**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eri and Expanded Sets
time_limit_ms: 3000
memory_limit_mb: 512
Description: Let there be a set that contains distinct positive integers. To expand the set to contain as many integers as possible, Eri can choose two integers $x\neq y$ from the set such that their average $\frac{x+y}2$ is still a positive integer and isn't contained in the set, and add it to the set. The integers $x$ and $y$ remain in the set.

Let's call the set of integers consecutive if, after the elements are sorted, the difference between any pair of adjacent elements is $1$. For example, sets $\\{2\\}$, $\\{2, 5, 4, 3\\}$, $\\{5, 6, 8, 7\\}$ are consecutive, while $\\{2, 4, 5, 6\\}$, $\\{9, 7\\}$ are not.

Eri likes consecutive sets. Suppose there is an array $b$, then Eri puts all elements in $b$ into the set. If after a finite number of operations described above, the set can become consecutive, the array $b$ will be called brilliant.

Note that if the same integer appears in the array multiple times, we only put it into the set once, as a set always contains distinct positive integers.

Eri has an array $a$ of $n$ positive integers. Please help him to count the number of pairs of integers $(l,r)$ such that $1 \leq l \leq r \leq n$ and the subarray $a_l, a_{l+1}, \ldots, a_r$ is brilliant.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 4 \cdot 10^5$) — length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $4 \cdot 10^5$.

Output: For each test case, output a single integer — the number of brilliant subarrays.

Examples:
input:
6
2
2 2
6
1 3 6 10 15 21
5
6 30 18 36 9
1
1000000000
6
1 1 4 5 1 4
12
70 130 90 90 90 108 612 500 451 171 193 193
output:
3
18
5
1
18
53

Note: In the first test case, the array $a = [2, 2]$ has $3$ subarrays: $[2]$, $[2]$, and $[2, 2]$. For all of them, the set only contains a single integer $2$, therefore it's always consecutive. All these subarrays are brilliant, so the answer is $3$.

In the second test case, let's consider the subarray $[3, 6, 10]$. We can do operations as follows:

$$\\{3,6,10\\} \xrightarrow{x=6,y=10} \\{3,6,8,10\\} \xrightarrow{x=6,y=8} \\{3,6,7,8,10\\} \xrightarrow{x=3,y=7} \\{3,5,6,7,8,10\\}$$ $$\xrightarrow{x=3,y=5} \\{3,4,5,6,7,8,10\\} \xrightarrow{x=8,y=10} \\{3,4,5,6,7,8,9,10\\}$$

$\\{3,4,5,6,7,8,9,10\\}$ is a consecutive set, so the subarray $[3, 6, 10]$ is brilliant.

Buggy Code:
powers_of_two = set()
for i in range(34):
    powers_of_two.add( 1<<i )
from math import gcd

ans = []
for _ in range(int(input())):
    n = int(input())
    a = [int(x) for x in input().split()]
    b = [a[i+1]-a[i] for i in range(n-1)]

    seg = [0]*(4*n)
    def gg(v,tl,tr,l,r):
        if l > r: return 0
        if l == tl and r == tr: return seg[v]
        tm = (tl+tr)//2
        return gcd(gg(v*2,tl,tm,l,min(r,tm)), gg(v*2+1,tm+1,tr,max(l,tm+1),r))
    def update(v,tl,tr,pos,new_val):
        if tl == tr: seg[v] = new_val
        else:
            tm = (tl+tr)//2
            if pos <= tm:
                update(v*2,tl,tm,pos,new_val)
            else:
                update(v*2+1,tm+1,tr,pos,new_val)
            seg[v] = gcd( seg[v*2], seg[v*2+1] )

    for i in range(n-1):
        update(1,0,n-2,i,b[i])

    def gcdd(L,R):
        if L >= R: return 0
        return gg(1,0,n-2,L,R-1)

    #print(gcdd(1,3))

    S = -1
    prev = -1
    run = 1
    a.append(-3)
    for e in a:
        if e == prev:
            run += 1
        else:
            prev = e
            S += (run+1)*run//2
            run = 1

    L = 0
    R = 1
    it = 0
    while True:
        it += 1
        while R < n and not gcdd(L,R) in powers_of_two:
            R += 1
        if R == n: break
        while gcdd(L,R) in powers_of_two:
            S += (n-R)
            L += 1

    ans.append(S)

for i in ans:
    print(i)

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

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

Think step by step.

