**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mirror Maze
time_limit_ms: 1000
memory_limit_mb: 1024
Description: You are given a grid of $$$R$$$ rows (numbered from $$$1$$$ to $$$R$$$ from north to south) and $$$C$$$ columns (numbered from $$$1$$$ to $$$C$$$ from west to east). Every cell in this grid is a square of the same size. The cell located at row $$$r$$$ and column $$$c$$$ is denoted as $$$(r, c)$$$. Each cell can either be empty or have a mirror in one of the cell's diagonals. Each mirror is represented by a line segment. The mirror is type $$$1$$$ if it is positioned diagonally from the southwest corner to the northeast corner of the cell, or type $$$2$$$ for the other diagonal.

These mirrors follow the law of reflection, that is, the angle of reflection equals the angle of incidence. Formally, for type $$$1$$$ mirror, if a beam of light comes from the north, south, west, or east of the cell, then it will be reflected to the west, east, north, and south of the cell, respectively. Similarly, for type $$$2$$$ mirror, if a beam of light comes from the north, south, west, or east of the cell, then it will be reflected to the east, west, south, and north of the cell, respectively.

You want to put a laser from outside the grid such that all mirrors are hit by the laser beam. There are $$$2 \cdot (R+C)$$$ possible locations to put the laser:

- from the north side of the grid at column $$$c$$$, for $$$1 \leq c \leq C$$$, shooting a laser beam to the south;
- from the south side of the grid at column $$$c$$$, for $$$1 \leq c \leq C$$$, shooting a laser beam to the north;
- from the east side of the grid at row $$$r$$$, for $$$1 \leq r \leq R$$$, shooting a laser beam to the west; and
- from the west side of the grid at row $$$r$$$, for $$$1 \leq r \leq R$$$, shooting a laser beam to the east.

Determine all possible locations for the laser such that all mirrors are hit by the laser beam.

Input: The first line consists of two integers $$$R$$$ $$$C$$$ ($$$1 \leq R, C \leq 200$$$).

Each of the next $$$R$$$ lines consists of a string $$$S_r$$$ of length $$$C$$$. The $$$c$$$-th character of string $$$S_r$$$ represents cell $$$(r, c)$$$. Each character can either be . if the cell is empty, / if the cell has type $$$1$$$ mirror, or \ if the cell has type $$$2$$$ mirror. There is at least one mirror in the grid.

Output: Output a single integer representing the number of possible locations for the laser such that all mirrors are hit by the laser beam. Denote this number as $$$k$$$.

If $$$k > 0$$$, then output $$$k$$$ space-separated strings representing the location of the laser. Each string consists of a character followed without any space by an integer. The character represents the side of the grid, which could be N, S, E, or W if you put the laser on the north, south, east, or west side of the grid, respectively. The integer represents the row/column number. You can output the strings in any order.

Examples:
input:
4 4
.//.
.\\.
.\/.
....
output:
2
N3 W2

input:
4 6
./..\.
.\...\
./../\
......
output:
2
E3 S2

input:
4 4
....
./\.
.\/.
....
output:
0

Note: Explanation for the sample input/output #1

The following illustration shows one of the solutions of this sample.

Explanation for the sample input/output #2

The following illustration shows one of the solutions of this sample.

Buggy Code:
import java.util.*;

public class Main{
    static int R, C;
    static char[][] grid;
    static Set<String> mirrors;
    static List<String> validStarts;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        R = sc.nextInt();
        C = sc.nextInt();
        grid = new char[R][C];
        mirrors = new HashSet<>();
        validStarts = new ArrayList<>();

        // Read the grid and identify mirror positions
        for (int r = 0; r < R; r++) {
            String line = sc.next();
            for (int c = 0; c < C; c++) {
                grid[r][c] = line.charAt(c);
                if (grid[r][c] == '/' || grid[r][c] == '\\') {
                    mirrors.add(r + "," + c);
                }
            }
        }

        // Iterate over all possible starting positions
        for (int c = 1; c <= C; c++) {
            simulateLaser("N" + c, 0, c - 1, 1, 0); // North side, beam to South
            simulateLaser("S" + c, R - 1, c - 1, -1, 0); // South side, beam to North
        }
        for (int r = 1; r <= R; r++) {
            simulateLaser("E" + r, r - 1, C - 1, 0, -1); // East side, beam to West
            simulateLaser("W" + r, r - 1, 0, 0, 1); // West side, beam to East
        }

        // Output the result
        System.out.println(validStarts.size());
        for (String pos : validStarts) {
            System.out.print(pos + " ");
        }
        sc.close();
    }

    private static void simulateLaser(String startPos, int r, int c, int dr, int dc) {
        Set<String> visited = new HashSet<>(mirrors);
        while (r >= 0 && r < R && c >= 0 && c < C) {
            String cell = r + "," + c;
            if (grid[r][c] == '/') {
                // Type 1 mirror
                int temp = dr;
                dr = -dc;
                dc = -temp;
                visited.remove(cell);
            } else if (grid[r][c] == '\\') {
                // Type 2 mirror
                int temp = dr;
                dr = dc;
                dc = temp;
                visited.remove(cell);
            }
            r += dr;
            c += dc;
        }
        if (visited.isEmpty()) {
            validStarts.add(startPos);
        }
    }
}

// ukdFRMV9IafswLGH8p7

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios that might cause the algorithm to fail. The primary issue in the code is that it only removes mirrors from the `visited` set when the laser hits them, but it does not check if the laser actually hits all mirrors in the correct sequence or if it gets stuck in a loop.

Here are some edge cases to consider:
1. **Mirrors forming a closed loop:** The laser could get stuck in a loop and never hit all mirrors.
2. **Mirrors at the edges:** The laser might not hit all mirrors if they are positioned in a way that the laser cannot reach them.
3. **Mirrors in a straight line:** The laser might not hit all mirrors if they are aligned in a way that the laser only hits a subset of them.

Let's create a small-scale test input that maximizes the failure probability:

```plaintext
4 4
/\/\
\//\
/\/\
\//\
```

**Explanation:**
- This grid is designed to have mirrors forming a closed loop.
- The mirrors are arranged in a way that the laser will get stuck in a loop and will not hit all mirrors.
- The laser will start from any side and will eventually return to its starting point without hitting all mirrors.

This test input should expose the bug in the code where the laser gets stuck in a loop and does not hit all mirrors, leading to incorrect results.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Mirror Maze
time_limit_ms: 1000
memory_limit_mb: 1024
Description: You are given a grid of $$$R$$$ rows (numbered from $$$1$$$ to $$$R$$$ from north to south) and $$$C$$$ columns (numbered from $$$1$$$ to $$$C$$$ from west to east). Every cell in this grid is a square of the same size. The cell located at row $$$r$$$ and column $$$c$$$ is denoted as $$$(r, c)$$$. Each cell can either be empty or have a mirror in one of the cell's diagonals. Each mirror is represented by a line segment. The mirror is type $$$1$$$ if it is positioned diagonally from the southwest corner to the northeast corner of the cell, or type $$$2$$$ for the other diagonal.

These mirrors follow the law of reflection, that is, the angle of reflection equals the angle of incidence. Formally, for type $$$1$$$ mirror, if a beam of light comes from the north, south, west, or east of the cell, then it will be reflected to the west, east, north, and south of the cell, respectively. Similarly, for type $$$2$$$ mirror, if a beam of light comes from the north, south, west, or east of the cell, then it will be reflected to the east, west, south, and north of the cell, respectively.

You want to put a laser from outside the grid such that all mirrors are hit by the laser beam. There are $$$2 \cdot (R+C)$$$ possible locations to put the laser:

- from the north side of the grid at column $$$c$$$, for $$$1 \leq c \leq C$$$, shooting a laser beam to the south;
- from the south side of the grid at column $$$c$$$, for $$$1 \leq c \leq C$$$, shooting a laser beam to the north;
- from the east side of the grid at row $$$r$$$, for $$$1 \leq r \leq R$$$, shooting a laser beam to the west; and
- from the west side of the grid at row $$$r$$$, for $$$1 \leq r \leq R$$$, shooting a laser beam to the east.

Determine all possible locations for the laser such that all mirrors are hit by the laser beam.

Input: The first line consists of two integers $$$R$$$ $$$C$$$ ($$$1 \leq R, C \leq 200$$$).

Each of the next $$$R$$$ lines consists of a string $$$S_r$$$ of length $$$C$$$. The $$$c$$$-th character of string $$$S_r$$$ represents cell $$$(r, c)$$$. Each character can either be . if the cell is empty, / if the cell has type $$$1$$$ mirror, or \ if the cell has type $$$2$$$ mirror. There is at least one mirror in the grid.

Output: Output a single integer representing the number of possible locations for the laser such that all mirrors are hit by the laser beam. Denote this number as $$$k$$$.

If $$$k > 0$$$, then output $$$k$$$ space-separated strings representing the location of the laser. Each string consists of a character followed without any space by an integer. The character represents the side of the grid, which could be N, S, E, or W if you put the laser on the north, south, east, or west side of the grid, respectively. The integer represents the row/column number. You can output the strings in any order.

Examples:
input:
4 4
.//.
.\\.
.\/.
....
output:
2
N3 W2

input:
4 6
./..\.
.\...\
./../\
......
output:
2
E3 S2

input:
4 4
....
./\.
.\/.
....
output:
0

Note: Explanation for the sample input/output #1

The following illustration shows one of the solutions of this sample.

Explanation for the sample input/output #2

The following illustration shows one of the solutions of this sample.

Buggy Code:
import java.io.*;
import java.util.*;
import java.util.function.BiFunction;
import java.util.function.Function;

public class CodeChef3 {

    static class FastWriter {
        private final BufferedWriter bw;

        public FastWriter() {
            this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
        }

        public void print(Object object) throws IOException {
            bw.append("").append(String.valueOf(object));
        }

        public void println(Object object) throws IOException {
            print(object);
            bw.append("\n");
        }

        public void close() throws IOException {
            bw.close();
        }
    }

    static int [][] dir = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};

    public static void main(String[] args) throws IOException {
        int n = in.nextInt();
        int m = in.nextInt();
        String [] arr = new String[n];
        Arrays.setAll(arr, i -> in.nextLine());
        int [][][] con = new int [n][m][];
        int tot = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                con[i][j] = getCodes(arr[i].charAt(j));
                if(arr[i].charAt(j) != '.') tot++;
            }
        }
        List<String> res = new ArrayList<>();
        for (int i = 0; i < m; i++) {
            if(check(con, tot, 0, i, 2)){
                res.add("N"+(i+1));
            }
        }
        for (int i = 0; i < n; i++) {
            if(check(con, tot, i, m-1, 3)){
                res.add("E"+(i+1));
            }
        }
        for (int i = 0; i < m; i++) {
            if(check(con, tot, n-1, i, 0)){
                res.add("S"+(i+1));
            }
        }
        for (int i = 0; i < n; i++) {
            if(check(con, tot, i, 0, 1)){
                res.add("W"+(i+1));
            }
        }
        if(res.isEmpty()){
            System.out.println(0);
            return;
        }
        System.out.println(res.size());
        for(String s : res){
            System.out.print(s + " ");
        }
        System.out.println();
    }

    static boolean check(int [][][] con, int tot, int i, int j, int code){
        int n = con.length, m = con[0].length;
        int cnt = 0;
        boolean [][][] vis = new boolean[n][m][4];
        while (i >= 0 && i < n && j >= 0 && j < m){
            if(con[i][j][0] != 0){
                boolean flag = false;
                for (int k = 0; k < 4 && !flag; k++) {
                    flag = vis[i][j][k];
                }
                if(!flag) cnt++;
            }
            if(vis[i][j][code]) break;
            vis[i][j][code] = true;
            code = con[i][j][code];
            i += dir[code][0];
            j += dir[code][1];
        }
        return cnt == tot;
    }

    static int [] getCodes(char c){
        if(c == '.') return new int [] {0, 1, 2, 3};
        if(c == '/'){
            return new int [] {1, 0, 3, 2};
        }
        // for \
        return new int [] {3, 2, 1, 0};
    }

    static long pow(long a, long b){
        long res = 1;
        while(b > 0){
            if(b%2 == 1){
                res = (res * a%mod)%mod;
            }
            a = (a%mod*a%mod)%mod;
            b >>= 1;
        }
        return res%mod;
    }

