**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum modulo equality
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$ and $$$q$$$ queries $$$l$$$, $$$r$$$.

For each query, find the maximum possible $$$m$$$, such that all elements $$$a_l$$$, $$$a_{l+1}$$$, ..., $$$a_r$$$ are equal modulo $$$m$$$. In other words, $$$a_l \bmod m = a_{l+1} \bmod m = \dots = a_r \bmod m$$$, where $$$a \bmod b$$$ — is the remainder of division $$$a$$$ by $$$b$$$. In particular, when $$$m$$$ can be infinite, print $$$0$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$q$$$ ($$$1 \le n, q \le 2\cdot 10^5$$$) — the length of the array and the number of queries.

The second line of each test case contains $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$) — the elements of the array.

In the following $$$q$$$ lines of each test case, two integers $$$l$$$, $$$r$$$ are provided ($$$1 \le l \le r \le n$$$) — the range of the query.

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

Output: For each query, output the maximum value $$$m$$$ described in the statement.

Examples:
input:
3
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
output:
3 1 4 1 0
0
1 6

Note: In the first query of the first sample, $$$6 \bmod 3 = 3 \bmod 3 = 0$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

In the third query of the first sample, $$$14 \bmod 4 = 2 \bmod 4 = 6 \bmod 4 = 2$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

Buggy Code:
from math import gcd

def solve():
    n, q = map(int, input().split())
    a = list(map(int, input().split()))

    if n == 1:
        for _ in range(q):
            l, r = map(int, input().split())
            print(0)
        return

    diff = [abs(a[i+1] - a[i]) for i in range(n-1)]
    for _ in range(q):
        l, r = map(int, input().split())
        l -= 1
        r -= 1
        if a[l] == a[r]:
            print(0, end=' ')
        else:
            res = diff[l]
            for i in range(l+1, r):
                res = gcd(res, diff[i])
            print(res, end=' ')
    print()

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 code, we need to focus on edge cases and scenarios that maximize the failure probability in faulty solutions. Here are the key points to consider:

1. **Single Element Arrays**: The code has a special case for arrays of length 1. Ensure that this case is handled correctly.
2. **Equal Elements in a Range**: The code has a special case for ranges where all elements are equal. This should return 0.
3. **Different Elements in a Range**: The code calculates the GCD of differences. Ensure that the GCD calculation is correct for ranges with different elements.
4. **Large Differences**: The code might fail if the differences between elements are large.
5. ** Queries on the Same Range**: Test multiple queries on the same range to ensure the code handles repeated queries correctly.

Given these points, let's create a small-scale test input:

```plaintext
3
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
3 3
5 1 2 3 4 5
1 5
1 1
2 2
3 3
4 4
5 5
```

### Explanation:
1. **First Test Case**:
   - Array: `[5, 14, 2, 6, 3]`
   - Queries:
     - `4 5`: Elements are `6` and `3`. The GCD of differences is `3`.
     - `1 4`: Elements are `5, 14, 2, 6`. The GCD of differences is `1`.
     - `2 4`: Elements are `14, 2, 6`. The GCD of differences is `4`.
     - `3 5`: Elements are `2, 6, 3`. The GCD of differences is `1`.
     - `1 1`: Single element, should return `0`.

2. **Second Test Case**:
   - Array: `[7]`
   - Queries:
     - `1 1`: Single element, should return `0`.
     - `3 2`: Invalid range, but should be handled gracefully. The code should handle this correctly.

3. **Third Test Case**:
   - Array: `[1, 7, 8, 2, 3, 4, 5]`
   - Queries:
     - `2 3`: Elements are `7, 8`. The GCD of differences is `1`.
     - `1 2`: Elements are `1, 7`. The GCD of differences is `6`.
     - `3 3`: Single element, should return `0`.
     - `5 1 2 3 4 5`: Elements are `1, 2, 3, 4, 5`. The GCD of differences is `1`.
     - `1 1`: Single element, should return `0`.
     - `2 2`: Single element, should return `0`.
     - `3 3`: Single element, should return `0`.
     - `4 4`: Single element, should return `0`.
     - `5 5`: Single element, should return `0`.

This test input covers a variety of scenarios, including single elements, equal elements in a range, different elements in a range, and repeated queries on the same range. It should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum modulo equality
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$ and $$$q$$$ queries $$$l$$$, $$$r$$$.

For each query, find the maximum possible $$$m$$$, such that all elements $$$a_l$$$, $$$a_{l+1}$$$, ..., $$$a_r$$$ are equal modulo $$$m$$$. In other words, $$$a_l \bmod m = a_{l+1} \bmod m = \dots = a_r \bmod m$$$, where $$$a \bmod b$$$ — is the remainder of division $$$a$$$ by $$$b$$$. In particular, when $$$m$$$ can be infinite, print $$$0$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$q$$$ ($$$1 \le n, q \le 2\cdot 10^5$$$) — the length of the array and the number of queries.

The second line of each test case contains $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$) — the elements of the array.

In the following $$$q$$$ lines of each test case, two integers $$$l$$$, $$$r$$$ are provided ($$$1 \le l \le r \le n$$$) — the range of the query.

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

Output: For each query, output the maximum value $$$m$$$ described in the statement.

Examples:
input:
3
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
output:
3 1 4 1 0
0
1 6

Note: In the first query of the first sample, $$$6 \bmod 3 = 3 \bmod 3 = 0$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

In the third query of the first sample, $$$14 \bmod 4 = 2 \bmod 4 = 6 \bmod 4 = 2$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

Buggy Code:
nTest = int(input())
for i in range(nTest):
    n, q = list(map(int, input().split()))
    lst = list(map(int, input().split()))
    for j in range(q):
        l, r = list(map(int, input().split()))
        if l==r:
            print(0, end=" ")
            break
        accessed = lst[l-1: r]
        maxVal = max(accessed)
        minVal = min(accessed)
        maxMod = maxVal
        modAvail = False
        rem = 0
        while not modAvail and maxMod >= minVal:
            modVal = True
            for k in range(len(accessed)):
                if accessed[k] % maxMod != rem:
                    modVal = False
                    break
            if modVal:
                modAvail = True
            else:
                maxMod -= 1
                rem = maxVal % maxMod
        if modAvail:
            print(maxMod, end=" ")
        else:
            print(1, end=" ")
    print()

**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 maximize the failure probability. Here are the key points to consider:

1. **Single Element Query**: The code has a special case for when the query range is a single element (`l == r`). This should always print `0`, but the code uses a `break` statement which could cause issues if there are multiple queries in a test case.

2. **All Elements Equal**: If all elements in the query range are equal, the maximum modulo value should be the maximum value in the range, which is the element itself.

3. **Different Elements**: If the elements in the query range are different, the code should find the maximum modulo value that makes all elements equal modulo that value.

4. **Edge Case with Large Values**: The code might struggle with large values and large ranges, so we should test with large numbers and large query ranges.

5. **Consecutive Queries**: The code should handle multiple queries within a single test case without breaking.

Given these considerations, let's construct a test input that exposes the bug:

```plaintext
3
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
10 10
1 2 3 4 5 6 7 8 9 10
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 10
```

This test input includes:
1. A test case with a small array and multiple queries, including a single-element query.
2. A test case with a small array and queries that test different elements and ranges.
3. A test case with a larger array and multiple single-element queries to ensure the `break` statement does not cause issues.

This should help expose the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum modulo equality
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$ and $$$q$$$ queries $$$l$$$, $$$r$$$.

For each query, find the maximum possible $$$m$$$, such that all elements $$$a_l$$$, $$$a_{l+1}$$$, ..., $$$a_r$$$ are equal modulo $$$m$$$. In other words, $$$a_l \bmod m = a_{l+1} \bmod m = \dots = a_r \bmod m$$$, where $$$a \bmod b$$$ — is the remainder of division $$$a$$$ by $$$b$$$. In particular, when $$$m$$$ can be infinite, print $$$0$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$q$$$ ($$$1 \le n, q \le 2\cdot 10^5$$$) — the length of the array and the number of queries.

The second line of each test case contains $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$) — the elements of the array.

In the following $$$q$$$ lines of each test case, two integers $$$l$$$, $$$r$$$ are provided ($$$1 \le l \le r \le n$$$) — the range of the query.

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

Output: For each query, output the maximum value $$$m$$$ described in the statement.

