**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^2$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define lowbit(x) (x & (-x))
#define lc p << 1
#define rc p << 1 | 1
typedef long long ll;
typedef pair<int,int> pii;
mt19937 rnd(time(0));
const int N = 2e5 + 7;
const int mod = 1e9 + 7;
int s[105],l[105],f[105],ans[105],a[105][105],c[105][105];
int m , n;
void cheng() {
    for(int i = 1;i <= m;i++) {
        for(int j = 1;j <= m;j++) {
            c[i][j] = 0;
            for(int k = 1;k <= m;k++) (c[i][j] += a[i][k] * a[k][j]) %= mod;
        }
    }
    for(int i = 1;i <= m;i++) {
        for(int j = 1;j <= m;j++) {
            a[i][j] = c[i][j];
        }
    }
}
void qpow(int n) {
    while(n) {
        if(n & 1) cheng();
        n >>= 1;
    }
}
void solve() {
    cin >> m >> n;
    for(int i = 1;i <= m;i++) cin >> s[i];
    for(int i = 1;i <= m;i++) cin >> l[i];
    for(int i = 1;i <= m;i++) {
        for(int j = 1;j <= m;j++) {
            a[i][j] = (s[i] * (s[j] + l[j]) + l[i] * s[j]) % mod;
        }
    }
    qpow(n);
    f[1] = 1;
    for(int i = 1;i <= m;i++) {
        for(int j = 1;j <= m;j++)
            (ans[i] += f[j] * a[j][i]) %= mod;
    }
    int res = 0;
    for(int i = 1;i <= m;i++) (res += ans[i]) %= mod;
    cout << res << "\n";
}
signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t = 1;
    // cin >> t;
    while(t--) solve();
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider the constraints and the logic of the problem. The key points to focus on are:

1. **Edge Cases**: Consider cases where the number of short and long trails is zero for some cabins.
2. **Large Values**: Use large values for `m` and `n` to stress the algorithm.
3. **Specific Combinations**: Create combinations that might cause issues in the matrix multiplication and exponentiation.

Given the constraints:
- \(1 \le m \le 10^2\)
- \(1 \le n \le 10^9\)
- \(0 \le s_i, l_i \le 10^3\)

Let's create a test input that includes a mix of cabins with no trails, cabins with only short trails, and cabins with only long trails. This will help expose any issues in the matrix multiplication and exponentiation logic.

Here is the test input:

```plaintext
5 10
1 0 2 0 3
0 1 0 2 0
```

Explanation:
- **m = 5**: There are 5 cabins.
- **n = 10**: Harry walks for 10 days.
- **s = [1, 0, 2, 0, 3]**: Cabin 1 has 1 short trail, cabin 2 has 0 short trails, cabin 3 has 2 short trails, cabin 4 has 0 short trails, and cabin 5 has 3 short trails.
- **l = [0, 1, 0, 2, 0]**: Cabin 1 has 0 long trails, cabin 2 has 1 long trail, cabin 3 has 0 long trails, cabin 4 has 2 long trails, and cabin 5 has 0 long trails.

This input includes:
- Cabins with no trails (cabin 2 and cabin 4).
- Cabins with only short trails (cabin 1 and cabin 3).
- Cabins with only long trails (cabin 2 and cabin 4).
- Cabins with both short and long trails (cabin 5).

This combination should help expose any issues in the matrix multiplication and exponentiation logic, particularly how the code handles zero values and large exponents.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^2$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
import os, sys
from io import BytesIO, IOBase
from array import array
from itertools import accumulate
import bisect
import math
from collections import deque

# from functools import cache
# cache cf需要自己提交 pypy3.9!
from copy import deepcopy

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, 8192))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, 8192))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

input = lambda: sys.stdin.readline().strip()
ints = lambda: list(map(int, input().split()))
Int = lambda: int(input())

def queryInteractive(a, b, c):
    print('? {} {} {}'.format(a, b, c))
    sys.stdout.flush()
    return int(input())

def answerInteractive(x1, x2):
    print('! {} {}'.format(x1, x2))
    sys.stdout.flush()

inf = float('inf')
#if "PyPy" in sys.version:
#    import pypyjit; pypyjit.set_param('max_unroll_recursion=-1')
MOD = int(1e9 + 7)
m,n = ints()
s = ints()
l = ints()

def solve_easy():
    dp = [[0]*m for _ in range(n+1)]
    dp[0][0] = 1

    for i in range(1,n+1):
        for j in range(m):
            for k in range(m):
                dp[i][j] = (dp[i][j] + dp[i-1][k]*(s[k]*s[j] + s[k]*l[j] + l[k]*s[j])%MOD)%MOD
    # print(dp)
    print(sum(dp[n])%MOD)

def mat_mul(X,Y):
    n,m = len(X),len(Y[0])
    res = [[0 for j in range(m)] for i in range(n)]
    for i in range(n):
        for j in range(m):
            for k in range(len(Y)):
                res[i][j] += X[i][k] * Y[k][j]
                res[i][j] %= MOD
    return res

def exp(x, i):
    ans = [[0]*m for _ in range(m)]
    for i in range(m):
        ans[i][i] = 1

    tmp = x
    while i:
        if i & 1:
            ans = mat_mul(ans , tmp)
        i >>= 1
        tmp = mat_mul(tmp , tmp)
    return ans

def solve_med():
    P = [[0]*m for _ in range(m)]
    for i in range(m):
        for j in range(m):
            P[i][j] = (s[i]*s[j] + s[i]*l[j] + l[i]*s[j])%MOD
    ans = exp(P,n)
    print(sum(ans[0])%MOD)

solve_med()

# import sys
# for line in sys.stdin:
#     a,b = map(int, line.split())
#     print(a+b)

# python A.py < i.txt > o.txt

**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 the process:

1. **Edge Cases for Input Constraints:**
   - Minimum and maximum values for `m` and `n`.
   - Zero values for `s_i` and `l_i`.
   - Large values for `s_i` and `l_i`.

2. **Scenarios to Test:**
   - Single cabin (`m = 1`).
   - Multiple cabins with varying numbers of short and long trails.
   - Cases where some cabins have no short trails or no long trails.

3. **Combining Edge Cases:**
   - Test with `m = 1` and `n = 10^9` to check if the code handles large `n` correctly.
   - Test with `m = 100` and `n = 1` to check if the code handles large `m` correctly.
   - Test with `m = 100` and `n = 10^9` to check if the code handles both large `m` and `n` correctly.

4. **Specific Test Case:**
   - Let's choose a small-scale test case that is complete and valid:
     - `m = 3`
     - `n = 3`
     - `s = [0, 1, 0]` (one cabin with no short trails)
     - `l = [1, 0, 1]` (one cabin with no long trails)

This test case will challenge the code to handle cabins with no short or long trails and to correctly compute the number of combinations over multiple days.

Here is the test input:

```plaintext
3 3
0 1 0
1 0 1
```

This input is small-scale and valid, and it should expose any issues in the code related to handling cabins with no short or long trails.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^2$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
from sys import stdin
from math import prod

def line(): return stdin.readline().strip()
def rd(converter=int): return converter(line())
def rl(converter=int,delimeter=None): return list(map(converter, line().split(delimeter)))
def rls(num_lines,converter=int): return [rd(converter) for _ in range(num_lines)]
def rg(num_lines,converter=int,delimeter=None): return [rl(converter,delimeter) for _ in range(num_lines)]

MULTIPLE_CASES = 0
mod = 10**9+7

def main():
    m,n = rl()
    S = rl()
    L = rl()

    dp = [0]*m
    dp[0] = 1
    matrix = [[S[c]*S[r] + S[c]*L[r] + L[c]*S[r] for c in range(m)] for r in range(m)]

    def multiply(m1,m2,mod): return [[sum(map(prod,zip(row,col)))%mod for col in zip(*m2)] for row in m1]

    while n:
        if n%2: dp = [sum(map(prod,zip(dp,col)))%mod for col in zip(*matrix)]
        matrix = multiply(matrix,matrix,mod)
        n >>= 1

    print(sum(dp)%mod)

for _ in range(rd() if MULTIPLE_CASES else 1): main()

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

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

Think step by step.

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

This test case is designed to expose potential issues in the given buggy code. Here's the reasoning:

1. **Small Scale**: The test case uses the smallest possible number of cabins (2) and only 1 day, which helps in making the input manageable and easy to verify manually.

2. **Edge Case for Trail Types**:
   - Cabin 1 has only short trails (1 short trail, 0 long trails).
   - Cabin 2 has only long trails (0 short trails, 1 long trail).

3. **Expected Behavior**:
   - Since Harry starts in Cabin 1 and can only take a short trail to the lake, he has only one option for the first part of the day.
   - From the lake, he can only go to Cabin 1 (since taking a long trail to Cabin 2 would violate the rule that at least one trail must be short).