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

    static long gcd(long a, long b) {
        while (b != 0) {
            long temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }

    static class Pair<T, U>  {
        T f;
        U s;

        public Pair(T first, U sec) {
            this.f = first;
            this.s = sec;
        }

        @Override
        public String toString() {
            return f + " " + s;
        }

        @Override
        public boolean equals(Object obj) {
            if(!(obj instanceof Pair<?,?>)) return false;
            Pair<T, U> pair = (Pair<T, U>) obj;
            return f.equals(pair.f) && s.equals(pair.s);
        }
    }

    static int mod = (int)1e9 + 7;

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

    static FastReader in = new FastReader();

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

        String nextLine() {
            String str = "";
            try {
                str = br.readLine().trim();
            } catch (Exception e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}

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

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

Think step by step.

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

1. **Identify Potential Bugs:**
   - The code uses a direction array to navigate the grid.
   - It uses a `getCodes` function to determine the reflection directions for each type of mirror.
   - The `check` function simulates the laser beam's path and counts the number of mirrors hit.

2. **Edge Cases:**
   - **Empty Grid:** The code should handle an empty grid correctly.
   - **Single Mirror:** The code should handle a grid with a single mirror correctly.
   - **Multiple Mirrors in a Line:** The code should handle a grid where mirrors are aligned in a way that they form a continuous path.
   - **Cycles:** The code should handle cases where the laser beam forms a cycle.
   - **Boundary Conditions:** The code should handle cases where the laser beam hits the boundary of the grid.

3. **Generate Test Input:**
   - **Single Mirror:**
     - A grid with a single mirror that should be hit by the laser beam.
   - **Multiple Mirrors in a Line:**
     - A grid where mirrors are aligned in a way that they form a continuous path.
   - **Cycle:**
     - A grid where the laser beam forms a cycle and should not hit all mirrors.
   - **Boundary Conditions:**
     - A grid where the laser beam hits the boundary of the grid.

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

```plaintext
3 3
/..
...\.
...
```

This test input includes:
- A grid with a single mirror in the first row and a mirror in the second row that forms a path.
- The laser beam should be able to hit both mirrors from the north or west side.

This test input will help expose any issues in the code's handling of multiple mirrors and their paths.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Segmentation Folds
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Peter loves folding segments. There is a segment on a number line occupying the interval $$$[\ell, r]$$$. Since it is the prime time for folding the segments, Peter decides to fold the segment carefully. In each step, he chooses one of the two following operations whenever possible:

1. Operation $$$\tt{LTR}$$$: he folds the segment from left to right, where $$$\ell$$$ coincides with a point $$$x$$$ ($$$\ell < x \le r$$$) such that $$$\ell+x$$$ is a prime number$$$^{\text{∗}}$$$. When Peter chooses this operation, he always chooses the largest possible value $$$x$$$. Note that the segment occupies the interval $$$[\frac{1}{2}(\ell+x), r]$$$ afterwards.
2. Operation $$$\tt{RTL}$$$: he folds the segment from right to left, where $$$r$$$ coincides with a point $$$x$$$ ($$$\ell \le x < r$$$) such that $$$r+x$$$ is a prime number. When Peter chooses this operation, he always chooses the smallest possible value $$$x$$$. Note that the segment occupies the interval $$$[\ell, \frac{1}{2}(r+x)]$$$ afterwards.

A folding sequence refers to a sequence of operations specified above. Peter wants to fold the segment several times, resulting in the shortest possible interval whose length that cannot be further reduced. The length of an interval $$$[\ell, r]$$$ is defined naturally to be $$$r-\ell$$$. Let's consider the following example. Suppose that we are folding a segment initially occupying the interval $$$[1, 30]$$$. There are three folding sequences that lead to the shortest possible resulting interval, as shown in the following figure.

Please help Peter determine the number of folding sequences such that the resulting interval has a shortest possible length. Output the number modulo $$$998244353$$$.

Input: The first line contains an integer $$$t$$$, denoting the number of test cases. In each of the following $$$t$$$ lines, there are two integers $$$\ell$$$ and $$$r$$$.

- $$$1 \le t \le 10$$$
- $$$1 \le \ell < r \le 10^{12}$$$
- $$$r - \ell \le 10^5$$$

Output: For each test case, please output a line denoting the number of ways to fold the given segment such that the resulting segment has the shortest possible length, modulo $$$998244353$$$.

Examples:
input:
3
1 30
16 18
142857 240135
output:
3
1
63

Note:

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

using namespace std;
// using ll = long long;
using i64 = long long;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)

i64 modpow(i64 a, i64 b, i64 mod) {
    i64 res = 1;
    while (b) {
        if (b & 1) res = (__int128_t)res * a % mod;
        a = (__int128_t)a * a % mod;
        b >>= 1;
    }
    return res;
}

bool MillerRabin(i64 n, vector<i64> A) {
    i64 s = 0, d = n - 1;
    while (d % 2 == 0) {
        ++s;
        d >>= 1;
    }
    for (auto a: A) {
        if (n <= a) return true;
        i64 t, x = modpow(a, d, n);
        if (x != 1) {
            for (t = 0; t < s; t++) {
                if (x == n - 1) break;
                x = (__int128_t)x * x % n;
            }
            if (t == s) return false;
        }
    }
    return true;
}

// bool is_prime(i64 n) {
//     if (n <= 1) return false;
//     if (n == 2) return true;
//     if (n % 2 == 0) return false;
//     return MillerRabin(n, {2, 325, 9375, 28178, 450775, 9780504, 1795265022});
// }

#include <queue>

vector<int> P;
const int M = 2000000;

void linear_sieve() {
    vector<int> T(M, 1);
    T[0] = T[1] = 0;
    for (int i = 2; i < M; i++) {
        if (T[i]) P.emplace_back(i);
        for (auto j: P) {
            if (i * j >= M) break;
            T[i * j] = 0;
        }
    }
}

void solve() {
    i64 l, r;
    cin >> l >> r;
    vector<int> T(2 * (r - l) + 10, 1);
    for (i64 p: P) {
        if (p * p > r * 2) break;
        for (i64 x = (2 * l + p - 1) / p * p; x <= (2*r) / p * p; x += p) {
            if (x == p) continue;
            T[x - l * 2] = 0;
        }
    }
    // rep(i, 31) {
    //     if (i) cout << ' ';
    //     cout << T[i];
    // }
    // cout << '\n';
    auto is_prime = [&](i64 x) -> bool {
      //  cout<<x<<" "<<x-l<<" "<<T.size()<<endl;
        return T[x - l * 2] == 1;
    };
    int len = 1e9, ans = 0;
    queue<pair<i64, i64>> que;
    que.emplace(l, r);
    while (que.size()) {
        auto [L, R] = que.front();
        que.pop();
        //cout << L << ' ' << R << endl;
        if (L == 1 && R == 1) continue;
        bool f = true;
        for (i64 x = R; x >= L + 1; x--) {
            if (is_prime(x + L)) {
                f = false;
                que.emplace((x + L + 1) / 2, R);
                break;
            }
        }
        for (int x = L; x < R; x++) {
            if (is_prime(x + R)) {
                f = false;
                que.emplace(L, (R + x) / 2);
                break;
            }
        }
        if (f) {
            if (R - L < len) {
                len = R - L;
                ans = 1;
            }
            else if (R - L == len) ans++;
        }
    }
    cout << ans << '\n';
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    linear_sieve();
    int t;
    cin >> t;
    while (t--) solve();
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Segmentation Folds
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Peter loves folding segments. There is a segment on a number line occupying the interval $$$[\ell, r]$$$. Since it is the prime time for folding the segments, Peter decides to fold the segment carefully. In each step, he chooses one of the two following operations whenever possible:

1. Operation $$$\tt{LTR}$$$: he folds the segment from left to right, where $$$\ell$$$ coincides with a point $$$x$$$ ($$$\ell < x \le r$$$) such that $$$\ell+x$$$ is a prime number$$$^{\text{∗}}$$$. When Peter chooses this operation, he always chooses the largest possible value $$$x$$$. Note that the segment occupies the interval $$$[\frac{1}{2}(\ell+x), r]$$$ afterwards.
2. Operation $$$\tt{RTL}$$$: he folds the segment from right to left, where $$$r$$$ coincides with a point $$$x$$$ ($$$\ell \le x < r$$$) such that $$$r+x$$$ is a prime number. When Peter chooses this operation, he always chooses the smallest possible value $$$x$$$. Note that the segment occupies the interval $$$[\ell, \frac{1}{2}(r+x)]$$$ afterwards.

A folding sequence refers to a sequence of operations specified above. Peter wants to fold the segment several times, resulting in the shortest possible interval whose length that cannot be further reduced. The length of an interval $$$[\ell, r]$$$ is defined naturally to be $$$r-\ell$$$. Let's consider the following example. Suppose that we are folding a segment initially occupying the interval $$$[1, 30]$$$. There are three folding sequences that lead to the shortest possible resulting interval, as shown in the following figure.

Please help Peter determine the number of folding sequences such that the resulting interval has a shortest possible length. Output the number modulo $$$998244353$$$.

Input: The first line contains an integer $$$t$$$, denoting the number of test cases. In each of the following $$$t$$$ lines, there are two integers $$$\ell$$$ and $$$r$$$.

- $$$1 \le t \le 10$$$
- $$$1 \le \ell < r \le 10^{12}$$$
- $$$r - \ell \le 10^5$$$

Output: For each test case, please output a line denoting the number of ways to fold the given segment such that the resulting segment has the shortest possible length, modulo $$$998244353$$$.

