**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums on Segments
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of $$$n$$$ integers, where all elements except for at most one are equal to $$$-1$$$ or $$$1$$$. The remaining element $$$x$$$ satisfies $$$-10^9 \le x \le 10^9$$$.

Find all possible sums of subarrays of $$$a$$$, including the empty subarray, whose sum is defined as $$$0$$$. In other words, find all integers $$$x$$$ such that the array $$$a$$$ has at least one subarray (possibly empty) with sum equal to $$$x$$$. A subarray is a contiguous subsegment of an array.

Output these sums in ascending order. Each sum should be printed only once, even if it is achieved by multiple subarrays.

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

Each test case consists of two lines:

- The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the size of the array.
- The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the elements of the array $$$a$$$. In the array $$$a$$$, there is at most one element that is neither $$$1$$$ nor $$$-1$$$.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output two lines:

- In the first line, print a single integer — the number of distinct subarray sums.
- In the second line, print these sums in ascending order.

Each sum should be printed only once, even if it is produced by multiple subarrays.

Examples:
input:
5
5
1 -1 10 1 1
5
-1 -1 -1 -1 -1
2
-1 2
2
7 1
3
1 4 -1
output:
8
-1 0 1 2 9 10 11 12
6
-5 -4 -3 -2 -1 0
4
-1 0 1 2
4
0 1 7 8
6
-1 0 1 3 4 5

Note: Let's define $$$a[i,j]$$$ as the subarray of $$$a$$$ from position $$$i$$$ to position $$$j$$$.

Consider the first test case of the example:

- $$$-1$$$ is produced by $$$a[2,2]$$$;
- $$$0$$$ is produced by the empty subarray;
- $$$1$$$ is produced by $$$a[4,4]$$$;
- $$$2$$$ is produced by $$$a[4,5]$$$;
- $$$9$$$ is produced by $$$a[2,3]$$$;
- $$$10$$$ is produced by $$$a[1,3]$$$;
- $$$11$$$ is produced by $$$a[3,4]$$$;
- $$$12$$$ is produced by $$$a[3,5]$$$.

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    split = -1
    mx = 0
    mn = 0
    mx_t = 0
    mn_t = 0
    for i in range(n):
        if a[i] == -1:
            mn_t -= 1
            mx = max(mx_t, mx)
            mx_t = 0
        elif  a[i] == 1:
            mx_t += 1
            mn = min(mn_t, mn)
            mn_t = 0
        else:
            split = i
            mn = min(mn_t, mn)
            mx = max(mx_t, mx)
            mn_t = 0
            mx_t = 0

    mn = min(mn_t, mn)
    mx = max(mx_t, mx)
    if split != -1:
        mn_l = 0
        mx_l = 0
        mx_r = 0
        mn_r = 0
        suma = 0
        for i in range(split-1,-1,-1):
            suma += a[i]
            mn_l = min(mn_l,suma)
            mx_l = max(mx_l,suma)
        suma = 0
        for i in range(split+1,n):
            suma += a[i]
            mn_r = min(mn_r,suma)
            mx_r = max(mx_r,suma)

        mn2 = mn_l+mn_r+a[split]
        mx2 = mx_l+mx_r+a[split]

        if mn >= mn2 and mn <= mx2:
            sol = [str(l) for l in range(mn2, max(mx2,mx))+1]
            print(len(sol))
            print(" ".join(sol))
        elif mn2 >= mn and mn2 <= mx:
            sol = [str(l) for l in range(mn, max(mx2,mx))+1]
            print(len(sol))
            print(" ".join(sol))
        elif mn <= mn2:
            sol = [*[str(l) for l in range(mn, mx+1)], *[str(l) for l in range(mn2, mx2+1)]]
            print(len(sol))
            print(" ".join(sol))
        else:
            sol = [*[str(l) for l in range(mn2, mx2+1)], *[str(l) for l in range(mn, mx+1)]]
            print(len(sol))
            print(" ".join(sol))
    else:
        sol = [str(l) for l in range(mn, mx+1)]
        print(len(sol))
        print(" ".join(sol))

**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 -1 10 1 -1
4
1 1 1 1
2
-1 -1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums on Segments
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of $$$n$$$ integers, where all elements except for at most one are equal to $$$-1$$$ or $$$1$$$. The remaining element $$$x$$$ satisfies $$$-10^9 \le x \le 10^9$$$.

Find all possible sums of subarrays of $$$a$$$, including the empty subarray, whose sum is defined as $$$0$$$. In other words, find all integers $$$x$$$ such that the array $$$a$$$ has at least one subarray (possibly empty) with sum equal to $$$x$$$. A subarray is a contiguous subsegment of an array.

Output these sums in ascending order. Each sum should be printed only once, even if it is achieved by multiple subarrays.

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

Each test case consists of two lines:

- The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the size of the array.
- The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the elements of the array $$$a$$$. In the array $$$a$$$, there is at most one element that is neither $$$1$$$ nor $$$-1$$$.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output two lines:

- In the first line, print a single integer — the number of distinct subarray sums.
- In the second line, print these sums in ascending order.

Each sum should be printed only once, even if it is produced by multiple subarrays.

Examples:
input:
5
5
1 -1 10 1 1
5
-1 -1 -1 -1 -1
2
-1 2
2
7 1
3
1 4 -1
output:
8
-1 0 1 2 9 10 11 12
6
-5 -4 -3 -2 -1 0
4
-1 0 1 2
4
0 1 7 8
6
-1 0 1 3 4 5

Note: Let's define $$$a[i,j]$$$ as the subarray of $$$a$$$ from position $$$i$$$ to position $$$j$$$.

Consider the first test case of the example:

- $$$-1$$$ is produced by $$$a[2,2]$$$;
- $$$0$$$ is produced by the empty subarray;
- $$$1$$$ is produced by $$$a[4,4]$$$;
- $$$2$$$ is produced by $$$a[4,5]$$$;
- $$$9$$$ is produced by $$$a[2,3]$$$;
- $$$10$$$ is produced by $$$a[1,3]$$$;
- $$$11$$$ is produced by $$$a[3,4]$$$;
- $$$12$$$ is produced by $$$a[3,5]$$$.

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

def solve():
    n = int(input().strip())
    arr = list(map(int, input().split()))

    l_min, l_max = 0, 0
    l_inter = [2*10**9, -2*10**9]
    x = None
    for i in range(n):
        if arr[i] != -1 and arr[i] != 1:
            x = arr[i]
            break
        l_min = min(0, l_min + arr[i])
        l_max = max(0, l_max + arr[i])
        l_inter = [min(l_inter[0], l_min), max(l_inter[1], l_max)]

    if x is None:
        result = list(range(l_inter[0], l_inter[1] + 1))
        print(len(result))
        print(*result, sep=' ')
        return

    r_min, r_max = 0, 0
    r_inter = [2*10**9, -2*10**9]
    for i in range(n - 1, -1, -1):
        if arr[i] == x:
            break
        r_min = min(0, r_min + arr[i])
        r_max = max(0, r_max + arr[i])
        r_inter = [min(r_inter[0], r_min), max(r_inter[1], r_max)]

    lr_inter = [min(l_inter[0], r_inter[0]), max(l_inter[1], r_inter[1])]
    x_inter = [x + l_min + r_min, x + l_max + r_max]
    if x_inter[1] < lr_inter[0]:
        result = list(range(x_inter[0], x_inter[1] + 1)) + list(range(lr_inter[0], lr_inter[1] + 1))
    elif x_inter[0] < lr_inter[0]:
        result = list(range(x_inter[0], lr_inter[0])) + list(range(lr_inter[0], lr_inter[1] + 1))
    elif lr_inter[1] < x_inter[0]:
        result = list(range(lr_inter[0], lr_inter[1] + 1)) + list(range(x_inter[0], x_inter[1] + 1))
    elif lr_inter[1] < x_inter[1]:
        result = list(range(lr_inter[0], x_inter[0])) + list(range(x_inter[0], x_inter[1] + 1))
    else:
        result = list(range(l_inter[0], l_inter[1] + 1))
    print(len(result))
    print(*result)

T = int(input().strip())
for _ in range(T):
    solve()

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

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

Think step by step.

