**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimize Equal Sum Subarrays
time_limit_ms: 1500
memory_limit_mb: 256
Description: You are given a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$.

Find a permutation $$$q$$$ of length $$$n$$$ that minimizes the number of pairs ($$$i, j$$$) ($$$1 \leq i \leq j \leq n$$$) such that $$$p_i + p_{i+1} + \ldots + p_j = q_i + q_{i+1} + \ldots + q_j$$$.

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

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

The following line contains $$$n$$$ space-separated integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \leq p_i \leq n$$$) — denoting the permutation $$$p$$$ of length $$$n$$$.

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 one line containing any permutation of length $$$n$$$ (the permutation $$$q$$$) such that $$$q$$$ minimizes the number of pairs.

Examples:
input:
3
2
1 2
5
1 2 3 4 5
7
4 7 5 1 2 6 3
output:
2 1
3 5 4 2 1
6 2 1 4 7 3 5

Note: For the first test, there exists only one pair ($$$i, j$$$) ($$$1 \leq i \leq j \leq n$$$) such that $$$p_i + p_{i+1} + \ldots + p_j = q_i + q_{i+1} + \ldots + q_j$$$, which is ($$$1, 2$$$). It can be proven that no such $$$q$$$ exists for which there are no pairs.

Buggy Code:
from math import *
def gcd(a, b):
    while b != 0:
        r = a % b
        a = b
        b = r
    return a

def nok(a, b):
    return a * b // gcd(a, b)

def bin_poisk(a, n, t):
    l = -1
    r = n
    while r - l > 1:
        m = (r + l) // 2
        if a[m] < t:
            l = m
        else:
            r = m
    if r < n and a[r] == t:
        return r
    elif l > -1 and a[l] == t:
        return l
    else:
        return -1

def lower_bound(a, n, t):
    l = -1
    r = n
    while r - l > 1:
        m = (r + l) // 2
        if a[m] < t:
            l = m
        else:
            r = m
    return r

def upper_bound(a, n, t):
    l = -1
    r = n
    while r - l > 1:
        m = (r + l) // 2
        if a[m] <= t:
            l = m
        else:
            r = m
    return l

def power(value, pow):
    result = 1
    while pow > 0:
        if pow % 2 == 1:
            result *= value
        value *= value
        pow //= 2
    return result

def prost(n):
    for i in range(2, n):
        if n % i == 0:
            return False
    return True

def inp():
    return map(int, input().split())

def inp_lst():
    return list(inp())

def inp_2lst(index):
    lst = []
    for i in range(index):
        lst.append(inp_lst())
    return lst

for i in range(int(input())):
    n = int(input())
    a = inp_lst()
    for j in range(1 + n % 2):
        for k in range(n - 1):
            a[k], a[k + 1] = a[k + 1], a[k]
    print(*a)

#1 2 3 4 5
#2 3 4 5 1
#3 4 5 1 2
#4 5 1 2 3
#5 1 2 3 4
#1 2 3 4 5

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

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

Think step by step.

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

1. **Edge Cases**: Small values of `n` and specific permutations that might cause the code to fail.
2. **Symmetry and Cycles**: Permutations that create symmetrical or cyclic patterns which might be incorrectly handled by the code.
3. **Large Values**: Although the output length limit restricts us from generating large inputs, we can still use the maximum allowed value for `n` in a small-scale test case.

Let's consider a permutation of length `n = 5` that is likely to cause issues. The code provided swaps elements in a specific pattern, which might not always minimize the number of equal sum subarrays.

### Test Input
```plaintext
3
5
1 2 3 4 5
5
5 4 3 2 1
5
2 1 4 3 5
```

### Explanation
1. **First Test Case**: `1 2 3 4 5`
   - This is a simple permutation where the elements are in increasing order. The code might not handle this correctly, as it only swaps adjacent elements, which might not minimize the number of equal sum subarrays.

2. **Second Test Case**: `5 4 3 2 1`
   - This is a permutation where the elements are in decreasing order. The code might also fail to handle this correctly due to the same reason as the first test case.

3. **Third Test Case**: `2 1 4 3 5`
   - This permutation has a mixed order and might expose issues in the code's logic for handling non-sequential patterns.

These test cases are designed to challenge the code by providing permutations that are likely to result in a higher number of equal sum subarrays if the code does not correctly minimize them.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimize Equal Sum Subarrays
time_limit_ms: 1500
memory_limit_mb: 256
Description: You are given a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$.

Find a permutation $$$q$$$ of length $$$n$$$ that minimizes the number of pairs ($$$i, j$$$) ($$$1 \leq i \leq j \leq n$$$) such that $$$p_i + p_{i+1} + \ldots + p_j = q_i + q_{i+1} + \ldots + q_j$$$.

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

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

The following line contains $$$n$$$ space-separated integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \leq p_i \leq n$$$) — denoting the permutation $$$p$$$ of length $$$n$$$.

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 one line containing any permutation of length $$$n$$$ (the permutation $$$q$$$) such that $$$q$$$ minimizes the number of pairs.

Examples:
input:
3
2
1 2
5
1 2 3 4 5
7
4 7 5 1 2 6 3
output:
2 1
3 5 4 2 1
6 2 1 4 7 3 5

Note: For the first test, there exists only one pair ($$$i, j$$$) ($$$1 \leq i \leq j \leq n$$$) such that $$$p_i + p_{i+1} + \ldots + p_j = q_i + q_{i+1} + \ldots + q_j$$$, which is ($$$1, 2$$$). It can be proven that no such $$$q$$$ exists for which there are no pairs.

Buggy Code:
from typing import List,Optional
from collections import deque
import sys,math
class Solution:
    def run(self):
        n = int(input().strip())
        p = list(map(int,input().strip().split()))

        if(n%2==1):
            ele = p.pop(n//2)
            print(ele,end =" ")

        for i in range(len(p)-1,-1,-1):
            print(p[i],end=" ")
        print()

# time complexity: O()
# space complexity: O()
if __name__ == "__main__":
    for _ in range(int(input().strip())):
        Solution().run()

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

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

Think step by step.

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

### Explanation:
1. **Input Structure**:
   - The first line contains the number of test cases, which is `1`.
   - The second line contains the length of the permutation, which is `3`.
   - The third line contains the permutation `1 2 3`.

2. **Expected Output**:
   - For the permutation `1 2 3`, a valid output that minimizes the number of pairs ($$i, j$$) such that the sum of elements in the subarray of `p` equals the sum of elements in the subarray of `q` should be a permutation that avoids creating such pairs as much as possible.

3. **Challenging the Buggy Code**:
   - The buggy code has a specific behavior when the length of the permutation `n` is odd. It pops the middle element and prints it first, then prints the remaining elements in reverse order.
   - For the input `1 2 3`, the middle element is `2`. The code will output `2 3 1`.
   - However, the correct output should be a permutation that minimizes the number of pairs. The output `2 3 1` creates a pair (1, 1) where the sum is `2`, and the same sum is also `2` in the original permutation `1 2 3`. This is not optimal.

4. **Why This Input is Challenging**:
   - This input tests the behavior of the buggy code when the permutation length is odd and the middle element is not the maximum or minimum value.
   - It forces the code to print the middle element first, which can create unnecessary pairs if not handled correctly.

By using this input, we can effectively expose the bug in the buggy code and highlight the need for a more sophisticated approach to minimize the number of equal sum subarrays.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimize Equal Sum Subarrays
time_limit_ms: 1500
memory_limit_mb: 256
Description: You are given a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$.

Find a permutation $$$q$$$ of length $$$n$$$ that minimizes the number of pairs ($$$i, j$$$) ($$$1 \leq i \leq j \leq n$$$) such that $$$p_i + p_{i+1} + \ldots + p_j = q_i + q_{i+1} + \ldots + q_j$$$.

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

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

The following line contains $$$n$$$ space-separated integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \leq p_i \leq n$$$) — denoting the permutation $$$p$$$ of length $$$n$$$.

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 one line containing any permutation of length $$$n$$$ (the permutation $$$q$$$) such that $$$q$$$ minimizes the number of pairs.

Examples:
input:
3
2
1 2
5
1 2 3 4 5
7
4 7 5 1 2 6 3
output:
2 1
3 5 4 2 1
6 2 1 4 7 3 5

Note: For the first test, there exists only one pair ($$$i, j$$$) ($$$1 \leq i \leq j \leq n$$$) such that $$$p_i + p_{i+1} + \ldots + p_j = q_i + q_{i+1} + \ldots + q_j$$$, which is ($$$1, 2$$$). It can be proven that no such $$$q$$$ exists for which there are no pairs.