Examples:
input:
3
1 30
16 18
142857 240135
output:
3
1
63

Note:

Buggy Code:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
long long T,prim[3000001],ans,m[2000001];
long long p[1000001],pp[1000001],kp;
long long kpb;
double l,r,pb[100001];
double mlen=100000000;
void dfs (double l,double r,long long kl,long long kr)
{
	//if (r-l==0.5) cout<<"AAAAAAAA"<<l<<" "<<r<<endl;
//	cout<<l<<" "<<r<<" "<<r-l<<endl;
	if (r<l) return;
	mlen=min(mlen,r-l);m[(long long)(r*2-l*2)]++;
	long long x1=lower_bound(pb+kl,pb+kr+1,r+l)-pb;
	long long x2=upper_bound(pb+kl,pb+kr,r+l)-pb-1;
//	if (r-l==1)
//	{
//	cout<<"l="<<l<<" r="<<r<<" pb[x1]="<<pb[x1]<<endl;
//	cout<<"l="<<l<<" r="<<r<<" pb[x2]="<<pb[x2]<<endl;
//	cout<<endl;
//	}
	if (pb[x1]*1.0/2!=r) dfs(l,pb[x1]*1.0/2,kl,x1);
	if (pb[x2]*1.0/2!=l&&pb[x2]!=0) dfs(pb[x2]*1.0/2,r,x2,kr);

}
int main(){
	prim[1]=0;
	for (long long i=2;i<=1200000;i++)
	{
		prim[i]=1;
		for (long long j=2;j*j<=i;j++)
		{
			if (i%j==0)
			{
				prim[i]=0;
				break;
			}
		}
		if (prim[i]) pp[++kp]=i;
	}
	cin>>T;
	while(T--)
	{
		cin>>l>>r;
		memset(m,0,sizeof(m));
		memset(p,0,sizeof(p));
		kpb=0;
		memset(pb,0,sizeof(pb));
		long long mid1=0,mid2=0;
		for (long long i=2*l;i<=2*r;i++)
		{
			//cout<<"i="<<i<<endl;
			p[(long long)(i-2*l)]=1;
			for (long long j=1;pp[j]*pp[j]<=i&&j<=kp;j++)
			{
				if (i%pp[j]==0)
				{
					p[(long long)(i-2*l)]=0;
					break;
				}
			}
			if (p[(long long)(i-2*l)]==1)
			{
				pb[++kpb]=i;
			}
		}
		mlen=10000000000;
		ans=0;
		//cout<<"AAAAAAAAAAA"<<endl;
		dfs(l,r,1,kpb);
		cout<<m[(long long)(mlen*2)]%998244353<<endl;

	}
	return 0;
}
/*
1
900000000000 900000100000
*/

**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 30
14 16
999999999999 1000000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Segmentation Folds
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Peter loves folding segments. There is a segment on a number line occupying the interval $$$[\ell, r]$$$. Since it is the prime time for folding the segments, Peter decides to fold the segment carefully. In each step, he chooses one of the two following operations whenever possible:

1. Operation $$$\tt{LTR}$$$: he folds the segment from left to right, where $$$\ell$$$ coincides with a point $$$x$$$ ($$$\ell < x \le r$$$) such that $$$\ell+x$$$ is a prime number$$$^{\text{∗}}$$$. When Peter chooses this operation, he always chooses the largest possible value $$$x$$$. Note that the segment occupies the interval $$$[\frac{1}{2}(\ell+x), r]$$$ afterwards.
2. Operation $$$\tt{RTL}$$$: he folds the segment from right to left, where $$$r$$$ coincides with a point $$$x$$$ ($$$\ell \le x < r$$$) such that $$$r+x$$$ is a prime number. When Peter chooses this operation, he always chooses the smallest possible value $$$x$$$. Note that the segment occupies the interval $$$[\ell, \frac{1}{2}(r+x)]$$$ afterwards.

A folding sequence refers to a sequence of operations specified above. Peter wants to fold the segment several times, resulting in the shortest possible interval whose length that cannot be further reduced. The length of an interval $$$[\ell, r]$$$ is defined naturally to be $$$r-\ell$$$. Let's consider the following example. Suppose that we are folding a segment initially occupying the interval $$$[1, 30]$$$. There are three folding sequences that lead to the shortest possible resulting interval, as shown in the following figure.

Please help Peter determine the number of folding sequences such that the resulting interval has a shortest possible length. Output the number modulo $$$998244353$$$.

Input: The first line contains an integer $$$t$$$, denoting the number of test cases. In each of the following $$$t$$$ lines, there are two integers $$$\ell$$$ and $$$r$$$.

- $$$1 \le t \le 10$$$
- $$$1 \le \ell < r \le 10^{12}$$$
- $$$r - \ell \le 10^5$$$

Output: For each test case, please output a line denoting the number of ways to fold the given segment such that the resulting segment has the shortest possible length, modulo $$$998244353$$$.

Examples:
input:
3
1 30
16 18
142857 240135
output:
3
1
63

Note:

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

using namespace std;

#define all(x) x.begin(), x.end()
#define ar array
#define pb push_back
#define ln '\n'
#define size(x) (int)(x).size()
#define int long long

using i64 = long long;

template <class F, class _S>
bool chmin(F &u, const _S &v){
    bool flag = false;
    if ( u > v ){
        u = v; flag |= true;
    }
    return flag;
}

template <class F, class _S>
bool chmax(F &u, const _S &v){
    bool flag = false;
    if ( u < v ){
        u = v; flag |= true;
    }
    return flag;
}

typedef pair <int,int> pi;

const int Mod = 998244353, inf = 1e9;

const int N = 1e6 + 5e5 + 7, M = 5e5 + 1;

bool p[N];

vector <int> pr;

int pre[M], nxt[M], L, R;

pi ans;

void chk_ans(pi &A, const pi &B){
	if ( A.first > B.first ) A = B;
	else if ( A.first == B.first ){
		A.second = (A.second + B.second) % Mod;
	}
}

int get_nxt(int l, int r){
	int m = (l + r) / 2 - L;

	return m >= 0 && pre[m] > l ? pre[m] : inf;
}

int get_pre(int l, int r){
	int m = (l + r + 1) / 2 - L;

	return nxt[m] != -1 && nxt[m] < r ? nxt[m] : -inf;
}

pi dfs(int l, int r){
	pi ret = {inf, 0};

	if ( l > r ) return ret;

	pi ans = {r - l, 1};

	chk_ans(ans, dfs(get_nxt(l, r), r));
	chk_ans(ans, dfs(l, get_pre(l, r)));

	return ans;
}

void solve(){
	int l, r; cin >> l >> r;

	L = l * 2, R = r * 2;

	vector <int> is(R - L + 1, 1);

	for ( auto &x: pr ){
		for ( int j = max(2LL, (L + x - 1) / x) * x; j <= R; j += x ) is[j - L] = 0;
	}

	memset(pre, -1, sizeof(pre));
	memset(nxt, -1, sizeof(nxt));

	int lst = -1;

	for ( int i = L; i <= R; i++ ){
		if ( is[i - L] ) lst = i;

		pre[i - L] = lst;
	} lst = -1;

	for ( int i = R; i >= L; i-- ){
		if ( is[i - L] ) lst = i;

		nxt[i - L] = lst;
	}

	ans = {inf, 0};

	cout << dfs(L, R).second << ln;
}

signed main(){
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    for ( int i = 2; i < N; i++ ){
		if ( !p[i] ) pr.pb(i);

		for ( int j = i * 2; j < N; j += i ) p[j] = 1;
	}

    int t; cin >> t;

    while ( t-- ) solve();

 	cout << '\n';
}

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

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

Think step by step.

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

1. **Prime Number Constraints**: The operations depend on finding prime numbers, so we need to test intervals where the sum of the endpoints with potential fold points is a prime number.
2. **Interval Length**: The interval length should be within the constraints but should be small enough to be manageable and testable.
3. **Edge Cases**: Test cases where the interval length is very small (e.g., 1 or 2) or where the initial interval is already the shortest possible.

Given these considerations, let's generate a test input that includes these edge cases:

```plaintext
3
1 2
1 10
9 10
```

### Explanation:
1. **Test Case 1: [1, 2]**
   - This is the smallest possible interval. The only possible operation is to fold it from left to right or right to left, but it cannot be folded further.
   - Expected output: 1 (since there is only one way to handle this interval).

2. **Test Case 2: [1, 10]**
   - This interval is larger and allows for multiple folding operations. The prime number constraints will be more significant here.
   - Expected output: The number of ways to fold the segment to its shortest possible length.

3. **Test Case 3: [9, 10]**
   - This is another small interval, but it is not the smallest possible. It should test the algorithm's ability to handle very short intervals.
   - Expected output: 1 (since there is only one way to handle this interval).

This test input will help in verifying the correctness of the algorithm, especially in handling small and edge case intervals.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Segmentation Folds
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Peter loves folding segments. There is a segment on a number line occupying the interval $$$[\ell, r]$$$. Since it is the prime time for folding the segments, Peter decides to fold the segment carefully. In each step, he chooses one of the two following operations whenever possible:

1. Operation $$$\tt{LTR}$$$: he folds the segment from left to right, where $$$\ell$$$ coincides with a point $$$x$$$ ($$$\ell < x \le r$$$) such that $$$\ell+x$$$ is a prime number$$$^{\text{∗}}$$$. When Peter chooses this operation, he always chooses the largest possible value $$$x$$$. Note that the segment occupies the interval $$$[\frac{1}{2}(\ell+x), r]$$$ afterwards.
2. Operation $$$\tt{RTL}$$$: he folds the segment from right to left, where $$$r$$$ coincides with a point $$$x$$$ ($$$\ell \le x < r$$$) such that $$$r+x$$$ is a prime number. When Peter chooses this operation, he always chooses the smallest possible value $$$x$$$. Note that the segment occupies the interval $$$[\ell, \frac{1}{2}(r+x)]$$$ afterwards.

A folding sequence refers to a sequence of operations specified above. Peter wants to fold the segment several times, resulting in the shortest possible interval whose length that cannot be further reduced. The length of an interval $$$[\ell, r]$$$ is defined naturally to be $$$r-\ell$$$. Let's consider the following example. Suppose that we are folding a segment initially occupying the interval $$$[1, 30]$$$. There are three folding sequences that lead to the shortest possible resulting interval, as shown in the following figure.

Please help Peter determine the number of folding sequences such that the resulting interval has a shortest possible length. Output the number modulo $$$998244353$$$.

Input: The first line contains an integer $$$t$$$, denoting the number of test cases. In each of the following $$$t$$$ lines, there are two integers $$$\ell$$$ and $$$r$$$.

- $$$1 \le t \le 10$$$
- $$$1 \le \ell < r \le 10^{12}$$$
- $$$r - \ell \le 10^5$$$

Output: For each test case, please output a line denoting the number of ways to fold the given segment such that the resulting segment has the shortest possible length, modulo $$$998244353$$$.

Examples:
input:
3
1 30
16 18
142857 240135
output:
3
1
63

Note:

Buggy Code:
print("3\n")
print("1\n")
print("63\n")

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

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

Think step by step.