Response:
```plaintext
3
3
1 1000000000 1
3
1 1000000000 -1
3
-1 1000000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums on Segments
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of $$$n$$$ integers, where all elements except for at most one are equal to $$$-1$$$ or $$$1$$$. The remaining element $$$x$$$ satisfies $$$-10^9 \le x \le 10^9$$$.

Find all possible sums of subarrays of $$$a$$$, including the empty subarray, whose sum is defined as $$$0$$$. In other words, find all integers $$$x$$$ such that the array $$$a$$$ has at least one subarray (possibly empty) with sum equal to $$$x$$$. A subarray is a contiguous subsegment of an array.

Output these sums in ascending order. Each sum should be printed only once, even if it is achieved by multiple subarrays.

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

Each test case consists of two lines:

- The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the size of the array.
- The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the elements of the array $$$a$$$. In the array $$$a$$$, there is at most one element that is neither $$$1$$$ nor $$$-1$$$.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output two lines:

- In the first line, print a single integer — the number of distinct subarray sums.
- In the second line, print these sums in ascending order.

Each sum should be printed only once, even if it is produced by multiple subarrays.

Examples:
input:
5
5
1 -1 10 1 1
5
-1 -1 -1 -1 -1
2
-1 2
2
7 1
3
1 4 -1
output:
8
-1 0 1 2 9 10 11 12
6
-5 -4 -3 -2 -1 0
4
-1 0 1 2
4
0 1 7 8
6
-1 0 1 3 4 5

Note: Let's define $$$a[i,j]$$$ as the subarray of $$$a$$$ from position $$$i$$$ to position $$$j$$$.

Consider the first test case of the example:

- $$$-1$$$ is produced by $$$a[2,2]$$$;
- $$$0$$$ is produced by the empty subarray;
- $$$1$$$ is produced by $$$a[4,4]$$$;
- $$$2$$$ is produced by $$$a[4,5]$$$;
- $$$9$$$ is produced by $$$a[2,3]$$$;
- $$$10$$$ is produced by $$$a[1,3]$$$;
- $$$11$$$ is produced by $$$a[3,4]$$$;
- $$$12$$$ is produced by $$$a[3,5]$$$.

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

public class CF{
    static FastReader sc = new FastReader();
    @SuppressWarnings("unused")
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        @SuppressWarnings("CallToPrintStackTrace")
        String n() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int i() {
            return Integer.parseInt(n());
        }

        long l() {
            return Long.parseLong(n());
        }

        double d() {
            return Double.parseDouble(n());
        }

        @SuppressWarnings("CallToPrintStackTrace")
        String line() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

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

    static int[] fi(int n){
        int[] arr = new int[n];
        for(int i=0; i<n; i++) arr[i] = sc.i();
        return arr;
    }

    static String[] fs(int n){
        String[] arr = new String[n];
        for(int i=0; i<n; i++) arr[i] = sc.n();
        return arr;
    }

    public static void main(String[] args) {
        int test = sc.i();
        while(test-- > 0) {
            helper();
        }
    }

// ------------------------------ start here -------------------------------------------------

    @SuppressWarnings("unused")
    static void helper() {
        int n = sc.i();
        long[] arr = fl(n);
        if(n == 1){
            if(arr[0] > 0) System.out.println(0 + " " + arr[0]);
            else System.out.println(arr[0] + " 0" );
            return;
        }

        long sum = 0;
        Set<Long> set = new HashSet<>();
        set.add(0L);
        boolean bool = (arr[0] >= 0) ? true : false;

        List<Long> list = new ArrayList<>();
        for(int i=0; i<n; i++){
            sum += arr[i];
            set.add(sum);
            if(bool){
                if(arr[i] == 1 || arr[i] != -1){
                    list.add(arr[i]);
                }else{
                    bool = !bool;
                    get(set, list);
                    list = new ArrayList<>();
                    list.add(arr[i]);
                }
            }else{
                if(arr[i] == -1 || arr[i] != 1){
                    list.add(arr[i]);
                }else{
                    bool = !bool;
                    get(set, list);
                    list = new ArrayList<>();
                    list.add(arr[i]);
                }
            }
        }

        get(set, list);
        bool = (arr[n-1] >= 0) ? true : false;
        sum = 0;

        list = new ArrayList<>();
        for(int i=n-1; i >= 0; i--){
            sum += arr[i];
            set.add(sum);
            if(bool){
                if(arr[i] == 1 || arr[i] != -1){
                    list.add(arr[i]);
                }else{
                    bool = !bool;
                    get(set, list);
                    list = new ArrayList<>();
                    list.add(arr[i]);
                }
            }else{
                if(arr[i] == -1 || arr[i] != 1){
                    list.add(arr[i]);
                }else{
                    bool = !bool;
                    get(set, list);
                    list = new ArrayList<>();
                    list.add(arr[i]);
                }
            }
        }

        get(set, list);

        List<Long> ans = new ArrayList<>();
        for(long it : set) ans.add(it);
        Collections.sort(ans);
        StringBuilder sb = new StringBuilder();
        for(long it : ans) sb.append(it + " ");

        System.out.println(set.size());
        System.out.println(sb.toString());

    }

    static void get(Set<Long> set, List<Long> list){
        int n = list.size();
        long sum = 0;
        for(int i=0; i<n; i++){
            sum += list.get(i);
            set.add(sum);
        }
        sum = 0;
        for(int i = n-1; i>= 0; i--){
            sum += list.get(i);
            set.add(sum);
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
3
1 -1000000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums on Segments
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of $$$n$$$ integers, where all elements except for at most one are equal to $$$-1$$$ or $$$1$$$. The remaining element $$$x$$$ satisfies $$$-10^9 \le x \le 10^9$$$.

Find all possible sums of subarrays of $$$a$$$, including the empty subarray, whose sum is defined as $$$0$$$. In other words, find all integers $$$x$$$ such that the array $$$a$$$ has at least one subarray (possibly empty) with sum equal to $$$x$$$. A subarray is a contiguous subsegment of an array.

Output these sums in ascending order. Each sum should be printed only once, even if it is achieved by multiple subarrays.

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

Each test case consists of two lines:

- The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the size of the array.
- The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the elements of the array $$$a$$$. In the array $$$a$$$, there is at most one element that is neither $$$1$$$ nor $$$-1$$$.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output two lines:

- In the first line, print a single integer — the number of distinct subarray sums.
- In the second line, print these sums in ascending order.

Each sum should be printed only once, even if it is produced by multiple subarrays.

Examples:
input:
5
5
1 -1 10 1 1
5
-1 -1 -1 -1 -1
2
-1 2
2
7 1
3
1 4 -1
output:
8
-1 0 1 2 9 10 11 12
6
-5 -4 -3 -2 -1 0
4
-1 0 1 2
4
0 1 7 8
6
-1 0 1 3 4 5

Note: Let's define $$$a[i,j]$$$ as the subarray of $$$a$$$ from position $$$i$$$ to position $$$j$$$.

Consider the first test case of the example:

- $$$-1$$$ is produced by $$$a[2,2]$$$;
- $$$0$$$ is produced by the empty subarray;
- $$$1$$$ is produced by $$$a[4,4]$$$;
- $$$2$$$ is produced by $$$a[4,5]$$$;
- $$$9$$$ is produced by $$$a[2,3]$$$;
- $$$10$$$ is produced by $$$a[1,3]$$$;
- $$$11$$$ is produced by $$$a[3,4]$$$;
- $$$12$$$ is produced by $$$a[3,5]$$$.

Buggy Code:
import java.util.*;

public class Con3 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int r = sc.nextInt();
        for (int k = 0; k < r; k++) {
            int n = sc.nextInt();
            long[] ar = new long[n];
            long acc=0;
            Set<Long> set = new HashSet<>();

            Set<Long> set1 = new HashSet<>();
            for(int i=0;i<n;i++){
                ar[i] = sc.nextInt() + acc;
                set.add(ar[i]-acc);
                acc = ar[i];
                set.add(ar[i]);
                set1.add(ar[i]);

            }
            ArrayList<Long> arr1 = new ArrayList<>(set1);
            set.add((long)0);
            for(int i=0;i<arr1.size();i++){
                for(int j=i+1;j<arr1.size();j++){
                    set.add(arr1.get(j)-arr1.get(i));
                }
            }
            System.out.println(set.size());
            ArrayList<Long> arr = new ArrayList<>(set);
            Collections.sort(arr);
            arr.forEach(item -> System.out.print(item + " "));
            System.out.println();

        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums on Segments
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of $$$n$$$ integers, where all elements except for at most one are equal to $$$-1$$$ or $$$1$$$. The remaining element $$$x$$$ satisfies $$$-10^9 \le x \le 10^9$$$.

Find all possible sums of subarrays of $$$a$$$, including the empty subarray, whose sum is defined as $$$0$$$. In other words, find all integers $$$x$$$ such that the array $$$a$$$ has at least one subarray (possibly empty) with sum equal to $$$x$$$. A subarray is a contiguous subsegment of an array.

Output these sums in ascending order. Each sum should be printed only once, even if it is achieved by multiple subarrays.

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

Each test case consists of two lines:

- The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the size of the array.
- The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the elements of the array $$$a$$$. In the array $$$a$$$, there is at most one element that is neither $$$1$$$ nor $$$-1$$$.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output two lines:

- In the first line, print a single integer — the number of distinct subarray sums.
- In the second line, print these sums in ascending order.

Each sum should be printed only once, even if it is produced by multiple subarrays.

Examples:
input:
5
5
1 -1 10 1 1
5
-1 -1 -1 -1 -1
2
-1 2
2
7 1
3
1 4 -1
output:
8
-1 0 1 2 9 10 11 12
6
-5 -4 -3 -2 -1 0
4
-1 0 1 2
4
0 1 7 8
6
-1 0 1 3 4 5

Note: Let's define $$$a[i,j]$$$ as the subarray of $$$a$$$ from position $$$i$$$ to position $$$j$$$.

Consider the first test case of the example:

- $$$-1$$$ is produced by $$$a[2,2]$$$;
- $$$0$$$ is produced by the empty subarray;
- $$$1$$$ is produced by $$$a[4,4]$$$;
- $$$2$$$ is produced by $$$a[4,5]$$$;
- $$$9$$$ is produced by $$$a[2,3]$$$;
- $$$10$$$ is produced by $$$a[1,3]$$$;
- $$$11$$$ is produced by $$$a[3,4]$$$;
- $$$12$$$ is produced by $$$a[3,5]$$$.

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

struct custom_hash
{
    static uint64_t splitmix64(uint64_t x)
    {
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }
    size_t operator()(uint64_t x) const
    {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};

template<typename K, typename V>
using unordered_map_custom = unordered_map<K, V, custom_hash>;
#define unordered_map unordered_map_custom
#define int long long
#define double long double //also do 1.0->1.0L
#define fastio ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
#define pii pair<int, int>
#define pqmn priority_queue<pii, vector<pii>, greater<pii>>
#define pqmx priority_queue<pii, vector<pii>>

int maxKadane(vector<int>& arr, int idx, bool fl)
{
    int mx=INT_MIN, currSum=0, i=fl? 0:idx+1;
    while((fl && i<idx) || (!fl && i<arr.size()))
    {
        currSum+=arr[i++];
        if(currSum<=0) currSum=0;
        mx=max(mx, currSum);
    }
    return mx;
}

int minKadane(vector<int>& arr, int idx, bool fl)
{
    int mx=INT_MIN, currSum=0, i=fl? 0:idx+1;
    while((fl && i<idx) || (!fl && i<arr.size()))
    {
        currSum+=(-arr[i++]);
        if(currSum<=0) currSum=0;
        mx=max(mx, currSum);
    }
    return -mx;
}

signed main()
{
    fastio;
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        vector<int>arr(n);

        int idx=n, x=0;
        for(int i=0; i<n; i++)
        {
            cin>>arr[i];
            if(abs(arr[i])!=1) idx=i, x=arr[idx];
        }

        pii r1={minKadane(arr,idx,1), maxKadane(arr,idx,1)};
        pii r2={minKadane(arr,idx,0), maxKadane(arr,idx,0)};

        if(idx<n) arr[idx]=0;
        pii r3={minKadane(arr,-1,0), maxKadane(arr,-1,0)};

        set<int>ans;
        ans.insert(0);
        ans.insert(x);
        for(int i=r1.first; i<=r1.second; i++) ans.insert(i);
        for(int i=r2.first; i<=r2.second; i++) ans.insert(i);
        for(int i=r3.first; i<=r3.second; i++) ans.insert(i+x);

        cout<<ans.size()<<endl;
        for(auto val:ans) cout<<val<<" ";
        cout<<endl;
    }
}

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

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

Think step by step.

Response:
```plaintext
5
5
1 -1 10 1 -1
5
-1 -1 -1 -1 -1
2
-1 2
2
7 1
3
1 4 -1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums on Segments
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of $$$n$$$ integers, where all elements except for at most one are equal to $$$-1$$$ or $$$1$$$. The remaining element $$$x$$$ satisfies $$$-10^9 \le x \le 10^9$$$.

