i = 1; i < numbers.size(); i++) {
            gcd = findGCD(gcd, numbers.get(i));
            if (gcd == 1) { // Early exit if GCD becomes 1
                return 1;
            }
        }

        return gcd;
    }

    // Method to find all divisors of a number
    public static ArrayList<Integer> findDivisors(int number) {
        ArrayList<Integer> divisors = new ArrayList<>();
        for (int i = 1; i <= Math.sqrt(number); i++) {
            if (number % i == 0) {
                divisors.add(i);
                if (i != number / i) {
                    divisors.add(number / i);
                }
            }
        }
        Collections.sort(divisors); // Optional: Sort the divisors
        return divisors;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt(); // Number of test cases

        while (t-- > 0) {
            long x = sc.nextInt();
            long m = sc.nextLong();
            // ArrayList<Integer> div = findDivisors(x);
            // HashSet<Integer> hs = new HashSet<>();
            int cnt = 0;
            if(x<m){
                for(long i=1;i<=Math.min(m,2*x+10);i++){
                    if( (((i^x)%i)==0 || ((i^x)%x)==0)) cnt++;
                }

            }
            else{
                for(long i=1;i<=m;i++){
                    if( (((i^x)%i)==0 || ((i^x)%x)==0)) cnt++;
                }
            }

            System.out.println(cnt);
        }
        sc.close();
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases that might cause the algorithm to fail. Here are the key points to consider:

1. **Small Values of `x` and `m`**: These can help us verify the basic logic.
2. **Large Values of `m`**: These can test the performance and correctness of the algorithm, especially when `m` is much larger than `x`.
3. **Edge Cases for `x` and `m`**: Values like `x = 1` or `m = 1` can reveal issues with boundary conditions.
4. **Values of `x` that have many divisors**: These can test the logic that checks divisibility.
5. **Values of `x` that are powers of 2**: These can test the bitwise operations and their effects on the results.

Given these considerations, let's construct a test input that includes a mix of these scenarios:

```plaintext
4
1 10
2 10
7 10
1024 1000000000000
```

### Explanation:
1. **Test Case 1: `1 10`**
   - `x = 1` and `m = 10` should yield 10 valid `y` values (all `y` from 1 to 10).
2. **Test Case 2: `2 10`**
   - `x = 2` and `m = 10` should yield 2 valid `y` values (2 and 10).
3. **Test Case 3: `7 10`**
   - `x = 7` and `m = 10` should yield 3 valid `y` values (1, 7, and 9).
4. **Test Case 4: `1024 1000000000000`**
   - `x = 1024` and `m = 1000000000000` should test the performance and correctness for large values. This case can expose issues with the loop and the logic for checking divisibility.

This test input is designed to cover a range of scenarios that can help identify potential bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$x \oplus y$$$ is divisible$$$^{\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

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

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
7 10
2 3
6 4
1 6
4 1
output:
3
2
2
6
1

Note: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.

- $$$y = 1$$$ is valid because $$$x \oplus y = 7 \oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 7 \oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.
- $$$y = 9$$$ is valid because $$$x \oplus y = 7 \oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.

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

using namespace std;
#define int long long
#define inf INT32_MAX
//#define x first
//#define y second
#define endl '\n'

void solve() {
    int n, m;
    cin >> n >> m;
    if (n == 1) {
        cout << m << endl;
        return;
    }
    int ans = 0;
    for (int i = 1; i <= min(n, m); i++) {//m作为因子
        if ((i ^ n) % i == 0 || (i ^ n) % n == 0) {
            ans++;
        }
    }
    int num = m | n;
    ans += max(num / n - 1, 0ll);
    for (int i = num / n * n; i >= 2 * n; i -= n) {
        if ((num ^ n) > m)ans--;
    }
    cout << ans << endl;
}

signed main() {
#ifdef ONLINE_JUDGE
#else
    freopen("test.in", "r", stdin);
    freopen("test.out", "w", stdout);
#endif
    ios::sync_with_stdio(0);
    cout.tie(0);
    cin.tie(0);
    int t = 1;
    cin >> t;
    while (t--)
        solve();
    return 0;
}

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

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

Think step by step.

Response:
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. Let's break down the potential issues and the corresponding test inputs:

1. **Edge Case: \( x = 1 \)**:
   - When \( x = 1 \), the code correctly handles this case by outputting \( m \). However, this is not a challenging case for the buggy code.

2. **Edge Case: \( m < n \)**:
   - If \( m \) is less than \( n \), the loop in the buggy code might not run as expected. This can lead to incorrect results.

3. **Edge Case: \( m \) is very large**:
   - Large values of \( m \) can cause the loop to run for a long time, potentially leading to performance issues or incorrect results due to the way the code handles large numbers.

4. **Edge Case: \( n \) is a power of 2**:
   - When \( n \) is a power of 2, the bitwise XOR operation has specific properties that might expose bugs in the code.

5. **Edge Case: \( n \) and \( m \) are both large and \( n \) is a multiple of \( m \)**:
   - This can lead to complex interactions in the loop and the bitwise XOR operation.

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

```plaintext
3
1 10
1024 2048
1000000 1000000
```

This test input includes:
- A simple case where \( x = 1 \) and \( m = 10 \).
- A case where \( x = 1024 \) (a power of 2) and \( m = 2048 \).
- A case where \( x \) and \( m \) are both large and \( x \) is a multiple of \( m \).

These test cases should help expose any bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$x \oplus y$$$ is divisible$$$^{\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

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

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
7 10
2 3
6 4
1 6
4 1
output:
3
2
2
6
1

Note: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.

- $$$y = 1$$$ is valid because $$$x \oplus y = 7 \oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 7 \oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.
- $$$y = 9$$$ is valid because $$$x \oplus y = 7 \oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.

