**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Polygonal Segments
time_limit_ms: 8000
memory_limit_mb: 512
Description: You are given an array $a$ of size $n$.

A segment $[l, r](1 \le l < r \le n)$ is called a polygonal segment only if the following conditions hold:

  * $(r-l+1) \geq 3$;    * Considering $a_l, a_{l+1}, \ldots, a_r$ as side lengths, these sides can form a polygon with $(r-l+1)$ sides.

Process $q$ queries of two types:

  * "1 l r": find the length of the longest segment among all polygonal segments $[l_0,r_0]$ satisfying $l \le l_0 \le r_0 \le r$. If there is no such polygonal segment, output $-1$ instead;    * "2 i x": assign $a_i := x$.

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

For each test case:

  * The first line of each testcase contains two integers $n$, $q$ ($4 \le n \le 2\cdot 10^5$, $1 \le q \le 10^5$);    * The second line of each testcase contains $n$ integers $a_1,a_2,\ldots, a_n$ ($1 \le a_i \le 10^{12}$);    * The following $q$ lines contain the description of queries. Each line is of one of two types:     * "1 l r" ($1 \le l < r \le n$, $r-l+1\ge 3$);      * "2 i x" ($1 \le i \le n$, $1 \le x \le 10^{12}$).

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

Output: For each query, if there is no suitable segment, output $-1$ in a new line. Otherwise, output the length of the longest segment satisfying the condition above in a new line.

Examples:
input:
2
5 6
3 1 2 2 8
1 1 3
1 1 4
1 1 5
2 1 5
1 1 4
1 1 5
4 10
500000000000 500000000000 1000000000000 500000000000
1 1 3
1 2 4
1 1 4
2 1 499999999999
2 3 999999999999
1 1 3
1 2 4
1 1 4
2 3 1000000000000
1 1 3
output:
-1
4
4
3
5
-1
-1
4
-1
3
4
-1

Note: In the first query of the first test case, there is no polygonal segment under the given condition. For example, considering segment $[1,3]$, you can not form a triangle with side lengths of $a_1=3$, $a_2=1$, and $a_3=2$.

In the second query of the first test case, the longest polygonal segment is $[1,4]$. You can form a quadrilateral with side lengths of $a_1=3$, $a_2=1$, $a_3=2$, and $a_4=2$.

![](CDN_BASE_URL/589caa1d808f5b5cc88c1eb5f51f2eaf) An example of a quadrilateral with side lengths of $3$, $1$, $2$, and $2$.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Codeforces1990F {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

    static class Ent {
        long sum;
        long max;
        int maxIndex;
    }

    static long[] readIntArray() throws IOException {
        String line = br.readLine();
        String[] linesplit = line.split(" ");
        long[] is = new long[linesplit.length];
        for (int i = 0; i < linesplit.length; i++) {
            is[i] = Long.parseLong(linesplit[i]);
        }
        return is;
    }

    static Ent[][] sums;

    public static void main(String[] args) throws NumberFormatException, IOException {
        int tests = Integer.parseInt(br.readLine());

        StringBuilder out = new StringBuilder();

        for (int t = 0; t < tests; t++) {
            long[] nqLine = readIntArray();
            int n = (int) nqLine[0];
            int q = (int) nqLine[1];

            long[] a = readIntArray();

            Ent[] ent = new Ent[n];
            for (int i = 0; i < n; i++) {
                Ent e = new Ent();
                e.sum = a[i];
                e.max = a[i];
                e.maxIndex = i;
                ent[i] = e;
            }

            sums = new Ent[20][];
            int sumIndex = 0;
            while (true) {
                sums[sumIndex] = ent;
                if (ent.length == 1) {
                    break;
                }
                sumIndex++;
                Ent[] nextent = new Ent[(ent.length + 1) / 2];
                for (int i = 0; i < ent.length; i++) {
                    if (nextent[i / 2] == null) {
                        Ent e = new Ent();
                        e.sum = ent[i].sum;
                        e.max = ent[i].max;
                        e.maxIndex = ent[i].maxIndex;
                        nextent[i / 2] = e;
                    } else {
                        nextent[i / 2].sum += ent[i].sum;
                        if (nextent[i / 2].max < ent[i].max) {
                            nextent[i / 2].max = ent[i].max;
                            nextent[i / 2].maxIndex = ent[i].maxIndex;
                        }
                    }
                }
                ent = nextent;
            }

            for (int i = 0; i < q; i++) {
                long[] qLine = readIntArray();
                if (qLine[0] == 1) {
                    int l = (int) qLine[1] - 1;
                    int r = (int) qLine[2] - 1;
                    int b = getBest(l, r);
                    out.append(b + "\r\n");
                } else {
                    int index = (int) qLine[1] - 1;
                    long x = qLine[2];
                    Ent other = sums[0][index];
                    other.sum = x;
                    other.max = x;
                    for (sumIndex = 1; sumIndex < sums.length; sumIndex++) {
                        if (sums[sumIndex] == null) {
                            break;
                        }
                        index /= 2;
                        Ent e = sums[sumIndex][index];
                        Ent o1 = sums[sumIndex - 1][index * 2];
                        Ent o2 = null;
                        if (index * 2 + 1 < sums[sumIndex - 1].length) {
                            o2 = sums[sumIndex - 1][index * 2 + 1];
                        }
                        e.sum = o1.sum;
                        e.max = o1.max;
                        e.maxIndex = o1.maxIndex;
                        if (o2 != null) {
                            e.sum += o2.sum;
                            if (e.max < o2.max) {
                                e.max = o2.max;
                                e.maxIndex = o2.maxIndex;
                            }
                        }
                    }
                }

            }
        }

        System.out.println(out);
    }

    public static int getBest(int l, int r) {
        if (r - l + 1 < 3) {
            return -1;
        }

        Ent e = getEnt(l, r);
        if (e.max * 2 < e.sum) {
            return r - l + 1;
        }

        int b1 = getBest(l, e.maxIndex - 1);
        int b2 = getBest(e.maxIndex + 1, r);
        return Math.max(b1, b2);
    }

    public static Ent getEnt(int l, int r) {
        //        System.out.print("SUM: " + l + "; " + r);
        Ent e = new Ent();
        while (l <= r) {
            int sumIndex = 0;
            int sumLength = 1;
            while (true) {
                if ((l & sumLength) == 0) {
                    if (sumLength * 2 <= r - l + 1) {
                        sumIndex++;
                        sumLength <<= 1;
                        continue;
                    }
                }
                break;
            }
            e.sum += sums[sumIndex][l >> sumIndex].sum;
            if (e.max < sums[sumIndex][l >> sumIndex].max) {
                e.max = sums[sumIndex][l >> sumIndex].max;
                e.maxIndex = sums[sumIndex][l >> sumIndex].maxIndex;
            }
            l += sumLength;
        }

        //        System.out.println(" = " + e.sum + "; " + e.max + "; " + e.maxIndex);

        return e;
    }
}

**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
5 4
1 2 3 4 5
1 1 5
2 3 6
1 1 5
1 2 4
```

### Explanation:
1. **Initial Array**: `[1, 2, 3, 4, 5]`
   - The first query `1 1 5` asks for the longest polygonal segment in the range `[1, 5]`.
   - The segment `[1, 5]` can form a polygon with side lengths `[1, 2, 3, 4, 5]`, so the output should be `5`.

2. **Update Query**: `2 3 6`
   - The array is updated to `[1, 2, 6, 4, 5]`.

3. **Second Query**: `1 1 5`
   - The segment `[1, 5]` now has side lengths `[1, 2, 6, 4, 5]`.
   - The maximum side length is `6`, and `6 * 2 > 1 + 2 + 4 + 5 = 12`, so the segment cannot form a polygon.
   - The output should be `-1`.

4. **Third Query**: `1 2 4`
   - The segment `[2, 4]` has side lengths `[2, 6, 4]`.
   - The maximum side length is `6`, and `6 * 2 > 2 + 4 = 6`, so the segment cannot form a polygon.
   - The output should be `-1`.

This test case is designed to check the handling of updates and the correct determination of polygonal segments, especially when the maximum side length in a segment is significantly larger than the sum of the other side lengths.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and an Infinite Sequence
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a sequence $a_0, a_1, a_2, \ldots$ of infinite length. Initially $a_i = i$ for every non-negative integer $i$.

After every second, each element of the sequence will simultaneously change. $a_i$ will change to $a_{i - 1} \mid a_i \mid a_{i + 1}$ for every positive integer $i$. $a_0$ will change to $a_0 \mid a_1$. Here, $|$ denotes [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR).

Turtle is asked to find the value of $a_n$ after $m$ seconds. In particular, if $m = 0$, then he needs to find the initial value of $a_n$. He is tired of calculating so many values, so please help him!

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

The first line of each test case contains two integers $n, m$ ($0 \le n, m \le 10^9$).

Output: For each test case, output a single integer — the value of $a_n$ after $m$ seconds.

Examples:
input:
9
0 0
0 1
0 2
1 0
5 2
10 1
20 3
1145 14
19198 10
output:
0
1
3
1
7
11
23
1279
19455

Note: After $1$ second, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[1, 3, 3, 7, 7, 7]$.

After $2$ seconds, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[3, 3, 7, 7, 7, 7]$.

Buggy Code:
#include<bits/stdc++.h>
#define int long long
#define dbg(a) cout << #a << " = " << a << '\n'
using namespace std;

const int INF = 1e18;
const int MOD = 1e9 + 7;

void solve()
{
    int n,m,power = 0;
    cin >> n >> m;
    int l = n,r = n,ans = n;
    while(m >= (1 << power))
    {
        ans|=(1 << power);
        power++;
    }
    ans|=(n + m);
    if(n - m > 0)
    {
        ans|=(n + m);
    }
    cout << ans << '\n';
}

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while(t > 0)
    {
        solve();
        t--;
    }
    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 produce incorrect results. Here are some considerations:

1. **Initial Values and Simple Cases:**
   - Test cases where \( n = 0 \) and \( m = 0 \) (initial value).
   - Test cases where \( n = 0 \) and \( m > 0 \) (first element after some seconds).

2. **Small Values of \( n \) and \( m \):**
   - Test cases where \( n \) and \( m \) are small to observe the pattern of changes.

3. **Large Values of \( n \) and \( m \):**
   - Test cases where \( n \) and \( m \) are large to ensure the code handles large inputs correctly.

4. **Edge Cases:**
   - Test cases where \( n \) is just greater than \( m \) to check the behavior of the code in such scenarios.
   - Test cases where \( n \) is much larger than \( m \) to ensure the code does not overflow or produce incorrect results.

5. **Boundary Conditions:**
   - Test cases where \( n \) and \( m \) are at the boundary of the allowed range.

Based on these considerations, here is a small-scale test input that includes a variety of cases:

```plaintext
10
0 0
0 1
0 2
1 0
5 2
10 1
20 3
1145 14
19198 10
1000000000 1000000000
```

This test input includes:
- Initial value for \( n = 0 \) and \( m = 0 \).
- Values for \( n = 0 \) with different \( m \) values to observe the first element after some seconds.
- Small values for \( n \) and \( m \) to check the pattern.
- Large values for \( n \) and \( m \) to ensure the code handles large inputs correctly.
- A boundary condition with the maximum allowed values for \( n \) and \( m \).

This should help in exposing 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: Turtle and an Infinite Sequence
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a sequence $a_0, a_1, a_2, \ldots$ of infinite length. Initially $a_i = i$ for every non-negative integer $i$.

After every second, each element of the sequence will simultaneously change. $a_i$ will change to $a_{i - 1} \mid a_i \mid a_{i + 1}$ for every positive integer $i$. $a_0$ will change to $a_0 \mid a_1$. Here, $|$ denotes [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR).

Turtle is asked to find the value of $a_n$ after $m$ seconds. In particular, if $m = 0$, then he needs to find the initial value of $a_n$. He is tired of calculating so many values, so please help him!

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

The first line of each test case contains two integers $n, m$ ($0 \le n, m \le 10^9$).

Output: For each test case, output a single integer — the value of $a_n$ after $m$ seconds.

Examples:
input:
9
0 0
0 1
0 2
1 0
5 2
10 1
20 3
1145 14
19198 10
output:
0
1
3
1
7
11
23
1279
19455

Note: After $1$ second, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[1, 3, 3, 7, 7, 7]$.

After $2$ seconds, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[3, 3, 7, 7, 7, 7]$.

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

#define ll long long
#define endl "\n"
#define pb push_back
#define ppb pop_back
#define ff first
#define ss second

#define MOD 1000000007

#define REP(j, a, b) for (int j = a; j < b; j++)
#define print(x) cout << x << endl
#define yep cout << "YES" << endl
#define nope cout << "NO" << endl

#define int long long

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

    int ans = n;

    for (int i = 0; i < 61; i++)
    {
        if ((n >> i) & 1LL != 0)
        {
            continue;
        }

        int fst = n;
        int snd = n;

        // In fst set the ith bit

        fst = (fst | (1LL << i));

        // unsetting the right bits:

        for (int j = i - 1; j >= 0; j--)
        {
            if ((fst >> j) & 1LL != 0)
            {
                fst = (fst ^ (1LL << j));
            }
        }

        if (abs(fst - n) <= m)
        {
            ans = (ans | (1LL << i));
        }

        for (int j = 61; j >= 0; j--)
        {
            if ((snd >> j) & 1LL != 0)
            {
                snd = (snd ^ (1LL << j));

                for (int k = j + 1; k >= 0; k--)
                {
                    snd = snd | (1LL << k);
                }

                break;
            }
        }

        snd = (snd | (1LL << i));

        if (abs(snd - n) <= m)
        {
            ans = (ans | (1LL << i));
        }
    }

    print(ans);

    return;
}

signed main()
{

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

    int t = 1;
    cin >> t;

    while (t--)
    {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
0 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and an Infinite Sequence
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a sequence $a_0, a_1, a_2, \ldots$ of infinite length. Initially $a_i = i$ for every non-negative integer $i$.

After every second, each element of the sequence will simultaneously change. $a_i$ will change to $a_{i - 1} \mid a_i \mid a_{i + 1}$ for every positive integer $i$. $a_0$ will change to $a_0 \mid a_1$. Here, $|$ denotes [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR).

Turtle is asked to find the value of $a_n$ after $m$ seconds. In particular, if $m = 0$, then he needs to find the initial value of $a_n$. He is tired of calculating so many values, so please help him!

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

The first line of each test case contains two integers $n, m$ ($0 \le n, m \le 10^9$).

Output: For each test case, output a single integer — the value of $a_n$ after $m$ seconds.

Examples:
input:
9
0 0
0 1
0 2
1 0
5 2
10 1
20 3
1145 14
19198 10
output:
0
1
3
1
7
11
23
1279
19455

Note: After $1$ second, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[1, 3, 3, 7, 7, 7]$.

After $2$ seconds, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[3, 3, 7, 7, 7, 7]$.

Buggy Code:
#include <iostream>
#include <vector>
using namespace std;

unsigned long long find_value(long long n, long long m) {
    //if (m == 0) return n;
    unsigned long long result = 0;

    // If m >= 60, the OR result will stabilize because it covers all bits up to that point
    if (m >= 60) {
        for (unsigned long long i = n; i <= n + 60; ++i) {
            result |= i;
        }
    } else {
        for (unsigned long long i = n; i <= n + m; ++i) {
            result |= i;
        }
    }
    return result;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t;
    cin >> t;
    vector<pair<long long, long long>> test_cases(t);

    for (int i = 0; i < t; ++i) {
        cin >> test_cases[i].first >> test_cases[i].second;
    }

    for (const auto& [n, m] : test_cases) {
        cout << find_value(n, m) << "\n";
    }

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code has a specific condition for `m >= 60` and a loop that performs bitwise OR operations. Let's consider the following scenarios:

1. **Initial Values (m = 0)**: The initial value of `a_n` should be `n`.
2. **Small Values of m**: The code should correctly compute the bitwise OR for small values of `m`.
3. **Large Values of m (m >= 60)**: The code assumes that the OR result will stabilize after 60 seconds, but we need to check if this assumption holds for all `n`.

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

1. **Initial Values**:
   - `n = 0, m = 0`
   - `n = 1, m = 0`
   - `n = 100, m = 0`

2. **Small Values of m**:
   - `n = 0, m = 1`
   - `n = 0, m = 2`
   - `n = 1, m = 1`
   - `n = 5, m = 2`
   - `n = 10, m = 1`

3. **Large Values of m (m >= 60)**:
   - `n = 0, m = 60`
   - `n = 1, m = 60`
   - `n = 100, m = 60`
   - `n = 1000, m = 100`

Here is the test input:

```plaintext
10
0 0
1 0
100 0
0 1
0 2
1 1
5 2
10 1
0 60
1 60
100 60
1000 100
```

This test input covers a range of values for `n` and `m`, including initial values, small values of `m`, and large values of `m` to ensure the code's assumptions are validated.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and an Infinite Sequence
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a sequence $a_0, a_1, a_2, \ldots$ of infinite length. Initially $a_i = i$ for every non-negative integer $i$.

After every second, each element of the sequence will simultaneously change. $a_i$ will change to $a_{i - 1} \mid a_i \mid a_{i + 1}$ for every positive integer $i$. $a_0$ will change to $a_0 \mid a_1$. Here, $|$ denotes [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR).

Turtle is asked to find the value of $a_n$ after $m$ seconds. In particular, if $m = 0$, then he needs to find the initial value of $a_n$. He is tired of calculating so many values, so please help him!

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

The first line of each test case contains two integers $n, m$ ($0 \le n, m \le 10^9$).

Output: For each test case, output a single integer — the value of $a_n$ after $m$ seconds.

Examples:
input:
9
0 0
0 1
0 2
1 0
5 2
10 1
20 3
1145 14
19198 10
output:
0
1
3
1
7
11
23
1279
19455

Note: After $1$ second, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[1, 3, 3, 7, 7, 7]$.

After $2$ seconds, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[3, 3, 7, 7, 7, 7]$.

Buggy Code:
def calculate_value(n, m, cache):
    key = (n, m)
    if key in cache:
        return cache[key]

    if m == 0:
        result = n
    elif n == 0:
        result = calculate_value(0, m - 1, cache) | abs(calculate_value(1, m - 1, cache))
    else:
        prev_m1 = m - 1
        val1 = calculate_value(n - 1, prev_m1, cache)
        val2 = abs(calculate_value(n, prev_m1, cache))
        val3 = calculate_value(n + 1, prev_m1, cache)

        result = val1 | val2 | val3

    cache[key] = result
    return result

def main():
    t = int(input())
    results = []
    for _ in range(t):
        n, m = map(int, input().split())
        results.append(calculate_value(n, m, {}))
    for result in results:
        print(result)

if __name__ == "__main__":
    main()

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

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

Think step by step.

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

1. **Initial Values**: The initial values of the sequence are straightforward, but we should still test them.
2. **Boundary Conditions**: The behavior at the boundaries, especially for `n = 0` and the first few values of `n`.
3. **Recursive Depth**: The recursion might lead to excessive depth or incorrect calculations, especially for larger values of `m`.
4. **Bitwise Operations**: Ensure that the bitwise OR operations are correctly applied.

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

- **Initial Values**: Test with `m = 0` to ensure the initial values are correctly returned.
- **Boundary Conditions**: Test with `n = 0` and small values of `n` and `m`.
- **Larger Values**: Test with larger values of `n` and `m` to check the recursive calculations and bitwise operations.

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