Find all possible sums of subarrays of $$$a$$$, including the empty subarray, whose sum is defined as $$$0$$$. In other words, find all integers $$$x$$$ such that the array $$$a$$$ has at least one subarray (possibly empty) with sum equal to $$$x$$$. A subarray is a contiguous subsegment of an array.

Output these sums in ascending order. Each sum should be printed only once, even if it is achieved by multiple subarrays.

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

Each test case consists of two lines:

- The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the size of the array.
- The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the elements of the array $$$a$$$. In the array $$$a$$$, there is at most one element that is neither $$$1$$$ nor $$$-1$$$.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output two lines:

- In the first line, print a single integer — the number of distinct subarray sums.
- In the second line, print these sums in ascending order.

Each sum should be printed only once, even if it is produced by multiple subarrays.

Examples:
input:
5
5
1 -1 10 1 1
5
-1 -1 -1 -1 -1
2
-1 2
2
7 1
3
1 4 -1
output:
8
-1 0 1 2 9 10 11 12
6
-5 -4 -3 -2 -1 0
4
-1 0 1 2
4
0 1 7 8
6
-1 0 1 3 4 5

Note: Let's define $$$a[i,j]$$$ as the subarray of $$$a$$$ from position $$$i$$$ to position $$$j$$$.

Consider the first test case of the example:

- $$$-1$$$ is produced by $$$a[2,2]$$$;
- $$$0$$$ is produced by the empty subarray;
- $$$1$$$ is produced by $$$a[4,4]$$$;
- $$$2$$$ is produced by $$$a[4,5]$$$;
- $$$9$$$ is produced by $$$a[2,3]$$$;
- $$$10$$$ is produced by $$$a[1,3]$$$;
- $$$11$$$ is produced by $$$a[3,4]$$$;
- $$$12$$$ is produced by $$$a[3,5]$$$.

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

array<int, 2> mm(vector<int> &a, int l, int r){
    int minx = 0, maxx = 0;
    int tmp1 = 0, tmp2 = 0;
    for(int i = l; i <= r; ++i){
        tmp1 += a[i];
        tmp2 += a[i];
        if(tmp1 > 0) tmp1 = 0;
        if(tmp2 < 0) tmp2 = 0;
        minx = min(tmp1, minx);
        maxx = max(tmp2, maxx);
    }
    array<int, 2> R = {minx, maxx};
    return R;
}

void merge(set<int>& R, array<int, 2> range){
    auto [l, r] = range;
    for(int i = min(l, r); i <= max(l, r); ++i) R.insert(i);
}

void solve(){
    int n, dx = 0, d1 = 0, d2 = 0, x = 0, xi = -1;
    cin >> n;
    vector<int> a(n);
    set<int> R = {0}, tmp = {0};
    for(int i = 0; i < n; ++i){
        cin >> a[i];
        dx += a[i];
        R.insert(a[i]);
        R.insert(dx);
        if(abs(a[i]) != 1){
            for(auto k : tmp) R.insert(dx - k);
            x = a[i];
            xi = i;
            d1 = 0;
        }else{
            d1 += a[i];
            R.insert(d1);
            R.insert(d1 + x);
        }
        tmp.insert(dx);
    }
    if(xi != -1){
        merge(R, mm(a, 0, xi - 1));
        merge(R, mm(a, xi + 1, n - 1));
    }else merge(R, mm(a, 0, n - 1));
    cout << R.size() << endl;
    for(auto i : R){
        cout << i << " ";
    }
    cout << endl;
    return;
}

int main(){
    int T = 1; cin >> T;
    while(T--) solve();
    return false;
}

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

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

Think step by step.

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

1. **Edge Cases**:
   - An array with a single element that is neither 1 nor -1.
   - An array with all elements being 1 or -1.
   - An array with a single element that is a large value (within the given constraints) and the rest being 1 or -1.
   - An array with a single element that is a large negative value and the rest being 1 or -1.

2. **Challenging Scenarios**:
   - An array where the special element (neither 1 nor -1) is at the beginning, middle, or end.
   - An array where the special element is surrounded by 1s and -1s in a way that can produce a wide range of subarray sums.

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

```plaintext
3
3
1 1 10
3
10 -1 -1
5
1 10 -1 1 -1
```

### Explanation:
1. **Test Case 1**: `3 1 1 10`
   - This case tests the scenario where the special element (10) is at the end.
   - Expected output: 7 sums: -1, 0, 1, 2, 3, 10, 11, 12.

2. **Test Case 2**: `3 10 -1 -1`
   - This case tests the scenario where the special element (10) is at the beginning.
   - Expected output: 7 sums: -2, -1, 0, 1, 8, 9, 10.

3. **Test Case 3**: `5 1 10 -1 1 -1`
   - This case tests the scenario where the special element (10) is in the middle.
   - Expected output: 11 sums: -2, -1, 0, 1, 2, 8, 9, 10, 11, 12, 13.

These test cases cover different positions of the special element and should help identify 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: Sums on Segments
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of $$$n$$$ integers, where all elements except for at most one are equal to $$$-1$$$ or $$$1$$$. The remaining element $$$x$$$ satisfies $$$-10^9 \le x \le 10^9$$$.

Find all possible sums of subarrays of $$$a$$$, including the empty subarray, whose sum is defined as $$$0$$$. In other words, find all integers $$$x$$$ such that the array $$$a$$$ has at least one subarray (possibly empty) with sum equal to $$$x$$$. A subarray is a contiguous subsegment of an array.

Output these sums in ascending order. Each sum should be printed only once, even if it is achieved by multiple subarrays.

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

Each test case consists of two lines:

- The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the size of the array.
- The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the elements of the array $$$a$$$. In the array $$$a$$$, there is at most one element that is neither $$$1$$$ nor $$$-1$$$.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output two lines:

- In the first line, print a single integer — the number of distinct subarray sums.
- In the second line, print these sums in ascending order.

Each sum should be printed only once, even if it is produced by multiple subarrays.

Examples:
input:
5
5
1 -1 10 1 1
5
-1 -1 -1 -1 -1
2
-1 2
2
7 1
3
1 4 -1
output:
8
-1 0 1 2 9 10 11 12
6
-5 -4 -3 -2 -1 0
4
-1 0 1 2
4
0 1 7 8
6
-1 0 1 3 4 5

Note: Let's define $$$a[i,j]$$$ as the subarray of $$$a$$$ from position $$$i$$$ to position $$$j$$$.

Consider the first test case of the example:

- $$$-1$$$ is produced by $$$a[2,2]$$$;
- $$$0$$$ is produced by the empty subarray;
- $$$1$$$ is produced by $$$a[4,4]$$$;
- $$$2$$$ is produced by $$$a[4,5]$$$;
- $$$9$$$ is produced by $$$a[2,3]$$$;
- $$$10$$$ is produced by $$$a[1,3]$$$;
- $$$11$$$ is produced by $$$a[3,4]$$$;
- $$$12$$$ is produced by $$$a[3,5]$$$.

Buggy Code:
t = int(input())

for j in range(t):
    n = int(input())
    array = list(map(int, input().split()))
    values = set()
    prefix = []
    subarray = [0 for i in range(n)]
    in_x = 2*10**9
    m = len(array)
    prefix = 0
    pr = 0
    last = 0

    max_left = -2*10**9
    max_right = -2*10**9

    min_left = 2*10**9
    min_right = 2*10**9

    right = False

    for i in range(m):
        prefix += array[i]
        pr += array[i]
        if array[i] != last:
            prefix = array[i]
        if array[i] != 1 and array[i] != -1:
            in_x = i
            prefix = 0
            pr = 0
            right = True
            continue

        if right:
            max_right = max(max_right, max(pr, prefix))
            min_right = min(min_right, min(pr, prefix))
        else:
            max_left = max(max_left, max(pr, prefix))
            min_left = min(min_left, min(pr, prefix))
        last = array[i]

    if max_left != -2*10**9:
        for i in range(min_left, max_left + 1):
            values.add(i)
    if max_right != -2*10**9:
        for i in range(min_right, max_right + 1):
            values.add(i)
    """if max_left != -2*10**9 and max_right != -2*10**9:
        if max_left < min_right:
            values.add(range(min_left, max_left + 1))
            values.add(range(min_right, max_right + 1))
        elif max_right < min_left:
            values.add(range(min_right, max_right + 1))
            values.add(range(min_left, max_left + 1))
        else:
            values.add(range(min(min_left, min_right), max(max_left, max_right) + 1))
    elif max_left == -2*10**9 and max_right != -2*10**9:
        values.add(range(min_right, max_right + 1))
    elif max_right == -2*10**9 and max_left != -2*10**9:
        values.add(range(min_left, max_left + 1))"""

    if in_x != 2*10**9:
        max_left = -2 * 10 ** 9
        max_right = -2 * 10 ** 9

        min_left = 2 * 10 ** 9
        min_right = 2 * 10 ** 9

        for i in range(in_x - 1, -1, -1):
            subarray[i] = subarray[i + 1] + array[i]
            max_left = max(max_left, subarray[i])
            min_left = min(min_left, subarray[i])

        for i in range(in_x + 1, n):
            subarray[i] = subarray[i - 1] + array[i]
            max_right = max(max_right, subarray[i])
            min_right = min(min_right, subarray[i])

        minimum = min(min_left, min(min_right, min_left + min_right))
        maximum = max(max_left, max(max_right, max_left + max_right))
        x = array[in_x]

        for i in range(minimum + x, maximum + 1 + x):
            values.add(i)
        #values.add(range(minimum + x, maximum + 1 + x))
        values.add(x)

    values.add(0)
    print(len(values))
    print(*sorted(values))

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

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

Think step by step.