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

public class Main {
    //    static long res;
//    private final static int m = (int)1e6;
//    private final static int[]prime = new int[78498];
//    private final static boolean[]isPrime = new boolean[m + 1];
//
//    static {
//        int cnt = 0;
//        for(int i = 2; i <= m; ++i) {
//            if(isPrime[i]) {
//                prime[cnt++] = i;
//                for(int j = i; j <= m / i; ++j) {
//                    isPrime[i * j] = true;
//                }
//            }
//        }
//    }
    static PrintWriter out;
    static FastRead read;
    static long resA, resB;

    static void solve() {

        int n = read.nextInt();
        int []arr = new int[n];
        for(int i = 0; i < n; ++i) arr[i] = read.nextInt();
        int []res = new int[n];
        int j = 0;
        if(n % 2 == 0 ) {
            int i = n - 1;
            while(i >= 0) {
                res[j] = arr[i];
                --i;
                ++j;
            }
        }else{
            res[j] = arr[n / 2];
            ++j;
            int i = n - 1;
            while(j < n) {
                if(i == n / 2) {
                    --i;
                    continue;
                }
                res[j] = arr[i];
                ++j;
                --i;
            }
        }
        for(int i = 0; i < n; ++i) {
            out.print(res[i] + " ");
        }
        out.println();
        out.flush();

    }

    public static void main(String[] args) {
        out = new PrintWriter(new OutputStreamWriter(System.out));
        read =  new FastRead();
        int t = 1;
        t = read.nextInt();
        while(t > 0) {
            --t;
//           Random random = new Random();
            solve();
        }
    }
    static long fastPow(long x, long n) {
        long res = 1;

        while(n > 0) {
            if((n & 1) == 1) {
                res = res * x;
            }
            x = x * x;
            n >>>= 1;
        }
        return res;
    }
    static String trans(String str) {
        str = str.replaceAll("\\[", "{");
        return str.replaceAll("]", "}");
    }
}

class FastRead {
    BufferedReader br;
    StringTokenizer st;
    public FastRead() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }

    String next(){
        try{
            while(st == null || !st.hasMoreElements()) {
                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());
    }
    String nextLine() {
        String res = "";
        try{
            if(st != null && st.hasMoreElements()) res = st.nextToken("\n");
            else res = br.readLine();
        }catch (IOException e) {
            e.printStackTrace();
        }
        return res;
    }
}
class Solution{
    public long minimumOperations(int[] nums, int[] target, int []tmp) {
        int n = nums.length;
        long res = 0;
        int []diff = new int[n];
//        for(int i = 0; i < n; ++i) {
//            diff[i] = nums[i] - target[i];
//            res += Math.abs(diff[i]);
//        }
        diff = tmp.clone();
        n = diff.length;
        int x = diff[0];
        long l = 0, r = 0;
        for(int i = 1; i < n; ++i) {
            int y = diff[i];
            if(y > x) l += y - x;
            else r += x - y;
            x = y;
        }
        return Math.abs(l - r) + Math.max(l, r);
    }
    public int[] maxNumber(int[] nums1, int[] nums2, int k) {
        int n = nums1.length, m = nums2.length;
        int []res = new int[k];
        for(int i = 0; i <= k && i <= n; ++i) {
            int j = k - i;
            if(j > m) continue;
            Deque<Integer> x = new ArrayDeque<>();
            Deque<Integer> y = new ArrayDeque<>();
            int []ans = new int[k];
            op(nums1, x, i, n);
            op(nums2, y, j, m);
            int now = 0;
            while(!x.isEmpty() && !y.isEmpty()) {
                if(x.peekLast() > y.peekLast()) {
                    ans[now] = x.pollLast();
                }else if(Objects.equals(x.peekLast(), y.peekLast())) {
                    if(check(x, y)) {
                        ans[now] = x.pollLast();
                    }else{
                        ans[now] = y.pollLast();
                    }
                }else{
                    ans[now] = y.pollLast();
                }
                ++now;
            }
            while(!x.isEmpty()) {
                ans[now] = x.pollLast();
                ++now;
            }
            while(!y.isEmpty()) {
                ans[now] = y.pollLast();
                ++now;
            }
            if(check(ans, res, k)) {
                res = ans.clone();
            }
        }
        return res;
    }
    private boolean check(Deque<Integer> x, Deque<Integer> y) {
        int num = x.peekLast();
        int cnt1 = 0, cnt2 = 0;
        for(int xx : x) {
            if(xx != num) {
                cnt1 = xx;
                break;
            }
        }
        for(int xx : y) {
            if(xx != num) {
                cnt2 = xx;
                break;
            }
        }
        System.out.println(cnt1 + " " + cnt2);
        return cnt1 > cnt2;
    }
    private boolean check(int []x, int []y, int k) {
        for (int i = 0; i < k; ++i) {
            if (x[i] > y[i]) return true;
            else if(x[i] < y[i]) return false;
        }
        return false;
    }
//    private void op(int []nums1, Deque<Integer> xx, int i, int n) {
//        for(int t = 0; t < n && i != 0; ++t) {
//            if(n - t - 1 < i - xx.size()) {
//                xx.offerFirst(nums1[t]);
//            }else if(xx.isEmpty()) {
//                xx.offerFirst(nums1[t]);
//            }else if(xx.peekFirst() <= nums1[t]) {
//                xx.pollFirst();
//                xx.offerFirst(nums1[t]);
//            }
//        }
//    }

    private void op(int []nums1, Deque<Integer> xx, int i, int n) {
        for(int t = 0; t < n && i != 0; ++t) {
            while(xx.size() >= i && xx.peekLast() < nums1[t] && n - t - 1 >= i - xx.size()) {
                xx.pollLast();
            }
            xx.offer(nums1[t]);
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimize Equal Sum Subarrays
time_limit_ms: 1500
memory_limit_mb: 256
Description: You are given a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$.

Find a permutation $$$q$$$ of length $$$n$$$ that minimizes the number of pairs ($$$i, j$$$) ($$$1 \leq i \leq j \leq n$$$) such that $$$p_i + p_{i+1} + \ldots + p_j = q_i + q_{i+1} + \ldots + q_j$$$.

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

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

The following line contains $$$n$$$ space-separated integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \leq p_i \leq n$$$) — denoting the permutation $$$p$$$ of length $$$n$$$.

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 one line containing any permutation of length $$$n$$$ (the permutation $$$q$$$) such that $$$q$$$ minimizes the number of pairs.

Examples:
input:
3
2
1 2
5
1 2 3 4 5
7
4 7 5 1 2 6 3
output:
2 1
3 5 4 2 1
6 2 1 4 7 3 5

Note: For the first test, there exists only one pair ($$$i, j$$$) ($$$1 \leq i \leq j \leq n$$$) such that $$$p_i + p_{i+1} + \ldots + p_j = q_i + q_{i+1} + \ldots + q_j$$$, which is ($$$1, 2$$$). It can be proven that no such $$$q$$$ exists for which there are no pairs.

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

public class Solution2
{
    public static void swap(int[] a, int i, int j) {
        int temp = a[i];
        a[i] = a[j];
        a[j] = temp;
    }
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(br.readLine().trim());
        while(t-- > 0) {
            int n = Integer.parseInt(br.readLine().trim());
            String[] pStr = br.readLine().trim().split("\\s+");
            // store reverse sorted in q
            int[] p = new int[pStr.length];
            int[] q = new int[pStr.length];
            for(int i = 0, j = p.length-1; i < pStr.length; i++, j--) {
                p[i] = Integer.parseInt(pStr[i]);
                q[j] = p[i];
            }
            // now swap mid and start elements in q to avoid middle subarray always having common
            if(q.length > 2) {
                int mid = q.length/2, start = 0;
                swap(q, start, mid);
                start++;

                // handle after-swap issues
                while(mid >= 0 && (q[mid] + q[mid+1] == p[mid] + p[mid+1])) {
                    swap(q, start, mid);
                    mid -= 2;
                    start++;
                }
            }
            for(int i = 0; i < q.length; i++) {
                System.out.print(q[i] + " ");
            }
            System.out.println();
        }
    }
}

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

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

Think step by step.

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

1. **Edge Cases**: Test the smallest and largest possible values for `n` and `t`.
2. **Specific Patterns**: Create permutations that might cause the algorithm to fail, especially in the swapping logic.
3. **Symmetry and Reversal**: Test with permutations that have symmetric or reversed properties to see if the algorithm handles them correctly.

Given the constraints, let's focus on a small-scale test input that can effectively expose the bug. The buggy code primarily fails in the way it handles the swapping of elements to avoid equal subarray sums. Let's create a test case where the permutation is such that the middle and start elements are the same or have a specific pattern that the algorithm might not handle correctly.

Here's a small-scale test input:

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

### Explanation:
1. **Test Case 1**: `n = 3`
   - Permutation: `1 2 3`
   - The algorithm should handle this small permutation correctly, but we can see if it fails to avoid equal subarray sums.

2. **Test Case 2**: `n = 4`
   - Permutation: `4 3 2 1`
   - This is a reverse permutation. The algorithm might fail to correctly swap elements to avoid equal subarray sums.

3. **Test Case 3**: `n = 5`
   - Permutation: `1 2 3 4 5`
   - This is a straightforward increasing permutation. The algorithm might fail to handle the middle and start elements correctly.

This test input should help identify if the buggy code fails to correctly handle the swapping logic and avoid equal subarray sums.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimize Equal Sum Subarrays
time_limit_ms: 1500
memory_limit_mb: 256
Description: You are given a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$.

Find a permutation $$$q$$$ of length $$$n$$$ that minimizes the number of pairs ($$$i, j$$$) ($$$1 \leq i \leq j \leq n$$$) such that $$$p_i + p_{i+1} + \ldots + p_j = q_i + q_{i+1} + \ldots + q_j$$$.

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

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

The following line contains $$$n$$$ space-separated integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \leq p_i \leq n$$$) — denoting the permutation $$$p$$$ of length $$$n$$$.

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 one line containing any permutation of length $$$n$$$ (the permutation $$$q$$$) such that $$$q$$$ minimizes the number of pairs.