```plaintext
4
0 0
0 1
1 2
10 5
```

This test input includes:
- `0 0`: Initial value of `a_0`.
- `0 1`: Value of `a_0` after 1 second.
- `1 2`: Value of `a_1` after 2 seconds.
- `10 5`: Value of `a_10` after 5 seconds, which will test the recursive depth and bitwise operations.

This should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and an Infinite Sequence
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a sequence $a_0, a_1, a_2, \ldots$ of infinite length. Initially $a_i = i$ for every non-negative integer $i$.

After every second, each element of the sequence will simultaneously change. $a_i$ will change to $a_{i - 1} \mid a_i \mid a_{i + 1}$ for every positive integer $i$. $a_0$ will change to $a_0 \mid a_1$. Here, $|$ denotes [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR).

Turtle is asked to find the value of $a_n$ after $m$ seconds. In particular, if $m = 0$, then he needs to find the initial value of $a_n$. He is tired of calculating so many values, so please help him!

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

The first line of each test case contains two integers $n, m$ ($0 \le n, m \le 10^9$).

Output: For each test case, output a single integer — the value of $a_n$ after $m$ seconds.

Examples:
input:
9
0 0
0 1
0 2
1 0
5 2
10 1
20 3
1145 14
19198 10
output:
0
1
3
1
7
11
23
1279
19455

Note: After $1$ second, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[1, 3, 3, 7, 7, 7]$.

After $2$ seconds, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[3, 3, 7, 7, 7, 7]$.

Buggy Code:
t=int(input())

for _ in range(t):
    n,m=map(int,input().split())
    count=0
    list_n=[]
    min_n=n-m
    zxc=0
    if min_n<0:
        min_n=0
        zxc=1
    max_n=n+m
    num_n=max_n-min_n
    for i1 in range(min_n,max_n+1):
        list_n.append(i1)
    while count!=m:
        count+=1
        list_m=[0]*(num_n+1)
        for i in range(num_n):
            if zxc==1:
                if i == 0:
                    a = list_n[i] | list_n[i + 1]
                    list_m[i] = a
                else:
                    a = list_n[i - 1] | list_n[i] | list_n[i + 1]
                    list_m[i] = a
            else:
                a = list_n[i - 1] | list_n[i] | list_n[i + 1]
                list_m[i] = a
        for o in range(num_n):
            list_n[o]=list_m[o]
    if zxc==1:
        print(list_n[n])
    else:
        print(list_n[m])

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

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

Think step by step.

Response:
To create a test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that highlight potential issues in the logic. Here are some considerations:

1. **Initial Values and Simple Cases**:
   - Test cases where `m = 0` to ensure the initial value is correctly returned.
   - Test cases where `n = 0` to check the special handling for the first element.

2. **Boundary Conditions**:
   - Test cases where `n` is very close to `0` and `m` is large to see how the code handles the boundary conditions for the first element.
   - Test cases where `n` is large and `m` is small to ensure the code can handle large initial values.

3. **Complex Cases**:
   - Test cases where `n` and `m` are both large to see how the code performs under more complex conditions.

4. **Edge Cases**:
   - Test cases where `n` is exactly `m` to check the handling of equal values.

Based on these considerations, here is a small-scale test input that includes various edge cases and scenarios:

```plaintext
5
0 0
0 1
1 1
10 0
10 10
```

This test input includes:
1. A simple case where `n = 0` and `m = 0` to check the initial value.
2. A simple case where `n = 0` and `m = 1` to check the first element after one second.
3. A simple case where `n = 1` and `m = 1` to check a regular element after one second.
4. A case where `n = 10` and `m = 0` to check a larger initial value.
5. A complex case where `n = 10` and `m = 10` to check a larger value after multiple seconds.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and an Infinite Sequence
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a sequence $a_0, a_1, a_2, \ldots$ of infinite length. Initially $a_i = i$ for every non-negative integer $i$.

After every second, each element of the sequence will simultaneously change. $a_i$ will change to $a_{i - 1} \mid a_i \mid a_{i + 1}$ for every positive integer $i$. $a_0$ will change to $a_0 \mid a_1$. Here, $|$ denotes [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR).

Turtle is asked to find the value of $a_n$ after $m$ seconds. In particular, if $m = 0$, then he needs to find the initial value of $a_n$. He is tired of calculating so many values, so please help him!

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

The first line of each test case contains two integers $n, m$ ($0 \le n, m \le 10^9$).

Output: For each test case, output a single integer — the value of $a_n$ after $m$ seconds.

Examples:
input:
9
0 0
0 1
0 2
1 0
5 2
10 1
20 3
1145 14
19198 10
output:
0
1
3
1
7
11
23
1279
19455

Note: After $1$ second, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[1, 3, 3, 7, 7, 7]$.

After $2$ seconds, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[3, 3, 7, 7, 7, 7]$.

Buggy Code:
def solve():
  n, m = [int(x) for x in input().split(" ")]
  print(n|(n+m)|(1 if m else 0))
for _ in range(int(input())):
  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
3
0 0
1 1
2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and an Infinite Sequence
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a sequence $a_0, a_1, a_2, \ldots$ of infinite length. Initially $a_i = i$ for every non-negative integer $i$.

After every second, each element of the sequence will simultaneously change. $a_i$ will change to $a_{i - 1} \mid a_i \mid a_{i + 1}$ for every positive integer $i$. $a_0$ will change to $a_0 \mid a_1$. Here, $|$ denotes [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR).

Turtle is asked to find the value of $a_n$ after $m$ seconds. In particular, if $m = 0$, then he needs to find the initial value of $a_n$. He is tired of calculating so many values, so please help him!

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

The first line of each test case contains two integers $n, m$ ($0 \le n, m \le 10^9$).

Output: For each test case, output a single integer — the value of $a_n$ after $m$ seconds.

Examples:
input:
9
0 0
0 1
0 2
1 0
5 2
10 1
20 3
1145 14
19198 10
output:
0
1
3
1
7
11
23
1279
19455

Note: After $1$ second, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[1, 3, 3, 7, 7, 7]$.

After $2$ seconds, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[3, 3, 7, 7, 7, 7]$.

Buggy Code:
import java.util.*;

public class Tais {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int a = sc.nextInt();
        long[] results = new long[a];

        for (int j = 0; j < a; j++) {
            long q = sc.nextLong();
            long p = sc.nextLong();
            results[j] = fvs(q, p);
        }

        for (long result : results) {
            System.out.println(result);
        }

        sc.close();
    }

    private static long fvs(long q, long p) {
        if (p == 0) {
            return q;
        }

        long result = q;
        for (long j = 1; j <= p; j++) {
            result |= (q + j);
        }

        return result;
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on scenarios where the code's logic fails. The buggy code has a fundamental issue in the way it calculates the sequence values after each second. Specifically, it incorrectly assumes that the value of \( a_i \) after \( p \) seconds can be calculated by performing a bitwise OR on the initial value \( q \) and the values \( q + j \) for \( j \) from 1 to \( p \). This is incorrect because the sequence values are updated based on the values of their neighbors, not just the initial value plus an offset.

Here are the steps to generate a test input that exposes this bug:

1. **Edge Case with Small \( n \) and \( m \)**: Test with small values of \( n \) and \( m \) to ensure the basic logic is correct.
2. **Edge Case with Large \( n \) and \( m \)**: Test with large values of \( n \) and \( m \) to ensure the code handles large inputs correctly.
3. **Edge Case with \( m = 0 \)**: Test with \( m = 0 \) to ensure the initial value is returned correctly.
4. **Edge Case with \( n = 0 \)**: Test with \( n = 0 \) to ensure the boundary condition is handled correctly.
5. **Edge Case with \( n \) and \( m \) such that the sequence values should change**: Test with values that should result in different values after each second to ensure the bitwise OR logic is applied correctly.

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