Response:
```plaintext
4
3
1 2 3
4
2 4 6 8
5
1 3 5 7 9
6
1 2 4 8 16 32
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eri and Expanded Sets
time_limit_ms: 3000
memory_limit_mb: 512
Description: Let there be a set that contains distinct positive integers. To expand the set to contain as many integers as possible, Eri can choose two integers $x\neq y$ from the set such that their average $\frac{x+y}2$ is still a positive integer and isn't contained in the set, and add it to the set. The integers $x$ and $y$ remain in the set.

Let's call the set of integers consecutive if, after the elements are sorted, the difference between any pair of adjacent elements is $1$. For example, sets $\\{2\\}$, $\\{2, 5, 4, 3\\}$, $\\{5, 6, 8, 7\\}$ are consecutive, while $\\{2, 4, 5, 6\\}$, $\\{9, 7\\}$ are not.

Eri likes consecutive sets. Suppose there is an array $b$, then Eri puts all elements in $b$ into the set. If after a finite number of operations described above, the set can become consecutive, the array $b$ will be called brilliant.

Note that if the same integer appears in the array multiple times, we only put it into the set once, as a set always contains distinct positive integers.

Eri has an array $a$ of $n$ positive integers. Please help him to count the number of pairs of integers $(l,r)$ such that $1 \leq l \leq r \leq n$ and the subarray $a_l, a_{l+1}, \ldots, a_r$ is brilliant.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 4 \cdot 10^5$) — length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $4 \cdot 10^5$.

Output: For each test case, output a single integer — the number of brilliant subarrays.

Examples:
input:
6
2
2 2
6
1 3 6 10 15 21
5
6 30 18 36 9
1
1000000000
6
1 1 4 5 1 4
12
70 130 90 90 90 108 612 500 451 171 193 193
output:
3
18
5
1
18
53

Note: In the first test case, the array $a = [2, 2]$ has $3$ subarrays: $[2]$, $[2]$, and $[2, 2]$. For all of them, the set only contains a single integer $2$, therefore it's always consecutive. All these subarrays are brilliant, so the answer is $3$.

In the second test case, let's consider the subarray $[3, 6, 10]$. We can do operations as follows:

$$\\{3,6,10\\} \xrightarrow{x=6,y=10} \\{3,6,8,10\\} \xrightarrow{x=6,y=8} \\{3,6,7,8,10\\} \xrightarrow{x=3,y=7} \\{3,5,6,7,8,10\\}$$ $$\xrightarrow{x=3,y=5} \\{3,4,5,6,7,8,10\\} \xrightarrow{x=8,y=10} \\{3,4,5,6,7,8,9,10\\}$$

$\\{3,4,5,6,7,8,9,10\\}$ is a consecutive set, so the subarray $[3, 6, 10]$ is brilliant.

Buggy Code:
import java.util.*;

public class Main {
    // GCD function using iterative method for better performance
    static long gcd(long a, long b) {
        while (b != 0) {
            long temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }

    // Function to check if a number is a power of 2
    static boolean isPow2(long n) {
        while (n % 2 == 0 && n > 0) n /= 2;
        return n == 1;
    }

    // Main solve function
    static void solve(long tcn, long tc, Scanner sc) {
        long n = sc.nextLong();
        long[] arr = new long[(int) n];

        for (int i = 0; i < n; i++) {
            arr[i] = sc.nextLong();
        }

        if (tcn == 9330){
            if (tc == 2741){
                System.out.println(n);
                for (int i = 0; i < n; i++) System.out.printf("%d ", arr[i]);
            }
        }

        else{
            long[] diff = new long[(int) (n - 1)];
            long[] no_duplicate = new long[(int) (n - 1)];
            Arrays.fill(no_duplicate, -1);

            long idx = -1;
            for (int i = 0; i < n - 1; i++) {
                diff[i] = Math.abs(arr[i + 1] - arr[i]);
                if (diff[i] == 0) {
                    no_duplicate[i] = idx;
                } else {
                    idx = i;
                    no_duplicate[i] = idx;
                }
            }

            // Prepare the powers of 2 array
            List<Long> pow2 = new ArrayList<>();
            long w = 1;
            int p = 0;
            while (w <= n) {
                pow2.add(w);
                w *= 2;
                p++;
            }

            // Build sparse table for GCD
            long[][] sp_table = new long[p][(int) (n - 1)];
            for (int i = 0; i < n - 1; i++) {
                sp_table[0][i] = diff[i];
            }

            for (int pow = 1; pow < p; pow++) {
                for (int i = 0; i + pow2.get(pow) <= n - 1; i++) {
                    sp_table[pow][i] = gcd(sp_table[pow - 1][i], sp_table[pow - 1][i + (int) (long) pow2.get(pow - 1)]);
                }
            }

            long ans = n;
            for (int i = 0; i < n - 1; i++) {
                long lo = -1;
                long hi = no_duplicate[i];
                long duplicate = i - hi;

                while (lo < hi) {
                    long mid = (lo + hi + 1) / 2;
                    int pow = Collections.binarySearch(pow2, hi - mid + 1);
                    if (pow < 0) pow = -pow - 2;

                    long val = gcd(sp_table[pow][(int) mid], sp_table[pow][i - (int) (long) pow2.get(pow) + 1]);
                    if (isPow2(val)) {
                        lo = mid;
                    } else {
                        hi = mid - 1;
                    }
                }
                ans += duplicate + lo + 1;
            }
            System.out.println(ans);
        }
    }

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

        long test_case_number = sc.nextLong();
        for (long test_case = 0; test_case < test_case_number; test_case++) {
            solve(test_case_number, test_case, sc);
        }

        sc.close();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that might cause the algorithm to fail. Here are some key points to consider:

1. **Duplicate Elements**: The code handles duplicates in a specific way, but it might not be robust enough.
2. **Large Differences**: Large differences between consecutive elements can affect the GCD calculations.
3. **Power of 2 Differences**: The code checks if the GCD is a power of 2, so we should include cases where the differences are not powers of 2.
4. **Small Subarrays**: Small subarrays can be tricky, especially when they contain only a few elements.
5. **Consecutive Elements**: Ensure that some subarrays are consecutive to test the correctness of the brilliant subarray detection.

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

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

### Explanation:
1. **Test Case 1**: `1 2 4 8`
   - Differences: `1, 2, 4`
   - GCD of differences: `1, 2, 4` (all are powers of 2)
   - This subarray should be brilliant because the differences are powers of 2.

2. **Test Case 2**: `1 3 5 7 9`
   - Differences: `2, 2, 2, 2`
   - GCD of differences: `2` (a power of 2)
   - This subarray should be brilliant because the differences are all 2, which is a power of 2.

3. **Test Case 3**: `2 4 6`
   - Differences: `2, 2`
   - GCD of differences: `2` (a power of 2)
   - This subarray should be brilliant because the differences are all 2, which is a power of 2.

This test input includes subarrays with differences that are powers of 2, which should help expose any issues in the GCD and power of 2 checks 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: Eri and Expanded Sets
time_limit_ms: 3000
memory_limit_mb: 512
Description: Let there be a set that contains distinct positive integers. To expand the set to contain as many integers as possible, Eri can choose two integers $x\neq y$ from the set such that their average $\frac{x+y}2$ is still a positive integer and isn't contained in the set, and add it to the set. The integers $x$ and $y$ remain in the set.

Let's call the set of integers consecutive if, after the elements are sorted, the difference between any pair of adjacent elements is $1$. For example, sets $\\{2\\}$, $\\{2, 5, 4, 3\\}$, $\\{5, 6, 8, 7\\}$ are consecutive, while $\\{2, 4, 5, 6\\}$, $\\{9, 7\\}$ are not.

Eri likes consecutive sets. Suppose there is an array $b$, then Eri puts all elements in $b$ into the set. If after a finite number of operations described above, the set can become consecutive, the array $b$ will be called brilliant.

Note that if the same integer appears in the array multiple times, we only put it into the set once, as a set always contains distinct positive integers.

Eri has an array $a$ of $n$ positive integers. Please help him to count the number of pairs of integers $(l,r)$ such that $1 \leq l \leq r \leq n$ and the subarray $a_l, a_{l+1}, \ldots, a_r$ is brilliant.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 4 \cdot 10^5$) — length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $4 \cdot 10^5$.

Output: For each test case, output a single integer — the number of brilliant subarrays.

Examples:
input:
6
2
2 2
6
1 3 6 10 15 21
5
6 30 18 36 9
1
1000000000
6
1 1 4 5 1 4
12
70 130 90 90 90 108 612 500 451 171 193 193
output:
3
18
5
1
18
53

Note: In the first test case, the array $a = [2, 2]$ has $3$ subarrays: $[2]$, $[2]$, and $[2, 2]$. For all of them, the set only contains a single integer $2$, therefore it's always consecutive. All these subarrays are brilliant, so the answer is $3$.

In the second test case, let's consider the subarray $[3, 6, 10]$. We can do operations as follows:

$$\\{3,6,10\\} \xrightarrow{x=6,y=10} \\{3,6,8,10\\} \xrightarrow{x=6,y=8} \\{3,6,7,8,10\\} \xrightarrow{x=3,y=7} \\{3,5,6,7,8,10\\}$$ $$\xrightarrow{x=3,y=5} \\{3,4,5,6,7,8,10\\} \xrightarrow{x=8,y=10} \\{3,4,5,6,7,8,9,10\\}$$

$\\{3,4,5,6,7,8,9,10\\}$ is a consecutive set, so the subarray $[3, 6, 10]$ is brilliant.

Buggy Code:
import java.util.*;

public class Main {
    // GCD function using iterative method for better performance
    static long gcd(long a, long b) {
        while (b != 0) {
            long temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }

    // Function to check if a number is a power of 2
    static boolean isPow2(long n) {
        while (n % 2 == 0 && n > 0) n /= 2;
        return n == 1;
    }

    // Main solve function
    static void solve(Scanner sc) {
        long n = sc.nextLong();
        long[] arr = new long[(int) n];

        for (int i = 0; i < n; i++) {
            arr[i] = sc.nextLong();
        }

        if (n == 1) {
            System.out.println(1);
        } else {
            long[] diff = new long[(int) (n - 1)];
            long[] no_duplicate = new long[(int) (n - 1)];
            Arrays.fill(no_duplicate, -1);

            long idx = -1;
            for (int i = 0; i < n - 1; i++) {
                diff[i] = Math.abs(arr[i + 1] - arr[i]);
                if (diff[i] == 0) {
                    no_duplicate[i] = idx;
                } else {
                    idx = i;
                    no_duplicate[i] = idx;
                }
            }

            // Prepare the powers of 2 array
            List<Long> pow2 = new ArrayList<>();
            long w = 1;
            int p = 0;
            while (w <= n) {
                pow2.add(w);
                w *= 2;
                p++;
            }

            // Build sparse table for GCD
            long[][] sp_table = new long[p][(int) (n - 1)];
            for (int i = 0; i < n - 1; i++) {
                sp_table[0][i] = diff[i];
            }

            for (int pow = 1; pow < p; pow++) {
                for (int i = 0; i + pow2.get(pow) <= n - 1; i++) {
                    sp_table[pow][i] = gcd(sp_table[pow - 1][i], sp_table[pow - 1][i + (int) (long) pow2.get(pow - 1)]);
                }
            }

            long ans = n;
            for (int i = 0; i < n - 1; i++) {
                long lo = -1;
                long hi = no_duplicate[i];
                long duplicate = i - hi;

                while (lo < hi) {
                    long mid = (lo + hi + 1) / 2;
                    int pow = Collections.binarySearch(pow2, hi - mid + 1);
                    if (pow < 0) pow = -pow - 2;

                    long val = gcd(sp_table[pow][(int) mid], sp_table[pow][i - (int) (long) pow2.get(pow) + 1]);
                    if (isPow2(val)) {
                        lo = mid;
                    } else {
                        hi = mid - 1;
                    }
                }
                ans += duplicate + lo + 1;
            }
            System.out.println(ans);
        }
    }

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

        long test_case_number = sc.nextLong();
        for (long test_case = 0; test_case < test_case_number; test_case++) {
            solve(sc);
        }

        sc.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
3
2
1 2
3
1 2 3
4
1 2 4 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eri and Expanded Sets
time_limit_ms: 3000
memory_limit_mb: 512
Description: Let there be a set that contains distinct positive integers. To expand the set to contain as many integers as possible, Eri can choose two integers $x\neq y$ from the set such that their average $\frac{x+y}2$ is still a positive integer and isn't contained in the set, and add it to the set. The integers $x$ and $y$ remain in the set.

Let's call the set of integers consecutive if, after the elements are sorted, the difference between any pair of adjacent elements is $1$. For example, sets $\\{2\\}$, $\\{2, 5, 4, 3\\}$, $\\{5, 6, 8, 7\\}$ are consecutive, while $\\{2, 4, 5, 6\\}$, $\\{9, 7\\}$ are not.

Eri likes consecutive sets. Suppose there is an array $b$, then Eri puts all elements in $b$ into the set. If after a finite number of operations described above, the set can become consecutive, the array $b$ will be called brilliant.

Note that if the same integer appears in the array multiple times, we only put it into the set once, as a set always contains distinct positive integers.

Eri has an array $a$ of $n$ positive integers. Please help him to count the number of pairs of integers $(l,r)$ such that $1 \leq l \leq r \leq n$ and the subarray $a_l, a_{l+1}, \ldots, a_r$ is brilliant.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 4 \cdot 10^5$) — length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $4 \cdot 10^5$.

Output: For each test case, output a single integer — the number of brilliant subarrays.

Examples:
input:
6
2
2 2
6
1 3 6 10 15 21
5
6 30 18 36 9
1
1000000000
6
1 1 4 5 1 4
12
70 130 90 90 90 108 612 500 451 171 193 193
output:
3
18
5
1
18
53

Note: In the first test case, the array $a = [2, 2]$ has $3$ subarrays: $[2]$, $[2]$, and $[2, 2]$. For all of them, the set only contains a single integer $2$, therefore it's always consecutive. All these subarrays are brilliant, so the answer is $3$.

In the second test case, let's consider the subarray $[3, 6, 10]$. We can do operations as follows:

$$\\{3,6,10\\} \xrightarrow{x=6,y=10} \\{3,6,8,10\\} \xrightarrow{x=6,y=8} \\{3,6,7,8,10\\} \xrightarrow{x=3,y=7} \\{3,5,6,7,8,10\\}$$ $$\xrightarrow{x=3,y=5} \\{3,4,5,6,7,8,10\\} \xrightarrow{x=8,y=10} \\{3,4,5,6,7,8,9,10\\}$$

$\\{3,4,5,6,7,8,9,10\\}$ is a consecutive set, so the subarray $[3, 6, 10]$ is brilliant.

Buggy Code:
def main():
    import sys
    import threading
    def solve():
        t = int(sys.stdin.readline())
        for _ in range(t):
            n = int(sys.stdin.readline())
            a = list(map(int, sys.stdin.readline().split()))
            total_subarrays = n * (n + 1) // 2
            bad_subarrays = 0
            max_len = 5  # We can process subarrays up to length 5
            for i in range(n):
                S = set()
                for j in range(i, min(n, i + max_len)):
                    S.add(a[j])
                    closure_set = set(S)
                    added = True
                    while added:
                        added = False
                        new_elements = set()
                        for x in closure_set:
                            for y in closure_set:
                                if x != y and (x + y) % 2 == 0:
                                    avg = (x + y) // 2
                                    if avg not in closure_set:
                                        new_elements.add(avg)
                        if new_elements:
                            closure_set.update(new_elements)
                            added = True
                    sorted_S = sorted(closure_set)
                    is_consecutive = all(sorted_S[k] + 1 == sorted_S[k + 1] for k in range(len(sorted_S) - 1))
                    if not is_consecutive:
                        bad_subarrays += 1
            good_subarrays = total_subarrays - bad_subarrays
            print(good_subarrays)
    threading.Thread(target=solve).start()
if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause the code to fail. The key points to consider are:

1. **Subarrays with repeated elements**: The code should handle subarrays with repeated elements correctly, as the set should only contain unique elements.
2. **Subarrays with a large number of elements**: The code should handle subarrays with a large number of elements without timing out or running out of memory.
3. **Subarrays that are already consecutive**: The code should correctly identify subarrays that are already consecutive.
4. **Subarrays that can become consecutive after operations**: The code should correctly identify subarrays that can become consecutive after the operations described.

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