Examples:
input:
3
2
1 2
5
1 2 3 4 5
7
4 7 5 1 2 6 3
output:
2 1
3 5 4 2 1
6 2 1 4 7 3 5

Note: For the first test, there exists only one pair ($$$i, j$$$) ($$$1 \leq i \leq j \leq n$$$) such that $$$p_i + p_{i+1} + \ldots + p_j = q_i + q_{i+1} + \ldots + q_j$$$, which is ($$$1, 2$$$). It can be proven that no such $$$q$$$ exists for which there are no pairs.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
int main(){
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        vector <int> v;
        for(int i=0;i<n;i++){
            int x;
            cin>>x;
            v.push_back(x);
        }

       reverse(v.begin(),v.end());
   if(v.size()>2){ for(int i=v.size()/2;i>0;i--){
        swap(v[i-1],v[i]);
    }
   }
    for(int i=0;i<v.size();i++){
        cout<<v[i]<<" ";
         }
        cout<<endl;

}
}

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

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

Think step by step.

Response:
```plaintext
3
3
1 2 3
4
1 2 3 4
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: Minimize Equal Sum Subarrays
time_limit_ms: 1500
memory_limit_mb: 256
Description: You are given a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$.

Find a permutation $$$q$$$ of length $$$n$$$ that minimizes the number of pairs ($$$i, j$$$) ($$$1 \leq i \leq j \leq n$$$) such that $$$p_i + p_{i+1} + \ldots + p_j = q_i + q_{i+1} + \ldots + q_j$$$.

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

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

The following line contains $$$n$$$ space-separated integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \leq p_i \leq n$$$) — denoting the permutation $$$p$$$ of length $$$n$$$.

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 one line containing any permutation of length $$$n$$$ (the permutation $$$q$$$) such that $$$q$$$ minimizes the number of pairs.

Examples:
input:
3
2
1 2
5
1 2 3 4 5
7
4 7 5 1 2 6 3
output:
2 1
3 5 4 2 1
6 2 1 4 7 3 5

Note: For the first test, there exists only one pair ($$$i, j$$$) ($$$1 \leq i \leq j \leq n$$$) such that $$$p_i + p_{i+1} + \ldots + p_j = q_i + q_{i+1} + \ldots + q_j$$$, which is ($$$1, 2$$$). It can be proven that no such $$$q$$$ exists for which there are no pairs.

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

using namespace std;

template < typename A, typename B > ostream & operator << (ostream & os,
    const pair < A, B > & p) {
    return os << '(' << p.first << ", " << p.second << ')';
}
template < typename T_container, typename T = typename enable_if < !is_same < T_container, string > ::value, typename T_container::value_type > ::type > ostream & operator << (ostream & os,
    const T_container & v) {
    os << '{';
    string sep;
    for (const T & x: v) os << sep << x, sep = ", ";
    return os << '}';
}
void dbg_out() {
    cerr << endl;
}
template < typename Head, typename...Tail > void dbg_out(Head H, Tail...T) {
    cerr << ' ' << H;
    dbg_out(T...);
}
#ifdef LOCAL
#define dbg(...) cerr << "(" << #__VA_ARGS__ << "):", dbg_out(__VA_ARGS__)
#else
#define dbg(...)
#endif

#define ar array
#define int long long
#define ll long long
#define ld long double
#define sza(x)((ll) x.size())
#define all(a)(a).begin(), (a).end()

const ll MAX_N = 1e5 + 5;
const ll MOD = 1e9 + 7;
const ll INF = 1e9;
const ld EPS = 1e-9;

vector < ll > factors(int x) {
    vector < ll > b;
    for (ll i = 2; i <= sqrt(x); i++) {
        if (x % i == 0) {
            b.push_back(i);

            if (i != x / i) {
                b.push_back(x / i);

            }
        }
    }
    sort(b.begin(), b.end());
    return b;
}

bool isprime(int n) {
    for (int i = 2; i <= sqrt(n); i++) {
        if (n % i == 0) return 0;
    }
    return 1;
}

bool oneone(int n) {
    return (n & (n + 1)) == 0;
}
bool prime[1000000];

void seive() {
    fill(prime, prime + 1000001, true);
    prime[0] = prime[1] = 0;
    for (int i = 2; i < 1000001; i++) {
        if (prime[i]) {
            for (int j = i * i; j < 1000001; j += i) {
                prime[j] = 0;
            }
        }
    }

}

void solve() {
    //Partik Malasi
    //Partik Malasi
    ll n,a,b,c;
    cin>>n;
    if(n==1){
      cout<<1<<endl;
      return;
    }
    vector<ll> v(n);
    vector<ll> ans(n);
    for(int i=0;i<n;i++){
        cin>>v[i];
    }
    for(int i=0;i<n;i++){
        ans[i]=(v[i])%n+1;

    }
    for(auto i:ans){

    cout<<i<<" ";
    }
    cout<<endl;

}

signed main() {
    ios_base::sync_with_stdio(0);
    ll temp;
    cin >> temp;

    cin.tie(0);
    cout.tie(0);

    for (ll i = 0; i < temp; i++) {
        solve();
    }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases or scenarios that maximize the failure probability. Let's analyze the code and identify potential issues:

1. **Permutation Handling**: The code generates the permutation `ans` by taking each element of `v` modulo `n` and adding 1. This means that the elements of `ans` will be in the range `[1, n]`.

2. **Edge Cases**:
   - **Single Element**: The code handles the case when `n == 1` correctly.
   - **Small Permutations**: Small permutations can help us quickly verify the correctness of the algorithm.
   - **Large Permutations**: Large permutations can test the performance and correctness of the algorithm under more complex conditions.

3. **Challenging Scenarios**:
   - **Reversed Permutation**: A reversed permutation can be challenging because the sums of subarrays can be very different from the original permutation.
   - **Sorted Permutation**: A sorted permutation can also be challenging because the sums of subarrays are more predictable.

Given the constraints and the need for a small-scale test input, let's focus on a small permutation that can expose the bug.

### Chosen Test Input
We will use a permutation of length 4, which is small enough to be manageable and large enough to show the issue. Let's choose the permutation `[4, 3, 2, 1]`.

### Explanation
- **Permutation**: `[4, 3, 2, 1]`
- **Expected Output**: A permutation that minimizes the number of pairs (`i, j`) such that the sum of subarrays is equal. For this permutation, a possible output could be `[1, 2, 3, 4]`.

### Test Input
```plaintext
1
4
4 3 2 1
```

This test input will help expose the bug in the provided code. The expected output for this test case should be a permutation that minimizes the number of equal sum subarrays. The code's current logic will produce `[2, 1, 4, 3]`, which is not optimal.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimize Equal Sum Subarrays
time_limit_ms: 1500
memory_limit_mb: 256
Description: You are given a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$.

Find a permutation $$$q$$$ of length $$$n$$$ that minimizes the number of pairs ($$$i, j$$$) ($$$1 \leq i \leq j \leq n$$$) such that $$$p_i + p_{i+1} + \ldots + p_j = q_i + q_{i+1} + \ldots + q_j$$$.

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

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

The following line contains $$$n$$$ space-separated integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \leq p_i \leq n$$$) — denoting the permutation $$$p$$$ of length $$$n$$$.

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 one line containing any permutation of length $$$n$$$ (the permutation $$$q$$$) such that $$$q$$$ minimizes the number of pairs.

Examples:
input:
3
2
1 2
5
1 2 3 4 5
7
4 7 5 1 2 6 3
output:
2 1
3 5 4 2 1
6 2 1 4 7 3 5

Note: For the first test, there exists only one pair ($$$i, j$$$) ($$$1 \leq i \leq j \leq n$$$) such that $$$p_i + p_{i+1} + \ldots + p_j = q_i + q_{i+1} + \ldots + q_j$$$, which is ($$$1, 2$$$). It can be proven that no such $$$q$$$ exists for which there are no pairs.

Buggy Code:
def li(): #list of integers
    return list(map(int,input().split()))
def iim(): #consecutive integers
    return map(int,input().split())
def si():  #str list
    return list(input())
def ii():  #integer
    return int(input())

from math import *
from collections import *

def solution():
    n = ii()
    lst = li()
    lst = lst[::-1]
    x = lst.pop((n//2))
    if n & 1:
        lst = [x]+lst
    else:lst.append(x)
    print(*lst)
for _ in range(int(input())):
    solution()

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios where the code's logic is likely to fail. The buggy code reverses the list and then manipulates it by popping an element and inserting it back in a specific position. This can lead to issues, especially when the input permutation has specific patterns or when the length of the permutation is small or has a specific parity.

### Step-by-Step Analysis:
1. **Reversing the List**: The list is reversed, which can change the relative positions of elements significantly.
2. **Popping and Inserting an Element**: The code pops an element from the middle of the list and then either inserts it at the beginning or appends it to the end based on the parity of the length of the list.
3. **Parity of `n`**: The behavior of the code changes based on whether `n` is odd or even.

### Edge Cases to Consider:
1. **Small `n`**: Test with the smallest possible `n` (e.g., `n = 1` and `n = 2`).
2. **Specific Permutations**: Test with permutations that have a specific pattern, such as all elements in ascending or descending order.
3. **Middle Element Manipulation**: Test with permutations where the middle element (or the element that gets popped) is crucial for the sum conditions.

### Chosen Test Input:
Let's choose a small `n` and a permutation that will highlight the issues with the code's logic.

#### Test Input:
```plaintext
3
1
1
2
1 2
4
1 2 3 4
```

### Explanation:
1. **Test Case 1 (`n = 1`)**:
   - Input: `1`
   - Expected Output: `1`
   - This is a trivial case, but it should help verify the basic functionality.

2. **Test Case 2 (`n = 2`)**:
   - Input: `1 2`
   - Expected Output: `2 1`
   - This is a simple case where the code should reverse the list. The expected output is `2 1`, but the buggy code will also output `2 1`, which is correct in this case. However, it's still a good test to ensure the code handles small `n` correctly.

3. **Test Case 3 (`n = 4`)**:
   - Input: `1 2 3 4`
   - Expected Output: `4 3 2 1` or any permutation that minimizes the number of pairs.
   - The buggy code will reverse the list to `4 3 2 1` and then pop the middle element `3` and append it to the end, resulting in `4 2 1 3`. This is not the optimal solution, as the optimal solution would be `4 3 2 1` or another permutation that minimizes the number of pairs.

### Final Output:
```plaintext
3
1
1
2
1 2
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: Minimize Equal Sum Subarrays
time_limit_ms: 1500
memory_limit_mb: 256
Description: You are given a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$.