4. **Expected Output**:
   - The number of possible combinations of trails is 1, as Harry can only take the short trail from Cabin 1 to the lake and then return to Cabin 1.

This test case is designed to check if the code correctly handles the constraints and rules, especially when there are limited trail options available.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^2$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

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

public class E_2_Trails_Medium {
    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 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;
        }
    }

    static FastReader vt = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        try {
            // int testCases = vt.nextInt();
            int testCases = 1;
            while (testCases-- > 0) {
                // write code here
                solve();
            }
            out.flush();
        } catch (Exception e) {
            return;
        }
    }

    static void solve() {
        int n = vt.nextInt();
        int m = vt.nextInt();
        int[][] ar = new int[n][2];
        for (int i = 0; i < n; i++) {
            ar[i][0] = vt.nextInt();
        }
        for (int i = 0; i < n; i++) {
            ar[i][1] = vt.nextInt();
        }
        long ans = 0;
        long[] dp = new long[n];
        long mod = 1000000007;
        for (int i = 0; i < n; i++) {
            dp[i] = ar[0][0] * ar[i][0];
            dp[i] += ar[0][0] * ar[i][1];
            dp[i] += ar[0][1] * ar[i][0];
            dp[i] %= mod;
        }
        long[][] mat = new long[n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                mat[i][j] = ar[i][0] * ar[j][0];
                mat[i][j] += ar[i][0] * ar[j][1];
                mat[i][j] += ar[i][1] * ar[j][0];
                mat[i][j] %= mod;
            }
        }

        computepowerofmatrix(mat, m - 1, n, mod);

        long[] temp = new long[n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                temp[i] += dp[j] * mat[i][j];
                temp[i] %= mod;
            }
        }
        for (int i = 0; i < n; i++) {
            ans += temp[i];
            ans %= mod;
        }
        out.println(ans);
    }

    static void computepowerofmatrix(long[][] mat, int power, int n, long mod) {
        long[][] res = new long[n][n];
        for (int i = 0; i < n; i++) {
            res[i][i] = 1;
        }
        while (power > 0) {
            if (power % 2 == 1) {
                res = multiply(res, mat, n, mod);
            }
            mat = multiply(mat, mat, n, mod);
            power /= 2;
        }
    }

    static long[][] multiply(long[][] a, long[][] b, int n, long mod) {
        long[][] res = new long[n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                for (int k = 0; k < n; k++) {
                    res[i][j] += a[i][k] * b[k][j];
                    res[i][j] %= mod;
                }
            }
        }
        return res;
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^2$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

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

public class TrailsMedium {
    static int mod = 1_000_000_007;
    static int[][] grid;
    static int m;
    public static void main(String[] args) throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
        StringTokenizer st = new StringTokenizer(br.readLine());
        m = Integer.parseInt(st.nextToken());
        int n = Integer.parseInt(st.nextToken());
        int[] small = new int[m];
        int[] big = new int[m];
        st = new StringTokenizer(br.readLine());
        for(int i = 0; i < m; i++){
            small[i] = Integer.parseInt(st.nextToken());
        }
        st = new StringTokenizer(br.readLine());
        for(int i = 0; i < m; i++){
            big[i] = Integer.parseInt(st.nextToken());
        }
        grid = new int[m][m];
        for(int i = 0; i < m; i++){
            for(int j = 0; j < m; j++){
                grid[i][j] = (small[i] + big[i]) * (small[j] + big[j]) - big[i] * big[j];
            }
        }
        power(n);
        int sum = 0;
        for(int i = 0; i < m; i++){
            sum += grid[0][i];
        }
        sum %= mod;
        pw.println(sum);
        pw.close();
    }
    public static void power(int n){
        int[][] temp = new int[m][m];
        int[][] ans = new int[m][m];
        for(int i = 0; i < m; i++){
            ans[i][i] = 1;
            temp[i][i] = 1;
        }
        for(int i = 0; i <= 31 - Integer.numberOfLeadingZeros(n); i++){
            temp = matrix(temp, grid);
            if((n & (1 << i)) != 0){
                ans = matrix(ans, temp);
            }

        }
        grid = ans;
    }

    public static int[][] matrix(int[][] a, int[][] b){
        int[][] ans = new int[m][m];
        for(int i = 0; i < m; i++){
            for(int j = 0; j < m; j++){
                for(int k = 0; k < m; k++){
                    ans[i][j] += (1L * a[i][k] * b[k][j]) % mod;
                    ans[i][j] %= mod;
                }
            }
        }
        return 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:
To expose the bug in the provided code, we need to focus on edge cases that might cause issues, such as:

1. **Single Cabin**: When there is only one cabin (i.e., `m = 1`), the code should handle this case correctly.
2. **No Short Trails**: If there are no short trails available, the code should still work correctly.
3. **No Long Trails**: If there are no long trails available, the code should still work correctly.
4. **High Number of Days**: Large values of `n` should be handled correctly, especially when `n` is close to the upper limit (`10^9`).

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

```plaintext
1 1000000000
1
0
```

This input tests the following:
- **Single Cabin**: Only one cabin is present.
- **High Number of Days**: The number of days is set to a very large value (`10^9`).
- **No Long Trails**: There are no long trails, only one short trail.

This input should force the code to handle the matrix exponentiation and the modulo operation correctly, even with a large number of days.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^2$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

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

#define f(i, n) for (ll i = 0; i < n; i++)
#define ia(a, n) \
    ll a[n];     \
    f(i, n) cin >> a[i]

#pragma GCC optimize("unroll-loops")
#pragma gcc optimize("Ofast")
#pragma GCC optimization("Ofast")
#pragma optimize(Ofast)
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")

typedef long long ll;
#define endl '\n'

const ll INF = 1000000000;
const ll MOD = 1000000007;
const ll N = 100005;

bool cmp1(pair<ll  , ll  > a, pair<ll  ,ll  > b)
{
  // Sort based on first element (ascending) and second element (descending)
  return (a.first < b.first) && ((a.first == b.first) && (a.second > b.second));
}

// Custom comparison function to compare based on the second element of pairs
bool cmp2(pair<ll, ll>& a, pair<ll, ll>& b) {
	if(a.second == b.second)
	{
		return a.first > b.first;
	}
    return a.second > b.second;
}

ll  lcm(ll  a , ll  b)
{
	return ((a*b)/(__gcd(a,b)));
}

void pv(vector <ll > v)
{
	for(ll i = 0 ; i < v.size() ; i++)
	{
		cout << v[i] << " ";
	}
	cout << endl;
}

void pvp(vector <pair <ll  , ll >> v)
{
	for(auto x: v)
	{
		cout << x.first << " " << x.second ;
		cout << endl;
	}
	cout << endl;
}

ll maxSubArraySum(vector <ll> a)
{
	ll size = a.size();
    ll max_so_far = INT_MIN , max_ending_here = 0;

    for (ll i = 0; i < size; i++) {
        max_ending_here = max_ending_here + a[i];
        if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;

        if (max_ending_here < 0)
            max_ending_here = 0;
    }
    return max_so_far;
}

ll ps(set <ll> &s)
{
	cout << endl;
	for(auto x : s)
	{
		cout << x << endl;
	}
	cout << endl;
}

ll pms(multiset <ll> &s)
{
	cout << endl;
	for(auto x : s)
	{
		cout << x << endl;
	}
	cout << endl;
}

bool cmp3(string a, string b)
{
	return a+b < b+a;
}

string toBinary(ll n)
{
	string ans = "";
	while(n > 0)
	{

		if(n%2 == 1)
		{

			ans.push_back('1');
		}
		else
		{
			ans.push_back('0');
		}
		n = n/2;
	}
	//reverse(ans.begin() , ans.end());  this would give binary in msb to lsb format
	return ans;