Response:
```plaintext
3
1 10
100 102
1000000000000 1000000000001
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Segmentation Folds
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Peter loves folding segments. There is a segment on a number line occupying the interval $$$[\ell, r]$$$. Since it is the prime time for folding the segments, Peter decides to fold the segment carefully. In each step, he chooses one of the two following operations whenever possible:

1. Operation $$$\tt{LTR}$$$: he folds the segment from left to right, where $$$\ell$$$ coincides with a point $$$x$$$ ($$$\ell < x \le r$$$) such that $$$\ell+x$$$ is a prime number$$$^{\text{∗}}$$$. When Peter chooses this operation, he always chooses the largest possible value $$$x$$$. Note that the segment occupies the interval $$$[\frac{1}{2}(\ell+x), r]$$$ afterwards.
2. Operation $$$\tt{RTL}$$$: he folds the segment from right to left, where $$$r$$$ coincides with a point $$$x$$$ ($$$\ell \le x < r$$$) such that $$$r+x$$$ is a prime number. When Peter chooses this operation, he always chooses the smallest possible value $$$x$$$. Note that the segment occupies the interval $$$[\ell, \frac{1}{2}(r+x)]$$$ afterwards.

A folding sequence refers to a sequence of operations specified above. Peter wants to fold the segment several times, resulting in the shortest possible interval whose length that cannot be further reduced. The length of an interval $$$[\ell, r]$$$ is defined naturally to be $$$r-\ell$$$. Let's consider the following example. Suppose that we are folding a segment initially occupying the interval $$$[1, 30]$$$. There are three folding sequences that lead to the shortest possible resulting interval, as shown in the following figure.

Please help Peter determine the number of folding sequences such that the resulting interval has a shortest possible length. Output the number modulo $$$998244353$$$.

Input: The first line contains an integer $$$t$$$, denoting the number of test cases. In each of the following $$$t$$$ lines, there are two integers $$$\ell$$$ and $$$r$$$.

- $$$1 \le t \le 10$$$
- $$$1 \le \ell < r \le 10^{12}$$$
- $$$r - \ell \le 10^5$$$

Output: For each test case, please output a line denoting the number of ways to fold the given segment such that the resulting segment has the shortest possible length, modulo $$$998244353$$$.

Examples:
input:
3
1 30
16 18
142857 240135
output:
3
1
63

Note:

Buggy Code:
# standard inputs
import itertools
from collections import *
from bisect import *
import heapq
from functools import *
import math
import sys
import random
from copy import deepcopy
#fin=open('i', 'r'); input = lambda: fin.readline().strip(); from icecream import ic; from h import SortedList, SortedKeyList
R = lambda: [*map(int, input().split())]; # read one line(turn to int array)
lines = lambda x: [R() for i in range(x)]; # read multiple input lines

#from icecream import ic
def generate_small_primes(limit):
    """
    Generate all prime numbers up to `limit` using the Sieve of Eratosthenes.
    """
    is_prime = [True] * (limit + 1)
    is_prime[0] = is_prime[1] = False
    primes = []

    for i in range(2, limit + 1):
        if is_prime[i]:
            primes.append(i)
            for j in range(i * i, limit + 1, i):
                is_prime[j] = False

    return primes

small_primes = generate_small_primes(int(3e6))

def find_primes_in_range(l, r):
    """
    Find all prime numbers in the range [l, r].
    """
    range_size = r - l + 1
    is_prime = [True] * range_size

    # Generate small primes up to sqrt(r)
    sqrt_r = int(math.sqrt(r)) + 1

    # Mark multiples of small primes in the range [l, r]
    for prime in small_primes:
        # Find the starting point for marking
        start = max(prime * prime, (l + prime - 1) // prime * prime)
        for j in range(start, r + 1, prime):
            is_prime[j - l] = False

    # Handle the case where l == 1 (1 is not a prime number)
    if l == 1:
        is_prime[0] = False

    # Collect all primes in the range
    primes_in_range = [l + i for i in range(range_size) if is_prime[i]]
    return primes_in_range
def solve():
  origL, origR = R()
  primes = find_primes_in_range(2*origL, 2*origR)
  graph = defaultdict(list)
  vis = set()

  def dfs(l, r):
    if (l, r) in vis:
      return

    vis.add((l, r))
    # LTR
    # get biggest prime p such that 2*l < p <= l+r
    idx = bisect(primes, l+r)-1
    p = primes[idx] if 0 <= idx < len(primes) else math.inf
    if 2*l < p <= l+r:
      graph[(p/2, r)].append((l, r))
      dfs(p/2, r)

    # RTL
    # get smallest prime p such that l+r <= p < 2*r
    idx = bisect_left(primes, r+l)
    p = primes[idx] if 0 <= idx < len(primes) else -math.inf

    if l+r <= p < 2*r:
      graph[(l, p/2)].append((l, r))
      dfs(l, p/2)
  dfs(origL, origR)

  # num ways to get to l, r
  @cache
  def dp(l, r):
    if (l, r) == (origL, origR):
      return 1

    ans = 0
    for p, q in graph[(l, r)]:
      ans += dp(p, q)
    return ans

  minDiff = min(r-l for l, r in vis)

  ans = 0
  for l, r in vis:
    if r-l == minDiff:
      ans += dp(l, r)

  print(ans)

for i in range(int(input())):
  solve()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Segmentation Folds
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Peter loves folding segments. There is a segment on a number line occupying the interval $$$[\ell, r]$$$. Since it is the prime time for folding the segments, Peter decides to fold the segment carefully. In each step, he chooses one of the two following operations whenever possible:

1. Operation $$$\tt{LTR}$$$: he folds the segment from left to right, where $$$\ell$$$ coincides with a point $$$x$$$ ($$$\ell < x \le r$$$) such that $$$\ell+x$$$ is a prime number$$$^{\text{∗}}$$$. When Peter chooses this operation, he always chooses the largest possible value $$$x$$$. Note that the segment occupies the interval $$$[\frac{1}{2}(\ell+x), r]$$$ afterwards.
2. Operation $$$\tt{RTL}$$$: he folds the segment from right to left, where $$$r$$$ coincides with a point $$$x$$$ ($$$\ell \le x < r$$$) such that $$$r+x$$$ is a prime number. When Peter chooses this operation, he always chooses the smallest possible value $$$x$$$. Note that the segment occupies the interval $$$[\ell, \frac{1}{2}(r+x)]$$$ afterwards.

A folding sequence refers to a sequence of operations specified above. Peter wants to fold the segment several times, resulting in the shortest possible interval whose length that cannot be further reduced. The length of an interval $$$[\ell, r]$$$ is defined naturally to be $$$r-\ell$$$. Let's consider the following example. Suppose that we are folding a segment initially occupying the interval $$$[1, 30]$$$. There are three folding sequences that lead to the shortest possible resulting interval, as shown in the following figure.

Please help Peter determine the number of folding sequences such that the resulting interval has a shortest possible length. Output the number modulo $$$998244353$$$.

Input: The first line contains an integer $$$t$$$, denoting the number of test cases. In each of the following $$$t$$$ lines, there are two integers $$$\ell$$$ and $$$r$$$.

- $$$1 \le t \le 10$$$
- $$$1 \le \ell < r \le 10^{12}$$$
- $$$r - \ell \le 10^5$$$

Output: For each test case, please output a line denoting the number of ways to fold the given segment such that the resulting segment has the shortest possible length, modulo $$$998244353$$$.

Examples:
input:
3
1 30
16 18
142857 240135
output:
3
1
63

Note:

Buggy Code:
import math

MAX = 15010
prime = [0]*(MAX+1)

for i in range(2,MAX+1):
    if prime[i]>1: continue
    for j in range(i,MAX+1,i):
        prime[j] += 1

primelist = []
for i in range(2,MAX+1):
    if prime[i]==1:  primelist.append(i)

dic = {}
def judgeprime(num):
    if num in dic:
        return dic[num]

    for p in primelist:
        if num%p==0:
            dic[num] = False
            return False
        if p*p > num:
            dic[num] = True
            return True
    dic[num] = True
    return True

M = 998244353

inf = 10**9

def getnext(l,r):

    bound = l + r

    bound1 = math.floor(bound)
    bound2 = math.ceil(bound)

    while not judgeprime(bound1):
        bound1 -= 1
        if bound1 <= 2*l:
            break

    while not judgeprime(bound2):
        bound2 += 1
        if bound2 >= 2*r:
            break

    res1,count1 = inf, 0
    res2,count2 = inf, 0

    if bound1 > 2*l + 0.00001:
        [res1,count1] = getnext(bound1*0.5,r)

    if bound2 < 2*r - 0.00001:
        [res2,count2] = getnext(l,bound2*0.5)

    if res1 < res2:
        res = [res1,count1]
    elif res1 > res2:
        res = [res2,count2]
    else:
        if res1 == inf:
            res = [r-l,1]
        else:
            res = [res1,(count1+count2)%M]

    return res

def main():

    l,r = map(int,input().split())

    [res,count] = getnext(l,r)

    print(count)

t = 1
T = int(input())
while t <= T:
    main()
    t += 1

p = 0
for key in dic:
    if dic[key]:
        p += 1

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

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

Think step by step.

Response:
```plaintext
3
1 2
999999999999 1000000000000
142857 142858
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Segmentation Folds
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Peter loves folding segments. There is a segment on a number line occupying the interval $$$[\ell, r]$$$. Since it is the prime time for folding the segments, Peter decides to fold the segment carefully. In each step, he chooses one of the two following operations whenever possible:

1. Operation $$$\tt{LTR}$$$: he folds the segment from left to right, where $$$\ell$$$ coincides with a point $$$x$$$ ($$$\ell < x \le r$$$) such that $$$\ell+x$$$ is a prime number$$$^{\text{∗}}$$$. When Peter chooses this operation, he always chooses the largest possible value $$$x$$$. Note that the segment occupies the interval $$$[\frac{1}{2}(\ell+x), r]$$$ afterwards.
2. Operation $$$\tt{RTL}$$$: he folds the segment from right to left, where $$$r$$$ coincides with a point $$$x$$$ ($$$\ell \le x < r$$$) such that $$$r+x$$$ is a prime number. When Peter chooses this operation, he always chooses the smallest possible value $$$x$$$. Note that the segment occupies the interval $$$[\ell, \frac{1}{2}(r+x)]$$$ afterwards.

A folding sequence refers to a sequence of operations specified above. Peter wants to fold the segment several times, resulting in the shortest possible interval whose length that cannot be further reduced. The length of an interval $$$[\ell, r]$$$ is defined naturally to be $$$r-\ell$$$. Let's consider the following example. Suppose that we are folding a segment initially occupying the interval $$$[1, 30]$$$. There are three folding sequences that lead to the shortest possible resulting interval, as shown in the following figure.

Please help Peter determine the number of folding sequences such that the resulting interval has a shortest possible length. Output the number modulo $$$998244353$$$.

Input: The first line contains an integer $$$t$$$, denoting the number of test cases. In each of the following $$$t$$$ lines, there are two integers $$$\ell$$$ and $$$r$$$.

- $$$1 \le t \le 10$$$
- $$$1 \le \ell < r \le 10^{12}$$$
- $$$r - \ell \le 10^5$$$

Output: For each test case, please output a line denoting the number of ways to fold the given segment such that the resulting segment has the shortest possible length, modulo $$$998244353$$$.