Examples:
input:
3
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
output:
3 1 4 1 0
0
1 6

Note: In the first query of the first sample, $$$6 \bmod 3 = 3 \bmod 3 = 0$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

In the third query of the first sample, $$$14 \bmod 4 = 2 \bmod 4 = 6 \bmod 4 = 2$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

Buggy Code:
import java.util.*;
public class Main {
    static int aLength = 0;
    static int bLength = 0;
    static int cLength = 0;
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int t = s.nextInt();
        for (int i = 0; i < t; i++) {
            int n = s.nextInt();
            int q = s.nextInt();
            int[] arr = new int[n];
            int[][] query = new int[q][2];
            takeInput(arr,query, s);
            for(int j=0; j < query.length; j++) {
                int maxValInQuery = getMax(query[j][0]-1, query[j][1]-1, arr);
                if(j == 0) {
                    System.out.print(getMaxModInQuery(arr, query[j], 1 , maxValInQuery));
                }
                else {
                    System.out.print(" "+getMaxModInQuery(arr, query[j], 1 , maxValInQuery));
                }
            }
            System.out.println();
        }
    }

    public static int getMaxModInQuery(int[] arr, int[] query, int start, int end) {
        if(start > end) return 0;
        int result = 0;
        int middle = start + (end-start)/2;
        int leftSol = getMaxModInQuery(arr,query, start, middle-1);
        int rightSol = getMaxModInQuery(arr,query, middle+1, end);
        for(int i= query[0]-1; i < query[1]-1; i++) {
            if((arr[i] % middle) !=(arr[i+1] % middle)) {
                result = 0;
                break;
            }
            else {
                result = middle;
            }
        }
        return Math.max(Math.max(leftSol, rightSol), result);
    }

    public static int getMax(int start, int end, int[] arr) {
        int maxVal = arr[start];
        for(int i=start+1; i<= end; i++) {
            maxVal = Math.max(arr[i], maxVal);
        }
        return maxVal;
    }

    public static void takeInput(int[] arr, int[][] query, Scanner s) {
        for(int i=0; i< arr.length; i++) {
            arr[i] = s.nextInt();
        }
        for(int i=0; i< query.length; i++) {
            query[i][0] = s.nextInt();
            query[i][1] = s.nextInt();
        }
    }
}

**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
5 5
1 2 3 4 5
1 2
2 3
3 4
4 5
1 5
1 1
10
1 1
3 2
1 1 2
1 2
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: Maximum modulo equality
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$ and $$$q$$$ queries $$$l$$$, $$$r$$$.

For each query, find the maximum possible $$$m$$$, such that all elements $$$a_l$$$, $$$a_{l+1}$$$, ..., $$$a_r$$$ are equal modulo $$$m$$$. In other words, $$$a_l \bmod m = a_{l+1} \bmod m = \dots = a_r \bmod m$$$, where $$$a \bmod b$$$ — is the remainder of division $$$a$$$ by $$$b$$$. In particular, when $$$m$$$ can be infinite, print $$$0$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$q$$$ ($$$1 \le n, q \le 2\cdot 10^5$$$) — the length of the array and the number of queries.

The second line of each test case contains $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$) — the elements of the array.

In the following $$$q$$$ lines of each test case, two integers $$$l$$$, $$$r$$$ are provided ($$$1 \le l \le r \le n$$$) — the range of the query.

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

Output: For each query, output the maximum value $$$m$$$ described in the statement.

Examples:
input:
3
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
output:
3 1 4 1 0
0
1 6

Note: In the first query of the first sample, $$$6 \bmod 3 = 3 \bmod 3 = 0$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

In the third query of the first sample, $$$14 \bmod 4 = 2 \bmod 4 = 6 \bmod 4 = 2$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

Buggy Code:
import java.io.*;
import java.util.ArrayList;
import java.util.Collections;
import java.util.InputMismatchException;
import java.util.List;
import java.util.function.BiFunction;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        OutputWriter out = new OutputWriter(outputStream);
        Problem solver = new Problem();
        int n = in.nextInt();
        for (int i = 1; i <= n; i++) {
            solver.solve(i, in, out);
        }
        out.close();
    }

    static class Problem {
        static final int K = 26;
        static final int MAX_N = 3_000_00;
        public void solve(int testNumber, InputReader in, OutputWriter out)
        {
            int n = in.nextInt();
            int q = in.nextInt();
            int [] arr = new int[n - 1];
            int cur = in.nextInt();
            for (int i = 0; i < n - 1; ++i)
            {
                int nxt = in.nextInt();
                arr[i] = Math.abs(cur - nxt);
                cur = nxt;
            }

            int[][] tb = buildSparseTable(arr);
            for (int i = 0; i < q; ++i)
            {
                int l = in.nextInt() - 1;
                int r = in.nextInt() - 2;
                if (i > 0)
                {
                    out.print(' ');
                }
                out.print(valueOnRange(tb, l, r));
            }
            out.println();
        }

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

        int valueOnRange(int[][] sparseTable, int l, int r)
        {
            int res = 0;
            for (int i = K - 1; i >= 0; --i)
            {
                if ((1 << i) <= r - l + 1)
                {
                    res = this.gcd(res, sparseTable[i][l]);
                    l += (1 << i);
                }
            }

            return res;
        }

        int[][] buildSparseTable(int[] arr)
        {
            int[][] sparseTable = new int[K][MAX_N];
            System.arraycopy(arr, 0, sparseTable[0], 0, arr.length);

            for (int pow = 1; pow < K; ++pow)
            {
                for (int start = 0; start + (1 << pow) < MAX_N; ++start)
                {
                    sparseTable[pow][start] = this.gcd(
                            sparseTable[pow - 1][start],
                            sparseTable[pow - 1][start + (1 << (pow - 1))]);
                }
            }
            return sparseTable;
        }
    }

    static class OutputWriter {
        private final PrintWriter writer;

        public OutputWriter(OutputStream outputStream) {
            writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
        }

        public OutputWriter(Writer writer) {
            this.writer = new PrintWriter(writer);
        }

        public void close() {
            writer.close();
        }

        public void println() {
            writer.println();
        }

        public void println(long l) {
            writer.println(l);
        }

        public void print(char l) {
            writer.print(l);
        }

        public void print(int l) {
            writer.print(l);
        }

        public void println(int i) {
            writer.println(i);
        }

        public void println(String s) {
            writer.println(s);
        }

    }

    static class InputReader {
        final private int BUFFER_SIZE = 1 << 16;
        private InputStream stream;
        private byte[] buf = new byte[BUFFER_SIZE];
        private int curChar;
        private int numChars;
        private InputReader.SpaceCharFilter filter;

        public InputReader(InputStream stream) {
            this.stream = stream;
        }

        public int read() {
            if (numChars == -1) {
                throw new InputMismatchException();
            }
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (numChars <= 0) {
                    return -1;
                }
            }
            return buf[curChar++];
        }

        public char nextChar() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            return (char)c;
        }

        public String nextString() {
            int c = read();
            while (isSpaceChar(c))
            {
                c = read();
            }
            StringBuilder sb = new StringBuilder();
            do {
                sb.append((char)c);
                c = read();
            } while (!isSpaceChar(c));

            return sb.toString();
        }

        public long nextLong() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            long sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            long res = 0;
            do {
                if (c < '0' || c > '9') {
                    throw new InputMismatchException();
                }
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        }

        public int nextInt() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            int res = 0;
            do {
                if (c < '0' || c > '9') {
                    throw new InputMismatchException();
                }
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        }

        public boolean isSpaceChar(int c) {
            if (filter != null) {
                return filter.isSpaceChar(c);
            }
            return isWhitespace(c);
        }

        public static boolean isWhitespace(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        public interface SpaceCharFilter {
            public boolean isSpaceChar(int ch);

