**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Destruction of the Universe (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, $n \leq 5000$. You can only make hacks if both versions of the problem are solved.

Orangutans are powerful beings—so powerful that they only need $1$ unit of time to destroy every vulnerable planet in the universe!

There are $n$ planets in the universe. Each planet has an interval of vulnerability $[l, r]$, during which it will be exposed to destruction by orangutans. Orangutans can also expand the interval of vulnerability of any planet by $1$ unit.

Specifically, suppose the expansion is performed on planet $p$ with interval of vulnerability $[l_p, r_p]$. Then, the resulting interval of vulnerability may be either $[l_p - 1, r_p]$ or $[l_p, r_p + 1]$.

Given a set of planets, orangutans can destroy all planets if the intervals of vulnerability of all planets in the set intersect at least one common point. Let the score of such a set denote the minimum number of expansions that must be performed.

Orangutans are interested in the sum of scores of all non-empty subsets of the planets in the universe. As the answer can be large, output it modulo $998\,244\,353$.

Input: The first line contains an integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 5000$) — the number of planets in the universe.

The following $n$ lines contain two integers $l_i$ and $r_i$ ($1 \leq l_i \leq r_i \leq n$) — the initial interval of vulnerability of the $i$-th planet.

It is guaranteed that the sum of $n$ does not exceed $5000$ over all test cases.

Output: For each test case, output an integer — the sum of scores to destroy all non- empty subsets of the planets in the universe, modulo $998\,244\,353$.

Examples:
input:
3
3
1 1
2 3
3 3
4
1 4
2 3
2 4
1 1
5
1 2
2 3
3 4
4 5
1 5
output:
5
6
24

Note: In the first testcase, there are seven non-empty subsets of planets we must consider:

  * For each of the subsets $\\{[1,1]\\}, \\{[2,3]\\}, \\{[3,3]\\}$, the score is $0$.    * For the subset $\\{[2,3], [3,3]\\}$, the score is $0$, because the point $3$ is already contained in both planets' interval of vulnerability.    * For the subset $\\{[1,1], [2,3]\\}$, the score is $1$. By using one operation on changing the interval of vulnerability of the second planet to be $[1,3]$, the two planets now both have the point $1$ in their interval.    * For the subset $\\{[1,1], [3,3]\\}$, the score is $2$. By using two operations on changing the interval of vulnerability of the first planet to be $[1,3]$, the two planets now both have the point $3$ in their interval.    * For the subset $\\{[1,1], [2,3], [3,3]\\}$, the score is $2$. By using one operation on changing the interval of vulnerability of the first planet to be $[1,2]$ and one operation on changing the interval of vulnerability of the third planet to $[2,3]$, all three planets will have the point $2$ in their interval.

The sum of scores of all non-empty subsets of the first testcase is $0 \cdot 4 + 1 \cdot 1 + 2\cdot2 = 5$.

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

public class Main {
    static final int MOD = 998244353;
    static final int MAX_N = 5000;
    static long[] pow2 = new long[MAX_N + 1];

    // Function to precompute powers of 2 modulo MOD
    static void precomputePowers() {
        pow2[0] = 1;
        for(int i = 1; i <= MAX_N; i++) {
            pow2[i] = (pow2[i-1] * 2) % MOD;
        }
    }

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

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(br.readLine());
        while(t-- >0){
            int n = Integer.parseInt(br.readLine());
            // Initialize counts[L][R] = number of planets with l_i = L and r_i = R
            int[][] counts = new int[n+2][n+2];
            for(int i =0; i < n; i++) {
                StringTokenizer st = new StringTokenizer(br.readLine());
                int l = Integer.parseInt(st.nextToken());
                int r = Integer.parseInt(st.nextToken());
                counts[l][r]++;
            }

            // Compute c[L][R] = number of planets with l_i <= L and r_i >= R
            long[][] c = new long[n+2][n+2];
            // Initialize difference array
            long[][] D = new long[n+2][n+2];
            for(int L=1; L<=n; L++) {
                for(int R=1; R<=n; R++) {
                    if(counts[L][R] >0){
                        D[L][1] = (D[L][1] + counts[L][R]) % MOD;
                        D[L][R+1] = (D[L][R+1] - counts[L][R] + MOD) % MOD;
                    }
                }
            }

            // Compute prefix sums along R
            for(int L=1; L<=n; L++) {
                for(int R=1; R<=n; R++) {
                    D[L][R] = (D[L][R] + D[L][R-1]) % MOD;
                }
            }

            // Compute prefix sums along L
            for(int L=1; L<=n; L++) {
                for(int R=1; R<=n; R++) {
                    if(L >1){
                        D[L][R] = (D[L][R] + D[L-1][R]) % MOD;
                    }
                    c[L][R] = D[L][R];
                }
            }

            // Compute sumA1[L][R] = number of planets with l_i = L and r_i >= R
            long[][] sumA1 = new long[n+2][n+2];
            for(int L=1; L<=n; L++) {
                for(int R=n; R>=1; R--){
                    sumA1[L][R] = counts[L][R];
                    if(R <n){
                        sumA1[L][R] = (sumA1[L][R] + sumA1[L][R+1]) % MOD;
                    }
                }
            }

            // Compute sumA2[L][R] = number of planets with r_i = R and l_i <= L
            long[][] sumA2 = new long[n+2][n+2];
            for(int R=1; R<=n; R++) {
                for(int L=1; L<=n; L++) {
                    sumA2[L][R] = counts[L][R];
                    if(L >1){
                        sumA2[L][R] = (sumA2[L][R] + sumA2[L-1][R]) % MOD;
                    }
                }
            }

            long answer =0;
            // Iterate over all L > R
            for(int L=1; L<=n; L++) {
                for(int R=1; R<L; R++) {
                    long totalC = c[L][R];
                    if(totalC ==0){
                        continue;
                    }
                    long A = sumA1[L][R];
                    long B = sumA2[L][R];
                    long C = counts[L][R];

                    // Calculate M(L, R) = 2^c - 2^{c-A} - 2^{c-B} + 2^{c -A - B + C}
                    long term1 = (totalC <= MAX_N) ? pow2[(int) totalC] : 0;
                    long term2 = (totalC >= A && (totalC - A) <= MAX_N) ? pow2[(int)(totalC - A)] : 0;
                    long term3 = (totalC >= B && (totalC - B) <= MAX_N) ? pow2[(int)(totalC - B)] : 0;
                    long exponent = totalC - A - B + C;
                    long term4 = (exponent >=0 && exponent <= MAX_N) ? pow2[(int)exponent] : 0;

                    // Calculate M with proper modular arithmetic
                    long M = (term1 - term2 - term3 + term4) % MOD;
                    if(M <0){
                        M += MOD;
                    }

                    // Calculate contribution
                    long contrib = ( (L - R) * M ) % MOD;
                    answer = (answer + contrib) % MOD;
                }
            }

            System.out.println(answer);
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Destruction of the Universe (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, $n \leq 5000$. You can only make hacks if both versions of the problem are solved.

Orangutans are powerful beings—so powerful that they only need $1$ unit of time to destroy every vulnerable planet in the universe!

There are $n$ planets in the universe. Each planet has an interval of vulnerability $[l, r]$, during which it will be exposed to destruction by orangutans. Orangutans can also expand the interval of vulnerability of any planet by $1$ unit.

Specifically, suppose the expansion is performed on planet $p$ with interval of vulnerability $[l_p, r_p]$. Then, the resulting interval of vulnerability may be either $[l_p - 1, r_p]$ or $[l_p, r_p + 1]$.

Given a set of planets, orangutans can destroy all planets if the intervals of vulnerability of all planets in the set intersect at least one common point. Let the score of such a set denote the minimum number of expansions that must be performed.

Orangutans are interested in the sum of scores of all non-empty subsets of the planets in the universe. As the answer can be large, output it modulo $998\,244\,353$.

Input: The first line contains an integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 5000$) — the number of planets in the universe.

The following $n$ lines contain two integers $l_i$ and $r_i$ ($1 \leq l_i \leq r_i \leq n$) — the initial interval of vulnerability of the $i$-th planet.

It is guaranteed that the sum of $n$ does not exceed $5000$ over all test cases.

Output: For each test case, output an integer — the sum of scores to destroy all non- empty subsets of the planets in the universe, modulo $998\,244\,353$.

Examples:
input:
3
3
1 1
2 3
3 3
4
1 4
2 3
2 4
1 1
5
1 2
2 3
3 4
4 5
1 5
output:
5
6
24

Note: In the first testcase, there are seven non-empty subsets of planets we must consider:

  * For each of the subsets $\\{[1,1]\\}, \\{[2,3]\\}, \\{[3,3]\\}$, the score is $0$.    * For the subset $\\{[2,3], [3,3]\\}$, the score is $0$, because the point $3$ is already contained in both planets' interval of vulnerability.    * For the subset $\\{[1,1], [2,3]\\}$, the score is $1$. By using one operation on changing the interval of vulnerability of the second planet to be $[1,3]$, the two planets now both have the point $1$ in their interval.    * For the subset $\\{[1,1], [3,3]\\}$, the score is $2$. By using two operations on changing the interval of vulnerability of the first planet to be $[1,3]$, the two planets now both have the point $3$ in their interval.    * For the subset $\\{[1,1], [2,3], [3,3]\\}$, the score is $2$. By using one operation on changing the interval of vulnerability of the first planet to be $[1,2]$ and one operation on changing the interval of vulnerability of the third planet to $[2,3]$, all three planets will have the point $2$ in their interval.