```plaintext
5
0 0
0 1
1 1
10 1
10 2
```

This test input includes:
- \( (0, 0) \): Initial value of \( a_0 \).
- \( (0, 1) \): Value of \( a_0 \) after 1 second.
- \( (1, 1) \): Value of \( a_1 \) after 1 second.
- \( (10, 1) \): Value of \( a_10 \) after 1 second.
- \( (10, 2) \): Value of \( a_10 \) after 2 seconds.

These cases should help identify the incorrect logic in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and an Infinite Sequence
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a sequence $a_0, a_1, a_2, \ldots$ of infinite length. Initially $a_i = i$ for every non-negative integer $i$.

After every second, each element of the sequence will simultaneously change. $a_i$ will change to $a_{i - 1} \mid a_i \mid a_{i + 1}$ for every positive integer $i$. $a_0$ will change to $a_0 \mid a_1$. Here, $|$ denotes [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR).

Turtle is asked to find the value of $a_n$ after $m$ seconds. In particular, if $m = 0$, then he needs to find the initial value of $a_n$. He is tired of calculating so many values, so please help him!

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

The first line of each test case contains two integers $n, m$ ($0 \le n, m \le 10^9$).

Output: For each test case, output a single integer — the value of $a_n$ after $m$ seconds.

Examples:
input:
9
0 0
0 1
0 2
1 0
5 2
10 1
20 3
1145 14
19198 10
output:
0
1
3
1
7
11
23
1279
19455

Note: After $1$ second, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[1, 3, 3, 7, 7, 7]$.

After $2$ seconds, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[3, 3, 7, 7, 7, 7]$.

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

public class Main{
    public static void main(String args[]){int t=sc.nextInt();while(t-->0){solve();}}
    static int mod=(int)1e9;
    static long ceil(long a, long b) {return ((long) Math.ceil(((double) (a) / b)));}
    static long abs(long a) {if (a < 0) return (-1 * a);return a;}
    static long log(long a) {return (long) (Math.log(a));}
    static long log2(long a) {return (long) (Math.log(a) / Math.log(2));}
    static boolean isSet(long x, long bit) {return ((x >> bit) & 1L) == 1L;}
    static long sum(ArrayList<Long> a) {long sum = 0; for(long ele : a) sum += ele; return sum;}
    static void sort(ArrayList<Long> a) {Collections.sort(a);}
    static void rsort(ArrayList<Long> a) {Collections.sort(a, Collections.reverseOrder());}
    public static long sum(int a[]){long sum=0;for(int i:a)sum+=i;return sum;}
    static int max(long a[]) {int in = 0;long m = a[0];for (int i = 1; i < a.length; i++) {if (a[i] > m) {m = a[i];in = i;}}return in;}
    static int min(long a[]) {int in = 0;long m = a[0];for (int i = 1; i < a.length; i++) {if (a[i] < m) {m = a[i];in = i;}}return in;}
    public static void sieve(int MAXN,int spf[]) {spf[1] = 1; for (int i = 2; i < MAXN; i++) spf[i] = i;for (int i = 4; i < MAXN; i += 2) spf[i] = 2;for (int i = 3; i * i < MAXN; i++) if (spf[i] == i) for (int j = i * i; j < MAXN; j += i) if (spf[j] == j) spf[j] = i;}
    public static int upper_bound(long a[], long val) {int start = 0;int end = a.length - 1;while (start <= end) {int mid = (start + end) / 2;if (a[mid] >= val) {if (mid == 0 || a[mid-1] < val)return mid;end = mid - 1;} else start = mid + 1;}return -1;}
    public static int upper_bound(ArrayList<Long> a, long val) {if (a.size() == 0)return -1;int start = 0;int end = a.size() - 1;while (start <= end) {int mid = (start + end) / 2;if (a.get(mid) >= val) {if (mid == 0 || a.get(mid-1) < val) return mid;end = mid - 1;} else start = mid + 1;}return -1;}
    public static int lower_bound(ArrayList<Long> a, long val) {if (a.size() == 0)return -1;int start = 0;int end = a.size() - 1;while (start <= end) {int mid = (start + end) / 2;if (a.get(mid) > val) {end = mid - 1;continue;} else {if (mid == a.size() - 1 || a.get(mid+1) > val) return mid; start = mid + 1;}}return -1;}
    public static int lower_bound(long a[], long val) {int start = 0;int end = a.length - 1;while (start <= end) {int mid = (start + end) / 2;if (a[mid] > val) {end = mid - 1;continue;} else {if (mid == a.length - 1 || a[mid + 1] > val) return mid; start = mid + 1;}}return -1;}
    public static long sum(long a[]){long sum=0;for(long i:a)sum+=i;return sum;}
    static Scanner sc=new Scanner(System.in);
    static int dir[][]=new int[][]{{1,0},{-1,0},{0,1},{0,-1}};
    static int dir8[][]=new int[][]{{1,0},{-1,0},{0,1},{0,-1},{1,-1},{-1,-1},{-1,1},{1,1}};
    public static int binarySearch(int i,int nums[]){int a=(int)1e9;int left=0,right=nums.length-1;while(left<=right){int mid=(left+right)/2;if(nums[mid]>nums[i]){right=mid-1;a=mid;} else left=mid+1;}return a;}
    public static void readInt(int a[],int n){for(int  i=0;i<n;i++)a[i]=sc.nextInt();}
    public static void readLong(long a[],int n){for(int  i=0;i<n;i++)a[i]=sc.nextLong();}
    public static void println(String s){System.out.println(s);}
    public static void print(int a[]) {for (int i : a) System.out.print(i + " ");};
    public static void print(String n){System.out.print(n+" ");}
    public static List<Integer> factors(int n){List<Integer> factors = new ArrayList<>();for (int i = 1; i <= Math.sqrt(n); i++) {if (n % i == 0) {factors.add(i);if (i != n / i) {factors.add(n / i);}}}return factors;}
    public static void println(){System.out.println();}
    public static int gcd(int a,int b){if(b==0)return a;return gcd(b,a%b);}
    public static int lcm(int a,int b){return (a/(gcd(a,b))*b);}
    public static boolean[] prime(int n){boolean a[]=new boolean[n+1];Arrays.fill(a,true);for(int i=2;i*i<=n;i++){if(a[i]==true){for(int j=i+i;j<=n;j+=i){a[i]=false;}}}return a;}
    public static boolean isPrime(int n) {if (n <= 1) {return false;}if (n <= 3) {return true;}if (n % 2 == 0 || n % 3 == 0) {return false;}for (int i = 5; i * i <= n; i = i + 6) {if (n % i == 0 || n % (i + 2) == 0) {return false;}}return true;}
    public static long factorial(int n) {if (n == 0) {return 1;}long fact = 1;for (int i = 1; i <= n; i++) {fact *= i;}return fact;}
    public static long power(long a, long b) {long result = 1;while (b > 0) {if ((b & 1) != 0) {result = result * a;}a = a * a;b >>= 1;}return result;}
    public static void read2DArray(int a[][],int n,int m){for(int i=0;i<n;i++){for(int j=0;j<m;j++)a[i][j]=sc.nextInt();}}
    public static void reverse(long []a){int i=0,j=a.length-1;while(i<j){swap(i,j,a);i++;j--;}}
    public static String intToBins(int n) {return Integer.toBinaryString(n);};
    public static void no(){System.out.println("NO");}
    public static void yes(){System.out.println("YES");}
    public static long powerUsingArray(int a[]){long ans=0;for(int i=0;i<a.length;i++){if(a[i]>=1){ans+=Math.pow(2,i);}}return ans;}
    public static void swap(int i,int j,long a[]){long temp=a[i];a[i]=a[j];a[j]=temp;}
    public static long cal(long n){return n*(n+1)/2;}
    //-----------------------------------------------------Dont modify above this line-------------------------------------------//
    //-----------------------------------------------------Dont modify above this line-------------------------------------------//

    static void solve() {
        long l=sc.nextLong(),r=sc.nextInt();

        long x=l+1;
        long ans=x-1;
        while(r>0){
            ans=ans|x;
            x++;
            r--;
        }

        if(l-1>=0)ans|=l-1;
        println(ans+" ");

    }
}

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

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

Think step by step.

Response:
```plaintext
1
0 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and an Infinite Sequence
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a sequence $a_0, a_1, a_2, \ldots$ of infinite length. Initially $a_i = i$ for every non-negative integer $i$.

After every second, each element of the sequence will simultaneously change. $a_i$ will change to $a_{i - 1} \mid a_i \mid a_{i + 1}$ for every positive integer $i$. $a_0$ will change to $a_0 \mid a_1$. Here, $|$ denotes [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR).

Turtle is asked to find the value of $a_n$ after $m$ seconds. In particular, if $m = 0$, then he needs to find the initial value of $a_n$. He is tired of calculating so many values, so please help him!

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

The first line of each test case contains two integers $n, m$ ($0 \le n, m \le 10^9$).

Output: For each test case, output a single integer — the value of $a_n$ after $m$ seconds.

Examples:
input:
9
0 0
0 1
0 2
1 0
5 2
10 1
20 3
1145 14
19198 10
output:
0
1
3
1
7
11
23
1279
19455

Note: After $1$ second, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[1, 3, 3, 7, 7, 7]$.

After $2$ seconds, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[3, 3, 7, 7, 7, 7]$.

Buggy Code:
#include <bits/stdc++.h>
#define int long long
#define sz(a) (int)a.size()
#define all(a) a.begin(), a.end()

using namespace std;

signed main() {
    ios_base::sync_with_stdio(false); cin.tie(nullptr);

    int t; cin >>t;
    while(t--){
        int n,m; cin >> n >> m;
        if(m>0) {
            vector<int> a(m*2+2);
            vector<int> pa(m*2+2,-1);
            for (int i =0; i <= m*2; i++)
            {
                pa[i]=max(0LL,i+(n-m));
            }
            for (int M = m; M >= 1; M--)
            {
                for (int i = m-M; i <= m+M; i++)
                {
                    a[i]=pa[i]|pa[i+1];
                    a[i]|=pa[i-1];
                }
                for (int i = m; i <= m+M; i++)
                {
                    pa[i]=a[i];
                }
            }
            cout << a[m] << "\n";
        }else if(m==0){
            cout << n << "\n";
        }else{
            int lg=(1<<(int)(log2(n+m)))-1;
            int out=(lg-1)|(lg+1)|lg;
            cout << out << "\n";
        }
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
5
0 0
1 1
2 2
3 3
4 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and an Infinite Sequence
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a sequence $a_0, a_1, a_2, \ldots$ of infinite length. Initially $a_i = i$ for every non-negative integer $i$.

After every second, each element of the sequence will simultaneously change. $a_i$ will change to $a_{i - 1} \mid a_i \mid a_{i + 1}$ for every positive integer $i$. $a_0$ will change to $a_0 \mid a_1$. Here, $|$ denotes [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR).

Turtle is asked to find the value of $a_n$ after $m$ seconds. In particular, if $m = 0$, then he needs to find the initial value of $a_n$. He is tired of calculating so many values, so please help him!