Buggy Code:
// Author : hejinming2012
#include <bits/stdc++.h>
#define int long long
#define endl '\n'
#define dbg(x) cout << #x " = " << (x) << endl
#define quickio ios::sync_with_stdio(false);
#define quickin cin.tie(0);
#define quickout cout.tie(0);

using namespace std;
inline int read() {
    int now = 0, nev = 1; char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') nev = -1; c = getchar(); }
    while(c >= '0' && c <= '9') { now = (now << 1) + (now << 3) + (c & 15); c = getchar(); }
    return now * nev;
}
void write(int x) {
    if(x < 0) putchar('-'), x = -x;
    if(x > 9) write(x / 10);
    putchar(x % 10 + '0');
}
vector <int> f(int x) {
    vector <int> res;
    for(int i = 1; i * i <= x; i++)
        if(x % i == 0) {
            res.push_back(i);
            if(x / i != i) res.push_back(x / i);
        }
    return res;
}
signed main() {
    quickio
    quickin
    quickout
    int T = read();
    while(T--) {
        int x = read(), m = read(), ans = 0;
        for(int i = 1; i <= 60; i++) {
            int j = x * (1ll << i - 1);
            if(j > m + x) break ;
        }
        for(int i = 1; i <= 60; i++) {
            if(x != 0 && i > m / x + 2) break ;
            int z = x * i, y = x ^ z;
            if(y >= 1 && y <= m) ans++;
        }
        vector <int> divs = f(x);
        for(auto i : divs) {
            if(i > m) continue ;
            int z = x ^ i;
            if(z % i == 0) ans++;
        }
        write(ans), putchar(10);
    }
    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 can maximize the failure probability. Here are the steps to consider:

1. **Edge Cases for `x` and `m`**:
   - `x = 1` and `m` is a large number.
   - `x` is a large number and `m` is small.
   - `x` is a large number and `m` is also large.
   - `x` is a small number and `m` is a large number.

2. **Special Values**:
   - `x` is a power of 2.
   - `x` is a prime number.
   - `x` is a composite number.

3. **Boundary Conditions**:
   - `m` is just slightly larger than `x`.
   - `m` is just slightly smaller than `x`.

4. **Multiple Test Cases**:
   - Include a mix of the above cases to ensure the code is thoroughly tested.

Given these considerations, a small-scale test input that includes a variety of edge cases is:

```plaintext
5
1 1000000000000
2 3
7 10
999999 1000000
1048576 1048576
```

This test input includes:
- A case where `x = 1` and `m` is very large.
- A case where `x` is a small prime number.
- A case where `x` is a small composite number.
- A case where `x` is a large number and `m` is just slightly larger.
- A case where `x` is a power of 2 and `m` is equal to `x`.

These cases should help expose any bugs in the algorithm, especially in the handling of large numbers and special values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$x \oplus y$$$ is divisible$$$^{\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

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

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
7 10
2 3
6 4
1 6
4 1
output:
3
2
2
6
1

Note: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.

- $$$y = 1$$$ is valid because $$$x \oplus y = 7 \oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 7 \oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.
- $$$y = 9$$$ is valid because $$$x \oplus y = 7 \oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.

Buggy Code:
def kq(x, m):
    t=0
    for y in range(1,min(m,2*x)+1):
        xor=x^y
        if (xor%x==0) or (y!=0 and xor%y==0):
            t+=1
    if m>2*x:
        t+=(m-2*x)//x
    return t
for _ in range(int(input())):
    x,m=map(int,input().split())
    print(kq(x,m))

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

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

Think step by step.

Response:
```plaintext
3
1 10
2 10
7 10
```

### Explanation:
1. **Test Case 1: `1 10`**
   - For `x = 1`, the possible values of `y` from 1 to 10 are all valid because `x ⊕ y` will always be a number that is divisible by 1.
   - Expected output: 10

2. **Test Case 2: `2 10`**
   - For `x = 2`, the valid values of `y` from 1 to 10 are 1, 2, 3, 6.
   - Expected output: 4
   - The buggy code might fail here because it does not correctly handle the case where `m` is much larger than `2*x`.

3. **Test Case 3: `7 10`**
   - For `x = 7`, the valid values of `y` from 1 to 10 are 1, 7, 9.
   - Expected output: 3
   - This test case is included to ensure the algorithm correctly handles the XOR operation and the divisibility check.

These test cases are designed to cover edge cases and scenarios that can expose the bug in the given code, especially where the range of `y` is larger than `2*x` and the divisibility conditions need to be checked accurately.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$x \oplus y$$$ is divisible$$$^{\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

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

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
7 10
2 3
6 4
1 6
4 1
output:
3
2
2
6
1

Note: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.

- $$$y = 1$$$ is valid because $$$x \oplus y = 7 \oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 7 \oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.
- $$$y = 9$$$ is valid because $$$x \oplus y = 7 \oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.

Buggy Code:
import math
t=int(input())

for _ in range(t):
    n,m=map(int,input().split())
    p=(m-n)//n

    j=math.ceil((m-n)/n)
    if(j<0):
        j=0
    if((n^(j*n))<=m):
       p+=1
    if((n^((j+1)*n))<=m):
       p+=1

    for x in range(1,min(n,m+1)):
        if(((x^n)%x)==0):
            p+=1
    print(p)

**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, let's consider the following points:

1. **Edge Cases:**
   - When \( x = 1 \): This is a special case because \( x \oplus y \) will always be \( y \), and \( y \) is always divisible by \( y \).
   - When \( m \) is very small: This can help in testing the boundary conditions.
   - When \( m \) is very large: This can test the performance and correctness of the algorithm.

2. **Key Scenarios:**
   - When \( m \) is close to \( n \): This can expose issues with the logic that checks the range of valid \( y \).
   - When \( m \) is a multiple of \( n \): This can test the logic for divisibility by \( n \).
   - When \( n \) is a power of 2: This can test the bitwise XOR operations and the divisibility conditions.

3. **Combining the Above:**
   - We will create a test case where \( x \) is a power of 2 and \( m \) is close to \( n \).
   - We will also create a test case where \( x \) is a small number and \( m \) is a large number.

Given these considerations, let's create the test input:

```plaintext
4
1 10
8 10
1000000 1000000
2 1000000000000
```

This test input includes:
- A test case where \( x = 1 \) and \( m = 10 \).
- A test case where \( x = 8 \) and \( m = 10 \).
- A test case where \( x = 1000000 \) and \( m = 1000000 \).
- A test case where \( x = 2 \) and \( m \) is a very large number.

These cases should help in exposing any bugs in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$x \oplus y$$$ is divisible$$$^{\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

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

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
7 10
2 3
6 4
1 6
4 1
output:
3
2
2
6
1

Note: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.