The sum of scores of all non-empty subsets of the first testcase is $0 \cdot 4 + 1 \cdot 1 + 2\cdot2 = 5$.

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

using namespace std;

const int MOD = 998244353;

int main() {
    int t;
    cin >> t;

    while (t--) {
        int n;
        cin >> n;

        vector<pair<int, int>> iv(n);
        for (int i = 0; i < n; ++i) {
            cin >> iv[i].first >> iv[i].second;
        }

        long long totalScore = 0;

        sort(iv.begin(), iv.end());

        for (int mask = 1; mask < (1 << n); ++mask) {
            int maxL = -1e9;
            int minR = 1e9;
            int expansionsNeeded = 0;

            for (int j = 0; j < n; ++j) {
                if (mask & (1 << j)) {
                    maxL = max(maxL, iv[j].first);
                    minR = min(minR, iv[j].second);
                }
            }

            if (maxL > minR) {
                expansionsNeeded = maxL - minR;
            }

            totalScore = (totalScore + expansionsNeeded) % MOD;
        }

        cout << totalScore << endl;
    }

    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Destruction of the Universe (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, $n \leq 5000$. You can only make hacks if both versions of the problem are solved.

Orangutans are powerful beings—so powerful that they only need $1$ unit of time to destroy every vulnerable planet in the universe!

There are $n$ planets in the universe. Each planet has an interval of vulnerability $[l, r]$, during which it will be exposed to destruction by orangutans. Orangutans can also expand the interval of vulnerability of any planet by $1$ unit.

Specifically, suppose the expansion is performed on planet $p$ with interval of vulnerability $[l_p, r_p]$. Then, the resulting interval of vulnerability may be either $[l_p - 1, r_p]$ or $[l_p, r_p + 1]$.

Given a set of planets, orangutans can destroy all planets if the intervals of vulnerability of all planets in the set intersect at least one common point. Let the score of such a set denote the minimum number of expansions that must be performed.

Orangutans are interested in the sum of scores of all non-empty subsets of the planets in the universe. As the answer can be large, output it modulo $998\,244\,353$.

Input: The first line contains an integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 5000$) — the number of planets in the universe.

The following $n$ lines contain two integers $l_i$ and $r_i$ ($1 \leq l_i \leq r_i \leq n$) — the initial interval of vulnerability of the $i$-th planet.

It is guaranteed that the sum of $n$ does not exceed $5000$ over all test cases.

Output: For each test case, output an integer — the sum of scores to destroy all non- empty subsets of the planets in the universe, modulo $998\,244\,353$.

Examples:
input:
3
3
1 1
2 3
3 3
4
1 4
2 3
2 4
1 1
5
1 2
2 3
3 4
4 5
1 5
output:
5
6
24

Note: In the first testcase, there are seven non-empty subsets of planets we must consider:

  * For each of the subsets $\\{[1,1]\\}, \\{[2,3]\\}, \\{[3,3]\\}$, the score is $0$.    * For the subset $\\{[2,3], [3,3]\\}$, the score is $0$, because the point $3$ is already contained in both planets' interval of vulnerability.    * For the subset $\\{[1,1], [2,3]\\}$, the score is $1$. By using one operation on changing the interval of vulnerability of the second planet to be $[1,3]$, the two planets now both have the point $1$ in their interval.    * For the subset $\\{[1,1], [3,3]\\}$, the score is $2$. By using two operations on changing the interval of vulnerability of the first planet to be $[1,3]$, the two planets now both have the point $3$ in their interval.    * For the subset $\\{[1,1], [2,3], [3,3]\\}$, the score is $2$. By using one operation on changing the interval of vulnerability of the first planet to be $[1,2]$ and one operation on changing the interval of vulnerability of the third planet to $[2,3]$, all three planets will have the point $2$ in their interval.

The sum of scores of all non-empty subsets of the first testcase is $0 \cdot 4 + 1 \cdot 1 + 2\cdot2 = 5$.

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

using namespace std;

const int MOD = 998244353;

long long calculateTotalScore(vector<pair<int, int>>& intervals) {
    long long totalScore = 0;
    int n = intervals.size();

    for (int mask = 1; mask < (1 << n); ++mask) {
        int maxL = -1e9;
        int minR = 1e9;
        int expansionsNeeded = 0;

        for (int j = 0; j < n; ++j) {
            if (mask & (1 << j)) {
                maxL = max(maxL, intervals[j].first);
                minR = min(minR, intervals[j].second);
            }
        }

        if (maxL > minR) {
            expansionsNeeded = maxL - minR;
        }

        totalScore = (totalScore + expansionsNeeded) % MOD;
    }

    return totalScore;
}

int main() {
    int t;
    cin >> t;

    while (t--) {
        int n;
        cin >> n;

        vector<pair<int, int>> intervals(n);
        for (int i = 0; i < n; ++i) {
            cin >> intervals[i].first >> intervals[i].second;
        }

        long long result = calculateTotalScore(intervals);
        cout << result << endl;
    }

    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Destruction of the Universe (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, $n \leq 5000$. You can only make hacks if both versions of the problem are solved.

Orangutans are powerful beings—so powerful that they only need $1$ unit of time to destroy every vulnerable planet in the universe!

There are $n$ planets in the universe. Each planet has an interval of vulnerability $[l, r]$, during which it will be exposed to destruction by orangutans. Orangutans can also expand the interval of vulnerability of any planet by $1$ unit.

Specifically, suppose the expansion is performed on planet $p$ with interval of vulnerability $[l_p, r_p]$. Then, the resulting interval of vulnerability may be either $[l_p - 1, r_p]$ or $[l_p, r_p + 1]$.

Given a set of planets, orangutans can destroy all planets if the intervals of vulnerability of all planets in the set intersect at least one common point. Let the score of such a set denote the minimum number of expansions that must be performed.

Orangutans are interested in the sum of scores of all non-empty subsets of the planets in the universe. As the answer can be large, output it modulo $998\,244\,353$.

Input: The first line contains an integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 5000$) — the number of planets in the universe.

The following $n$ lines contain two integers $l_i$ and $r_i$ ($1 \leq l_i \leq r_i \leq n$) — the initial interval of vulnerability of the $i$-th planet.

It is guaranteed that the sum of $n$ does not exceed $5000$ over all test cases.

Output: For each test case, output an integer — the sum of scores to destroy all non- empty subsets of the planets in the universe, modulo $998\,244\,353$.

Examples:
input:
3
3
1 1
2 3
3 3
4
1 4
2 3
2 4
1 1
5
1 2
2 3
3 4
4 5
1 5
output:
5
6
24

Note: In the first testcase, there are seven non-empty subsets of planets we must consider:

  * For each of the subsets $\\{[1,1]\\}, \\{[2,3]\\}, \\{[3,3]\\}$, the score is $0$.    * For the subset $\\{[2,3], [3,3]\\}$, the score is $0$, because the point $3$ is already contained in both planets' interval of vulnerability.    * For the subset $\\{[1,1], [2,3]\\}$, the score is $1$. By using one operation on changing the interval of vulnerability of the second planet to be $[1,3]$, the two planets now both have the point $1$ in their interval.    * For the subset $\\{[1,1], [3,3]\\}$, the score is $2$. By using two operations on changing the interval of vulnerability of the first planet to be $[1,3]$, the two planets now both have the point $3$ in their interval.    * For the subset $\\{[1,1], [2,3], [3,3]\\}$, the score is $2$. By using one operation on changing the interval of vulnerability of the first planet to be $[1,2]$ and one operation on changing the interval of vulnerability of the third planet to $[2,3]$, all three planets will have the point $2$ in their interval.

The sum of scores of all non-empty subsets of the first testcase is $0 \cdot 4 + 1 \cdot 1 + 2\cdot2 = 5$.

Buggy Code:
MOD = 998244353

def min_expansions(intervals):
    n = len(intervals)
    dp = [[float('inf')] * (n + 1) for _ in range(1 << n)]

    # Initialize base cases
    for i in range(n):
        l, r = intervals[i]
        for p in range(l, r + 1):
            dp[1 << i][p] = 0

    # Fill DP table
    for mask in range(1 << n):
        for i in range(n):
            if mask & (1 << i):
                continue
            new_mask = mask | (1 << i)
            l, r = intervals[i]
            for p in range(1, n + 1):
                if dp[mask][p] == float('inf'):
                    continue
                # Expand left boundary
                if p <= r:
                    dp[new_mask][max(p, l)] = min(dp[new_mask][max(p, l)], dp[mask][p] + max(0, l - p))
                # Expand right boundary
                if p >= l:
                    dp[new_mask][min(p, r)] = min(dp[new_mask][min(p, r)], dp[mask][p] + max(0, p - r))

    # Compute total score for all non-empty subsets
    total_score = 0
    for mask in range(1, 1 << n):
        min_cost = min(dp[mask])
        total_score += min_cost
        total_score %= MOD

    return total_score

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

    index = 0
    t = int(data[index])
    index += 1
    results = []

    for _ in range(t):
        n = int(data[index])
        index += 1
        intervals = []

        for __ in range(n):
            l = int(data[index])
            r = int(data[index + 1])
            intervals.append((l, r))
            index += 2

        result = min_expansions(intervals)
        results.append(result)

    for res in results:
        print(res)

if __name__ == "__main__":
    main()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Destruction of the Universe (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, $n \leq 5000$. You can only make hacks if both versions of the problem are solved.

Orangutans are powerful beings—so powerful that they only need $1$ unit of time to destroy every vulnerable planet in the universe!