```plaintext
4
5
1 2 3 4 5
6
1 1 2 2 3 3
7
1 3 6 10 15 21 28
8
1 2 4 5 7 8 10 11
```

### Explanation:
1. **Test Case 1**: `1 2 3 4 5`
   - This is a consecutive subarray. The code should correctly identify all subarrays as brilliant.
2. **Test Case 2**: `1 1 2 2 3 3`
   - This subarray contains repeated elements. The code should handle the repeated elements correctly and only consider unique elements in the set.
3. **Test Case 3**: `1 3 6 10 15 21 28`
   - This subarray can be expanded to become consecutive. The code should correctly identify the subarrays that can become consecutive.
4. **Test Case 4**: `1 2 4 5 7 8 10 11`
   - This subarray has gaps that can be filled to become consecutive. The code should correctly identify the subarrays that can become consecutive.

This test input should help expose any bugs in the provided code by covering a variety of scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eri and Expanded Sets
time_limit_ms: 3000
memory_limit_mb: 512
Description: Let there be a set that contains distinct positive integers. To expand the set to contain as many integers as possible, Eri can choose two integers $x\neq y$ from the set such that their average $\frac{x+y}2$ is still a positive integer and isn't contained in the set, and add it to the set. The integers $x$ and $y$ remain in the set.

Let's call the set of integers consecutive if, after the elements are sorted, the difference between any pair of adjacent elements is $1$. For example, sets $\\{2\\}$, $\\{2, 5, 4, 3\\}$, $\\{5, 6, 8, 7\\}$ are consecutive, while $\\{2, 4, 5, 6\\}$, $\\{9, 7\\}$ are not.

Eri likes consecutive sets. Suppose there is an array $b$, then Eri puts all elements in $b$ into the set. If after a finite number of operations described above, the set can become consecutive, the array $b$ will be called brilliant.

Note that if the same integer appears in the array multiple times, we only put it into the set once, as a set always contains distinct positive integers.

Eri has an array $a$ of $n$ positive integers. Please help him to count the number of pairs of integers $(l,r)$ such that $1 \leq l \leq r \leq n$ and the subarray $a_l, a_{l+1}, \ldots, a_r$ is brilliant.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 4 \cdot 10^5$) — length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $4 \cdot 10^5$.

Output: For each test case, output a single integer — the number of brilliant subarrays.

Examples:
input:
6
2
2 2
6
1 3 6 10 15 21
5
6 30 18 36 9
1
1000000000
6
1 1 4 5 1 4
12
70 130 90 90 90 108 612 500 451 171 193 193
output:
3
18
5
1
18
53

Note: In the first test case, the array $a = [2, 2]$ has $3$ subarrays: $[2]$, $[2]$, and $[2, 2]$. For all of them, the set only contains a single integer $2$, therefore it's always consecutive. All these subarrays are brilliant, so the answer is $3$.

In the second test case, let's consider the subarray $[3, 6, 10]$. We can do operations as follows:

$$\\{3,6,10\\} \xrightarrow{x=6,y=10} \\{3,6,8,10\\} \xrightarrow{x=6,y=8} \\{3,6,7,8,10\\} \xrightarrow{x=3,y=7} \\{3,5,6,7,8,10\\}$$ $$\xrightarrow{x=3,y=5} \\{3,4,5,6,7,8,10\\} \xrightarrow{x=8,y=10} \\{3,4,5,6,7,8,9,10\\}$$

$\\{3,4,5,6,7,8,9,10\\}$ is a consecutive set, so the subarray $[3, 6, 10]$ is brilliant.

Buggy Code:
def main():
    import sys
    import threading
    def solve():
        import sys
        sys.setrecursionlimit(1 << 25)
        t = int(sys.stdin.readline())
        for _ in range(t):
            n = int(sys.stdin.readline())
            a = list(map(int, sys.stdin.readline().split()))
            total_subarrays = n * (n + 1) // 2
            bad_subarrays = 0

            max_len = 6  # We can process subarrays up to length 6

            for i in range(n):
                S = set()
                for j in range(i, min(n, i + max_len)):
                    S.add(a[j])
                    closure_set = set(S)
                    added = True
                    while added:
                        added = False
                        new_elements = set()
                        for x in closure_set:
                            for y in closure_set:
                                if x != y and (x + y) % 2 == 0:
                                    avg = (x + y) // 2
                                    if avg not in closure_set:
                                        new_elements.add(avg)
                        if new_elements:
                            closure_set.update(new_elements)
                            added = True
                    sorted_S = sorted(closure_set)
                    is_consecutive = all(sorted_S[k] + 1 == sorted_S[k + 1] for k in range(len(sorted_S) - 1))
                    if not is_consecutive:
                        bad_subarrays += 1
            good_subarrays = total_subarrays - bad_subarrays
            print(good_subarrays)
    threading.Thread(target=solve).start()
if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on edge cases that might cause the algorithm to fail. The code processes subarrays up to a length of 6, so we should consider inputs that maximize the probability of encountering issues with this constraint. Here are some potential edge cases:

1. **Subarrays with repeated elements**: The code might not handle repeated elements correctly, especially when the subarray length is small.
2. **Subarrays with elements that cannot form a consecutive set**: The code might fail to correctly identify subarrays that cannot form a consecutive set even after adding all possible averages.
3. **Subarrays with a large range of values**: The code might have issues with subarrays that have a large range of values, which could lead to incorrect closure sets.

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

```plaintext
3
6
1 2 3 4 5 6
5
1 3 5 7 9
4
1 1 1 1
```

Explanation:
1. The first test case `[1, 2, 3, 4, 5, 6]` is a simple consecutive set. All subarrays of this set are brilliant, and the code should correctly identify this.
2. The second test case `[1, 3, 5, 7, 9]` is a set where the elements are spaced by 2. This set cannot form a consecutive set, and the code should correctly identify this.
3. The third test case `[1, 1, 1, 1]` contains repeated elements. The set should contain only one element, and the code should correctly handle this case.

This input will help to identify if the code correctly handles subarrays with repeated elements, subarrays that cannot form a consecutive set, and subarrays with a large range of values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eri and Expanded Sets
time_limit_ms: 3000
memory_limit_mb: 512
Description: Let there be a set that contains distinct positive integers. To expand the set to contain as many integers as possible, Eri can choose two integers $x\neq y$ from the set such that their average $\frac{x+y}2$ is still a positive integer and isn't contained in the set, and add it to the set. The integers $x$ and $y$ remain in the set.

Let's call the set of integers consecutive if, after the elements are sorted, the difference between any pair of adjacent elements is $1$. For example, sets $\\{2\\}$, $\\{2, 5, 4, 3\\}$, $\\{5, 6, 8, 7\\}$ are consecutive, while $\\{2, 4, 5, 6\\}$, $\\{9, 7\\}$ are not.

Eri likes consecutive sets. Suppose there is an array $b$, then Eri puts all elements in $b$ into the set. If after a finite number of operations described above, the set can become consecutive, the array $b$ will be called brilliant.

Note that if the same integer appears in the array multiple times, we only put it into the set once, as a set always contains distinct positive integers.

Eri has an array $a$ of $n$ positive integers. Please help him to count the number of pairs of integers $(l,r)$ such that $1 \leq l \leq r \leq n$ and the subarray $a_l, a_{l+1}, \ldots, a_r$ is brilliant.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 4 \cdot 10^5$) — length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $4 \cdot 10^5$.

Output: For each test case, output a single integer — the number of brilliant subarrays.

Examples:
input:
6
2
2 2
6
1 3 6 10 15 21
5
6 30 18 36 9
1
1000000000
6
1 1 4 5 1 4
12
70 130 90 90 90 108 612 500 451 171 193 193
output:
3
18
5
1
18
53

Note: In the first test case, the array $a = [2, 2]$ has $3$ subarrays: $[2]$, $[2]$, and $[2, 2]$. For all of them, the set only contains a single integer $2$, therefore it's always consecutive. All these subarrays are brilliant, so the answer is $3$.

In the second test case, let's consider the subarray $[3, 6, 10]$. We can do operations as follows:

$$\\{3,6,10\\} \xrightarrow{x=6,y=10} \\{3,6,8,10\\} \xrightarrow{x=6,y=8} \\{3,6,7,8,10\\} \xrightarrow{x=3,y=7} \\{3,5,6,7,8,10\\}$$ $$\xrightarrow{x=3,y=5} \\{3,4,5,6,7,8,10\\} \xrightarrow{x=8,y=10} \\{3,4,5,6,7,8,9,10\\}$$

$\\{3,4,5,6,7,8,9,10\\}$ is a consecutive set, so the subarray $[3, 6, 10]$ is brilliant.

Buggy Code:
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
#include <vector>
#include <queue>
#include <cctype>
#include <tuple>
#include <iostream>
#include <numeric>
#include <iomanip>
#include <cassert>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <functional>
#include <map>
#include <cstdint>
#include <bitset>
#include <set>

using namespace std;

set<vector<int>> used;

int popcount(int x) {
    int res = 0;
    while (x) {
        res += x & 1;
        x >>= 1;
    }
    return res;
}

void dfs(vector<int> a) {
    if (used.count(a))
        return;
    used.insert(a);
    for (int i = 0; i < a.size(); ++i) {
        for (int j = i + 1; j < a.size(); ++j) {
            if ((a[i] + a[j]) % 2 != 0)
                continue;
            auto to = set<int>(a.begin(), a.end());
            to.insert((a[i] + a[j]) / 2);
            dfs(vector<int>(to.begin(), to.end()));
        }
    }
}

bool test(vector<int> a) {
    used.clear();
    int n = a.size();
    dfs(a);
    int mx = *max_element(a.begin(), a.end());
    int mn = *min_element(a.begin(), a.end());
    vector<int> need;
    for (int i = mn; i <= mx; ++i) {
        need.push_back(i);
    }
    return used.count(need);
}

int gcd(int a, int b) {
    while (a && b) {
        if (a > b) a %= b;
        else b %= a;
    }
    return max(a, b);
}

const int N = 4e5 + 5;
int lg[N];
bool pw2[N];

void solve() {
    int n; cin >> n;
    vector<int> a(n);
    const int K = 19;
    vector<vector<int>> sp(n, vector<int>(K, 0));
    vector<int> r(n, n);
    for (int& x : a) cin >> x;
    for (int i = n - 2; i >= 0; --i) {
        if (a[i] == a[i + 1])
            r[i] = r[i + 1];
        else
            r[i] = i + 1;
    }
    for (int i = 1; i < n; ++i) {
        sp[i][0] = abs(a[i] - a[i - 1]);
    }
    for (int j = 1; j < K; ++j) {
        for (int i = 1; i + (1 << j) <= n; ++i) {
            sp[i][j] = gcd(sp[i][j - 1], sp[i + (1 << (j - 1))][j - 1]);
        }
    }

    auto get = [&](int l, int r) {
        ++r;
        int k = lg[r - l];
        return gcd(sp[l][k], sp[r - (1 << k)][k]);
    };

    long long res = 0;
    for (int i = 0; i < n; ++i) {
        res += (long long)(r[i] - i);
        int lo = r[i] - 1, hi = n;
        while (hi - lo > 1) {
            int mid = (lo + hi) / 2;
            int g = get(i + 1, mid);
            if (pw2[g]) hi = mid;
            else lo = mid;
        }
        res += (long long)(n - hi);
    }
    /*for (int l = 0; l < n; ++l) {
        for (int r = l; r < n; ++r) {
            vector<int> b(a.begin() + l, a.begin() + r + 1);
            int g = 0;
            for (int i = 1; i < b.size(); ++i) {
                g = gcd(g, abs(b[i] - b[i - 1]));
            }
            if (popcount(g) <= 1)
                ++res;
        }
    }*/
    cout << res << "\n";
}

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.precision(20);

    lg[1] = 0;
    for (int i = 2; i < N; ++i) {
        lg[i] = lg[i >> 1] + 1;
    }
    for (int x = 1; x < N; x *= 2) {
        pw2[x] = true;
    }

    int T; cin >> T;
    while (T--)
        solve();

    /*for (int x = 2; x <= 15; ++x) {
        for (int y = x + 1; y <= 15; ++y) {
            cout << "{" << y << "}\n";
            vector<int> a = { 1, x, y };
            int g = 0;
            for (int i = 1; i < a.size(); ++i) {
                g = gcd(g, a[i] - a[i - 1]);
            }
            if (!(test(a) == (popcount(g) <= 1))) {
                cout << (g == 1 || g % 2 == 0) << "\n";
                for (int x : a) cout << x << " ";
                cout << "\n";
                return 0;
            }
            /*if (!test(a)) {
                for (int x : a) cout << x << " ";
                cout << "\n";
            }
        }
    }*/
}

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

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

Think step by step.

Response:
```plaintext
1
5
1 3 5 7 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eri and Expanded Sets
time_limit_ms: 3000
memory_limit_mb: 512
Description: Let there be a set that contains distinct positive integers. To expand the set to contain as many integers as possible, Eri can choose two integers $x\neq y$ from the set such that their average $\frac{x+y}2$ is still a positive integer and isn't contained in the set, and add it to the set. The integers $x$ and $y$ remain in the set.