- $$$y = 1$$$ is valid because $$$x \oplus y = 7 \oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 7 \oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.
- $$$y = 9$$$ is valid because $$$x \oplus y = 7 \oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.

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

public class Main {
boolean MULTI_CASE = true;
boolean ALWAYS_FLUSH = false;

void go() {
    int x = nextInt();
    long m = nextLong();
//    long r1 = 0;
//    for (int i = 1; i <= m; i++) {
//        int t = x ^ i;
//        if (t % i == 0 || t % x == 0) {
//            r1++;
//        }
//    }
//    H.out(r1);
    long res = 0;
    long top = highBit1(x) << 2;
    if (m < top * 2L) {
        for (int i = 1; i <= m; i++) {
            int t = x ^ i;
            if (t % i == 0 || t % x == 0) {
                res++;
            }
        }
    } else {
        long t1 = top / 2 / x;
//        long lo1 = 0x3f3f3f3f3f3f3f3fL;
//        long hi1 = 0;
        for (int i = 1; i < top; i++) {
            int t = x ^ i;
            if (t % i == 0) {
                res++;
            } else if (t % x == 0) {
                if (t <= t1) res++;
//                lo1=Math.min(lo1, t/x);
//                hi1 = Math.max(hi1, t/x);
            }
        }
        long t2 = (m - top / 2) / x;
//        long lo2 = 0x3f3f3f3f3f3f3f3fL;
//        long hi2 = 0;
        for (long i = m - top; i <= m; i++) {
            long t = x ^ i;
            if (t % i == 0) {
                res++;
            } else if (t % x == 0) {
                if (t >= t2) res++;
//                lo2=Math.min(lo2, t/x);
//                hi2 = Math.max(hi2, t/x);
            }
        }
        res += t2 - t1 - 1;
    }
    sl(res);
}

long highBit1(int x) {
    // 11010 -> 10000
    x |= x >> 1;
    x |= x >> 2;
    x |= x >> 4;
    x |= x >> 8;
    x |= x >> 16;
    x = (x + 1) >> 1;
    return x;
}

/**
 * Input.
 */
InputStream inStream;
byte[] inBuff = new byte[1024];
int inBuffCursor = 0, inBuffLen = 0;

boolean isVisibleChar(int c) {
    return 33 <= c && c <= 126;
}

boolean hasNextByte() {
    if (inBuffCursor < inBuffLen) return true;
    inBuffCursor = 0;
    try {
        inBuffLen = inStream.read(inBuff);
    } catch (Exception e) {
        e.printStackTrace(System.out);
    }
    return inBuffLen > 0;
}

boolean hasNext() {
    while (hasNextByte() && !isVisibleChar(inBuff[inBuffCursor])) inBuffCursor++;
    return hasNextByte();
}

int nextByte() {
    return hasNextByte() ? inBuff[inBuffCursor++] : -1;
}

String next() {
    if (!hasNext()) throw new RuntimeException("no next.");
    StringBuilder sb = new StringBuilder();
    int b = nextByte();
    while (isVisibleChar(b)) {
        sb.appendCodePoint(b);
        b = nextByte();
    }
    return sb.toString();
}

long nextLong() {
    if (!hasNext()) throw new RuntimeException("no next.");
    long result = 0;
    boolean negative = false;
    int b = nextByte();
    if (b < '0') {
        if (b == '-') negative = true;
        else if (b != '+') throw new RuntimeException("long number must start with +/-.");
        b = nextByte();
    }
    while (isVisibleChar(b)) {
        if (b < '0' || b > '9') throw new RuntimeException("wrong digit in long:" + (char) b);
        // NOTE: may overflow here, even if it is a valid Long, eg.-(1<<63).
        result = result * 10 + (b - '0');
        b = nextByte();
    }
    return negative ? -result : result;
}

int nextInt() {
    long x = nextLong();
    if (x < Integer.MIN_VALUE || x > Integer.MAX_VALUE)
        throw new RuntimeException("int overflow:" + x);
    return (int) x;
}

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

/**
 * Output.
 */
PrintWriter printOut = new PrintWriter(System.out);

void so(Object obj, boolean newLine) {
    if (newLine) printOut.println(obj);
    else printOut.print(obj);
    if (ALWAYS_FLUSH) printOut.flush();
}

void so(Object obj) {
    so(obj, false);
}

void sl(Object obj) {
    so(obj, true);
}

void sl() {
    sl("");
}

/**
 * Main.
 */
void mainGo() {
    try {
        inStream = new FileInputStream("src/main.in");
        ALWAYS_FLUSH = true;
    } catch (Exception e) {
        inStream = System.in;
    }
    while (hasNext()) {
        if (MULTI_CASE) {
            int T = nextInt();
            if (T == 0) break;
            for (int i = 0; i < T; ++i) {
                go();
            }
        } else {
            go();
        }
    }
    printOut.flush();
}

public static void main(String[] args) throws Exception {
    new Main().mainGo();
}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code has a complex structure with multiple conditions and loops, so we need to carefully choose inputs that can expose issues in these conditions.

Here are some key points to consider:
1. **Small values of \( x \) and \( m \)**: These can help us verify the correctness of the brute-force part of the code.
2. **Large values of \( m \)**: These can help us verify the correctness of the optimized part of the code.
3. **Special values of \( x \)**: Values like powers of 2, and values that are close to powers of 2, can expose issues in the bitwise operations.
4. **Values of \( m \) that are just below or just above the threshold for the optimized part**.

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

```plaintext
5
1 10
2 10
7 10
1048576 1048576
1048575 1048576
```

Explanation:
1. **Test Case 1: \( x = 1 \), \( m = 10 \)**:
   - This is a simple case where the brute-force part should work correctly.
2. **Test Case 2: \( x = 2 \), \( m = 10 \)**:
   - This is a simple case where the brute-force part should work correctly.
3. **Test Case 3: \( x = 7 \), \( m = 10 \)**:
   - This is a case from the example, which should help verify the correctness of the brute-force part.
4. **Test Case 4: \( x = 1048576 \), \( m = 1048576 \)**:
   - This is a large value of \( x \) and \( m \), which should trigger the optimized part of the code.
5. **Test Case 5: \( x = 1048575 \), \( m = 1048576 \)**:
   - This is a large value of \( m \) just above a power of 2, which can expose issues in the bitwise operations.