        }

    }
}

**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 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
```

### Explanation:
This test case is designed to expose several potential issues in the buggy code:

1. **Array Length and Query Length**: The array length is 5, and there are 5 queries. This is a manageable size to ensure the test case is not too large but still significant enough to test the algorithm's correctness.

2. **Query Ranges**:
   - **Query 1 (4 5)**: The elements are 6 and 3. The maximum modulo value is 3 because 6 % 3 = 3 % 3 = 0.
   - **Query 2 (1 4)**: The elements are 5, 14, 2, 6. The maximum modulo value is 1 because the greatest common divisor (GCD) of the differences (9, 12, 4) is 1.
   - **Query 3 (2 4)**: The elements are 14, 2, 6. The maximum modulo value is 4 because 14 % 4 = 2 % 4 = 6 % 4 = 2.
   - **Query 4 (3 5)**: The elements are 2, 6, 3. The maximum modulo value is 1 because the GCD of the differences (4, 3) is 1.
   - **Query 5 (1 1)**: The elements are 5. The maximum modulo value is 0 because the query length is 1.

3. **Edge Cases**:
   - **Single Element Query (1 1)**: This tests the case where the query range is a single element, which should always return 0.
   - **Different Ranges (1 4, 2 4, 3 5)**: These ranges test different subarrays to ensure the algorithm correctly handles varying lengths and positions.

This test case should help identify issues in the buggy code, particularly with the handling of query ranges and the calculation of the GCD.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum modulo equality
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$ and $$$q$$$ queries $$$l$$$, $$$r$$$.

For each query, find the maximum possible $$$m$$$, such that all elements $$$a_l$$$, $$$a_{l+1}$$$, ..., $$$a_r$$$ are equal modulo $$$m$$$. In other words, $$$a_l \bmod m = a_{l+1} \bmod m = \dots = a_r \bmod m$$$, where $$$a \bmod b$$$ — is the remainder of division $$$a$$$ by $$$b$$$. In particular, when $$$m$$$ can be infinite, print $$$0$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$q$$$ ($$$1 \le n, q \le 2\cdot 10^5$$$) — the length of the array and the number of queries.

The second line of each test case contains $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$) — the elements of the array.

In the following $$$q$$$ lines of each test case, two integers $$$l$$$, $$$r$$$ are provided ($$$1 \le l \le r \le n$$$) — the range of the query.

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

Output: For each query, output the maximum value $$$m$$$ described in the statement.

Examples:
input:
3
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
output:
3 1 4 1 0
0
1 6

Note: In the first query of the first sample, $$$6 \bmod 3 = 3 \bmod 3 = 0$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

In the third query of the first sample, $$$14 \bmod 4 = 2 \bmod 4 = 6 \bmod 4 = 2$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

Buggy Code:
def solve(a, l, r):
         if l == r:
                  return 0
         res = abs(a[l]-a[l+1])
         for i in range(l+1, r):
                  diff = abs(a[i]-a[i+1])
                  if diff % res:
                           res = gcd(diff, res)
                  if res == 1:
                           break
         return res

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

t = int(input())
for i in range(t):
    n, k = tuple([int(i) for i in input().split()])
    a = [int(i) for i in input().split()]
    for j in range(k):
             l, r = tuple([int(i) for i in input().split()])
             print(solve(a, l-1, r-1), end=' ')
    print('')

**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
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
```

### Explanation:
1. **Test Case 1:**
   - **Array:** `[5, 14, 2, 6, 3]`
   - **Queries:**
     - `[4, 5]`: Elements are `6` and `3`. The maximum `m` is `3` because `6 % 3 = 3 % 3 = 0`.
     - `[1, 4]`: Elements are `5, 14, 2, 6`. The maximum `m` is `1` because `5 % 1 = 14 % 1 = 2 % 1 = 6 % 1 = 0`.
     - `[2, 4]`: Elements are `14, 2, 6`. The maximum `m` is `4` because `14 % 4 = 2 % 4 = 6 % 4 = 2`.
     - `[3, 5]`: Elements are `2, 6, 3`. The maximum `m` is `1` because `2 % 1 = 6 % 1 = 3 % 1 = 0`.
     - `[1, 1]`: Single element `5`. The maximum `m` is `0` because there's only one element.

2. **Test Case 2:**
   - **Array:** `[7]`
   - **Query:**
     - `[1, 1]`: Single element `7`. The maximum `m` is `0` because there's only one element.

3. **Test Case 3:**
   - **Array:** `[1, 7, 8]`
   - **Queries:**
     - `[2, 3]`: Elements are `7, 8`. The maximum `m` is `1` because `7 % 1 = 8 % 1 = 0`.
     - `[1, 2]`: Elements are `1, 7`. The maximum `m` is `6` because `1 % 6 = 7 % 6 = 1`.

This test input includes a variety of scenarios, including single-element queries, ranges with different lengths, and cases where the elements are not all the same. These scenarios are designed to expose potential 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: Maximum modulo equality
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$ and $$$q$$$ queries $$$l$$$, $$$r$$$.

For each query, find the maximum possible $$$m$$$, such that all elements $$$a_l$$$, $$$a_{l+1}$$$, ..., $$$a_r$$$ are equal modulo $$$m$$$. In other words, $$$a_l \bmod m = a_{l+1} \bmod m = \dots = a_r \bmod m$$$, where $$$a \bmod b$$$ — is the remainder of division $$$a$$$ by $$$b$$$. In particular, when $$$m$$$ can be infinite, print $$$0$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$q$$$ ($$$1 \le n, q \le 2\cdot 10^5$$$) — the length of the array and the number of queries.

The second line of each test case contains $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$) — the elements of the array.

In the following $$$q$$$ lines of each test case, two integers $$$l$$$, $$$r$$$ are provided ($$$1 \le l \le r \le n$$$) — the range of the query.

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

Output: For each query, output the maximum value $$$m$$$ described in the statement.

Examples:
input:
3
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
output:
3 1 4 1 0
0
1 6

Note: In the first query of the first sample, $$$6 \bmod 3 = 3 \bmod 3 = 0$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

In the third query of the first sample, $$$14 \bmod 4 = 2 \bmod 4 = 6 \bmod 4 = 2$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

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

public class Main {

    static int mod = 998__244______________________________________________________353;

    static int inf = (int) 2e9;

    static long infL = (long) 9e18;
/*

3
3 8
2 5
4 5

3
2 5
1 4
3 4
*/
    private static final int N = (int) (1e6+10);
    static int []min = new int[N],ans = new int[N];
    static int [][]cnt = new int[N][2];
    static int []f = new int[N];
    static int []las = new int[N];
    static int []pre = new int[N];
    static int []a = new int[N];

    static Pair<Integer,Integer> []pair = new Pair[N];
    static int n,k;

    static int[][] st;
    static int[] dx = new int[]{-1,1,0,0};
    static int[] dy = new int[]{0,0,-1,1};
    static int [][]type;

    private static void solve(){
        n = in();k = in();
        pre[0] = 0;
        for(int i=1;i<=n;i++){
            f[i] = in();
        }
        for(int i=1;i<n;i++){//9 12 4 3
            f[i] = Math.abs(f[i] - f[i+1]);
        }
        GcdST gcdST = new GcdST();
        gcdST.init(f,n-1);

        while(k-- > 0){
            int l = in();int r = in();
            if(l==r){
                print(0);
            }else {
                print(gcdST.get(l,r-1));
            }
            print(" ");
        }
        println();
    }

    public static void main(String[] args) {
        int t = 1;
        t = in();
        while (t-- > 0) {
            solve();
        }
    }

    private static class GcdST{

        private final int[][] st = new int[200020][35];

        public void init(int[] f,int n){
            for(int k=0;k<=25;k++){
                for(int i=1;i<=n;i++){
                    if(k==0)st[i][k] = f[i];
                    else {
                        if(i + (1<<k) - 1 > n)break;
                        st[i][k] = gcd(st[i][k-1],st[i+(1<<(k-1))][k-1]);
                    }
                }
            }
        }

        public int get(int l,int r){
            int x = r - l + 1;
            int res = 0,c = 0,base = 1;
            while(x > 0){
                if((x & 1) == 1){
                    res = gcd(res,st[l][c]);
                    l += base; // x x x x
                }
                x /= 2;
                c++;
                base *= 2;
            }
//            for (int k = 25;k>=0;k--){
//                if(((x >> k) & 1) == 1){
//                    res = gcd(res,st[l][k]);
//                    l += 1<<k; // x x x x
//                }
//            }
            return res;
        }

    }

    private static class SegTree{

        private static class Node{
            int l,r;
            long max,min;
        }
        final Node []tr;
        public SegTree(int n){
            tr = new Node[n << 2];
            for (int i = 0; i < tr.length; i++) {
                tr[i] = new Node(); // 初始化所有节点
            }
        }
        public void init(int idx,int l,int r){
            tr[idx].l = l;
            tr[idx].r = r;
            if (l == r) { // 叶节点
                tr[idx].max = -inf;
                tr[idx].min = inf;
                return;
            }
            int mid = (l + r) >> 1;
            init(idx * 2, l, mid ); // 左子树
            init(idx * 2 + 1, mid + 1, r); // 右子树
            pushUp(idx); // 合并子树信息
        }
        // 合并子节点信息
        private void pushUp(int idx) {
            tr[idx].max = Math.max(tr[idx * 2].max, tr[idx * 2 + 1].max);
            tr[idx].min = Math.min(tr[idx * 2].min, tr[idx * 2 + 1].min);
        }

        // 区间查询最大值和最小值
        public long[] query(int l, int r, int idx) {
            if (tr[idx].l >= l && tr[idx].r <= r) { // 当前节点区间完全包含在查询区间中
                return new long[]{tr[idx].max, tr[idx].min};
            }
            int mid = (tr[idx].l + tr[idx].r) >> 1;
            long maxVal = Long.MIN_VALUE;
            long minVal = Long.MAX_VALUE;
            if (l <= mid) { // 查询左子树
                long[] leftRes = query(l, r, idx * 2);
                maxVal = leftRes[0];
                minVal = leftRes[1];
            }
            if (r > mid) { // 查询右子树
                long[] rightRes = query(l, r, idx * 2 + 1);
                maxVal = Math.max(maxVal, rightRes[0]);
                minVal = Math.min(minVal, rightRes[1]);
            }
            return new long[]{maxVal, minVal};
        }