Let's call the set of integers consecutive if, after the elements are sorted, the difference between any pair of adjacent elements is $1$. For example, sets $\\{2\\}$, $\\{2, 5, 4, 3\\}$, $\\{5, 6, 8, 7\\}$ are consecutive, while $\\{2, 4, 5, 6\\}$, $\\{9, 7\\}$ are not.

Eri likes consecutive sets. Suppose there is an array $b$, then Eri puts all elements in $b$ into the set. If after a finite number of operations described above, the set can become consecutive, the array $b$ will be called brilliant.

Note that if the same integer appears in the array multiple times, we only put it into the set once, as a set always contains distinct positive integers.

Eri has an array $a$ of $n$ positive integers. Please help him to count the number of pairs of integers $(l,r)$ such that $1 \leq l \leq r \leq n$ and the subarray $a_l, a_{l+1}, \ldots, a_r$ is brilliant.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 4 \cdot 10^5$) — length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $4 \cdot 10^5$.

Output: For each test case, output a single integer — the number of brilliant subarrays.

Examples:
input:
6
2
2 2
6
1 3 6 10 15 21
5
6 30 18 36 9
1
1000000000
6
1 1 4 5 1 4
12
70 130 90 90 90 108 612 500 451 171 193 193
output:
3
18
5
1
18
53

Note: In the first test case, the array $a = [2, 2]$ has $3$ subarrays: $[2]$, $[2]$, and $[2, 2]$. For all of them, the set only contains a single integer $2$, therefore it's always consecutive. All these subarrays are brilliant, so the answer is $3$.

In the second test case, let's consider the subarray $[3, 6, 10]$. We can do operations as follows:

$$\\{3,6,10\\} \xrightarrow{x=6,y=10} \\{3,6,8,10\\} \xrightarrow{x=6,y=8} \\{3,6,7,8,10\\} \xrightarrow{x=3,y=7} \\{3,5,6,7,8,10\\}$$ $$\xrightarrow{x=3,y=5} \\{3,4,5,6,7,8,10\\} \xrightarrow{x=8,y=10} \\{3,4,5,6,7,8,9,10\\}$$

$\\{3,4,5,6,7,8,9,10\\}$ is a consecutive set, so the subarray $[3, 6, 10]$ is brilliant.

Buggy Code:
from math import gcd

t = int(input())

class Node():
    def __init__(self, arr, i, j):
        assert i != j

        self.i = i
        self.j = j

        if j-i == 1:
            self.val = 0
            self.left = None
            self.right = None
        else:
            m = (i + j) // 2
            self.left = Node(arr, i, m)
            self.right = Node(arr, m, j)

            diff = abs(arr[i] - arr[m])
            v1 = self.left.val
            v2 = self.right.val

            self.val = gcd(gcd(v1, v2), diff)

    def print(self):
        print(self.i, self.j, self.val)

        if not self.left is None:
            self.left.print()
        if not self.right is None:
            self.right.print()

    def query(self, arr, i, j):
        if self.i >= i and self.j <= j:
            return self.val
        if self.i >= j or self.j <= i:
            return 0

        m = (self.i + self.j) // 2
        if m >= j:
            return self.left.query(arr, i, j)
        elif m <= i:
            return self.right.query(arr, i, j)
        else:
            diff = abs(arr[i] - arr[m])
            v1 = self.left.query(arr, i, j)
            v2 = self.right.query(arr, i, j)

            return gcd(gcd(v1, v2), diff)

def check(x):
    if x == 0:
        return True
    if x % 2 == 1:
        return x == 1
    else:
        return check(x // 2)

outs = []
for _ in range(t):
    n = int(input())

    arr = input().split(" ")
    arr = [int(x) for x in arr]

    tree = Node(arr, 0, n)
    # print()
    # tree.print()
    # print()

    # print()
    # for i in range(n):
    #     for j in range(i+1, n+1):
    #         print(i, j, tree.query(arr, i, j))
    # print()

    # out = 0
    # for i in range(n):
    #     for j in range(i+1, n+1):
    #         v = tree.query(arr, i, j)
    #         if v in powers:
    #             out += 1

    counts = {}
    nonzero = 0
    def add(x):
        global nonzero
        if not x in counts or counts[x] == 0:
            nonzero += 1
            counts[x] = 0
        counts[x] += 1
    def remove(x):
        global nonzero
        counts[x] -= 1
        if counts[x] == 0:
            nonzero -= 1

    out = 0
    i = 0
    j = 0
    while i < n and j <= n:
        # print(i, j, out)
        if i == j:
            add(arr[j])
            j += 1
            continue

        v = tree.query(arr, i, j)
        if check(v):
            if nonzero > 1:
                out += n - j + 1
                remove(arr[i])
                i += 1
            else:
                if j == n:
                    break
                add(arr[j])
                j += 1
        else:
            if j == n:
                break
            add(arr[j])
            j += 1

    val = None
    count = 0
    for x in arr:
        if x != val:
            val = x
            count = 1
            out += 1
        else:
            count += 1
            out += count

    outs.append(out)

for out in outs:
    print(out)

**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
3
1 2 3
4
1 2 4 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eri and Expanded Sets
time_limit_ms: 3000
memory_limit_mb: 512
Description: Let there be a set that contains distinct positive integers. To expand the set to contain as many integers as possible, Eri can choose two integers $x\neq y$ from the set such that their average $\frac{x+y}2$ is still a positive integer and isn't contained in the set, and add it to the set. The integers $x$ and $y$ remain in the set.

Let's call the set of integers consecutive if, after the elements are sorted, the difference between any pair of adjacent elements is $1$. For example, sets $\\{2\\}$, $\\{2, 5, 4, 3\\}$, $\\{5, 6, 8, 7\\}$ are consecutive, while $\\{2, 4, 5, 6\\}$, $\\{9, 7\\}$ are not.

Eri likes consecutive sets. Suppose there is an array $b$, then Eri puts all elements in $b$ into the set. If after a finite number of operations described above, the set can become consecutive, the array $b$ will be called brilliant.

Note that if the same integer appears in the array multiple times, we only put it into the set once, as a set always contains distinct positive integers.

Eri has an array $a$ of $n$ positive integers. Please help him to count the number of pairs of integers $(l,r)$ such that $1 \leq l \leq r \leq n$ and the subarray $a_l, a_{l+1}, \ldots, a_r$ is brilliant.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 4 \cdot 10^5$) — length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $4 \cdot 10^5$.

Output: For each test case, output a single integer — the number of brilliant subarrays.

Examples:
input:
6
2
2 2
6
1 3 6 10 15 21
5
6 30 18 36 9
1
1000000000
6
1 1 4 5 1 4
12
70 130 90 90 90 108 612 500 451 171 193 193
output:
3
18
5
1
18
53

Note: In the first test case, the array $a = [2, 2]$ has $3$ subarrays: $[2]$, $[2]$, and $[2, 2]$. For all of them, the set only contains a single integer $2$, therefore it's always consecutive. All these subarrays are brilliant, so the answer is $3$.

In the second test case, let's consider the subarray $[3, 6, 10]$. We can do operations as follows:

$$\\{3,6,10\\} \xrightarrow{x=6,y=10} \\{3,6,8,10\\} \xrightarrow{x=6,y=8} \\{3,6,7,8,10\\} \xrightarrow{x=3,y=7} \\{3,5,6,7,8,10\\}$$ $$\xrightarrow{x=3,y=5} \\{3,4,5,6,7,8,10\\} \xrightarrow{x=8,y=10} \\{3,4,5,6,7,8,9,10\\}$$

$\\{3,4,5,6,7,8,9,10\\}$ is a consecutive set, so the subarray $[3, 6, 10]$ is brilliant.