There are $n$ planets in the universe. Each planet has an interval of vulnerability $[l, r]$, during which it will be exposed to destruction by orangutans. Orangutans can also expand the interval of vulnerability of any planet by $1$ unit.

Specifically, suppose the expansion is performed on planet $p$ with interval of vulnerability $[l_p, r_p]$. Then, the resulting interval of vulnerability may be either $[l_p - 1, r_p]$ or $[l_p, r_p + 1]$.

Given a set of planets, orangutans can destroy all planets if the intervals of vulnerability of all planets in the set intersect at least one common point. Let the score of such a set denote the minimum number of expansions that must be performed.

Orangutans are interested in the sum of scores of all non-empty subsets of the planets in the universe. As the answer can be large, output it modulo $998\,244\,353$.

Input: The first line contains an integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 5000$) — the number of planets in the universe.

The following $n$ lines contain two integers $l_i$ and $r_i$ ($1 \leq l_i \leq r_i \leq n$) — the initial interval of vulnerability of the $i$-th planet.

It is guaranteed that the sum of $n$ does not exceed $5000$ over all test cases.

Output: For each test case, output an integer — the sum of scores to destroy all non- empty subsets of the planets in the universe, modulo $998\,244\,353$.

Examples:
input:
3
3
1 1
2 3
3 3
4
1 4
2 3
2 4
1 1
5
1 2
2 3
3 4
4 5
1 5
output:
5
6
24

Note: In the first testcase, there are seven non-empty subsets of planets we must consider:

  * For each of the subsets $\\{[1,1]\\}, \\{[2,3]\\}, \\{[3,3]\\}$, the score is $0$.    * For the subset $\\{[2,3], [3,3]\\}$, the score is $0$, because the point $3$ is already contained in both planets' interval of vulnerability.    * For the subset $\\{[1,1], [2,3]\\}$, the score is $1$. By using one operation on changing the interval of vulnerability of the second planet to be $[1,3]$, the two planets now both have the point $1$ in their interval.    * For the subset $\\{[1,1], [3,3]\\}$, the score is $2$. By using two operations on changing the interval of vulnerability of the first planet to be $[1,3]$, the two planets now both have the point $3$ in their interval.    * For the subset $\\{[1,1], [2,3], [3,3]\\}$, the score is $2$. By using one operation on changing the interval of vulnerability of the first planet to be $[1,2]$ and one operation on changing the interval of vulnerability of the third planet to $[2,3]$, all three planets will have the point $2$ in their interval.

The sum of scores of all non-empty subsets of the first testcase is $0 \cdot 4 + 1 \cdot 1 + 2\cdot2 = 5$.

Buggy Code:
def sum_of_scores(planets):
    MOD = 998244353
    n = len(planets)
    total_score = 0

    # Iterate through all non-empty subsets using bitmask
    for mask in range(1, 1 << n):
        L = float('-inf')
        R = float('inf')

        # Calculate L and R for the current subset
        for i in range(n):
            if mask & (1 << i):  # If the i-th planet is included in the subset
                l_i, r_i = planets[i]
                L = max(L, l_i)
                R = min(R, r_i)

        # Calculate the number of expansions needed
        if L > R:
            expansions = L - R
        else:
            expansions = 0

        # Add the score of this subset to the total score
        total_score = (total_score + expansions) % MOD

    return total_score

t = int(input())
for i in range(t):
    n = int(input())
    planets = []
    for j in range(n):
        planets.append([int(x) for x in input().split()])
    result = sum_of_scores(planets)
    print(result)  # Output the 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
1
3
1 2
2 3
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Destruction of the Universe (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, $n \leq 5000$. You can only make hacks if both versions of the problem are solved.

Orangutans are powerful beings—so powerful that they only need $1$ unit of time to destroy every vulnerable planet in the universe!

There are $n$ planets in the universe. Each planet has an interval of vulnerability $[l, r]$, during which it will be exposed to destruction by orangutans. Orangutans can also expand the interval of vulnerability of any planet by $1$ unit.

Specifically, suppose the expansion is performed on planet $p$ with interval of vulnerability $[l_p, r_p]$. Then, the resulting interval of vulnerability may be either $[l_p - 1, r_p]$ or $[l_p, r_p + 1]$.

Given a set of planets, orangutans can destroy all planets if the intervals of vulnerability of all planets in the set intersect at least one common point. Let the score of such a set denote the minimum number of expansions that must be performed.

Orangutans are interested in the sum of scores of all non-empty subsets of the planets in the universe. As the answer can be large, output it modulo $998\,244\,353$.

Input: The first line contains an integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 5000$) — the number of planets in the universe.

The following $n$ lines contain two integers $l_i$ and $r_i$ ($1 \leq l_i \leq r_i \leq n$) — the initial interval of vulnerability of the $i$-th planet.

It is guaranteed that the sum of $n$ does not exceed $5000$ over all test cases.

Output: For each test case, output an integer — the sum of scores to destroy all non- empty subsets of the planets in the universe, modulo $998\,244\,353$.

Examples:
input:
3
3
1 1
2 3
3 3
4
1 4
2 3
2 4
1 1
5
1 2
2 3
3 4
4 5
1 5
output:
5
6
24

Note: In the first testcase, there are seven non-empty subsets of planets we must consider:

  * For each of the subsets $\\{[1,1]\\}, \\{[2,3]\\}, \\{[3,3]\\}$, the score is $0$.    * For the subset $\\{[2,3], [3,3]\\}$, the score is $0$, because the point $3$ is already contained in both planets' interval of vulnerability.    * For the subset $\\{[1,1], [2,3]\\}$, the score is $1$. By using one operation on changing the interval of vulnerability of the second planet to be $[1,3]$, the two planets now both have the point $1$ in their interval.    * For the subset $\\{[1,1], [3,3]\\}$, the score is $2$. By using two operations on changing the interval of vulnerability of the first planet to be $[1,3]$, the two planets now both have the point $3$ in their interval.    * For the subset $\\{[1,1], [2,3], [3,3]\\}$, the score is $2$. By using one operation on changing the interval of vulnerability of the first planet to be $[1,2]$ and one operation on changing the interval of vulnerability of the third planet to $[2,3]$, all three planets will have the point $2$ in their interval.

The sum of scores of all non-empty subsets of the first testcase is $0 \cdot 4 + 1 \cdot 1 + 2\cdot2 = 5$.

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

public class Main {
    static final int MOD = 998244353;
    static final int MAX_N = 5000;
    static long[] pow2 = new long[MAX_N + 1];

    public static void main(String[] args) throws IOException {
        // Precompute powers of 2 modulo MOD
        pow2[0] = 1;
        for(int i=1;i<=MAX_N;i++) {
            pow2[i] = (pow2[i-1] * 2) % MOD;
        }

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(br.readLine());
        while(t-- >0){
            int n = Integer.parseInt(br.readLine());
            // Initialize counts
            // counts[L][R] = number of planets with l_i = L and r_i = R
            int[][] counts = new int[n+2][n+2];
            for(int i=0;i<n;i++){
                String[] parts = br.readLine().split(" ");
                int l = Integer.parseInt(parts[0]);
                int r = Integer.parseInt(parts[1]);
                counts[l][r]++;
            }

            // Precompute sum2D[L][R] = number of planets with l_i <= L and r_i >= R
            long[][] sum2D = new long[n+2][n+2];
            for(int L=1; L<=n; L++){
                for(int R=n; R>=1; R--){
                    sum2D[L][R] = counts[L][R];
                    if(L >1){
                        sum2D[L][R] = (sum2D[L][R] + sum2D[L-1][R]) % MOD;
                    }
                    if(R <n){
                        sum2D[L][R] = (sum2D[L][R] + sum2D[L][R+1]) % MOD;
                    }
                    if(L >1 && R <n){
                        sum2D[L][R] = (sum2D[L][R] - sum2D[L-1][R+1] + MOD) % MOD;
                    }
                }
            }

            // Precompute sumA1[L][R] = number of planets with l_i = L and r_i >= R
            long[][] sumA1 = new long[n+2][n+2];
            for(int L=1; L<=n; L++){
                for(int R=n; R>=1; R--){
                    sumA1[L][R] = counts[L][R];
                    if(R <n){
                        sumA1[L][R] = (sumA1[L][R] + sumA1[L][R+1]) % MOD;
                    }
                }
            }

            // Precompute sumA2[L][R] = number of planets with r_i = R and l_i <= L
            long[][] sumA2 = new long[n+2][n+2];
            for(int R=1; R<=n; R++){
                for(int L=1; L<=n; L++){
                    sumA2[L][R] = counts[L][R];
                    if(L >1){
                        sumA2[L][R] = (sumA2[L][R] + sumA2[L-1][R]) % MOD;
                    }
                }
            }

            // Now, iterate over all L > R
            long answer =0;
            for(int L=1; L<=n; L++){
                for(int R=1; R<L; R++){
                    long N = sum2D[L][R];
                    if(N ==0){
                        continue;
                    }
                    long NA1 = sumA1[L][R];
                    long NA2 = sumA2[L][R];
                    long NA12 = counts[L][R];

                    // If either NA1 or NA2 is zero, there are no valid subsets
                    if(NA1 ==0 || NA2 ==0){
                        continue;
                    }

                    // Compute exponent parts, ensure that exponents are within 0 to MAX_N
                    // pow2[k] precomputed
                    // M(L,R) = (pow2[N] -1) - (pow2[N - NA1] -1) - (pow2[N - NA2] -1) + (pow2[N - NA1 - NA2 + NA12] -1)
                    long term1 = (pow2[(int)N] -1 + MOD) % MOD;
                    long term2 = (pow2[(int)(N - NA1)] -1 + MOD) % MOD;
                    long term3 = (pow2[(int)(N - NA2)] -1 + MOD) % MOD;
                    long term4 = 0;
                    if((int)(N - NA1 - NA2 + NA12) >=0){
                        term4 = (pow2[(int)(N - NA1 - NA2 + NA12)] -1 + MOD) % MOD;
                    }
                    long M = (term1 - term2 - term3 + term4) % MOD;
                    if(M <0){
                        M += MOD;
                    }
                    // (L - R) * M
                    long contrib = ((long)(L - R) * M) % MOD;
                    answer = (answer + contrib) % MOD;
                }
            }

            System.out.println(answer);
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Destruction of the Universe (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, $n \leq 5000$. You can only make hacks if both versions of the problem are solved.