Find a permutation $$$q$$$ of length $$$n$$$ that minimizes the number of pairs ($$$i, j$$$) ($$$1 \leq i \leq j \leq n$$$) such that $$$p_i + p_{i+1} + \ldots + p_j = q_i + q_{i+1} + \ldots + q_j$$$.

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

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

The following line contains $$$n$$$ space-separated integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \leq p_i \leq n$$$) — denoting the permutation $$$p$$$ of length $$$n$$$.

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 one line containing any permutation of length $$$n$$$ (the permutation $$$q$$$) such that $$$q$$$ minimizes the number of pairs.

Examples:
input:
3
2
1 2
5
1 2 3 4 5
7
4 7 5 1 2 6 3
output:
2 1
3 5 4 2 1
6 2 1 4 7 3 5

Note: For the first test, there exists only one pair ($$$i, j$$$) ($$$1 \leq i \leq j \leq n$$$) such that $$$p_i + p_{i+1} + \ldots + p_j = q_i + q_{i+1} + \ldots + q_j$$$, which is ($$$1, 2$$$). It can be proven that no such $$$q$$$ exists for which there are no pairs.

Buggy Code:
for _ in range(int(input())):
    n = int(input())
    res = list(map(int,input().split()))
    if n & 1:
        ans = [res[n // 2]]
        for i in range(n - 1, -1, -1):
            if i != n // 2: ans.append(res[i])
        print(*ans)
    else:
        ans = [res[n // 2]]
        for i in range(n - 1, -1, -1):
            if i != n // 2: ans.append(res[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
2
3
1 3 2
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: Minimize Equal Sum Subarrays
time_limit_ms: 1500
memory_limit_mb: 256
Description: You are given a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$.

Find a permutation $$$q$$$ of length $$$n$$$ that minimizes the number of pairs ($$$i, j$$$) ($$$1 \leq i \leq j \leq n$$$) such that $$$p_i + p_{i+1} + \ldots + p_j = q_i + q_{i+1} + \ldots + q_j$$$.

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

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

The following line contains $$$n$$$ space-separated integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \leq p_i \leq n$$$) — denoting the permutation $$$p$$$ of length $$$n$$$.

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 one line containing any permutation of length $$$n$$$ (the permutation $$$q$$$) such that $$$q$$$ minimizes the number of pairs.

Examples:
input:
3
2
1 2
5
1 2 3 4 5
7
4 7 5 1 2 6 3
output:
2 1
3 5 4 2 1
6 2 1 4 7 3 5

Note: For the first test, there exists only one pair ($$$i, j$$$) ($$$1 \leq i \leq j \leq n$$$) such that $$$p_i + p_{i+1} + \ldots + p_j = q_i + q_{i+1} + \ldots + q_j$$$, which is ($$$1, 2$$$). It can be proven that no such $$$q$$$ exists for which there are no pairs.

Buggy Code:
import java.util.*;

public class ques2 {
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-- > 0){
            int n = sc.nextInt();
            int a[] = new int[n];
            for(int i = 0; i<n; i++){
                a[i]=sc.nextInt();
            }
            if(n==2){
                System.out.println(a[1]+" "+a[0]);
                continue;
            }
            for(int i = 0; i<=n-3; i++){
                int a1 = a[i];
                int a2 = a[i+1];
                int a3 = a[i+2];
                a[i]=a3;
                a[i+1]=a1;
                a[i+2]=a2;
            }
            for(int i = 0; i<n; i++){
                System.out.print(a[i]+" ");
            }
            System.out.println();
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to consider edge cases and scenarios where the code's logic might fail. The code attempts to minimize the number of equal sum subarrays by rearranging elements in a specific way. However, the logic used in the code is not robust and can fail in certain scenarios.

### Key Points to Consider:
1. **Small Test Cases**: The code has a special case for \( n = 2 \). We should test with \( n = 3 \) and other small values to see if the code handles them correctly.
2. **Special Permutations**: Test with permutations that have specific patterns, such as sequences with repeating sums or sequences that are already sorted.
3. **Edge Cases**: Test with the maximum and minimum values of \( n \) within the constraints.

### Chosen Test Case:
Let's choose \( n = 5 \) and a permutation that is already sorted. This will test the code's ability to handle a sorted sequence, which might not be optimally handled by the current logic.

### Test Input:
```plaintext
1
5
1 2 3 4 5
```

### Explanation:
- **Test Case 1**: \( n = 5 \) and the permutation is \( [1, 2, 3, 4, 5] \).
- **Expected Output**: The code should ideally output a permutation that minimizes the number of equal sum subarrays. For this specific input, a valid output could be \( [3, 5, 4, 2, 1] \), but the code's current logic might not produce an optimal result.

By providing this test input, we can observe if the code's logic fails to produce a permutation that minimizes the number of equal sum subarrays.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimize Equal Sum Subarrays
time_limit_ms: 1500
memory_limit_mb: 256
Description: You are given a permutation$$$^{\text{∗}}$$$ $$$p$$$ of length $$$n$$$.

Find a permutation $$$q$$$ of length $$$n$$$ that minimizes the number of pairs ($$$i, j$$$) ($$$1 \leq i \leq j \leq n$$$) such that $$$p_i + p_{i+1} + \ldots + p_j = q_i + q_{i+1} + \ldots + q_j$$$.

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

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

The following line contains $$$n$$$ space-separated integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \leq p_i \leq n$$$) — denoting the permutation $$$p$$$ of length $$$n$$$.

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 one line containing any permutation of length $$$n$$$ (the permutation $$$q$$$) such that $$$q$$$ minimizes the number of pairs.

Examples:
input:
3
2
1 2
5
1 2 3 4 5
7
4 7 5 1 2 6 3
output:
2 1
3 5 4 2 1
6 2 1 4 7 3 5

Note: For the first test, there exists only one pair ($$$i, j$$$) ($$$1 \leq i \leq j \leq n$$$) such that $$$p_i + p_{i+1} + \ldots + p_j = q_i + q_{i+1} + \ldots + q_j$$$, which is ($$$1, 2$$$). It can be proven that no such $$$q$$$ exists for which there are no pairs.

Buggy Code:
import java.util.*;
import java.lang.*;
import java.io.*;
/**
 * B_Minimize_Equal_Sum_Subarrays
 */
public class B_Minimize_Equal_Sum_Subarrays {

    public static void main(String[] args) throws java.lang.Exception {
        FastWriter out = new FastWriter();
        int t = in.nextInt();

        while (t != 0) {
            t--;
            int n=in.nextInt();
            int a[]=ai(n);
            int ans[]=new int[n];
            for(int i=0;i<n;i++){
                if(a[i]+1 <= n){
                    ans[i]=a[i]+1;
                }else{
                    ans[i]=(a[i]+1)%(n);

                }
            }
            for(int i=0;i<n;i++){
                out.print(ans[i]+" ");
            }
            out.println("");

        }
        out.close();
    }

    // GCD Method
    static long gcd(long a, long b) {
        while (b != 0) {
            long t = b;
            b = a % b;
            a = t;
        }
        return a;
    }

    // LCM Method
    static long lcm(long a, long b) {
        return (a / gcd(a, b)) * b;
    }

    // Power Calculation Method
    static long pow(long base, int exp) {
        long result = 1;
        while (exp > 0) {
            if (exp % 2 == 1) {
                result *= base;
            }
            base *= base;
            exp /= 2;
        }
        return result;
    }

    // Sieve of Eratosthenes Method
    static boolean[] sieve(int n) {
        boolean[] isPrime = new boolean[n + 1];
        Arrays.fill(isPrime, true);
        isPrime[0] = isPrime[1] = false;
        for (int i = 2; i * i <= n; i++) {
            if (isPrime[i]) {
                for (int j = i * i; j <= n; j += i) {
                    isPrime[j] = false;
                }
            }
        }
        return isPrime;
    }

    // Prime Factorization Method using Smallest Prime Factor (SPF)
    public static List<Long> primeFactorization(long number) {
        int max = (int)Math.sqrt(number) + 1;
        long[] spf = new long[max];
        Arrays.fill(spf, -1);
        for (int i = 2; i < max; i++) {
            if (spf[i] == -1) {
                for (int j = i; j < max; j += i) {
                    if (spf[j] == -1) {
                        spf[j] = i;
                    }
                }
            }
        }

        List<Long> primeFactors = new ArrayList<>();
        while (number > 1) {
            if (number < max && spf[(int)number] != -1) {
                primeFactors.add(spf[(int)number]);
                number /= spf[(int)number];
            } else {
                boolean found = false;
                for (int i = 2; i < max; i++) {
                    if (number % i == 0) {
                        primeFactors.add((long)i);
                        number /= i;
                        found = true;
                        break;
                    }
                }
                if (!found) {
                    primeFactors.add(number);
                    break;
                }
            }
        }

        return primeFactors;
    }

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

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

    // get array max
    static int aMax(int[] arr) {
        int a = Integer.MIN_VALUE;
        for (int i = 0; i < arr.length; i++) a = Math.max(a, arr[i]);
        return a;
    }

    // get array min
    static int aMin(int[] arr) {
        int a = Integer.MAX_VALUE;
        for (int i = 0; i < arr.length; i++) a = Math.min(a, arr[i]);
        return a;
    }

    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 class FastWriter {
        private final BufferedWriter bw;

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

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

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

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

    static class Pair {
        long f;
        long s;
        public Pair(long first, long second) {
            this.f = first;
            this.s = second;
        }
    }

    static class Triplet {
        long f;
        long s;
        long t;
        public Triplet(long first, long second, long third) {
            this.f = first;
            this.s = second;
            this.t = third;
        }
    }
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided code, we need to understand the logic and potential issues in the code. The buggy code seems to generate a permutation `q` by incrementing each element of the input permutation `p` and wrapping around if it exceeds `n`. This approach is likely to produce a permutation that is not optimal for minimizing the number of equal sum subarrays.