        // 单点更新
        public void update(int idx, int pos, int value) {
            if (tr[idx].l == tr[idx].r) { // 找到目标叶节点
                tr[idx].max = tr[idx].min = value;
                return;
            }
            int mid = (tr[idx].l + tr[idx].r) >> 1;
            if (pos <= mid) { // 更新左子树
                update(idx * 2,pos, value);
            } else { // 更新右子树
                update(idx * 2 + 1,pos, value);
            }
            pushUp(idx); // 更新当前节点
        }

    }

    //  -===-      MATH       -===-  //

    static int fact(int n) {
        int r = 1;
        for (int i = 1; i <= n; i++) {
            r = (int) (((long) r * i) % mod);
        }
        return r;
    }

    static int C(int n, int k) {
        return (int) (((((long) fact(n) * pow(fact(k), mod - 2)) % mod) * pow(fact(n - k), mod - 2)) % mod);
    }

    static int gcd(int i1, int i2) {
        if (i2 == 0) return i1;
        return gcd(i2, i1 % i2);
    }

    static long gcd(long i1, long i2) {
        if (i2 == 0) return i1;
        return gcd(i2, i1 % i2);
    }

    static int pow(int x, int y) {
        x = x % mod;
        if (x == 0) return 0;
        int r = 1;
        while (y > 0) {
            if ((y & 1) != 0) r = (int) (((long) r * x) % mod);
            y /= 2;
            x = (int) (((long) x * x) % mod);
        }
        return r % mod;
    }

    static void swap(int[] a, int i, int j) {
        int tmp = a[i];
        a[i] = a[j];
        a[j] = tmp;
    }

    static int digitSum(long n) {
        int r = 0;
        while (n != 0) {
            r += (int) (n % 10);
            n /= 10;
        }
        return r;
    }

    //  -===- DATA STRUCTURES -===-  //

    static class UnionFind {

        int[] p, rank, min, max, sz;
        int n, c;

        UnionFind(int n) {
            this.n = n;
            p = new int[n];
            rank = new int[n];
            min = new int[n];
            max = new int[n];
            sz = new int[n];
            for (int i = 0; i < p.length; i++) {
                p[i] = i;
                min[i] = i;
                max[i] = i;
                sz[i] = 1;
            }
        }

        int find(int a) {
            return p[a] = (p[a] == a ? a : find(p[a]));
        }

        void union(int a, int b) {
            a = find(a);
            b = find(b);
            if (a == b) return;
            if (rank[a] > rank[b]) {
                p[b] = a;
                max[a] = Math.max(max[a], max[b]);
                min[a] = Math.min(min[a], min[b]);
                sz[a] += sz[b];
            } else {
                p[a] = b;
                max[b] = Math.max(max[b], max[a]);
                min[b] = Math.min(min[b], min[a]);
                sz[b] += sz[a];
                if (rank[a] == rank[b]) {
                    rank[b]++;
                }
            }
            c--;
        }
    }

    private static class Pair<F, S> {

        int index;

        F first;
        S second;

        public Pair(F f, S s) {
            this.first = f;
            this.second = s;
        }

        public F getFirst() {
            return first;
        }

        public S getSecond() {
            return second;
        }

        public void setFirst(F first) {
            this.first = first;
        }

        public void setSecond(S second) {
            this.second = second;
        }

        public int getIndex() {
            return index;
        }

        public void setIndex(Integer index) {
            this.index = index;
        }

        @Override
        public String toString(){
            return first.toString() + ',' + second.toString();
        }

    }

    //  -===-      IN      -===-  //
    static StringTokenizer STT;

    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

    static PrintStream out = System.out;

    static String inStr() {
        while (STT == null || !STT.hasMoreElements()) {
            try {
                STT = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return STT.nextToken();
    }

    static int in() {
        return Integer.parseInt(inStr());
    }

    static long inLong() {
        return Long.parseLong(inStr());
    }

    static double inDouble() {
        return Double.parseDouble(inStr());
    }

    static String inLine() {
        String s;
        try {
            if (STT != null && STT.hasMoreTokens()) {
                s = STT.nextToken("\n");
            } else {
                s = br.readLine();
            }
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
        return s;
    }

    static long[] inArrLong() {
        String[] split = inLine().split(" ");
        long[] a = new long[split.length];
        for (int i = 0; i < a.length; i++) {
            a[i] = Long.parseLong(split[i]);
        }
        return a;
    }

    static int[] inArr() {
        String s = inLine();
        String[] split = s.split(" ");
        int[] a = new int[split.length];
        for (int i = 0; i < a.length; i++) {
            a[i] = Integer.parseInt(split[i]);
        }
        return a;
    }

    //  -===-      PRINT      -===-  //

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

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

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

    static void println(int a) {
        out.println(a);
    }

    static void print(int a) {
        out.print(a);
    }

    static void println(long a) {
        out.println(a);
    }

    static void print(long a) {
        out.print(a);
    }

    static void println(double a) {
        out.println(a);
    }

    static void print(double a) {
        out.print(a);
    }

    static void println(String a) {
        out.println(a);
    }

    static void print(String a) {
        out.print(a);
    }

    static void print(int[] a) {
        for (int i = 0; i < a.length - 1; i++) {
            out.print(a[i] + " ");
        }
        if (a.length > 0) {
            out.println(a[a.length - 1]);
        }
    }

    static void print(List<?> l) {
        for (int i = 0; i < l.size() - 1; i++) {
            out.print(l.get(i) + " ");
        }
        if (!l.isEmpty()) {
            out.println(l.get(l.size() - 1));
        }
    }

    public static class D {
        private final int[] dArray; // 离散化后的数组

        public int index;

        // 初始化方法
        public D(int[] oArray,int n) {
            // 1. 去重
            Set<Integer> set = new HashSet<>();
            for(int i=1;i<=n;i++){
                set.add(oArray[i]);
            }

            // 2. 转换为数组并排序
            dArray = new int[set.size()+1];
            index = 0;
            for (int num : set) {
                dArray[++index] = num;
            }
            Arrays.sort(dArray,1,index+1);
        }

        public int getOld(int x){
            return dArray[x];
        }

        // 获取离散化后的值（二分查找实现）
        public int get(int target) {
            int left = 1;
            int right = index;

            while (left <= right) {
                int mid = left + (right - left) / 2;
                if (dArray[mid] == target) {
                    return mid; // 返回下标 + 1（假设下标从 1 开始）
                } else if (dArray[mid] < target) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            }

            return -1; // 如果未找到，返回 -1
        }

    }

}

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

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

Think step by step.

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

1. **Understand the Problem**: The problem requires finding the maximum possible modulo value `m` such that all elements in a given range `[l, r]` of the array are equal modulo `m`.

2. **Identify Potential Bugs**:
   - **Edge Cases**: Queries where `l == r`, arrays with a single element, and arrays with all elements being the same.
   - **GCD Calculation**: The GCD calculation in the `GcdST` class might fail for certain edge cases.
   - **Array Initialization and Indexing**: Ensure that the array indices are correctly handled, especially in the `GcdST` class.

3. **Construct Test Input**:
   - **Single Element Array**: Test with an array that has only one element.
   - **All Elements Equal**: Test with an array where all elements are the same.
   - **Mixed Elements**: Test with an array that has a mix of elements to ensure the GCD calculation is correct.
   - **Large Range Queries**: Test with large range queries to ensure the GCD calculation is efficient and correct.

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

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

Explanation:
- **Test Case 1**: An array with a single element `[7]` and a query `[1, 1]`. The expected output is `0` because the range has only one element.
- **Test Case 2**: An array with three elements, all the same `[7, 7, 7]`, and queries `[1, 3]` and `[3, 3]`. The expected output is `7` for both queries because all elements are equal.
- **Test Case 3**: An array with five elements `[1, 2, 3, 4, 5]` and queries `[1, 5]`, `[2, 4]`, and `[3, 5]`. The expected output is `1` for all queries because the GCD of any range in this array is `1`.

This test input 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: Maximum modulo equality
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$ and $$$q$$$ queries $$$l$$$, $$$r$$$.

For each query, find the maximum possible $$$m$$$, such that all elements $$$a_l$$$, $$$a_{l+1}$$$, ..., $$$a_r$$$ are equal modulo $$$m$$$. In other words, $$$a_l \bmod m = a_{l+1} \bmod m = \dots = a_r \bmod m$$$, where $$$a \bmod b$$$ — is the remainder of division $$$a$$$ by $$$b$$$. In particular, when $$$m$$$ can be infinite, print $$$0$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$q$$$ ($$$1 \le n, q \le 2\cdot 10^5$$$) — the length of the array and the number of queries.

The second line of each test case contains $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$) — the elements of the array.

In the following $$$q$$$ lines of each test case, two integers $$$l$$$, $$$r$$$ are provided ($$$1 \le l \le r \le n$$$) — the range of the query.