}

ll num_of_digs(ll n)
{
	ll c = 0;
	while( n!= 0)
	{
		n = n/10;
		c++;
	}
	return c;
}

ll binexBest(ll a , ll b , ll mod)
{

	ll result = 1;
	b = b % (MOD-1);
	while(b > 0)
	{
		if(b%2 == 1)
		{
			result = (result%mod * a%mod)%mod ;
		}
		a = (a%mod*a%mod)%mod;
		b = b/2;
	}
	return result;
}

ll invUsingBinexBest(ll a, ll p)
{
	return (binexBest(a , p-2 , p) );
}

ll num_of_div(vector <pair <ll , ll>> &v , ll mod)
{

	ll result = 1;
	for(ll i = 0 ; i < v.size() ; i++)
	{
		result = (result % mod * (v[i].second + 1)) % mod;
	}
	return result;
}

ll sum_of_div(vector <pair <ll , ll>> &v , ll mod)
{
	vector <ll> for_mul;
	for(ll i = 0 ; i < v.size() ; i++)
	{
		ll p = v[i].first;
		ll a = v[i].second;
		for_mul.push_back( ((binexBest(p,a+1,mod) - 1)*invUsingBinexBest(p-1 , mod))%mod );
	}

	ll result = 1;
	for(ll i = 0 ; i < for_mul.size() ; i++)
	{
		result = (result%mod * for_mul[i]%mod)%mod;
	}
	return result;
}

ll prod_of_div(vector <pair <ll , ll>> v , ll mod)
{
	ll divs = 1 , b = 0;
	for(ll i = 0 ; i < v.size() ; i++)
	{
		if(b == 0 && v[i].second % 2 == 1)
		{
			b = 1;
			divs = (divs % (mod-1) * ((v[i].second + 1)/2)) % (mod-1);
			continue;
		}
		divs = (divs % (mod-1) * (v[i].second + 1)) % (mod-1);
	}
	vector <ll> pows;
	if(b == 1)
	{
		for(ll i = 0 ; i < v.size() ; i++)
		{
			ll p = v[i].first;
			ll a = v[i].second;
			pows.push_back( ( a * divs ) % (mod-1) );
		}
	}
	else
	{
		for(ll i = 0 ; i < v.size() ; i++)
		{
			ll p = v[i].first;
			ll a = v[i].second;
			pows.push_back( ( a/2 * divs ) % (mod-1) );
		}
	}
	ll result = 1;
	for(ll i = 0 ; i < pows.size() ; i++)
	{
		ll p = v[i].first;
		result = (result * binexBest(p , pows[i] , mod)) % mod;
	}
	return result;

}

vector <ll> FibGlobal = {0 , 1};
void FibGen(ll n , ll mod)
{
	for(ll i = 2 ; i <= n ; i++)
	{
		FibGlobal.push_back((FibGlobal[i-1] + FibGlobal[i-2]) % mod);
	}
}

vector <vector <ll>> mat_mul(vector <vector<ll>> &v , vector <vector<ll>> &p , ll mod )
{

	ll n = v.size();

    vector<vector<ll>> result(n, vector<ll>(n, 0));

    for(ll i = 0 ; i < n ; i++)
    {
    	for(ll j = 0 ; j < n ; j++)
    	{
    		ll temp = 0;
    		for(ll k = 0 ; k < n ; k++)
    		{

    			temp = (temp + (v[i][k] * p[k][j]));
    		}
    		result[i][j] = temp;
    	}
    }
    return result;
}

ll diophantine_using_extended_euler(ll a, ll b, ll& x, ll& y)  //ax + by == gcd(a,b)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return 0;
    }
    ll x1, y1;
    diophantine_using_extended_euler(b, a % b, x1, y1);
    x = y1;
    y = x1 - y1 * (a / b);
    return 0;
}

pair <ll,ll> chinese_remainder_theoram(vector <pair<ll,ll>> &v) // x = r % m -> {r , m}
{
	for(ll i = 0 ; i < v.size() ; i++)
	{
		swap(v[i].first , v[i].second);
	}
	ll product = 1;
	for(ll i = 0 ; i < v.size() ; i++)
	{
		product = product * v[i].first;
	}

	pair <ll,ll> ans = {0,product};
	for(ll i = 0 ; i < v.size() ; i++)
	{
		ll temp = product/v[i].first;
		ans.first = (ans.first + (v[i].second*temp*invUsingBinexBest(temp , v[i].first)))%product;
	}

	cout << ans.first << " mod " << ans.second;
	return ans;

}

vector <vector <ll>> mat_exp(vector <vector <ll>> &v , ll n , ll mod)
{
	vector<vector<ll>> ans(v.size(), vector<ll>(v.size(), 0));

	for(ll i = 0 ; i < v.size() ; i++)
	{
		ans[i][i] = 1;
	}

	while(n>0)
	{
		if(n%2 == 1)
		{
			ans = mat_mul(ans , v , mod);
		}
		v = mat_mul(v , v , mod);
		n = n/2;
	}
	return ans;
}

vector<ll> sieve(ll n)
{
    vector<ll> sieveArr(n + 1, 1);
    sieveArr[0] = sieveArr[1] = 0;

    for(ll i = 2; i * i <= n; i++)
    {
        if(sieveArr[i] == 1)
        {
            for(ll j = i * i; j <= n; j += i)
            {
                sieveArr[j] = 0;
            }
        }
    }

    //return sieveArr;

    vector<ll> primes;
    for(ll i = 2; i <= n; i++)
    {
        if(sieveArr[i] == 1)
        {
            primes.push_back(i);
        }
    }

    return primes;
}

vector <ll> betterSieve(ll n)
{
	vector<ll> lp(n+1); //gives the smalles prime factor for every element
	vector<ll> pr;
	ll counter = 0 ;
	for (ll i=2; i <= n; ++i)
	{
	    if (lp[i] == 0)
	    {
	        lp[i] = i;
	        pr.push_back(i);
	    }
	    for (ll j = 0; i * pr[j] <= n; ++j)
	    {
	    	counter++;
	        lp[i * pr[j]] = pr[j];
	        if (pr[j] == lp[i])
	        {
	            break;
	        }
	    }
	}
	//return pr;
	return lp;
}

string pad(string s , ll n) //works for lsb to msb
{
	for(ll i = s.size() ; i < n ; i++)
	{
		s.push_back('0');
	}
	return s;
}
void modulo(vector<vector<ll>> &v)
{
    for (auto &x : v)
    {
        for (auto &y : x)
        {
            y %= MOD;
        }
    }
}

vector<vector<ll>> power(vector<vector<ll>> v, ll b)
{
    ll n = v.size();
    modulo(v);
    vector<vector<ll>> ans(n, vector<ll>(n, 0));
    f(i, n)
    {
        ans[i][i] = 1;
    }
    while (b)
    {
        if (b & 1)
        {
            ans = mat_mul(ans, v , MOD);
            modulo(ans);
        }
        v = mat_mul(v, v , MOD);
        modulo(v);
        b >>= 1;
    }
    return ans;
}

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

    ll m , n;
    cin >> m >> n;
    vector <ll> s , l;
    for(ll i = 0 ; i < m ; i++)
    {
    	ll temp;
    	cin >> temp;
    	s.push_back(temp);
    }
    for(ll i = 0 ; i < m ; i++)
    {
    	ll temp;
    	cin >> temp;
    	l.push_back(temp);
    }

    vector<vector<ll>> A(m, vector<ll>(m, 0));

    for(ll i = 0 ; i < m ; i++)
    {
    	for(ll j = 0 ; j < m ; j++)
    	{
    		A[i][j] = s[i]*l[j] + l[i]*s[j] + s[i]*s[j];
    	}
    }

    vector <vector <ll>> An = power(A, n-1);

    ll ans = 0;
    for(ll i = 0 ; i < m ; i++)
    {
        for(ll j = 0 ; j < m ; j++)
        {
            ans += An[i][j] * (s[0] * l[j] + l[0] * s[j] + s[0] * s[j]);
            ans %= MOD;
        }
    }
    cout << 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:
To generate a challenging test input that exposes potential bugs in the given code, we need to focus on edge cases and scenarios that might cause the code to fail. Here are the steps to consider:

1. **Edge Cases:**
   - Minimum and maximum values for `m` and `n`.
   - Cases where all trails are short or all trails are long.
   - Cases where there are no short trails or no long trails.

2. **Complex Cases:**
   - Cases where the number of short and long trails is balanced.
   - Cases with a mix of high and low values for `s_i` and `l_i`.

3. **Small-Scale Input:**
   - Given the constraints, a small-scale input with `m = 3` and `n = 2` or `n = 3` can be sufficient to test the algorithm.

Let's construct a test input that combines these considerations:

- `m = 3` and `n = 2` to keep the input small.
- `s_i` and `l_i` values that include both zero and non-zero values to test edge cases.

Here is the test input:

```plaintext
3 2
1 0 1
0 1 1
```

This input includes:
- One cabin with only short trails.
- One cabin with no trails.
- One cabin with both short and long trails.