Orangutans are powerful beings—so powerful that they only need $1$ unit of time to destroy every vulnerable planet in the universe!

There are $n$ planets in the universe. Each planet has an interval of vulnerability $[l, r]$, during which it will be exposed to destruction by orangutans. Orangutans can also expand the interval of vulnerability of any planet by $1$ unit.

Specifically, suppose the expansion is performed on planet $p$ with interval of vulnerability $[l_p, r_p]$. Then, the resulting interval of vulnerability may be either $[l_p - 1, r_p]$ or $[l_p, r_p + 1]$.

Given a set of planets, orangutans can destroy all planets if the intervals of vulnerability of all planets in the set intersect at least one common point. Let the score of such a set denote the minimum number of expansions that must be performed.

Orangutans are interested in the sum of scores of all non-empty subsets of the planets in the universe. As the answer can be large, output it modulo $998\,244\,353$.

Input: The first line contains an integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 5000$) — the number of planets in the universe.

The following $n$ lines contain two integers $l_i$ and $r_i$ ($1 \leq l_i \leq r_i \leq n$) — the initial interval of vulnerability of the $i$-th planet.

It is guaranteed that the sum of $n$ does not exceed $5000$ over all test cases.

Output: For each test case, output an integer — the sum of scores to destroy all non- empty subsets of the planets in the universe, modulo $998\,244\,353$.

Examples:
input:
3
3
1 1
2 3
3 3
4
1 4
2 3
2 4
1 1
5
1 2
2 3
3 4
4 5
1 5
output:
5
6
24

Note: In the first testcase, there are seven non-empty subsets of planets we must consider:

  * For each of the subsets $\\{[1,1]\\}, \\{[2,3]\\}, \\{[3,3]\\}$, the score is $0$.    * For the subset $\\{[2,3], [3,3]\\}$, the score is $0$, because the point $3$ is already contained in both planets' interval of vulnerability.    * For the subset $\\{[1,1], [2,3]\\}$, the score is $1$. By using one operation on changing the interval of vulnerability of the second planet to be $[1,3]$, the two planets now both have the point $1$ in their interval.    * For the subset $\\{[1,1], [3,3]\\}$, the score is $2$. By using two operations on changing the interval of vulnerability of the first planet to be $[1,3]$, the two planets now both have the point $3$ in their interval.    * For the subset $\\{[1,1], [2,3], [3,3]\\}$, the score is $2$. By using one operation on changing the interval of vulnerability of the first planet to be $[1,2]$ and one operation on changing the interval of vulnerability of the third planet to $[2,3]$, all three planets will have the point $2$ in their interval.

The sum of scores of all non-empty subsets of the first testcase is $0 \cdot 4 + 1 \cdot 1 + 2\cdot2 = 5$.

Buggy Code:
import java.util.*;

public class Main {
    public static final int MOD = 998244353;

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

        while (t-- > 0) {
            int n = sc.nextInt();
            int arr[][] = new int[n][2];

            for (int i = 0; i < n; i++) {
                arr[i][0] = sc.nextInt();
                arr[i][1] = sc.nextInt();
            }
            System.out.println(sumOfScores(n, arr));
        }
    }

    public static long sumOfScores(int n, int[][] arr) {
        long totalScore = 0;

        for (int j = 1; j < (1 << n); j++) {
            int lMax = Integer.MIN_VALUE;
            int rMin = Integer.MAX_VALUE;
            int expansionsNeeded = 0;

            for (int i = 0; i < n; i++) {
                if ((j & (1 << i)) != 0) {
                    lMax = Math.max(lMax, arr[i][0]);
                    rMin = Math.min(rMin, arr[i][1]);
                }
            }
            if (lMax > rMin) {
                expansionsNeeded = lMax - rMin;
            }
            totalScore = (totalScore + expansionsNeeded) % MOD;
        }

        return totalScore;
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Destruction of the Universe (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, $n \leq 5000$. You can only make hacks if both versions of the problem are solved.

Orangutans are powerful beings—so powerful that they only need $1$ unit of time to destroy every vulnerable planet in the universe!

There are $n$ planets in the universe. Each planet has an interval of vulnerability $[l, r]$, during which it will be exposed to destruction by orangutans. Orangutans can also expand the interval of vulnerability of any planet by $1$ unit.

Specifically, suppose the expansion is performed on planet $p$ with interval of vulnerability $[l_p, r_p]$. Then, the resulting interval of vulnerability may be either $[l_p - 1, r_p]$ or $[l_p, r_p + 1]$.

Given a set of planets, orangutans can destroy all planets if the intervals of vulnerability of all planets in the set intersect at least one common point. Let the score of such a set denote the minimum number of expansions that must be performed.

Orangutans are interested in the sum of scores of all non-empty subsets of the planets in the universe. As the answer can be large, output it modulo $998\,244\,353$.

Input: The first line contains an integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 5000$) — the number of planets in the universe.

The following $n$ lines contain two integers $l_i$ and $r_i$ ($1 \leq l_i \leq r_i \leq n$) — the initial interval of vulnerability of the $i$-th planet.

It is guaranteed that the sum of $n$ does not exceed $5000$ over all test cases.

Output: For each test case, output an integer — the sum of scores to destroy all non- empty subsets of the planets in the universe, modulo $998\,244\,353$.

Examples:
input:
3
3
1 1
2 3
3 3
4
1 4
2 3
2 4
1 1
5
1 2
2 3
3 4
4 5
1 5
output:
5
6
24

Note: In the first testcase, there are seven non-empty subsets of planets we must consider:

  * For each of the subsets $\\{[1,1]\\}, \\{[2,3]\\}, \\{[3,3]\\}$, the score is $0$.    * For the subset $\\{[2,3], [3,3]\\}$, the score is $0$, because the point $3$ is already contained in both planets' interval of vulnerability.    * For the subset $\\{[1,1], [2,3]\\}$, the score is $1$. By using one operation on changing the interval of vulnerability of the second planet to be $[1,3]$, the two planets now both have the point $1$ in their interval.    * For the subset $\\{[1,1], [3,3]\\}$, the score is $2$. By using two operations on changing the interval of vulnerability of the first planet to be $[1,3]$, the two planets now both have the point $3$ in their interval.    * For the subset $\\{[1,1], [2,3], [3,3]\\}$, the score is $2$. By using one operation on changing the interval of vulnerability of the first planet to be $[1,2]$ and one operation on changing the interval of vulnerability of the third planet to $[2,3]$, all three planets will have the point $2$ in their interval.

The sum of scores of all non-empty subsets of the first testcase is $0 \cdot 4 + 1 \cdot 1 + 2\cdot2 = 5$.

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

const int MOD = 998244353;

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

    while (t--) {
        int n;
        cin >> n;

        vector<pair<int, int>> planets(n);
        for (int i = 0; i < n; i++) {
            cin >> planets[i].first >> planets[i].second;
        }

        // Result for this test case
        long long total_score = 0;

        // Process each subset
        for (int mask = 1; mask < (1 << n); mask++) {
            int l_max = 1, r_min = n;
            int expansions = 0;
            for (int i = 0; i < n; i++) {
                if (mask & (1 << i)) {  // If the i-th planet is in the subset
                    l_max = max(l_max, planets[i].first);
                    r_min = min(r_min, planets[i].second);
                }
            }
            if (l_max <= r_min) {
                // No expansions needed if the intervals overlap
                total_score = (total_score + expansions) % MOD;
            } else {
                // Expansion needed to make intervals overlap
                expansions = l_max - r_min;
                total_score = (total_score + expansions) % MOD;
            }
        }

        cout << total_score << endl;
    }
}

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

    solve();
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Destruction of the Universe (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, $n \leq 5000$. You can only make hacks if both versions of the problem are solved.

Orangutans are powerful beings—so powerful that they only need $1$ unit of time to destroy every vulnerable planet in the universe!

There are $n$ planets in the universe. Each planet has an interval of vulnerability $[l, r]$, during which it will be exposed to destruction by orangutans. Orangutans can also expand the interval of vulnerability of any planet by $1$ unit.

Specifically, suppose the expansion is performed on planet $p$ with interval of vulnerability $[l_p, r_p]$. Then, the resulting interval of vulnerability may be either $[l_p - 1, r_p]$ or $[l_p, r_p + 1]$.

Given a set of planets, orangutans can destroy all planets if the intervals of vulnerability of all planets in the set intersect at least one common point. Let the score of such a set denote the minimum number of expansions that must be performed.

Orangutans are interested in the sum of scores of all non-empty subsets of the planets in the universe. As the answer can be large, output it modulo $998\,244\,353$.