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

Output: For each query, output the maximum value $$$m$$$ described in the statement.

Examples:
input:
3
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
output:
3 1 4 1 0
0
1 6

Note: In the first query of the first sample, $$$6 \bmod 3 = 3 \bmod 3 = 0$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

In the third query of the first sample, $$$14 \bmod 4 = 2 \bmod 4 = 6 \bmod 4 = 2$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

Buggy Code:
#include <iostream>
#include <algorithm>
#include <vector>
#include <stack>
#include <map>
#include <queue>
#include <math.h>
#include <set>
#include <string>
#include <cassert>
#include <deque>
#include <functional>
#define ll long long
#define pii pair<ll, ll>
#define inf 1000000000000000LL
#define mod 1000000007
#define vi vector<ll>
using namespace std;
ll gcd(ll a, ll b) {
    while (b != 0) {
        if (b > a) swap(a, b);
        a -= b;
    }
    return a;
}
void solve() {
    ll n, q, temp;
    cin >> n >> q;
    if (n == 1) {
        ll x; cin >> x;
        for (int i = 0; i < q; i++) { ll a, b; cin >> a >> b; cout << 0 << ' '; }
    }
    else {
        vector<ll> v(n - 1);
        for (int i = 0; i < n - 1; i++) cin >> v[i];
        cin >> temp;
        for (int i = 0; i < n - 2; i++) v[i] = abs(v[i] - v[i + 1]);
        n--;
        v[n - 1] = abs(temp - v[n - 1]);
        vector<vi> T(19, vi(n));
        for (int i = 0; i < n; i++) {
            T[0][i] = v[i];
        }
        ll p = 1;
        for (int k = 1; k < 19; k++) {
            for (int i = 0; i < n; i++) {
                T[k][i] = gcd(T[k - 1][i], T[k - 1][fmin(n - 1, p + i)]);
            }
            p *= 2;
        }
        for (int i = 0; i < q; i++) {
            ll l, r, p = 1, k = -1;
            cin >> l >> r;
            l--; r -= 2;
            if (r < l) cout << 0 << ' '; else {
                while (r - l + 1 > p) {
                    p *= 2; k++;
                }
                if (p != 1)
                    p /= 2;
                if (k < 0) k++;
                ll R = gcd(T[k][l], T[k][r - p + 1]);
                cout << R << ' ';
            }
        }
    }
    cout << endl;
}
int main()
{
    //int t = 1;
    int t;  cin >> t;
    while (t--)
    solve();
}

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

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

Think step by step.

Response:
```plaintext
2
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum modulo equality
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$ and $$$q$$$ queries $$$l$$$, $$$r$$$.

For each query, find the maximum possible $$$m$$$, such that all elements $$$a_l$$$, $$$a_{l+1}$$$, ..., $$$a_r$$$ are equal modulo $$$m$$$. In other words, $$$a_l \bmod m = a_{l+1} \bmod m = \dots = a_r \bmod m$$$, where $$$a \bmod b$$$ — is the remainder of division $$$a$$$ by $$$b$$$. In particular, when $$$m$$$ can be infinite, print $$$0$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$q$$$ ($$$1 \le n, q \le 2\cdot 10^5$$$) — the length of the array and the number of queries.

The second line of each test case contains $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$) — the elements of the array.

In the following $$$q$$$ lines of each test case, two integers $$$l$$$, $$$r$$$ are provided ($$$1 \le l \le r \le n$$$) — the range of the query.

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

Output: For each query, output the maximum value $$$m$$$ described in the statement.

Examples:
input:
3
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
output:
3 1 4 1 0
0
1 6

Note: In the first query of the first sample, $$$6 \bmod 3 = 3 \bmod 3 = 0$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

In the third query of the first sample, $$$14 \bmod 4 = 2 \bmod 4 = 6 \bmod 4 = 2$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

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

int findMaxM(const vector<int>& subarray) {
    int n = subarray.size();
    if (n == 1) return 0;

    int minVal = *min_element(subarray.begin(), subarray.end());
    int maxVal = *max_element(subarray.begin(), subarray.end());

    if (minVal == maxVal) return 0;

    for (int m = maxVal; m > 1; --m) {
        bool valid = true;
        int modValue = subarray[0] % m;

        for (int i = 1; i < n; i++) {
            if (subarray[i] % m != modValue) {
                valid = false;
                break;
            }
        }

        if (valid) return m;
    }

    return 1;
}

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

    int t;
    cin >> t;
    while (t--) {
        int n, q;
        cin >> n >> q;
        vector<int> a(n);
        for (int i = 0; i < n; ++i) {
            cin >> a[i];
        }

        vector<int> answers;
        for (int i = 0; i < q; ++i) {
            int l, r;
            cin >> l >> r;
            --l; --r;

            vector<int> subarray(a.begin() + l, a.begin() + r + 1);
            answers.push_back(findMaxM(subarray));
        }

        for (int i = 0; i < answers.size(); ++i) {
            cout << answers[i];
            if (i < answers.size() - 1) cout << " ";
        }
        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
1
3 3
1 2 3
1 2
2 3
1 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum modulo equality
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$ and $$$q$$$ queries $$$l$$$, $$$r$$$.

For each query, find the maximum possible $$$m$$$, such that all elements $$$a_l$$$, $$$a_{l+1}$$$, ..., $$$a_r$$$ are equal modulo $$$m$$$. In other words, $$$a_l \bmod m = a_{l+1} \bmod m = \dots = a_r \bmod m$$$, where $$$a \bmod b$$$ — is the remainder of division $$$a$$$ by $$$b$$$. In particular, when $$$m$$$ can be infinite, print $$$0$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$q$$$ ($$$1 \le n, q \le 2\cdot 10^5$$$) — the length of the array and the number of queries.

The second line of each test case contains $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$) — the elements of the array.

In the following $$$q$$$ lines of each test case, two integers $$$l$$$, $$$r$$$ are provided ($$$1 \le l \le r \le n$$$) — the range of the query.

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

Output: For each query, output the maximum value $$$m$$$ described in the statement.

Examples:
input:
3
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
output:
3 1 4 1 0
0
1 6

Note: In the first query of the first sample, $$$6 \bmod 3 = 3 \bmod 3 = 0$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

In the third query of the first sample, $$$14 \bmod 4 = 2 \bmod 4 = 6 \bmod 4 = 2$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

Buggy Code:
for _ in range(int(input())):
    n, q = map(int, input().split(" "))
    a = list(map(int, input().split(" ")))
    def isequal(array):
        for i in array:
            if i != array[0]:
                return False
        return True
    query = ""
    for __ in range(q):
        l, r = map(int, input().split(" "))
        if isequal(a[l-1: r]):
            query += "0 "
        else:
            m = 1
            for i in sorted(range(2,max(a[l-1: r]) + 1), reverse=True):
                if isequal([z % i for z in a[l-1: r]]):
                    m = i
                    break
            query += str(m) + " "

    print(query)

**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
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
```

### Explanation:
1. **Test Case 1:**
   - Array: `[5, 14, 2, 6, 3]`
   - Queries:
     - `4 5`: The subarray is `[6, 3]`. The maximum `m` is `3` because `6 % 3 = 3 % 3 = 0`.
     - `1 4`: The subarray is `[5, 14, 2, 6]`. The maximum `m` is `1` because no greater `m` can make all elements equal modulo `m`.
     - `2 4`: The subarray is `[14, 2, 6]`. The maximum `m` is `4` because `14 % 4 = 2 % 4 = 6 % 4 = 2`.
     - `3 5`: The subarray is `[2, 6, 3]`. The maximum `m` is `1` because no greater `m` can make all elements equal modulo `m`.
     - `1 1`: The subarray is `[5]`. The maximum `m` is `0` because a single element is always equal to itself modulo any `m`.

2. **Test Case 2:**
   - Array: `[7]`
   - Query:
     - `1 1`: The subarray is `[7]`. The maximum `m` is `0` because a single element is always equal to itself modulo any `m`.

3. **Test Case 3:**
   - Array: `[1, 7, 8]`
   - Queries:
     - `2 3`: The subarray is `[7, 8]`. The maximum `m` is `1` because `7 % 1 = 8 % 1 = 0`.
     - `1 2`: The subarray is `[1, 7]`. The maximum `m` is `6` because `1 % 6 = 7 % 6 = 1`.

This test input is designed to cover various edge cases, including single-element subarrays, subarrays with different elements, and subarrays that require checking multiple possible values of `m`. It should help expose the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum modulo equality
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$ and $$$q$$$ queries $$$l$$$, $$$r$$$.