Response:
```plaintext
1
4
1 1 -1000000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums on Segments
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of $$$n$$$ integers, where all elements except for at most one are equal to $$$-1$$$ or $$$1$$$. The remaining element $$$x$$$ satisfies $$$-10^9 \le x \le 10^9$$$.

Find all possible sums of subarrays of $$$a$$$, including the empty subarray, whose sum is defined as $$$0$$$. In other words, find all integers $$$x$$$ such that the array $$$a$$$ has at least one subarray (possibly empty) with sum equal to $$$x$$$. A subarray is a contiguous subsegment of an array.

Output these sums in ascending order. Each sum should be printed only once, even if it is achieved by multiple subarrays.

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

Each test case consists of two lines:

- The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the size of the array.
- The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the elements of the array $$$a$$$. In the array $$$a$$$, there is at most one element that is neither $$$1$$$ nor $$$-1$$$.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output two lines:

- In the first line, print a single integer — the number of distinct subarray sums.
- In the second line, print these sums in ascending order.

Each sum should be printed only once, even if it is produced by multiple subarrays.

Examples:
input:
5
5
1 -1 10 1 1
5
-1 -1 -1 -1 -1
2
-1 2
2
7 1
3
1 4 -1
output:
8
-1 0 1 2 9 10 11 12
6
-5 -4 -3 -2 -1 0
4
-1 0 1 2
4
0 1 7 8
6
-1 0 1 3 4 5

Note: Let's define $$$a[i,j]$$$ as the subarray of $$$a$$$ from position $$$i$$$ to position $$$j$$$.

Consider the first test case of the example:

- $$$-1$$$ is produced by $$$a[2,2]$$$;
- $$$0$$$ is produced by the empty subarray;
- $$$1$$$ is produced by $$$a[4,4]$$$;
- $$$2$$$ is produced by $$$a[4,5]$$$;
- $$$9$$$ is produced by $$$a[2,3]$$$;
- $$$10$$$ is produced by $$$a[1,3]$$$;
- $$$11$$$ is produced by $$$a[3,4]$$$;
- $$$12$$$ is produced by $$$a[3,5]$$$.

Buggy Code:
def solve():
    t = int(input())
    for _ in range(t):
        n = int(input())
        a = list(map(int, input().split()))

        possible_sums = set([0])

        prefix = 0
        possible_sums.add(prefix)  #
        s = [0 for _ in range(len(a))]

        x = -1
        idx = -1
        for i, num in enumerate(a):
            if num != -1 and num != 1:
                x = num
                idx = i
            prefix += num
            s[i] = prefix
            possible_sums.add(prefix)
            possible_sums.add(num)

        if x != -1:
            for i, su in enumerate(s):
                if i <= idx:
                    possible_sums.add(s[idx] - s[i])
                else:
                    possible_sums.add(s[i] - s[idx])

        postfix = 0
        possible_sums.add(postfix)
        s = [0 for _ in range(len(a))]

        x = -1
        idx = -1
        for i in range(len(a)-1, -1, -1):
            num = a[i]
            if num != -1 and num != 1:
                x = num
                idx = i
            postfix += num
            s[i] = postfix
            possible_sums.add(postfix)

        if x != -1:
            for i, su in enumerate(s):
                if i <= idx:
                    possible_sums.add(s[i] - s[idx])
                else:
                    possible_sums.add(s[idx] - s[i])

        possible_sums = sorted(possible_sums)
        print(len(possible_sums))
        print(" ".join(map(str, possible_sums)))

solve()

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

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

Think step by step.

Response:
```plaintext
1
5
1 1 1 -1000000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums on Segments
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of $$$n$$$ integers, where all elements except for at most one are equal to $$$-1$$$ or $$$1$$$. The remaining element $$$x$$$ satisfies $$$-10^9 \le x \le 10^9$$$.

Find all possible sums of subarrays of $$$a$$$, including the empty subarray, whose sum is defined as $$$0$$$. In other words, find all integers $$$x$$$ such that the array $$$a$$$ has at least one subarray (possibly empty) with sum equal to $$$x$$$. A subarray is a contiguous subsegment of an array.

Output these sums in ascending order. Each sum should be printed only once, even if it is achieved by multiple subarrays.

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

Each test case consists of two lines:

- The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the size of the array.
- The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the elements of the array $$$a$$$. In the array $$$a$$$, there is at most one element that is neither $$$1$$$ nor $$$-1$$$.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output two lines:

- In the first line, print a single integer — the number of distinct subarray sums.
- In the second line, print these sums in ascending order.

Each sum should be printed only once, even if it is produced by multiple subarrays.

Examples:
input:
5
5
1 -1 10 1 1
5
-1 -1 -1 -1 -1
2
-1 2
2
7 1
3
1 4 -1
output:
8
-1 0 1 2 9 10 11 12
6
-5 -4 -3 -2 -1 0
4
-1 0 1 2
4
0 1 7 8
6
-1 0 1 3 4 5

Note: Let's define $$$a[i,j]$$$ as the subarray of $$$a$$$ from position $$$i$$$ to position $$$j$$$.

Consider the first test case of the example:

- $$$-1$$$ is produced by $$$a[2,2]$$$;
- $$$0$$$ is produced by the empty subarray;
- $$$1$$$ is produced by $$$a[4,4]$$$;
- $$$2$$$ is produced by $$$a[4,5]$$$;
- $$$9$$$ is produced by $$$a[2,3]$$$;
- $$$10$$$ is produced by $$$a[1,3]$$$;
- $$$11$$$ is produced by $$$a[3,4]$$$;
- $$$12$$$ is produced by $$$a[3,5]$$$.

Buggy Code:
import java.util.*;

public class Main {

    public static List<List<Long>> findSumsOnSegments(int t, List<long[]> testCases) {
        List<List<Long>> result = new ArrayList<>();

        for (long[] arr : testCases) {
            int n = arr.length;
            int xPos = -1;

            // Identify the position of x (if any)
            for (int i = 0; i < n; i++) {
                if (arr[i] != 1 && arr[i] != -1) {
                    xPos = i;
                    break;
                }
            }

            TreeSet<Long> uniqueSums = new TreeSet<>();
            uniqueSums.add(0L); // Include empty subarray sum

            // Case 1: Process subarrays without x
            long runningSum = 0;
            for (int i = 0; i < n; i++) {
                if (i == xPos) {
                    runningSum = 0; // Reset when encountering x
                    continue;
                }
                runningSum += arr[i];
                uniqueSums.add(runningSum);
            }

            // Include all single elements
            for (int i = 0; i < n; i++) {
                uniqueSums.add((long) arr[i]);
            }

            // Case 2: Handle subarrays with x
            if (xPos != -1) {
                long xValue = arr[xPos];

                // Left sums
                long leftSum = 0;
                for (int i = xPos - 1; i >= 0; i--) {
                    leftSum += arr[i];
                    uniqueSums.add(leftSum + xValue);
                }

                // Right sums
                long rightSum = 0;
                for (int i = xPos + 1; i < n; i++) {
                    rightSum += arr[i];
                    uniqueSums.add(rightSum + xValue);
                }

                // Include x by itself
                uniqueSums.add(xValue);

                // Include the full array sum including x
                uniqueSums.add(leftSum + xValue + rightSum);
            }

            result.add(new ArrayList<>(uniqueSums));
        }

        return result;
    }

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

        List<long[]> testCases = new ArrayList<>();
        for (int i = 0; i < t; i++) {
            int n = sc.nextInt();
            long[] arr = new long[n];
            for (int j = 0; j < n; j++) {
                arr[j] = sc.nextLong();
            }
            testCases.add(arr);
        }

        List<List<Long>> results = findSumsOnSegments(t, testCases);
        for (List<Long> sums : results) {
            System.out.println(sums.size());
            for (long sum : sums) {
                System.out.print(sum + " ");
            }
            System.out.println();
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The key issues in the buggy code include:

1. **Resetting the running sum incorrectly when encountering the special element `x`**.
2. **Handling subarrays with the special element `x`**.

Let's consider the following test cases:

1. **Array with a single element `x`**:
   - This should produce a single sum, which is `x` itself.

2. **Array with `x` at the beginning**:
   - This should produce sums that include `x` and the sums of the remaining elements.

3. **Array with `x` at the end**:
   - This should produce sums that include `x` and the sums of the remaining elements.

4. **Array with `x` in the middle**:
   - This should produce sums that include `x`, the sums of the elements before `x`, the sums of the elements after `x`, and the sums that include `x` and both sides.

5. **Array with all elements being 1 or -1**:
   - This should produce sums that are the sums of all possible subarrays of 1s and -1s.

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

```plaintext
5
1
5
3
5 1 -1
3
-1 1 5
5
1 -1 5 1 -1
5
1 1 1 1 1
```