Input: The first line contains an integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 5000$) — the number of planets in the universe.

The following $n$ lines contain two integers $l_i$ and $r_i$ ($1 \leq l_i \leq r_i \leq n$) — the initial interval of vulnerability of the $i$-th planet.

It is guaranteed that the sum of $n$ does not exceed $5000$ over all test cases.

Output: For each test case, output an integer — the sum of scores to destroy all non- empty subsets of the planets in the universe, modulo $998\,244\,353$.

Examples:
input:
3
3
1 1
2 3
3 3
4
1 4
2 3
2 4
1 1
5
1 2
2 3
3 4
4 5
1 5
output:
5
6
24

Note: In the first testcase, there are seven non-empty subsets of planets we must consider:

  * For each of the subsets $\\{[1,1]\\}, \\{[2,3]\\}, \\{[3,3]\\}$, the score is $0$.    * For the subset $\\{[2,3], [3,3]\\}$, the score is $0$, because the point $3$ is already contained in both planets' interval of vulnerability.    * For the subset $\\{[1,1], [2,3]\\}$, the score is $1$. By using one operation on changing the interval of vulnerability of the second planet to be $[1,3]$, the two planets now both have the point $1$ in their interval.    * For the subset $\\{[1,1], [3,3]\\}$, the score is $2$. By using two operations on changing the interval of vulnerability of the first planet to be $[1,3]$, the two planets now both have the point $3$ in their interval.    * For the subset $\\{[1,1], [2,3], [3,3]\\}$, the score is $2$. By using one operation on changing the interval of vulnerability of the first planet to be $[1,2]$ and one operation on changing the interval of vulnerability of the third planet to $[2,3]$, all three planets will have the point $2$ in their interval.

The sum of scores of all non-empty subsets of the first testcase is $0 \cdot 4 + 1 \cdot 1 + 2\cdot2 = 5$.

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

using namespace std;
const int MOD = 998244353;

// Function to ensure non-negative modulo result
inline int mod(int a, int m) {
    return (a % m + m) % m;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    int t;
    cin >> t;

    while (t--) {
        int n;
        cin >> n;

        vector<pair<int, int>> planets(n);
        for (int i = 0; i < n; ++i) {
            cin >> planets[i].first >> planets[i].second;
        }

        // To store the sum of scores for all subsets
        int totalScore = 0;

        // Iterate over all subsets using bitmasking
        for (int mask = 1; mask < (1 << n); ++mask) {
            int lmax = 1, rmin = n;  // Initialize lmax and rmin

            for (int i = 0; i < n; ++i) {
                if (mask & (1 << i)) {
                    lmax = max(lmax, planets[i].first);
                    rmin = min(rmin, planets[i].second);
                }
            }

            // Check if the subset has a valid range
            if (lmax <= rmin) {
                // No expansions needed, so expansions = 0
                totalScore = mod(totalScore, MOD);
            } else {
                // Calculate expansions
                int expansions = lmax - rmin;
                totalScore = mod(totalScore + expansions, MOD);
            }
        }

        cout << totalScore << endl;
    }

    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Destruction of the Universe (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, $n \leq 5000$. You can only make hacks if both versions of the problem are solved.

Orangutans are powerful beings—so powerful that they only need $1$ unit of time to destroy every vulnerable planet in the universe!

There are $n$ planets in the universe. Each planet has an interval of vulnerability $[l, r]$, during which it will be exposed to destruction by orangutans. Orangutans can also expand the interval of vulnerability of any planet by $1$ unit.

Specifically, suppose the expansion is performed on planet $p$ with interval of vulnerability $[l_p, r_p]$. Then, the resulting interval of vulnerability may be either $[l_p - 1, r_p]$ or $[l_p, r_p + 1]$.

Given a set of planets, orangutans can destroy all planets if the intervals of vulnerability of all planets in the set intersect at least one common point. Let the score of such a set denote the minimum number of expansions that must be performed.

Orangutans are interested in the sum of scores of all non-empty subsets of the planets in the universe. As the answer can be large, output it modulo $998\,244\,353$.

Input: The first line contains an integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 5000$) — the number of planets in the universe.

The following $n$ lines contain two integers $l_i$ and $r_i$ ($1 \leq l_i \leq r_i \leq n$) — the initial interval of vulnerability of the $i$-th planet.

It is guaranteed that the sum of $n$ does not exceed $5000$ over all test cases.

Output: For each test case, output an integer — the sum of scores to destroy all non- empty subsets of the planets in the universe, modulo $998\,244\,353$.

Examples:
input:
3
3
1 1
2 3
3 3
4
1 4
2 3
2 4
1 1
5
1 2
2 3
3 4
4 5
1 5
output:
5
6
24

Note: In the first testcase, there are seven non-empty subsets of planets we must consider:

  * For each of the subsets $\\{[1,1]\\}, \\{[2,3]\\}, \\{[3,3]\\}$, the score is $0$.    * For the subset $\\{[2,3], [3,3]\\}$, the score is $0$, because the point $3$ is already contained in both planets' interval of vulnerability.    * For the subset $\\{[1,1], [2,3]\\}$, the score is $1$. By using one operation on changing the interval of vulnerability of the second planet to be $[1,3]$, the two planets now both have the point $1$ in their interval.    * For the subset $\\{[1,1], [3,3]\\}$, the score is $2$. By using two operations on changing the interval of vulnerability of the first planet to be $[1,3]$, the two planets now both have the point $3$ in their interval.    * For the subset $\\{[1,1], [2,3], [3,3]\\}$, the score is $2$. By using one operation on changing the interval of vulnerability of the first planet to be $[1,2]$ and one operation on changing the interval of vulnerability of the third planet to $[2,3]$, all three planets will have the point $2$ in their interval.

The sum of scores of all non-empty subsets of the first testcase is $0 \cdot 4 + 1 \cdot 1 + 2\cdot2 = 5$.

Buggy Code:
from itertools import combinations
def scoremin(arr):
    # print(arr)
    xleft=min(arr,key=lambda x: x[0])[0]
    xright=max(arr,key=lambda x: x[1])[1]
    # print(xleft,xright)
    scoreleft=sum(min(abs(xleft-i[0]),abs(xleft-i[1])) if ((xleft-i[0])*(xleft-i[1]))>0 else 0 for i in arr)
    scoreright=sum(min(abs(xright-i[0]),abs(xright-i[1])) if ((xright-i[0])*(xright-i[1]))>0 else 0 for i in arr)
    if scoreleft==0:

        # print(xleft)

        return 0
    elif scoreright==0:
        # print(xright)
        return 0
    while scoreleft!=scoreright:
        # print(scoreleft,scoreright,xleft,xright)
        if scoreleft==0:
            return 0
        elif scoreright==0:
            return 0

        elif scoreleft>scoreright:
            xleft=(xright+xleft+1)//2
            scoreleft = sum(
                min(abs(xleft - i[0]), abs(xleft - i[1])) if ((xleft - i[0]) * (xleft - i[1])) > 0 else 0 for i in arr)

        else:
            xright=(xright+xleft+1)//2
            scoreright = sum(
                min(abs(xright - i[0]), abs(xright - i[1])) if ((xright - i[0]) * (xright - i[1])) > 0 else 0 for i in
                arr)
    # print('scorel:',scoreleft, 'scorer:',scoreright, 'xleft:',xleft,'xright:', xright)

    x=(xleft+xright+1)//2
    # print('optimal x:',x)
    scoremin = sum(
        min(abs(x - i[0]), abs(x - i[1])) if ((x - i[0]) * (x - i[1])) > 0 else 0 for i in arr)
    # print('scoremin:',scoremin)
    return scoremin

for _ in range(int(input())):
    arr=[]
    score = 0
    for _ in range(int(input())):
        a,b=map(int,input().split())
        arr.append((a,b))
    arr.sort()
    for i in range(2,len(arr)+1):
        for j in combinations(arr,i):
            j = list(j)
            # j.sort()
            # print(j)
            # print(j)
            score+=scoremin(j)

    print(score%998244353)
# arr=[[1,6],[2,6],[4,6],[1,6],[5,6]]
# arr.sort()
# print(arr)

# print(scoremin(arr))

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Destruction of the Universe (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, $n \leq 5000$. You can only make hacks if both versions of the problem are solved.

Orangutans are powerful beings—so powerful that they only need $1$ unit of time to destroy every vulnerable planet in the universe!

There are $n$ planets in the universe. Each planet has an interval of vulnerability $[l, r]$, during which it will be exposed to destruction by orangutans. Orangutans can also expand the interval of vulnerability of any planet by $1$ unit.

Specifically, suppose the expansion is performed on planet $p$ with interval of vulnerability $[l_p, r_p]$. Then, the resulting interval of vulnerability may be either $[l_p - 1, r_p]$ or $[l_p, r_p + 1]$.

Given a set of planets, orangutans can destroy all planets if the intervals of vulnerability of all planets in the set intersect at least one common point. Let the score of such a set denote the minimum number of expansions that must be performed.

Orangutans are interested in the sum of scores of all non-empty subsets of the planets in the universe. As the answer can be large, output it modulo $998\,244\,353$.

Input: The first line contains an integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 5000$) — the number of planets in the universe.

The following $n$ lines contain two integers $l_i$ and $r_i$ ($1 \leq l_i \leq r_i \leq n$) — the initial interval of vulnerability of the $i$-th planet.

It is guaranteed that the sum of $n$ does not exceed $5000$ over all test cases.