For each query, find the maximum possible $$$m$$$, such that all elements $$$a_l$$$, $$$a_{l+1}$$$, ..., $$$a_r$$$ are equal modulo $$$m$$$. In other words, $$$a_l \bmod m = a_{l+1} \bmod m = \dots = a_r \bmod m$$$, where $$$a \bmod b$$$ — is the remainder of division $$$a$$$ by $$$b$$$. In particular, when $$$m$$$ can be infinite, print $$$0$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$q$$$ ($$$1 \le n, q \le 2\cdot 10^5$$$) — the length of the array and the number of queries.

The second line of each test case contains $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$) — the elements of the array.

In the following $$$q$$$ lines of each test case, two integers $$$l$$$, $$$r$$$ are provided ($$$1 \le l \le r \le n$$$) — the range of the query.

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

Output: For each query, output the maximum value $$$m$$$ described in the statement.

Examples:
input:
3
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
output:
3 1 4 1 0
0
1 6

Note: In the first query of the first sample, $$$6 \bmod 3 = 3 \bmod 3 = 0$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

In the third query of the first sample, $$$14 \bmod 4 = 2 \bmod 4 = 6 \bmod 4 = 2$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

Buggy Code:
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt(), q = sc.nextInt();
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
            }
            int m = n - 1;
            int[] b = new int[m];
            for (int i = 0; i < m; i++) {
                b[i] = Math.abs(a[i] - a[i + 1]);
            }
            int[][] f = new int[m][20];
            for (int i = 0; i < m; i++) {
                f[i][0] = b[i];
            }
            for (int i = 0; i < m; i++) {
                for (int j = 1; j < 20 && i + (1 << j) - 1 < m; j++) {
                    f[i][j] = gcd(f[i][j - 1], f[i + (1 << j - 1)][j - 1]);
                }
            }
            StringBuffer sb = new StringBuffer();
            for (int i = 0; i < q; i++) {
                int l = sc.nextInt() - 1, r = sc.nextInt() - 2;
                if (l > r) {
                    sb.append("0 ");
                    continue;
                }
                if (l == r) {
                    sb.append(b[l]).append(' ');
                    continue;
                }
                int len = r - l + 1;
                int j = 1;
                while ((1 << j + 1) < len) j++;
//                System.out.println("l = " + l);
//                System.out.println("r = " + r);
//                System.out.println("j = " + j);
                int g = gcd(f[l][j], f[r - (1 << j) + 1][j]);
                sb.append(g).append(' ');
            }
            System.out.println(sb);
        }
    }
    static int gcd(int a, int b) {
        return b == 0? a: gcd(b, a % b);
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum modulo equality
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$ and $$$q$$$ queries $$$l$$$, $$$r$$$.

For each query, find the maximum possible $$$m$$$, such that all elements $$$a_l$$$, $$$a_{l+1}$$$, ..., $$$a_r$$$ are equal modulo $$$m$$$. In other words, $$$a_l \bmod m = a_{l+1} \bmod m = \dots = a_r \bmod m$$$, where $$$a \bmod b$$$ — is the remainder of division $$$a$$$ by $$$b$$$. In particular, when $$$m$$$ can be infinite, print $$$0$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$q$$$ ($$$1 \le n, q \le 2\cdot 10^5$$$) — the length of the array and the number of queries.

The second line of each test case contains $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$) — the elements of the array.

In the following $$$q$$$ lines of each test case, two integers $$$l$$$, $$$r$$$ are provided ($$$1 \le l \le r \le n$$$) — the range of the query.

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

Output: For each query, output the maximum value $$$m$$$ described in the statement.

Examples:
input:
3
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
output:
3 1 4 1 0
0
1 6

Note: In the first query of the first sample, $$$6 \bmod 3 = 3 \bmod 3 = 0$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

In the third query of the first sample, $$$14 \bmod 4 = 2 \bmod 4 = 6 \bmod 4 = 2$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

Buggy Code:
#include <bits/stdc++.h>
#ifdef LOCAL
#include "bits/debug.h"
#else
#define debug(...) 0
#endif
using namespace std;
using ll = long long;
using pii = pair<int, int>;
#define sz(x) signed(size(x))
#define fi first
#define se second

void Atarayo() {
    int n, q;
    cin >> n >> q;

    vector<int> a(n);
    vector table(n, vector<int>(15));
    for (int& i : a)
        cin >> i;

    for (int i = 1; i < n; ++i)
        table[i][0] = abs(a[i] - a[i - 1]);

    for (int j = 1; j < 15; ++j)
        for (int i = 0; i < n; ++i) {
            int nxt = min(n - 1, i + (1 << (j - 1)));
            table[i][j] = gcd(table[i][j - 1], table[nxt][j - 1]);
        }

    auto query = [&](int l, int r) {
        if (l == r) return 0;
        ++l;
        int len = r - l + 1;
        int k = __lg(len);
        return gcd(table[l][k], table[r - (1 << k) + 1][k]);
    };

    while (q--) {
        int l, r;
        cin >> l >> r;
        cout << query(l - 1, r - 1) << " \n"[q == 0];
    }
}

signed main() {
    cin.tie(0)->sync_with_stdio(0);
    int t = 1;
    cin >> t;
    while (t--)
        Atarayo();
}

**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 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum modulo equality
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$ and $$$q$$$ queries $$$l$$$, $$$r$$$.

For each query, find the maximum possible $$$m$$$, such that all elements $$$a_l$$$, $$$a_{l+1}$$$, ..., $$$a_r$$$ are equal modulo $$$m$$$. In other words, $$$a_l \bmod m = a_{l+1} \bmod m = \dots = a_r \bmod m$$$, where $$$a \bmod b$$$ — is the remainder of division $$$a$$$ by $$$b$$$. In particular, when $$$m$$$ can be infinite, print $$$0$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$q$$$ ($$$1 \le n, q \le 2\cdot 10^5$$$) — the length of the array and the number of queries.

The second line of each test case contains $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$) — the elements of the array.

In the following $$$q$$$ lines of each test case, two integers $$$l$$$, $$$r$$$ are provided ($$$1 \le l \le r \le n$$$) — the range of the query.

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

Output: For each query, output the maximum value $$$m$$$ described in the statement.

Examples:
input:
3
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
output:
3 1 4 1 0
0
1 6

Note: In the first query of the first sample, $$$6 \bmod 3 = 3 \bmod 3 = 0$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

In the third query of the first sample, $$$14 \bmod 4 = 2 \bmod 4 = 6 \bmod 4 = 2$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

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

using namespace std;

// INPUT
#ifndef ONLINE_JUDGE
#define IO freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout)
#define time  cout << "\n"; auto end = chrono::steady_clock::now();auto diff = end - start; cout << chrono::duration<double, milli>(diff).count() << " ms" << endl;
#else
#define IO
#define time
#endif
#define fastio ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ll long long
#define testcase int t; cin>>t; while(t--)

// typo
#define Endl endl
#define itn int

// Loops
#define fori(i,n) for(int i=0;i<n;i++)
#define rfor(i,n) for(int i=n-1;i>=0;i--)
#define foran(i,a,n) for(int i=a;i<n;i++)

// Vector
#define vi vector<int>
#define vll vector<ll>
#define pii pair<int, int>
#define vpii vector<pair<int, int>>
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define vin(p) for(auto &i:p) cin >> i
#define vout(p) for(auto i:p) cout << i << " "; cout<<endl

auto start = chrono::steady_clock::now();

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

void solve()
{
    int n, q;
    cin>>n>>q;
    vi v(n);
    vin(v);
    vi a(n-1);
    fori(i, n-1)
    {
        a[i] = abs(v[i]-v[i+1]);
    }
    vector<vi> b(n, vi(n));

    for(int i=0;i<n-1;i++)
    {
        b[i][0] = a[i];
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=0;j<n-1 && i+j<n-1;j++)
        {
            b[j][i] = gcd(b[j][i-1], b[j+1][i-1]);
        }
    }
    // fori(i, b.size())
    // {
    //     fori(j, b[0].size())
    //     {
    //         cout<<b[i][j]<<" ";
    //     }
    //     cout<<endl;
    // }
    // cout<<Endl;

    int l, r;
    while(q--)
    {
        cin>>l>>r;
        if(l==r)
        {
            cout<<0<<" ";
            continue;
        }
        int ans;
        ans = b[l-1][r-l-1];

        cout<<ans<<" ";
    }
    cout<<endl;
    return ;
}

int main()
{
    IO;
    fastio;

    testcase
    {
        solve();
    }
    time

    return 0;
}

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

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

Think step by step.

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

1. **Understand the Bug**: The bug in the code is related to the way the GCD is calculated and stored in the `b` array. The code assumes that the GCD of a range can be efficiently computed using a precomputed table, but it doesn't handle cases where the range is very small or where all elements are the same.

2. **Edge Cases**:
   - **Single Element Query**: When the query range has only one element, the output should be 0.
   - **Identical Elements**: When all elements in the query range are the same, the output should be a large number (in this case, we can use 0 as per the problem statement).
   - **Consecutive Elements with Small Differences**: When the elements in the query range have small differences, the GCD might be large, but the code might not handle this correctly.