Examples:
input:
3
1 30
16 18
142857 240135
output:
3
1
63

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>

// using namespace __gnu_pbds;
// #define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
// // order_of_key(k) : no. of elements < k
// // *find_by_order(i) : value at index i (0-based)
#define int long long int
#define vi vector<int>
#define mi map<int,int>
#define si set<int>
#define mv map<int,vector<int>>
#define pii pair<int,int>
const int inf = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9+7;
const int N = 2e6+5;
int primes[N];
map<long double, int> fin ;
vi vv;
vi ans;
void doo(){
    for(int i = 2 ; i<N ; i++){
        if(!primes[i]){
            vv.push_back(i);
            int j = 2*i;
            while(j<N){
                primes[j] = 1;
                j+=i;
            }
        }
    }
}
void func(long double l, long double r){
    // cerr<<r<<" "<<l<<endl;
    fin[(r - l)]++;
    int temp = r+l;
    auto tt = upper_bound(ans.begin(),ans.end(),temp);
    if(tt!=ans.begin()){
        tt--;
        int tk = *tt;
        if(tk>2*l){
            long double kk = (long double)tk;
            func(kk/2,r);
        }
    }
    temp = r + l + 0.5;
    tt = lower_bound(ans.begin(),ans.end(),temp);
    if(tt!=ans.end() && *tt < 2*r){
        int tk = *tt;
        long double kk = (long double)tk;
        func(l,kk/2);
    }
}
void solve(){
    fin.clear();
    ans.clear();
    int l,r;
    cin>>l>>r;
    int tt = sqrt(2*r+1);
    mi anss;
    for(auto i:vv){
        if(i>tt)break;
        int tk = 2*l/i;
        tk = max(tk,2ll);
        while(tk*i < 2*r){
            anss[tk*i] = 1;
            tk++;
        }
    }
    for(int i = 2*l ; i <= 2*r ; i++){
        if(!anss[i]){
            ans.push_back(i);
        }
    }
    func((long double)l, (long double)r);
    cout<<fin.begin()->second<<endl;
}
int32_t main(){
ios::sync_with_stdio(0);
cin.tie(0);
int t;
cin>>t;
doo();
while(t--){
solve();
}
}

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

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

Think step by step.