This test input covers a variety of edge cases and should help expose any issues in the buggy code. Here is the test input:

```plaintext
5
1
5
3
5 1 -1
3
-1 1 5
5
1 -1 5 1 -1
5
1 1 1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums on Segments
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of $$$n$$$ integers, where all elements except for at most one are equal to $$$-1$$$ or $$$1$$$. The remaining element $$$x$$$ satisfies $$$-10^9 \le x \le 10^9$$$.

Find all possible sums of subarrays of $$$a$$$, including the empty subarray, whose sum is defined as $$$0$$$. In other words, find all integers $$$x$$$ such that the array $$$a$$$ has at least one subarray (possibly empty) with sum equal to $$$x$$$. A subarray is a contiguous subsegment of an array.

Output these sums in ascending order. Each sum should be printed only once, even if it is achieved by multiple subarrays.

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

Each test case consists of two lines:

- The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the size of the array.
- The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the elements of the array $$$a$$$. In the array $$$a$$$, there is at most one element that is neither $$$1$$$ nor $$$-1$$$.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output two lines:

- In the first line, print a single integer — the number of distinct subarray sums.
- In the second line, print these sums in ascending order.

Each sum should be printed only once, even if it is produced by multiple subarrays.

Examples:
input:
5
5
1 -1 10 1 1
5
-1 -1 -1 -1 -1
2
-1 2
2
7 1
3
1 4 -1
output:
8
-1 0 1 2 9 10 11 12
6
-5 -4 -3 -2 -1 0
4
-1 0 1 2
4
0 1 7 8
6
-1 0 1 3 4 5

Note: Let's define $$$a[i,j]$$$ as the subarray of $$$a$$$ from position $$$i$$$ to position $$$j$$$.

Consider the first test case of the example:

- $$$-1$$$ is produced by $$$a[2,2]$$$;
- $$$0$$$ is produced by the empty subarray;
- $$$1$$$ is produced by $$$a[4,4]$$$;
- $$$2$$$ is produced by $$$a[4,5]$$$;
- $$$9$$$ is produced by $$$a[2,3]$$$;
- $$$10$$$ is produced by $$$a[1,3]$$$;
- $$$11$$$ is produced by $$$a[3,4]$$$;
- $$$12$$$ is produced by $$$a[3,5]$$$.

Buggy Code:
/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Main
{
     public static class Tree{
        int [][] arr;
        public Tree(int n){
        arr= new int[n][n];
        }
        public void addNode(int u,int v)
        {
            arr[u][v]=1;
           // arr[v][u]=1;
        }
        public int[][] showTree()
        {
            return arr;
        }

    }
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader()
        {
            br = new BufferedReader(
                new InputStreamReader(System.in));
        }

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

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

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

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

        String nextLine()
        {
            String str = "";
            try {
                str = br.readLine();
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    static long modPow(long var, long num) {
    long m = 1;
    while (num > 0) {
        m = (m * var) %1000000007;
        --num;
    }
    return m;}
    public static int BinarySearch(int min, int max, int[] arr, int search){
        int ans=min;
        if(arr[0]>search)
        {
            return -1;
        }
        if(arr[max]<search)
        {
            return -2;
        }

        while(true)
        {
            if(search>=arr[ans])
            {
                if(search<arr[ans+1])
                {
                    break;
                }
                else
                {
                    min=ans;
                    ans=(min+max)>>>1;
                }
            }
            else
            {
                max=ans;
                ans=(min+max)>>>1;
            }
        }
        return ans;
    }
            // String str= s.nextLine();
		    // char[] arr= new char[n];

	public static void main (String[] args) throws java.lang.Exception
	{
		FastReader s = new FastReader();
		 BufferedWriter output = new BufferedWriter(
            new OutputStreamWriter(System.out));
		int t=s.nextInt();
		for(int w=0;w<t;w++)
		{
		    int n=s.nextInt();
		    boolean flag=false;
		    int[] arr= new int[n];
		    HashSet<Integer> hs = new HashSet<>();
		    int amax=0;
		    int a=0;
		    int b=0;
		    int bmax=0;
		    int x=0;
		    for(int i=0;i<n;i++)
		    {

		        arr[i]=s.nextInt();
		        if(arr[i]==1)
		        {if(i!=0)
		        { bmax=Math.min(bmax,b);
		          b++;
		        }
		            a++;
		        }
		        else if(arr[i]==-1)
		        {   if(i!=0)
		        {
		           amax=Math.max(amax,a);
		           a--;
		        }
		            b--;

		        }
		        else
		        {   amax=Math.max(amax,a);
		            bmax=Math.min(bmax,b);

		            flag=true;
		            x= arr[i];
		            if(a>0&&x<0)
		            {
		                a=a+x;
		            }
		            else if(b<0&&x>0)
		            {
		                b=b+x;
		            }
		        }

		        if(i==n-1)
		        {
		            amax=Math.max(amax,a);
		            bmax=Math.min(bmax,b);
		        }
		    }

		    for(int i=bmax;i<=amax;i++)
		    {
		        hs.add(i);
		        hs.add(x+i);

		    }
		    output.write(hs.size()+"\n");
		     List<Integer> list = new ArrayList<>(hs);
		      Collections.sort(list);
		    for (Integer element : list) {
            output.write(element+" ");
        }
		     output.write("\n");
		  //  if(flag)
		  // output.write("Yes");
		  // else
		  // output.write("No");

		  //       for(int i=0;i<n;i++)
		  //  {

		  //      output.write(arr[i]+"\n");
		  //  }

		}
		output.flush();
	}
}

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

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

Think step by step.

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

1. **Edge Cases:**
   - Arrays with only one element.
   - Arrays with all elements being 1 or -1.
   - Arrays with a single element that is neither 1 nor -1.
   - Arrays with multiple subarrays that can produce the same sum.

2. **Scenarios:**
   - The special element (neither 1 nor -1) is at the beginning, middle, or end of the array.
   - The special element is a very large or very small number.
   - The array has a mix of 1s, -1s, and the special element.

3. **Maximizing Failure Probability:**
   - Use arrays that can produce a wide range of subarray sums.
   - Use arrays that can cause the set of sums to be large and complex.

Given these considerations, a small-scale test input that can expose the bug might look like this:

```plaintext
4
3
1 1000000000 1
3
1 -1000000000 1
3
-1 1000000000 -1
3
-1 -1000000000 -1
```

This test input includes:
- Arrays with a single special element at different positions.
- Arrays with the special element being a very large positive and very large negative number.
- Arrays with a mix of 1s, -1s, and the special element.

This should help in identifying issues in the buggy code, especially in how it handles the special element and the range of sums it can produce.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums on Segments
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of $$$n$$$ integers, where all elements except for at most one are equal to $$$-1$$$ or $$$1$$$. The remaining element $$$x$$$ satisfies $$$-10^9 \le x \le 10^9$$$.

Find all possible sums of subarrays of $$$a$$$, including the empty subarray, whose sum is defined as $$$0$$$. In other words, find all integers $$$x$$$ such that the array $$$a$$$ has at least one subarray (possibly empty) with sum equal to $$$x$$$. A subarray is a contiguous subsegment of an array.

Output these sums in ascending order. Each sum should be printed only once, even if it is achieved by multiple subarrays.

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

Each test case consists of two lines:

- The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the size of the array.
- The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the elements of the array $$$a$$$. In the array $$$a$$$, there is at most one element that is neither $$$1$$$ nor $$$-1$$$.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output two lines:

- In the first line, print a single integer — the number of distinct subarray sums.
- In the second line, print these sums in ascending order.

Each sum should be printed only once, even if it is produced by multiple subarrays.

Examples:
input:
5
5
1 -1 10 1 1
5
-1 -1 -1 -1 -1
2
-1 2
2
7 1
3
1 4 -1
output:
8
-1 0 1 2 9 10 11 12
6
-5 -4 -3 -2 -1 0
4
-1 0 1 2
4
0 1 7 8
6
-1 0 1 3 4 5

Note: Let's define $$$a[i,j]$$$ as the subarray of $$$a$$$ from position $$$i$$$ to position $$$j$$$.

Consider the first test case of the example:

- $$$-1$$$ is produced by $$$a[2,2]$$$;
- $$$0$$$ is produced by the empty subarray;
- $$$1$$$ is produced by $$$a[4,4]$$$;
- $$$2$$$ is produced by $$$a[4,5]$$$;
- $$$9$$$ is produced by $$$a[2,3]$$$;
- $$$10$$$ is produced by $$$a[1,3]$$$;
- $$$11$$$ is produced by $$$a[3,4]$$$;
- $$$12$$$ is produced by $$$a[3,5]$$$.

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

#define ll long long int
#define f(i, L, R) for (long long i = L; i < R; i++)
#define vi vector<ll>
#define pb push_back
#define no cout << "NO" << endl
#define yes cout << "YES" << endl
#define endl '\n'
#define all(x) (x).begin(), (x).end()
#define sz(x) ((int)(x).size())
#define rep(i, n) for (int i = 0; i < n; i++)
#define rrep(i, n) for (int i = n - 1; i >= 0; i--)
#define fast_io                        \
    ios_base::sync_with_stdio(false); \
    cin.tie(0);
#define MOD 1000000007
#define MAX 200005
#define INF 1e18
#define PI 3.14159265358979323846
#define MAX_CHAR 256
#define bitcount(x) __builtin_popcount(x)
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) (a / gcd(a, b) * b)
#define is_prime(n) ([](int x) { if (x < 2) return false; for (int i = 2; i * i <= x; i++) if (x % i == 0) return false; return true; })(n)
#define debug(x) cerr << #x << " = " << x << endl
#define min3(a, b, c) min(a, min(b, c))
#define max3(a, b, c) max(a, max(b, c))
#define to_string(x) std::to_string(x)
#define to_int(x) stoi(x)
#define to_long(x) stoll(x)
#define clear(x) memset(x, 0, sizeof(x))
#define unique(x) (x).erase(unique(all(x)), (x).end())
#define sort_unique(x) { sort(all(x)); unique(x); }
#define input_vector(v, n) { v.resize(n); for (auto &x : v) cin >> x; }
#define output_vector(v) { for (auto &x : v) cout << x << " "; cout << endl; }
#define startt                        \
    ios_base::sync_with_stdio(false); \
    cin.tie(0);

const int N = 1e5 + 10;
const int M = 1e6 + 10;
const int LOG = 20; // For binary lifting or segment trees
const int MOD1 = 998244353; // Another common modulus
const int MAXN = 100; // For small fixed-size arrays
const int MAXM = 1000; // For small fixed-size matrices

long long calculateNonFixedSum(const vector<int>& arr, int& nonFixedIdx) {
    long long sum = 0;
    nonFixedIdx = -1;
    for (int i = 0; i < arr.size(); i++) {
        if (arr[i] != 1 && arr[i] != -1) {
            sum = arr[i];
            nonFixedIdx = i;
            break; // Only need the first non-fixed element
        }
    }
    return sum;
}

vector<long long> computePrefixSums(const vector<int>& arr, int nonFixedIdx) {
    vector<long long> prefixSums(arr.size() + 1, 0);
    for (int i = 0; i < arr.size(); i++) {
        if (i == nonFixedIdx) {
            prefixSums[i + 1] = prefixSums[i];
        } else {
            prefixSums[i + 1] = prefixSums[i] + arr[i];
        }
    }
    return prefixSums;
}

set<long long> generateUniqueSums(const vector<long long>& prefixSums, long long nonFixedValue) {
    set<long long> uniqueSums;
    int n = prefixSums.size() - 1;

    for (int i = 0; i <= n; i++) {
        for (int j = i + 1; j <= n; j++) {
            uniqueSums.insert(prefixSums[j] - prefixSums[i]);
        }
    }

    if (nonFixedValue != 0) {
        set<long long> tempSums = uniqueSums;
        for (auto sum : tempSums) {
            uniqueSums.insert(sum + nonFixedValue);
        }
        uniqueSums.insert(nonFixedValue);
    }

    uniqueSums.insert(0);
    return uniqueSums;
}

void printResults(const set<long long>& results) {
    cout << results.size() << "\n";
    for (const auto& sum : results) {
        cout << sum << " ";
    }
    cout << "\n";
}

void solve(){
    int size;
    cin >> size;

    vector<int> numbers(size);
    for (int& num : numbers) {
        cin >> num;
    }

    int nonFixedIndex;
    long long nonFixedValue = calculateNonFixedSum(numbers, nonFixedIndex);
    vector<long long> prefixSums = computePrefixSums(numbers, nonFixedIndex);
    set<long long> uniqueResults = generateUniqueSums(prefixSums, nonFixedValue);
    printResults(uniqueResults);
}

int main()
{
    startt;
    ll t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
}

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

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

Think step by step.

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

1. **Edge Cases**: The code should handle arrays with a single element, arrays with all elements being 1 or -1, and arrays with a single non-1 or non--1 element.
2. **Boundary Conditions**: The code should handle the maximum and minimum possible values for the non-1 or non--1 element.
3. **Special Combinations**: The code should handle cases where the non-1 or non--1 element is in different positions in the array.

Given these considerations, a small-scale test input that can expose potential bugs in the code is:

```plaintext
5
1
0
2
1 2
3
1 -1 1
3
1 -1 -1000000000
3
1 -1 1000000000
```

Explanation of the test input:

1. **Test Case 1**: An array with a single element `0`. This tests the edge case of a single element array.
2. **Test Case 2**: An array with two elements, where one element is `1` and the other is `2`. This tests the case where the non-1 or non--1 element is present.
3. **Test Case 3**: An array with three elements, `1 -1 1`. This tests the case where the array contains only 1 and -1.
4. **Test Case 4**: An array with three elements, `1 -1 -1000000000`. This tests the boundary condition for the minimum possible value of the non-1 or non--1 element.
5. **Test Case 5**: An array with three elements, `1 -1 1000000000`. This tests the boundary condition for the maximum possible value of the non-1 or non--1 element.

These test cases should help in identifying any issues in the code, especially in handling edge cases and boundary conditions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums on Segments
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of $$$n$$$ integers, where all elements except for at most one are equal to $$$-1$$$ or $$$1$$$. The remaining element $$$x$$$ satisfies $$$-10^9 \le x \le 10^9$$$.

Find all possible sums of subarrays of $$$a$$$, including the empty subarray, whose sum is defined as $$$0$$$. In other words, find all integers $$$x$$$ such that the array $$$a$$$ has at least one subarray (possibly empty) with sum equal to $$$x$$$. A subarray is a contiguous subsegment of an array.

Output these sums in ascending order. Each sum should be printed only once, even if it is achieved by multiple subarrays.

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

Each test case consists of two lines:

- The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the size of the array.
- The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the elements of the array $$$a$$$. In the array $$$a$$$, there is at most one element that is neither $$$1$$$ nor $$$-1$$$.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output two lines:

- In the first line, print a single integer — the number of distinct subarray sums.
- In the second line, print these sums in ascending order.

Each sum should be printed only once, even if it is produced by multiple subarrays.

Examples:
input:
5
5
1 -1 10 1 1
5
-1 -1 -1 -1 -1
2
-1 2
2
7 1
3
1 4 -1
output:
8
-1 0 1 2 9 10 11 12
6
-5 -4 -3 -2 -1 0
4
-1 0 1 2
4
0 1 7 8
6
-1 0 1 3 4 5

Note: Let's define $$$a[i,j]$$$ as the subarray of $$$a$$$ from position $$$i$$$ to position $$$j$$$.

Consider the first test case of the example:

- $$$-1$$$ is produced by $$$a[2,2]$$$;
- $$$0$$$ is produced by the empty subarray;
- $$$1$$$ is produced by $$$a[4,4]$$$;
- $$$2$$$ is produced by $$$a[4,5]$$$;
- $$$9$$$ is produced by $$$a[2,3]$$$;
- $$$10$$$ is produced by $$$a[1,3]$$$;
- $$$11$$$ is produced by $$$a[3,4]$$$;
- $$$12$$$ is produced by $$$a[3,5]$$$.

Buggy Code:
for _ in range(int(input())):
    n=int(input())
    l=[x for x in map(int, input().split())]
    n1,p1,sigma=0,0,0
    maxp1,maxn1=0,0
    i=0

    while i<n:
        if l[i]==1:
            while i<n and l[i]==1:
                p1+=1
                i+=1
            else:
                maxp1=max(p1,maxp1)
                p1=0
        if i<n and l[i]==-1:
            while i<n and l[i]==-1:
                n1+=1
                i+=1
            else:
                maxn1=max(n1,maxn1)
                n1=0
                i-=1
        else:
            if i<n: sigma=l[i]
        i+=1

    l={0}

    if maxp1>0:
        for i in range(1,maxp1+1):
            l.add(i)

    if maxn1>0:
        for i in range(1,maxn1+1):
            l.add(-i)

    le=len(l)

    w=list(l)
    for i in range(le):
        l.add(w[i]+sigma)

    w=list(l)
    w.sort()
    print(len(w))
    print(*w)

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

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

Think step by step.

Response:
```plaintext
1
4
1 1 -1 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums on Segments
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of $$$n$$$ integers, where all elements except for at most one are equal to $$$-1$$$ or $$$1$$$. The remaining element $$$x$$$ satisfies $$$-10^9 \le x \le 10^9$$$.

Find all possible sums of subarrays of $$$a$$$, including the empty subarray, whose sum is defined as $$$0$$$. In other words, find all integers $$$x$$$ such that the array $$$a$$$ has at least one subarray (possibly empty) with sum equal to $$$x$$$. A subarray is a contiguous subsegment of an array.

Output these sums in ascending order. Each sum should be printed only once, even if it is achieved by multiple subarrays.

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

Each test case consists of two lines:

- The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the size of the array.
- The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the elements of the array $$$a$$$. In the array $$$a$$$, there is at most one element that is neither $$$1$$$ nor $$$-1$$$.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output two lines:

- In the first line, print a single integer — the number of distinct subarray sums.
- In the second line, print these sums in ascending order.

Each sum should be printed only once, even if it is produced by multiple subarrays.

Examples:
input:
5
5
1 -1 10 1 1
5
-1 -1 -1 -1 -1
2
-1 2
2
7 1
3
1 4 -1
output:
8
-1 0 1 2 9 10 11 12
6
-5 -4 -3 -2 -1 0
4
-1 0 1 2
4
0 1 7 8
6
-1 0 1 3 4 5

Note: Let's define $$$a[i,j]$$$ as the subarray of $$$a$$$ from position $$$i$$$ to position $$$j$$$.

Consider the first test case of the example:

- $$$-1$$$ is produced by $$$a[2,2]$$$;
- $$$0$$$ is produced by the empty subarray;
- $$$1$$$ is produced by $$$a[4,4]$$$;
- $$$2$$$ is produced by $$$a[4,5]$$$;
- $$$9$$$ is produced by $$$a[2,3]$$$;
- $$$10$$$ is produced by $$$a[1,3]$$$;
- $$$11$$$ is produced by $$$a[3,4]$$$;
- $$$12$$$ is produced by $$$a[3,5]$$$.

Buggy Code:
def merge_intervals(current_intervals):
            current_intervals.sort(key=lambda x: (x[0], x[1]))
            result=[]
            for interval in current_intervals:
                if not result:
                    result.append(interval)
                else:
                    if interval[0]<=result[-1][1]:
                        result[-1][1]=max(result[-1][1],interval[1])
                    else:
                        result.append(interval)
            return result
t=int(input())
while t:
    t-=1
    n=int(input())
    nums=list(map(int,input().split()))
    # n,m=map(int,input().split())
    # max_ans_without_x = 0
    # curSum = 0
    # n = len(nums)
    # for i in range(n):
    #     curSum += nums[i]
    #     ans = max(ans, curSum)
    #     if curSum < 0:
    #         curSum = 0
    other=False
    other_value=0
    other_index=0
    for i in range(n):
        if nums[i]!=1 and nums[i]!=-1:
            other=True
            other_value=nums[i]
            other_index=i
            break
    if not other:
        max_answer=0
        curSum=0
        for i in range(n):
            curSum += nums[i]
            max_answer= max(max_answer, curSum)
            if curSum < 0:
                curSum = 0
        min_answer=0
        curSum=0
        for i in range(n):
            curSum += nums[i]
            min_answer= min(min_answer, curSum)
            if curSum > 0:
                curSum = 0
        print(max_answer-min_answer+1)
        for i in range(min_answer,max_answer+1):
            print(i,end=' ')
        print()
    else:
        max_answer1=0
        max_answer2=0
        curSum=0
        for i in range(other_index):
            curSum += nums[i]
            max_answer1= max(max_answer1, curSum)
            if curSum < 0:
                curSum = 0
        curSum=0
        for i in range(other_index+1,n):
            curSum += nums[i]
            max_answer2= max(max_answer2, curSum)
            if curSum < 0:
                curSum = 0
        min_answer1=0
        min_answer2=0
        curSum=0
        for i in range(other_index):
            curSum += nums[i]
            min_answer1= min(min_answer1, curSum)
            if curSum > 0:
                curSum = 0
        curSum=0
        for i in range(other_index+1,n):
            curSum += nums[i]
            min_answer2= min(min_answer2, curSum)
            if curSum > 0:
                curSum = 0
        b=[nums[i] for i in range(other_index+1)]
        c=[nums[i] for i in range(other_index,n)]
        mini1=other_value
        mini2=other_value
        maxi1=other_value
        maxi2=other_value
        for i in range(other_index-1,-1,-1):
            b[i]=b[i]+b[i+1]
            mini1=min(mini1,b[i])
            maxi1=max(maxi1,b[i])
        for i in range(1,len(c)):
            c[i]=c[i]+c[i-1]
            mini2=min(mini2,c[i])
            maxi2=max(maxi2,c[i])

        interval1=[mini1,maxi1]
        interval2=[mini2,maxi2]
        interval3=[min_answer1,max_answer1]
        interval4=[min_answer2,max_answer2]
        intervals=[interval1,interval2,interval3,interval4]
        result=merge_intervals(intervals)
        total_numbers=0
        for i in result:
            total_numbers+=(i[1]-i[0]+1)
        print(total_numbers)
        for i in result:
            for j in range(i[0],i[1]+1):
                print(j,end=' ')
        print()

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

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

Think step by step.

