**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums of Segments
time_limit_ms: 4000
memory_limit_mb: 512
Description: You are given a sequence of integers $[a_1, a_2, \dots, a_n]$. Let $s(l,r)$ be the sum of elements from $a_l$ to $a_r$ (i. e. $s(l,r) = \sum\limits_{i=l}^{r} a_i$).

Let's construct another sequence $b$ of size $\frac{n(n+1)}{2}$ as follows: $b = [s(1,1), s(1,2), \dots, s(1,n), s(2,2), s(2,3), \dots, s(2,n), s(3,3), \dots, s(n,n)]$.

For example, if $a = [1, 2, 5, 10]$, then $b = [1, 3, 8, 18, 2, 7, 17, 5, 15, 10]$.

You are given $q$ queries. During the $i$-th query, you are given two integers $l_i$ and $r_i$, and you have to calculate $\sum \limits_{j=l_i}^{r_i} b_j$.

Input: The first line contains one integer $n$ ($1 \le n \le 3 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-10 \le a_i \le 10$).

The third line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$).

Then $q$ lines follow, the $i$-th of them contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le \frac{n(n+1)}{2}$).

Output: Print $q$ integers, the $i$-th of which should be equal to $\sum \limits_{j=l_i}^{r_i} b_j$.

Examples:
input:
4
1 2 5 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
output:
1
4
12
30
32
86
56
54
60
26
82
57
9
2
61

Note:

Buggy Code:
def main():

    n = 4
    a = [1, 2, 5, 10]
    q = 15
    queries = [
        (1, 1),
        (1, 2),
        (1, 3),
        (1, 4),
        (1, 5),
        (1, 10),
        (5, 10),
        (6, 10),
        (2, 8),
        (3, 4),
        (3, 10),
        (3, 8),
        (5, 6),
        (5, 5),
        (1, 8)
    ]

    prefix_a = [0] * (n + 1)
    for i in range(1, n + 1):
        prefix_a[i] = prefix_a[i - 1] + a[i - 1]

    b = []
    for l in range(1, n + 1):
        for r in range(l, n + 1):
            b.append(prefix_a[r] - prefix_a[l - 1])

    m = len(b)
    prefix_b = [0] * (m + 1)
    for i in range(1, m + 1):
        prefix_b[i] = prefix_b[i - 1] + b[i - 1]

    result = []
    for l, r in queries:
        result.append(prefix_b[r] - prefix_b[l - 1])

    print('\n'.join(map(str, result)))

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that could potentially cause issues. Here are the steps to consider:

1. **Edge Cases for Sequence Length:**
   - Test with the minimum and maximum allowed sequence lengths (1 and 300,000).
   - Test with sequences of length 2 and 3 to check boundary conditions.

2. **Edge Cases for Element Values:**
   - Test with sequences containing all positive, all negative, and mixed values.
   - Test with sequences containing the minimum and maximum allowed values (-10 and 10).

3. **Edge Cases for Queries:**
   - Test with queries that span the entire range of the sequence.
   - Test with queries that span the first and last few elements.
   - Test with queries that span the middle of the sequence.

4. **Special Cases:**
   - Test with a sequence of all zeros.
   - Test with a sequence where all elements are the same.

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

```plaintext
4
-10 0 10 -10
5
1 1
1 6
3 6
4 4
1 10
```

Explanation:
- **Sequence Length:** 4 (small but non-trivial)
- **Element Values:** Includes the minimum, maximum, and zero values
- **Queries:**
  - `(1, 1)`: Single element
  - `(1, 6)`: Spanning the first half of the sequence
  - `(3, 6)`: Spanning the middle of the sequence
  - `(4, 4)`: Single element
  - `(1, 10)`: Spanning the entire sequence

This test input should help expose any issues in the buggy code, particularly with handling edge cases and large ranges.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums of Segments
time_limit_ms: 4000
memory_limit_mb: 512
Description: You are given a sequence of integers $[a_1, a_2, \dots, a_n]$. Let $s(l,r)$ be the sum of elements from $a_l$ to $a_r$ (i. e. $s(l,r) = \sum\limits_{i=l}^{r} a_i$).

Let's construct another sequence $b$ of size $\frac{n(n+1)}{2}$ as follows: $b = [s(1,1), s(1,2), \dots, s(1,n), s(2,2), s(2,3), \dots, s(2,n), s(3,3), \dots, s(n,n)]$.

For example, if $a = [1, 2, 5, 10]$, then $b = [1, 3, 8, 18, 2, 7, 17, 5, 15, 10]$.

You are given $q$ queries. During the $i$-th query, you are given two integers $l_i$ and $r_i$, and you have to calculate $\sum \limits_{j=l_i}^{r_i} b_j$.

Input: The first line contains one integer $n$ ($1 \le n \le 3 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-10 \le a_i \le 10$).

The third line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$).

Then $q$ lines follow, the $i$-th of them contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le \frac{n(n+1)}{2}$).

Output: Print $q$ integers, the $i$-th of which should be equal to $\sum \limits_{j=l_i}^{r_i} b_j$.

Examples:
input:
4
1 2 5 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
output:
1
4
12
30
32
86
56
54
60
26
82
57
9
2
61

Note:

Buggy Code:
n = 4
a = [1, 2, 5, 10]
q = 15
queries = [
    (1, 1),
    (1, 2),
    (1, 3),
    (1, 4),
    (1, 5),
    (1, 10),
    (5, 10),
    (6, 10),
    (2, 8),
    (3, 4),
    (3, 10),
    (3, 8),
    (5, 6),
    (5, 5),
    (1, 8)
]

# Step 1: Calculate the prefix sums of `a`
prefix_a = [0] * (n + 1)
for i in range(1, n + 1):
    prefix_a[i] = prefix_a[i - 1] + a[i - 1]

# Step 2: Construct the sequence `b`
b = []
for l in range(1, n + 1):
    for r in range(l, n + 1):
        b.append(prefix_a[r] - prefix_a[l - 1])

# Step 3: Calculate the prefix sums of `b`
m = len(b)
prefix_b = [0] * (m + 1)
for i in range(1, m + 1):
    prefix_b[i] = prefix_b[i - 1] + b[i - 1]

# Step 4: Process the queries
result = []
for l, r in queries:
    result.append(prefix_b[r] - prefix_b[l - 1])

# Output the results
print('\n'.join(map(str, result)))

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

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

Think step by step.

Response:
```plaintext
4
1 2 5 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums of Segments
time_limit_ms: 4000
memory_limit_mb: 512
Description: You are given a sequence of integers $[a_1, a_2, \dots, a_n]$. Let $s(l,r)$ be the sum of elements from $a_l$ to $a_r$ (i. e. $s(l,r) = \sum\limits_{i=l}^{r} a_i$).

Let's construct another sequence $b$ of size $\frac{n(n+1)}{2}$ as follows: $b = [s(1,1), s(1,2), \dots, s(1,n), s(2,2), s(2,3), \dots, s(2,n), s(3,3), \dots, s(n,n)]$.

For example, if $a = [1, 2, 5, 10]$, then $b = [1, 3, 8, 18, 2, 7, 17, 5, 15, 10]$.

You are given $q$ queries. During the $i$-th query, you are given two integers $l_i$ and $r_i$, and you have to calculate $\sum \limits_{j=l_i}^{r_i} b_j$.

Input: The first line contains one integer $n$ ($1 \le n \le 3 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-10 \le a_i \le 10$).

The third line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$).

Then $q$ lines follow, the $i$-th of them contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le \frac{n(n+1)}{2}$).

Output: Print $q$ integers, the $i$-th of which should be equal to $\sum \limits_{j=l_i}^{r_i} b_j$.

Examples:
input:
4
1 2 5 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
output:
1
4
12
30
32
86
56
54
60
26
82
57
9
2
61

Note:

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

typedef unsigned long long ull;

int main() {
    ull n, q;
    cin >> n;

    vector<ull> a(n);
    for (ull i = 0; i < n; ++i) {
        cin >> a[i];
    }

    vector<ull> b;
    for (ull i = 0; i < n; ++i) {
        ull sum = 0;
        for (ull j = i; j < n; ++j) {
            sum += a[j];
            b.push_back(sum);
        }
    }

    ull m = b.size();
    vector<ull> pb(m + 1, 0);
    for (ull i = 0; i < m; ++i) {
        pb[i + 1] = pb[i] + b[i];
    }

    cin >> q;
    for (ull i = 0; i < q; ++i) {
        ull li, ri;
        cin >> li >> ri;

        li--, ri--;
        cout << pb[ri + 1] - pb[li] << endl;
    }

    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases and scenarios that can maximize the failure probability. The primary issue in the code is the use of `unsigned long long` (ULL) for variables that can potentially be negative, such as the elements of the array `a`.

Here are the steps to generate a challenging test input:

1. **Choose a small `n` to keep the test input manageable.**
2. **Include negative values in the array `a` to test the handling of negative sums.**
3. **Include a query that spans the entire range of `b` to ensure the prefix sum calculation is correct.**
4. **Include a query that spans a small range to test the correctness of the prefix sum differences.**

Let's construct a test input based on these criteria:

```plaintext
4
-10 -5 0 5
3
1 6
2 5
3 4
```

Explanation:
- `n = 4` to keep the input small.
- `a = [-10, -5, 0, 5]` includes negative values to test the sum calculations.
- `q = 3` to have multiple queries.
- The queries are:
  - `1 6`: This query spans the first 6 elements of `b`.
  - `2 5`: This query spans a smaller range to test the prefix sum differences.
  - `3 4`: This query spans a very small range to test the individual elements.

This test input will help expose any issues with handling negative values and the correctness of the prefix sum calculations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums of Segments
time_limit_ms: 4000
memory_limit_mb: 512
Description: You are given a sequence of integers $[a_1, a_2, \dots, a_n]$. Let $s(l,r)$ be the sum of elements from $a_l$ to $a_r$ (i. e. $s(l,r) = \sum\limits_{i=l}^{r} a_i$).

Let's construct another sequence $b$ of size $\frac{n(n+1)}{2}$ as follows: $b = [s(1,1), s(1,2), \dots, s(1,n), s(2,2), s(2,3), \dots, s(2,n), s(3,3), \dots, s(n,n)]$.

For example, if $a = [1, 2, 5, 10]$, then $b = [1, 3, 8, 18, 2, 7, 17, 5, 15, 10]$.

You are given $q$ queries. During the $i$-th query, you are given two integers $l_i$ and $r_i$, and you have to calculate $\sum \limits_{j=l_i}^{r_i} b_j$.

Input: The first line contains one integer $n$ ($1 \le n \le 3 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-10 \le a_i \le 10$).

The third line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$).

Then $q$ lines follow, the $i$-th of them contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le \frac{n(n+1)}{2}$).

Output: Print $q$ integers, the $i$-th of which should be equal to $\sum \limits_{j=l_i}^{r_i} b_j$.

Examples:
input:
4
1 2 5 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
output:
1
4
12
30
32
86
56
54
60
26
82
57
9
2
61

Note:

Buggy Code:
//package com.aakash.Contest.E171;

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

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

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

            StringTokenizer s2
                    = new StringTokenizer(br.readLine());
            int n = Integer.parseInt(s2.nextToken());
            if(n==100) {
                long[] arr = new long[n+1];
                StringTokenizer s3
                        = new StringTokenizer(br.readLine());
                for (int i = 1; i <=n; i++) arr[i] = Integer.parseInt(s3.nextToken());
                int i=1;
                while(i<=n) {
                    int c=0;
                    while(c<20) {
                        System.out.print(arr[i++]+" ");
                        c++;
                    }
                }
            } else {
                long[] arr = new long[n+1];
                StringTokenizer s3
                        = new StringTokenizer(br.readLine());
                for (int i = 1; i <=n; i++) arr[i] = Integer.parseInt(s3.nextToken());
                StringTokenizer s4
                        = new StringTokenizer(br.readLine());
                int q = Integer.parseInt(s4.nextToken());
                long[][] que=new long[q][2];
                for(int i=0;i<q;i++) {
                    StringTokenizer s5
                            = new StringTokenizer(br.readLine());
                    long a=Integer.parseInt(s5.nextToken());
                    long b=Integer.parseInt(s5.nextToken());
                    que[i][0]=a;
                    que[i][1]=b;
                }
                solve(arr,que,n,q);
            }

    }
    static long[] res;
    static long[] sum;
    static long[] indsum;
    static long[] fisum;
    static long[] bisum;
    static long[] bsum;
    static long[] tsum;
    static long total;
    static void solve(long[] arr,long[][] que, int n,int q) {
        res=new long[q];
        sum=new long[n+2];
        tsum=new long[n+2];
        fisum=new long[n+2];
        bisum=new long[n+2];
        bsum=new long[n+2];
        indsum=new long[n+2];
        fill(arr,n);
        int i=0;
        for(long[] temp:que) {
            long L=temp[0],R=temp[1];
            int s=bss(L,n),e=bse(R,n);
           // s=Math.min(s,e);
            long mp=mp(n,s,e);
            long fp=fp(n,s-1,L-1);
            if(mp==0) fp=0;
            long bp=fp(n,e,R);
            long curr=mp-fp+bp;
            res[i++]=curr;
            //System.out.println(s+" "+e);
            //System.out.println(mp+" "+fp+" "+bp);
        }
        for(i=0;i<q;i++) System.out.println(res[i]+" ");
    }
    static long fp(int n,int s,long L) {

        long prev=n-s;
        long curr=(long)n*(n+1)/2;
        curr-=prev*(prev+1)/2;
        long diff=L-curr;
        int a=s+1,b=s+(int)diff;
        return bisum(bsum,sum,n,a,b);
    }

    static long mp(int n,int s,int e) {
        if(e<s) return 0;
        long range=e-s+1;
        long first=range*total;
        long rem1=indsum[s-1]*range;
        long rem2=bisum(bisum,indsum,n,s,e-1);
        return first-rem1-rem2;
    }
    static int bse(long p,int n) {
        int s=1,e=n;
        while(s<=e) {
            int m=s+(e-s)/2;
            long prev=n-m;
            long curr=(long)n*(n+1)/2;
            curr-=prev*(prev+1)/2;
            if(curr<=p) s=m+1;
            else e=m-1;
        }
        return e;
    }
    static int bss(long p,int n) {
        int s=1,e=n;
        while(s<=e) {
            int m=s+(e-s)/2;
            long prev=n-m;
            long curr=(long)n*(n+1)/2;
            curr-=prev*(prev+1)/2;
            if(curr<p) s=m+1;
            else e=m-1;
        }
        return s;
    }
    static void fill(long[] arr,int n) {
        long add=0;
        for(int i=1;i<=n;i++) {
            add+=arr[i];
            sum[i]=add;
        }
        add=0;
        for(int i=1;i<=n;i++) {
            int t=n-i+1;
            long curr=arr[i]*t*t;
            add+=curr;
            fisum[i]=add;
        }
        for(int i=n;i>=1;i--) {
            int t=n-i+1;
            long curr=arr[i]*t*t;
            add+=curr;
            bisum[i]=add;
        }
        total=0;
        for(int i=1;i<=n;i++) {
            int t=n-i+1;
            total+=arr[i]*t;
            indsum[i]=total;
        }

        add=0;
        for(int i=n;i>=1;i--) {
            int t=n-i+1;
            long curr=arr[i]*t;
            add+=curr;
            bsum[i]=add;
        }
    }
    static long bisum(long[] indarr,long[] sumarr,int n,int s,int e)  {
        if(e<s||s==0) return 0;
        long res=indarr[s]-indarr[e+1];
        long sum=sumarr[e]-sumarr[s-1];
        long rem=sum*(n-e);
        return res-rem;
    }
    static long fisum(long[] indarr,long[] sumarr,int s,int e) {
        if(e<s) return 0;
        long res=indarr[e]-indarr[s-1];
        long sum=sumarr[e]-sumarr[s-1];
        long rem=sum*(s-1);
        return res-rem;
    }
}

**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
100
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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 of Segments
time_limit_ms: 4000
memory_limit_mb: 512
Description: You are given a sequence of integers $[a_1, a_2, \dots, a_n]$. Let $s(l,r)$ be the sum of elements from $a_l$ to $a_r$ (i. e. $s(l,r) = \sum\limits_{i=l}^{r} a_i$).

Let's construct another sequence $b$ of size $\frac{n(n+1)}{2}$ as follows: $b = [s(1,1), s(1,2), \dots, s(1,n), s(2,2), s(2,3), \dots, s(2,n), s(3,3), \dots, s(n,n)]$.

For example, if $a = [1, 2, 5, 10]$, then $b = [1, 3, 8, 18, 2, 7, 17, 5, 15, 10]$.

You are given $q$ queries. During the $i$-th query, you are given two integers $l_i$ and $r_i$, and you have to calculate $\sum \limits_{j=l_i}^{r_i} b_j$.

Input: The first line contains one integer $n$ ($1 \le n \le 3 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-10 \le a_i \le 10$).

The third line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$).

Then $q$ lines follow, the $i$-th of them contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le \frac{n(n+1)}{2}$).

Output: Print $q$ integers, the $i$-th of which should be equal to $\sum \limits_{j=l_i}^{r_i} b_j$.

Examples:
input:
4
1 2 5 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
output:
1
4
12
30
32
86
56
54
60
26
82
57
9
2
61

Note:

Buggy Code:
import java.util.*;

public class Codechef {
    static int MOD = 1000000007;

    public static void main(String[] args) {
        Scanner ab = new Scanner(System.in);
        int T = 1;
        while (T-- > 0) {
            int N = ab.nextInt();
            int[] arr = new int[N+1];
            for(int i=1; i<=N; i++) arr[i] = ab.nextInt();

            int[][] dp = new int[N+1][N+1];
            for(int i=1; i<=N; i++) {
                int sum = 0;
                for(int j=i; j<=N; j++) {
                    sum += arr[j];
                    dp[i][j] = sum;
                }
            }

            int[] cache = new int[N+1];
            for(int i=1; i<=N; i++) {
                int sum = 0;
                for(int j=i; j<=N; j++) sum += dp[i][j];
                cache[i] = sum;
            }

            int Q = ab.nextInt();
            while(Q-- > 0) {
                int l = ab.nextInt();
                int r = ab.nextInt();
                int x = seg(l, N);
                int y = seg(r, N);
                int ans = 0;
                int a = l - f(x, N);
                int b = r - f(y, N);
                if(x != y) {
//                    System.out.print("no: ");
                    for(int i=x+1; i<y; i++) ans += cache[i];
                    for(int i=a; i<=N+1-x; i++) ans += dp[x][i+x-1];
                    for(int i=1; i<=b; i++) ans += dp[y][i+y-1];
                } else {
//                    System.out.print("yes: ");
                    for(int i=a; i<=b; i++) ans += dp[x][i+x-1];
                }
                System.out.println(ans);
            }
        }
    }
    static int seg(int i, int N) {
        int low = 1;
        int high = N;
        int ans = N;
        while(low <= high) {
            int mid = (low + high) / 2;
            if(f(mid, N) <= i) {
                ans = mid;
                low = mid + 1;
            } else high = mid - 1;
        }
        return ans;
    }
    static int f(int seg, int N) {
        return (seg - 1)*N - (seg - 1)*(seg - 2) / 2;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. The primary issues in the code seem to be related to the segmentation and the calculation of the sums for the segments. Here are some steps to consider:

1. **Edge Cases for Small `n`**: Test with small values of `n` to ensure the code handles the base cases correctly.
2. **Edge Cases for Large `n`**: Test with large values of `n` to ensure the code handles large inputs efficiently.
3. **Queries Spanning Multiple Segments**: Ensure the code can handle queries that span multiple segments correctly.
4. **Negative and Positive Values**: Include both negative and positive values in the array to test the sum calculations.

Given the constraints and the need for a small-scale test input, we can focus on a scenario where `n` is small but the queries are designed to expose potential issues in the segment calculations.

### Challenging Test Input
```plaintext
4
1 2 5 10
5
1 1
1 2
1 10
5 10
1 8
```

### Explanation
- **`n = 4`**: A small value to keep the input manageable.
- **Array `[1, 2, 5, 10]`**: A simple array with both small and large values.
- **Queries**:
  - **`1 1`**: Single element sum, should return `1`.
  - **`1 2`**: Sum of the first two elements, should return `3`.
  - **`1 10`**: Sum of all elements in `b`, should return `86`.
  - **`5 10`**: Sum of the last six elements in `b`, should return `56`.
  - **`1 8`**: Sum of the first eight elements in `b`, should return `61`.

This input is designed to test the code's ability to handle single-element sums, sums of multiple elements, and sums spanning multiple segments. It also includes a mix of small and large values to ensure the sum calculations are correct.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums of Segments
time_limit_ms: 4000
memory_limit_mb: 512
Description: You are given a sequence of integers $[a_1, a_2, \dots, a_n]$. Let $s(l,r)$ be the sum of elements from $a_l$ to $a_r$ (i. e. $s(l,r) = \sum\limits_{i=l}^{r} a_i$).

Let's construct another sequence $b$ of size $\frac{n(n+1)}{2}$ as follows: $b = [s(1,1), s(1,2), \dots, s(1,n), s(2,2), s(2,3), \dots, s(2,n), s(3,3), \dots, s(n,n)]$.

For example, if $a = [1, 2, 5, 10]$, then $b = [1, 3, 8, 18, 2, 7, 17, 5, 15, 10]$.

You are given $q$ queries. During the $i$-th query, you are given two integers $l_i$ and $r_i$, and you have to calculate $\sum \limits_{j=l_i}^{r_i} b_j$.

Input: The first line contains one integer $n$ ($1 \le n \le 3 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-10 \le a_i \le 10$).

The third line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$).

Then $q$ lines follow, the $i$-th of them contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le \frac{n(n+1)}{2}$).

Output: Print $q$ integers, the $i$-th of which should be equal to $\sum \limits_{j=l_i}^{r_i} b_j$.

Examples:
input:
4
1 2 5 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
output:
1
4
12
30
32
86
56
54
60
26
82
57
9
2
61

Note:

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

long long get_b_sum(int l, int r, const vector<long long>& prefix) {
    long long sum = 0;
    for (int i = l; i <= r; ++i) {
        sum += prefix[r] - prefix[i - 1];
    }
    return sum;
}

int main() {
    int n;
    cin >> n;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
    }

    vector<long long> prefix(n + 1);
    for (int i = 1; i <= n; ++i) {
        prefix[i] = prefix[i - 1] + a[i];
    }

    vector<long long> bprefix((n * (n + 1)) / 2 + 1);
    int ind = 1;

    for (int l = 1; l <= n; ++l) {
        for (int r = l; r <= n; ++r) {
            bprefix[ind] = prefix[r] - prefix[l - 1];
            ind++;
        }
    }

    int q;
    cin >> q;
    for (int i = 0; i < q; ++i) {
        int li, ri;
        cin >> li >> ri;

        long long sum = 0;
        for (int j = li; j <= ri; ++j) {
            sum += bprefix[j];
        }

        cout << sum << '\n';
    }

    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on scenarios where the code might fail. The primary issue in the code is in the `get_b_sum` function, which incorrectly calculates the sum of the segments. However, the main loop in the `main` function also has an inefficiency that can be exploited.

Here are some key points to consider for generating a test input:
1. **Edge Cases**: Small values of `n` and `q` to ensure the test is manageable.
2. **Specific Queries**: Queries that span large ranges in the `b` array to ensure the sum calculation is thoroughly tested.
3. **Negative and Positive Values**: Mix of negative and positive values in the input array to test the sum calculation with different sign combinations.

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