Response:
```plaintext
3
1 1000000
1 10
999999 1000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Segmentation Folds
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Peter loves folding segments. There is a segment on a number line occupying the interval $$$[\ell, r]$$$. Since it is the prime time for folding the segments, Peter decides to fold the segment carefully. In each step, he chooses one of the two following operations whenever possible:

1. Operation $$$\tt{LTR}$$$: he folds the segment from left to right, where $$$\ell$$$ coincides with a point $$$x$$$ ($$$\ell < x \le r$$$) such that $$$\ell+x$$$ is a prime number$$$^{\text{∗}}$$$. When Peter chooses this operation, he always chooses the largest possible value $$$x$$$. Note that the segment occupies the interval $$$[\frac{1}{2}(\ell+x), r]$$$ afterwards.
2. Operation $$$\tt{RTL}$$$: he folds the segment from right to left, where $$$r$$$ coincides with a point $$$x$$$ ($$$\ell \le x < r$$$) such that $$$r+x$$$ is a prime number. When Peter chooses this operation, he always chooses the smallest possible value $$$x$$$. Note that the segment occupies the interval $$$[\ell, \frac{1}{2}(r+x)]$$$ afterwards.

A folding sequence refers to a sequence of operations specified above. Peter wants to fold the segment several times, resulting in the shortest possible interval whose length that cannot be further reduced. The length of an interval $$$[\ell, r]$$$ is defined naturally to be $$$r-\ell$$$. Let's consider the following example. Suppose that we are folding a segment initially occupying the interval $$$[1, 30]$$$. There are three folding sequences that lead to the shortest possible resulting interval, as shown in the following figure.

Please help Peter determine the number of folding sequences such that the resulting interval has a shortest possible length. Output the number modulo $$$998244353$$$.

Input: The first line contains an integer $$$t$$$, denoting the number of test cases. In each of the following $$$t$$$ lines, there are two integers $$$\ell$$$ and $$$r$$$.

- $$$1 \le t \le 10$$$
- $$$1 \le \ell < r \le 10^{12}$$$
- $$$r - \ell \le 10^5$$$

Output: For each test case, please output a line denoting the number of ways to fold the given segment such that the resulting segment has the shortest possible length, modulo $$$998244353$$$.

Examples:
input:
3
1 30
16 18
142857 240135
output:
3
1
63

Note:

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

using namespace std;

void solve()
{
    map<long long, bool> notP;
    long long l, r;
    cin>>l>>r;

    int up = sqrt(2*r);
    for(int i=2; i<=up; i++)
    {
        for(long long j = max(2ll, 2*l / i); j<=2*r / i; j++) notP[j*i] = 1;
    }

    vector<int> pr;
    pr.push_back(0);
    for(long long i = 2*l; i <= 2*r; i++) if(!notP[i]) pr.push_back(i);
    // for(auto i = pr.begin(); i!=pr.end(); i++) cout<<(*i)<<"\n";

    long double mn = (r-l)*2; int cnt = 0;

    function<void(long long, long long)> dfs = [&] (long long L, long long R) -> void
    {
        bool flag = false;

        long long rOfL = (L+R)/2, lOfR = (L+R+1)/2;

        auto p = lower_bound(pr.begin(), pr.end(), rOfL + 1);
        p--;
        if((*p)>L)
        {
            flag = 1;
            dfs((*p), R);
        }
        p = lower_bound(pr.begin(), pr.end(), lOfR);
        if(p!=pr.end() && (*p)<R)
        {
            flag = 1;
            dfs(L, (*p));
        }

        if(!flag)
        {
            // cout<<L<<" "<<R<<" "<<mn<<"\n";
            if(R-L<mn)
            {
                mn = R-L;
                cnt = 0;
            }
            if(R-L == mn)
            {
                cnt++;
            }
        }
        return;
    };

    dfs(2*l, 2*r);

    cout<<cnt<<"\n";

    return;
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);

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

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
1 2
1 100000
1 1000000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Segmentation Folds
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Peter loves folding segments. There is a segment on a number line occupying the interval $$$[\ell, r]$$$. Since it is the prime time for folding the segments, Peter decides to fold the segment carefully. In each step, he chooses one of the two following operations whenever possible:

1. Operation $$$\tt{LTR}$$$: he folds the segment from left to right, where $$$\ell$$$ coincides with a point $$$x$$$ ($$$\ell < x \le r$$$) such that $$$\ell+x$$$ is a prime number$$$^{\text{∗}}$$$. When Peter chooses this operation, he always chooses the largest possible value $$$x$$$. Note that the segment occupies the interval $$$[\frac{1}{2}(\ell+x), r]$$$ afterwards.
2. Operation $$$\tt{RTL}$$$: he folds the segment from right to left, where $$$r$$$ coincides with a point $$$x$$$ ($$$\ell \le x < r$$$) such that $$$r+x$$$ is a prime number. When Peter chooses this operation, he always chooses the smallest possible value $$$x$$$. Note that the segment occupies the interval $$$[\ell, \frac{1}{2}(r+x)]$$$ afterwards.

A folding sequence refers to a sequence of operations specified above. Peter wants to fold the segment several times, resulting in the shortest possible interval whose length that cannot be further reduced. The length of an interval $$$[\ell, r]$$$ is defined naturally to be $$$r-\ell$$$. Let's consider the following example. Suppose that we are folding a segment initially occupying the interval $$$[1, 30]$$$. There are three folding sequences that lead to the shortest possible resulting interval, as shown in the following figure.

Please help Peter determine the number of folding sequences such that the resulting interval has a shortest possible length. Output the number modulo $$$998244353$$$.

Input: The first line contains an integer $$$t$$$, denoting the number of test cases. In each of the following $$$t$$$ lines, there are two integers $$$\ell$$$ and $$$r$$$.

- $$$1 \le t \le 10$$$
- $$$1 \le \ell < r \le 10^{12}$$$
- $$$r - \ell \le 10^5$$$

Output: For each test case, please output a line denoting the number of ways to fold the given segment such that the resulting segment has the shortest possible length, modulo $$$998244353$$$.

Examples:
input:
3
1 30
16 18
142857 240135
output:
3
1
63

Note:

Buggy Code:
//People who believe in miracles are as amazing as miracles themselves.
#include<bits/stdc++.h>
#define ll long long
using namespace std;
inline ll read(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-') f=-f;ch=getchar();}
    while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
    return x*f;
}

const int N=1e5+10,P=998244353;

ll L,R,ans,mn; vector<int> vc={2,3,5,7,11,13,17,19,23,29,31,37},pr;

map<pair<ll,ll>,int> mp,num;

inline ll ksm(ll a,ll k,ll p){
    ll res=1; a%=p;
    for(;k;a=(a*a)%p,k>>=1)
        if(k&1) res=(res*a)%p;
    return res;
}

inline bool check(ll p,ll x){
    if(!(x%p)) return 0;
    ll s=x-1,num=0;
    while(!(s&1)) ++num,s>>=1;
    ll v=ksm(p,s,x);
    if(v==1) return 1;
    while(num--){
        ll las=v;
        v=(v*v)%x;
        if(v==1) return las==x-1;
    }
    return 0;
}

inline bool check(ll x){
    if(x<=100){
        if(x<=1) return 0;
        for(int i=2;i*i<=x;++i)
            if(!(x%i)) return 0;
        return 1;
    }
    for(auto p:vc) if(!check(p,x)) return 0;
    return 1;
}

ll solve(ll L,ll R){
    auto p=make_pair(L,R);
    if(mp.count(p)) return mp[p];
    ll xL=-1,xR=-1,res=R-L;

    auto it=upper_bound(pr.begin(),pr.end(),(L+R)/2);
    if(it!=pr.begin()){
        --it;
        xL=(*it)*2-L;
        if(!(L<xL&&xL<=R)) xL=-1;
    }

    it=lower_bound(pr.begin(),pr.end(),(L+R+1)/2);
    if(it!=pr.end()){
        xR=(*it)*2-R;
        if(!(L<=xR&&xR<R)) xR=-1;
    }

    if(xL!=-1) res=min(res,solve((L+xL)/2,R));
    if(xR!=-1) res=min(res,solve(L,(xR+R)/2));
    return mp[p]=res;
}

ll get_ans(ll L,ll R){
    if(solve(L,R)!=mn) return 0;
    if(R-L==mn) return 1;
    auto p=make_pair(L,R);
    if(num.count(p)) return num[p];
    ll xL=-1,xR=-1,res=0;

    auto it=upper_bound(pr.begin(),pr.end(),(L+R)/2);
    if(it!=pr.begin()){
        --it;
        xL=(*it)*2-L;
        if(!(L<xL&&xL<=R)) xL=-1;
    }

    it=lower_bound(pr.begin(),pr.end(),(L+R+1)/2);
    if(it!=pr.end()){
        xR=(*it)*2-R;
        if(!(L<=xR&&xR<R)) xR=-1;
    }

    if(xL!=-1) res=(res+get_ans((L+xL)/2,R))%P;
    if(xR!=-1) res=(res+get_ans(L,(xR+R)/2))%P;
    return num[p]=res;
}

inline void solve(){
    L=read(),R=read(),mp.clear(),num.clear(),pr.clear();
    for(ll x=2*L;x<=2*R;++x) if(check(x)) pr.emplace_back(x);
    mn=solve(L*2,R*2),ans=get_ans(L*2,R*2),printf("%lld\n",ans);
}

signed main(){
#ifndef ONLINE_JUDGE
    freopen("../1.in","r",stdin);
    freopen("../1.out","w",stdout);
#endif
    for(int T=read();T;--T) solve();
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
1 100
1 101
1 200
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Segmentation Folds
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Peter loves folding segments. There is a segment on a number line occupying the interval $$$[\ell, r]$$$. Since it is the prime time for folding the segments, Peter decides to fold the segment carefully. In each step, he chooses one of the two following operations whenever possible:

1. Operation $$$\tt{LTR}$$$: he folds the segment from left to right, where $$$\ell$$$ coincides with a point $$$x$$$ ($$$\ell < x \le r$$$) such that $$$\ell+x$$$ is a prime number$$$^{\text{∗}}$$$. When Peter chooses this operation, he always chooses the largest possible value $$$x$$$. Note that the segment occupies the interval $$$[\frac{1}{2}(\ell+x), r]$$$ afterwards.
2. Operation $$$\tt{RTL}$$$: he folds the segment from right to left, where $$$r$$$ coincides with a point $$$x$$$ ($$$\ell \le x < r$$$) such that $$$r+x$$$ is a prime number. When Peter chooses this operation, he always chooses the smallest possible value $$$x$$$. Note that the segment occupies the interval $$$[\ell, \frac{1}{2}(r+x)]$$$ afterwards.

A folding sequence refers to a sequence of operations specified above. Peter wants to fold the segment several times, resulting in the shortest possible interval whose length that cannot be further reduced. The length of an interval $$$[\ell, r]$$$ is defined naturally to be $$$r-\ell$$$. Let's consider the following example. Suppose that we are folding a segment initially occupying the interval $$$[1, 30]$$$. There are three folding sequences that lead to the shortest possible resulting interval, as shown in the following figure.

Please help Peter determine the number of folding sequences such that the resulting interval has a shortest possible length. Output the number modulo $$$998244353$$$.

Input: The first line contains an integer $$$t$$$, denoting the number of test cases. In each of the following $$$t$$$ lines, there are two integers $$$\ell$$$ and $$$r$$$.

- $$$1 \le t \le 10$$$
- $$$1 \le \ell < r \le 10^{12}$$$
- $$$r - \ell \le 10^5$$$

Output: For each test case, please output a line denoting the number of ways to fold the given segment such that the resulting segment has the shortest possible length, modulo $$$998244353$$$.

Examples:
input:
3
1 30
16 18
142857 240135
output:
3
1
63

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define int long long
ll l,r,vis[1000005];
const int p = 998244353;
vector<int> pri;
void isprime(){
    for(int i = 2;i <= 1000000;i++){
        if(!vis[i])pri.push_back(i);
        for(auto j:pri){
            if(i * j > 1000000)break;
            vis[i * j]=1;
            if(i % j == 0)break;
        }
    }
}

void solve(){
    cin >> l >> r;
    vector<int> a(2 * (r - l) + 1,1),b(2 * (r - l) + 1),c(2 * (r - l) + 1);
    vector<ll> v;
    for(auto i:pri){
        if(i > 2ll * r)break;
        ll k = (2ll * l + i - 1) / i * i;
        if(k == i)k += i;
        for(ll j = k;j < 2 * r;j += i){
            a[j - 2ll * l] = 0;
        }
    }

    int tmp = 1;
    for(int i = 1;i < 2 * (r - l);i++) {
        if(a[i])tmp = i;
        b[i] = tmp;
        //cout << b[i] << " \n"[i == 2 * (r - l) - 1];
    }
    tmp = (r - l) * 2;
    for(int i = (r - l) * 2 - 1;i >= 1;i--){
        if(a[i])
            tmp = i;
        c[i] = tmp;
        //cout << c[i] << " \n"[i == 1];
    }
    int ans = 1e8,cnt = 0;
    function<void(int,int,int,int,int)> dfs = [&](int l,int f1,int r,int f2,int num) {
        if(l == r)return;
        int vl = l * 2 + f1,rl = r * 2 + f2;
        int x = l + r + (f1 * f2);
        int l1 = (l + 1) * 2 + f1,r1 = (r + 1) * 2 + f2;
        //cout << vl << " " << rl << " " << x << " "<< b[x] << " " << c[x] << "\n";
        //cout << l << " " << r << " " << f1 << " " << f2 << "  " << num << "\n";
        if(b[x] == vl && c[x] == rl){
            if(ans > r1 - l1)ans = r1 - l1,cnt = 1;
            else if(ans == r1 - l1)cnt++;
            //cout << ans << " " << cnt << "\n";
            return;
        }else{
            if(b[x] != vl)dfs(b[x] / 2,1,r,f2,num + 1);
            if(c[x] != rl)dfs(l,f1,c[x] / 2,1,num + 1);
            return;
        }
    };
    dfs(0,0,r - l,0,1);
    cout  << cnt << "\n";
}

signed main() {
    isprime();
    int T;cin >> T;
    while(T--) solve();
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
1 100000
1 2
99999 100001
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Segmentation Folds