Output: For each test case, output an integer — the sum of scores to destroy all non- empty subsets of the planets in the universe, modulo $998\,244\,353$.

Examples:
input:
3
3
1 1
2 3
3 3
4
1 4
2 3
2 4
1 1
5
1 2
2 3
3 4
4 5
1 5
output:
5
6
24

Note: In the first testcase, there are seven non-empty subsets of planets we must consider:

  * For each of the subsets $\\{[1,1]\\}, \\{[2,3]\\}, \\{[3,3]\\}$, the score is $0$.    * For the subset $\\{[2,3], [3,3]\\}$, the score is $0$, because the point $3$ is already contained in both planets' interval of vulnerability.    * For the subset $\\{[1,1], [2,3]\\}$, the score is $1$. By using one operation on changing the interval of vulnerability of the second planet to be $[1,3]$, the two planets now both have the point $1$ in their interval.    * For the subset $\\{[1,1], [3,3]\\}$, the score is $2$. By using two operations on changing the interval of vulnerability of the first planet to be $[1,3]$, the two planets now both have the point $3$ in their interval.    * For the subset $\\{[1,1], [2,3], [3,3]\\}$, the score is $2$. By using one operation on changing the interval of vulnerability of the first planet to be $[1,2]$ and one operation on changing the interval of vulnerability of the third planet to $[2,3]$, all three planets will have the point $2$ in their interval.

The sum of scores of all non-empty subsets of the first testcase is $0 \cdot 4 + 1 \cdot 1 + 2\cdot2 = 5$.

Buggy Code:
import itertools as it

def calc_distance(x, y):
    if (x, y) in memo.values():
        return memo[(x, y)]
    distance = min(x[-1], y[-1]) - max(x[0], y[0])
    val = max(0, -distance)
    memo[(x, y)] = val
    return val

num_tests = int(input())

for i in range(0, num_tests):
    try:
        memo = {}
        score = 0
        num_planets = int(input())
        if num_planets < 2:
            input()
            print(score)
            continue
        planet_tuples = [(tuple(int(x) for x in input().split(' ')))
            for _ in range(0, num_planets)]
        for j in range(2, num_planets + 1):
            combinations = it.combinations(planet_tuples, j)
            score += sum([max(set(it.starmap(calc_distance,
                              it.combinations(x, 2)))) for x in combinations])
    except EOFError:
        continue
    print(score)

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. Here are some key points to consider:

1. **Single Planet Test Case**: The code should handle a single planet correctly.
2. **Overlapping Intervals**: Test cases with overlapping intervals to ensure the code correctly calculates the minimum number of expansions.
3. **Non-Overlapping Intervals**: Test cases with non-overlapping intervals to ensure the code correctly calculates the required number of expansions.
4. **Edge Cases**: Test cases with intervals that are very close to each other or have minimal overlap.

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

```plaintext
3
1
1 1
2
1 2
3 4
3
1 2
2 3
3 4
```

### Explanation:

1. **Test Case 1**:
   - **Number of Planets**: 1
   - **Interval**: [1, 1]
   - **Expected Output**: 0
   - **Reason**: There is only one planet, and no expansion is needed.

2. **Test Case 2**:
   - **Number of Planets**: 2
   - **Intervals**: [1, 2] and [3, 4]
   - **Expected Output**: 2
   - **Reason**: The intervals do not overlap. One expansion is needed for each planet to make them overlap.

3. **Test Case 3**:
   - **Number of Planets**: 3
   - **Intervals**: [1, 2], [2, 3], [3, 4]
   - **Expected Output**: 3
   - **Reason**: The intervals are close to each other but do not overlap. Two expansions are needed to make all intervals overlap.

This test input covers a range of scenarios that should help expose any bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Destruction of the Universe (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, $n \leq 5000$. You can only make hacks if both versions of the problem are solved.

Orangutans are powerful beings—so powerful that they only need $1$ unit of time to destroy every vulnerable planet in the universe!

There are $n$ planets in the universe. Each planet has an interval of vulnerability $[l, r]$, during which it will be exposed to destruction by orangutans. Orangutans can also expand the interval of vulnerability of any planet by $1$ unit.

Specifically, suppose the expansion is performed on planet $p$ with interval of vulnerability $[l_p, r_p]$. Then, the resulting interval of vulnerability may be either $[l_p - 1, r_p]$ or $[l_p, r_p + 1]$.

Given a set of planets, orangutans can destroy all planets if the intervals of vulnerability of all planets in the set intersect at least one common point. Let the score of such a set denote the minimum number of expansions that must be performed.

Orangutans are interested in the sum of scores of all non-empty subsets of the planets in the universe. As the answer can be large, output it modulo $998\,244\,353$.

Input: The first line contains an integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 5000$) — the number of planets in the universe.

The following $n$ lines contain two integers $l_i$ and $r_i$ ($1 \leq l_i \leq r_i \leq n$) — the initial interval of vulnerability of the $i$-th planet.

It is guaranteed that the sum of $n$ does not exceed $5000$ over all test cases.

Output: For each test case, output an integer — the sum of scores to destroy all non- empty subsets of the planets in the universe, modulo $998\,244\,353$.

Examples:
input:
3
3
1 1
2 3
3 3
4
1 4
2 3
2 4
1 1
5
1 2
2 3
3 4
4 5
1 5
output:
5
6
24

Note: In the first testcase, there are seven non-empty subsets of planets we must consider:

  * For each of the subsets $\\{[1,1]\\}, \\{[2,3]\\}, \\{[3,3]\\}$, the score is $0$.    * For the subset $\\{[2,3], [3,3]\\}$, the score is $0$, because the point $3$ is already contained in both planets' interval of vulnerability.    * For the subset $\\{[1,1], [2,3]\\}$, the score is $1$. By using one operation on changing the interval of vulnerability of the second planet to be $[1,3]$, the two planets now both have the point $1$ in their interval.    * For the subset $\\{[1,1], [3,3]\\}$, the score is $2$. By using two operations on changing the interval of vulnerability of the first planet to be $[1,3]$, the two planets now both have the point $3$ in their interval.    * For the subset $\\{[1,1], [2,3], [3,3]\\}$, the score is $2$. By using one operation on changing the interval of vulnerability of the first planet to be $[1,2]$ and one operation on changing the interval of vulnerability of the third planet to $[2,3]$, all three planets will have the point $2$ in their interval.

The sum of scores of all non-empty subsets of the first testcase is $0 \cdot 4 + 1 \cdot 1 + 2\cdot2 = 5$.

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

public class CF_G1_DestructionOfUniverse_Easy {
    static final int MOD = 998244353;

    public static void main(String[] args) throws Exception {
        FastReader in = new FastReader();
        PrintWriter out = new PrintWriter(System.out);
        int t = in.nextInt();
        while (t-- > 0) {
            int n = in.nextInt();
            int[] l = new int[n];
            int[] r = new int[n];
            int minL = Integer.MAX_VALUE;
            int maxR = Integer.MIN_VALUE;
            for (int i = 0; i < n; i++) {
                l[i] = in.nextInt();
                r[i] = in.nextInt();
                minL = Math.min(minL, l[i]);
                maxR = Math.max(maxR, r[i]);
            }

            long totalSum = 0;
            int totalSubsets = (1 << n) - 1;
            for (int mask = 1; mask <= totalSubsets; mask++) {
                int[] indices = new int[n];
                int idx = 0;
                for (int i = 0; i < n; i++) {
                    if (((mask >> i) & 1) == 1) {
                        indices[idx++] = i;
                    }
                }
                int size = idx;
                long minExpansions = Long.MAX_VALUE;
                for (int x = 1; x <= n; x++) {
                    long expansions = 0;
                    for (int i = 0; i < size; i++) {
                        int planet = indices[i];
                        long cost = 0;
                        if (x < l[planet]) {
                            cost = l[planet] - x;
                        } else if (x > r[planet]) {
                            cost = x - r[planet];
                        }
                        expansions += cost;
                    }
                    minExpansions = Math.min(minExpansions, expansions);
                }
                totalSum = (totalSum + minExpansions) % MOD;
            }
            out.println(totalSum);
        }
        out.flush();
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

Think step by step.

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

1. **Understand the Bug**: The given code calculates the minimum number of expansions required for each subset of planets to have a common point. The main issue might arise when the code incorrectly handles the minimum and maximum values of the intervals or the subset calculations.

2. **Edge Cases**:
   - **Single Planet**: Test cases with a single planet.
   - **Overlapping Intervals**: Test cases where intervals overlap in various ways.
   - **Non-Overlapping Intervals**: Test cases where intervals do not overlap.
   - **Maximum and Minimum Values**: Test cases where intervals are at the extreme values of the range.

3. **Small-Scale Test Input**:
   - A small number of planets to ensure the test input is valid and complete.
   - A mix of overlapping and non-overlapping intervals to test the code's handling of different scenarios.

Given these considerations, a challenging test input could be:

```plaintext
1
3
1 1
2 2
3 3
```

This test input has three planets with non-overlapping intervals. The expected output for this test case is the sum of scores for all non-empty subsets of these planets. Let's break it down:

- **Subsets and Scores**:
  - `{[1,1]}`: Score = 0
  - `{[2,2]}`: Score = 0
  - `{[3,3]}`: Score = 0
  - `{[1,1], [2,2]}`: Score = 1 (Expand `[1,1]` to `[1,2]` or `[2,2]` to `[1,2]`)
  - `{[1,1], [3,3]}`: Score = 2 (Expand `[1,1]` to `[1,3]` or `[3,3]` to `[1,3]`)
  - `{[2,2], [3,3]}`: Score = 1 (Expand `[2,2]` to `[2,3]` or `[3,3]` to `[2,3]`)
  - `{[1,1], [2,2], [3,3]}`: Score = 2 (Expand `[1,1]` to `[1,2]` and `[2,2]` to `[2,3]` or `[1,1]` to `[1,3]` and `[3,3]` to `[2,3]`)

The sum of scores is:
\[ 0 + 0 + 0 + 1 + 2 + 1 + 2 = 6 \]

Therefore, the expected output for this test input is `6`.