```plaintext
4
1 -2 3 -4
3
1 10
5 10
3 8
```

### Explanation:
- **n = 4**: A small value to keep the input manageable.
- **a = [1, -2, 3, -4]**: A mix of positive and negative values to test the sum calculation.
- **q = 3**: A small number of queries to ensure the test is concise.
- **Queries**:
  - **1 10**: This query spans the entire `b` array, ensuring the sum of all segments is calculated.
  - **5 10**: This query spans a middle portion of the `b` array.
  - **3 8**: This query spans another portion of the `b` array to test different segments.

This test input should help expose any issues in the sum calculation and the efficiency of the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums of Segments
time_limit_ms: 4000
memory_limit_mb: 512
Description: You are given a sequence of integers $[a_1, a_2, \dots, a_n]$. Let $s(l,r)$ be the sum of elements from $a_l$ to $a_r$ (i. e. $s(l,r) = \sum\limits_{i=l}^{r} a_i$).

Let's construct another sequence $b$ of size $\frac{n(n+1)}{2}$ as follows: $b = [s(1,1), s(1,2), \dots, s(1,n), s(2,2), s(2,3), \dots, s(2,n), s(3,3), \dots, s(n,n)]$.

For example, if $a = [1, 2, 5, 10]$, then $b = [1, 3, 8, 18, 2, 7, 17, 5, 15, 10]$.

You are given $q$ queries. During the $i$-th query, you are given two integers $l_i$ and $r_i$, and you have to calculate $\sum \limits_{j=l_i}^{r_i} b_j$.

Input: The first line contains one integer $n$ ($1 \le n \le 3 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-10 \le a_i \le 10$).

The third line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$).

Then $q$ lines follow, the $i$-th of them contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le \frac{n(n+1)}{2}$).

Output: Print $q$ integers, the $i$-th of which should be equal to $\sum \limits_{j=l_i}^{r_i} b_j$.

Examples:
input:
4
1 2 5 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
output:
1
4
12
30
32
86
56
54
60
26
82
57
9
2
61

Note:

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

vector<long long> computePrefixSums(const vector<int>& a) {
    int n = a.size();
    vector<long long> prefixSum(n + 1, 0);
    for (int i = 1; i <= n; ++i) {
        prefixSum[i] = prefixSum[i - 1] + a[i - 1];
    }
    return prefixSum;
}

vector<long long> handleQueries(const vector<long long>& prefixSum, int n, const vector<pair<int, int>>& queries) {
    vector<long long> results;
    for (const auto& query : queries) {
        long long l = query.first;
        long long r = query.second;
        long long sum = 0;

        for (int i = 1; i <= n; ++i) {
            for (int j = i; j <= n; ++j) {
                long long idx = (i - 1) * n - (i - 1) * (i - 2) / 2 + j - i + 1;
                if (idx >= l && idx <= r) {
                    sum += prefixSum[j] - prefixSum[i - 1];
                }
            }
        }

        results.push_back(sum);
    }
    return results;
}

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

    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
    }

    vector<long long> prefixSum = computePrefixSums(a);

    int q;
    cin >> q;
    vector<pair<int, int>> queries(q);
    for (int i = 0; i < q; ++i) {
        int li, ri;
        cin >> li >> ri;
        queries[i] = {li, ri};
    }

    vector<long long> results = handleQueries(prefixSum, n, queries);
    for (const auto& result : results) {
        cout << result << "\n";
    }

    return 0;
}

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

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

Think step by step.

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

1. **Edge Cases**: Small values of `n` and `q` to ensure the test case is small-scale.
2. **Complex Queries**: Queries that span a wide range of indices in the `b` array to ensure the code handles them correctly.
3. **Negative and Positive Values**: Including both negative and positive values in the array `a` to test the sum calculations.

Let's construct a test case with `n = 4` and `q = 5` to ensure it is small-scale and complete.

### Test Input
```plaintext
4
1 -2 3 -4
5
1 1
1 10
5 10
3 8
2 8
```

### Explanation
- **Array `a`**: `[1, -2, 3, -4]`
  - This array includes both positive and negative values to test the sum calculations.
- **Queries**:
  - `1 1`: This query should return the sum of the first element in `b`, which is `1`.
  - `1 10`: This query should return the sum of all elements in `b`.
  - `5 10`: This query should return the sum of the last six elements in `b`.
  - `3 8`: This query should return the sum of the middle elements in `b`.
  - `2 8`: This query should return the sum of a subset of elements in `b`.

This test input is designed to ensure that the code handles the sum calculations correctly, especially for queries that span a wide range of indices in the `b` array.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums of Segments
time_limit_ms: 4000
memory_limit_mb: 512
Description: You are given a sequence of integers $[a_1, a_2, \dots, a_n]$. Let $s(l,r)$ be the sum of elements from $a_l$ to $a_r$ (i. e. $s(l,r) = \sum\limits_{i=l}^{r} a_i$).

Let's construct another sequence $b$ of size $\frac{n(n+1)}{2}$ as follows: $b = [s(1,1), s(1,2), \dots, s(1,n), s(2,2), s(2,3), \dots, s(2,n), s(3,3), \dots, s(n,n)]$.

For example, if $a = [1, 2, 5, 10]$, then $b = [1, 3, 8, 18, 2, 7, 17, 5, 15, 10]$.

You are given $q$ queries. During the $i$-th query, you are given two integers $l_i$ and $r_i$, and you have to calculate $\sum \limits_{j=l_i}^{r_i} b_j$.

Input: The first line contains one integer $n$ ($1 \le n \le 3 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-10 \le a_i \le 10$).

The third line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$).

Then $q$ lines follow, the $i$-th of them contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le \frac{n(n+1)}{2}$).

Output: Print $q$ integers, the $i$-th of which should be equal to $\sum \limits_{j=l_i}^{r_i} b_j$.

Examples:
input:
4
1 2 5 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
output:
1
4
12
30
32
86
56
54
60
26
82
57
9
2
61

Note:

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

    idx = 0

    # Read n
    n = int(data[idx])
    idx += 1

    # Read array a
    a = list(map(int, data[idx:idx + n]))
    idx += n

    # Read number of queries
    q = int(data[idx])
    idx += 1

    # Read queries
    queries = []
    for _ in range(q):
        l_i = int(data[idx])
        r_i = int(data[idx + 1])
        queries.append((l_i, r_i))
        idx += 2

    # Step 1: Compute prefix sums of a
    prefix_a = [0] * (n + 1)
    for i in range(1, n + 1):
        prefix_a[i] = prefix_a[i - 1] + a[i - 1]

    # Step 2: Construct sequence b
    b = []
    for l in range(1, n + 1):
        for r in range(l, n + 1):
            b.append(prefix_a[r] - prefix_a[l - 1])

    # Step 3: Compute cumulative sum array for b
    cumulative_b = [0] * (len(b) + 1)
    for i in range(1, len(b) + 1):
        cumulative_b[i] = cumulative_b[i - 1] + b[i - 1]

    # Step 4: Answer each query
    result = []
    for l_i, r_i in queries:
        result.append(cumulative_b[r_i] - cumulative_b[l_i - 1])

    print("\n".join(map(str, result)))

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider edge cases and scenarios that can potentially cause issues. Here are the steps to think about:

1. **Boundary Conditions**: Test with the smallest and largest possible values for `n` and `q`.
2. **Edge Cases for `a`**: Include both positive and negative values, and values at the boundary of the range (-10 to 10).
3. **Query Boundaries**: Test queries that cover the entire range of `b`, as well as queries that are just at the boundaries.

Given the constraints:
- \(1 \le n \le 3 \cdot 10^5\)
- \(-10 \le a_i \le 10\)
- \(1 \le q \le 3 \cdot 10^5\)

Let's create a small-scale test input that is complete and valid:

```plaintext
3
1 -2 3
5
1 1
1 3
3 6
1 6
6 6
```

### Explanation:
- **n = 3**: This is a small value to keep the input manageable.
- **a = [1, -2, 3]**: This array includes both positive and negative values within the range.
- **q = 5**: A small number of queries to test different scenarios.
- **Queries**:
  - **(1, 1)**: Sum of the first element in `b`.
  - **(1, 3)**: Sum of the first three elements in `b`.
  - **(3, 6)**: Sum of elements in the middle of `b`.
  - **(1, 6)**: Sum of all elements in `b`.
  - **(6, 6)**: Sum of the last element in `b`.

This test input should help in identifying any issues with the code, especially in handling the construction of the `b` array and the cumulative sum calculations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums of Segments
time_limit_ms: 4000
memory_limit_mb: 512
Description: You are given a sequence of integers $[a_1, a_2, \dots, a_n]$. Let $s(l,r)$ be the sum of elements from $a_l$ to $a_r$ (i. e. $s(l,r) = \sum\limits_{i=l}^{r} a_i$).

Let's construct another sequence $b$ of size $\frac{n(n+1)}{2}$ as follows: $b = [s(1,1), s(1,2), \dots, s(1,n), s(2,2), s(2,3), \dots, s(2,n), s(3,3), \dots, s(n,n)]$.

For example, if $a = [1, 2, 5, 10]$, then $b = [1, 3, 8, 18, 2, 7, 17, 5, 15, 10]$.

You are given $q$ queries. During the $i$-th query, you are given two integers $l_i$ and $r_i$, and you have to calculate $\sum \limits_{j=l_i}^{r_i} b_j$.

Input: The first line contains one integer $n$ ($1 \le n \le 3 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-10 \le a_i \le 10$).

The third line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$).

Then $q$ lines follow, the $i$-th of them contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le \frac{n(n+1)}{2}$).

Output: Print $q$ integers, the $i$-th of which should be equal to $\sum \limits_{j=l_i}^{r_i} b_j$.

Examples:
input:
4
1 2 5 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
output:
1
4
12
30
32
86
56
54
60
26
82
57
9
2
61

Note:

Buggy Code:
n=int(input())
a=list(map(int,input().split()))
q=int(input())
w=[a[0]]
for i in range(1,n):
    w.append(a[i]+w[i-1])
#print(w)#
b=[]
for i in range(n):
    for j in range(i,n):
        # if j+1==n:
        #     b.append(w[j]-w[i-1])
        if i==0:
            b.append(w[j])
        else:
            b.append(w[j]-w[i-1])
#print(b)#
bb=[b[0]]
for i in range(1,len(b)):
    bb.append(b[i]+bb[i-1])
#print(bb)#
for i in range(q):
    l,r=map(int,input().split())
    l-=1
    r-=1
    # if r>=n:
    if l==0:
        print(bb[r])
    else:
        print(bb[r]-bb[l-1])
    #print()

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

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

Think step by step.

Response:
To 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. Here are the steps to create such a test input:

1. **Edge Cases for Sequence Length (n):**
   - Test with the minimum value of `n` (i.e., 1).
   - Test with a value of `n` that is large enough to trigger potential issues in the code (e.g., 10 or 20).

2. **Edge Cases for Element Values (a_i):**
   - Test with all positive values.
   - Test with all negative values.
   - Test with a mix of positive and negative values.

3. **Edge Cases for Queries (q):**
   - Test with the minimum value of `q` (i.e., 1).
   - Test with a value of `q` that is large enough to trigger potential issues in the code (e.g., 10 or 20).

4. **Edge Cases for Query Ranges (l_i, r_i):**
   - Test with queries that cover the entire range of `b`.
   - Test with queries that cover a single element in `b`.
   - Test with queries that cover the start and end of `b`.

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

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

### Explanation:
- **n = 4**: This is a small value that is easy to verify manually.
- **a = [1, 2, 5, 10]**: A sequence with positive values to avoid issues with negative sums.
- **q = 4**: A small number of queries to keep the input manageable.
- **Queries:**
  - **1 1**: Single element query.
  - **1 10**: Full range query.
  - **1 4**: Partial range query.
  - **4 10**: Another partial range query.

This test input will help expose potential issues in the buggy code, such as incorrect handling of the prefix sums and the query ranges.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums of Segments
time_limit_ms: 4000
memory_limit_mb: 512
Description: You are given a sequence of integers $[a_1, a_2, \dots, a_n]$. Let $s(l,r)$ be the sum of elements from $a_l$ to $a_r$ (i. e. $s(l,r) = \sum\limits_{i=l}^{r} a_i$).

Let's construct another sequence $b$ of size $\frac{n(n+1)}{2}$ as follows: $b = [s(1,1), s(1,2), \dots, s(1,n), s(2,2), s(2,3), \dots, s(2,n), s(3,3), \dots, s(n,n)]$.

For example, if $a = [1, 2, 5, 10]$, then $b = [1, 3, 8, 18, 2, 7, 17, 5, 15, 10]$.

You are given $q$ queries. During the $i$-th query, you are given two integers $l_i$ and $r_i$, and you have to calculate $\sum \limits_{j=l_i}^{r_i} b_j$.

Input: The first line contains one integer $n$ ($1 \le n \le 3 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-10 \le a_i \le 10$).

The third line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$).

Then $q$ lines follow, the $i$-th of them contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le \frac{n(n+1)}{2}$).

Output: Print $q$ integers, the $i$-th of which should be equal to $\sum \limits_{j=l_i}^{r_i} b_j$.

Examples:
input:
4
1 2 5 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
output:
1
4
12
30
32
86
56
54
60
26
82
57
9
2
61

Note:

Buggy Code:
import java.util.Scanner;

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

        int n = sc.nextInt();
        long[] a = new long[n + 1];
        long[] prefixSum = new long[n + 1];

        for (int i = 1; i <= n; i++) {
            a[i] = sc.nextLong();
            prefixSum[i] = prefixSum[i - 1] + a[i];
        }

        int q = sc.nextInt();
        StringBuilder output = new StringBuilder();

        while (q-- > 0) {
            int l = sc.nextInt();
            int r = sc.nextInt();

            long sum = 0;

            int start = 1, end = 1, currentIndex = 1;
            while (currentIndex <= r) {
                if (currentIndex >= l) {
                    sum += prefixSum[end] - prefixSum[start - 1];
                }
                currentIndex++;
                if (end == n) {
                    start++;
                    end = start;
                } else {
                    end++;
                }
            }

            output.append(sum).append("\n");
        }

        System.out.print(output);
        sc.close();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability. The code has a loop that constructs the sum of segments, and it might fail when dealing with specific boundary conditions or large ranges.

### Step-by-Step Reasoning:

1. **Boundary Conditions:**
   - Test with the smallest possible input size (`n = 1`).
   - Test with the maximum possible input size (`n = 300000`).

2. **Large Range Queries:**
   - Test with queries that cover a large range in the `b` array, especially those that might cross segment boundaries.

3. **Negative and Positive Values:**
   - Test with a mix of negative and positive values to ensure the prefix sum calculations are correct.

4. **Specific Edge Cases:**
   - Test with a single element array.
   - Test with an array where all elements are the same.
   - Test with an array where elements are in a specific pattern (e.g., alternating positive and negative values).

### Chosen Test Input:
- **n = 5** (a small but non-trivial size)
- **a = [1, -2, 3, -4, 5]**
- **q = 3**
- **Queries:**
  - **l1 = 1, r1 = 15** (covers a large range)
  - **l2 = 5, r2 = 10** (covers a smaller range but still crosses segments)
  - **l3 = 1, r3 = 3** (covers the first few segments)

### Test Input:
```plaintext
5
1 -2 3 -4 5
3
1 15
5 10
1 3
```

### Explanation:
- **n = 5**: A small but non-trivial size.
- **a = [1, -2, 3, -4, 5]**: A mix of positive and negative values.
- **q = 3**: Three queries to test different scenarios.
  - **1 15**: Covers a large range, including multiple segments.
  - **5 10**: Covers a smaller range but still crosses segments.
  - **1 3**: Covers the first few segments to test basic functionality.

This test input should help expose any issues in the loop that constructs the sum of segments and the handling of segment boundaries.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums of Segments
time_limit_ms: 4000
memory_limit_mb: 512
Description: You are given a sequence of integers $[a_1, a_2, \dots, a_n]$. Let $s(l,r)$ be the sum of elements from $a_l$ to $a_r$ (i. e. $s(l,r) = \sum\limits_{i=l}^{r} a_i$).

Let's construct another sequence $b$ of size $\frac{n(n+1)}{2}$ as follows: $b = [s(1,1), s(1,2), \dots, s(1,n), s(2,2), s(2,3), \dots, s(2,n), s(3,3), \dots, s(n,n)]$.

For example, if $a = [1, 2, 5, 10]$, then $b = [1, 3, 8, 18, 2, 7, 17, 5, 15, 10]$.

You are given $q$ queries. During the $i$-th query, you are given two integers $l_i$ and $r_i$, and you have to calculate $\sum \limits_{j=l_i}^{r_i} b_j$.

Input: The first line contains one integer $n$ ($1 \le n \le 3 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-10 \le a_i \le 10$).

The third line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$).

Then $q$ lines follow, the $i$-th of them contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le \frac{n(n+1)}{2}$).

Output: Print $q$ integers, the $i$-th of which should be equal to $\sum \limits_{j=l_i}^{r_i} b_j$.

Examples:
input:
4
1 2 5 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
output:
1
4
12
30
32
86
56
54
60
26
82
57
9
2
61

Note:

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

public class Main {

    static class SegmentTree {
        private long[] tree;
        private int n;

        public SegmentTree(int[] arr) {
            n = arr.length;
            tree = new long[n * 4];
            build(arr, 0, 0, n - 1);
        }

        private void build(int[] arr, int node, int start, int end) {
            if (start == end) {
                tree[node] = arr[start];
            } else {
                int mid = (start + end) / 2;
                build(arr, 2 * node + 1, start, mid);
                build(arr, 2 * node + 2, mid + 1, end);
                tree[node] = tree[2 * node + 1] + tree[2 * node + 2];
            }
        }

        public long query(int l, int r) {
            return query(0, 0, n - 1, l, r);
        }

        private long query(int node, int start, int end, int l, int r) {
            if (r < start || end < l) {
                return 0;
            }
            if (l <= start && end <= r) {
                return tree[node];
            }
            int mid = (start + end) / 2;
            long leftSum = query(2 * node + 1, start, mid, l, r);
            long rightSum = query(2 * node + 2, mid + 1, end, l, r);
            return leftSum + rightSum;
        }
    }

    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());

        int n = Integer.parseInt(st.nextToken());
        int[] arr = new int[n];
        st = new StringTokenizer(br.readLine());

        for (int i = 0; i < n; i++) {
            arr[i] = Integer.parseInt(st.nextToken());
        }

        SegmentTree segmentTree = new SegmentTree(arr);

        int bSize = (n * (n + 1)) / 2;
        int[] b = new int[bSize];

        int index = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                b[index++] = (int) segmentTree.query(i, j);
            }
        }

        long[] prefixB = new long[bSize + 1];
        for (int i = 1; i <= bSize; i++) {
            prefixB[i] = prefixB[i - 1] + b[i - 1];
        }

        int q = Integer.parseInt(br.readLine());
        StringBuilder result = new StringBuilder();
        for (int i = 0; i < q; i++) {
            st = new StringTokenizer(br.readLine());
            int l = Integer.parseInt(st.nextToken());
            int r = Integer.parseInt(st.nextToken());
            // Answer the query using the prefix sums
            long sum = prefixB[r] - prefixB[l - 1];
            result.append(sum).append("\n");
        }

        System.out.print(result);
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums of Segments
time_limit_ms: 4000
memory_limit_mb: 512
Description: You are given a sequence of integers $[a_1, a_2, \dots, a_n]$. Let $s(l,r)$ be the sum of elements from $a_l$ to $a_r$ (i. e. $s(l,r) = \sum\limits_{i=l}^{r} a_i$).

Let's construct another sequence $b$ of size $\frac{n(n+1)}{2}$ as follows: $b = [s(1,1), s(1,2), \dots, s(1,n), s(2,2), s(2,3), \dots, s(2,n), s(3,3), \dots, s(n,n)]$.

For example, if $a = [1, 2, 5, 10]$, then $b = [1, 3, 8, 18, 2, 7, 17, 5, 15, 10]$.

You are given $q$ queries. During the $i$-th query, you are given two integers $l_i$ and $r_i$, and you have to calculate $\sum \limits_{j=l_i}^{r_i} b_j$.

Input: The first line contains one integer $n$ ($1 \le n \le 3 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-10 \le a_i \le 10$).

The third line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$).

Then $q$ lines follow, the $i$-th of them contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le \frac{n(n+1)}{2}$).

Output: Print $q$ integers, the $i$-th of which should be equal to $\sum \limits_{j=l_i}^{r_i} b_j$.

Examples:
input:
4
1 2 5 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
output:
1
4
12
30
32
86
56
54
60
26
82
57
9
2
61

Note:

Buggy Code:
n=int(input())
a=list(map(int,input().split()))
b=[]
for i in range(0,n):
    s=0
    for j in range(i,n):
        s+=a[j]
        b.append(s)
for _ in range(int(input())):
    s,e=list(map(int,input().split()))
    i=s-1
    s=0
    while(i<e):
        s+=b[i]
        i+=1
    print(s)

**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 and scenarios that can maximize the failure probability. The code provided has a nested loop that constructs the sequence `b`, and another nested loop that processes the queries. The primary issues with this code are:

1. **Efficiency**: The code is not optimized for large inputs, as it uses a linear search for each query, which can be very slow for large `n` and `q`.
2. **Edge Cases**: The code might fail for edge cases where the sequence `a` contains negative values or the queries span large ranges.

Given the constraints:
- \(1 \le n \le 3 \cdot 10^5\)
- \(-10 \le a_i \le 10\)
- \(1 \le q \le 3 \cdot 10^5\)
- \(1 \le l_i \le r_i \le \frac{n(n+1)}{2}\)

We can create a small-scale test input that is complete and valid. Let's focus on a small `n` and a few queries that span different ranges to ensure the code is tested thoroughly.

Here is a small-scale test input:

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

Explanation:
- **Sequence `a`**: [1, -2, 3, -4, 5]
- **Sequence `b`**: [1, -1, 2, -2, 3, -1, 2, 1, 6, 2, 7, 3, 8, 4, 5]
- **Queries**:
  - (1, 1): Sum of `b[0]` = 1
  - (1, 5): Sum of `b[0] + b[1] + b[2] + b[3] + b[4]` = 1 - 1 + 2 - 2 + 3 = 3
  - (2, 5): Sum of `b[1] + b[2] + b[3] + b[4]` = -1 + 2 - 2 + 3 = 2
  - (3, 10): Sum of `b[2] + b[3] + b[4] + b[5] + b[6] + b[7] + b[8] + b[9]` = 2 - 2 + 3 - 1 + 2 + 1 + 6 + 2 = 13
  - (1, 10): Sum of all elements in `b` = 1 - 1 + 2 - 2 + 3 - 1 + 2 + 1 + 6 + 2 + 7 + 3 + 8 + 4 + 5 = 38

This test input should expose any issues with the code, especially in handling negative values and large ranges.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums of Segments
time_limit_ms: 4000
memory_limit_mb: 512
Description: You are given a sequence of integers $[a_1, a_2, \dots, a_n]$. Let $s(l,r)$ be the sum of elements from $a_l$ to $a_r$ (i. e. $s(l,r) = \sum\limits_{i=l}^{r} a_i$).

Let's construct another sequence $b$ of size $\frac{n(n+1)}{2}$ as follows: $b = [s(1,1), s(1,2), \dots, s(1,n), s(2,2), s(2,3), \dots, s(2,n), s(3,3), \dots, s(n,n)]$.

For example, if $a = [1, 2, 5, 10]$, then $b = [1, 3, 8, 18, 2, 7, 17, 5, 15, 10]$.

You are given $q$ queries. During the $i$-th query, you are given two integers $l_i$ and $r_i$, and you have to calculate $\sum \limits_{j=l_i}^{r_i} b_j$.

Input: The first line contains one integer $n$ ($1 \le n \le 3 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-10 \le a_i \le 10$).

The third line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$).

Then $q$ lines follow, the $i$-th of them contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le \frac{n(n+1)}{2}$).

Output: Print $q$ integers, the $i$-th of which should be equal to $\sum \limits_{j=l_i}^{r_i} b_j$.

Examples:
input:
4
1 2 5 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
output:
1
4
12
30
32
86
56
54
60
26
82
57
9
2
61

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
using ll=long long;
using ull=unsigned long long int;
/*#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template <typename T>
 // Ordered Set Tree
using ordered_set= tree<T, null_type, less<T>,
             rb_tree_tag,
             tree_order_statistics_node_update>;*/
ll cmod(ll a,ll b){
if(a%b==0)return a/b;
else return a/b+1;}
ll fact(ll d,ll MOD){
    ll f=1;
    if(d==0)return f;
    while(d>=1){
        f=f*d%MOD;
        d--;
    }
    return f;
}
ll binpow(ll a,ll b,ll MODe){
ll res=1;
a=a;
while(b>0){
    if(b&1){
        res=res*a%MODe;
    }
    a=a*a%MODe;
    b>>=1;}
    return res;
}
bool check(vector<ll> a,ll m,ll k,ll q){
    if(m==0)return false;
ll z=m*k,c=1,d=0;
for(int i=0;i<a.size();i++){
   if(a[i]>z)return false;
   d+=a[i];
   if(d>z){
    d=a[i];
    c++;
   }
}
if(c<=q)return true;
else return false;
}
vector<ll> sieve(ll limit) {
    vector<bool> is_prime(limit + 1, true);
    vector<ll> primes;
    is_prime[0] = is_prime[1] = false;
    for (ll i = 2; i <= limit; ++i) {
        if (is_prime[i]) {
            primes.push_back(i);
            for (ll j = 2 * i; j <= limit; j += i) {
                is_prime[j] = false;
            }
        }
    }
    return primes;
}
int query(int i,int j){
cout<<"? "<<i<<" "<<j<<endl;
cout.flush();
int p;
cin>>p;
return p;}
void code(){
cout<<"Y"<<endl;}
//check for binary search and dp and greedy
//check t input lena hain yah nhi
//check while binary searching for l and r base case
//check for ordered set
//check if something is empty
void solve(){
ll n;
cin>>n;
vector<ll> a(n+1);
for(ll i=1;i<=n;i++)cin>>a[i];
ll q;
cin>>q;
while(q--){
    ll ll,rr;
    cin>>ll>>rr;
    long long p,q,r,s;
    long long u=0,v=1;
    for(long long i=n;i>0;i--){
        if(ll<=u+i){
            p=v;
            q=ll-u;
            break;
        }
        u+=i;
        v++;
    }
    u=0,v=1;
    for(long long i=n;i>0;i--){
        if(rr<=u+i){
            r=v;
            s=rr-u;
            break;
        }
        u+=i;
        v++;
    }
    q+=p-1;
    s+=r-1;
    //cout<<p<<" "<<q<<" "<<r<<" "<<s<<endl;
    long long ans=0;
    for(long long i=p;i<=n;i++){
      long long k=0;
      if(p!=r){
      if(i==p)k+=n+1-q;
      if(i>p){
        if(i>=q)k+=n+1-i;
        else k+=n+1-q;
        k+=(min(r-1,i)-p)*(n+1-i);
        if(i>=r&&i<=s)k+=(s-i+1);
      }}
      else{
        if(i>=p&&i<q)k+=s-q+1;
        if(i>=q)k+=s-i+1;
      }
      if(k>0)ans+=k*a[i];
      //cout<<i<<" "<<k<<"  ";
    }//cout<<endl;
    cout<<ans<<endl;
}
}
int main(){
      ios_base::sync_with_stdio(false);
      cin.tie(NULL);
int 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 given buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. Here are the steps to consider:

1. **Edge Case for Small `n`**:
   - Test with the smallest possible value of `n` (1) to see if the code handles a single-element array correctly.

2. **Edge Case for Large `n`**:
   - Test with a large value of `n` (close to the upper limit) to see if the code handles large arrays and large queries efficiently.

3. **Edge Case for Negative and Positive Values**:
   - Test with arrays containing both negative and positive values to ensure the code handles mixed values correctly.

4. **Edge Case for Queries**:
   - Test with queries that span the entire range of `b` to ensure the code handles large ranges correctly.
   - Test with queries that are single-element ranges to ensure the code handles small ranges correctly.

5. **Edge Case for Boundary Values**:
   - Test with queries that start or end at the boundaries of the `b` array to ensure the code handles boundary conditions correctly.

Given the constraints and the need to keep the test input small, we will focus on a combination of the above cases. Here is a small-scale test input that includes a mix of these edge cases:

```plaintext
4
-10 10 -10 10
3
1 1
1 10
5 6
```

This test input includes:
- A small `n` value (4).
- A mix of negative and positive values in the array.
- Queries that:
  - Cover a single element (`1 1`).
  - Cover the entire range of `b` (`1 10`).
  - Cover a small range in the middle (`5 6`).

This should help in exposing 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 of Segments
time_limit_ms: 4000
memory_limit_mb: 512
Description: You are given a sequence of integers $[a_1, a_2, \dots, a_n]$. Let $s(l,r)$ be the sum of elements from $a_l$ to $a_r$ (i. e. $s(l,r) = \sum\limits_{i=l}^{r} a_i$).

Let's construct another sequence $b$ of size $\frac{n(n+1)}{2}$ as follows: $b = [s(1,1), s(1,2), \dots, s(1,n), s(2,2), s(2,3), \dots, s(2,n), s(3,3), \dots, s(n,n)]$.

For example, if $a = [1, 2, 5, 10]$, then $b = [1, 3, 8, 18, 2, 7, 17, 5, 15, 10]$.

You are given $q$ queries. During the $i$-th query, you are given two integers $l_i$ and $r_i$, and you have to calculate $\sum \limits_{j=l_i}^{r_i} b_j$.

Input: The first line contains one integer $n$ ($1 \le n \le 3 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-10 \le a_i \le 10$).

The third line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$).

Then $q$ lines follow, the $i$-th of them contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le \frac{n(n+1)}{2}$).

Output: Print $q$ integers, the $i$-th of which should be equal to $\sum \limits_{j=l_i}^{r_i} b_j$.

Examples:
input:
4
1 2 5 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
output:
1
4
12
30
32
86
56
54
60
26
82
57
9
2
61

Note:

Buggy Code:
#include <bits/stdc++.h>
#define F first
#define S second
#define ld long double
#define pii pair<int, int>
#define int long long
#define pb push_back
#define endl "\n"
#define setbits(x) __builtin_popcount(int num);
using namespace std;
int n;

int getF(int k,int all, int add[], int sub[], int pf[]){
	//cout<<"HERE "<<k<<' '<<all<<' '<<add[all+k]<<' '<<sub[all]<<' '<<(n-all-k)<<' '<<pf[all]<<endl;
	int sum=add[all+k]-sub[all]+(n-all-k)*pf[all];
	return sum;
}

int sol(int k, int n, int psum[],int add[], int sub[], int pf[]){
	int low=0,high=n;
	int all=0;
	int took;
	while(low<=high){
		int mid=(low+high);
		int S=(mid*(2*n+(1-mid)))/2;
		if(S<=k){
			all=mid;
			low=mid+1;
			took=S;
		}
		else high=mid-1;
	}
	int total=psum[all];
	//cout<<"Come "<<k<<' '<<all<<' '<<total<<' '<<k-took<<endl;
	k-=took;
	if(k==0)all--;
	int xtra=getF(k,all,add,sub,pf);
	//cout<<"NUM "<<xtra<<endl;
	return total+xtra;
}

void solve(){
	cin>>n;
	int a[n+1]={0};
	int count[n+1]={0};
	int sum=0;
	int add[n+1]={0},sub[n+1]={0},pf[n+1]={0},psum[n+1]={0};
	int tsum=0;
	for(int i=1;i<=n;i++){
		cin>>a[i];
		sum+=a[i];
		add[i]=sum;
		tsum+=add[i];
		add[i]+=add[i-1];
		sub[i]=(n-i+1)*a[i]+sub[i-1];
		pf[i]=sum;

	}
	for(int i=1;i<=n;i++){
		psum[i]=tsum+psum[i-1];
		tsum-=(n-i+1)*a[i];
	}

	// for(int i=0;i<=n;i++){
		// cout<<add[i]<<' '<<sub[i]<<endl;
	// }
	int q;
	cin>>q;
	for(int i=0;i<q;i++){
		int l,r;
		cin>>l>>r;
		l--;
		int R=sol(r,n,psum,add,sub,pf);
		int L=0;
		if(l>0)L=sol(l,n,psum,add,sub,pf);
		//cout<<"FINAL "<<R<<' '<<L<<endl;
		cout<<R-L<<endl;

	}

}

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int t;
    t = 1;
    //cin>>t;
    while (t--)
    {
        solve();
    }
    return 0;
}

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

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

Think step by step.

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

1. **Edge Cases for `n`**:
   - Minimum value of `n`: 1
   - Maximum value of `n`: 300,000 (though we will use a smaller value for a small-scale test)

2. **Edge Cases for `a`**:
   - All elements are the same (e.g., all 1s or all -1s)
   - All elements are at the extreme values (e.g., all 10s or all -10s)
   - Mixed values with both positive and negative numbers

3. **Edge Cases for Queries**:
   - Single element queries (e.g., l = r)
   - Full range queries (e.g., l = 1, r = n(n+1)/2)
   - Queries that span multiple segments

4. **Combination of Edge Cases**:
   - A combination of the above to ensure the algorithm handles all possible scenarios.

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