This test input should help identify any bugs in the provided code by covering a range of edge cases and scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$x \oplus y$$$ is divisible$$$^{\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

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

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
7 10
2 3
6 4
1 6
4 1
output:
3
2
2
6
1

Note: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.

- $$$y = 1$$$ is valid because $$$x \oplus y = 7 \oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 7 \oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.
- $$$y = 9$$$ is valid because $$$x \oplus y = 7 \oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.

Buggy Code:
//package com.aakash.Contest.C2039;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class C2 {
    static BufferedReader br = new BufferedReader(
            new InputStreamReader(System.in));

    public static void main(String[] args) throws IOException {

        StringTokenizer s1
                = new StringTokenizer(br.readLine());
        int t = Integer.parseInt(s1.nextToken());
        while (t-- > 0) {
            StringTokenizer s2
                    = new StringTokenizer(br.readLine());
            long x = Integer.parseInt(s2.nextToken());
            long m=Long.parseLong(s2.nextToken());
            solve(x,m);
        }

    }

    static void solve(long x,long m) {
        long tx=x,tm=m;
        long pow=1;
        while(tx!=0) {
            tx=tx>>1;
            if((pow&tm)!=0) tm-=pow;
            pow=pow<<1;
        }
        long c=(tm-1)/x;
        for(int i=0;i<=x;i++) {
            long xor=tm+i,y=x^xor;
            if(y<=m &&y!=0 &&xor%x==0) c++;
        }
        for(int i=1;i<=Math.min(x,m);i++) {
            if(i==x) continue;
            long y=i,xor=i^x;
            if((xor%y)==0) c++;
        }
        System.out.println(c);
    }
}

**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 or scenarios that can cause the algorithm to fail. Let's analyze the code and identify potential issues:

1. **Edge Case with `x = 1`**:
   - When `x = 1`, the code should return `m` because every `y` from 1 to `m` will satisfy the condition `x ⊕ y` being divisible by `y` (since `1 ⊕ y` is always `y`).

2. **Edge Case with `x = 0`**:
   - The code does not handle `x = 0` explicitly, but the problem constraints specify `1 ≤ x ≤ 10^6`, so this case is not applicable.

3. **Edge Case with `m = 1`**:
   - When `m = 1`, the only possible `y` is 1. The code should return 1.

4. **Edge Case with `x = m`**:
   - When `x = m`, the only possible `y` is `x` itself because `x ⊕ x = 0` which is divisible by both `x` and `y`.

5. **Edge Case with `x` being a power of 2**:
   - When `x` is a power of 2, the behavior of the XOR operation can be different. For example, `x = 2` and `m = 3` should return 2 because `2 ⊕ 1 = 3` and `2 ⊕ 2 = 0`.

6. **Edge Case with large `m`**:
   - When `m` is very large, the code should still handle it correctly. However, the loop that iterates from 0 to `x` and from 1 to `min(x, m)` can be a potential issue if not optimized.

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

```plaintext
5
1 10
2 3
3 3
4 4
7 10
```

This test input includes:
- `x = 1` and `m = 10` to test the case where `x = 1`.
- `x = 2` and `m = 3` to test the case where `x` is a power of 2.
- `x = 3` and `m = 3` to test the case where `x = m`.
- `x = 4` and `m = 4` to test another case where `x = m`.
- `x = 7` and `m = 10` to test a general case as provided in the example.

This test input should help expose any issues in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$x \oplus y$$$ is divisible$$$^{\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

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

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
7 10
2 3
6 4
1 6
4 1
output:
3
2
2
6
1

Note: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.

- $$$y = 1$$$ is valid because $$$x \oplus y = 7 \oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 7 \oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.
- $$$y = 9$$$ is valid because $$$x \oplus y = 7 \oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
using lint = long long;
using pi = array<lint, 2>;
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
#define cr(v, n) (v).clear(), (v).resize(n);
int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int tc;
	cin >> tc;
	while (tc--) {
		lint x, m;
		cin >> x >> m;
		lint cnt = 0;
		for (lint y = 1; y <= m; y++) {
			lint xorv = x ^ y;
			if (xorv % x != 0 && xorv % y == 0) {
				cnt++;
			}
		}
		m++;
		for (int i = 60; i >= 0; i--) {
			if ((m >> i) & 1) {
				lint z = ((m >> i) << i) - (1ll << i);
				z ^= ((x >> i) << i);
				cnt += (z + (1ll << i) + x - 1) / x - (z + x - 1) / x;
			}
		}
		cnt--;
		cout << cnt << "\n";
	}
}

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

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

Think step by step.

Response:
```plaintext
1
1048576 1048576
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$x \oplus y$$$ is divisible$$$^{\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

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

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
7 10
2 3
6 4
1 6
4 1
output:
3
2
2
6
1

Note: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.

- $$$y = 1$$$ is valid because $$$x \oplus y = 7 \oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 7 \oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.
- $$$y = 9$$$ is valid because $$$x \oplus y = 7 \oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.

Buggy Code:
#include <bits/stdc++.h>
#define sz(a) int((a).size())
using namespace std;

using ll = long long;
const int inf = 1e9 + 7;

int main() {
    #ifndef ONLINE_JUDGE
        freopen("in.txt", "r", stdin);
    #endif
    ios::sync_with_stdio(0),cin.tie(NULL),cout.tie(NULL);

    int t;
    cin >> t;
    while (t--) {
        ll n, m;
        cin >> n>> m;
        ll answer = 0;

        for (int i = 1; i <= min(m, n); i++) {
            if (((n ^ i) % n == 0 || (n ^ i) % i == 0)) {
                ++answer;
            }
        }

        for (int i = n * 2; i <= m + n; i += n) {
            if ((n ^ i) <= m) {
                ++answer;
            }
        }

        cout << answer <<"\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 consider edge cases and scenarios where the code might fail. The code provided has a few potential issues:

1. **Loop Limits**: The first loop iterates from 1 to `min(m, n)`, which might miss some valid `y` values if `m` is much larger than `n`.
2. **Second Loop**: The second loop starts from `n * 2` and increments by `n`, but it checks if `(n ^ i) <= m` which might be incorrect or unnecessary.

Given these potential issues, let's create a test case that maximizes the failure probability:

1. **Large `m` and small `n`**: This will test the first loop's limit.
2. **Special values of `n` and `m`**: Values where `n` is a power of 2 or where `m` is a multiple of `n` can expose issues in the second loop.

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