Buggy Code:
from math import gcd

t = int(input())

class Node():
    def __init__(self, arr, i, j):
        assert i != j

        self.i = i
        self.j = j

        if j-i == 1:
            self.val = 0
            self.left = None
            self.right = None
        else:
            m = (i + j) // 2
            self.left = Node(arr, i, m)
            self.right = Node(arr, m, j)

            diff = abs(arr[i] - arr[m])
            v1 = self.left.val
            v2 = self.right.val

            self.val = gcd(gcd(v1, v2), diff)

    def print(self):
        print(self.i, self.j, self.val)

        if not self.left is None:
            self.left.print()
        if not self.right is None:
            self.right.print()

    def query(self, arr, i, j):
        if self.i >= i and self.j <= j:
            return self.val
        if self.i >= j or self.j <= i:
            return 0

        m = (self.i + self.j) // 2
        if m >= j:
            return self.left.query(arr, i, j)
        elif m <= i:
            return self.right.query(arr, i, j)
        else:
            diff = abs(arr[i] - arr[m])
            v1 = self.left.query(arr, i, j)
            v2 = self.right.query(arr, i, j)

            return gcd(gcd(v1, v2), diff)

powers = [2 ** k for k in range(40)]
powers.append(0)
powers = set(powers)

outs = []
for _ in range(t):
    n = int(input())

    arr = input().split(" ")
    arr = [int(x) for x in arr]

    tree = Node(arr, 0, n)
    # print()
    # tree.print()
    # print()

    # print()
    # for i in range(n):
    #     for j in range(i+1, n+1):
    #         print(i, j, tree.query(arr, i, j))
    # print()

    # out = 0
    # for i in range(n):
    #     for j in range(i+1, n+1):
    #         v = tree.query(arr, i, j)
    #         if v in powers:
    #             out += 1

    counts = {}
    nonzero = 0
    def add(x):
        global nonzero
        if not x in counts or counts[x] == 0:
            nonzero += 1
            counts[x] = 0
        counts[x] += 1
    def remove(x):
        global nonzero
        counts[x] -= 1
        if counts[x] == 0:
            nonzero -= 1

    out = 0
    i = 0
    j = 0
    while i < n and j <= n:
        # print(i, j, out)
        if i == j:
            add(arr[j])
            j += 1
            continue

        v = tree.query(arr, i, j)
        if v in powers:
            if nonzero > 1:
                out += n - j + 1
                remove(arr[i])
                i += 1
            else:
                if j == n:
                    break
                add(arr[j])
                j += 1
        else:
            if j == n:
                break
            add(arr[j])
            j += 1

    val = None
    count = 0
    for x in arr:
        if x != val:
            val = x
            count = 1
            out += 1
        else:
            count += 1
            out += count

    outs.append(out)

for out in outs:
    print(out)

**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
1 2 4 8 16
3
1 1 2
4
1 3 5 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eri and Expanded Sets
time_limit_ms: 3000
memory_limit_mb: 512
Description: Let there be a set that contains distinct positive integers. To expand the set to contain as many integers as possible, Eri can choose two integers $x\neq y$ from the set such that their average $\frac{x+y}2$ is still a positive integer and isn't contained in the set, and add it to the set. The integers $x$ and $y$ remain in the set.

Let's call the set of integers consecutive if, after the elements are sorted, the difference between any pair of adjacent elements is $1$. For example, sets $\\{2\\}$, $\\{2, 5, 4, 3\\}$, $\\{5, 6, 8, 7\\}$ are consecutive, while $\\{2, 4, 5, 6\\}$, $\\{9, 7\\}$ are not.

Eri likes consecutive sets. Suppose there is an array $b$, then Eri puts all elements in $b$ into the set. If after a finite number of operations described above, the set can become consecutive, the array $b$ will be called brilliant.

Note that if the same integer appears in the array multiple times, we only put it into the set once, as a set always contains distinct positive integers.

Eri has an array $a$ of $n$ positive integers. Please help him to count the number of pairs of integers $(l,r)$ such that $1 \leq l \leq r \leq n$ and the subarray $a_l, a_{l+1}, \ldots, a_r$ is brilliant.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 4 \cdot 10^5$) — length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $4 \cdot 10^5$.

Output: For each test case, output a single integer — the number of brilliant subarrays.

Examples:
input:
6
2
2 2
6
1 3 6 10 15 21
5
6 30 18 36 9
1
1000000000
6
1 1 4 5 1 4
12
70 130 90 90 90 108 612 500 451 171 193 193
output:
3
18
5
1
18
53

Note: In the first test case, the array $a = [2, 2]$ has $3$ subarrays: $[2]$, $[2]$, and $[2, 2]$. For all of them, the set only contains a single integer $2$, therefore it's always consecutive. All these subarrays are brilliant, so the answer is $3$.

In the second test case, let's consider the subarray $[3, 6, 10]$. We can do operations as follows:

$$\\{3,6,10\\} \xrightarrow{x=6,y=10} \\{3,6,8,10\\} \xrightarrow{x=6,y=8} \\{3,6,7,8,10\\} \xrightarrow{x=3,y=7} \\{3,5,6,7,8,10\\}$$ $$\xrightarrow{x=3,y=5} \\{3,4,5,6,7,8,10\\} \xrightarrow{x=8,y=10} \\{3,4,5,6,7,8,9,10\\}$$

$\\{3,4,5,6,7,8,9,10\\}$ is a consecutive set, so the subarray $[3, 6, 10]$ is brilliant.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define Loop(x,l,r) for(ll x=ll(l);x<ll(r);x++)
#define LoopR(x,l,r) for(ll x=ll(r)-1;x>=ll(l);x--)
//#define endl "\n"
typedef long long ll;
#define int ll
typedef pair<int,int>pii;
typedef pair<ll,ll>pll;
const int mod=1e9+7;
const int lg=19,kaf=(1<<lg);

long long mypow(ll m,ll y){
    if(y==0){
        return 1;
    }
    ll p=mypow(m,(y>>1));
    p*=p;
    p%=mod;
    if(y&1){
        p*=m;
        p%=mod;
    }
    return p;
}
int n;
struct segment{
    int seg[(2<<lg)];
    void add(int i,int w){
        i+=kaf;
        seg[i]=w;
        i>>=1;
        while(i>0){
            seg[i]=gcd(seg[(i<<1)],seg[(i<<1)^1]);
            i>>=1;
        }
        return ;
    }
    int pors(int i,int l,int r,int tl,int tr){
        if(l>r||l>tr||r<tl||tl>tr){
            return 0;
        }
        if(l>=tl&&r<=tr){
            return seg[i];
        }
        int m=(l+r)>>1;
        return gcd(pors((i<<1),l,m,tl,tr),pors((i<<1)^1,m+1,r,tl,tr));
    }
    void cl(int i){
        i+=kaf;
        while(i>0){
            seg[i]=0;
            i>>=1;
        }
    }
}seg;

void clear(){
    for(int i=0;i<=n;i++){
        seg.cl(i);
    }
}

void solve(){
    clear();
    cin>>n;
    vector<int>all(n);
    for(int i=0;i<n;i++){
        cin>>all[i];
    }
    for(int i=1;i<=n;i++){
        seg.add(i,abs(all[i]-all[i-1]));
    }
    int res=(n*(n-1))/2;
    res+=n;
    int last=0;
    for(int i=n-1;i>=0;i--){
        if(all[i]==all[i+1]){
            res-=last;
            continue;
        }
        int low=i,high=n,mid;
        while(high-low>1){
            mid=(high+low)>>1;
            int z=seg.pors(1,0,kaf-1,i+1,mid);
            //cout<<i<<" "<<mid<<" "<<z<<endl;
            if(__builtin_popcount(z)==1){
                high=mid;
            }else{
                low=mid;
            }
        }
        res-=high-i-1;
        last=high-i-1;
    }
    cout<<res<<"\n";
}

signed main(){
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int t;
    cin>>t;
    Loop(asd,0,t){
        solve();
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases that can cause the algorithm to fail. The code is designed to count the number of brilliant subarrays, and it uses a segment tree to manage the differences between adjacent elements. Here are some edge cases to consider:

1. **Single Element Array**: The code should handle single-element arrays correctly.
2. **All Elements are the Same**: The code should handle arrays where all elements are the same.
3. **Consecutive Elements**: The code should handle arrays where the elements are already consecutive.
4. **Non-Consecutive Elements with Possible Expansion**: The code should handle arrays where the elements can be expanded to become consecutive.
5. **Large Range of Elements**: The code should handle arrays with a large range of elements.

Let's generate a test input that includes these edge cases:

```plaintext
6
1
10
2
2 2
3
1 2 3
4
1 3 5 7
5
1 2 4 8 16
```

Explanation:
1. **Single Element Array**: The first test case with a single element `10` should be considered brilliant.
2. **All Elements are the Same**: The second test case with two elements `2 2` should be considered brilliant.
3. **Consecutive Elements**: The third test case with elements `1 2 3` is already consecutive.
4. **Non-Consecutive Elements with Possible Expansion**: The fourth test case with elements `1 3 5 7` can be expanded to become consecutive.
5. **Large Range of Elements**: The fifth test case with elements `1 2 4 8 16` cannot be expanded to become consecutive.

This test input is small-scale and complete, designed to cover various edge cases that can expose potential bugs in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eri and Expanded Sets
time_limit_ms: 3000
memory_limit_mb: 512
Description: Let there be a set that contains distinct positive integers. To expand the set to contain as many integers as possible, Eri can choose two integers $x\neq y$ from the set such that their average $\frac{x+y}2$ is still a positive integer and isn't contained in the set, and add it to the set. The integers $x$ and $y$ remain in the set.

Let's call the set of integers consecutive if, after the elements are sorted, the difference between any pair of adjacent elements is $1$. For example, sets $\\{2\\}$, $\\{2, 5, 4, 3\\}$, $\\{5, 6, 8, 7\\}$ are consecutive, while $\\{2, 4, 5, 6\\}$, $\\{9, 7\\}$ are not.

Eri likes consecutive sets. Suppose there is an array $b$, then Eri puts all elements in $b$ into the set. If after a finite number of operations described above, the set can become consecutive, the array $b$ will be called brilliant.