### Analysis of the Buggy Code
1. **Increment and Wrap Around**: The code increments each element of `p` and wraps around if the result exceeds `n`. This means that `p[i]` becomes `p[i] + 1` if `p[i] + 1 <= n`, and `(p[i] + 1) % n` otherwise.
2. **Potential Issues**:
   - The generated permutation `q` might not be a valid permutation (i.e., it might contain duplicates or missing elements).
   - The generated permutation `q` might not minimize the number of equal sum subarrays.

### Challenging Test Input
To expose the bug, we need a permutation `p` that, when incremented and wrapped around, results in a permutation `q` that is not optimal and might contain duplicates or missing elements. A good candidate for this is a permutation where the elements are already close to `n`.

### Example Test Input
Let's consider a small permutation `p` where all elements are close to `n`:

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

### Explanation
- **Input**:
  - `t = 1` (one test case)
  - `n = 4` (length of the permutation)
  - `p = [3, 4, 1, 2]` (the permutation)

- **Expected Output**:
  - A permutation `q` that minimizes the number of equal sum subarrays.

- **Buggy Code Output**:
  - The buggy code will increment each element of `p`:
    - `p[0] = 3` becomes `4`
    - `p[1] = 4` becomes `1` (since `(4 + 1) % 4 = 1`)
    - `p[2] = 1` becomes `2`
    - `p[3] = 2` becomes `3`
  - The resulting permutation `q` will be `[4, 1, 2, 3]`.

- **Analysis**:
  - The resulting permutation `q` is not optimal for minimizing the number of equal sum subarrays.
  - For example, the subarray sums for `p` and `q`:
    - `p[0] + p[1] = 3 + 4 = 7`
    - `q[0] + q[1] = 4 + 1 = 5`
    - `p[2] + p[3] = 1 + 2 = 3`
    - `q[2] + q[3] = 2 + 3 = 5`
  - The subarray sums do not match, but the permutation `q` generated by the buggy code is not optimal.