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

The first line of each test case contains two integers $n, m$ ($0 \le n, m \le 10^9$).

Output: For each test case, output a single integer — the value of $a_n$ after $m$ seconds.

Examples:
input:
9
0 0
0 1
0 2
1 0
5 2
10 1
20 3
1145 14
19198 10
output:
0
1
3
1
7
11
23
1279
19455

Note: After $1$ second, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[1, 3, 3, 7, 7, 7]$.

After $2$ seconds, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[3, 3, 7, 7, 7, 7]$.

Buggy Code:
#include<iostream>
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define rep(i,a,b) for(long long i=a;i<b;i++)
#define rrep(i,a,b) for(long long i=a;i>=b;i--)
#define repin rep(i,0,n)
#define di(a) ll a;cin>>a;
#define precise(i) cout<<fixed<<setprecision(i)
#define vl vector<ll>
#define sl set<ll>
#define mll map<ll,ll>
#define take(a,n) for(ll j=0;j<n;j++) cin>>a[j];
#define give(a,n) for(ll j=0;j<n;j++) cout<<a[j]<<' ';
#define vpll vector<pair<ll,ll>>
#define sis string s;
#define sin string s;cin>>s;
#define db double
#define be(x) x.begin(),x.end()
#define pii pair<ll,ll>
#define pb push_back
#define pob pop_back
#define ff first
#define ss second
#define lb lower_bound
#define ub upper_bound
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp>
#include <functional> // for less
#include <iostream>
using namespace __gnu_pbds;
using namespace std;
typedef tree<ll, null_type, less<ll>, rb_tree_tag,
            tree_order_statistics_node_update>
    ordered_set;
const long long int M = 1e9+7;
ll power(ll a, ll b){
    ll res = 1;
    while(b>0){
        if(b&1){
            res = ((res%M)*(a%M))%M;
        }
        a = ((a%M)*(a%M))%M;
        b >>= 1;
    }
    return res;
}
void batman(){
    ll n,m;
    cin>>n>>m;
    ll bit[31]={0};
    // cout<<m<<"\n";
    for(ll i=0;i<=30;i++){
        bit[i]=((n>>i)&1);
    }
    ll ans=0;
    // cout<<m<<"\n";
    ll tot=0;
    for(ll i=30;i>=0;i--){
        if(bit[i]){ans+=(1ll<<i);tot+=(1ll<<i);}
        else{
            ll to=(1ll<<i);
            ll val=(to+tot)-n;
            ll val2=tot-n;
            for(ll j=0;j<=i;j++)val2+=(1ll<<j);
            // if(i==2)cout<<tot<<"\n";
            // if(i==0)cout<<(1ll<<i)<<"\n";
            // cout<<m<<"\n";
            if(min(val,val2)<=m){ans+=(1ll<<i);}
        }
    }
    cout<<ans<<"\n";
}
int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
     ll t=1;
     cin>>t;
     while(t--){
       batman();
     }
    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
0 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and an Infinite Sequence
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a sequence $a_0, a_1, a_2, \ldots$ of infinite length. Initially $a_i = i$ for every non-negative integer $i$.

After every second, each element of the sequence will simultaneously change. $a_i$ will change to $a_{i - 1} \mid a_i \mid a_{i + 1}$ for every positive integer $i$. $a_0$ will change to $a_0 \mid a_1$. Here, $|$ denotes [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR).

Turtle is asked to find the value of $a_n$ after $m$ seconds. In particular, if $m = 0$, then he needs to find the initial value of $a_n$. He is tired of calculating so many values, so please help him!

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

The first line of each test case contains two integers $n, m$ ($0 \le n, m \le 10^9$).

Output: For each test case, output a single integer — the value of $a_n$ after $m$ seconds.

Examples:
input:
9
0 0
0 1
0 2
1 0
5 2
10 1
20 3
1145 14
19198 10
output:
0
1
3
1
7
11
23
1279
19455

Note: After $1$ second, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[1, 3, 3, 7, 7, 7]$.

After $2$ seconds, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[3, 3, 7, 7, 7, 7]$.

Buggy Code:
# import math

def solve():
    n, m = list(map(int, input().split()))
    x = n.bit_length()
    mask = (1 << m) - 1
    result = n | mask
    if result > 2**x + 2:
        y = result.bit_length()
        mask = (1 << (2 * x - y)) - 1
        result = n | mask
    print(result)

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

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that might cause the code to fail. The buggy code has a condition that checks if the result is greater than `2**x + 2` and then modifies the mask. This suggests that the code might not handle certain edge cases correctly, especially when `m` is large and the bitwise operations produce unexpected results.

Here are some key scenarios to consider:
1. **Initial Values**: Test with `m = 0` to ensure the initial value is correctly returned.
2. **Small Values of `n` and `m`**: Test with small values to ensure the basic logic works.
3. **Large Values of `n` and `m`**: Test with large values to check how the code handles large bitwise operations.
4. **Edge Cases**: Test with values of `n` and `m` that are powers of 2 or just below/above powers of 2.

Based on these considerations, here is a small-scale test input that covers a range of scenarios:

```plaintext
5
0 0
1 1
10 1
1000 10
1073741823 30
```

This test input includes:
- A case where `m = 0` to check the initial value.
- A simple case with small `n` and `m` to verify basic logic.
- A case with a moderate `n` and `m` to check the bitwise operations.
- A case with a large `n` and `m` to stress the code and check for potential overflow or incorrect mask handling.
- A case with `n` and `m` at the upper limit of the input constraints to ensure the code can handle extreme values.

This should help in exposing 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: Turtle and an Infinite Sequence
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a sequence $a_0, a_1, a_2, \ldots$ of infinite length. Initially $a_i = i$ for every non-negative integer $i$.

After every second, each element of the sequence will simultaneously change. $a_i$ will change to $a_{i - 1} \mid a_i \mid a_{i + 1}$ for every positive integer $i$. $a_0$ will change to $a_0 \mid a_1$. Here, $|$ denotes [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR).

Turtle is asked to find the value of $a_n$ after $m$ seconds. In particular, if $m = 0$, then he needs to find the initial value of $a_n$. He is tired of calculating so many values, so please help him!

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

The first line of each test case contains two integers $n, m$ ($0 \le n, m \le 10^9$).

Output: For each test case, output a single integer — the value of $a_n$ after $m$ seconds.

Examples:
input:
9
0 0
0 1
0 2
1 0
5 2
10 1
20 3
1145 14
19198 10
output:
0
1
3
1
7
11
23
1279
19455

Note: After $1$ second, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[1, 3, 3, 7, 7, 7]$.

After $2$ seconds, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[3, 3, 7, 7, 7, 7]$.

Buggy Code:
test=int(input())
for i in range(test):
    st1=list(map(int,input().strip().split()))
    x,t=st1
    ans=0
    if x==0:
            for i in range(t+1):
                ans=(ans|(x+i))
            print(ans)
    else:
        for i in range(t+1):
            ans=(ans|(x-i))
        for i in range(t+1):
            ans=(ans|(x+i))
        print(ans)

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

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

Think step by step.

Response:
```plaintext
3
0 1000
1 1000
1000 1000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and an Infinite Sequence
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a sequence $a_0, a_1, a_2, \ldots$ of infinite length. Initially $a_i = i$ for every non-negative integer $i$.

After every second, each element of the sequence will simultaneously change. $a_i$ will change to $a_{i - 1} \mid a_i \mid a_{i + 1}$ for every positive integer $i$. $a_0$ will change to $a_0 \mid a_1$. Here, $|$ denotes [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR).

Turtle is asked to find the value of $a_n$ after $m$ seconds. In particular, if $m = 0$, then he needs to find the initial value of $a_n$. He is tired of calculating so many values, so please help him!

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

The first line of each test case contains two integers $n, m$ ($0 \le n, m \le 10^9$).

Output: For each test case, output a single integer — the value of $a_n$ after $m$ seconds.

Examples:
input:
9
0 0
0 1
0 2
1 0
5 2
10 1
20 3
1145 14
19198 10
output:
0
1
3
1
7
11
23
1279
19455

Note: After $1$ second, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[1, 3, 3, 7, 7, 7]$.

After $2$ seconds, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[3, 3, 7, 7, 7, 7]$.

Buggy Code:
import java.util.*;
import java.util.Map;
import java.util.Scanner;
import java.util.Collection;
public class Main {
    static long gcd(int a, int b)
    {
        int i;
        if (a < b)
            i = a;
        else
            i = b;
        for (i = i; i > 1; i--) {
            if (a % i == 0 && b % i == 0)
                return i;
        }
        return 1;
    }
    public static long lcm(int a, int b) {
        long j1= (long) a *b;
        // return j1;
        return (j1/ gcd(a, b));
    }
    public static double log2(double x) {
        return Math.log(x) / Math.log(2);
    }
    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        int t;
        t=in.nextInt();
        while(t>0){
            int n,m;
            n=in.nextInt();
            m=in.nextInt();
            int l1=n;
            int[] arr=new int[65];
            for(int i=0;i<65;++i)arr[i]=0;
            int pos=0;
            while(l1!=0){
                if((l1&1)==1){
                    arr[pos]=1;
                }
                l1>>=1;
                pos++;
            }
            long sum=0;
            int p1=-1;
            for(int i=0;i<65;++i){
                if(arr[i]==1)sum+=(1L<<i);
                // if((arr[i]==1 && arr[i+1]==1))continue;
                if((arr[i]==0)){
                    if(m>=((1L<<i)-sum)){
                        p1=Math.max(p1,i);
                        // m-=(1L<<i)-sum;
                    }else{
                        break;
                    }
                }
            }
            for(int i=p1;i>-1;i--){
                arr[i]=1;
            }
            long  ans=0;
            // System.out.println(ans);
            for(int i=0;i<65;++i){
                if(arr[i]==1)ans+=(1L<<i);
            }
            System.out.println(ans);
            t--;
        }
    }
}

**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. The buggy code seems to have issues with how it handles the bitwise operations and the transformation of the sequence over time. Here are the steps to think through the test case:

1. **Initial Values and Zero Seconds**: The code should handle the initial values correctly, so a test case with `m = 0` should work fine.
2. **Single Step Transformations**: Test the sequence after one second to see if the bitwise OR operations are applied correctly.
3. **Multiple Step Transformations**: Test the sequence after multiple seconds to see if the cumulative effects of the bitwise OR operations are handled correctly.
4. **Large Values of `n` and `m`**: Test with large values to ensure the code can handle large inputs without performance issues or incorrect logic.
5. **Edge Cases**: Test with edge cases like `n = 0` and `m = 1`, `n = 1` and `m = 1`, and so on.

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