Note that if the same integer appears in the array multiple times, we only put it into the set once, as a set always contains distinct positive integers.

Eri has an array $a$ of $n$ positive integers. Please help him to count the number of pairs of integers $(l,r)$ such that $1 \leq l \leq r \leq n$ and the subarray $a_l, a_{l+1}, \ldots, a_r$ is brilliant.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 4 \cdot 10^5$) — length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $4 \cdot 10^5$.

Output: For each test case, output a single integer — the number of brilliant subarrays.

Examples:
input:
6
2
2 2
6
1 3 6 10 15 21
5
6 30 18 36 9
1
1000000000
6
1 1 4 5 1 4
12
70 130 90 90 90 108 612 500 451 171 193 193
output:
3
18
5
1
18
53

Note: In the first test case, the array $a = [2, 2]$ has $3$ subarrays: $[2]$, $[2]$, and $[2, 2]$. For all of them, the set only contains a single integer $2$, therefore it's always consecutive. All these subarrays are brilliant, so the answer is $3$.

In the second test case, let's consider the subarray $[3, 6, 10]$. We can do operations as follows:

$$\\{3,6,10\\} \xrightarrow{x=6,y=10} \\{3,6,8,10\\} \xrightarrow{x=6,y=8} \\{3,6,7,8,10\\} \xrightarrow{x=3,y=7} \\{3,5,6,7,8,10\\}$$ $$\xrightarrow{x=3,y=5} \\{3,4,5,6,7,8,10\\} \xrightarrow{x=8,y=10} \\{3,4,5,6,7,8,9,10\\}$$

$\\{3,4,5,6,7,8,9,10\\}$ is a consecutive set, so the subarray $[3, 6, 10]$ is brilliant.

Buggy Code:
import java.util.*;

public class Main {
    // GCD function
    static long gcd(long a, long b) {
        if (a < b) return gcd(b, a);
        if (b == 0) return a;
        return gcd(b, a % b);
    }

    // Function to check if a number is a power of 2
    static boolean isPow2(long n) {
        if (n == 0) return true;
        while (n % 2 == 0) n /= 2;
        return n == 1;
    }

    // Function to calculate gcd of a segment from a segment tree
    static long calculate(long[] segtree, long w, long L, long R) {
        long left = w + L;
        long right = w + R + 1;
        long res = 0;

        while (left < right) {
            if (left % 2 == 1) {
                res = gcd(res, segtree[(int) left]);
                left++;
            }

            if (right % 2 == 1) {
                res = gcd(res, segtree[(int) (right - 1)]);
                right--;
            }

            left /= 2;
            right /= 2;
        }

        return res;
    }

    // Main solve function
    static void solve(Scanner sc) {
        if (!sc.hasNextLong()) return; // Prevents NoSuchElementException
        long n = sc.nextLong();

        long[] arr = new long[(int) n];
        for (int i = 0; i < n; i++) {
            arr[i] = sc.nextLong();
        }

        if (n == 1) {
            // Special case when n == 1, we can directly print 1
            System.out.print(1);
        } else {
            long[] diff = new long[(int) (n - 1)];
            long[] no_duplicate = new long[(int) (n - 1)];
            Arrays.fill(no_duplicate, -1);

            long idx = -1;
            for (int i = 0; i < n - 1; i++) {
                diff[i] = Math.abs(arr[i + 1] - arr[i]);

                if (diff[i] == 0) {
                    no_duplicate[i] = idx;
                } else {
                    idx = i;
                    no_duplicate[i] = idx;
                }
            }

            // Build a segment tree to calculate gcd(diff[l], ..., diff[r])
            long w = 1;
            while (w < n) w *= 2;
            long[] segtree = new long[(int) (2 * w)];

            for (int i = 0; i < n - 1; i++) {
                segtree[(int) (w + i)] = diff[i];
            }

            for (int i = (int) w - 1; i > 0; i--) {
                segtree[i] = gcd(segtree[2 * i], segtree[2 * i + 1]);
            }

            long ans = n;
            for (int i = 0; i < n - 1; i++) {
                long lo = -1, hi = no_duplicate[i];
                long res = i - hi;

                while (lo < hi) {
                    long mid = (lo + hi + 1) / 2;
                    if (isPow2(calculate(segtree, w, mid, i))) lo = mid;
                    else hi = mid - 1;
                }

                ans += res + lo + 1;
            }

            System.out.print(ans);
        }
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long test_case_number = sc.nextLong();

        for (long test_case = 0; test_case < test_case_number; test_case++) {
            solve(sc);
            System.out.println();
        }

        sc.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
6
2
2 2
6
1 3 6 10 15 21
5
6 30 18 36 9
1
1000000000
6
1 1 4 5 1 4
12
70 130 90 90 90 108 612 500 451 171 193 193
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eri and Expanded Sets
time_limit_ms: 3000
memory_limit_mb: 512
Description: Let there be a set that contains distinct positive integers. To expand the set to contain as many integers as possible, Eri can choose two integers $x\neq y$ from the set such that their average $\frac{x+y}2$ is still a positive integer and isn't contained in the set, and add it to the set. The integers $x$ and $y$ remain in the set.

Let's call the set of integers consecutive if, after the elements are sorted, the difference between any pair of adjacent elements is $1$. For example, sets $\\{2\\}$, $\\{2, 5, 4, 3\\}$, $\\{5, 6, 8, 7\\}$ are consecutive, while $\\{2, 4, 5, 6\\}$, $\\{9, 7\\}$ are not.

Eri likes consecutive sets. Suppose there is an array $b$, then Eri puts all elements in $b$ into the set. If after a finite number of operations described above, the set can become consecutive, the array $b$ will be called brilliant.

Note that if the same integer appears in the array multiple times, we only put it into the set once, as a set always contains distinct positive integers.

Eri has an array $a$ of $n$ positive integers. Please help him to count the number of pairs of integers $(l,r)$ such that $1 \leq l \leq r \leq n$ and the subarray $a_l, a_{l+1}, \ldots, a_r$ is brilliant.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 4 \cdot 10^5$) — length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $4 \cdot 10^5$.

Output: For each test case, output a single integer — the number of brilliant subarrays.

Examples:
input:
6
2
2 2
6
1 3 6 10 15 21
5
6 30 18 36 9
1
1000000000
6
1 1 4 5 1 4
12
70 130 90 90 90 108 612 500 451 171 193 193
output:
3
18
5
1
18
53

Note: In the first test case, the array $a = [2, 2]$ has $3$ subarrays: $[2]$, $[2]$, and $[2, 2]$. For all of them, the set only contains a single integer $2$, therefore it's always consecutive. All these subarrays are brilliant, so the answer is $3$.

In the second test case, let's consider the subarray $[3, 6, 10]$. We can do operations as follows:

$$\\{3,6,10\\} \xrightarrow{x=6,y=10} \\{3,6,8,10\\} \xrightarrow{x=6,y=8} \\{3,6,7,8,10\\} \xrightarrow{x=3,y=7} \\{3,5,6,7,8,10\\}$$ $$\xrightarrow{x=3,y=5} \\{3,4,5,6,7,8,10\\} \xrightarrow{x=8,y=10} \\{3,4,5,6,7,8,9,10\\}$$

$\\{3,4,5,6,7,8,9,10\\}$ is a consecutive set, so the subarray $[3, 6, 10]$ is brilliant.

Buggy Code:
import java.util.*;
import java.io.*;
public class C {
  public static void main(String[] args) throws IOException {
    BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);
    int[] two = new int[30];
    two[0] = 1;
    for(int i = 1; i < 30; i++) {
      two[i] = two[i - 1] * 2;
    }

    StringTokenizer st = new StringTokenizer(r.readLine());
    int T = Integer.parseInt(st.nextToken());
    while(T-- > 0) {
      st = new StringTokenizer(r.readLine());
      int n = Integer.parseInt(st.nextToken());
      st = new StringTokenizer(r.readLine());
      int[] arr = new int[n];
      for(int i = 0; i < n; i++) {
        arr[i] = Integer.parseInt(st.nextToken());
      }
      long ans = 0;
      int prev = -1;
      long freq = 0;
      for(int i = 0; i < n; i++) {
        if(arr[i] == prev) {freq++;}
        else {ans += freq*(freq + 1)/2; freq = 1; prev = arr[i];}
      }
      ans += freq*(freq + 1)/2;
      int[] diff = new int[n - 1];
      for(int i = 0; i < n - 1; i++) {
        diff[i] = Math.abs(arr[i] - arr[i + 1]);
      }
      gcdSegTree tree = new gcdSegTree(diff);
      //System.out.println(Arrays.toString(tree.first));
      //System.out.println(Arrays.toString(tree.last));
      //System.out.println(Arrays.toString(tree.tree));
      for(int i = 0; i < n; i++) {
        int left = i - 1; //bad
        int right = n - 1; //good
        while(right - left > 1) {
          int index = (left + right)/2;
          int g = tree.f(1, i, index);
          if(Arrays.binarySearch(two, g) >= 0) {
            right = index;
          } else {
            left = index;
          }
        }
        //System.out.println(i + " | " + left + ", " + right);
        ans += n - 1 - right;
      }
      pw.println(ans);
    }

    pw.close();
  }

  static class gcdSegTree {
    int n;
    int[] first, last, tree;

    public gcdSegTree(int size) {
      n = 1;
      while(size > 0) {
        n *= 2;
        size /= 2;
      }
      first = new int[2*n];
      last = new int[2*n];
      tree = new int[2*n];
      for(int i = n; i < 2*n; i++) {
        first[i] = last[i] = i-n;
      }
      for(int i = n-1; i > 0; i--) {
        first[i] = first[2*i];
        last[i] = last[2*i+1];
      }
    }

    public gcdSegTree(int[] orig) {
      //constructor for initializing segment tree with values
      this(orig.length);
      for(int i = 0; i < orig.length; i++) {
        tree[i+n]=orig[i];
      }
      //min (OPERATOR)
      for(int i = n-1; i > 0; i--) {
        tree[i] = gcd(tree[2*i], tree[2*i+1]);
      }
    }

    void update(int i, int index, int value) {//ALWAYS PUT i = 1
      if(i >= n) {
        //Base case: we are at the node containing only index in the original array
        tree[i] = value;
        return;
      }
      //recursive case
      if(first[2*i] <= index && last[2*i] >= index) {
        //the node that contains index is in the left child
        update(2*i, index, value);
      } else {
        //the node that contains index is in the right child
        update(2*i+1, index, value);
      }
      //update the current node min (OPERATOR)
      tree[i] = gcd(tree[2*i], tree[2*i+1]);
    }

    int f(int i, int left, int right) {//ALWAYS PUT i = 1
      if(first[i]==left && last[i]==right) {
        //base case: interval of node is exactly interval we want. Simply return tree value.
        return tree[i];
      }
      int result = 0; //STARTING VALUE
      if(first[2*i]<=right && last[2*i]>=left) {
        //interval overlaps with interval of left child node
        //Math.min is used to make sure the interval we put in the argument is a subset of the
        //interval of the node
        //min (OPERATOR)
        result = gcd(result, f(2*i, left, Math.min(right, last[2*i])));
      }
      if(first[2*i+1]<=right && last[2*i+1]>=left) {
        //interval overlaps with interval of right child node
        //min (OPERATOR)
        result = gcd(result, f(2*i+1, Math.max(left, first[2*i+1]), right));
      }
      return result;
    }
  }

  public static int gcd(int a, int b) {
    while(a != 0 && b != 0) {
      if(a > b) {
        a %= b;
      } else {
        b %= a;
      }
    }
    if(a == 0) {return b;}
    return a;
  }
}

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

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

Think step by step.