```plaintext
4
-10 10 -10 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
```

This test input:
- Uses a small `n` (4) to keep the input manageable.
- Uses extreme values for `a` (both positive and negative).
- Includes a variety of queries, including single element queries, full range queries, and queries that span multiple segments.

This should help expose any bugs in the algorithm, especially those related to handling edge cases and large values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums of Segments
time_limit_ms: 4000
memory_limit_mb: 512
Description: You are given a sequence of integers $[a_1, a_2, \dots, a_n]$. Let $s(l,r)$ be the sum of elements from $a_l$ to $a_r$ (i. e. $s(l,r) = \sum\limits_{i=l}^{r} a_i$).

Let's construct another sequence $b$ of size $\frac{n(n+1)}{2}$ as follows: $b = [s(1,1), s(1,2), \dots, s(1,n), s(2,2), s(2,3), \dots, s(2,n), s(3,3), \dots, s(n,n)]$.

For example, if $a = [1, 2, 5, 10]$, then $b = [1, 3, 8, 18, 2, 7, 17, 5, 15, 10]$.

You are given $q$ queries. During the $i$-th query, you are given two integers $l_i$ and $r_i$, and you have to calculate $\sum \limits_{j=l_i}^{r_i} b_j$.

Input: The first line contains one integer $n$ ($1 \le n \le 3 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-10 \le a_i \le 10$).

The third line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$).

Then $q$ lines follow, the $i$-th of them contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le \frac{n(n+1)}{2}$).

Output: Print $q$ integers, the $i$-th of which should be equal to $\sum \limits_{j=l_i}^{r_i} b_j$.

Examples:
input:
4
1 2 5 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
output:
1
4
12
30
32
86
56
54
60
26
82
57
9
2
61

Note:

Buggy Code:
# sums of segments

import sys, math
# sys.stdin = open("2026D.in", "r")

n = int(input())
a = list(map(int, input().split()))
pa = [0]*(n+1)
ppa = [0]*(n+1)
pb = [0]*(n+1)
istart = [0]*(n+1)
for i in range(1, n+1):
    pa[i] = pa[i-1] + a[i-1]
    ppa[i] = ppa[i-1] + pa[i]
for x in range(1, n+1):
    pb[x] = pb[x-1] + ppa[n] - ppa[x-1] - (n-x+1) * pa[x-1]

istart[1] = 1
for i in range(2, n+1):
    istart[i] = istart[i-1] + n - i + 2

# print (a)
# print (pa)
# print (ppa)
# print (istart)

def geti(x, lo, hi):
    # print (x, lo, hi)
    if lo < hi:
        mid = math.ceil((lo + hi)/2)
        if istart[mid] <= x:
            lo = mid
        else:
            hi = mid-1
        return geti(x, lo, hi)
    else:
        return lo

# for i in range(1, n*n):
#     print (i, geti(i, 1, n))

q = int(input())
for i in range(q):
    l, r = map(int, input().split())
    il = geti(l, 1, n)
    jl = l - istart[il] + il
    ir = geti(r, 1, n)
    jr = r - istart[ir] + ir
    # print (l, il, jl, ',', r, ir, jr)
    if ir == il:
        ans = ppa[jr] - ppa[jl - 1] - (jr - jl + 1) * pa[il - 1]
    else:
        ans = ppa[n] - ppa[jl - 1] - (n - jl + 1) * pa[il - 1]
        # for x in range(il + 1, ir):
        #     ans += ppa[n] - ppa[x-1] - (n-x+1) * pa[x-1]
        ans += pb[ir-1] - pb[il]
        ans += ppa[jr] - ppa[ir-1] - (jr - ir + 1) * pa[ir - 1]
    print (ans)

# print (l)
# print (r)

# n = 6
# for i in range(1, n+1):
#     for j in range(i, n+1):
#         # l = int((i-1)*(2*n+2-i)/2)+(j-i)+1
#         l = n * (i-1) + j - int(i*(i-1)/2)
#         l = int((2*n - i)*(i-1)/2) + j
#         print (i, j, l)

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

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

Think step by step.

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

1. **Edge Cases for `n` and `a`**:
   - Use the smallest possible value for `n` to simplify the problem and ensure the input is small.
   - Use values for `a` that can cause issues, such as negative numbers or zeros.

2. **Queries**:
   - Use queries that span different segments of the sequence `b` to test the correctness of the segment sum calculation.
   - Include queries that test the edge cases, such as the first and last elements of `b`.

3. **Expected Output**:
   - Ensure the test input has a clear and simple expected output to make it easy to verify the correctness of the solution.

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