This test input should expose any issues in the code's handling of non-overlapping intervals and the calculation of the minimum number of expansions required.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: The Destruction of the Universe (Easy Version)
time_limit_ms: 4000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, $n \leq 5000$. You can only make hacks if both versions of the problem are solved.

Orangutans are powerful beings—so powerful that they only need $1$ unit of time to destroy every vulnerable planet in the universe!

There are $n$ planets in the universe. Each planet has an interval of vulnerability $[l, r]$, during which it will be exposed to destruction by orangutans. Orangutans can also expand the interval of vulnerability of any planet by $1$ unit.

Specifically, suppose the expansion is performed on planet $p$ with interval of vulnerability $[l_p, r_p]$. Then, the resulting interval of vulnerability may be either $[l_p - 1, r_p]$ or $[l_p, r_p + 1]$.

Given a set of planets, orangutans can destroy all planets if the intervals of vulnerability of all planets in the set intersect at least one common point. Let the score of such a set denote the minimum number of expansions that must be performed.

Orangutans are interested in the sum of scores of all non-empty subsets of the planets in the universe. As the answer can be large, output it modulo $998\,244\,353$.

Input: The first line contains an integer $t$ ($1 \leq t \leq 1000$) — the number of test cases.

The first line of each test case contains an integer $n$ ($1 \leq n \leq 5000$) — the number of planets in the universe.

The following $n$ lines contain two integers $l_i$ and $r_i$ ($1 \leq l_i \leq r_i \leq n$) — the initial interval of vulnerability of the $i$-th planet.

It is guaranteed that the sum of $n$ does not exceed $5000$ over all test cases.

Output: For each test case, output an integer — the sum of scores to destroy all non- empty subsets of the planets in the universe, modulo $998\,244\,353$.

Examples:
input:
3
3
1 1
2 3
3 3
4
1 4
2 3
2 4
1 1
5
1 2
2 3
3 4
4 5
1 5
output:
5
6
24

Note: In the first testcase, there are seven non-empty subsets of planets we must consider:

  * For each of the subsets $\\{[1,1]\\}, \\{[2,3]\\}, \\{[3,3]\\}$, the score is $0$.    * For the subset $\\{[2,3], [3,3]\\}$, the score is $0$, because the point $3$ is already contained in both planets' interval of vulnerability.    * For the subset $\\{[1,1], [2,3]\\}$, the score is $1$. By using one operation on changing the interval of vulnerability of the second planet to be $[1,3]$, the two planets now both have the point $1$ in their interval.    * For the subset $\\{[1,1], [3,3]\\}$, the score is $2$. By using two operations on changing the interval of vulnerability of the first planet to be $[1,3]$, the two planets now both have the point $3$ in their interval.    * For the subset $\\{[1,1], [2,3], [3,3]\\}$, the score is $2$. By using one operation on changing the interval of vulnerability of the first planet to be $[1,2]$ and one operation on changing the interval of vulnerability of the third planet to $[2,3]$, all three planets will have the point $2$ in their interval.

The sum of scores of all non-empty subsets of the first testcase is $0 \cdot 4 + 1 \cdot 1 + 2\cdot2 = 5$.

Buggy Code:
import java.util.*;
import java.io.*;
public class PlanetDestructionScorer{
    static final int MOD = 998244353;
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(br.readLine().trim());
        while (t-- > 0) {
            int n = Integer.parseInt(br.readLine().trim());
            int[][] intervals = new int[n][2];
            for (int i = 0; i < n; i++) {
                String[] input = br.readLine().trim().split(" ");
                intervals[i][0] = Integer.parseInt(input[0]);
                intervals[i][1] = Integer.parseInt(input[1]);
            }
            System.out.println(calculateTotalScore(intervals));
        }
    }
    private static long calculateTotalScore(int[][] intervals) {
        int n = intervals.length;
        long totalScore = 0;
            for (int mask = 1; mask < (1 << n); mask++) {
            int left = 1, right = 5000;
            for (int i = 0; i < n; i++) {
                if ((mask & (1 << i)) != 0) {
                    left = Math.max(left, intervals[i][0]);
                    right = Math.min(right, intervals[i][1]);
                }
            }
            int score = Math.max(0, left - right);
            totalScore = (totalScore + score) % MOD;
        }
        return totalScore;
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inhabitant of the Deep Sea
time_limit_ms: 2000
memory_limit_mb: 256
Description: $$$n$$$ ships set out to explore the depths of the ocean. The ships are numbered from $$$1$$$ to $$$n$$$ and follow each other in ascending order; the $$$i$$$-th ship has a durability of $$$a_i$$$.

The Kraken attacked the ships $$$k$$$ times in a specific order. First, it attacks the first of the ships, then the last, then the first again, and so on.

Each attack by the Kraken reduces the durability of the ship by $$$1$$$. When the durability of the ship drops to $$$0$$$, it sinks and is no longer subjected to attacks (thus the ship ceases to be the first or last, and the Kraken only attacks the ships that have not yet sunk). If all the ships have sunk, the Kraken has nothing to attack and it swims away.

For example, if $$$n=4$$$, $$$k=5$$$, and $$$a=[1, 2, 4, 3]$$$, the following will happen:

1. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [2, 4, 3]$$$;
2. The Kraken attacks the last ship, now $$$a = [2, 4, 2]$$$;
3. The Kraken attacks the first ship, now $$$a = [1, 4, 2]$$$;
4. The Kraken attacks the last ship, now $$$a = [1, 4, 1]$$$;
5. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [4, 1]$$$.

How many ships were sunk after the Kraken's attack?

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le k \le 10^{15}$$$) — the number of ships and how many times the Kraken will attack the ships.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the durability of the ships.

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

Output: For each test case, output the number of ships sunk by the Kraken on a separate line.

Examples:
input:
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
output:
2
3
5
0
2
2

Note:

Buggy Code:
#include <stdio.h>

int main()
{
    int t,sum=0;
    scanf("%d", &t);

    while (t--)
    {
        int n, b;
        scanf("%d %d", &n, &b);
        sum=0;
        int arr[n];
        for (int i = 0; i < n; i++)
        {
            scanf("%d", &arr[i]);
            sum+=arr[i];
        }
        if(sum<=b)
        {
            printf("%d\n",n);
            continue;
        }
        int count = 0;
        int c=0,d=n-1;
        for (int i = 0; i < b; i++)
        {
            if (i % 2 == 0)
            {
                if (arr[c] > 0)
                {
                    arr[c]--;
                    if(arr[c]==0)
                    {
                        count++;
                        c++;
                    }
                }
            }
            else
            {
                if (arr[d] > 0)
                {
                    arr[n - 1]--;
                    if(arr[d]==0)
                    {
                        count++;
                        d--;
                    }
                }
            }
        }
        printf("%d\n", count);
    }
}

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

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

Think step by step.

Response:
```plaintext
1
5 10
1 1 1 1 1
```

### Explanation:
This test case is designed to expose the bug in the given code. Here's why it's challenging:

1. **Small `n` and `k`**: The number of ships (`n = 5`) and the number of attacks (`k = 10`) are small, making it easy to manually verify the expected behavior.

2. **Uniform Durability**: All ships have a durability of `1`. This means that each attack will immediately sink a ship, and the Kraken will attack the next ship in the sequence.

3. **Even Number of Attacks**: The total number of attacks (`k = 10`) is exactly twice the number of ships (`n = 5`). This means that the Kraken will attack each ship exactly twice, which should sink all the ships.

### Expected Behavior:
- The Kraken will attack the first ship, sinking it.
- The Kraken will attack the last ship, sinking it.
- This process will continue until all ships are sunk.

### Expected Output:
```plaintext
5
```

### Why This Test Case Exposes the Bug:
- The code should correctly count the number of ships that are sunk. However, the given code has a potential issue with how it handles the indices `c` and `d` when a ship is sunk. This test case will help identify if the code correctly updates these indices and counts the sunk ships accurately.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inhabitant of the Deep Sea
time_limit_ms: 2000
memory_limit_mb: 256
Description: $$$n$$$ ships set out to explore the depths of the ocean. The ships are numbered from $$$1$$$ to $$$n$$$ and follow each other in ascending order; the $$$i$$$-th ship has a durability of $$$a_i$$$.

The Kraken attacked the ships $$$k$$$ times in a specific order. First, it attacks the first of the ships, then the last, then the first again, and so on.

Each attack by the Kraken reduces the durability of the ship by $$$1$$$. When the durability of the ship drops to $$$0$$$, it sinks and is no longer subjected to attacks (thus the ship ceases to be the first or last, and the Kraken only attacks the ships that have not yet sunk). If all the ships have sunk, the Kraken has nothing to attack and it swims away.

For example, if $$$n=4$$$, $$$k=5$$$, and $$$a=[1, 2, 4, 3]$$$, the following will happen:

1. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [2, 4, 3]$$$;
2. The Kraken attacks the last ship, now $$$a = [2, 4, 2]$$$;
3. The Kraken attacks the first ship, now $$$a = [1, 4, 2]$$$;
4. The Kraken attacks the last ship, now $$$a = [1, 4, 1]$$$;
5. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [4, 1]$$$.