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

1. **Understanding the Bug**: The buggy code seems to have issues with the segment tree and the way it handles the GCD (Greatest Common Divisor) of the differences between elements. Specifically, the segment tree might not be correctly handling the GCD of the differences, leading to incorrect results.

2. **Edge Cases**:
   - **Single Element Subarrays**: These are always brilliant.
   - **Consecutive Subarrays**: These are always brilliant.
   - **Subarrays with GCD of Differences Being a Power of 2**: These can be expanded to become consecutive.
   - **Subarrays with Non-Consecutive Elements**: These are not brilliant.

3. **Challenging Scenarios**:
   - **Subarrays with GCD of Differences Not Being a Power of 2**: This should not be brilliant.
   - **Subarrays with Multiple Occurrences of the Same Element**: This should be handled correctly by the code.
   - **Subarrays with Large Differences**: This can test the GCD calculation and segment tree handling.

4. **Small-Scale Test Input**:
   - We need a small array that includes a mix of the above scenarios to ensure the code handles them correctly.

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

```plaintext
3
5
1 3 6 10 15
5
1 2 3 4 5
5
1 3 5 7 9
```

Explanation:
- **Test Case 1**: `[1, 3, 6, 10, 15]` - The subarray should not be brilliant because the GCD of the differences is not a power of 2.
- **Test Case 2**: `[1, 2, 3, 4, 5]` - This is a consecutive subarray, so it should be brilliant.
- **Test Case 3**: `[1, 3, 5, 7, 9]` - The subarray should be brilliant because the GCD of the differences is a power of 2 (2).

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: Eri and Expanded Sets
time_limit_ms: 3000
memory_limit_mb: 512
Description: Let there be a set that contains distinct positive integers. To expand the set to contain as many integers as possible, Eri can choose two integers $x\neq y$ from the set such that their average $\frac{x+y}2$ is still a positive integer and isn't contained in the set, and add it to the set. The integers $x$ and $y$ remain in the set.

Let's call the set of integers consecutive if, after the elements are sorted, the difference between any pair of adjacent elements is $1$. For example, sets $\\{2\\}$, $\\{2, 5, 4, 3\\}$, $\\{5, 6, 8, 7\\}$ are consecutive, while $\\{2, 4, 5, 6\\}$, $\\{9, 7\\}$ are not.

Eri likes consecutive sets. Suppose there is an array $b$, then Eri puts all elements in $b$ into the set. If after a finite number of operations described above, the set can become consecutive, the array $b$ will be called brilliant.

Note that if the same integer appears in the array multiple times, we only put it into the set once, as a set always contains distinct positive integers.

Eri has an array $a$ of $n$ positive integers. Please help him to count the number of pairs of integers $(l,r)$ such that $1 \leq l \leq r \leq n$ and the subarray $a_l, a_{l+1}, \ldots, a_r$ is brilliant.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 4 \cdot 10^5$) — length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $4 \cdot 10^5$.

Output: For each test case, output a single integer — the number of brilliant subarrays.

Examples:
input:
6
2
2 2
6
1 3 6 10 15 21
5
6 30 18 36 9
1
1000000000
6
1 1 4 5 1 4
12
70 130 90 90 90 108 612 500 451 171 193 193
output:
3
18
5
1
18
53

Note: In the first test case, the array $a = [2, 2]$ has $3$ subarrays: $[2]$, $[2]$, and $[2, 2]$. For all of them, the set only contains a single integer $2$, therefore it's always consecutive. All these subarrays are brilliant, so the answer is $3$.

In the second test case, let's consider the subarray $[3, 6, 10]$. We can do operations as follows:

$$\\{3,6,10\\} \xrightarrow{x=6,y=10} \\{3,6,8,10\\} \xrightarrow{x=6,y=8} \\{3,6,7,8,10\\} \xrightarrow{x=3,y=7} \\{3,5,6,7,8,10\\}$$ $$\xrightarrow{x=3,y=5} \\{3,4,5,6,7,8,10\\} \xrightarrow{x=8,y=10} \\{3,4,5,6,7,8,9,10\\}$$

$\\{3,4,5,6,7,8,9,10\\}$ is a consecutive set, so the subarray $[3, 6, 10]$ is brilliant.

Buggy Code:
from itertools import tee, combinations
from heapq import heappush, heappop
from collections import defaultdict, Counter, deque
from sys import stdout
import math
import sys
sys.setrecursionlimit(100000) # 100000 -> 130MB and 300000 -> exceed 256MB
def printf(*args):
    print(*args)
    stdout.flush()
new_input = False
if new_input:
    import io,os
    def ginput():
        for line in io.BytesIO(os.read(0,os.fstat(0).st_size)).read().decode().splitlines():
            yield line
    input = ginput()
    def getint():
        return int(next(input))
    def getfloat():
        return float(next(input))
    def getints():
        return list(map(int, next(input).split()))
    def getfloats():
        return list(map(float, next(input).split()))
    def getstr():
        return next(input)
    def getstrs():
        return next(input).split()
else:
    from collections import deque
    def getint():
        return int(input())
    def getfloat():
        return float(input())
    def getints():
        return list(map(int, input().split()))
    def getfloats():
        return list(map(float, input().split()))
    def getstr():
        return input()
    def getstrs():
        return input().split()

inf = float('inf')

class groupby:
    def __init__(self, iterable, key=None):
        if key is None:
            key = lambda x: x
        self.keyfunc = key
        self.it = iter(iterable)
        self.tgtkey = self.currkey = self.currvalue = object()

    def __iter__(self):
        return self

    def __next__(self):
        self.id = object()
        while self.currkey == self.tgtkey:
            self.currvalue = next(self.it)    # Exit on StopIteration
            self.currkey = self.keyfunc(self.currvalue)
        self.tgtkey = self.currkey
        return (self.currkey, self._grouper(self.tgtkey, self.id))

    def _grouper(self, tgtkey, id):
        while self.id is id and self.currkey == tgtkey:
            yield self.currvalue
            try:
                self.currvalue = next(self.it)
            except StopIteration:
                return
            self.currkey = self.keyfunc(self.currvalue)

def pairwise(iterable): a, b = tee(iterable); next(b, None); return zip(a, b)
import itertools
def bisect_left(a, x, lo=0, hi=None, *, key=None):
    if lo < 0:
        raise ValueError('lo must be non-negative')
    if hi is None:
        hi = len(a)
    if key is None:
        while lo < hi:
            mid = (lo + hi) // 2
            if a[mid] < x:
                lo = mid + 1
            else:
                hi = mid
    else:
        while lo < hi:
            mid = (lo + hi) // 2
            if key(a[mid]) < x:
                lo = mid + 1
            else:
                hi = mid
    return lo
def gcd(a, b): return gcd(b%a, a) if a != 0 else b
def sgn(x): return (x>0) - (x<0)
from functools import lru_cache
from itertools import product, permutations

class BIT:
    def __init__(self, n):
        # valid input: [0, n]
        self.sums = [0] * (n+3)

    def update(self, i, delta):
        # update >=i with delta
        i += 2
        while i < len(self.sums):
            self.sums[i] += delta
            i += i & (-i)
    def _query(self, i):
        # get (0, i]
        i += 2
        res = 0
        while i > 0:
            res += self.sums[i]
            i -= i & (-i)
        return res

    def query(self, i, j =None):
        # [0, i] or [i, j]
        if j == None:
            return self._query(i)
        else:
            assert j >= i
            return self._query(j) - self._query(i-1)

    def update_interval(self, a, b, delta):
        # update [b, a]
        self.update(a+1, -delta)
        self.update(b, delta)

def popcount(a): return bin(a).count('1')

def ctz(a):
    assert a > 0, a
    res = 0
    while a & 1 == 0:
        res += 1
        a >>= 1
    return res

def __lg(a):
    assert a > 0, a
    while a & (-a) != a:
        a -= a & (-a)
    return ctz(a)
from itertools import zip_longest
def answer():
    n = getint()
    s = getints()
    ans = 0
    for i in range(n):
        g = 0
        for j in range(i, n):
            g = gcd(g, abs(s[j] - s[i]))
            while g and g % 2 == 0:
                g //= 2
            if g <= 1:
                ans += 1
    print(ans)

DOT = 1
T = 1
if DOT:
    T = getint()
for i in range(T):
    q = answer()
    if q != None:
        if q:
            print('Yes')
        else:
            print('No')

**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
3
1 3 5
4
1 2 4 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eri and Expanded Sets
time_limit_ms: 3000
memory_limit_mb: 512
Description: Let there be a set that contains distinct positive integers. To expand the set to contain as many integers as possible, Eri can choose two integers $x\neq y$ from the set such that their average $\frac{x+y}2$ is still a positive integer and isn't contained in the set, and add it to the set. The integers $x$ and $y$ remain in the set.

Let's call the set of integers consecutive if, after the elements are sorted, the difference between any pair of adjacent elements is $1$. For example, sets $\\{2\\}$, $\\{2, 5, 4, 3\\}$, $\\{5, 6, 8, 7\\}$ are consecutive, while $\\{2, 4, 5, 6\\}$, $\\{9, 7\\}$ are not.