```plaintext
3
1 1000000000000000000
2 1000000000000000000
4 1000000000000000000
```

### Explanation:
1. **Test Case 1: \( n = 1 \) and \( m = 10^{18} \)**:
   - For \( n = 1 \), \( x \oplus y \) is always \( y \), which is divisible by \( y \). Therefore, all \( y \) from 1 to \( m \) are valid.
   - The expected output is \( 10^{18} \).

2. **Test Case 2: \( n = 2 \) and \( m = 10^{18} \)**:
   - For \( n = 2 \), \( x \oplus y \) can be \( y - 2 \) or \( y + 2 \) depending on the parity of \( y \).
   - The code might miss some valid \( y \) values if it doesn't handle the large \( m \) correctly.
   - The expected output is a large number, but the code might fail to count all valid \( y \) values.

3. **Test Case 3: \( n = 4 \) and \( m = 10^{18} \)**:
   - For \( n = 4 \), \( x \oplus y \) can be \( y - 4 \), \( y + 4 \), or other values depending on the binary representation of \( y \).
   - The code might miss some valid \( y \) values if it doesn't handle the large \( m \) correctly.
   - The expected output is a large number, but the code might fail to count all valid \( y \) values.

This test input should help identify the issues in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$x \oplus y$$$ is divisible$$$^{\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

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

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
7 10
2 3
6 4
1 6
4 1
output:
3
2
2
6
1

Note: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.

- $$$y = 1$$$ is valid because $$$x \oplus y = 7 \oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 7 \oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.
- $$$y = 9$$$ is valid because $$$x \oplus y = 7 \oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.

Buggy Code:
def count_valid_y(x, m):
    valid_count = 0

    for y in range(1, m + 1):
        xor_result = x ^ y
        if (xor_result % x == 0) or (xor_result % y == 0):
            valid_count += 1
        if valid_count >= m:
            break

    return valid_count

def main():
    import sys
    input = sys.stdin.read
    data = input().split()

    t = int(data[0])
    index = 1
    results = []

    for _ in range(t):
        x = int(data[index])
        m = int(data[index + 1])
        index += 2
        results.append(count_valid_y(x, m))

    for result in results:
        print(result)

if __name__ == '__main__':
    main()

# hclzCd1yTQ5SOK02n4a

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. The bug in the code is that it incorrectly breaks the loop when `valid_count` reaches `m`, which is not a valid condition for breaking the loop. This can lead to incorrect results.

Here are the steps to generate the test input:

1. **Edge Cases:**
   - When `x` is 1, every `y` from 1 to `m` should be valid.
   - When `x` is a large number and `m` is small, the loop should not break early.
   - When `x` is a power of 2, the XOR results have specific patterns that need to be checked.

2. **Challenging Scenarios:**
   - When `m` is very large compared to `x`, the loop should run for a long time.
   - When `x` is a small number and `m` is large, the loop should run for a long time.

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

```plaintext
5
1 100
1000000 100
64 1000000
7 1000000
1 1000000000000
```

### Explanation:
1. **Test Case 1: `1 100`**
   - For `x = 1`, every `y` from 1 to 100 should be valid. This tests the scenario where `x` is 1.

2. **Test Case 2: `1000000 100`**
   - For `x = 1000000`, `m` is much smaller. This tests the scenario where `x` is large and `m` is small.

3. **Test Case 3: `64 1000000`**
   - For `x = 64`, `m` is large. This tests the scenario where `x` is a power of 2 and `m` is large.

4. **Test Case 4: `7 1000000`**
   - For `x = 7`, `m` is large. This tests the scenario where `x` is a small number and `m` is large.

5. **Test Case 5: `1 1000000000000`**
   - For `x = 1`, `m` is very large. This tests the scenario where `m` is extremely large.

This test input should help expose the bug in the code and ensure that the loop runs for the correct number of iterations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$x \oplus y$$$ is divisible$$$^{\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

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

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
7 10
2 3
6 4
1 6
4 1
output:
3
2
2
6
1

Note: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.

- $$$y = 1$$$ is valid because $$$x \oplus y = 7 \oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 7 \oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.
- $$$y = 9$$$ is valid because $$$x \oplus y = 7 \oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.

Buggy Code:
import sys
input = lambda: sys.stdin.readline().rstrip("\r\n")
def ii(): return int(input())
def mii(): return map(int, input().split())
def lii(): return list(map(int, input().split()))

mod = 10**9+7
maxn = 2*10**5+10

t=ii()
for _ in range(t):
    x,m=mii()
    if x==1:
        print(m)
        continue
    ans=set()
    for i in range(1,min(m,x)+1):
        if (i^x)%i==0:
            ans.add(i)
    # x ke jo bhi multiples hai
    mm=10**8//t
    for mul in range(1,2*m//x+10):
        y=x^(mul*x)
        if y!=0 and 1<=y<=m:
            ans.add(y)
    print(len(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 focus on edge cases and scenarios that can maximize the failure probability. Here are some key considerations:

1. **Edge Case for \( x = 1 \)**:
   - The code handles \( x = 1 \) separately by directly printing \( m \). This is correct, but we should still include it to ensure it works as expected.

2. **Edge Case for \( x \) being a power of 2**:
   - Powers of 2 have a unique binary representation (e.g., 2, 4, 8, 16, etc.). This can affect the behavior of the XOR operation.

3. **Edge Case for \( x \) being a prime number**:
   - Prime numbers have fewer multiples, which can make the loop more efficient but also more prone to missing valid \( y \) values.

4. **Edge Case for \( x \) being a composite number**:
   - Composite numbers have more multiples, which can make the loop more complex and potentially miss valid \( y \) values.

5. **Edge Case for large \( m \)**:
   - Large values of \( m \) can push the limits of the code, especially in the loop that checks multiples.

6. **Edge Case for \( x \) and \( m \) being the same**:
   - This can simplify the problem but also test the edge condition where \( y = x \).

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