How many ships were sunk after the Kraken's attack?

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le k \le 10^{15}$$$) — the number of ships and how many times the Kraken will attack the ships.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the durability of the ships.

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

Output: For each test case, output the number of ships sunk by the Kraken on a separate line.

Examples:
input:
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
output:
2
3
5
0
2
2

Note:

Buggy Code:
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
LL k;
int n;
const int N = 2e5 + 10;
int a[N];
int ss[N],se[N];

int main()
{
    int t;
    scanf("%d", &t);
    while(t -- )
    {
        int last = 1;
        scanf("%d%lld", &n, &k);
        for(int i = 1;i <= n; ++ i) scanf("%d", &a[i]);
        for(int i = 1;i <= n; ++ i) ss[i] = ss[i - 1] + a[i];
        for(int i = n;i >= 1;-- i) se[i] = se[i + 1] + a[i];

        int i = 1,j = n;
        while(i <= j)
        {
            if(last == 1)
            {
                last = 0;
                if(k < a[i]) {printf("%d\n",i - 1 + n - j);break;}
                else if(k >= a[i])
                    k -= a[i++];
            }
            else
            {
                last = 1;
                if(k < a[j]) {printf("%d\n",i - 1 + n - j);break;}
                else if(k >= a[j])
                    k -= a[j--];
            }
        }
        if(i > j) printf("%d\n",n);
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
5 10
1 2 3 4 5
2 10
1 1
4 100
1 10 1 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inhabitant of the Deep Sea
time_limit_ms: 2000
memory_limit_mb: 256
Description: $$$n$$$ ships set out to explore the depths of the ocean. The ships are numbered from $$$1$$$ to $$$n$$$ and follow each other in ascending order; the $$$i$$$-th ship has a durability of $$$a_i$$$.

The Kraken attacked the ships $$$k$$$ times in a specific order. First, it attacks the first of the ships, then the last, then the first again, and so on.

Each attack by the Kraken reduces the durability of the ship by $$$1$$$. When the durability of the ship drops to $$$0$$$, it sinks and is no longer subjected to attacks (thus the ship ceases to be the first or last, and the Kraken only attacks the ships that have not yet sunk). If all the ships have sunk, the Kraken has nothing to attack and it swims away.

For example, if $$$n=4$$$, $$$k=5$$$, and $$$a=[1, 2, 4, 3]$$$, the following will happen:

1. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [2, 4, 3]$$$;
2. The Kraken attacks the last ship, now $$$a = [2, 4, 2]$$$;
3. The Kraken attacks the first ship, now $$$a = [1, 4, 2]$$$;
4. The Kraken attacks the last ship, now $$$a = [1, 4, 1]$$$;
5. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [4, 1]$$$.

How many ships were sunk after the Kraken's attack?

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le k \le 10^{15}$$$) — the number of ships and how many times the Kraken will attack the ships.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the durability of the ships.

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

Output: For each test case, output the number of ships sunk by the Kraken on a separate line.

Examples:
input:
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
output:
2
3
5
0
2
2

Note:

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

using namespace std;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        long long k;
        cin >> n >> k;
        vector <long long> a(n);
        for (int i = 0; i < n; ++i)
            cin >> a[i];
        long long sum = 0;
        int l = 0;
        int r = n - 1;
        while (l < r)
        {
            long long tempSum = a[l] + a[r];
            if (sum + tempSum >= k)
            {
                ///вопросик какие уцелели
                if (a[l] <= a[r])
                {
                    if (sum + 2 * a[l] - 2 > k)
                    {
                        ///уцелели оба
                        break;
                    }
                    else
                    {
                        ///первый потопили
                        a[r] -= a[l];
                        sum += 2 * a[l];
                        a[l] = 0;
                        l++;
                        if (!a[r])
                            r--;
                    }
                }
                else
                {
                    ///a[r] < a[l]
                    if (sum + 2 * a[r] - 1 > k)
                    {
                        ///уцелели оба
                        break;
                    }
                    else
                    {
                        ///последний потопили
                        a[l] -= a[r];
                        sum += 2 * a[r];
                        a[r] = 0;
                        r--;
                    }
                }
            }
            else
            {
                ///оба корабля уничтожены
                a[l] = 0;
                a[r] = 0;
                l++; r--;
                sum += tempSum;
            }
        }
        if (l == r && sum + a[l] <= k) ///все абсолютно уничтожены
            l++;
        cout << n - (r - l + 1) << '\n';
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
3 6
1 2 1
```

### Explanation:
This test case is designed to expose potential issues in the algorithm when the Kraken attacks a small number of ships with specific durabilities that can lead to edge cases.

- **n = 3**: There are 3 ships.
- **k = 6**: The Kraken attacks 6 times.
- **a = [1, 2, 1]**: The durabilities of the ships are 1, 2, and 1.

### Expected Behavior:
1. **First attack**: The first ship (durability 1) is attacked and sinks. Remaining ships: [2, 1].
2. **Second attack**: The last ship (durability 1) is attacked and sinks. Remaining ships: [2].
3. **Third attack**: The first ship (durability 2) is attacked, reducing its durability to 1. Remaining ships: [1].
4. **Fourth attack**: The last ship (durability 1) is attacked and sinks. Remaining ships: [1].
5. **Fifth attack**: The first ship (durability 1) is attacked and sinks. Remaining ships: [].
6. **Sixth attack**: There are no ships left to attack, so the Kraken swims away.

### Expected Output:
The number of ships sunk should be **3**.

This test case is designed to check if the algorithm correctly handles the scenario where the Kraken attacks the ships in such a way that all ships eventually sink. The specific durabilities and the number of attacks are chosen to ensure that the algorithm must correctly manage the sequence of attacks and the state of the ships.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Inhabitant of the Deep Sea
time_limit_ms: 2000
memory_limit_mb: 256
Description: $$$n$$$ ships set out to explore the depths of the ocean. The ships are numbered from $$$1$$$ to $$$n$$$ and follow each other in ascending order; the $$$i$$$-th ship has a durability of $$$a_i$$$.

The Kraken attacked the ships $$$k$$$ times in a specific order. First, it attacks the first of the ships, then the last, then the first again, and so on.

Each attack by the Kraken reduces the durability of the ship by $$$1$$$. When the durability of the ship drops to $$$0$$$, it sinks and is no longer subjected to attacks (thus the ship ceases to be the first or last, and the Kraken only attacks the ships that have not yet sunk). If all the ships have sunk, the Kraken has nothing to attack and it swims away.

For example, if $$$n=4$$$, $$$k=5$$$, and $$$a=[1, 2, 4, 3]$$$, the following will happen:

1. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [2, 4, 3]$$$;
2. The Kraken attacks the last ship, now $$$a = [2, 4, 2]$$$;
3. The Kraken attacks the first ship, now $$$a = [1, 4, 2]$$$;
4. The Kraken attacks the last ship, now $$$a = [1, 4, 1]$$$;
5. The Kraken attacks the first ship, its durability becomes zero and now $$$a = [4, 1]$$$.

How many ships were sunk after the Kraken's attack?

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le k \le 10^{15}$$$) — the number of ships and how many times the Kraken will attack the ships.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the durability of the ships.

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

Output: For each test case, output the number of ships sunk by the Kraken on a separate line.

Examples:
input:
6
4 5
1 2 4 3
4 6
1 2 4 3
5 20
2 7 1 8 2
2 2
3 2
2 15
1 5
2 7
5 2
output:
2
3
5
0
2
2

Note:

Buggy Code:
t = int(input())
def minium(a):
    y =[a[0],a[-1]]
    x = min(y)
    p = y.index(x)
    if p==0: return p
    else:return -p

for _ in range(t):
    n,k = map(int,input().split())
    b = list(map(int,input().split()))
    c=0
    while(k>0 and len(b)>0):
        if len(b)==1:
            if k>=b[0]:c+=1
            break
        #print(b,k)
        i = minium(b)
        if i ==0:j=-1
        else:j=0
        n = 2*b[i]
        if 2*b[i]-1>k:break
        if k>n:
            if b[0]==b[-1]:
                k-=n
                c+=2
                del b[0],b[-1]
            else:
                k-=n
                c+=1
                b[j]-=b[i]
                del b[i]
        elif k>=n-1:
            if b[0]==b[-1]:
                if k ==2*b[0]:c+=2
                else: c+=1
                break
            elif b[-1]>b[0]:
                c+=1
                break
            else:
                break
    print(c)

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

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

Think step by step.

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

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

1. **Initial Durability Distribution**: The ships have durabilities `[1, 2, 3, 4, 5]`. This ensures that the first ship will sink quickly, and the remaining ships will have different durabilities, making the pattern of attacks more complex.

2. **Number of Attacks**: The Kraken attacks 10 times, which is enough to sink multiple ships but not all of them. This tests the logic of handling multiple attacks and the correct order of attacks.

3. **Edge Case for First and Last Ships**: The first ship has the lowest durability, so it will be sunk first. The last ship has the highest durability, so it will take more attacks to sink. This tests the logic of alternating attacks between the first and last ships.

4. **Intermediate Durability Checks**: The intermediate durabilities (2, 3, 4) ensure that the code must correctly handle the reduction of durabilities and the removal of ships from the list.

This test case should help identify any issues in the logic for handling the attacks and the removal of ships.