```plaintext
3
-1 0 1
4
1 1
1 3
2 3
3 3
```

### Explanation:
- **Input**:
  - `n = 3`: This is the smallest non-trivial value for `n` to ensure the input is small but still meaningful.
  - `a = [-1, 0, 1]`: This sequence includes negative, zero, and positive values to test various edge cases.
  - `q = 4`: We have four queries to cover different scenarios.
  - `l, r` values:
    - `1 1`: This tests the sum of the first element of `b`.
    - `1 3`: This tests the sum of the first three elements of `b`.
    - `2 3`: This tests the sum of the second and third elements of `b`.
    - `3 3`: This tests the sum of the third element of `b`.

- **Expected Output**:
  - For `1 1`, `b[1] = s(1,1) = -1`.
  - For `1 3`, `b[1] = -1`, `b[2] = -1 + 0 = -1`, `b[3] = -1 + 0 + 1 = 0`.
    - Sum: `-1 + (-1) + 0 = -2`.
  - For `2 3`, `b[2] = -1`, `b[3] = 0`.
    - Sum: `-1 + 0 = -1`.
  - For `3 3`, `b[3] = 0`.

This test input is small and complete, and it should help expose any bugs in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums of Segments
time_limit_ms: 4000
memory_limit_mb: 512
Description: You are given a sequence of integers $[a_1, a_2, \dots, a_n]$. Let $s(l,r)$ be the sum of elements from $a_l$ to $a_r$ (i. e. $s(l,r) = \sum\limits_{i=l}^{r} a_i$).

Let's construct another sequence $b$ of size $\frac{n(n+1)}{2}$ as follows: $b = [s(1,1), s(1,2), \dots, s(1,n), s(2,2), s(2,3), \dots, s(2,n), s(3,3), \dots, s(n,n)]$.

For example, if $a = [1, 2, 5, 10]$, then $b = [1, 3, 8, 18, 2, 7, 17, 5, 15, 10]$.

You are given $q$ queries. During the $i$-th query, you are given two integers $l_i$ and $r_i$, and you have to calculate $\sum \limits_{j=l_i}^{r_i} b_j$.

Input: The first line contains one integer $n$ ($1 \le n \le 3 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-10 \le a_i \le 10$).

The third line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$).

Then $q$ lines follow, the $i$-th of them contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le \frac{n(n+1)}{2}$).

Output: Print $q$ integers, the $i$-th of which should be equal to $\sum \limits_{j=l_i}^{r_i} b_j$.

Examples:
input:
4
1 2 5 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
output:
1
4
12
30
32
86
56
54
60
26
82
57
9
2
61

Note:

Buggy Code:
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.TreeSet;

public class Main {

	static class Scanner {
		Scanner(InputStream in) { this.in = in; } InputStream in;
		byte[] bb = new byte[200000 * 19]; int i, n;
		byte getc() {
			if (i == n) {
				i = n = 0;
				try { n = in.read(bb); } catch (IOException e) {}
			}
			return i < n ? bb[i++] : 0;
		}
		int nextInt() {
			byte c = 0; while (c <= ' ') c = getc();
			boolean negate = false;
			if (c == '-') {
				negate = true;
				c = getc();
			}
			int a = 0; while (c > ' ') { a = a * 10 + c - '0'; c = getc(); }
			if (negate) a = -a;
			return a;
		}

		long nextLong() {
			byte c = 0; while (c <= ' ') c = getc();
			long a = 0; while (c > ' ') { a = a * 10 + c - '0'; c = getc(); }
			return a;
		}
	}

	private static BufferedWriter out;

	private static long f(int IDX, int N, int[] A, long[] suffixSum) throws Exception {
		if (IDX == 0)
			return 0;

		//out.write("IDX = " + IDX + "\n");

		long count = 0;
		int id=1;
		for (id=1; id<=N; id++) {
			count += (N-id+1);
			if (count >= IDX)
				break;
		}

		//out.write("id= " + id + "\n");

		long result = 0;
		for (int i=1; i<id; i++) {
			result += suffixSum[i];
		}

		//out.write("result = " + result + "\n");

		long counts = 0;
		for (int i=1; i<id; i++) {
			counts += (N-i+1);
		}

		//out.write("result = " + result + "\n");
		long current = 0;
		//out.write("counts = " + counts + "\n");
		for (int i=0; i<IDX - counts; i++) {
			current += A[id + i];
			result += current;
			//result += A[id+i];
		}

		return result;
	}

	public static void main(String[] args) throws Exception {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in), 200000 * 20);
		out = new BufferedWriter(new OutputStreamWriter(System.out), 50000000);

		int N = Integer.parseInt(in.readLine());
		String[] line = in.readLine().split(" ");
		int[] A = new int[N + 1];

		for (int i=1; i<=N; i++) {
			A[i] = Integer.parseInt(line[i-1]);
		}

		long[] prefixSum = new long[N+1];
		for (int i=1; i<=N; i++) {
			prefixSum[i] = prefixSum[i-1] + A[i];
		}

		long[] dp = new long[N+1];
		dp[N] = A[N];
		for (int i=N-1; i>=1; i--) {
			dp[i] = 1l * (N-i+1) * A[i] + dp[i+1];
		}

		/*for (int i=1; i<=N; i++) {
			dp[i] = dp[i+1]
			for (int j=i; j<=N; j++) {
				dp[i] += (prefixSum[j] - prefixSum[i-1]);
			}
		}*/

		int Q = Integer.parseInt(in.readLine());

		for (int q=1; q<=Q; q++) {
			line = in.readLine().split(" ");
			int L = Integer.parseInt(line[0]);
			int R = Integer.parseInt(line[1]);

			long result = f(R, N, A, dp) - f(L-1, N, A, dp);
			out.write(result + "\n");

		}

		in.close();
		out.close();
	}
}

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

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

Think step by step.

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

1. **Small Scale Input**: Since we need a small-scale test input, we will choose a small value for \( n \) and \( q \).

2. **Edge Cases**:
   - **Single Element Array**: Test with an array that has only one element.
   - **Negative and Positive Elements**: Test with an array that includes both negative and positive elements.
   - **Query Edge Cases**: Test queries that cover the entire range of possible indices in the \( b \) array.

3. **Complexity**: Ensure the test input is complex enough to challenge the logic of the code, especially the part that calculates the sum of segments.

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

```plaintext
4
-1 2 -3 4
5
1 1
1 6
5 10
1 10
3 8
```

### Explanation:
- **Array Length (n = 4)**: Small enough to be manageable but large enough to have multiple segments.
- **Array Elements (a = [-1, 2, -3, 4])**: Includes both negative and positive values to test the sum calculations.
- **Number of Queries (q = 5)**: Sufficient to cover different scenarios.
- **Queries**:
  - **(1, 1)**: Single element sum.
  - **(1, 6)**: Sum of the first 6 elements in the \( b \) array.
  - **(5, 10)**: Sum of the last 6 elements in the \( b \) array.
  - **(1, 10)**: Sum of all elements in the \( b \) array.
  - **(3, 8)**: Sum of a middle range of elements in the \( b \) array.

This test input should help expose any issues in the code, particularly in the logic for calculating the sum of segments.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sums of Segments
time_limit_ms: 4000
memory_limit_mb: 512
Description: You are given a sequence of integers $[a_1, a_2, \dots, a_n]$. Let $s(l,r)$ be the sum of elements from $a_l$ to $a_r$ (i. e. $s(l,r) = \sum\limits_{i=l}^{r} a_i$).

Let's construct another sequence $b$ of size $\frac{n(n+1)}{2}$ as follows: $b = [s(1,1), s(1,2), \dots, s(1,n), s(2,2), s(2,3), \dots, s(2,n), s(3,3), \dots, s(n,n)]$.

For example, if $a = [1, 2, 5, 10]$, then $b = [1, 3, 8, 18, 2, 7, 17, 5, 15, 10]$.

You are given $q$ queries. During the $i$-th query, you are given two integers $l_i$ and $r_i$, and you have to calculate $\sum \limits_{j=l_i}^{r_i} b_j$.

Input: The first line contains one integer $n$ ($1 \le n \le 3 \cdot 10^5$).

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($-10 \le a_i \le 10$).

The third line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$).

Then $q$ lines follow, the $i$-th of them contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le \frac{n(n+1)}{2}$).

Output: Print $q$ integers, the $i$-th of which should be equal to $\sum \limits_{j=l_i}^{r_i} b_j$.

Examples:
input:
4
1 2 5 10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
output:
1
4
12
30
32
86
56
54
60
26
82
57
9
2
61

Note:

Buggy Code:
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.TreeSet;

public class Main {

	static class Scanner {
		Scanner(InputStream in) { this.in = in; } InputStream in;
		byte[] bb = new byte[200000 * 19]; int i, n;
		byte getc() {
			if (i == n) {
				i = n = 0;
				try { n = in.read(bb); } catch (IOException e) {}
			}
			return i < n ? bb[i++] : 0;
		}
		int nextInt() {
			byte c = 0; while (c <= ' ') c = getc();
			boolean negate = false;
			if (c == '-') {
				negate = true;
				c = getc();
			}
			int a = 0; while (c > ' ') { a = a * 10 + c - '0'; c = getc(); }
			if (negate) a = -a;
			return a;
		}

		long nextLong() {
			byte c = 0; while (c <= ' ') c = getc();
			long a = 0; while (c > ' ') { a = a * 10 + c - '0'; c = getc(); }
			return a;
		}
	}

	private static BufferedWriter out;

	private static long f(int IDX, int N, int[] A, long[] suffixSum) throws Exception {
		if (IDX == 0)
			return 0;

		//out.write("IDX = " + IDX + "\n");

		long count = 0;
		int id=1;
		for (id=1; id<=N; id++) {
			count += (N-id+1);
			if (count >= IDX)
				break;
		}

		//out.write("id= " + id + "\n");

		long result = 0;
		for (int i=1; i<id; i++) {
			result += suffixSum[i];
		}

		//out.write("result = " + result + "\n");

		long counts = 0;
		for (int i=1; i<id; i++) {
			counts += (N-i+1);
		}

		//out.write("result = " + result + "\n");
		long current = 0;
		//out.write("counts = " + counts + "\n");
		for (int i=0; i<IDX - counts; i++) {
			current += A[id + i];
			result += current;
			//result += A[id+i];
		}

		return result;
	}

	public static void main(String[] args) throws Exception {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in), 200000 * 20);
		out = new BufferedWriter(new OutputStreamWriter(System.out), 50000000);

		int N = Integer.parseInt(in.readLine());
		String[] line = in.readLine().split(" ");
		int[] A = new int[N + 1];

		for (int i=1; i<=N; i++) {
			A[i] = Integer.parseInt(line[i-1]);
		}

		long[] prefixSum = new long[N+1];
		for (int i=1; i<=N; i++) {
			prefixSum[i] = prefixSum[i-1] + A[i];
		}

		long[] dp = new long[N+1];
		for (int i=1; i<=N; i++) {
			for (int j=i; j<=N; j++) {
				dp[i] += (prefixSum[j] - prefixSum[i-1]);
			}
		}

		long[] suffixSum = new long[N+1];
		suffixSum[N] = A[N];
		for (int i=N-1; i>=1; i--) {
			suffixSum[i] = suffixSum[i+1] + A[i];
		}

		//long[] sufSuf = new long[N+1];

		int Q = Integer.parseInt(in.readLine());

		for (int q=1; q<=Q; q++) {
			line = in.readLine().split(" ");
			int L = Integer.parseInt(line[0]);
			int R = Integer.parseInt(line[1]);

			long result = f(R, N, A, dp) - f(L-1, N, A, dp);
			out.write(result + "\n");

		}

		in.close();
		out.close();
	}
}

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

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

Think step by step.

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

1. **Boundary Conditions**: Test with the smallest and largest possible values for `n` and `q`.
2. **Edge Cases for Array Values**: Test with arrays that have all positive, all negative, and mixed values.
3. **Query Range**: Test with queries that span the entire range of `b`, as well as queries that focus on specific segments.

Given the constraints and the need to keep the test input small, we can focus on a small but comprehensive test case. Here's a step-by-step approach to generate the test input:

1. **Small `n` and `q`**: Start with a small `n` and `q` to ensure the test case is manageable.
2. **Mixed Values in `a`**: Include both positive and negative values to test the sum calculations.
3. **Various Query Ranges**: Include queries that cover different segments of `b`.

Let's generate a test input based on these considerations:

```plaintext
4
1 -2 5 -10
15
1 1
1 2
1 3
1 4
1 5
1 10
5 10
6 10
2 8
3 4
3 10
3 8
5 6
5 5
1 8
```

### Explanation:
- **n = 4**: A small array size to keep the test case manageable.
- **a = [1, -2, 5, -10]**: Includes both positive and negative values to test sum calculations.
- **q = 15**: A reasonable number of queries to test various segments of `b`.
- **Queries**:
  - **1 1**: Single element.
  - **1 2**: Small segment.
  - **1 3**: Larger segment.
  - **1 4**: Larger segment.
  - **1 5**: Larger segment.
  - **1 10**: Full segment.
  - **5 10**: Middle segment.
  - **6 10**: End segment.
  - **2 8**: Large segment.
  - **3 4**: Small segment.
  - **3 10**: Large segment.
  - **3 8**: Large segment.
  - **5 6**: Small segment.
  - **5 5**: Single element.
  - **1 8**: Large segment.

This test input should help expose any issues in the buggy code by covering a variety of edge cases and query ranges.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: MEXimize the Score
time_limit_ms: 2000
memory_limit_mb: 256
Description: Suppose we partition the elements of an array $b$ into any number $k$ of non-empty multisets $S_1, S_2, \ldots, S_k$, where $k$ is an arbitrary positive integer. Define the score of $b$ as the maximum value of $\operatorname{MEX}(S_1)$$^{\text{∗}}$$ + \operatorname{MEX}(S_2) + \ldots + \operatorname{MEX}(S_k)$ over all possible partitions of $b$ for any integer $k$.

Envy is given an array $a$ of size $n$. Since he knows that calculating the score of $a$ is too easy for you, he instead asks you to calculate the sum of scores of all $2^n - 1$ non-empty subsequences of $a$.$^{\text{†}}$ Since this answer may be large, please output it modulo $998\,244\,353$.

$^{\text{∗}}$$\operatorname{MEX}$ of a collection of integers $c_1, c_2, \ldots, c_k$ is defined as the smallest non-negative integer $x$ that does not occur in the collection $c$. For example, $\operatorname{MEX}([0,1,2,2]) = 3$ and $\operatorname{MEX}([1,2,2]) = 0$

$^{\text{†}}$A sequence $x$ is a subsequence of a sequence $y$ if $x$ can be obtained from $y$ by deleting several (possibly, zero or all) elements.

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

The first line of each test case contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of $a$.

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

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

Output: For each test case, output the answer, modulo $998\,244\,353$.

Examples:
input:
4
3
0 0 1
4
0 0 1 1
5
0 0 1 2 2
4
1 1 1 1
output:
11
26
53
0

Note: In the first testcase, we must consider seven subsequences:

  * $[0]$: The score is $1$.    * $[0]$: The score is $1$.    * $[1]$: The score is $0$.    * $[0,0]$: The score is $2$.    * $[0,1]$: The score is $2$.    * $[0,1]$: The score is $2$.    * $[0,0,1]$: The score is $3$.

The answer for the first testcase is $1+1+2+2+2+3=11$.

In the last testcase, all subsequences have a score of $0$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main()
{
    int t; cin>>t;
    if(t==4)
    {
        printf("11\n26\n53\n0");
    }
    else
    {
        printf("11\n26\n53\n0");
    }
    cin>>t;
}

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

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

Think step by step.

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

1. **Edge Cases**: Test cases with minimal or maximal values for `n` and `a_i`.
2. **Special Patterns**: Test cases with specific patterns in the array that might cause the faulty solution to fail.
3. **Multiple Test Cases**: Ensure the test input includes multiple test cases to cover different scenarios.

The provided buggy code simply prints the same output for any input, which is clearly incorrect. We need to create a test input that will demonstrate this bug.

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

1. **Test Case with Minimal `n`**: Use `n = 1` to see how the code handles a single element.
2. **Test Case with Maximal `n`**: Use `n` close to the upper limit to see how the code handles a large array.
3. **Test Case with Specific Patterns**: Use an array with all elements the same or an array with a specific pattern that might cause the faulty solution to fail.
4. **Test Case with Mixed Values**: Use an array with a mix of values to see how the code handles a more general case.

Here is the test input:

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

This test input includes:
- A test case with `n = 1` to check the minimal case.
- A test case with `n = 2` to check a small but non-trivial case.
- A test case with `n = 3` to check a slightly larger case.
- A test case with `n = 4` to check a larger case.
- A test case with `n = 5` to check a larger case with a specific pattern.

This should expose the bug in the provided code, as the code will print the same output for all test cases, which is incorrect.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: MEXimize the Score
time_limit_ms: 2000
memory_limit_mb: 256
Description: Suppose we partition the elements of an array $b$ into any number $k$ of non-empty multisets $S_1, S_2, \ldots, S_k$, where $k$ is an arbitrary positive integer. Define the score of $b$ as the maximum value of $\operatorname{MEX}(S_1)$$^{\text{∗}}$$ + \operatorname{MEX}(S_2) + \ldots + \operatorname{MEX}(S_k)$ over all possible partitions of $b$ for any integer $k$.

Envy is given an array $a$ of size $n$. Since he knows that calculating the score of $a$ is too easy for you, he instead asks you to calculate the sum of scores of all $2^n - 1$ non-empty subsequences of $a$.$^{\text{†}}$ Since this answer may be large, please output it modulo $998\,244\,353$.

$^{\text{∗}}$$\operatorname{MEX}$ of a collection of integers $c_1, c_2, \ldots, c_k$ is defined as the smallest non-negative integer $x$ that does not occur in the collection $c$. For example, $\operatorname{MEX}([0,1,2,2]) = 3$ and $\operatorname{MEX}([1,2,2]) = 0$

$^{\text{†}}$A sequence $x$ is a subsequence of a sequence $y$ if $x$ can be obtained from $y$ by deleting several (possibly, zero or all) elements.

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

The first line of each test case contains an integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the length of $a$.

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

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

Output: For each test case, output the answer, modulo $998\,244\,353$.

Examples:
input:
4
3
0 0 1
4
0 0 1 1
5
0 0 1 2 2
4
1 1 1 1
output:
11
26
53
0

Note: In the first testcase, we must consider seven subsequences:

  * $[0]$: The score is $1$.    * $[0]$: The score is $1$.    * $[1]$: The score is $0$.    * $[0,0]$: The score is $2$.    * $[0,1]$: The score is $2$.    * $[0,1]$: The score is $2$.    * $[0,0,1]$: The score is $3$.

The answer for the first testcase is $1+1+2+2+2+3=11$.

In the last testcase, all subsequences have a score of $0$.

Buggy Code:
#include <bits/stdc++.h>
#define MAX (ll) 2e5+5
#define mod (ll) 998244353
#define all(x) x.begin(),x.end()
typedef long long int ll;
#define pr pair<ll,ll>
typedef long double dl;
using namespace std;
ll n,x,y,p,k,m,s,i,j;
ll fact[MAX],inv[MAX];
ll sum[2][MAX],freq[2][MAX];
ll power(int a,int b){
    if(b==0) return 1;
    ll ans=power(a,b/2);
    if(b%2) return 1ll*ans*ans%mod*a%mod;
    return 1ll*ans*ans%mod;
}
void get_fact(){
    fact[0]=1;
    for(int i=1;i<MAX;i++) fact[i]=1ll*fact[i-1]*i%mod;
    inv[MAX-1]=power(fact[MAX-1],mod-2);
    for(int i=MAX-2;i>=0;i--) inv[i]=1ll*(i+1)*inv[i+1]%mod;
}
ll ncr(int n,int r){
    // if(!r) return 0ll;
    return 1ll*fact[n]*inv[r]%mod*inv[n-r]%mod;
}
int getMod(int i){
    return (i+2)%2;
}
void solve(){
    cin>>n;
    vector<int> v(n),hash(n+1);
    for(int &i:v) cin>>i,hash[i]++;
    for(int i=0;i<=hash[0];i++){
        sum[0][i]=ncr(hash[0],i)*i;
        freq[0][i]=ncr(hash[0],i);
        // cout<<i<<"("<<sum[0][i]-sum[0][i+1]<<" "<<freq[0][i]-freq[0][i+1]<<") ";
    }
    // cout<<endl;
    for(int i=hash[0]-1;i>=0;i--){
        sum[0][i]+=sum[0][i+1];
        freq[0][i]+=freq[0][i+1];
    }
    ll total=sum[0][0],prev=0;
    for(int i=1;i<=n;i++){
        // if(hash[i]==0) break;
        // cout<<i<<"->";
        prev=0;
        sum[getMod(i)][hash[i]+1]=freq[getMod(i)][hash[i]+1]=0;
        for(int j=hash[i];j>=0;j--){
            // cout<<"prev is "<<prev<<endl;
            // freq[getMod(i)][j]=1ll*ncr(hash[i],j)*freq[getMod(i-1)][j]+freq[getMod(i)][j+1];
            // sum[getMod(i)][j]=1ll*j*ncr(hash[i],j)*freq[getMod(i-1)][j]+(hash[i-1]>=j?sum[getMod(i-1)][j]:0)*ncr(hash[i],j)+sum[getMod(i)][j+1];
            // cout<<j<<"("<<sum[getMod(i)][j]<<" "<<sum[getMod(i)][j+1]<<" "<<(sum[getMod(i-1)][j])<<" "<<(sum[getMod(i-1)][j+1])<<" "<<getMod(i)<<" "<<getMod(i-1)<<") ";
            // sum[getMod(i)][j]+=1ll*prev*(freq[getMod(i-1)][j]-freq[getMod(i-1)][j+1])*j+(((hash[i-1]>=j?sum[getMod(i-1)][j]:0)-(hash[i-1]>=j+1?sum[getMod(i-1)][j+1]:0)))*prev;
            // freq[getMod(i)][j]+=1ll*prev*((hash[i-1]>=j?freq[getMod(i-1)][j]:0)-(hash[i-1]>=j?freq[getMod(i-1)][j+1]:0));
            // cout<<j<<"("<<sum[getMod(i)][j]-sum[getMod(i)][j+1]<<" "<<freq[getMod(i)][j]-freq[getMod(i)][j+1]<<") ";
            sum[getMod(i)][j]=((hash[i-1]>=j?sum[getMod(i-1)][j]:0)+j*(hash[i-1]>=j?freq[getMod(i-1)][j]:0))*ncr(hash[i],j)+sum[getMod(i)][j+1];
            freq[getMod(i)][j]=1ll*ncr(hash[i],j)*freq[getMod(i-1)][j]+freq[getMod(i)][j+1];
            sum[getMod(i)][j]+=((hash[i-1]>=j?sum[getMod(i-1)][j]:0)-(hash[i-1]>=j+1?sum[getMod(i-1)][j+1]:0)+j*((hash[i-1]>=j?freq[getMod(i-1)][j]:0)-(hash[i-1]>=j+1?freq[getMod(i-1)][j+1]:0)))*prev;//+sum[getMod(i)][j+1];
            freq[getMod(i)][j]+=1ll*prev*((hash[i-1]>=j?freq[getMod(i-1)][j]:0)-(hash[i-1]>=j+1?freq[getMod(i-1)][j+1]:0));
            prev+=ncr(hash[i],j);
        }
        total=sum[getMod(i)][0];
    }
    // cout<<"str"<<endl;
    // for(int i=0;i<=1;i++){
    //     cout<<i<<"->";
    //     for(int j=0;j<=n;j++){
    //         cout<<j<<"("<<sum[i][j]-sum[i][j+1]<<" "<<freq[i][j]-freq[i][j+1]<<") ";
    //     }
    //     cout<<endl;
    // }
    cout<<total<<endl;
}
int main(){
    int t=1;
    // std::ios::sync_with_stdio(false);
    // std::cin.tie(nullptr);
    get_fact();
    cin>>t;
    // memset(dp,-1,sizeof(dp));
    while(t--) solve();
    return 0;
}

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

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

Think step by step.

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