This setup should help expose any issues in the code, especially in handling zero values and ensuring the correct combination of trails.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^2$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
#include<bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp> // Common file
// #include <ext/pb_ds/tree_policy.hpp>
// #include <functional> // for less

// #include<iostream>
// #include<string>
// #include<vector>
// #include<fstream>
using namespace std;

// using namespace __gnu_pbds;
// typedef tree<int, null_type, less<int>, rb_tree_tag,
//              tree_order_statistics_node_update>
//     ordered_set;

typedef long long ll;
// const ll mod=998244353;
const ll hmod=1e9+7;
const int intinfi=1e9+10;
const ll infi=1e18+10;
#define pb push_back
#define vec vector<int>
#define llv vector<ll>
// #define st set<int>
#define fl(i,a,b) for(int i=a;i<=b ;i++)
#define rl(i,a,b) for(int i=b;i>=a;i--)
#define inarr(arr,n) int arr[n]; fl(i,0,n-1){cin>>arr[i];}
#define setarr(arr,n,x) fl(i,0,n-1){arr[i]=x;}
#define ff first
#define sec second
#define mp make_pair
#define all(v) v.begin(),v.end()
#define pii pair<int,int>
#define endl '\n'
#define int ll
/*****************************************FUNCTIONS*****************************************/
bool is_powtwo(ll n) {return !(n & (n - 1));}
int fgcd(int a,int b){if (a==0) return b ; if(b==0)return a; if (a>=b) return fgcd(b,a%b); return fgcd(a,b%a);}
int flcm(int a,int b){if(a==0||b==0) return 0; return (a/fgcd(a,b))*b;}
ll longlcm(ll a,ll b){if(a==0||b==0) return 0; return (a/fgcd(a,b))*b;}
// ll ncr(int n,int r){ if (r>n || r<0 || n<0) return 0; if(r==0||n==1||r==n) return 1; return (n*ncr(n-1,r-1))/r;}
ll power(ll mod,int a,int n){ ll mario=1; while(n--) mario=(mario*a)%mod; return mario;}
int modInverse(int A){
    int m0 = hmod;
    int y = 0, x = 1;
    if (m0 == 1)
        return 0;
    while (A > 1) {
        int q = A / m0;
        int t = m0;
        m0 = A % m0, A = t;
        t = y;
        y = x - q * y;
        x = t;
    }

    // Make x positive
    if (x < 0)
        x += hmod;

    return x;
}

int t,n,m,i,j,k,l,store,flg,a,b,c,d,e;
ll sum,prod,xx,appu;
float woo;
string str,final;

int nc2(int a,int b, vec &inv){
    int ca=1;
    fl(i,1,a){
        ca=(ca*i)%hmod;
    }
    fl(i,1,b){
        ca=(ca*inv[i])%hmod;
    }
    fl(i,1,a-b){
        ca=(ca*inv[i])%hmod;
    }
    return ca;

}

// void dfs(int x,vec &bad, vec &ids, vector<vec> &children){
//     bad.pb(x);
//     ids[x]=bad.size()-1;
//     for(auto u: children[x]){
//         dfs(u,bad,ids,children);
//     }
// }
int get_par(int index, vec &bad,vec &ids,vec &par){
    return ids[par[bad[index]]];
}
int get_c(int index,vec &bad, vec &ids, int num){
    int a=bad[index]*2+num;
    // cout<<index<<" "<<a<<endl;
    // cout<<n<<endl;
    if(a>n) return -1;
    return ids[bad[index]*2+num];
}
bool check_par(int a, vec &perm,vec &bad, vec &ids, vec &par){
    if(perm[get_par(a,bad,ids,par)]==perm[a]/2) return true;
    return false;
}
bool check_child(int a, int c, vec &perm,vec &bad, vec&ids){
    if(perm[c]/2!=perm[a]){
        return false;
    }
    // if(perm[get_c(a,bad,ids,1)]/2!=perm[a]) return false;
    return true;
}
void update(int c1, vec& perm, vec &bad, vec &ids, vec &wrngp,int &sum,vec &par){
    if(check_par(c1,perm,bad,ids,par)){
        if(wrngp[perm[c1]]){
            wrngp[perm[c1]]=0;
            sum--;
        }
    }else{
        if(wrngp[perm[c1]]==0){
            wrngp[perm[c1]]=1;
            sum++;
        }
    }
}
int solve(int a,int b,int c,int d){
    if(a<c && d<b){
        return d-c+2;
    }
    if(c<a && b<d) return b-a+2;
    if(a==c && b==d) return b-a;
    if(a==c) return min(b,d)-a+1;
    if(b==d) return b-max(a,c)+1;
    if(a<c && b>=c && b<d) return b-c+2;
    if(c<a && d>=a && d<b) return d-a+2;
    if(c>b) return 1;
    return 1;

}

// void dfs(int a, int b, vec &vis, vector<vec> &adj){
//     if(a==b) return;
//     if(vis[a]) return;
//     int c=b,d=b;
//     while(true){
//         cout<<"? "<<a<<" "<<b<<endl;
//         fflush(stdout);
//         cin>>c;
//         if(c==a){
//             vis[a]=1;
//             adj[a].pb(b);
//             dfs(b,d,vis,adj);
//             break;
//         }
//         b=c;
//     }
// }
// bool check(int n,string str,int k){
//     vec arr(n+1),cng(n+2,0);
//     fl(i,1,n) arr[i]=str[i-1]-'0';
//     fl(i,1,n){
//         cng[i]=cng[i-1]+cng[i];
//         arr[i]=(arr[i]+cng[i])%2;
//         if(arr[i]==0 && i<=n-k+1){
//             arr[i]=1;
//             cng[i]++;
//             cng[k+i]--;
//         }

//     }

//     fl(i,1,n){
//         if(arr[i]!=1) return false;
//     }
//     return true;

// }
// void dfs(int x,int y,vector<vec> &ex, vector<vec>&vis, vector<vec>&arr){
//     if(vis[x][y]) return;
//     vis[x][y]=1;
//     if(x==1||y==1){
//         ex[x][y]=1;
//     }
//     if(x-1>=1 && arr[x-1][y]){
//         dfs(x-1,y,ex,vis,arr);
//         if(ex[x-1][y]) ex[x][y]=1;
//     }
//     if(y-1>=1 && arr[x][y-1]){
//         dfs(x,y-1,ex,vis,arr);
//         if(ex[x][y-1]) ex[x][y]=1;
//     }
// }
// bool chk(string &a,string &b,int l1,int l2){
//     if(l2==b.length()) return true;
//     if(l1>=a.length()) return false;
//     if(a[l1]==b[l2]) return chk(a,b,l1+1,l2+1);
//     return chk(a,b,l1+2,l2);
// }

// void dfs(int x,map<int,vec> &ad,map<int,int> &cmp, map<int,int> &vis,map<int,int> &cmp_max,map<int,int> &cmp_min){
//     if(vis[x]) return;
//     vis[x]=1;
//     cmp_max[cmp[x]]=max(cmp_max[cmp[x]],x);
//     cmp_min[cmp[x]]=min(cmp_min[cmp[x]],x);
//     for(auto u:ad[x]){
//         cmp[u]=cmp[x];
//         dfs(u,ad,cmp,vis,cmp_max,cmp_min);
//     }
// }
int fastpr(int a,int n,int mod){
    if(n==0) return 1;
    int b=fastpr(a,n/2,mod);
    if(n%2){
        return (a*((b*b)%mod))%mod;
    }
    return (b*b)%mod;
}
int ff(int a,int b1,int b2){
    a+=min(b1,b2);
    int p=min(b1,b2);
    b1-=p;
    b2-=p;
    if(max(b1,b2)%2==0 && a>=max(b1,b2)) return 1;
    return 0;
}
pair<int,int> op(int x1,int x2,int a,int b){
    // if(x2<a||b<x1) return {0,-1};
    // return {max(a,x1),min(x2,b)};
    if(a>=x1 && b<=x2) return {a,b};
    if(x1>=a && x2<=b) return {x1,x2};
    if(x1<=a && a<=x2) return {a,x2};
    if(x1<=b && x2>=b) return {x1,b};
    return {0,-1};
}
int clr(int a,int b,int c,int d,int wht){
    int p=c-a+1;
    int q=d-b+1;
    int cutu=(p*q)/2;
    if((p*q)%2 && ((a+b)%2)==0) cutu++;
    if(wht) return cutu;
    return p*q-cutu;
}
vector<long long> fac(500001,1),invfac(500001,1);
long long mod=1e9+7;

int mi(long long a,long long n){
    if(n==0) return 1;
    long long b=mi(a,n/2);
    if(n%2){
        return (a*(((b*b)%mod)))%mod;
    }
    return (b*b)%mod;
}

long long ncr(long long n,long long r){
    return (((fac[n]*invfac[n-r])%mod) * invfac[r])%mod;
}
void dfs(int x, vector<vec> &adj,vec &vis,vec &par,vec &color, vec &bcc){
    vis[x]=1;
    for(auto u:adj[x]){
        if(vis[u]){continue;}
        if(color[u]) bcc[x]++;
        par[u]=x;
        dfs(u,adj,vis,par,color,bcc);
    }
}

int solve(int n, vector<vec> &conn,int a,int b){
    if(n==1) return conn[a][b];
    int plop=0;
    int j;
    if(n%2){
        n=n/2;
        fl(j,1,m){
            int c=solve(n,conn,a,j);
            int d=solve(n,conn,j,b);
            plop=(plop+c*d)%hmod;
        }
        return plop;
    }else{
        fl(j,1,m){
            int c=solve(n-1,conn,a,j);
            plop=(plop+c*conn[j][b])%hmod;
        }
        return plop;
    }
}
/*****************************************CODE HERE*****************************************/
int32_t main(){
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    // cin>>t;
    t=1;
    while(t--){
        cin>>m>>n;
        vec st(m+1),lt(m+1);
        k=log2(n)+1;
        vector<vec> conn(m+1,vec(m+1));
        fl(i,1,m) cin>>st[i];
        fl(i,1,m) cin>>lt[i];
        fl(i,1,m){
            fl(j,1,m){
                conn[i][j]=(st[i]*st[j]+st[i]*lt[j]+lt[i]*st[j])%hmod;
            }
        }
        // fl(i,1,m){
        //     fl(j,1,m){
        //         cout<<conn[i][j]<<" ";
        //     }
        //     cout<<endl;
        // }
        sum=0;
        fl(i,1,m){
            a=solve(n,conn,1,i);
            sum=(sum+a)%hmod;
            // cout<<a<<endl;
        }
        cout<<sum;

        // cout<<endl;

    }

}

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

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

Think step by step.