```plaintext
5
1 10
2 10
7 10
15 10
10 10
```

Explanation:
- The first test case (\( x = 1 \), \( m = 10 \)) tests the special case where \( x = 1 \).
- The second test case (\( x = 2 \), \( m = 10 \)) tests \( x \) being a power of 2.
- The third test case (\( x = 7 \), \( m = 10 \)) tests \( x \) being a prime number.
- The fourth test case (\( x = 15 \), \( m = 10 \)) tests \( x \) being a composite number.
- The fifth test case (\( x = 10 \), \( m = 10 \)) tests the edge condition where \( x = m \).

This input should help expose any bugs in the code by covering a variety of edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$x \oplus y$$$ is divisible$$$^{\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

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

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
7 10
2 3
6 4
1 6
4 1
output:
3
2
2
6
1

Note: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.

- $$$y = 1$$$ is valid because $$$x \oplus y = 7 \oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 7 \oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.
- $$$y = 9$$$ is valid because $$$x \oplus y = 7 \oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.

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

public class Main {
		static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		static StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
		static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		public static void main(String[] args) throws Exception {
				long t=Integer.parseInt(br.readLine());
				while (t-->0) {
						String[] s = br.readLine().split(" ");
						long x = Long.parseLong(s[0]);
						long m = Long.parseLong(s[1]);
						HashSet<Long> set = new HashSet<>();
						for(long i=1;i<=Math.min(m,x<<1);i++){
								long cur=x^i;
								if(cur%i==0){
										set.add(i);
								}
						}
						for(long i=x;i<=(m<<1);i+=x){
								if((x^i)<=m&&(x^i)!=0){
										set.add(x^i);
								}
						}
						out.println(set.size());
				}
				out.close();
		}
		static int nextInt() throws IOException {
				in.nextToken();
				return (int) in.nval;
		}
}

**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
10 1000000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves XOR (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard version of the problem. The differences between the two versions are highlighted in bold. You can only make hacks if both versions of the problem are solved.

Shohag has two integers $$$x$$$ and $$$m$$$. Help him count the number of integers $$$1 \le y \le m$$$ such that $$$x \oplus y$$$ is divisible$$$^{\text{∗}}$$$ by either $$$x$$$, $$$y$$$, or both. Here $$$\oplus$$$ is the bitwise XOR operator.

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

The first and only line of each test case contains two space-separated integers $$$x$$$ and $$$m$$$ ($$$1 \le x \le 10^6$$$, $$$1 \le m \le 10^{18}$$$).

It is guaranteed that the sum of $$$x$$$ over all test cases does not exceed $$$10^7$$$.

Output: For each test case, print a single integer — the number of suitable $$$y$$$.

Examples:
input:
5
7 10
2 3
6 4
1 6
4 1
output:
3
2
2
6
1

Note: In the first test case, for $$$x = 7$$$, there are $$$3$$$ valid values for $$$y$$$ among the integers from $$$1$$$ to $$$m = 10$$$, and they are $$$1$$$, $$$7$$$, and $$$9$$$.

- $$$y = 1$$$ is valid because $$$x \oplus y = 7 \oplus 1 = 6$$$ and $$$6$$$ is divisible by $$$y = 1$$$.
- $$$y = 7$$$ is valid because $$$x \oplus y = 7 \oplus 7 = 0$$$ and $$$0$$$ is divisible by both $$$x = 7$$$ and $$$y = 7$$$.
- $$$y = 9$$$ is valid because $$$x \oplus y = 7 \oplus 9 = 14$$$ and $$$14$$$ is divisible by $$$x = 7$$$.

Buggy Code:
import java.util.Scanner;

public class Main{
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();

        while (t-- > 0) {
            long x = scanner.nextLong();
            long m = scanner.nextLong();

            long result = countValidY(x, m);
            System.out.println(result);
        }

        scanner.close();
    }

    private static long countValidY(long x, long m) {
        long count = 0;

        // Check all y from 1 to m
        for (long y = 1; y <= m; y++) {
            if ((x ^ y) % x == 0 || (x ^ y) % y == 0) {
                count++;
            }
        }

        return count;
    }
}

// 31fFrgVuoSIz7wCTYDX

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

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

Think step by step.

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

1. **Large Values of \( m \)**: The code iterates from 1 to \( m \), which can be as large as \( 10^{18} \). This can lead to performance issues, especially for large values of \( m \).

2. **Small Values of \( x \)**: When \( x \) is small, there are more potential values of \( y \) that could satisfy the condition \( (x \oplus y) \) being divisible by \( x \) or \( y \).

3. **Edge Cases for \( x \) and \( m \)**: Values like \( x = 1 \) or \( x = 2 \) can have many valid \( y \) values, which can help in exposing any off-by-one errors or logical flaws.

4. **Specific Values of \( y \)**: Values of \( y \) that are powers of 2 or close to powers of 2 can be particularly interesting because the XOR operation can produce interesting results.

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