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Peter loves folding segments. There is a segment on a number line occupying the interval $$$[\ell, r]$$$. Since it is the prime time for folding the segments, Peter decides to fold the segment carefully. In each step, he chooses one of the two following operations whenever possible:

1. Operation $$$\tt{LTR}$$$: he folds the segment from left to right, where $$$\ell$$$ coincides with a point $$$x$$$ ($$$\ell < x \le r$$$) such that $$$\ell+x$$$ is a prime number$$$^{\text{∗}}$$$. When Peter chooses this operation, he always chooses the largest possible value $$$x$$$. Note that the segment occupies the interval $$$[\frac{1}{2}(\ell+x), r]$$$ afterwards.
2. Operation $$$\tt{RTL}$$$: he folds the segment from right to left, where $$$r$$$ coincides with a point $$$x$$$ ($$$\ell \le x < r$$$) such that $$$r+x$$$ is a prime number. When Peter chooses this operation, he always chooses the smallest possible value $$$x$$$. Note that the segment occupies the interval $$$[\ell, \frac{1}{2}(r+x)]$$$ afterwards.

A folding sequence refers to a sequence of operations specified above. Peter wants to fold the segment several times, resulting in the shortest possible interval whose length that cannot be further reduced. The length of an interval $$$[\ell, r]$$$ is defined naturally to be $$$r-\ell$$$. Let's consider the following example. Suppose that we are folding a segment initially occupying the interval $$$[1, 30]$$$. There are three folding sequences that lead to the shortest possible resulting interval, as shown in the following figure.

Please help Peter determine the number of folding sequences such that the resulting interval has a shortest possible length. Output the number modulo $$$998244353$$$.

Input: The first line contains an integer $$$t$$$, denoting the number of test cases. In each of the following $$$t$$$ lines, there are two integers $$$\ell$$$ and $$$r$$$.

- $$$1 \le t \le 10$$$
- $$$1 \le \ell < r \le 10^{12}$$$
- $$$r - \ell \le 10^5$$$

Output: For each test case, please output a line denoting the number of ways to fold the given segment such that the resulting segment has the shortest possible length, modulo $$$998244353$$$.

Examples:
input:
3
1 30
16 18
142857 240135
output:
3
1
63

Note:

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

using namespace std;

const int N = 2e6+5;

vector<int> prime;
bool mark[N];

void solve()
{
    map<long long, bool> notP;
    long long l, r;
    cin>>l>>r;

    int up = sqrt(2*r);
    for(auto i : prime)
    {
        if(i>up) break;
        for(long long j = max(2ll, 2*l / i); j<=2*r / i; j++) notP[j*i] = 1;
    }

    vector<long long> pr;
    pr.push_back(0);
    for(long long i = 2*l; i <= 2*r; i++) if(!notP[i]) pr.push_back(i);
    // for(auto i = pr.begin(); i!=pr.end(); i++) cout<<(*i)<<"\n";

    long double mn = (r-l)*2; int cnt = 0;

    function<void(long long, long long)> dfs = [&] (long long L, long long R) -> void
    {
        bool flag = false;

        long long rOfL = (L+R)/2, lOfR = (L+R+1)/2;

        auto p = lower_bound(pr.begin(), pr.end(), rOfL + 1);
        p--;
        if((*p)>L)
        {
            flag = 1;
            dfs((*p), R);
        }
        p = lower_bound(pr.begin(), pr.end(), lOfR);
        if(p!=pr.end() && (*p)<R)
        {
            flag = 1;
            dfs(L, (*p));
        }

        if(!flag)
        {
            // cout<<L<<" "<<R<<" "<<mn<<"\n";
            if(R-L<mn)
            {
                mn = R-L;
                cnt = 0;
            }
            if(R-L == mn)
            {
                cnt++;
            }
        }
        return;
    };

    dfs(2*l, 2*r);

    cout<<cnt<<"\n";

    return;
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);

    int up = N;
    for(int i=2; i<up; i++)
    {
        if(mark[i]) continue;
        prime.push_back(i);
        for(int j=2*i; j<up; j+=i) mark[j] = 1;
    }

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

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
1 2
999999999998 999999999999
1 100000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Segmentation Folds
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Peter loves folding segments. There is a segment on a number line occupying the interval $$$[\ell, r]$$$. Since it is the prime time for folding the segments, Peter decides to fold the segment carefully. In each step, he chooses one of the two following operations whenever possible:

1. Operation $$$\tt{LTR}$$$: he folds the segment from left to right, where $$$\ell$$$ coincides with a point $$$x$$$ ($$$\ell < x \le r$$$) such that $$$\ell+x$$$ is a prime number$$$^{\text{∗}}$$$. When Peter chooses this operation, he always chooses the largest possible value $$$x$$$. Note that the segment occupies the interval $$$[\frac{1}{2}(\ell+x), r]$$$ afterwards.
2. Operation $$$\tt{RTL}$$$: he folds the segment from right to left, where $$$r$$$ coincides with a point $$$x$$$ ($$$\ell \le x < r$$$) such that $$$r+x$$$ is a prime number. When Peter chooses this operation, he always chooses the smallest possible value $$$x$$$. Note that the segment occupies the interval $$$[\ell, \frac{1}{2}(r+x)]$$$ afterwards.

A folding sequence refers to a sequence of operations specified above. Peter wants to fold the segment several times, resulting in the shortest possible interval whose length that cannot be further reduced. The length of an interval $$$[\ell, r]$$$ is defined naturally to be $$$r-\ell$$$. Let's consider the following example. Suppose that we are folding a segment initially occupying the interval $$$[1, 30]$$$. There are three folding sequences that lead to the shortest possible resulting interval, as shown in the following figure.

Please help Peter determine the number of folding sequences such that the resulting interval has a shortest possible length. Output the number modulo $$$998244353$$$.

Input: The first line contains an integer $$$t$$$, denoting the number of test cases. In each of the following $$$t$$$ lines, there are two integers $$$\ell$$$ and $$$r$$$.

- $$$1 \le t \le 10$$$
- $$$1 \le \ell < r \le 10^{12}$$$
- $$$r - \ell \le 10^5$$$

Output: For each test case, please output a line denoting the number of ways to fold the given segment such that the resulting segment has the shortest possible length, modulo $$$998244353$$$.

Examples:
input:
3
1 30
16 18
142857 240135
output:
3
1
63

Note:

Buggy Code:
# pragma GCC optimize ("O3")
# pragma GCC optimize ("Ofast")
# pragma GCC optimize ("unroll-loops")
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
 ull mul(ull a, ull b, ull mod){ return (__uint128_t)a * b % mod; }
 bool isp[1101010];
 vector<int> prime;
 void sieve(){
 memset(isp, 1, sizeof isp);
 isp[0] = isp[1] = 0;
 for(ll i=2; i<=1000000; i++){
 if(isp[i]) prime.push_back(i);
 for(auto j : prime){
 if(i*j > 1000000) break;
 isp[i*j] = 0;
 if(i % j == 0) break;
 }
 }
 }

 ull pw(ull a,ull b,ull mod) {
     if (b==0) {
         return 1;
     }
     if (b&1) {
         return mul(pw(a,b-1,mod),a,mod);
     }
     ull ret=pw(a,b/2,mod);
     return mul(ret,ret,mod);
 }
 // 32bit : 2, 7, 61
 // 64bit : 2, 325, 9375, 28178, 450775, 9780504, 1795265022
 bool MR(ull n, ull a){ // Miller Rabin
 if(a % n == 0) return 1;
 int cnt = __builtin_ctzll(n-1);
 ull d = n >> cnt;
 ull p = pw(a, d, n);
 if(p == 1 || p == n- 1) return 1;
 while(cnt--){
 p = mul(p, p, n);
 if(p == n-1) return 1;
 }
 return 0;
 }
 bool isPrime(ll n){
 if(n < 1000001) return isp[n];
 if(n <= 2) return n == 2;
 if(!(n & 1)) return 0;
 if(n%3 == 0 || n%5 == 0 || n%7 == 0 || n%11 == 0) return 0;
 for(int p : {2,325,9375,28178,450775}){
 if(!MR(n, p)) return 0;
 }
 return 1;
 }

typedef pair<int,int> P;
vector<long long> pv;
unordered_map<long long,P> mp;
const int mod=998244353;
long long BIG=1e7;
long long st;

long long fh(P x) {
    return (x.first-st)*BIG+(x.second-st);
}

P solve(long long l,long long r) {
    if (mp.find(fh(P(l,r)))!=mp.end()) {
        return mp[fh(P(l,r))];
    }
    int ind=lower_bound(pv.begin(),pv.end(),(l+r+1)/2)-pv.begin();
    long long two=pv[ind];
    long long one;
    if (two*2==l+r) {
        one=two;
    }
    else {
        one=pv[ind-1];
    }
    P got1=P(r-l,0);
    P got2=P(r-l,0);
    if (one>l) {
        got1=solve(one,r);
    }
    if (two<r) {
        got2=solve(l,two);
    }
    if (got1.second==0||got2.second==0) {
        if (got1.second==0&&got2.second==0) {
            return mp[fh(P(l,r))]=P(r-l,1);
        }
        if (got1.second==0) {
            return mp[fh(P(l,r))]=got2;
        }
        if (got2.second==0) {
            return mp[fh(P(l,r))]=got1;
        }
    }
    if (got1.first<got2.first) {
        return mp[fh(P(l,r))]=got1;
    }
    if (got1.first>got2.first) {
        return mp[fh(P(l,r))]=got2;
    }
    return mp[fh(P(l,r))]=P(got1.first,got1.second+got2.second);
}

int main(void) {
    int tc;
    scanf("%d",&tc);
    sieve();
    for(int t=0;t<tc;t++) {
        //mp.clear();
        long long l,r;
        scanf("%lld %lld",&l,&r);
        l*=2;
        r*=2;
        st=l;
        pv.clear();
        pv.push_back(-1e14);
        for(ll i=l+1;i<r;i+=2) {
            if (isPrime(i)) {
                pv.push_back(i);
            }
        }
        pv.push_back(1e14);
        P ret=solve(l,r);
        printf("%lld\n",ret.second);
    }
}

**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 30
16 18
1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Segmentation Folds
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Peter loves folding segments. There is a segment on a number line occupying the interval $$$[\ell, r]$$$. Since it is the prime time for folding the segments, Peter decides to fold the segment carefully. In each step, he chooses one of the two following operations whenever possible:

1. Operation $$$\tt{LTR}$$$: he folds the segment from left to right, where $$$\ell$$$ coincides with a point $$$x$$$ ($$$\ell < x \le r$$$) such that $$$\ell+x$$$ is a prime number$$$^{\text{∗}}$$$. When Peter chooses this operation, he always chooses the largest possible value $$$x$$$. Note that the segment occupies the interval $$$[\frac{1}{2}(\ell+x), r]$$$ afterwards.
2. Operation $$$\tt{RTL}$$$: he folds the segment from right to left, where $$$r$$$ coincides with a point $$$x$$$ ($$$\ell \le x < r$$$) such that $$$r+x$$$ is a prime number. When Peter chooses this operation, he always chooses the smallest possible value $$$x$$$. Note that the segment occupies the interval $$$[\ell, \frac{1}{2}(r+x)]$$$ afterwards.

A folding sequence refers to a sequence of operations specified above. Peter wants to fold the segment several times, resulting in the shortest possible interval whose length that cannot be further reduced. The length of an interval $$$[\ell, r]$$$ is defined naturally to be $$$r-\ell$$$. Let's consider the following example. Suppose that we are folding a segment initially occupying the interval $$$[1, 30]$$$. There are three folding sequences that lead to the shortest possible resulting interval, as shown in the following figure.

Please help Peter determine the number of folding sequences such that the resulting interval has a shortest possible length. Output the number modulo $$$998244353$$$.

Input: The first line contains an integer $$$t$$$, denoting the number of test cases. In each of the following $$$t$$$ lines, there are two integers $$$\ell$$$ and $$$r$$$.

- $$$1 \le t \le 10$$$
- $$$1 \le \ell < r \le 10^{12}$$$
- $$$r - \ell \le 10^5$$$

Output: For each test case, please output a line denoting the number of ways to fold the given segment such that the resulting segment has the shortest possible length, modulo $$$998244353$$$.

Examples:
input:
3
1 30
16 18
142857 240135
output:
3
1
63

Note:

Buggy Code:
M=10**6
pre_is_prime=[True]*(M+1)
pre_is_prime[0]=False
pre_is_prime[1]=False
for p in range(M+1):
    if pre_is_prime[p]:
        for i in range(2*p,M+1,p):
            pre_is_prime[i]=False
mod=998244353
T=int(input())
for t in range(T):
    L,R=map(int,input().split())
    L*=2
    R*=2
    is_prime=[True]*(R-L+1)
    for p in range(2,R-L+1):
        if pre_is_prime[p]:
            for i in range((-L)%p,R-L+1,p):
                if i+L!=p:
                    is_prime[i]=False
    LR=[]
    queue=[(L,R)]
    while queue:
        l,r=queue.pop()
        end=True
        P=[]
        for p in range(l+1,(l+r)//2+1):
            if is_prime[p-L]:
                P.append(p)
        if P:
            queue.append((max(P),r))
            end=False
        P=[]
        for p in range((l+r+1)//2,r):
            if is_prime[p-L]:
                P.append(p)
        if P:
            queue.append((l,min(P)))
            end=False
        if end:
            LR.append((l,r))
    mi=min(r-l for l,r in LR)
    ans=0
    for l,r in LR:
        if r-l==mi:
            ans+=1
    print(ans)

**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
1 100
1 2
999999999999 1000000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Segmentation Folds
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Peter loves folding segments. There is a segment on a number line occupying the interval $$$[\ell, r]$$$. Since it is the prime time for folding the segments, Peter decides to fold the segment carefully. In each step, he chooses one of the two following operations whenever possible:

1. Operation $$$\tt{LTR}$$$: he folds the segment from left to right, where $$$\ell$$$ coincides with a point $$$x$$$ ($$$\ell < x \le r$$$) such that $$$\ell+x$$$ is a prime number$$$^{\text{∗}}$$$. When Peter chooses this operation, he always chooses the largest possible value $$$x$$$. Note that the segment occupies the interval $$$[\frac{1}{2}(\ell+x), r]$$$ afterwards.