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

1. **Edge Cases:**
   - Single cabin (`m = 1`).
   - Maximum number of cabins (`m = 100`).
   - No short trails available.
   - No long trails available.
   - All cabins have both short and long trails.

2. **Scenarios:**
   - Large number of days (`n = 10^9`).
   - Small number of days (`n = 1`).
   - Intermediate number of days (`n = 5`).

3. **Combination of Trails:**
   - Cabins with a large number of short trails and a small number of long trails.
   - Cabins with a small number of short trails and a large number of long trails.
   - Cabins with equal numbers of short and long trails.

Given these considerations, let's create a small-scale test input that combines some of these elements to maximize the failure probability in the faulty solution.

### Test Input
```plaintext
3 3
2 1 0
0 1 2
```

### Explanation
- **m = 3**: Three cabins.
- **n = 3**: Harry walks for 3 days.
- **s_i**: Number of short trails from each cabin.
  - Cabin 1: 2 short trails.
  - Cabin 2: 1 short trail.
  - Cabin 3: 0 short trails.
- **l_i**: Number of long trails from each cabin.
  - Cabin 1: 0 long trails.
  - Cabin 2: 1 long trail.
  - Cabin 3: 2 long trails.

This test input:
- Includes a cabin with no short trails (Cabin 3).
- Includes a cabin with no long trails (Cabin 1).
- Includes a cabin with both short and long trails (Cabin 2).
- Has a moderate number of days (3) to ensure the algorithm has to handle multiple days of walking.

This setup should expose any issues in the algorithm's handling of different combinations of short and long trails over multiple days.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^2$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
from sys import stdin
input=lambda :stdin.readline()[:-1]

MOD = 10**9 + 7

def solve():
    m, n = map(int, input().split())
    s = list(map(int, input().split()))
    l = list(map(int, input().split()))
    total = [s[i] + l[i] for i in range(m)]
    dp = [[0]*m for _ in range(n+1)]
    dp[0][0] = 1

    for i in range(1, n+1):
        for j in range(m):
            for k in range(m):
                dp[i][k] += ((total[j] * total[k] - l[k] * l[j]) * dp[i-1][j]) % MOD
                dp[i][k] %= MOD

    ans = dp[n]
    print(sum(ans)%MOD)

for __ in range(1):
    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
2 1
1 0
0 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^2$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
MOD = 10**9 + 7

def matrix_mult(A, B):
    n = len(A)
    result = [[0] * n for _ in range(n)]
    for i in range(n):
        for j in range(n):
            result[i][j] = sum(A[i][k] * B[k][j] for k in range(n)) % MOD
    return result

def matrix_power(M, power):
    n = len(M)
    result = [[1 if i == j else 0 for j in range(n)] for i in range(n)]
    base = M
    while power > 0:
        if power % 2 == 1:
            result = matrix_mult(result, base)
        base = matrix_mult(base, base)
        power //= 2
    return result

n, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))

M = [[0] * n for _ in range(n)]
for i in range(n):
    for j in range(n):
        M[i][j] = ((a[i] + b[i]) * (a[j] + b[j]) - b[i] * b[j]) % MOD

M = matrix_power(M, m)

v0 = [1] + [0] * (n - 1)
vn = [sum(M[i][j] * v0[j] for j in range(n)) % MOD for i in range(n)]

print(sum(vn) % MOD)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^2$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

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

public class Main {
    private static final int MOD = 1000000007;
    //private static final int MOD = 998244353;
    private static int add(long a, long b) { return (int) ((a+b)%MOD); }
    private static int multiply(long a, long b) { return (int) ((a*b)%MOD); }
    private static int[][] multiplyMat(int[][] a, int[][] b) {
        int[][] c = new int[a.length][b[0].length];
        for(int i = 0; i < a.length; i++) {
            for(int j = 0; j < b[0].length; j++) {
                for(int k = 0; k < a[0].length; k++) {
                    c[i][j] = add(c[i][j], multiply(a[i][k], b[k][j]));
                }
            }
        }
        return c;
    }
    private static int[][] powerMat(int[][] a, long b) {
        int[][] c = new int[a.length][a.length];
        for(int i = 0; i < a.length; i++) {
            c[i][i] = 1;
        }
        while(b > 0) {
            if((b&1) > 0) {
                c = multiplyMat(c, a);
            }
            a = multiplyMat(a, a);
            b >>= 1;
        }
        return c;
    }
    public static void main(String[] args) throws IOException {
        //BufferedReader f = new BufferedReader(new FileReader("uva.in"));
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        StringTokenizer st = new StringTokenizer(f.readLine());
        int m = Integer.parseInt(st.nextToken());
        int n = Integer.parseInt(st.nextToken());
        st = new StringTokenizer(f.readLine());
        int[] s = new int[m];
        for(int i = 0; i < m; i++) {
            s[i] = Integer.parseInt(st.nextToken());
        }
        st = new StringTokenizer(f.readLine());
        int[] l = new int[m];
        for(int i = 0; i < m; i++) {
            l[i] = Integer.parseInt(st.nextToken());
        }
        int[][] matrix = new int[m][m];
        for(int i = 0; i < m; i++) {
            for(int j = 0; j < m; j++) {
                matrix[i][j] = s[i]*l[j]+s[j]*l[i]+s[i]*s[j];
            }
        }
        matrix = powerMat(matrix, n);
        int ans = 0;
        for(int i = 0; i < m; i++) {
            ans = add(ans, matrix[0][i]);
        }
        out.println(ans);
        f.close();
        out.close();
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to consider edge cases that might cause the algorithm to fail. Here are some potential edge cases:

1. **Single Cabin (m = 1)**: This should be a simple case, but it can help verify if the code handles the base case correctly.
2. **No Short Trails (s_i = 0 for all i)**: This should force the algorithm to only use long trails, which might expose issues in the trail combination logic.
3. **No Long Trails (l_i = 0 for all i)**: This should force the algorithm to only use short trails, which might expose issues in the trail combination logic.
4. **Large Number of Days (n = 10^9)**: This can test the efficiency and correctness of the matrix exponentiation.
5. **Mixed Trails with Specific Patterns**: This can test the logic for combining trails in more complex scenarios.

Let's choose a small-scale test input that combines some of these edge cases to maximize the probability of exposing the bug:

### Test Input
```plaintext
3 1000000000
0 1 0
1 0 1
```

### Explanation
- **m = 3**: Three cabins.
- **n = 1000000000**: A very large number of days to test the matrix exponentiation.
- **s = [0, 1, 0]**: Cabin 1 and Cabin 3 have no short trails, Cabin 2 has one short trail.
- **l = [1, 0, 1]**: Cabin 1 and Cabin 3 have one long trail, Cabin 2 has no long trails.

This input will:
- Test the handling of cabins with no short trails.
- Test the handling of cabins with no long trails.
- Test the matrix exponentiation with a very large exponent.

By using this input, we can see if the code correctly handles these edge cases and if the matrix exponentiation works as expected.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Trails (Medium)
time_limit_ms: 1000
memory_limit_mb: 256
Description: Harry Potter is hiking in the Alps surrounding Lake Geneva. In this area there are $$$m$$$ cabins, numbered 1 to $$$m$$$. Each cabin is connected, with one or more trails, to a central meeting point next to the lake. Each trail is either short or long. Cabin $$$i$$$ is connected with $$$s_i$$$ short trails and $$$l_i$$$ long trails to the lake.