```plaintext
4
0 1
1 1
2 1
10 2
```

This test input includes:
- The initial value of `a_0` after one second.
- The value of `a_1` after one second.
- The value of `a_2` after one second.
- The value of `a_10` after two seconds.

These cases cover the initial transformation, single-step transformations, and a multi-step transformation, which should help identify any issues in the bitwise OR operations and the logic of the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and an Infinite Sequence
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a sequence $a_0, a_1, a_2, \ldots$ of infinite length. Initially $a_i = i$ for every non-negative integer $i$.

After every second, each element of the sequence will simultaneously change. $a_i$ will change to $a_{i - 1} \mid a_i \mid a_{i + 1}$ for every positive integer $i$. $a_0$ will change to $a_0 \mid a_1$. Here, $|$ denotes [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR).

Turtle is asked to find the value of $a_n$ after $m$ seconds. In particular, if $m = 0$, then he needs to find the initial value of $a_n$. He is tired of calculating so many values, so please help him!

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

The first line of each test case contains two integers $n, m$ ($0 \le n, m \le 10^9$).

Output: For each test case, output a single integer — the value of $a_n$ after $m$ seconds.

Examples:
input:
9
0 0
0 1
0 2
1 0
5 2
10 1
20 3
1145 14
19198 10
output:
0
1
3
1
7
11
23
1279
19455

Note: After $1$ second, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[1, 3, 3, 7, 7, 7]$.

After $2$ seconds, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[3, 3, 7, 7, 7, 7]$.

Buggy Code:
import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;

public class Solution2 {
    static int bitwiseOROverRange(int ll , int ul){
        int Ans = ll;
        while (Ans<ul) {
            Ans|= (Ans+1);
        }
        return Ans;
    }
    public static void main(String[] args) {
        FastScanner fs = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int t = fs.nextInt();
        while (t--!=0) {
            int n = fs.nextInt();
            int m = fs.nextInt();
            int Ans = 0;
            if (n==0) {
                Ans = bitwiseOROverRange(0,m);
            }
            else{
                Ans = bitwiseOROverRange(n-m, n+m);
            }
            out.println(Ans);
        }
        out.close();
    }
}

class Pair {
    int x, y;
    public Pair(int x, int y) {
        this.x = x;
        this.y = y;
    }
}

//Template stolen from Former IGM SecondThread
class FastScanner {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st = new StringTokenizer("");

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

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

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

// 9
// 0 0
// 0 1
// 0 2
// 1 0
// 5 2
// 10 1
// 20 3
// 1145 14
// 19198 10

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

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

Think step by step.

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

1. **Edge Case for \( n = 0 \):**
   - The code uses `bitwiseOROverRange(0, m)` for \( n = 0 \). We need to ensure that this function behaves correctly for small values of \( m \).

2. **Edge Case for \( n \neq 0 \):**
   - The code uses `bitwiseOROverRange(n-m, n+m)` for \( n \neq 0 \). We need to ensure that this function handles cases where \( n-m \) or \( n+m \) are small or large.

3. **Large Values of \( m \):**
   - The function `bitwiseOROverRange` might fail for large values of \( m \) due to the loop condition `while (Ans<ul)`. We need to test with large \( m \) values to see if the loop runs indefinitely or produces incorrect results.

4. **Small Values of \( n \) and \( m \):**
   - Test with small values of \( n \) and \( m \) to ensure the function behaves as expected.

5. **Large Values of \( n \):**
   - Test with large values of \( n \) to ensure the function handles large ranges correctly.

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