3. **Small-Scale Test Input**:
   - We will create a test case with a small array and a few queries to cover the edge cases mentioned above.

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

```plaintext
3
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
3
7
1 1
3 2
1 7 8
2 3
1 2
```

Explanation:
- **Test Case 1**:
  - Array: `[5, 14, 2, 6, 3]`
  - Queries:
    - `4 5`: Elements are `[6, 3]`, GCD is 3.
    - `1 4`: Elements are `[5, 14, 2, 6]`, GCD is 1.
    - `2 4`: Elements are `[14, 2, 6]`, GCD is 4.
    - `3 5`: Elements are `[2, 6, 3]`, GCD is 1.
    - `1 1`: Single element, output is 0.
- **Test Case 2**:
  - Array: `[7]`
  - Queries:
    - `1 1`: Single element, output is 0.
- **Test Case 3**:
  - Array: `[1, 7, 8]`
  - Queries:
    - `2 3`: Elements are `[7, 8]`, GCD is 1.
    - `1 2`: Elements are `[1, 7]`, GCD is 6.

This test input should help expose the bug in the given buggy code by covering various edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum modulo equality
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$ and $$$q$$$ queries $$$l$$$, $$$r$$$.

For each query, find the maximum possible $$$m$$$, such that all elements $$$a_l$$$, $$$a_{l+1}$$$, ..., $$$a_r$$$ are equal modulo $$$m$$$. In other words, $$$a_l \bmod m = a_{l+1} \bmod m = \dots = a_r \bmod m$$$, where $$$a \bmod b$$$ — is the remainder of division $$$a$$$ by $$$b$$$. In particular, when $$$m$$$ can be infinite, print $$$0$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$q$$$ ($$$1 \le n, q \le 2\cdot 10^5$$$) — the length of the array and the number of queries.

The second line of each test case contains $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$) — the elements of the array.

In the following $$$q$$$ lines of each test case, two integers $$$l$$$, $$$r$$$ are provided ($$$1 \le l \le r \le n$$$) — the range of the query.

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

Output: For each query, output the maximum value $$$m$$$ described in the statement.

Examples:
input:
3
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
output:
3 1 4 1 0
0
1 6

Note: In the first query of the first sample, $$$6 \bmod 3 = 3 \bmod 3 = 0$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

In the third query of the first sample, $$$14 \bmod 4 = 2 \bmod 4 = 6 \bmod 4 = 2$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

Buggy Code:
def GCD(num1, num2):
    if num1 == 0:
        return num2
    else:
        return GCD(num2%num1, num1)

t = int(input())
for tt in range(0, t):
    n, q = map(int, input().split())
    vals = list(map(int, input().split()))
    res = []
    for qq in range(0, q):
        l, r = map(int, input().split())
        if l == r:
            res.append(0)
        else:
            gcd = 0
            w = list(set(vals[(l-1):r]))
            for i in range(0, len(w) - 1):
                gcd = GCD(gcd, abs(w[i] - w[i+1]))
                if gcd == 1:
                    break
            res.append(gcd)
    print(*res)

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can highlight the issues in the algorithm. The main issue in the buggy code is that it does not correctly handle the case where all elements in the range are the same. Additionally, it does not handle the case where the range is a single element correctly.

Here's a step-by-step approach to creating the test input:

1. **Single Element Range**: The code should return 0 for a single element range, but it might not handle this case correctly.
2. **All Elements Equal**: The code should return the maximum possible \( m \) which is the difference between the maximum and minimum values in the range, but it might not handle this case correctly.
3. **Different Elements**: The code should return the greatest common divisor (GCD) of the differences between the elements in the range, but it might not handle this case correctly.

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

```plaintext
3
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
```

This input includes:
1. A test case with a single element range.
2. A test case with all elements equal.
3. A test case with different elements.

Let's break it down:

1. **Test Case 1**:
   - Array: `[5, 14, 2, 6, 3]`
   - Queries:
     - `4 5`: Elements are `6` and `3`. The GCD of `3` is `3`.
     - `1 4`: Elements are `5, 14, 2, 6`. The GCD of differences is `1`.
     - `2 4`: Elements are `14, 2, 6`. The GCD of differences is `4`.
     - `3 5`: Elements are `2, 6, 3`. The GCD of differences is `1`.
     - `1 1`: Single element. The code should return `0`.

2. **Test Case 2**:
   - Array: `[7]`
   - Queries:
     - `1 1`: Single element. The code should return `0`.

3. **Test Case 3**:
   - Array: `[1, 7, 8]`
   - Queries:
     - `2 3`: Elements are `7, 8`. The GCD of `1` is `1`.
     - `1 2`: Elements are `1, 7`. The GCD of `6` is `6`.

This test input should help expose 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: Maximum modulo equality
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$ and $$$q$$$ queries $$$l$$$, $$$r$$$.

For each query, find the maximum possible $$$m$$$, such that all elements $$$a_l$$$, $$$a_{l+1}$$$, ..., $$$a_r$$$ are equal modulo $$$m$$$. In other words, $$$a_l \bmod m = a_{l+1} \bmod m = \dots = a_r \bmod m$$$, where $$$a \bmod b$$$ — is the remainder of division $$$a$$$ by $$$b$$$. In particular, when $$$m$$$ can be infinite, print $$$0$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$q$$$ ($$$1 \le n, q \le 2\cdot 10^5$$$) — the length of the array and the number of queries.

The second line of each test case contains $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$) — the elements of the array.

In the following $$$q$$$ lines of each test case, two integers $$$l$$$, $$$r$$$ are provided ($$$1 \le l \le r \le n$$$) — the range of the query.

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

Output: For each query, output the maximum value $$$m$$$ described in the statement.

Examples:
input:
3
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
output:
3 1 4 1 0
0
1 6

Note: In the first query of the first sample, $$$6 \bmod 3 = 3 \bmod 3 = 0$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

In the third query of the first sample, $$$14 \bmod 4 = 2 \bmod 4 = 6 \bmod 4 = 2$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

Buggy Code:
def GCD(num1, num2):
    if num1 == 0:
        return num2
    elif num2 == 0:
        return num1
    else:
        if num1 >= num2:
            num1 %= num2
            if num1 == 0:
                return num2
            else:
                return GCD(num1, num2)
        else:
            num2 %= num1
            if num2 == 0:
                return num1
            else:
                return GCD(num1, num2)

t = int(input())
for tt in range(0, t):
    n, q = map(int, input().split())
    vals = list(map(int, input().split()))
    res = []
    for qq in range(0, q):
        l, r = map(int, input().split())
        if l == r:
            res.append(0)
        else:
            gcd = 0
            w = list(set(vals[(l-1):r]))
            for i in range(0, len(w) - 1):
                gcd = GCD(gcd, abs(w[i] - w[i+1]))
            res.append(gcd)
    print(*res)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code has issues with handling the GCD calculation and the set of unique values in the range. Here are the steps to create a suitable test input:

1. **Edge Case: Single Element Queries**:
   - The code correctly handles single element queries by outputting 0. We don't need to focus on this.

2. **Edge Case: Identical Elements in Range**:
   - If all elements in the range are identical, the maximum possible \( m \) should be the element itself. The code might fail if it doesn't handle this correctly.

3. **Edge Case: Consecutive Elements with Small Differences**:
   - If the elements in the range have small differences, the GCD calculation might be incorrect. For example, if the elements are 1, 2, 3, the GCD should be 1.

4. **Edge Case: Large Differences in Elements**:
   - If the elements in the range have large differences, the GCD calculation might be incorrect. For example, if the elements are 1, 1000, 1000000, the GCD should be 1.

5. **Edge Case: Mixed Elements**:
   - If the elements in the range are mixed with both small and large values, the GCD calculation might be incorrect. For example, if the elements are 1, 2, 1000, 1000000, the GCD should be 1.

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

```plaintext
3
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
3 3
1 1 1
1 1
3 3
3 5
1 7 8
2 3
1 2
```

This test input includes:
- A case with identical elements in the range.
- A case with consecutive elements with small differences.
- A case with large differences in elements.
- A case with mixed elements.

Each of these scenarios should help in identifying the issues in the GCD calculation and the handling of unique values in the range.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum modulo equality
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$ and $$$q$$$ queries $$$l$$$, $$$r$$$.

For each query, find the maximum possible $$$m$$$, such that all elements $$$a_l$$$, $$$a_{l+1}$$$, ..., $$$a_r$$$ are equal modulo $$$m$$$. In other words, $$$a_l \bmod m = a_{l+1} \bmod m = \dots = a_r \bmod m$$$, where $$$a \bmod b$$$ — is the remainder of division $$$a$$$ by $$$b$$$. In particular, when $$$m$$$ can be infinite, print $$$0$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$q$$$ ($$$1 \le n, q \le 2\cdot 10^5$$$) — the length of the array and the number of queries.