Each day, Harry walks a trail from the cabin where he currently is to Lake Geneva, and then from there he walks a trail to any of the $$$m$$$ cabins (including the one he started in). However, as he has to finish the hike in a day, at least one of the two trails has to be short.

How many possible combinations of trails can Harry take if he starts in cabin 1 and walks for $$$n$$$ days?

Give the answer modulo $$$10^9 + 7$$$.

Input: The first line contains the integers $$$m$$$ and $$$n$$$.

The second line contains $$$m$$$ integers, $$$s_1, \dots, s_m$$$, where $$$s_i$$$ is the number of short trails between cabin $$$i$$$ and Lake Geneva.

The third and last line contains $$$m$$$ integers, $$$l_1, \dots, l_m$$$, where $$$l_i$$$ is the number of long trails between cabin $$$i$$$ and Lake Geneva.

We have the following constraints:

$$$0 \le s_i, l_i \le 10^3$$$.

$$$1 \le m \le 10^2$$$.

$$$1 \le n \le 10^9$$$.

Output: The number of possible combinations of trails, modulo $$$10^9 + 7$$$.

Examples:
input:
3 2
1 0 1
0 1 1
output:
18

Note:

Buggy Code:
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;

public class trails {
	static long mod = (long) 1e9 + 7;
	public static void main(String[] args) throws Exception {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		int[] buf = Arrays.stream(in.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
		int m = buf[0];
		int n = buf[1];
		long[] s = Arrays.stream(in.readLine().split(" ")).mapToLong(Long::parseLong).toArray();
		long[] l = Arrays.stream(in.readLine().split(" ")).mapToLong(Long::parseLong).toArray();

		long[][] trans = new long[m][m]; //i -> j
		for (int i = 0; i < m; i++) {
			for (int j = 0; j < m; j++) {
				trans[i][j] = s[i]*((s[j] + l[j])%mod)%mod + l[i] * s[j] %mod;
				trans[i][j] %= mod;
			}

		}

		long[][] dp = new long[m][n+1];
		for (int rem = 0; rem <= n; rem++) {
			for (int i = 0; i < m; i++) {
				if (rem == 0) dp[i][rem] = 1;
				else {
					for (int j = 0; j < m; j++) {
						dp[i][rem] += trans[i][j] * dp[j][rem-1] % mod;
						dp[i][rem] %= mod;
					}
				}
			}
		}
		System.out.println(dp[0][n]);

		in.close();
	}
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases or scenarios that can maximize the failure probability. Here are the key points to consider:

1. **Edge Cases:**
   - Cabins with no trails (both short and long).
   - Cabins with only short trails.
   - Cabins with only long trails.
   - Large values for `s_i` and `l_i`.

2. **Constraints:**
   - `1 <= m <= 100`
   - `1 <= n <= 10^9`
   - `0 <= s_i, l_i <= 10^3`

3. **Test Input:**
   - We need to create a small-scale input that is complete and valid.
   - The input should be challenging enough to expose the bug in the code.

Given these considerations, let's create a test input where the bug in the transition matrix calculation can be more easily observed. A good candidate is a scenario where there are cabins with both short and long trails, and the number of days is relatively large.

```plaintext
3 100
1 0 1
0 1 1
```

### Explanation:
- **m = 3**: There are 3 cabins.
- **n = 100**: Harry walks for 100 days.
- **s = [1, 0, 1]**: Cabin 1 has 1 short trail, Cabin 2 has 0 short trails, and Cabin 3 has 1 short trail.
- **l = [0, 1, 1]**: Cabin 1 has 0 long trails, Cabin 2 has 1 long trail, and Cabin 3 has 1 long trail.

This input is designed to test the following:
- The transition from Cabin 1 to Cabin 2 and back.
- The transition from Cabin 1 to Cabin 3 and back.
- The transition from Cabin 2 to Cabin 3 and back.

The large value of `n` (100) ensures that the dynamic programming part of the code is also tested for a significant number of days.

By using this input, we can observe if the code correctly handles the transitions and the modulo operations, especially when dealing with large numbers and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Parity and Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an array $a$ of $n$ positive integers.

In one operation, you can pick any pair of indexes $(i, j)$ such that $a_i$ and $a_j$ have distinct parity, then replace the smaller one with the sum of them. More formally:

  * If $a_i < a_j$, replace $a_i$ with $a_i + a_j$;    * Otherwise, replace $a_j$ with $a_i + a_j$.

Find the minimum number of operations needed to make all elements of the array have the same parity.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$).

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

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

Output: For each test case, output a single integer — the minimum number of operations required.

Examples:
input:
7
5
1 3 5 7 9
4
4 4 4 4
3
2 3 4
4
3 2 2 8
6
4 3 6 1 2 1
6
3 6 1 2 1 2
5
999999996 999999997 999999998 999999999 1000000000
output:
0
0
2
4
3
3
3

Note: In the first test case, all integers already have the same parity. Therefore, no operation is needed.

In the third test case, we can perform two operations $(1, 2)$ and $(1, 3)$. The array $a$ transforms as follows: $a = [\color{red}2, \color{red}3, 4] \longrightarrow [\color{red}5, 3, \color{red}4] \longrightarrow [5, 3, 9]$.

In the fourth test case, an example of an optimal sequence of operations is $(1, 2)$, $(1, 3)$, $(1, 4)$, and $(1, 4)$. The array $a$ transforms as follows: $a = [\color{red}3, \color{red}2, 2, 8] \longrightarrow [\color{red}3, 5, \color{red}2, 8] \longrightarrow [\color{red}3, 5, 5, \color{red}8] \longrightarrow [\color{red}{11}, 5, 5, \color{red}8] \longrightarrow [11, 5, 5, 19]$.

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

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

        while(t--){
            int x;
            long long temp;
            cin>>x;
            multiset<long long> evenItems;
            long long maxOdd=0;
            int count=0;
            for(int i=0;i<x;i++){

                cin >> temp;
                if(temp%2==0){
                    evenItems.insert(temp);
                }
                else if(maxOdd<temp){
                    maxOdd=temp;
                }
            }
            if(maxOdd==0){
                cout<<0<<endl;

            }
            else{
                for(auto it:evenItems){
                    if(maxOdd<it){
                        count+=2;
                        maxOdd=maxOdd+it;
                    }
                    else{
                        count++;
                        maxOdd=maxOdd+it;
                    }
                }
                cout<<count<<endl;
            }

        }

    }

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's consider the following aspects:

1. **Edge Cases:**
   - All elements have the same parity (either all even or all odd).
   - Only one element is of different parity.
   - The array has a mix of even and odd elements, but the largest odd element is smaller than the smallest even element.

2. **Scenarios to Maximize Failure Probability:**
   - The code incorrectly handles the case where the largest odd element is smaller than the smallest even element.
   - The code might not handle the case where there are multiple even elements correctly.

Given these considerations, a challenging test input could be an array where the largest odd element is smaller than the smallest even element. This will test the logic of the code in handling such a scenario.

Here is the test input:

```plaintext
1
5
2 4 6 8 1
```

Explanation:
- The array has one odd element (1) and four even elements (2, 4, 6, 8).
- The largest odd element (1) is smaller than the smallest even element (2).
- This input should expose the bug in the code's logic for handling the case where the largest odd element is smaller than the smallest even element.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Parity and Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an array $a$ of $n$ positive integers.

In one operation, you can pick any pair of indexes $(i, j)$ such that $a_i$ and $a_j$ have distinct parity, then replace the smaller one with the sum of them. More formally:

  * If $a_i < a_j$, replace $a_i$ with $a_i + a_j$;    * Otherwise, replace $a_j$ with $a_i + a_j$.

Find the minimum number of operations needed to make all elements of the array have the same parity.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$).

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

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

Output: For each test case, output a single integer — the minimum number of operations required.

Examples:
input:
7
5
1 3 5 7 9
4
4 4 4 4
3
2 3 4
4
3 2 2 8
6
4 3 6 1 2 1
6
3 6 1 2 1 2
5
999999996 999999997 999999998 999999999 1000000000
output:
0
0
2
4
3
3
3

Note: In the first test case, all integers already have the same parity. Therefore, no operation is needed.

In the third test case, we can perform two operations $(1, 2)$ and $(1, 3)$. The array $a$ transforms as follows: $a = [\color{red}2, \color{red}3, 4] \longrightarrow [\color{red}5, 3, \color{red}4] \longrightarrow [5, 3, 9]$.

In the fourth test case, an example of an optimal sequence of operations is $(1, 2)$, $(1, 3)$, $(1, 4)$, and $(1, 4)$. The array $a$ transforms as follows: $a = [\color{red}3, \color{red}2, 2, 8] \longrightarrow [\color{red}3, 5, \color{red}2, 8] \longrightarrow [\color{red}3, 5, 5, \color{red}8] \longrightarrow [\color{red}{11}, 5, 5, \color{red}8] \longrightarrow [11, 5, 5, 19]$.