```plaintext
4
1 1000000000000000000
2 1000000000000000000
7 1000000000000000000
1000000 1000000000000000000
```

### Explanation:
1. **Test Case 1: \( x = 1, m = 10^{18} \)**: This is a large \( m \) and a small \( x \). For \( x = 1 \), \( (x \oplus y) \) is always \( y \), which is divisible by \( y \). This should return \( m \) valid \( y \) values.
2. **Test Case 2: \( x = 2, m = 10^{18} \)**: This is another large \( m \) with a small \( x \). The valid \( y \) values will be more numerous, and this can help in exposing any issues with the loop or condition.
3. **Test Case 3: \( x = 7, m = 10^{18} \)**: This is a moderate \( x \) with a large \( m \). The valid \( y \) values will be fewer but still numerous, which can help in exposing any performance issues.
4. **Test Case 4: \( x = 10^6, m = 10^{18} \)**: This is a large \( x \) with a large \( m \). The valid \( y \) values will be fewer, but the large \( m \) can still cause performance issues.

This test input should be challenging and help in identifying 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: Turtle and Paths on a Tree
time_limit_ms: 4000
memory_limit_mb: 1024
Description: Note the unusual definition of $\text{MEX}$ in this problem.

Piggy gave Turtle a binary tree$^{\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.

If a set of paths $P = \\{(x_i, y_i)\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.

Turtle defines the value of a set of paths as $\sum\limits_{(x, y) \in P} f(x, y)$, where $f(x, y)$ denotes the $\text{MEX}^{\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).

Turtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!

$^{\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.

$^{\ddagger}\text{MEX}$ of a collection of integers $c_1, c_2, \ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\text{MEX}$ of $[2, 3]$ is $1$.

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 a single integer $n$ ($2 \le n \le 2.5 \cdot 10^4$) — the number of vertices in the tree.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the sequence $a$.

The third line of each test case contains $n - 1$ integers $p_2, p_3, \ldots, p_n$ ($1 \le p_i < i$) — the parent of each vertex in the tree.

Additional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.

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

Output: For each test case, output a single integer — the minimum value over all good sets of paths.

Examples:
input:
5
5
3 2 2 1 1
1 1 2 2
5
3 2 1 1 1
1 1 2 2
6
1 2 1 2 1 3
1 2 3 3 4
7
2 1 2 3 1 2 1
1 1 2 2 3 3
10
1 2 2 1 4 2 3 1 2 1
1 1 2 2 3 3 4 5 5
output:
4
6
6
6
7

Note: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:

![](CDN_BASE_URL/45a1635df0fcab1b711d2165163f0dfb)

The good set of paths with the minimum value is $\\{(2, 3), (4, 5)\\}$.

Note that in this test case $\\{(4, 5)\\}$ and $\\{(3, 4), (4, 5)\\}$ are not good sets of paths because each edge should be covered exactly once.

In the second test case, the tree is as follows:

![](CDN_BASE_URL/a41b8a69adf0471a8353a08785fa6375)

The set of good paths with the minimum value is $\\{(1, 2), (1, 3), (4, 5)\\}$.

In the third test case, the tree is as follows:

![](CDN_BASE_URL/7e62b6f397447336b5c5732c0d915ebd)

The set of good paths with the minimum value is $\\{(1, 6), (3, 5)\\}$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int INF = 1e18, MOD = 1e9 + 7;
const int mxN = 25e3 + 7;
int dp[mxN];
array<int,2> child[mxN];
int a[mxN];
int ndp[mxN];
int n;
int dfs(int pos)
{
	//cout << pos << " ";
	int x = INF, y = INF;
	int dp1[mxN][2];
	int ans = INF;
	if(child[pos][0] != -1)
	{
		dfs(child[pos][0]);
		for(int i = 1;n + 1 >= i;i++) dp1[i][0] = dp[i];
		for(int i = 1;n + 1 >= i;i++)
		{
			if(a[pos] != i) x = min(x, dp1[i][0] + i);
		}
	}
	if(child[pos][1] != -1)
	{
		dfs(child[pos][1]);
		for(int i = 1;n + 1 >= i;i++) dp1[i][1] = dp[i];
		for(int i = 1;n + 1 >= i;i++)
		{
			if(a[pos] != i) y = min(y, dp1[i][1] + i),ans = min(ans, dp1[i][0] + dp1[i][1] + i);
		}
		ans = min(ans, x + y);
	}
	else ans = min(ans, x);
	for(int i = 0;n + 1 >= i;i++)
	{
		if(a[pos] != i)
			ndp[i] = 0;
		else
			ndp[i] = INF;
	}
	if(x == INF && y == INF)
	{
		for(int i = 0;n >= i;i++) dp[i] = ndp[i];
		/*cout << pos << endl;
		for(int i = 1;n >= i;i++) cout << dp[i] << " ";
		cout << endl;*/
		return 0;
	}

	for(int i = 0;n + 1 >= i;i++)
	{
		//cout << i << " " << ndp[i] << endl;
		if(a[pos] == i) continue;
	//	cout << i << endl;
		if(y == INF)
		{
			ndp[i] = min(x, dp1[i][0]);
		}
		else
		{
			ndp[i] = min({dp1[i][0] + y, dp1[i][1] + x,ans});
		}
	}
	for(int i = 0;n + 1 >= i;i++) dp[i] = ndp[i];
	/*cout << pos << endl;
	for(int i = 1;n >= i;i++) cout << dp[i] << " ";
	cout << endl;
	cout << "SO: " << ans << endl;*/
	return ans;
}
void solve()
{
	//int n;
	cin >> n;
	for(int i = 0;n > i;i++) cin >> a[i], child[i] = {-1,-1};
	for(int i = 0;n - 1 > i;i++)
	{
		int x;
		cin >> x;
		x--;
		if(child[x][0] == -1) child[x][0] = i + 1;
		else child[x][1] = i + 1;
	}
	for(int i = 0;n >= i;i++) ndp[i] = INF, dp[i] = INF;
	cout << dfs(0) << endl;
}

int32_t main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	int tt;
	cin >> tt;
	while(tt--) solve();
}

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

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

Think step by step.

Response:
```plaintext
1
4
1 2 3 4
1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and Paths on a Tree
time_limit_ms: 4000
memory_limit_mb: 1024
Description: Note the unusual definition of $\text{MEX}$ in this problem.

Piggy gave Turtle a binary tree$^{\dagger}$ with $n$ vertices and a sequence $a_1, a_2, \ldots, a_n$ on his birthday. The binary tree is rooted at vertex $1$.

If a set of paths $P = \\{(x_i, y_i)\\}$ in the tree covers each edge exactly once, then Turtle will think that the set of paths is good. Note that a good set of paths can cover a vertex twice or more.

Turtle defines the value of a set of paths as $\sum\limits_{(x, y) \in P} f(x, y)$, where $f(x, y)$ denotes the $\text{MEX}^{\ddagger}$ of all $a_u$ such that vertex $u$ is on the simple path from $x$ to $y$ in the tree (including the starting vertex $x$ and the ending vertex $y$).

Turtle wonders the minimum value over all good sets of paths. Please help him calculate the answer!

$^{\dagger}$A binary tree is a tree where every non-leaf vertex has at most $2$ sons.

$^{\ddagger}\text{MEX}$ of a collection of integers $c_1, c_2, \ldots, c_k$ is defined as the smallest positive integer $x$ which does not occur in the collection $c$. For example, $\text{MEX}$ of $[3, 3, 1, 4]$ is $2$, $\text{MEX}$ of $[2, 3]$ is $1$.