### Output
```plaintext
1
4
3 4 1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Min-Fund Prison (Easy)
time_limit_ms: 1500
memory_limit_mb: 256
Description: In the easy version, $$$m = n-1$$$ and there exists a path between $$$u$$$ and $$$v$$$ for all $$$u, v$$$ ($$$1 \leq u, v \leq n$$$).

After a worker's strike organized by the Dementors asking for equal rights, the prison of Azkaban has suffered some damage. After settling the spirits, the Ministry of Magic is looking to renovate the prison to ensure that the Dementors are kept in check. The prison consists of $$$n$$$ prison cells and $$$m$$$ bi-directional corridors. The $$$i^{th}$$$ corridor is from cells $$$u_i$$$ to $$$v_i$$$. A subset of these cells $$$S$$$ is called a complex if any cell in $$$S$$$ is reachable from any other cell in $$$S$$$. Formally, a subset of cells $$$S$$$ is a complex if $$$x$$$ and $$$y$$$ are reachable from each other for all $$$x, y \in S$$$, using only cells from $$$S$$$ on the way. The funding required for a complex $$$S$$$ consisting of $$$k$$$ cells is defined as $$$k^2$$$.

As part of your Intro to Magical Interior Design course at Hogwarts, you have been tasked with designing the prison. The Ministry of Magic has asked that you divide the prison into $$$2$$$ complexes with $$$\textbf{exactly one corridor}$$$ connecting them, so that the Dementors can't organize union meetings. For this purpose, you are allowed to build bi-directional corridors. The funding required to build a corridor between any $$$2$$$ cells is $$$c$$$.

Due to budget cuts and the ongoing fight against the Death Eaters, you must find the $$$\textbf{minimum total funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Note: The total funding is the sum of the funding required for the $$$2$$$ complexes and the corridors built. If after the division, the two complexes have $$$x$$$ and $$$y$$$ cells respectively and you have built a total of $$$a$$$ corridors, the total funding will be $$$x^2 + y^2 + c \times a$$$. Note that $$$x+y=n$$$.

Input: The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow.

The first line of each test case consists of three integers $$$n, m$$$ and $$$c$$$ ($$$2 \leq n \leq 10^5$$$, $$$m = n - 1$$$, $$$1 \leq c \leq 10^9$$$)

$$$m$$$ lines follow, each consisting of 2 integers — $$$u_i, v_i$$$ indicating a corridor is present between cells $$$u_i$$$ and $$$v_i$$$ ($$$1 \leq u_i, v_i \leq n$$$, $$$u_i \neq v_i$$$)

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

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

It is guaranteed that there exists at most one corridor between any two cells.

Output: Print the $$$\textbf{minimum funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Examples:
input:
2
2 1 3
1 2
8 7 76
3 1
3 2
2 4
2 5
4 6
4 7
7 8
output:
2
32

Note:

Buggy Code:
// Permanent
#include <iostream>
#include <vector>
#include <algorithm>
#include <climits>

// Temporary
#include <cmath>
#include <numeric>
#include <cstdlib>
#include <map>
#include <set>
#include <queue>

using namespace std;

#define int long long
#define bochi_orz cin.tie(0);cin.sync_with_stdio(0);

signed main(){
    bochi_orz
    int t;
    cin >> t;
    for (int i = 0; i < t; i++){
        int n, m, c;
        cin >> n >> m >> c;
        for (int i = 0; i < m; i++){
            int a, b;
            cin >> a >> b;
        }
        cout << (n / 2) * (n / 2) + ((n + 1) / 2) * ((n + 1) / 2) << "\n";
    }
}

/*
 ____   ___    ___     ____  _____          ____    ____   ___
|      |   |  |   \   |        |     /|    |       |      |   \
|      |   |  |    |  |____    |      |    |   _   |____  |___/
|      |   |  |    |  |        |      |    |    |  |      |  \
|____  |___|  |___/   |____    |    __|__  |____|  |____  |   \

*/

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Min-Fund Prison (Easy)
time_limit_ms: 1500
memory_limit_mb: 256
Description: In the easy version, $$$m = n-1$$$ and there exists a path between $$$u$$$ and $$$v$$$ for all $$$u, v$$$ ($$$1 \leq u, v \leq n$$$).

After a worker's strike organized by the Dementors asking for equal rights, the prison of Azkaban has suffered some damage. After settling the spirits, the Ministry of Magic is looking to renovate the prison to ensure that the Dementors are kept in check. The prison consists of $$$n$$$ prison cells and $$$m$$$ bi-directional corridors. The $$$i^{th}$$$ corridor is from cells $$$u_i$$$ to $$$v_i$$$. A subset of these cells $$$S$$$ is called a complex if any cell in $$$S$$$ is reachable from any other cell in $$$S$$$. Formally, a subset of cells $$$S$$$ is a complex if $$$x$$$ and $$$y$$$ are reachable from each other for all $$$x, y \in S$$$, using only cells from $$$S$$$ on the way. The funding required for a complex $$$S$$$ consisting of $$$k$$$ cells is defined as $$$k^2$$$.

As part of your Intro to Magical Interior Design course at Hogwarts, you have been tasked with designing the prison. The Ministry of Magic has asked that you divide the prison into $$$2$$$ complexes with $$$\textbf{exactly one corridor}$$$ connecting them, so that the Dementors can't organize union meetings. For this purpose, you are allowed to build bi-directional corridors. The funding required to build a corridor between any $$$2$$$ cells is $$$c$$$.

Due to budget cuts and the ongoing fight against the Death Eaters, you must find the $$$\textbf{minimum total funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Note: The total funding is the sum of the funding required for the $$$2$$$ complexes and the corridors built. If after the division, the two complexes have $$$x$$$ and $$$y$$$ cells respectively and you have built a total of $$$a$$$ corridors, the total funding will be $$$x^2 + y^2 + c \times a$$$. Note that $$$x+y=n$$$.

Input: The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow.

The first line of each test case consists of three integers $$$n, m$$$ and $$$c$$$ ($$$2 \leq n \leq 10^5$$$, $$$m = n - 1$$$, $$$1 \leq c \leq 10^9$$$)

$$$m$$$ lines follow, each consisting of 2 integers — $$$u_i, v_i$$$ indicating a corridor is present between cells $$$u_i$$$ and $$$v_i$$$ ($$$1 \leq u_i, v_i \leq n$$$, $$$u_i \neq v_i$$$)

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

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

It is guaranteed that there exists at most one corridor between any two cells.

Output: Print the $$$\textbf{minimum funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Examples:
input:
2
2 1 3
1 2
8 7 76
3 1
3 2
2 4
2 5
4 6
4 7
7 8
output:
2
32

Note:

Buggy Code:
#include <iostream>
#include <set>
#include <algorithm>
#include <vector>
#include <cassert>
#include <numeric>
#include <map>
#include <queue>
#include <bitset>
#include <stack>
#include <cmath>
#include <string>
#define sz(x) (int)(x).size()
using namespace std;
#define ll long long
#define pi pair<int, int>
int mod = 1000000007;
vector<char> alphabet = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r','s', 't', 'u', 'v', 'w', 'x', 'y','z'};
void setIO(string name = "") { // name is nonempty for USACO file I/O
	ios_base::sync_with_stdio(0); cin.tie(0); // see Fast Input & Output
	if(sz(name)){
		freopen((name+".in").c_str(), "r", stdin); // see Input & Output
		freopen((name+".out").c_str(), "w", stdout);
	}
}
ll nextpow(ll x) {
    ll res = 1;
    while (res < x) {
        res *= 2;
    }
    return res;
}

template<typename T>
void print(vector<T> &vec) {
    for (auto v: vec) {
        cout << v << " ";
    }
    cout << endl;
}

void dfs(int u, vector<vector<int>> &adj, vector<ll> &dp, vector<bool>& visited) {
    visited[u] = true;
    dp[u] = 1;
    for (auto v: adj[u]) {
        if (!visited[v]) {
            dfs(v, adj, dp, visited);
            dp[u] += dp[v];
        }
    }
}

int main() {
    int t; cin >> t;
    while (t--) {
        int n, m, c; cin >> n >> m >> c;
        vector<ll> dp(n + 1);
        vector<vector<int>> adj(n + 1);
        vector<pair<int, int>> edges;
        vector<bool> visited(n + 1);
        for (int i = 0; i < m; i ++) {
            int u, v; cin >> u >> v;
            //cout << "U " << u << " V . "<< v << endl;
            edges.push_back({u, v});
            adj[u].push_back(v);
            adj[v].push_back(u);
        }
        dfs(1, adj, dp, visited);
        ll res = LLONG_MAX;
        for (auto [u, v] : edges) {
            res = min(res, (dp[v]*dp[v]) + (n-dp[v])*(n-dp[v]));
        }
        cout << res << endl;
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Min-Fund Prison (Easy)
time_limit_ms: 1500
memory_limit_mb: 256
Description: In the easy version, $$$m = n-1$$$ and there exists a path between $$$u$$$ and $$$v$$$ for all $$$u, v$$$ ($$$1 \leq u, v \leq n$$$).

After a worker's strike organized by the Dementors asking for equal rights, the prison of Azkaban has suffered some damage. After settling the spirits, the Ministry of Magic is looking to renovate the prison to ensure that the Dementors are kept in check. The prison consists of $$$n$$$ prison cells and $$$m$$$ bi-directional corridors. The $$$i^{th}$$$ corridor is from cells $$$u_i$$$ to $$$v_i$$$. A subset of these cells $$$S$$$ is called a complex if any cell in $$$S$$$ is reachable from any other cell in $$$S$$$. Formally, a subset of cells $$$S$$$ is a complex if $$$x$$$ and $$$y$$$ are reachable from each other for all $$$x, y \in S$$$, using only cells from $$$S$$$ on the way. The funding required for a complex $$$S$$$ consisting of $$$k$$$ cells is defined as $$$k^2$$$.

As part of your Intro to Magical Interior Design course at Hogwarts, you have been tasked with designing the prison. The Ministry of Magic has asked that you divide the prison into $$$2$$$ complexes with $$$\textbf{exactly one corridor}$$$ connecting them, so that the Dementors can't organize union meetings. For this purpose, you are allowed to build bi-directional corridors. The funding required to build a corridor between any $$$2$$$ cells is $$$c$$$.

Due to budget cuts and the ongoing fight against the Death Eaters, you must find the $$$\textbf{minimum total funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Note: The total funding is the sum of the funding required for the $$$2$$$ complexes and the corridors built. If after the division, the two complexes have $$$x$$$ and $$$y$$$ cells respectively and you have built a total of $$$a$$$ corridors, the total funding will be $$$x^2 + y^2 + c \times a$$$. Note that $$$x+y=n$$$.

Input: The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow.

The first line of each test case consists of three integers $$$n, m$$$ and $$$c$$$ ($$$2 \leq n \leq 10^5$$$, $$$m = n - 1$$$, $$$1 \leq c \leq 10^9$$$)

$$$m$$$ lines follow, each consisting of 2 integers — $$$u_i, v_i$$$ indicating a corridor is present between cells $$$u_i$$$ and $$$v_i$$$ ($$$1 \leq u_i, v_i \leq n$$$, $$$u_i \neq v_i$$$)

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

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

It is guaranteed that there exists at most one corridor between any two cells.

Output: Print the $$$\textbf{minimum funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Examples:
input:
2
2 1 3
1 2
8 7 76
3 1
3 2
2 4
2 5
4 6
4 7
7 8
output:
2
32

Note:

Buggy Code:
#include <locale.h>
#include <iostream>
#include <fstream>
#include <algorithm>
#include <vector>
#include <numeric>
#include <iterator>
#include <sstream>
#include <map>
#include <math.h>
#include <stdio.h>
#include <stack>
#include <random>
#include <ctime>
#include <queue>
#include <set>
using namespace std;
using ld = long double;
using ll = long long;

ll mod = 1e9 + 7;
int const N = 3e5 + 100;

mt19937 rnd;

ll  random(ll  limit)
{
	uniform_int_distribution <ll > uniform_dist(0, limit);
	return uniform_dist(rnd);
}

ll bpow(ll a, ll p) {
	ll res = 1;
	while (p > 0) {
		if (p % 2) {
			res = res * a % mod;
		}
		a = a * a % mod;
		p /= 2;
	}
	return res;
}

ll inv(ll a) {
	return bpow(a, mod - 2);
}

void solve() {
	ll n, m, c;
	cin >> n >> m >> c;
	ll a = n / 2;
	ll b = n - a;
	cout << a * a + b * b <<'\n';
	for (int i = 0; i < n - 1; i++)
		cin >> m >> c;
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	rnd = mt19937(956231);
	int t;
	cin >> t;
	//t = 1;
	while (t--) {
		solve();
	}

}

/*

*/

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to create a test input that highlights the issue with the algorithm. The buggy code simply divides the prison into two complexes of equal (or nearly equal) size without considering the actual structure of the corridors. This approach can lead to incorrect results if the division does not form valid complexes.

### Steps to Create the Test Input

1. **Understand the Bug**: The bug is that the code does not check if the division forms valid complexes. It assumes that dividing the cells into two groups of `a` and `b` (where `a = n/2` and `b = n - a`) will always result in valid complexes, which is not true.

2. **Create a Valid Graph**: We need a graph where dividing the cells into two groups of `a` and `b` does not form valid complexes. This can be achieved by creating a tree (since `m = n - 1`) where the division results in one complex being disconnected from the other.

3. **Example Graph**: Consider a simple example with 4 cells and 3 corridors. If we divide the cells into two groups of 2, the division should result in one group being disconnected from the other. For instance, if the corridors are:
   - 1 -> 2
   - 2 -> 3
   - 3 -> 4

   Dividing the cells into groups {1, 2} and {3, 4} will result in the group {1, 2} being disconnected from {3, 4}.

### Test Input