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

#define int long long int
#define ll long long
#define MOD 1000000007
#define F first
#define S second
#define fast_io ios::sync_with_stdio(0);cin.tie(0);
int gcd(int a,int b){if(b==0)return a;else return gcd(b,a%b);}
int lcm(int a,int b){return a*b/gcd(a,b);}

void solve() {
    int even = 0;
    int odd = 0;
    int maxEven = 0;
    int n;
    cin>>n;
    vector<int> v;
    for(int i = 0; i < n; i++){
        int ele;
        cin>>ele;
        v.emplace_back(ele);
        if((ele % 2) == 0) even++;
        else odd++;
        maxEven = max(maxEven, ele);
    }

    if(odd == 0 || even == 0){
        cout<<0<<endl;
        return;
    }
    int ans = 0;
    ans += even;
    int a = maxEven;
    for(auto & e : v){
        if(e == a) continue;
        maxEven -= e;
        if(maxEven <= 0) break;
    }
    if(maxEven > 0) ans++;
    cout<<ans<<endl;

}

int32_t main()
{
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif
    fast_io;
    int t = 1;
    cin>>t;
    while(t--) solve();
    return 0;
}

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

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

Think step by step.

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

### Explanation:
1. **Test Case 1:**
   - **Input:** `1` (single element array)
   - **Output:** `0` (since there is only one element, all elements have the same parity)
   - **Purpose:** This tests the base case where the array has only one element.

2. **Test Case 2:**
   - **Input:** `2 1 2`
   - **Output:** `1` (one operation is needed to make all elements have the same parity)
   - **Purpose:** This tests the simplest case with two elements of different parities.

3. **Test Case 3:**
   - **Input:** `4 1 2 3 4`
   - **Output:** `2` (two operations are needed to make all elements have the same parity)
   - **Purpose:** This tests a case where the array has both even and odd elements, and multiple operations are needed to make all elements have the same parity.

These test cases are designed to cover basic scenarios and edge cases that can expose potential bugs in the code, such as handling single-element arrays and arrays with mixed parities.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Parity and Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an array $a$ of $n$ positive integers.

In one operation, you can pick any pair of indexes $(i, j)$ such that $a_i$ and $a_j$ have distinct parity, then replace the smaller one with the sum of them. More formally:

  * If $a_i < a_j$, replace $a_i$ with $a_i + a_j$;    * Otherwise, replace $a_j$ with $a_i + a_j$.

Find the minimum number of operations needed to make all elements of the array have the same parity.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$).

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

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

Output: For each test case, output a single integer — the minimum number of operations required.

Examples:
input:
7
5
1 3 5 7 9
4
4 4 4 4
3
2 3 4
4
3 2 2 8
6
4 3 6 1 2 1
6
3 6 1 2 1 2
5
999999996 999999997 999999998 999999999 1000000000
output:
0
0
2
4
3
3
3

Note: In the first test case, all integers already have the same parity. Therefore, no operation is needed.

In the third test case, we can perform two operations $(1, 2)$ and $(1, 3)$. The array $a$ transforms as follows: $a = [\color{red}2, \color{red}3, 4] \longrightarrow [\color{red}5, 3, \color{red}4] \longrightarrow [5, 3, 9]$.

In the fourth test case, an example of an optimal sequence of operations is $(1, 2)$, $(1, 3)$, $(1, 4)$, and $(1, 4)$. The array $a$ transforms as follows: $a = [\color{red}3, \color{red}2, 2, 8] \longrightarrow [\color{red}3, 5, \color{red}2, 8] \longrightarrow [\color{red}3, 5, 5, \color{red}8] \longrightarrow [\color{red}{11}, 5, 5, \color{red}8] \longrightarrow [11, 5, 5, 19]$.

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

using namespace std;

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

    vector<int> numbers;
    int ops = 0;
    long long max_odd = -2147483647;

    for (int i = 0; i < n; i++) {
        int in = 0;
        cin >> in;
        if (in%2 == 0) {
            // cout << "We are inputting: " << in << endl;
            numbers.push_back(in);
        }
        if (in > max_odd && in%2 != 0) {
            max_odd = in;
        }
    }

    if (numbers.size() == n) {
        cout << 0 << endl;
        return;
    }

    int i = 0;

    // cout << "This is the size of numbers:" << endl;
    // cout << numbers.size() << endl;

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

    while(i<numbers.size()) {
        if (numbers[i] < max_odd) {
            // cout << "In if" << endl;
            // cout << numbers[i] << " " << max_odd << endl;
            max_odd = max_odd + numbers[i];
            ops++;
            i++;
            // cout << "Max odd is now: " << max_odd << endl;
        }
        else {
            // cout << "In else" << endl;
            // cout << numbers[i] << " " << max_odd << endl;
            ops++;
            max_odd = numbers[i] + max_odd;
            // i++;
        }
    }

    cout << ops << endl;

}

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

    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
3
2 3 4
4
3 2 2 8
5
999999996 999999997 999999998 999999999 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Parity and Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an array $a$ of $n$ positive integers.

In one operation, you can pick any pair of indexes $(i, j)$ such that $a_i$ and $a_j$ have distinct parity, then replace the smaller one with the sum of them. More formally:

  * If $a_i < a_j$, replace $a_i$ with $a_i + a_j$;    * Otherwise, replace $a_j$ with $a_i + a_j$.

Find the minimum number of operations needed to make all elements of the array have the same parity.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$).

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

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

Output: For each test case, output a single integer — the minimum number of operations required.

Examples:
input:
7
5
1 3 5 7 9
4
4 4 4 4
3
2 3 4
4
3 2 2 8
6
4 3 6 1 2 1
6
3 6 1 2 1 2
5
999999996 999999997 999999998 999999999 1000000000
output:
0
0
2
4
3
3
3

Note: In the first test case, all integers already have the same parity. Therefore, no operation is needed.

In the third test case, we can perform two operations $(1, 2)$ and $(1, 3)$. The array $a$ transforms as follows: $a = [\color{red}2, \color{red}3, 4] \longrightarrow [\color{red}5, 3, \color{red}4] \longrightarrow [5, 3, 9]$.

In the fourth test case, an example of an optimal sequence of operations is $(1, 2)$, $(1, 3)$, $(1, 4)$, and $(1, 4)$. The array $a$ transforms as follows: $a = [\color{red}3, \color{red}2, 2, 8] \longrightarrow [\color{red}3, 5, \color{red}2, 8] \longrightarrow [\color{red}3, 5, 5, \color{red}8] \longrightarrow [\color{red}{11}, 5, 5, \color{red}8] \longrightarrow [11, 5, 5, 19]$.

Buggy Code:
for _ in range(int(input())):
    n = int(input())
    l0 = list(map(int, input().split()))
    l1 = [i for i in l0 if i % 2 == 0]
    l1.sort()
    if len(l1) == len(l0) or len(l1) == 0:
        print(0)
    else:
        ans = 0
        m = max([i for i in l0 if i % 2 == 1])
        for i in l1:
            if i > m:
                ans += 2
            else:
                ans += 1
            m += i
        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
1
5
1 2 3 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Parity and Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an array $a$ of $n$ positive integers.

In one operation, you can pick any pair of indexes $(i, j)$ such that $a_i$ and $a_j$ have distinct parity, then replace the smaller one with the sum of them. More formally:

  * If $a_i < a_j$, replace $a_i$ with $a_i + a_j$;    * Otherwise, replace $a_j$ with $a_i + a_j$.

Find the minimum number of operations needed to make all elements of the array have the same parity.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$).

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

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

Output: For each test case, output a single integer — the minimum number of operations required.

Examples:
input:
7
5
1 3 5 7 9
4
4 4 4 4
3
2 3 4
4
3 2 2 8
6
4 3 6 1 2 1
6
3 6 1 2 1 2
5
999999996 999999997 999999998 999999999 1000000000
output:
0
0
2
4
3
3
3

Note: In the first test case, all integers already have the same parity. Therefore, no operation is needed.

In the third test case, we can perform two operations $(1, 2)$ and $(1, 3)$. The array $a$ transforms as follows: $a = [\color{red}2, \color{red}3, 4] \longrightarrow [\color{red}5, 3, \color{red}4] \longrightarrow [5, 3, 9]$.

In the fourth test case, an example of an optimal sequence of operations is $(1, 2)$, $(1, 3)$, $(1, 4)$, and $(1, 4)$. The array $a$ transforms as follows: $a = [\color{red}3, \color{red}2, 2, 8] \longrightarrow [\color{red}3, 5, \color{red}2, 8] \longrightarrow [\color{red}3, 5, 5, \color{red}8] \longrightarrow [\color{red}{11}, 5, 5, \color{red}8] \longrightarrow [11, 5, 5, 19]$.