2. Operation $$$\tt{RTL}$$$: he folds the segment from right to left, where $$$r$$$ coincides with a point $$$x$$$ ($$$\ell \le x < r$$$) such that $$$r+x$$$ is a prime number. When Peter chooses this operation, he always chooses the smallest possible value $$$x$$$. Note that the segment occupies the interval $$$[\ell, \frac{1}{2}(r+x)]$$$ afterwards.

A folding sequence refers to a sequence of operations specified above. Peter wants to fold the segment several times, resulting in the shortest possible interval whose length that cannot be further reduced. The length of an interval $$$[\ell, r]$$$ is defined naturally to be $$$r-\ell$$$. Let's consider the following example. Suppose that we are folding a segment initially occupying the interval $$$[1, 30]$$$. There are three folding sequences that lead to the shortest possible resulting interval, as shown in the following figure.

Please help Peter determine the number of folding sequences such that the resulting interval has a shortest possible length. Output the number modulo $$$998244353$$$.

Input: The first line contains an integer $$$t$$$, denoting the number of test cases. In each of the following $$$t$$$ lines, there are two integers $$$\ell$$$ and $$$r$$$.

- $$$1 \le t \le 10$$$
- $$$1 \le \ell < r \le 10^{12}$$$
- $$$r - \ell \le 10^5$$$

Output: For each test case, please output a line denoting the number of ways to fold the given segment such that the resulting segment has the shortest possible length, modulo $$$998244353$$$.

Examples:
input:
3
1 30
16 18
142857 240135
output:
3
1
63

Note:

Buggy Code:
M=2*10**6
pre_is_prime=[True]*(M+1)
pre_is_prime[0]=False
pre_is_prime[1]=False
for p in range(M+1):
    if pre_is_prime[p]:
        for i in range(2*p,M+1,p):
            pre_is_prime[i]=False
mod=998244353
T=int(input())
for t in range(T):
    L,R=map(int,input().split())
    L*=2
    R*=2
    is_prime=[True]*(R-L+1)
    for p in range(2,R-L+1):
        if pre_is_prime[p]:
            for i in range((-L)%p,R-L+1,p):
                if i+L!=p:
                    is_prime[i]=False
    LR=[]
    queue=[(L,R)]
    while queue:
        l,r=queue.pop()
        end=True
        P=[]
        for p in range(l+1,(l+r)//2+1):
            if is_prime[p-L]:
                P.append(p)
        if P:
            queue.append((max(P),r))
            end=False
        P=[]
        for p in range((l+r+1)//2,r):
            if is_prime[p-L]:
                P.append(p)
        if P:
            queue.append((l,min(P)))
            end=False
        if end:
            LR.append((l,r))
    mi=min(r-l for l,r in LR)
    ans=0
    for l,r in LR:
        if r-l==mi:
            ans+=1
    print(ans)

**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 10
10 20
999999999991 999999999999
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Segmentation Folds
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Peter loves folding segments. There is a segment on a number line occupying the interval $$$[\ell, r]$$$. Since it is the prime time for folding the segments, Peter decides to fold the segment carefully. In each step, he chooses one of the two following operations whenever possible:

1. Operation $$$\tt{LTR}$$$: he folds the segment from left to right, where $$$\ell$$$ coincides with a point $$$x$$$ ($$$\ell < x \le r$$$) such that $$$\ell+x$$$ is a prime number$$$^{\text{∗}}$$$. When Peter chooses this operation, he always chooses the largest possible value $$$x$$$. Note that the segment occupies the interval $$$[\frac{1}{2}(\ell+x), r]$$$ afterwards.
2. Operation $$$\tt{RTL}$$$: he folds the segment from right to left, where $$$r$$$ coincides with a point $$$x$$$ ($$$\ell \le x < r$$$) such that $$$r+x$$$ is a prime number. When Peter chooses this operation, he always chooses the smallest possible value $$$x$$$. Note that the segment occupies the interval $$$[\ell, \frac{1}{2}(r+x)]$$$ afterwards.

A folding sequence refers to a sequence of operations specified above. Peter wants to fold the segment several times, resulting in the shortest possible interval whose length that cannot be further reduced. The length of an interval $$$[\ell, r]$$$ is defined naturally to be $$$r-\ell$$$. Let's consider the following example. Suppose that we are folding a segment initially occupying the interval $$$[1, 30]$$$. There are three folding sequences that lead to the shortest possible resulting interval, as shown in the following figure.

Please help Peter determine the number of folding sequences such that the resulting interval has a shortest possible length. Output the number modulo $$$998244353$$$.

Input: The first line contains an integer $$$t$$$, denoting the number of test cases. In each of the following $$$t$$$ lines, there are two integers $$$\ell$$$ and $$$r$$$.

- $$$1 \le t \le 10$$$
- $$$1 \le \ell < r \le 10^{12}$$$
- $$$r - \ell \le 10^5$$$

Output: For each test case, please output a line denoting the number of ways to fold the given segment such that the resulting segment has the shortest possible length, modulo $$$998244353$$$.

Examples:
input:
3
1 30
16 18
142857 240135
output:
3
1
63

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
using ui = unsigned int;
using ull = unsigned long long;
using ll = long long;
#define endl '\n'
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const int maxn = 1e6 + 10;
const int mod = 1000000007;
#define inl inline
#define fr(i, a, b) for (int i = a; i <= b; i++)
#define ford(i, a, b) for (int i = a; i >= b; i--)
#define forall(i, a) for (auto &i : a)

/**
   ____         ___ _____
  / ___| _   _ / _ \___ /
  \___ \| | | | | | ||_ \
   ___) | |_| | |_| |__) |
  |____/ \__, |\___/____/
         |___/
*/
istream &operator>>(istream &in, vector<int> &v)
{
    for (auto &i : v)
        in >> i;
    return in;
}
ostream &operator<<(ostream &out, vector<int> &v)
{
    for (auto &i : v)
        out << i << " ";
    return out;
}
bool _output = 1;

#define int ll
bool vis[maxn];
vector<int> primes;
int pf[maxn];
void init()
{
    vis[1] = vis[0] = 1;
    for (int i = 2; i < maxn; i++)
    {
        if (!vis[i])
        {
            primes.push_back(i);
            pf[i] = i;
        }
        for (auto p : primes)
        {
            if (i * p >= maxn)
                break;
            vis[i * p] = 1;
            pf[i * p] = p;
            if (i % p == 0)
                break;
        }
    }
}
int L, R;
int len = 1e18;
int ans = 0;
vector<int> ps;
void make(int ls, int rs)
{
    // cout << ls << " " << rs << endl;
    if (rs - ls < len)
    {
        len = rs - ls;
        ans = 0;
        // return;
    }
    if (rs - ls == len)
    {
        ans++;
        // return;
    }
    int mid = (ls + rs) / 2;
    auto f = lower_bound(ps.begin(), ps.end(), mid);
    // cout << ls << " " << rs << " ";
    // cout << *f << endl;
    auto nl = f, nr = f;

    while (1)
    {
        if (nl == ps.end())
            break;
        if (*nl <= ls)
            break;
        if (*nl - ls <= rs - *nl)
        {
            make(*nl, rs);
            break;
        }
        if (nl == ps.begin())
            break;
        nl--;
    }

    while (1)
    {
        if (nr == ps.end())
            break;
        if (*nr >= rs)
            break;
        if (*nr - ls >= rs - *nr)
        {
            make(ls, *nr);
            break;
        }
        nr++;
    }
    // cout << *nl << " " << *nr << " " << ls << " " << rs << endl;
    // make(*nr, rs);
    // make(ls, *nl);
}
void solve()
{
    len = 1e18;
    ans = 0;
    ps.clear();
    cin >> L >> R;
    L *= 2, R *= 2;
    unordered_set<int> st;
    for (auto p : primes)
    {
        int x = (L + p - 1) / p * p;
        if (x <= p)
            x = 2 * p;
        for (int i = x; i <= R; i += p)
        {
            st.insert(i);
        }
    }
    fr(i, L, R) if (!st.count(i))
    {
        ps.push_back(i);
        // cout<<i<<endl;
    }
    make(L, R);
    // cout << len << "\n";
    cout << ans << endl;
}
signed main()
{
    init();
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int _ = 1;
    if (_output)
        cin >> _;
    while (_--)
        solve();
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
1 2
1 1000000000000
999999999999 1000000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Segmentation Folds
time_limit_ms: 1000
memory_limit_mb: 1024
Description: Peter loves folding segments. There is a segment on a number line occupying the interval $$$[\ell, r]$$$. Since it is the prime time for folding the segments, Peter decides to fold the segment carefully. In each step, he chooses one of the two following operations whenever possible:

1. Operation $$$\tt{LTR}$$$: he folds the segment from left to right, where $$$\ell$$$ coincides with a point $$$x$$$ ($$$\ell < x \le r$$$) such that $$$\ell+x$$$ is a prime number$$$^{\text{∗}}$$$. When Peter chooses this operation, he always chooses the largest possible value $$$x$$$. Note that the segment occupies the interval $$$[\frac{1}{2}(\ell+x), r]$$$ afterwards.
2. Operation $$$\tt{RTL}$$$: he folds the segment from right to left, where $$$r$$$ coincides with a point $$$x$$$ ($$$\ell \le x < r$$$) such that $$$r+x$$$ is a prime number. When Peter chooses this operation, he always chooses the smallest possible value $$$x$$$. Note that the segment occupies the interval $$$[\ell, \frac{1}{2}(r+x)]$$$ afterwards.

A folding sequence refers to a sequence of operations specified above. Peter wants to fold the segment several times, resulting in the shortest possible interval whose length that cannot be further reduced. The length of an interval $$$[\ell, r]$$$ is defined naturally to be $$$r-\ell$$$. Let's consider the following example. Suppose that we are folding a segment initially occupying the interval $$$[1, 30]$$$. There are three folding sequences that lead to the shortest possible resulting interval, as shown in the following figure.

Please help Peter determine the number of folding sequences such that the resulting interval has a shortest possible length. Output the number modulo $$$998244353$$$.

Input: The first line contains an integer $$$t$$$, denoting the number of test cases. In each of the following $$$t$$$ lines, there are two integers $$$\ell$$$ and $$$r$$$.

- $$$1 \le t \le 10$$$
- $$$1 \le \ell < r \le 10^{12}$$$
- $$$r - \ell \le 10^5$$$

Output: For each test case, please output a line denoting the number of ways to fold the given segment such that the resulting segment has the shortest possible length, modulo $$$998244353$$$.

Examples:
input:
3
1 30
16 18
142857 240135
output:
3
1
63

Note:

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

typedef long long ll;
typedef unsigned long long ull;
 ull mul(ull a, ull b, ull mod){ return (__uint128_t)a * b % mod; }
 bool isp[10101010];
 vector<int> prime;
 void sieve(){
 memset(isp, 1, sizeof isp);
 isp[0] = isp[1] = 0;
 for(ll i=2; i<=10000000; i++){
 if(isp[i]) prime.push_back(i);
 for(auto j : prime){
 if(i*j > 10000000) break;
 isp[i*j] = 0;
 if(i % j == 0) break;
 }
 }
 }

 ull pw(ull a,ull b,ull mod) {
     if (b==0) {
         return 1;
     }
     if (b&1) {
         return mul(pw(a,b-1,mod),a,mod);
     }
     ull ret=pw(a,b/2,mod);
     return mul(ret,ret,mod);
 }
 // 32bit : 2, 7, 61
 // 64bit : 2, 325, 9375, 28178, 450775, 9780504, 1795265022
 bool MR(ull n, ull a){ // Miller Rabin
 if(a % n == 0) return 1;
 int cnt = __builtin_ctzll(n-1);
 ull d = n >> cnt;
 ull p = pw(a, d, n);
 if(p == 1 || p == n- 1) return 1;
 while(cnt--){
 p = mul(p, p, n);
 if(p == n-1) return 1;
 }
 return 0;
 }
 bool isPrime(ll n){
 if(n < 10000001) return isp[n];
 if(n <= 2) return n == 2;
 if(!(n & 1)) return 0;
 if(n%3 == 0 || n%5 == 0 || n%7 == 0 || n%11 == 0) return 0;
 for(int p : {2,325,9375,28178,450775,9780504,1795265022}){
 if(!MR(n, p)) return 0;
 }
 return 1;
 }

typedef pair<long long,long long> P;
vector<long long> pv;
map<P,P> mp;
const int mod=998244353;

P solve(long long l,long long r) {
    int ind=lower_bound(pv.begin(),pv.end(),(l+r+1)/2)-pv.begin();
    long long two=pv[ind];
    long long one;
    if (two*2==l+r) {
        one=two;
    }
    else {
        one=pv[ind-1];
    }
    P got1=P(r-l,0);
    P got2=P(r-l,0);
    if (one>l) {
        got1=solve(one,r);
    }
    if (two<r) {
        got2=solve(l,two);
    }
    if (got1.second==0||got2.second==0) {
        if (got1.second==0&&got2.second==0) {
            return mp[P(l,r)]=P(r-l,1);
        }
        if (got1.second==0) {
            return mp[P(l,r)]=got2;
        }
        if (got2.second==0) {
            return mp[P(l,r)]=got1;
        }
    }
    if (got1.first<got2.first) {
        return mp[P(l,r)]=got1;
    }
    if (got1.first>got2.first) {
        return mp[P(l,r)]=got2;
    }
    long long mn=got1.first;
    long long val=(got1.second+got2.second)%mod;
    return mp[P(l,r)]=P(mn,val);
}

int main(void) {
    int tc;
    scanf("%d",&tc);
    sieve();
    for(int t=0;t<tc;t++) {
        long long l,r;
        scanf("%lld %lld",&l,&r);
        l*=2;
        r*=2;
        pv.clear();
        pv.push_back(-1e14);
        for(ll i=l;i<=r;i++) {
            if (isPrime(i)) {
                pv.push_back(i);
            }
        }
        pv.push_back(1e14);
        P ret=solve(l,r);
        printf("%lld\n",ret.second);
    }
}

**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 2
9 11
142857 142858
```