Eri likes consecutive sets. Suppose there is an array $b$, then Eri puts all elements in $b$ into the set. If after a finite number of operations described above, the set can become consecutive, the array $b$ will be called brilliant.

Note that if the same integer appears in the array multiple times, we only put it into the set once, as a set always contains distinct positive integers.

Eri has an array $a$ of $n$ positive integers. Please help him to count the number of pairs of integers $(l,r)$ such that $1 \leq l \leq r \leq n$ and the subarray $a_l, a_{l+1}, \ldots, a_r$ is brilliant.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 4 \cdot 10^5$) — length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $4 \cdot 10^5$.

Output: For each test case, output a single integer — the number of brilliant subarrays.

Examples:
input:
6
2
2 2
6
1 3 6 10 15 21
5
6 30 18 36 9
1
1000000000
6
1 1 4 5 1 4
12
70 130 90 90 90 108 612 500 451 171 193 193
output:
3
18
5
1
18
53

Note: In the first test case, the array $a = [2, 2]$ has $3$ subarrays: $[2]$, $[2]$, and $[2, 2]$. For all of them, the set only contains a single integer $2$, therefore it's always consecutive. All these subarrays are brilliant, so the answer is $3$.

In the second test case, let's consider the subarray $[3, 6, 10]$. We can do operations as follows:

$$\\{3,6,10\\} \xrightarrow{x=6,y=10} \\{3,6,8,10\\} \xrightarrow{x=6,y=8} \\{3,6,7,8,10\\} \xrightarrow{x=3,y=7} \\{3,5,6,7,8,10\\}$$ $$\xrightarrow{x=3,y=5} \\{3,4,5,6,7,8,10\\} \xrightarrow{x=8,y=10} \\{3,4,5,6,7,8,9,10\\}$$

$\\{3,4,5,6,7,8,9,10\\}$ is a consecutive set, so the subarray $[3, 6, 10]$ is brilliant.

Buggy Code:
def II():
    return(int(input()))
def LMI():
    return(list(map(int,input().split())))
def I():
    return(input())
def MII():
    return(map(int,input().split()))
import sys
input=sys.stdin.readline
# import io,os
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# from collections import Counter
# int(math.log(len(L)))
# import math
# from collections import defaultdict
# mod=10**9+7
# from collections import deque
import  math
class SegmentTreeSum():
    def __init__(self, X):
        L = [0] + X
        n = int(math.ceil(math.log(len(L)) / math.log(2)))
        self.seg = [0] * (2 ** n - 1) + L + [0] * (2 ** n - len(L))
        for i in range(2 ** n - 2, -1, -1):
            self.seg[i] = math.gcd(self.seg[2 * i + 1] , self.seg[2 * i + 2])
    def update(self, m, n):
        x = m + len(self.seg) // 2
        self.seg[x] = n
        x = (x - 1) // 2
        while x > 0:
            self.seg[x] = self.seg[2 * x + 1] + self.seg[2 * x + 2]
            x = (x - 1) // 2
        self.seg[0]=math.gcd(self.seg[1],self.seg[2])

    def su(self, l, r):
        ans = 0
        a = l + len(self.seg) // 2
        b = r + len(self.seg) // 2
        while a < b:
            if a % 2 == 0:
                ans = math.gcd(ans,self.seg[a])
                a //= 2
            else:
                a //= 2
            if b % 2 == 0:
                b -= 1
                b //= 2
            else:
                ans = math.gcd(ans,self.seg[b])
                b -= 2
                b //= 2
        if a==b:
            if a % 2 == 0:
                ans = math.gcd(ans,self.seg[a])
                a //= 2
            else:
                a //= 2
            if b % 2 == 0:
                b -= 1
                b //= 2
            else:
                ans = math.gcd(ans,self.seg[b])
                b -= 2
                b //= 2

        return ans

# from functools import cache
# @cache
def t():
    n=II()
    L=LMI()
    diff=[0]
    for i in range(1,n):
        diff.append(abs(L[i]-L[i-1]))
    X=SegmentTreeSum(diff)
    ans=n
    for i in range(n-1):
        l=i+1
        h=n-1
        res=-1
        while l<=h:
            mid=(l+h)//2
            alp=X.su(i+2,mid+1)
            if alp!=0 and (math.log(alp)/math.log(2))%1==0:
                res=mid
                h=mid-1
            else:
                l=mid+1

        if res!=-1:
            ans+=n-res
        l=i+1
        h=n-1
        res=-1
        while l<=h:
            mid=(l+h)//2
            alp=X.su(i+2,mid+1)
            if alp==0:
                res=mid
                l=mid+1
            else:
                h=mid-1
        if res!=-1:
            ans+=res-i

    print(ans)

    return

if __name__=="__main__":

    for _ in range(II()):
        t()
    # t()

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

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

Think step by step.

Response:
```plaintext
6
2
2 2
6
1 3 6 10 15 21
5
6 30 18 36 9
1
1000000000
6
1 1 4 5 1 4
12
70 130 90 90 90 108 612 500 451 171 193 193
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Eri and Expanded Sets
time_limit_ms: 3000
memory_limit_mb: 512
Description: Let there be a set that contains distinct positive integers. To expand the set to contain as many integers as possible, Eri can choose two integers $x\neq y$ from the set such that their average $\frac{x+y}2$ is still a positive integer and isn't contained in the set, and add it to the set. The integers $x$ and $y$ remain in the set.

Let's call the set of integers consecutive if, after the elements are sorted, the difference between any pair of adjacent elements is $1$. For example, sets $\\{2\\}$, $\\{2, 5, 4, 3\\}$, $\\{5, 6, 8, 7\\}$ are consecutive, while $\\{2, 4, 5, 6\\}$, $\\{9, 7\\}$ are not.

Eri likes consecutive sets. Suppose there is an array $b$, then Eri puts all elements in $b$ into the set. If after a finite number of operations described above, the set can become consecutive, the array $b$ will be called brilliant.

Note that if the same integer appears in the array multiple times, we only put it into the set once, as a set always contains distinct positive integers.

Eri has an array $a$ of $n$ positive integers. Please help him to count the number of pairs of integers $(l,r)$ such that $1 \leq l \leq r \leq n$ and the subarray $a_l, a_{l+1}, \ldots, a_r$ is brilliant.

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

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 4 \cdot 10^5$) — length of the array $a$.

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

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $4 \cdot 10^5$.

Output: For each test case, output a single integer — the number of brilliant subarrays.

Examples:
input:
6
2
2 2
6
1 3 6 10 15 21
5
6 30 18 36 9
1
1000000000
6
1 1 4 5 1 4
12
70 130 90 90 90 108 612 500 451 171 193 193
output:
3
18
5
1
18
53

Note: In the first test case, the array $a = [2, 2]$ has $3$ subarrays: $[2]$, $[2]$, and $[2, 2]$. For all of them, the set only contains a single integer $2$, therefore it's always consecutive. All these subarrays are brilliant, so the answer is $3$.

In the second test case, let's consider the subarray $[3, 6, 10]$. We can do operations as follows:

$$\\{3,6,10\\} \xrightarrow{x=6,y=10} \\{3,6,8,10\\} \xrightarrow{x=6,y=8} \\{3,6,7,8,10\\} \xrightarrow{x=3,y=7} \\{3,5,6,7,8,10\\}$$ $$\xrightarrow{x=3,y=5} \\{3,4,5,6,7,8,10\\} \xrightarrow{x=8,y=10} \\{3,4,5,6,7,8,9,10\\}$$

$\\{3,4,5,6,7,8,9,10\\}$ is a consecutive set, so the subarray $[3, 6, 10]$ is brilliant.

Buggy Code:
def II():
    return(int(input()))
def LMI():
    return(list(map(int,input().split())))
def I():
    return(input())
def MII():
    return(map(int,input().split()))
import sys
input=sys.stdin.readline
# import io,os
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# from collections import Counter
# int(math.log(len(L)))
# import math
# from collections import defaultdict
# mod=10**9+7
# from collections import deque
import  math

class SegmentTreeSum():
    def __init__(self, X):
        L = [0] + X
        n = int(math.ceil(math.log(len(L)) / math.log(2)))
        self.seg = [0] * (2 ** n - 1) + L + [0] * (2 ** n - len(L))
        for i in range(2 ** n - 2, -1, -1):
            self.seg[i] = math.gcd(self.seg[2 * i + 1] , self.seg[2 * i + 2])
    def update(self, m, n):
        x = m + len(self.seg) // 2
        self.seg[x] = n
        x = (x - 1) // 2
        while x > -1:
            self.seg[x] =math.gcd( self.seg[2 * x + 1] , self.seg[2 * x + 2])
            x = (x - 1) // 2
    def su(self, l, r):
        ans = 0
        a = l + len(self.seg) // 2
        b = r + len(self.seg) // 2
        while a <= b:
            if a % 2 == 0:
                ans = math.gcd(ans,self.seg[a])
                a //= 2
            else:
                a //= 2
            if b % 2 == 0:
                b -= 1
                b //= 2
            else:
                ans = math.gcd(ans,self.seg[b])
                b -= 2
                b //= 2
        return ans

# from functools import cache
for _ in range(II()):
    n=II()
    L=LMI()
    diff=[0]
    for i in range(1,n):
        diff.append(abs(L[i]-L[i-1]))
    X=SegmentTreeSum(diff)
    ans=n
    for i in range(n-1):
        l=i+1
        h=n-1
        res=-1
        while l<=h:
            mid=(l+h)//2
            alp=X.su(i+2,mid+1)
            if alp!=0 and (math.log(alp)/math.log(2))%1==0:
                res=mid
                h=mid-1
            else:
                l=mid+1

        if res!=-1:
            ans+=n-res
        l=i+1
        h=n-1
        res=-1
        while l<=h:
            mid=(l+h)//2
            alp=X.su(i+2,mid+1)
            if alp==0:
                res=mid
                l=mid+1
            else:
                h=mid-1
        if res!=-1:
            ans+=res-i

    print(ans)

# if __name__=="__main__":

#     for _ in range(II()):
#         t()
    # t()

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

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

Think step by step.

Response:
```plaintext
6
2
2 2
6
1 3 6 10 15 21
5
6 30 18 36 9
1
1000000000
6
1 1 4 5 1 4
12
70 130 90 90 90 108 612 500 451 171 193 193
```