The second line of each test case contains $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$) — the elements of the array.

In the following $$$q$$$ lines of each test case, two integers $$$l$$$, $$$r$$$ are provided ($$$1 \le l \le r \le n$$$) — the range of the query.

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

Output: For each query, output the maximum value $$$m$$$ described in the statement.

Examples:
input:
3
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
output:
3 1 4 1 0
0
1 6

Note: In the first query of the first sample, $$$6 \bmod 3 = 3 \bmod 3 = 0$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

In the third query of the first sample, $$$14 \bmod 4 = 2 \bmod 4 = 6 \bmod 4 = 2$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

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

public class Main{

    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));

        int t = Integer.parseInt(reader.readLine());
        while (t-- > 0) {
            StringTokenizer st = new StringTokenizer(reader.readLine());
            int n = Integer.parseInt(st.nextToken());
            int q = Integer.parseInt(st.nextToken());

            int[] a = new int[n];
            st = new StringTokenizer(reader.readLine());
            for (int i = 0; i < n; i++) {
                a[i] = Integer.parseInt(st.nextToken());
            }

            // Precompute prefix GCDs for efficiency
            int[] prefixGCD = new int[n];
            for (int i = 1; i < n; i++) {
                prefixGCD[i] = gcd(prefixGCD[i - 1], Math.abs(a[i] - a[i - 1]));
            }

            while (q-- > 0) {
                st = new StringTokenizer(reader.readLine());
                int l = Integer.parseInt(st.nextToken()) - 1;
                int r = Integer.parseInt(st.nextToken()) - 1;

                if (l == r) {
                    writer.write("0 ");
                    continue;
                }

                // Check if all elements are the same
                boolean allSame = true;
                for (int i = l + 1; i <= r; i++) {
                    if (a[i] != a[l]) {
                        allSame = false;
                        break;
                    }
                }
                if (allSame) {
                    writer.write("0 ");
                    continue;
                }

                // Compute difference between the first element and rest
                int m = 0;
                for (int i = l + 1; i <= r; i++) {
                    m = gcd(m, Math.abs(a[i] - a[l]));
                }
                writer.write(m + " ");
            }
            writer.newLine();
        }
        writer.flush();
        writer.close();
    }

    private static int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}

// 1zliwbj6fNBxKPRUGVp

**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
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
```

### Explanation:
1. **Test Case 1:**
   - Array: `[5, 14, 2, 6, 3]`
   - Queries:
     - Query 1: `(4, 5)` - The elements are `[6, 3]`, and the maximum modulo is `3` (since `6 % 3 == 3 % 3 == 0`).
     - Query 2: `(1, 4)` - The elements are `[5, 14, 2, 6]`, and the maximum modulo is `1` (since `5 % 1 == 14 % 1 == 2 % 1 == 6 % 1 == 0`).
     - Query 3: `(2, 4)` - The elements are `[14, 2, 6]`, and the maximum modulo is `4` (since `14 % 4 == 2 % 4 == 6 % 4 == 2`).
     - Query 4: `(3, 5)` - The elements are `[2, 6, 3]`, and the maximum modulo is `1` (since `2 % 1 == 6 % 1 == 3 % 1 == 0`).
     - Query 5: `(1, 1)` - The elements are `[5]`, and the maximum modulo is `0` (since a single element is always equal to itself modulo any number).

2. **Test Case 2:**
   - Array: `[7]`
   - Queries:
     - Query 1: `(1, 1)` - The elements are `[7]`, and the maximum modulo is `0` (since a single element is always equal to itself modulo any number).

3. **Test Case 3:**
   - Array: `[1, 7, 8]`
   - Queries:
     - Query 1: `(2, 3)` - The elements are `[7, 8]`, and the maximum modulo is `1` (since `7 % 1 == 8 % 1 == 0`).
     - Query 2: `(1, 2)` - The elements are `[1, 7]`, and the maximum modulo is `6` (since `1 % 6 == 7 % 6 == 1`).

This test input includes various edge cases such as single-element queries, queries with all elements being the same, and queries with different elements, which can help expose bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum modulo equality
time_limit_ms: 5000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of length $$$n$$$ and $$$q$$$ queries $$$l$$$, $$$r$$$.

For each query, find the maximum possible $$$m$$$, such that all elements $$$a_l$$$, $$$a_{l+1}$$$, ..., $$$a_r$$$ are equal modulo $$$m$$$. In other words, $$$a_l \bmod m = a_{l+1} \bmod m = \dots = a_r \bmod m$$$, where $$$a \bmod b$$$ — is the remainder of division $$$a$$$ by $$$b$$$. In particular, when $$$m$$$ can be infinite, print $$$0$$$.

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

The first line of each test case contains two integers $$$n$$$, $$$q$$$ ($$$1 \le n, q \le 2\cdot 10^5$$$) — the length of the array and the number of queries.

The second line of each test case contains $$$n$$$ integers $$$a_i$$$ ($$$1 \le a_i \le 10^9$$$) — the elements of the array.

In the following $$$q$$$ lines of each test case, two integers $$$l$$$, $$$r$$$ are provided ($$$1 \le l \le r \le n$$$) — the range of the query.

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

Output: For each query, output the maximum value $$$m$$$ described in the statement.

Examples:
input:
3
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
output:
3 1 4 1 0
0
1 6

Note: In the first query of the first sample, $$$6 \bmod 3 = 3 \bmod 3 = 0$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

In the third query of the first sample, $$$14 \bmod 4 = 2 \bmod 4 = 6 \bmod 4 = 2$$$. It can be shown that for greater $$$m$$$, the required condition will not be fulfilled.

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

public class CodeForcesSolution {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    static StringTokenizer st;
    static int MOD = (int) 998244353, n, q, a[];
    public static void main(String[] args) throws IOException {
        int t = readInt();
        while (t-- > 0) {
            solve();
        }
    }
    static void solve() throws IOException {
        n = readInt(); q = readInt(); a = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            a[i] = readInt();
        }
        while (q-->0) {
            int l = readInt(), r = readInt();
            int min = 1<<30;
            List<Integer> val = new ArrayList<>();
            for (int i = l; i <= r; i++) {
                min = Math.min(min, a[i]);
                val.add(a[i]);
            }
            int g = 0;
            for (int v : val) {
                g = gcd(g, v - min);
            }
            System.out.print(g + " ");
        }
        System.out.println();
    }
    static int gcd(int a, int b) {
        if (b == 0) return a;
        return gcd(b, a % b);
    }
    static void printArray(int[] arr, int size) {
        for(int i = 0; i < size; i++) {
            System.out.print(arr[i]);
            if (i < size - 1) System.out.print(" ");
        }
        System.out.println();
    }
    static String next() throws IOException {
        while (st == null || !st.hasMoreTokens())
            st = new StringTokenizer(br.readLine().trim());
        return st.nextToken();
    }
    static long readLong() throws IOException {
        return Long.parseLong(next());
    }
    static int readInt() throws IOException {
        return Integer.parseInt(next());
    }
    static double readDouble() throws IOException {
        return Double.parseDouble(next());
    }
    static char readCharacter() throws IOException {
        return next().charAt(0);
    }
    static String readLine() throws IOException {
        return br.readLine().trim();
    }
    static int readALotInt() throws IOException{
        int x = 0, c;
        while((c = br.read()) != ' ' && c != '\n')
            x = x * 10 + (c - '0');
        return x;
    }
}

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

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

Think step by step.

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

1. **Single Element Queries**: Test queries where the range `[l, r]` contains only one element. This should always return `0` because the modulo condition is trivially satisfied for any `m`.

2. **Identical Elements in Range**: Test queries where all elements in the range `[l, r]` are identical. The maximum `m` should be `0` because all elements are equal, and the modulo condition is trivially satisfied for any `m`.

3. **Distinct Elements in Range**: Test queries where the elements in the range `[l, r]` are distinct. The maximum `m` should be the greatest common divisor (GCD) of the differences between the elements and the minimum element in the range.

4. **Large Range Queries**: Test queries with large ranges to ensure the algorithm can handle large inputs efficiently.

5. **Mixed Scenarios**: Combine the above scenarios in a single test case to ensure the algorithm handles different types of inputs correctly.

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

```plaintext
3
5 5
5 14 2 6 3
4 5
1 4
2 4
3 5
1 1
1 1
7
1 1
3 2
1 7 8
2 3
1 2
10 5
1 2 3 4 5 6 7 8 9 10
1 10
2 9
3 8
4 7
5 6
```

This test input includes:
1. A test case with a small array and various queries, including single element, identical elements, and distinct elements.
2. A test case with a single element in the array and a single element query.
3. A test case with a larger array and mixed queries to test the algorithm's performance and correctness.

This should help expose any bugs in the given buggy code.