```plaintext
6
0 0
0 1
0 2
1 0
1 1
1 2
```

This input includes:
- \( n = 0 \) with \( m = 0, 1, 2 \) to test the initial and subsequent changes for the first element.
- \( n = 1 \) with \( m = 0, 1, 2 \) to test the changes for a non-zero element.

This input is small enough to be manageable and should help identify any issues in the `bitwiseOROverRange` function.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and an Infinite Sequence
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a sequence $a_0, a_1, a_2, \ldots$ of infinite length. Initially $a_i = i$ for every non-negative integer $i$.

After every second, each element of the sequence will simultaneously change. $a_i$ will change to $a_{i - 1} \mid a_i \mid a_{i + 1}$ for every positive integer $i$. $a_0$ will change to $a_0 \mid a_1$. Here, $|$ denotes [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR).

Turtle is asked to find the value of $a_n$ after $m$ seconds. In particular, if $m = 0$, then he needs to find the initial value of $a_n$. He is tired of calculating so many values, so please help him!

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

The first line of each test case contains two integers $n, m$ ($0 \le n, m \le 10^9$).

Output: For each test case, output a single integer — the value of $a_n$ after $m$ seconds.

Examples:
input:
9
0 0
0 1
0 2
1 0
5 2
10 1
20 3
1145 14
19198 10
output:
0
1
3
1
7
11
23
1279
19455

Note: After $1$ second, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[1, 3, 3, 7, 7, 7]$.

After $2$ seconds, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[3, 3, 7, 7, 7, 7]$.

Buggy Code:
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n, m;
        cin >> n >> m;
        if (m == 0) {
            cout << n << endl;
        } else {
            int decimal_number = n + m - 1;
            int result;
            if (decimal_number == 0) {
                result = 0;
            } else if (decimal_number == 1) {
                result = 1;
            } else {
                int num_bits = static_cast<int>(log2(decimal_number)) + 1;
                result = 1 << (num_bits - 1);
            }

            int a = 0;
            for (int i = max(0, max(n - m + 1, result)); i < n + m; i++) {
                if (i == 0) {
                    a = max(a, i | (i + 1));
                } else {
                    a = max(a, (i - 1) | i | (i + 1));
                }
            }
            cout << a << endl;
        }
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
0 2
1 2
2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and an Infinite Sequence
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a sequence $a_0, a_1, a_2, \ldots$ of infinite length. Initially $a_i = i$ for every non-negative integer $i$.

After every second, each element of the sequence will simultaneously change. $a_i$ will change to $a_{i - 1} \mid a_i \mid a_{i + 1}$ for every positive integer $i$. $a_0$ will change to $a_0 \mid a_1$. Here, $|$ denotes [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR).

Turtle is asked to find the value of $a_n$ after $m$ seconds. In particular, if $m = 0$, then he needs to find the initial value of $a_n$. He is tired of calculating so many values, so please help him!

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

The first line of each test case contains two integers $n, m$ ($0 \le n, m \le 10^9$).

Output: For each test case, output a single integer — the value of $a_n$ after $m$ seconds.

Examples:
input:
9
0 0
0 1
0 2
1 0
5 2
10 1
20 3
1145 14
19198 10
output:
0
1
3
1
7
11
23
1279
19455

Note: After $1$ second, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[1, 3, 3, 7, 7, 7]$.

After $2$ seconds, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[3, 3, 7, 7, 7, 7]$.

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

typedef long long ll;
typedef vector<ll> v64;
typedef vector<int> v32;
#define loop(i, a, b) for (ll i = a; i < b; ++i) // Use pre-increment for clarity
#define all(x) (x).begin(), (x).end()

void solve() {
    ll n, t;
    cin >> n >> t;
    ll ans = 0;

    if (n <= (t + 1)) {
        ll k = n + t;
        for (ll i = k; i >= 0; i--) {
            ans |= i;
            if (((i+1) & (i)) == 0) {
                break;
            }
        }
    }
    else {
        for (ll i = n + t; i >= n - t; i--) { // Track iterations
            ans |= i;
            if (((i+1) & (i)) == 0) {
                break;
            }
        }
    }

    cout << ans;
}

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

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

    int test;
    cin >> test;

    while (test--) {
        solve();
        cout << "\n";
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
0 0
0 1
1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and an Infinite Sequence
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a sequence $a_0, a_1, a_2, \ldots$ of infinite length. Initially $a_i = i$ for every non-negative integer $i$.

After every second, each element of the sequence will simultaneously change. $a_i$ will change to $a_{i - 1} \mid a_i \mid a_{i + 1}$ for every positive integer $i$. $a_0$ will change to $a_0 \mid a_1$. Here, $|$ denotes [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR).

Turtle is asked to find the value of $a_n$ after $m$ seconds. In particular, if $m = 0$, then he needs to find the initial value of $a_n$. He is tired of calculating so many values, so please help him!

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

The first line of each test case contains two integers $n, m$ ($0 \le n, m \le 10^9$).

Output: For each test case, output a single integer — the value of $a_n$ after $m$ seconds.

Examples:
input:
9
0 0
0 1
0 2
1 0
5 2
10 1
20 3
1145 14
19198 10
output:
0
1
3
1
7
11
23
1279
19455

Note: After $1$ second, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[1, 3, 3, 7, 7, 7]$.

After $2$ seconds, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[3, 3, 7, 7, 7, 7]$.

Buggy Code:
for _ in range(int(input())):
    n,m=map(int,input().split())

    left,right=max(0,n-m),n+m
    def orr(low,high):
        if low==high:
            return low
        mid=(low+high)//2
        val1=orr(low,mid)
        val2=orr(mid+1,high)
        return val1|val2
    print(orr(left,right))

**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
0 0
1 1
2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and an Infinite Sequence
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a sequence $a_0, a_1, a_2, \ldots$ of infinite length. Initially $a_i = i$ for every non-negative integer $i$.

After every second, each element of the sequence will simultaneously change. $a_i$ will change to $a_{i - 1} \mid a_i \mid a_{i + 1}$ for every positive integer $i$. $a_0$ will change to $a_0 \mid a_1$. Here, $|$ denotes [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR).

Turtle is asked to find the value of $a_n$ after $m$ seconds. In particular, if $m = 0$, then he needs to find the initial value of $a_n$. He is tired of calculating so many values, so please help him!

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

The first line of each test case contains two integers $n, m$ ($0 \le n, m \le 10^9$).

Output: For each test case, output a single integer — the value of $a_n$ after $m$ seconds.

Examples:
input:
9
0 0
0 1
0 2
1 0
5 2
10 1
20 3
1145 14
19198 10
output:
0
1
3
1
7
11
23
1279
19455

Note: After $1$ second, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[1, 3, 3, 7, 7, 7]$.

After $2$ seconds, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[3, 3, 7, 7, 7, 7]$.

Buggy Code:
import sys
input = sys.stdin.read

def process_cases(cases):
    results = []
    for l in cases:
        imax = l[0]
        m = l[1]
        left, right = imax, imax
        soln = imax

        while m > 0:
            right += 1
            if left > 0:
                left -= 1
                soln |= left
            soln |= right
            m -= 1

        results.append(soln)
    return results

def main():
    data = input().split()
    t = int(data[0])
    index = 1
    cases = []
    for _ in range(t):
        imax = int(data[index])
        m = int(data[index + 1])
        cases.append([imax, m])
        index += 2

    results = process_cases(cases)
    for result in results:
        print(result)

if __name__ == "__main__":
    main()

**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
4
0 1000000000
1 1000000000
1000000000 0
1000000000 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and an Infinite Sequence
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a sequence $a_0, a_1, a_2, \ldots$ of infinite length. Initially $a_i = i$ for every non-negative integer $i$.

After every second, each element of the sequence will simultaneously change. $a_i$ will change to $a_{i - 1} \mid a_i \mid a_{i + 1}$ for every positive integer $i$. $a_0$ will change to $a_0 \mid a_1$. Here, $|$ denotes [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR).

Turtle is asked to find the value of $a_n$ after $m$ seconds. In particular, if $m = 0$, then he needs to find the initial value of $a_n$. He is tired of calculating so many values, so please help him!

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

The first line of each test case contains two integers $n, m$ ($0 \le n, m \le 10^9$).

Output: For each test case, output a single integer — the value of $a_n$ after $m$ seconds.

Examples:
input:
9
0 0
0 1
0 2
1 0
5 2
10 1
20 3
1145 14
19198 10
output:
0
1
3
1
7
11
23
1279
19455

Note: After $1$ second, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[1, 3, 3, 7, 7, 7]$.

After $2$ seconds, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[3, 3, 7, 7, 7, 7]$.

Buggy Code:
import java.util.Scanner;

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

        int testCases = scanner.nextInt();

        StringBuilder outputBuilder = new StringBuilder();

        for (int testCase = 0; testCase < testCases; testCase++) {
            long firstNum = scanner.nextLong();
            long secondNum = scanner.nextLong();

            long finalResult = calculateResult(firstNum, secondNum);

            outputBuilder.append(finalResult).append("\n");
        }

        System.out.print(outputBuilder.toString());

        scanner.close();
    }

    private static long calculateResult(long firstNum, long secondNum) {
        long startingPoint = Math.max(0, firstNum - secondNum);
        long endingPoint = firstNum + secondNum;

        long finalResult = 0;
        for (long i = startingPoint; i <= endingPoint; i++) {
            finalResult |= i;
        }

        return finalResult;
    }
}

**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 buggy code calculates the result by performing a bitwise OR over a range of numbers. The range is determined by `startingPoint` and `endingPoint`, which are derived from `firstNum` and `secondNum`.

### Key Points to Consider:
1. **Initial Values**: Test cases where `m = 0` to ensure the initial value of `a_n` is correctly returned.
2. **Boundary Conditions**: Test cases where `n` is close to `0` or `m` is close to `0`.
3. **Large Values**: Test cases with large values of `n` and `m` to check the performance and correctness of the bitwise OR operation.
4. **Special Cases**: Test cases where `n` and `m` are such that the range `startingPoint` to `endingPoint` is minimized or maximized.

### Chosen Test Input:
- **Test Case 1**: `n = 0, m = 0` (Initial value)
- **Test Case 2**: `n = 0, m = 1` (Boundary condition)
- **Test Case 3**: `n = 1, m = 1` (Simple case)
- **Test Case 4**: `n = 10, m = 10` (Large values)
- **Test Case 5**: `n = 1000000000, m = 1000000000` (Extreme large values)

### Test Input:
```plaintext
5
0 0
0 1
1 1
10 10
1000000000 1000000000
```

This test input includes a mix of simple and complex cases, ensuring that the code is tested for both small and large values, as well as boundary conditions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Turtle and an Infinite Sequence
time_limit_ms: 1000
memory_limit_mb: 256
Description: There is a sequence $a_0, a_1, a_2, \ldots$ of infinite length. Initially $a_i = i$ for every non-negative integer $i$.

After every second, each element of the sequence will simultaneously change. $a_i$ will change to $a_{i - 1} \mid a_i \mid a_{i + 1}$ for every positive integer $i$. $a_0$ will change to $a_0 \mid a_1$. Here, $|$ denotes [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR).

Turtle is asked to find the value of $a_n$ after $m$ seconds. In particular, if $m = 0$, then he needs to find the initial value of $a_n$. He is tired of calculating so many values, so please help him!

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

The first line of each test case contains two integers $n, m$ ($0 \le n, m \le 10^9$).

Output: For each test case, output a single integer — the value of $a_n$ after $m$ seconds.

Examples:
input:
9
0 0
0 1
0 2
1 0
5 2
10 1
20 3
1145 14
19198 10
output:
0
1
3
1
7
11
23
1279
19455

Note: After $1$ second, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[1, 3, 3, 7, 7, 7]$.

After $2$ seconds, $[a_0, a_1, a_2, a_3, a_4, a_5]$ will become $[3, 3, 7, 7, 7, 7]$.

Buggy Code:
import java.util.Scanner;
public class q2949 {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int testcases = input.nextInt();
        for (int i = 0; i < testcases; i++) {
            int n = input.nextInt();
            int m = input.nextInt();

            System.out.println(msecan(n,m));
        }
    }
    public static int msecan(int n, int m){
        int left = Math.max(0, n-m);
        int or1 = 0;

        for(int i = left; i<=n+m; i++){
            or1|=i;
        }
        return or1;

    }

}

**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 scenarios where the sequence transformation logic is not correctly handled. The buggy code seems to have an issue with the range of values it considers for the bitwise OR operation. Specifically, it uses a range from `n-m` to `n+m`, which might not correctly reflect the actual sequence transformation after multiple seconds.

### Key Points to Consider:
1. **Initial Values**: The initial values of the sequence are `a_i = i` for every non-negative integer `i`.
2. **Transformation**: After each second, each element `a_i` changes to `a_{i-1} | a_i | a_{i+1}` for positive `i`, and `a_0` changes to `a_0 | a_1`.
3. **Edge Cases**:
   - **Small `m`**: The sequence changes significantly after just a few seconds.
   - **Large `n` and `m`**: The range of values to consider can be very large, which might cause issues with the current logic.
   - **Boundary Conditions**: Edge cases where `n` is small or `m` is large can expose issues.

### Test Input:
To test the code, we can use a combination of small and large values for `n` and `m` to see if the code handles the transformations correctly.