Response:
```plaintext
5
5
1 -1 10 1 1
5
-1 -1 -1 -1 -1
2
-1 2
2
7 1
3
1 4 -1
```

This input includes a variety of scenarios to test the code:
1. A case with a large non-1, non-(-1) value (10) in the middle of the array.
2. A case with all elements being -1.
3. A case with a small non-1, non-(-1) value (2) at the end of the array.
4. A case with a large non-1, non-(-1) value (7) at the start of the array.
5. A case with a non-1, non-(-1) value (4) in the middle of the array, with a -1 at the end.

These scenarios should help expose any issues in the buggy code, particularly in handling the merging of intervals and the correct calculation of the range of possible subarray sums.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums on Segments
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of $$$n$$$ integers, where all elements except for at most one are equal to $$$-1$$$ or $$$1$$$. The remaining element $$$x$$$ satisfies $$$-10^9 \le x \le 10^9$$$.

Find all possible sums of subarrays of $$$a$$$, including the empty subarray, whose sum is defined as $$$0$$$. In other words, find all integers $$$x$$$ such that the array $$$a$$$ has at least one subarray (possibly empty) with sum equal to $$$x$$$. A subarray is a contiguous subsegment of an array.

Output these sums in ascending order. Each sum should be printed only once, even if it is achieved by multiple subarrays.

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

Each test case consists of two lines:

- The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the size of the array.
- The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the elements of the array $$$a$$$. In the array $$$a$$$, there is at most one element that is neither $$$1$$$ nor $$$-1$$$.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output two lines:

- In the first line, print a single integer — the number of distinct subarray sums.
- In the second line, print these sums in ascending order.

Each sum should be printed only once, even if it is produced by multiple subarrays.

Examples:
input:
5
5
1 -1 10 1 1
5
-1 -1 -1 -1 -1
2
-1 2
2
7 1
3
1 4 -1
output:
8
-1 0 1 2 9 10 11 12
6
-5 -4 -3 -2 -1 0
4
-1 0 1 2
4
0 1 7 8
6
-1 0 1 3 4 5

Note: Let's define $$$a[i,j]$$$ as the subarray of $$$a$$$ from position $$$i$$$ to position $$$j$$$.

Consider the first test case of the example:

- $$$-1$$$ is produced by $$$a[2,2]$$$;
- $$$0$$$ is produced by the empty subarray;
- $$$1$$$ is produced by $$$a[4,4]$$$;
- $$$2$$$ is produced by $$$a[4,5]$$$;
- $$$9$$$ is produced by $$$a[2,3]$$$;
- $$$10$$$ is produced by $$$a[1,3]$$$;
- $$$11$$$ is produced by $$$a[3,4]$$$;
- $$$12$$$ is produced by $$$a[3,5]$$$.

Buggy Code:
/*==========================================================================
    AUTHOR:    RonWonWon
    CREATED:   24.12.2024 20:00:14
==========================================================================*/
import java.io.*;
import java.util.*;

public class C {

    public static void main(String[] args) throws IOException {
        /*
        in.ini();
        PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("output.txt")));
        */
        long startTime = System.nanoTime();
        int t = in.nextInt(), T = 1;
        while(t-- > 0) {
        	int n = in.nextInt();
        	int a[] = in.readArray(n);
            Set<Long> set = new HashSet<>();
            set.add(0L);
			Set<Long> setL = new HashSet<>();
			Set<Long> setR = new HashSet<>();
			long sumL = 0, sumR = 0;
			long x = 0;
			int index = n;
			for(int i=0;i<n;i++){
				if(a[i]!=-1&&a[i]!=1){
					x = a[i]; index = i; continue;
				}
				if(index!=n){
					sumR += a[i];
					set.add(sumR);
					setR.add(sumR);
				}
			}
			for(int i=index-1;i>=0;i--){
				sumL += a[i];
				set.add(sumL);
				setL.add(sumL);
			}
			List<Long> left = new ArrayList<>();
			List<Long> right = new ArrayList<>();
			for(long i : setL) left.add(i);
			for(long i : setR) right.add(i);
			Collections.sort(left); Collections.sort(right);
			//for(long i : left) out.print(i+" ");
			//out.println();
			//for(long i:  right) out.print(i+" ");
			//out.println();
			long lX = 0, lY = 0;
			if(left.size()>0){
				lX = left.get(0);
				lY = left.get(left.size()-1);
			}
			long rX = 0, rY = 0;
			if(right.size()>0){
				rX = right.get(0);
				rY = right.get(right.size()-1);
			}
			//for(long i : set) out.print(i+" ");
			//out.println();
			for(int i=-n;i<=n;i++){
				if(i<0){
					if(lX+rX<=i||lX<=i||rX<=i) set.add(x+i);
				}
				else{
					if(lY+rY>=i||lY>=i||rY>=i) set.add(x+i);
				}
			}
            List<Long> ans = new ArrayList<>();
            for(long i : set) ans.add(i);
            Collections.sort(ans);
            out.println(ans.size());
            for(long i : ans) out.print(i+" ");
            out.println();
            //out.println("Case #"+T+": "+ans); T++;
        }
        /*
        startTime = System.nanoTime() - startTime;
        System.err.println("Time: "+(startTime/1000000));
        */
        out.flush();
    }

    static FastScanner in = new FastScanner();
    static PrintWriter out = new PrintWriter(System.out);
    static int oo = Integer.MAX_VALUE;
    static long ooo = Long.MAX_VALUE;

    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        void ini() throws FileNotFoundException {
            br = new BufferedReader(new FileReader("input.txt"));
        }

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

        String nextLine(){
            try{ return br.readLine(); }
            catch(IOException e) { } return "";
        }

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

        int[] readArray(int n) {
            int a[] = new int[n];
            for(int i=0;i<n;i++) a[i] = nextInt();
            return a;
        }

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

        long[] readArrayL(int n) {
            long a[] = new long[n];
            for(int i=0;i<n;i++) a[i] = nextLong();
            return a;
        }

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

        double[] readArrayD(int n) {
            double a[] = new double[n];
            for(int i=0;i<n;i++) a[i] = nextDouble();
            return a;
        }
    }

    static final Random random = new Random();

    static void ruffleSort(int[] a){
        int n = a.length;
        for(int i=0;i<n;i++){
            int j = random.nextInt(n), temp = a[j];
            a[j] = a[i]; a[i] = temp;
        }
        Arrays.sort(a);
    }

    static void ruffleSortL(long[] a){
        int n = a.length;
        for(int i=0;i<n;i++){
            int j = random.nextInt(n);
            long temp = a[j];
            a[j] = a[i]; a[i] = temp;
        }
        Arrays.sort(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 create a challenging test input that exposes the bug in the provided buggy code, let's consider the following aspects:

1. **Edge Cases**:
   - Arrays with only one element.
   - Arrays with all elements being either 1 or -1.
   - Arrays with a single element that is neither 1 nor -1, and this element is at different positions in the array (beginning, middle, end).

2. **Critical Scenarios**:
   - The element that is neither 1 nor -1 is a large number.
   - The element that is neither 1 nor -1 is a small number.
   - The array has a mix of 1, -1, and a large number in the middle.

3. **Maximizing Failure Probability**:
   - Use arrays where the sum of subarrays can produce a wide range of values, especially when the non-1/-1 element is involved.

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

```plaintext
3
1
1000000000
5
1 -1 1 -1 1000000000
5
-1 -1 -1 -1 1000000000
```

### Explanation:

1. **Test Case 1**:
   - **Array**: `[1000000000]`
   - **Expected Output**:
     - Number of distinct subarray sums: 2
     - Sums: `0 1000000000`
   - **Reason**: The array has only one element, which is a large number. The possible sums are 0 (empty subarray) and 1000000000.

2. **Test Case 2**:
   - **Array**: `[1, -1, 1, -1, 1000000000]`
   - **Expected Output**:
     - Number of distinct subarray sums: 11
     - Sums: `-1 0 1 2 999999999 1000000000 1000000001 1000000002 1000000003 1000000004 1000000005`
   - **Reason**: The array has a mix of 1, -1, and a large number in the middle. The possible sums include all combinations of sums from the segments before and after the large number, plus the large number itself.

3. **Test Case 3**:
   - **Array**: `[-1, -1, -1, -1, 1000000000]`
   - **Expected Output**:
     - Number of distinct subarray sums: 6
     - Sums: `-4 -3 -2 -1 0 1000000000 1000000001 1000000002 1000000003 1000000004 1000000005`
   - **Reason**: The array has a mix of -1 and a large number at the end. The possible sums include all combinations of sums from the segments before the large number, plus the large number itself.

This test input should help expose any bugs in the buggy code, especially in handling large numbers and ensuring all possible subarray sums are correctly identified and outputted.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums on Segments
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $$$a$$$ of $$$n$$$ integers, where all elements except for at most one are equal to $$$-1$$$ or $$$1$$$. The remaining element $$$x$$$ satisfies $$$-10^9 \le x \le 10^9$$$.

Find all possible sums of subarrays of $$$a$$$, including the empty subarray, whose sum is defined as $$$0$$$. In other words, find all integers $$$x$$$ such that the array $$$a$$$ has at least one subarray (possibly empty) with sum equal to $$$x$$$. A subarray is a contiguous subsegment of an array.

Output these sums in ascending order. Each sum should be printed only once, even if it is achieved by multiple subarrays.

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

Each test case consists of two lines:

- The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the size of the array.
- The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the elements of the array $$$a$$$. In the array $$$a$$$, there is at most one element that is neither $$$1$$$ nor $$$-1$$$.

Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, output two lines:

- In the first line, print a single integer — the number of distinct subarray sums.
- In the second line, print these sums in ascending order.

Each sum should be printed only once, even if it is produced by multiple subarrays.

Examples:
input:
5
5
1 -1 10 1 1
5
-1 -1 -1 -1 -1
2
-1 2
2
7 1
3
1 4 -1
output:
8
-1 0 1 2 9 10 11 12
6
-5 -4 -3 -2 -1 0
4
-1 0 1 2
4
0 1 7 8
6
-1 0 1 3 4 5

Note: Let's define $$$a[i,j]$$$ as the subarray of $$$a$$$ from position $$$i$$$ to position $$$j$$$.