Buggy Code:
tc = int(input())
from heapq import heappush, heappop, heapify

for _ in range(tc):
    n = int(input())
    a = list(map(int, input().split()))
    all_odd = all([i % 2 == 1 for i in a])
    all_even = all([i % 2 == 0 for i in a])
    if all_odd or all_even:
        print(0)
    else:
        evens = [i for i in a if i % 2 == 0]
        odds = [-i for i in a if i % 2 == 1]
        heapify(odds)
        evens.sort()
        a = 0
        for x in evens:
            odd = -heappop(odds)
            while odd < x:
                temp = odd + x
                heappush(odds, -temp)
                odd = -heappop(odds)
                a += 1
            heappush(odds, -(odd+x))
            a += 1
        print(a)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Parity and Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an array $a$ of $n$ positive integers.

In one operation, you can pick any pair of indexes $(i, j)$ such that $a_i$ and $a_j$ have distinct parity, then replace the smaller one with the sum of them. More formally:

  * If $a_i < a_j$, replace $a_i$ with $a_i + a_j$;    * Otherwise, replace $a_j$ with $a_i + a_j$.

Find the minimum number of operations needed to make all elements of the array have the same parity.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$).

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

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

Output: For each test case, output a single integer — the minimum number of operations required.

Examples:
input:
7
5
1 3 5 7 9
4
4 4 4 4
3
2 3 4
4
3 2 2 8
6
4 3 6 1 2 1
6
3 6 1 2 1 2
5
999999996 999999997 999999998 999999999 1000000000
output:
0
0
2
4
3
3
3

Note: In the first test case, all integers already have the same parity. Therefore, no operation is needed.

In the third test case, we can perform two operations $(1, 2)$ and $(1, 3)$. The array $a$ transforms as follows: $a = [\color{red}2, \color{red}3, 4] \longrightarrow [\color{red}5, 3, \color{red}4] \longrightarrow [5, 3, 9]$.

In the fourth test case, an example of an optimal sequence of operations is $(1, 2)$, $(1, 3)$, $(1, 4)$, and $(1, 4)$. The array $a$ transforms as follows: $a = [\color{red}3, \color{red}2, 2, 8] \longrightarrow [\color{red}3, 5, \color{red}2, 8] \longrightarrow [\color{red}3, 5, 5, \color{red}8] \longrightarrow [\color{red}{11}, 5, 5, \color{red}8] \longrightarrow [11, 5, 5, 19]$.

Buggy Code:
n = int(input())
for _ in range(n):
    t = int(input())
    l = list(map(int,input().split()))
    l.sort()
    odd = []
    even = []
    for i in l:
        if i%2 == 0:
            even.append(i)
        else:
            odd.append(i)
    if len(odd) == 0 or len(even) == 0:
        print(0)
    else:
        count = 0
        for i in range(len(even)):
            if even[i] < odd[-1]:
                odd.append(even[i]+odd[-1])
                count += 1
            elif even[i] > odd[-1]:
                odd[-1] = odd[-1] + 2*even[i]
                count += 2
        print(count)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Parity and Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an array $a$ of $n$ positive integers.

In one operation, you can pick any pair of indexes $(i, j)$ such that $a_i$ and $a_j$ have distinct parity, then replace the smaller one with the sum of them. More formally:

  * If $a_i < a_j$, replace $a_i$ with $a_i + a_j$;    * Otherwise, replace $a_j$ with $a_i + a_j$.

Find the minimum number of operations needed to make all elements of the array have the same parity.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$).

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

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

Output: For each test case, output a single integer — the minimum number of operations required.

Examples:
input:
7
5
1 3 5 7 9
4
4 4 4 4
3
2 3 4
4
3 2 2 8
6
4 3 6 1 2 1
6
3 6 1 2 1 2
5
999999996 999999997 999999998 999999999 1000000000
output:
0
0
2
4
3
3
3

Note: In the first test case, all integers already have the same parity. Therefore, no operation is needed.

In the third test case, we can perform two operations $(1, 2)$ and $(1, 3)$. The array $a$ transforms as follows: $a = [\color{red}2, \color{red}3, 4] \longrightarrow [\color{red}5, 3, \color{red}4] \longrightarrow [5, 3, 9]$.

In the fourth test case, an example of an optimal sequence of operations is $(1, 2)$, $(1, 3)$, $(1, 4)$, and $(1, 4)$. The array $a$ transforms as follows: $a = [\color{red}3, \color{red}2, 2, 8] \longrightarrow [\color{red}3, 5, \color{red}2, 8] \longrightarrow [\color{red}3, 5, 5, \color{red}8] \longrightarrow [\color{red}{11}, 5, 5, \color{red}8] \longrightarrow [11, 5, 5, 19]$.

Buggy Code:
import java.util.*;

public class Solution{
    static Scanner sc = new Scanner(System.in);

    static void helper(){
        int n = sc.nextInt();
        long[] arr = new long[n];
        Queue<Long> even = new PriorityQueue<>();
        Queue<Long> odd = new PriorityQueue<>(Collections.reverseOrder());
        for(int i=0; i<n; i++){
            arr[i] = sc.nextInt();
            if(arr[i] %2 == 0) even.offer(arr[i]);
            else{
                odd.offer(arr[i]);
            }

        }

        if(even.size() == n || odd.size() == n){
            System.out.println(0);
            return;
        }

        long chances = 0;

        while(!even.isEmpty()){
            long e = even.poll();
            long o = odd.peek();
            if(e < o){
                chances++;
                odd.offer(o+e);
            }else{
                chances++;
                odd.offer(o+e);
                even.offer(e);
            }
        }

        System.out.println(chances);

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

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Parity and Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an array $a$ of $n$ positive integers.

In one operation, you can pick any pair of indexes $(i, j)$ such that $a_i$ and $a_j$ have distinct parity, then replace the smaller one with the sum of them. More formally:

  * If $a_i < a_j$, replace $a_i$ with $a_i + a_j$;    * Otherwise, replace $a_j$ with $a_i + a_j$.

Find the minimum number of operations needed to make all elements of the array have the same parity.

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

The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$).

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

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

Output: For each test case, output a single integer — the minimum number of operations required.

Examples:
input:
7
5
1 3 5 7 9
4
4 4 4 4
3
2 3 4
4
3 2 2 8
6
4 3 6 1 2 1
6
3 6 1 2 1 2
5
999999996 999999997 999999998 999999999 1000000000
output:
0
0
2
4
3
3
3

Note: In the first test case, all integers already have the same parity. Therefore, no operation is needed.

In the third test case, we can perform two operations $(1, 2)$ and $(1, 3)$. The array $a$ transforms as follows: $a = [\color{red}2, \color{red}3, 4] \longrightarrow [\color{red}5, 3, \color{red}4] \longrightarrow [5, 3, 9]$.

In the fourth test case, an example of an optimal sequence of operations is $(1, 2)$, $(1, 3)$, $(1, 4)$, and $(1, 4)$. The array $a$ transforms as follows: $a = [\color{red}3, \color{red}2, 2, 8] \longrightarrow [\color{red}3, 5, \color{red}2, 8] \longrightarrow [\color{red}3, 5, 5, \color{red}8] \longrightarrow [\color{red}{11}, 5, 5, \color{red}8] \longrightarrow [11, 5, 5, 19]$.

Buggy Code:
import java.util.*;

public class ProblemB {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-- > 0)
        {
            int n = sc.nextInt();
            int[] arr = new int[n];
            int index = 0;
            int maxOdd = 0;
            int e = 0;
            int o = 0;
            for(int i = 0; i<n; i++)
            {
                arr[i] = sc.nextInt();
                if(arr[i] % 2 != 0)
                {
                    o++;
                    maxOdd = Math.max(arr[i], maxOdd);
                    if(arr[i] > maxOdd)
                    {
                        index = i;
                    }
                }
                else e++;
            }
            if(o==0 || e==0){
                System.out.println(0);
                continue;
            }
            Arrays.sort(arr);
            int c = 0;
            for(int i = 0; i<n; i++)
            {
                if(arr[i] % 2 == 0)
                {
                    if(arr[i] < maxOdd)
                    {
                        arr[i] += maxOdd;
                        maxOdd = arr[i];
                        index = i;
                        c++;
                    }
                    else
                    {
                        arr[index] += arr[i];
                        maxOdd = arr[index];
                        c++;
                        if(arr[i] < maxOdd)
                        {
                            arr[i] += maxOdd;
                            maxOdd = arr[i];
                            index = i;
                            c++;
                        }
                    }
                }
            }
            System.out.println(c);
        }

    }

}

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

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

Think step by step.

Response:
```plaintext
1
5
2 3 4 5 6
```