```plaintext
5
0 0
1 1
2 2
10 3
10000 10
```

### Explanation:
1. **Test Case 1**: `0 0`
   - Initial value of `a_0` is `0`.
   - Expected output: `0`

2. **Test Case 2**: `1 1`
   - After 1 second, `a_1` changes to `a_0 | a_1 | a_2` which is `0 | 1 | 2 = 3`.
   - Expected output: `3`

3. **Test Case 3**: `2 2`
   - After 1 second, `a_2` changes to `a_1 | a_2 | a_3` which is `1 | 2 | 3 = 3`.
   - After 2 seconds, `a_2` changes to `a_1 | a_2 | a_3` which is `3 | 3 | 7 = 7`.
   - Expected output: `7`

4. **Test Case 4**: `10 3`
   - After 1 second, `a_10` changes to `a_9 | a_10 | a_11` which is `9 | 10 | 11 = 15`.
   - After 2 seconds, `a_10` changes to `a_9 | a_10 | a_11` which is `15 | 15 | 15 = 15`.
   - After 3 seconds, `a_10` changes to `a_9 | a_10 | a_11` which is `15 | 15 | 15 = 15`.
   - Expected output: `15`

5. **Test Case 5**: `10000 10`
   - After 10 seconds, the value of `a_10000` will have been transformed multiple times.
   - Expected output: `10239` (This is a more complex case, but the current logic might fail here due to the large range).

This test input should help expose any issues with the range and transformation logic in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cat, Fox and the Lonely Array
time_limit_ms: 2000
memory_limit_mb: 256
Description: Today, Cat and Fox found an array $a$ consisting of $n$ non-negative integers.

Define the loneliness of $a$ as the smallest positive integer $k$ ($1 \le k \le n$) such that for any two positive integers $i$ and $j$ ($1 \leq i, j \leq n - k +1$), the following holds: $$a_i | a_{i+1} | \ldots | a_{i+k-1} = a_j | a_{j+1} | \ldots | a_{j+k-1},$$ where $x | y$ denotes the [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR) of $x$ and $y$. In other words, for every $k$ consecutive elements, their bitwise OR should be the same. Note that the loneliness of $a$ is well-defined, because for $k = n$ the condition is satisfied.

Cat and Fox want to know how lonely the array $a$ is. Help them calculate the loneliness of the found array.

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

The first line of each test case contains one integer $n$ ($1 \leq n \leq 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i < 2^{20}$) — the elements of the array.

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $10^5$.

Output: For each test case, print one integer  — the loneliness of the given array.

Examples:
input:
7
1
0
3
2 2 2
3
1 0 2
5
3 0 1 4 2
5
2 0 4 0 2
7
0 0 0 0 1 2 4
8
0 1 3 2 2 1 0 3
output:
1
1
3
4
4
7
3

Note: In the first example, the loneliness of an array with a single element is always $1$, so the answer is $1$.

In the second example, the OR of each subarray of length $k = 1$ is $2$, so the loneliness of the whole array is $1$.

In the seventh example, it's true that $(0 | 1 | 3) = (1 | 3 | 2) = (3 | 2 | 2) = (2 | 2 | 1) = (2 | 1 | 0) = (1 | 0 | 3) = 3$, so the condition is satisfied for $k = 3$. We can verify that the condition is not true for any smaller $k$, so the answer is indeed $3$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define py cout << "YES" << endl;
#define pn cout << "NO" << endl;
#define nl cout << endl;
#define all(p) p.begin(), p.end()
#define pyr                    \
    {                          \
        cout << "YES" << endl; \
        return;                \
    }
#define pnr                   \
    {                         \
        cout << "NO" << endl; \
        return;               \
    }

void breathe()
{
    ll n;
    cin >> n;
    vector<ll> arr(n);
    for (auto &&i : arr)
    {
        cin >> i;
    }

    map<ll, vector<ll>> m;

    for (ll i = 0; i < n; i++)
    {
        for (ll j = 0; j < 21; j++)
        {
            if (arr[i] & (1 << j))
            {
                m[i].push_back(j);
            }
        }
    }

    map<ll, ll> freq;

    for (ll k = 1; k <= n; k++)
    {
        ll orr = 0;
        for (ll i = 0; i < k; i++)
        {
            orr |= arr[i];
            for (ll j = 0; j < 21; j++)
            {
                if (arr[i] & (1 << j))
                {
                    freq[j] += 1;
                }
            }
        }

        ll check = orr;
        bool flag = 0;
        for (ll i = k; i < n; i++)
        {
            orr |= arr[i];
            for (ll j = 0; j < 21; j++)
            {
                if (arr[i] & (1 << j))
                {
                    freq[j] += 1;
                }
            }

            for (ll j = 0; j < 21; j++)
            {
                if (arr[i - k] & (1 << j))
                {
                    freq[j] -= 1;
                    if (freq[j] == 0)
                    {
                        check -= (1 << j);
                    }
                }
            }

            if (orr != check)
            {
                flag = 1;
                break;
            }
        }

        if (flag == 0)
        {
            cout << k << endl;
            return;
        }
    }

    return;
}

//

signed main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    ll tt = 1;
    cin >> tt;
    while (tt--)
    {
        breathe();
    }
    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 buggy code, we need to focus on edge cases and scenarios that can highlight the issues in the algorithm. The buggy code seems to have issues with correctly calculating the bitwise OR for subarrays and comparing them. Here are some steps to think through the process:

1. **Single Element Array**: This is a straightforward case where the loneliness is always 1.
2. **All Elements Identical**: If all elements are the same, the loneliness should be 1.
3. **All Elements Zero**: If all elements are zero, the loneliness should be 1.
4. **Alternating Elements**: This can create a pattern where the bitwise OR of subarrays might not be consistent.
5. **Mixed Values**: A mix of values that have different bitwise OR results for different subarray lengths.
6. **Edge Case with Large k**: Ensure that the algorithm correctly handles cases where the smallest k is close to n.

Given these considerations, let's create a small-scale test input that can expose the bug:

```plaintext
5
1
0
3
2 2 2
3
1 0 2
5
3 0 1 4 2
5
2 0 4 0 2
```

This test input includes:
1. A single element array.
2. An array with all elements identical.
3. An array with mixed values where the bitwise OR of subarrays might not be consistent.
4. An array where the smallest k is close to n.

These cases should help in identifying the issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Cat, Fox and the Lonely Array
time_limit_ms: 2000
memory_limit_mb: 256
Description: Today, Cat and Fox found an array $a$ consisting of $n$ non-negative integers.

Define the loneliness of $a$ as the smallest positive integer $k$ ($1 \le k \le n$) such that for any two positive integers $i$ and $j$ ($1 \leq i, j \leq n - k +1$), the following holds: $$a_i | a_{i+1} | \ldots | a_{i+k-1} = a_j | a_{j+1} | \ldots | a_{j+k-1},$$ where $x | y$ denotes the [bitwise OR](https://en.wikipedia.org/wiki/Bitwise_operation#OR) of $x$ and $y$. In other words, for every $k$ consecutive elements, their bitwise OR should be the same. Note that the loneliness of $a$ is well-defined, because for $k = n$ the condition is satisfied.

Cat and Fox want to know how lonely the array $a$ is. Help them calculate the loneliness of the found array.

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

The first line of each test case contains one integer $n$ ($1 \leq n \leq 10^5$) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i < 2^{20}$) — the elements of the array.

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $10^5$.

Output: For each test case, print one integer  — the loneliness of the given array.

Examples:
input:
7
1
0
3
2 2 2
3
1 0 2
5
3 0 1 4 2
5
2 0 4 0 2
7
0 0 0 0 1 2 4
8
0 1 3 2 2 1 0 3
output:
1
1
3
4
4
7
3

Note: In the first example, the loneliness of an array with a single element is always $1$, so the answer is $1$.

In the second example, the OR of each subarray of length $k = 1$ is $2$, so the loneliness of the whole array is $1$.

In the seventh example, it's true that $(0 | 1 | 3) = (1 | 3 | 2) = (3 | 2 | 2) = (2 | 2 | 1) = (2 | 1 | 0) = (1 | 0 | 3) = 3$, so the condition is satisfied for $k = 3$. We can verify that the condition is not true for any smaller $k$, so the answer is indeed $3$.

Buggy Code:
#include<bits/stdc++.h>
#define N 100005
using namespace std;
int n;
int a[N],lst[N];
template<class T=int> inline T rd(){
    T res=0;char c=getchar();bool p=0;
    while(c<'0'||c>'9') {if(c=='-')p=1;c=getchar();};
    while(c>='0'&&c<='9')   res=res*10+c-'0',c=getchar();
    return p?-res:res;
}
int main(void){
    int T=rd();
    while(T--){
        n=rd();int ans=0;
        for(int i=1;i<=n;i++)   a[i]=rd();
        for(int i=0;i<=20;i++){
            int lst=0,mx=0;
            for(int k=1;k<=n;k++){
                if(a[k]&(1<<i)){
                    mx=max(mx,k-lst);
                    lst=k;
                }
            }
            if(mx==0)   continue;
            ans=max(ans,mx);
        }
        printf("%i\n",max(1,ans));
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider the following aspects:

1. **Edge Cases**: Test cases that push the boundaries of the input constraints.
2. **Specific Patterns**: Test cases that have specific patterns in the array which might not be handled correctly by the buggy code.
3. **Maximum Constraints**: Test cases that use the maximum allowed values for `n` and `a_i`.

The buggy code has a logic issue in how it calculates the maximum distance between elements with a specific bit set. It only considers the maximum distance for each bit position and then takes the maximum of these distances. This approach might miss cases where the bitwise OR of subarrays of a certain length is the same, but the distances between elements with specific bits set are not the determining factor.

Let's create a test case that exposes this issue:

### Test Case
```plaintext
1
6
1 2 3 4 5 6
```

### Explanation
- **Array**: `[1, 2, 3, 4, 5, 6]`
- **Bitwise OR Analysis**:
  - For `k = 1`: Each element is unique, so the condition is not satisfied.
  - For `k = 2`:
    - `1 | 2 = 3`
    - `2 | 3 = 3`
    - `3 | 4 = 7`
    - `4 | 5 = 7`
    - `5 | 6 = 7`
    - The condition is not satisfied for all pairs.
  - For `k = 3`:
    - `1 | 2 | 3 = 3`
    - `2 | 3 | 4 = 7`
    - `3 | 4 | 5 = 7`
    - `4 | 5 | 6 = 7`
    - The condition is not satisfied for all triplets.
  - For `k = 4`:
    - `1 | 2 | 3 | 4 = 7`
    - `2 | 3 | 4 | 5 = 7`
    - `3 | 4 | 5 | 6 = 7`
    - The condition is satisfied for all quadruplets.
  - For `k = 5` and `k = 6`, the condition is also satisfied.

The correct loneliness of the array is `4`. The buggy code might not correctly handle this case because it only considers the maximum distance between elements with specific bits set, which might not be the determining factor for the smallest `k` that satisfies the condition.

This test case should expose the bug in the provided code.