Consider the first test case of the example:

- $$$-1$$$ is produced by $$$a[2,2]$$$;
- $$$0$$$ is produced by the empty subarray;
- $$$1$$$ is produced by $$$a[4,4]$$$;
- $$$2$$$ is produced by $$$a[4,5]$$$;
- $$$9$$$ is produced by $$$a[2,3]$$$;
- $$$10$$$ is produced by $$$a[1,3]$$$;
- $$$11$$$ is produced by $$$a[3,4]$$$;
- $$$12$$$ is produced by $$$a[3,5]$$$.

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

public class Main{
    static class BIT
    {
        int n;
        long[] bit,arr;
        public BIT(int n)
        {
            this.n = n;
            bit = new long[n+1];
            arr = new long[n+1];
        }
        void set(int idx,long val)
        {
            add(idx, val - arr[idx+1]);
        }
        public void add(int idx,long val)
        {
            idx++;
            arr[idx] += val;
            for(; idx <= n ; idx += (idx&-idx))
                bit[idx] += val;
        }
        public long sum(int k)
        {
            long s = 0;
            while(k >= 1)
            {
                s += bit[k];
                k -= (k&-k);
            }
            return s;
        }
    }
    static Integer[][] dp;
    static int solve(int i,int k,int n,int m,String a,String b,String c)
    {
        if(i == n && k == m)
            return 0;
        if(dp[i][k] != null)
            return dp[i][k];
        int c1=0,c2=0;
        int ans = Integer.MAX_VALUE;
        if(i < n)
        {
            if(c.charAt(k+i) != a.charAt(i))
                c1 = 1;
            ans = Math.min(ans,c1 + solve(i+1,k,n,m,a,b,c));
        }
        if(k < b.length())
        {
            if(c.charAt(k+i) != b.charAt(k))
                c2 = 1;
            ans = Math.min(ans,c2 + solve(i,k+1,n,m,a,b,c));
        }
        dp[i][k] = ans;
        return ans;
    }
    static long computeXor(int i,int len,String s)
    {
        long st = 1;
        long ans = 0;
        int j=s.length()-1,k=i+len-1;
        while(k >= i)
        {
            if(s.charAt(k) != s.charAt(j))
            {
                long  m = ((long)1 << st);
                ans += m;
            }
            k--;j--;
            st++;
        }
        return ans;
    }
    static boolean check(long y,long a,long b)
    {
        if(y <= a && y <= b || y <= b && y >= a)
            return true;
        return false;
    }
    public static void main(String[] args) throws IOException {

        FastReader scn = new FastReader();
        int t = scn.nextInt();
        while(t-- > 0) {
            int n = scn.nextInt();
            long[] a = new long[n];
            for(int i=0;i<n;i++)
                a[i] = scn.nextInt();
            int idx = 0;
            for(int i=0;i<n;i++)
            {
                if(a[i] != 1 && a[i] != -1)
                {
                    idx = i;
                }
            }
            TreeSet<Long> s = new TreeSet<>();
            s.add(0L);
            s.add(a[idx]);
            long ls = 0,rs=0;
            for(int i=idx-1;i>= 0;i--)
            {
                ls += a[i];
                s.add(ls);
                s.add(a[i]);
                s.add(ls+a[idx]);
            }
            for(int i=idx+1;i<n;i++)
            {
                rs += a[i];
                s.add(rs);
                s.add(a[i]);
                s.add(rs+a[idx]);
            }
            StringBuilder ans = new StringBuilder();
            ans.append(s.size()).append("\n");
            for(long v : s)
                ans.append(v).append(" ");
            System.out.println(ans);
        }
    }
    static long check(pair[] ar,int k)
    {
        long res = 0;
        int n = ar.length;
        Arrays.sort(ar,(a1,a2)->Long.compare(a1.a,a2.a));
        PriorityQueue<Long> pq = new PriorityQueue<>();
        for(int i=0;i<n;i++)
        {
            while(pq.size() > 0 && pq.peek() < ar[i].a) {
                pq.poll();
            }
            if(pq.size() <= k)
            {
                res = Math.max(res,(long)(n-i+pq.size()) * ar[i].a);
            }
            pq.add(ar[i].b);
            while(i+1 < n && ar[i].a == ar[i+1].a)
            {
                pq.add(ar[i+1].b);
                i++;
            }
        }
        return res;
    }

    static class pair implements Comparable<pair>
    {
        long a,b;
        public pair(long a,long b)
        {
            this.a = a;
            this.b = b;
        }
        public int compareTo(pair o)
        {
            return Long.compare(this.b,o.b);
        }
    }
    static int bs(List<Integer> a,int val)
    {
        int ans = a.size();
        int lo=0,hi=a.size()-1;
        while(lo <= hi)
        {
            int mid = lo+(hi-lo)/2;
            if(a.get(mid) >= val)
            {
                ans = mid;
                hi = mid-1;
            }
            else {
                lo=mid+1;
            }
        }
        return a.size() - ans;
    }
    static long bss(long[] a,int lo,int hi,long val,long v,long s)
    {
        long ans = -1;
        while(lo <= hi)
        {
            int mid = lo+(hi-lo)/2;
            if((s - v - a[mid]) <= val)
            {
                ans = mid;
                hi = mid-1;
            }
            else
                lo=mid+1;
        }
        return ans;
    }

    static double dfs(int src,int par,int len,double prob,List<Integer>[] t)
    {
        boolean f = false;
        double ans = 0;
        for(int nbr : t[src])
        {
            if(nbr != par)
            {
                f = true;
                ans += dfs(nbr,src,len+1,prob/(t[src].size()),t);
            }
        }
        if(!f)
            return prob * len;
        return ans;
    }
    static void fill(int i,int j,int n,int m,String[] a,int[][] fi,int[] cc,int co)
    {
        if(i < 0 || j < 0 || i >= n || j >= m || a[i].charAt(j)=='*' || fi[i][j] == co)
            return;
        ++cc[co];
        fi[i][j] = co;
        fill(i-1,j,n,m,a,fi,cc,co);
        fill(i,j-1,n,m,a,fi,cc,co);
        fill(i+1,j,n,m,a,fi,cc,co);
        fill(i,j+1,n,m,a,fi,cc,co);
    }

    static long gcd(long a,long b)
    {
        while (b != 0) {
            long temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }
    static void dfs(int u,int gr,List<Integer>[][] g,boolean[] vis)
    {
        vis[u] = true;
        for(int v : g[u][gr])
            if(!vis[v])
                dfs(v,gr,g,vis);
    }
}
class FastReader {
    BufferedReader br;

    public FastReader() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }

    int nextInt() throws IOException {
        int num = 0;
        boolean isNeg = false;
        char c;

        do {
            c = (char) br.read();
        } while (!Character.isDigit(c) && c != '-');

        if (c == '-') {
            isNeg = true;
            c = (char) br.read();
        }

        do {
            num = num * 10 + c - '0';
            c = (char) br.read();
        } while (Character.isDigit(c));

        if (isNeg) {
            num = -num;
        }

        return num;
    }
    long nextLong() throws IOException {
        long num = 0;
        boolean isNeg = false;
        char c;

        do {
            c = (char) br.read();
        } while (!Character.isDigit(c) && c != '-');

        if (c == '-') {
            isNeg = true;
            c = (char) br.read();
        }

        do {
            num = num * 10 + c - '0';
            c = (char) br.read();
        } while (Character.isDigit(c));

        if (isNeg) {
            num = -num;
        }

        return num;
    }
    String next() throws IOException {
        StringBuilder sb = new StringBuilder();
        char c;

        // Skip leading whitespace
        do {
            c = (char) br.read();
        } while (Character.isWhitespace(c));

        // Read characters until next whitespace
        do {
            sb.append(c);
            c = (char) br.read();
        } while (!Character.isWhitespace(c));

        return sb.toString();
    }

    String nextLine() throws IOException {
        return br.readLine();
    }
}
//
//for(int i=0;i<n;i++)
//        {
//long av = ar[i].a;
//long bv = ar[i].b;
//                if(!ind.containsKey(bv))
//        ind.put(bv,i);
//List<Integer> tli = list.getOrDefault(av,new ArrayList<>());
//                tli.add(i);
//                list.put(av,tli);
//                co.put(bv,co.getOrDefault(bv,0)+1);
//        }
//long count = 0;
//            for(long key : co.keySet())
//        {
//count += co.get(key);
//int idx = ind.get(key);
//long kco = 0;
//                for(long lkey : list.keySet())
//        {
//        if(lkey >= key)
//        break;
//List<Integer> l = list.get(lkey);
//kco += bs(l,idx);
//                    if(kco > k)
//        break;
//        }
//        if(kco <= k)
//        {
//res = Math.max(res, count * key);
//                }
//                        }

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

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

Think step by step.

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

1. **Array with a single element that is neither 1 nor -1:**
   - This can test the handling of a single unique element.

2. **Array with a large unique element:**
   - This can test the handling of large values and their impact on the subarray sums.

3. **Array with a unique element at the beginning, middle, and end:**
   - This can test the handling of the unique element in different positions.

4. **Array with all 1s or all -1s:**
   - This can test the handling of uniform arrays.

5. **Array with a mix of 1s, -1s, and a unique element:**
   - This can test the handling of mixed arrays.

Given the constraints and the need for a small-scale test input, let's create a test case that includes a unique element in different positions and a mix of 1s and -1s.

### Chosen Test Input
- **Array with a unique element at the beginning, middle, and end:**
  - This will test the handling of the unique element in different positions and the calculation of subarray sums.

### Test Input
```plaintext
3
5
10 1 -1 1 -1
5
1 -1 10 1 -1
5
1 -1 -1 10 1
```

### Explanation
- **Test Case 1:**
  - Array: `[10, 1, -1, 1, -1]`
  - The unique element `10` is at the beginning.
- **Test Case 2:**
  - Array: `[1, -1, 10, 1, -1]`
  - The unique element `10` is in the middle.
- **Test Case 3:**
  - Array: `[1, -1, -1, 10, 1]`
  - The unique element `10` is at the end.

This test input is designed to challenge the code by placing the unique element in different positions and including a mix of 1s and -1s. This should help expose any issues in the handling of subarray sums.