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 a single integer $n$ ($2 \le n \le 2.5 \cdot 10^4$) — the number of vertices in the tree.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the sequence $a$.

The third line of each test case contains $n - 1$ integers $p_2, p_3, \ldots, p_n$ ($1 \le p_i < i$) — the parent of each vertex in the tree.

Additional constraint on the input: the given tree is a binary tree, that is, every non-leaf vertex has at most $2$ sons.

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

Output: For each test case, output a single integer — the minimum value over all good sets of paths.

Examples:
input:
5
5
3 2 2 1 1
1 1 2 2
5
3 2 1 1 1
1 1 2 2
6
1 2 1 2 1 3
1 2 3 3 4
7
2 1 2 3 1 2 1
1 1 2 2 3 3
10
1 2 2 1 4 2 3 1 2 1
1 1 2 2 3 3 4 5 5
output:
4
6
6
6
7

Note: In the first test case, the tree is as follows. The number in brackets denotes the weight of the vertex:

![](CDN_BASE_URL/45a1635df0fcab1b711d2165163f0dfb)

The good set of paths with the minimum value is $\\{(2, 3), (4, 5)\\}$.

Note that in this test case $\\{(4, 5)\\}$ and $\\{(3, 4), (4, 5)\\}$ are not good sets of paths because each edge should be covered exactly once.

In the second test case, the tree is as follows:

![](CDN_BASE_URL/a41b8a69adf0471a8353a08785fa6375)

The set of good paths with the minimum value is $\\{(1, 2), (1, 3), (4, 5)\\}$.

In the third test case, the tree is as follows:

![](CDN_BASE_URL/7e62b6f397447336b5c5732c0d915ebd)

The set of good paths with the minimum value is $\\{(1, 6), (3, 5)\\}$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
const int N = 5;
int mex[(1<<N)];
void dfs(int i, vector<vector<int>>& e, vector<int>& a, vector<vector<int>>& dp){
    for(int j: e[i])dfs(j, e, a, dp);
    for(int j=0; j<(1<<N); j++)dp[j][i]=1e9;
    if(e[i].size()==1){
        int mask = (a[i]<=N?(1<<(a[i]-1)):0);
        int x = e[i].back();
        for(int j=0; j<(1<<N)-1; j++)dp[j|mask][i]=min(dp[j|mask][i], dp[j][x]+(i==0?mex[j|mask]:0));
        if(a[x]<=N)mask|=1<<(a[x]-1);
        int low = 1e9;
        for(int j=0; j<(1<<N)-1; j++)low=min(dp[j][x]+mex[j], low);
        dp[mask][i]=min(dp[mask][i], low+(i==0?mex[mask]:0));
    }
    else if(e[i].size()==0){
        if(a[i]<=N){
            dp[1<<(a[i]-1)][i]=0;
        }
        else{
            dp[0][i]=0;
        }
    }
    else{
        int mask = (a[i]<=N?(1<<(a[i]-1)):0);
        int x = e[i].back();
        e[i].pop_back();
        int y = e[i].back();
        for(int j=0; j<(1<<N)-1; j++){
            for(int w=0; w<(1<<N)-1; w++){
                if((w|j|mask)==(1<<N)-1)continue;
                dp[mask][i]=min(dp[mask][i], dp[j][x]+dp[w][y]+mex[w|j|mask]);
                dp[mask][i]=min(dp[mask][i], dp[j][x]+mex[j]+dp[w][y]+mex[w|mask]);
                dp[mask][i]=min(dp[mask][i], dp[w][y]+mex[w]+dp[j][x]+mex[j|mask]);
            }
        }
        int low = 1e9;
        mask|=(a[x]<=N?(1<<(a[x]-1)):0);
        for(int j=0; j<(1<<N)-1; j++){
            low=min(low, mex[mask]+dp[j][x]+mex[j]);
            if((j|mask)!=(1<<N)-1)low=min(low, dp[j][x]+mex[j|mask]);
        }
        mask = (a[i]<=N?(1<<(a[i]-1)):0);
        for(int j=0; j<(1<<N)-1; j++){
            dp[mask|j][i]=min(dp[mask|j][i], dp[j][y]+low+(i==0?mex[mask|j]:0));
        }
        mask|=(a[y]<=N?(1<<(a[y]-1)):0);
        int low2 = 1e9;
        for(int j=0; j<(1<<N)-1; j++)low2=min(low2, dp[j][y]+mex[j]+(i==0?mex[mask]:0));
        dp[mask][i]=min(dp[mask][i], low+low2);
        swap(x, y);
        low = 1e9;
        mask|=(a[x]<=N?(1<<(a[x]-1)):0);
        for(int j=0; j<(1<<N)-1; j++){
            low=min(low, mex[mask]+dp[j][x]+mex[j]);
            if((j|mask)!=(1<<N)-1)low=min(low, dp[j][x]+mex[j|mask]);
        }
        mask = (a[i]<=N?(1<<(a[i]-1)):0);
        for(int j=0; j<(1<<N)-1; j++){
            dp[mask|j][i]=min(dp[mask|j][i], dp[j][y]+low+(i==0?mex[mask|j]:0));
        }
        mask|=(a[y]<=N?(1<<(a[y]-1)):0);
        low2 = 1e9;
        for(int j=0; j<(1<<N)-1; j++)low2=min(low2, dp[j][y]+mex[j]+(i==0?mex[mask]:0));
        dp[mask][i]=min(dp[mask][i], low+low2);
    }
}
void solve(){
    int n; cin >> n;
    vector<vector<int>> e(n);
    vector<int> a(n);
    for(int& i: a)cin >> i;
    for(int i=1, u; i<n; i++){
        cin >> u;
        e[u-1].push_back(i);
    }
    vector<vector<int>> dp((1<<N), vector(n, 0));
    dfs(0, e, a, dp);
    int ans = 1e9;
    for(int i=0; i<(1<<N)-1; i++)ans=min(ans, dp[i][0]);
    cout << ans << '\n';
}
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    for(int i=0; i<(1<<N)-1; i++){
        mex[i]++;
        if(i&1){
            mex[i]++;
            if(i&2){
                mex[i]++;
                if(i&4)mex[i]++;
            }
        }
    }
    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:
```plaintext
1
7
1 2 3 4 5 6 7
1 1 2 2 3 3
```