```plaintext
1
4 3 10
1 2
2 3
3 4
```

### Explanation

- **Test Case Count**: 1
- **Number of Cells (n)**: 4
- **Number of Corridors (m)**: 3 (which is `n - 1`)
- **Cost to Build a Corridor (c)**: 10
- **Corridors**:
  - 1 -> 2
  - 2 -> 3
  - 3 -> 4

### Expected Output

The correct output for this test case should be `-1` because there is no way to divide the prison into two complexes with exactly one corridor connecting them. The buggy code will output `5` (since `2^2 + 2^2 = 4 + 1 = 5`), which is incorrect.

This test input should expose the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Min-Fund Prison (Easy)
time_limit_ms: 1500
memory_limit_mb: 256
Description: In the easy version, $$$m = n-1$$$ and there exists a path between $$$u$$$ and $$$v$$$ for all $$$u, v$$$ ($$$1 \leq u, v \leq n$$$).

After a worker's strike organized by the Dementors asking for equal rights, the prison of Azkaban has suffered some damage. After settling the spirits, the Ministry of Magic is looking to renovate the prison to ensure that the Dementors are kept in check. The prison consists of $$$n$$$ prison cells and $$$m$$$ bi-directional corridors. The $$$i^{th}$$$ corridor is from cells $$$u_i$$$ to $$$v_i$$$. A subset of these cells $$$S$$$ is called a complex if any cell in $$$S$$$ is reachable from any other cell in $$$S$$$. Formally, a subset of cells $$$S$$$ is a complex if $$$x$$$ and $$$y$$$ are reachable from each other for all $$$x, y \in S$$$, using only cells from $$$S$$$ on the way. The funding required for a complex $$$S$$$ consisting of $$$k$$$ cells is defined as $$$k^2$$$.

As part of your Intro to Magical Interior Design course at Hogwarts, you have been tasked with designing the prison. The Ministry of Magic has asked that you divide the prison into $$$2$$$ complexes with $$$\textbf{exactly one corridor}$$$ connecting them, so that the Dementors can't organize union meetings. For this purpose, you are allowed to build bi-directional corridors. The funding required to build a corridor between any $$$2$$$ cells is $$$c$$$.

Due to budget cuts and the ongoing fight against the Death Eaters, you must find the $$$\textbf{minimum total funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Note: The total funding is the sum of the funding required for the $$$2$$$ complexes and the corridors built. If after the division, the two complexes have $$$x$$$ and $$$y$$$ cells respectively and you have built a total of $$$a$$$ corridors, the total funding will be $$$x^2 + y^2 + c \times a$$$. Note that $$$x+y=n$$$.

Input: The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow.

The first line of each test case consists of three integers $$$n, m$$$ and $$$c$$$ ($$$2 \leq n \leq 10^5$$$, $$$m = n - 1$$$, $$$1 \leq c \leq 10^9$$$)

$$$m$$$ lines follow, each consisting of 2 integers — $$$u_i, v_i$$$ indicating a corridor is present between cells $$$u_i$$$ and $$$v_i$$$ ($$$1 \leq u_i, v_i \leq n$$$, $$$u_i \neq v_i$$$)

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

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

It is guaranteed that there exists at most one corridor between any two cells.

Output: Print the $$$\textbf{minimum funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Examples:
input:
2
2 1 3
1 2
8 7 76
3 1
3 2
2 4
2 5
4 6
4 7
7 8
output:
2
32

Note:

Buggy Code:
print (2); print (32);

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases or scenarios that can reveal the issues in the faulty solution. The buggy code simply prints the results for two specific test cases without any actual computation. Therefore, we should create a test case that is different from the provided examples and requires actual logic to compute the correct answer.

### Step-by-Step Reasoning:
1. **Test Case Structure**: Each test case should have the format:
   - First line: `n m c`
   - Next `m` lines: `u_i v_i`

2. **Edge Cases**:
   - **Small Number of Cells**: Test with a small number of cells to ensure the logic works for simple cases.
   - **Large Number of Cells**: Test with a large number of cells to ensure the logic handles larger inputs.
   - **Specific Corridor Configurations**: Test with specific corridor configurations that might expose issues in the algorithm.

3. **Example Test Case**:
   - **Small Case**: Use a small number of cells and corridors to ensure the logic is correct.
   - **Large Case**: Use a larger number of cells and corridors to ensure the logic handles larger inputs efficiently.

### Chosen Test Case:
- **Small Case**:
  - `n = 4`, `m = 3`, `c = 5`
  - Corridors: `(1, 2)`, `(2, 3)`, `(3, 4)`
  - Expected Output: `20` (since the only possible division is `1, 2, 3` and `4`, and the funding required is `3^2 + 1^2 + 5 * 1 = 9 + 1 + 5 = 15`)

- **Large Case**:
  - `n = 100`, `m = 99`, `c = 1000`
  - Corridors: `(1, 2)`, `(2, 3)`, ..., `(99, 100)`
  - Expected Output: `10001` (since the only possible division is `1, 2, ..., 50` and `51, 52, ..., 100`, and the funding required is `50^2 + 50^2 + 1000 * 1 = 2500 + 2500 + 1000 = 6000`)

### Final Test Input:
```plaintext
2
4 3 5
1 2
2 3
3 4
100 99 1000
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
10 11
11 12
12 13
13 14
14 15
15 16
16 17
17 18
18 19
19 20
20 21
21 22
22 23
23 24
24 25
25 26
26 27
27 28
28 29
29 30
30 31
31 32
32 33
33 34
34 35
35 36
36 37
37 38
38 39
39 40
40 41
41 42
42 43
43 44
44 45
45 46
46 47
47 48
48 49
49 50
50 51
51 52
52 53
53 54
54 55
55 56
56 57
57 58
58 59
59 60
60 61
61 62
62 63
63 64
64 65
65 66
66 67
67 68
68 69
69 70
70 71
71 72
72 73
73 74
74 75
75 76
76 77
77 78
78 79
79 80
80 81
81 82
82 83
83 84
84 85
85 86
86 87
87 88
88 89
89 90
90 91
91 92
92 93
93 94
94 95
95 96
96 97
97 98
98 99
99 100
```

This test input includes both a small and a large case to ensure the solution handles different scales of input correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Min-Fund Prison (Easy)
time_limit_ms: 1500
memory_limit_mb: 256
Description: In the easy version, $$$m = n-1$$$ and there exists a path between $$$u$$$ and $$$v$$$ for all $$$u, v$$$ ($$$1 \leq u, v \leq n$$$).

After a worker's strike organized by the Dementors asking for equal rights, the prison of Azkaban has suffered some damage. After settling the spirits, the Ministry of Magic is looking to renovate the prison to ensure that the Dementors are kept in check. The prison consists of $$$n$$$ prison cells and $$$m$$$ bi-directional corridors. The $$$i^{th}$$$ corridor is from cells $$$u_i$$$ to $$$v_i$$$. A subset of these cells $$$S$$$ is called a complex if any cell in $$$S$$$ is reachable from any other cell in $$$S$$$. Formally, a subset of cells $$$S$$$ is a complex if $$$x$$$ and $$$y$$$ are reachable from each other for all $$$x, y \in S$$$, using only cells from $$$S$$$ on the way. The funding required for a complex $$$S$$$ consisting of $$$k$$$ cells is defined as $$$k^2$$$.

As part of your Intro to Magical Interior Design course at Hogwarts, you have been tasked with designing the prison. The Ministry of Magic has asked that you divide the prison into $$$2$$$ complexes with $$$\textbf{exactly one corridor}$$$ connecting them, so that the Dementors can't organize union meetings. For this purpose, you are allowed to build bi-directional corridors. The funding required to build a corridor between any $$$2$$$ cells is $$$c$$$.

Due to budget cuts and the ongoing fight against the Death Eaters, you must find the $$$\textbf{minimum total funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Note: The total funding is the sum of the funding required for the $$$2$$$ complexes and the corridors built. If after the division, the two complexes have $$$x$$$ and $$$y$$$ cells respectively and you have built a total of $$$a$$$ corridors, the total funding will be $$$x^2 + y^2 + c \times a$$$. Note that $$$x+y=n$$$.

Input: The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow.

The first line of each test case consists of three integers $$$n, m$$$ and $$$c$$$ ($$$2 \leq n \leq 10^5$$$, $$$m = n - 1$$$, $$$1 \leq c \leq 10^9$$$)

$$$m$$$ lines follow, each consisting of 2 integers — $$$u_i, v_i$$$ indicating a corridor is present between cells $$$u_i$$$ and $$$v_i$$$ ($$$1 \leq u_i, v_i \leq n$$$, $$$u_i \neq v_i$$$)

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

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

It is guaranteed that there exists at most one corridor between any two cells.

Output: Print the $$$\textbf{minimum funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Examples:
input:
2
2 1 3
1 2
8 7 76
3 1
3 2
2 4
2 5
4 6
4 7
7 8
output:
2
32

Note:

Buggy Code:
import   sys, math
finp   = sys.stdin.readline
MOD    = 10**9 + 7
njoin  = lambda x: '\n'.join(map(str, x))
ljoin  = lambda x: ' '.join(map(str, x))
sjoin  = lambda x: '\n'.join(x)
initl  = lambda x,y: [y]*x
INF    = float('inf')
NINF   = float('-inf')
ilist  = lambda: list(map(int, finp().split()))
slist  = lambda: list(input())
slists = lambda: input().split()
ivars  = lambda: map(int, finp().split())
ipt    = lambda: int(finp())
out    = []
#=============================================#
def sol():
    n,m,c = ivars()
    adj = {}
    for _ in range(m):
        u,v = ivars()
        if u in adj:
            adj[u].append(v)
        else:
            adj[u] = [v]
        if v in adj:
            adj[v].append(u)
        else:
            adj[v] = [u]
    m = (n+1)//2
    for i in adj:
        if len(adj[i]) > m:
            m = adj[i] + 1
    print(m**2 + (n-m)**2)

#=============================================#

if __name__ == "__main__":
    for _ in range(ipt()): sol()
    #sol()
    pass

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, let's analyze the code and identify potential issues:

1. **Graph Representation and Adjacency List**:
   - The code uses an adjacency list to represent the graph.
   - The code correctly reads the input and constructs the adjacency list.

2. **Logic for Finding Complexes**:
   - The code attempts to find a way to divide the prison into two complexes.
   - The variable `m` is incorrectly used to store the midpoint of the number of cells, which is not the correct approach for finding the division.
   - The code then iterates over the adjacency list and checks the length of the adjacency list for each cell, which is not the correct way to determine the division.

3. **Output**:
   - The code outputs the sum of the squares of the sizes of the two complexes, but it does not consider the cost of building the corridor between the two complexes.

To expose this bug, we need a test case where the division into two complexes is not straightforward and the incorrect logic in the code will lead to an incorrect result. Specifically, we need a test case where the graph is not trivially divisible into two equal parts.

Here is a small-scale test input that should expose the bug:

```plaintext
2
5 4 10
1 2
2 3
3 4
4 5
6 5 100
1 2
2 3
3 4
4 5
5 6
```

### Explanation:
1. **Test Case 1**:
   - **n = 5, m = 4, c = 10**
   - **Edges**: (1, 2), (2, 3), (3, 4), (4, 5)
   - This forms a linear chain of cells. The correct division would be to split the chain into two parts, e.g., {1, 2} and {3, 4, 5}. The total funding would be \(2^2 + 3^2 + 10 = 4 + 9 + 10 = 23\).

2. **Test Case 2**:
   - **n = 6, m = 5, c = 100**
   - **Edges**: (1, 2), (2, 3), (3, 4), (4, 5), (5, 6)
   - This also forms a linear chain of cells. The correct division would be to split the chain into two parts, e.g., {1, 2, 3} and {4, 5, 6}. The total funding would be \(3^2 + 3^2 + 100 = 9 + 9 + 100 = 118\).

The buggy code will likely fail to correctly determine the division and will output an incorrect result for these test cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Min-Fund Prison (Easy)
time_limit_ms: 1500
memory_limit_mb: 256
Description: In the easy version, $$$m = n-1$$$ and there exists a path between $$$u$$$ and $$$v$$$ for all $$$u, v$$$ ($$$1 \leq u, v \leq n$$$).

After a worker's strike organized by the Dementors asking for equal rights, the prison of Azkaban has suffered some damage. After settling the spirits, the Ministry of Magic is looking to renovate the prison to ensure that the Dementors are kept in check. The prison consists of $$$n$$$ prison cells and $$$m$$$ bi-directional corridors. The $$$i^{th}$$$ corridor is from cells $$$u_i$$$ to $$$v_i$$$. A subset of these cells $$$S$$$ is called a complex if any cell in $$$S$$$ is reachable from any other cell in $$$S$$$. Formally, a subset of cells $$$S$$$ is a complex if $$$x$$$ and $$$y$$$ are reachable from each other for all $$$x, y \in S$$$, using only cells from $$$S$$$ on the way. The funding required for a complex $$$S$$$ consisting of $$$k$$$ cells is defined as $$$k^2$$$.

As part of your Intro to Magical Interior Design course at Hogwarts, you have been tasked with designing the prison. The Ministry of Magic has asked that you divide the prison into $$$2$$$ complexes with $$$\textbf{exactly one corridor}$$$ connecting them, so that the Dementors can't organize union meetings. For this purpose, you are allowed to build bi-directional corridors. The funding required to build a corridor between any $$$2$$$ cells is $$$c$$$.

Due to budget cuts and the ongoing fight against the Death Eaters, you must find the $$$\textbf{minimum total funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Note: The total funding is the sum of the funding required for the $$$2$$$ complexes and the corridors built. If after the division, the two complexes have $$$x$$$ and $$$y$$$ cells respectively and you have built a total of $$$a$$$ corridors, the total funding will be $$$x^2 + y^2 + c \times a$$$. Note that $$$x+y=n$$$.

Input: The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow.

The first line of each test case consists of three integers $$$n, m$$$ and $$$c$$$ ($$$2 \leq n \leq 10^5$$$, $$$m = n - 1$$$, $$$1 \leq c \leq 10^9$$$)

$$$m$$$ lines follow, each consisting of 2 integers — $$$u_i, v_i$$$ indicating a corridor is present between cells $$$u_i$$$ and $$$v_i$$$ ($$$1 \leq u_i, v_i \leq n$$$, $$$u_i \neq v_i$$$)

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

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

It is guaranteed that there exists at most one corridor between any two cells.

Output: Print the $$$\textbf{minimum funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Examples:
input:
2
2 1 3
1 2
8 7 76
3 1
3 2
2 4
2 5
4 6
4 7
7 8
output:
2
32

Note:

Buggy Code:
class MinFund:
    def __init__(self):
        self.test_case_count = None
        self.cell_count = []
        self.read_input()
        for cell in self.cell_count:
            out = self.min_complex(cell)
            print(out)

    def read_input(self):
        self.test_case_count = int(input())
        for _ in range(self.test_case_count):
            line = input()
            n,m,_ = list(map(int, line.split(" ")))
            self.cell_count.append(n)
            for i in range(m):
                _ = input()

    def min_complex(self, cell_count):
        if cell_count % 2 == 0:
            return (cell_count ** 2) // 2
        else:
            return 2*(cell_count // 2)*(cell_count // 2 + 1) + 1

if __name__ == "__main__":
    fund = MinFund()

**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
2 1 1
1 2
3 2 1
1 2
2 3
4 3 1
1 2
2 3
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Min-Fund Prison (Easy)
time_limit_ms: 1500
memory_limit_mb: 256
Description: In the easy version, $$$m = n-1$$$ and there exists a path between $$$u$$$ and $$$v$$$ for all $$$u, v$$$ ($$$1 \leq u, v \leq n$$$).

After a worker's strike organized by the Dementors asking for equal rights, the prison of Azkaban has suffered some damage. After settling the spirits, the Ministry of Magic is looking to renovate the prison to ensure that the Dementors are kept in check. The prison consists of $$$n$$$ prison cells and $$$m$$$ bi-directional corridors. The $$$i^{th}$$$ corridor is from cells $$$u_i$$$ to $$$v_i$$$. A subset of these cells $$$S$$$ is called a complex if any cell in $$$S$$$ is reachable from any other cell in $$$S$$$. Formally, a subset of cells $$$S$$$ is a complex if $$$x$$$ and $$$y$$$ are reachable from each other for all $$$x, y \in S$$$, using only cells from $$$S$$$ on the way. The funding required for a complex $$$S$$$ consisting of $$$k$$$ cells is defined as $$$k^2$$$.

As part of your Intro to Magical Interior Design course at Hogwarts, you have been tasked with designing the prison. The Ministry of Magic has asked that you divide the prison into $$$2$$$ complexes with $$$\textbf{exactly one corridor}$$$ connecting them, so that the Dementors can't organize union meetings. For this purpose, you are allowed to build bi-directional corridors. The funding required to build a corridor between any $$$2$$$ cells is $$$c$$$.

Due to budget cuts and the ongoing fight against the Death Eaters, you must find the $$$\textbf{minimum total funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Note: The total funding is the sum of the funding required for the $$$2$$$ complexes and the corridors built. If after the division, the two complexes have $$$x$$$ and $$$y$$$ cells respectively and you have built a total of $$$a$$$ corridors, the total funding will be $$$x^2 + y^2 + c \times a$$$. Note that $$$x+y=n$$$.

Input: The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow.

The first line of each test case consists of three integers $$$n, m$$$ and $$$c$$$ ($$$2 \leq n \leq 10^5$$$, $$$m = n - 1$$$, $$$1 \leq c \leq 10^9$$$)

$$$m$$$ lines follow, each consisting of 2 integers — $$$u_i, v_i$$$ indicating a corridor is present between cells $$$u_i$$$ and $$$v_i$$$ ($$$1 \leq u_i, v_i \leq n$$$, $$$u_i \neq v_i$$$)

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

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

It is guaranteed that there exists at most one corridor between any two cells.

Output: Print the $$$\textbf{minimum funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Examples:
input:
2
2 1 3
1 2
8 7 76
3 1
3 2
2 4
2 5
4 6
4 7
7 8
output:
2
32

Note:

Buggy Code:
// package Graph;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class MinFundingEasy {
    public static void main(String[] args)throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int test=Integer.parseInt(br.readLine());
        StringBuilder print=new StringBuilder();
        while(test--!=0){
            StringTokenizer st=new StringTokenizer(br.readLine());
            int n=Integer.parseInt(st.nextToken());
            int m=Integer.parseInt(st.nextToken());
            int c=Integer.parseInt(st.nextToken());
            for(int i=1;i<=m;i++){
                st=new StringTokenizer(br.readLine());
            }
            long x=n/2;
            long y=n-x;
            long ans=x*x+y*y;
            print.append(ans+"\n");
        }
        System.out.println(print);
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Min-Fund Prison (Easy)
time_limit_ms: 1500
memory_limit_mb: 256
Description: In the easy version, $$$m = n-1$$$ and there exists a path between $$$u$$$ and $$$v$$$ for all $$$u, v$$$ ($$$1 \leq u, v \leq n$$$).

After a worker's strike organized by the Dementors asking for equal rights, the prison of Azkaban has suffered some damage. After settling the spirits, the Ministry of Magic is looking to renovate the prison to ensure that the Dementors are kept in check. The prison consists of $$$n$$$ prison cells and $$$m$$$ bi-directional corridors. The $$$i^{th}$$$ corridor is from cells $$$u_i$$$ to $$$v_i$$$. A subset of these cells $$$S$$$ is called a complex if any cell in $$$S$$$ is reachable from any other cell in $$$S$$$. Formally, a subset of cells $$$S$$$ is a complex if $$$x$$$ and $$$y$$$ are reachable from each other for all $$$x, y \in S$$$, using only cells from $$$S$$$ on the way. The funding required for a complex $$$S$$$ consisting of $$$k$$$ cells is defined as $$$k^2$$$.

As part of your Intro to Magical Interior Design course at Hogwarts, you have been tasked with designing the prison. The Ministry of Magic has asked that you divide the prison into $$$2$$$ complexes with $$$\textbf{exactly one corridor}$$$ connecting them, so that the Dementors can't organize union meetings. For this purpose, you are allowed to build bi-directional corridors. The funding required to build a corridor between any $$$2$$$ cells is $$$c$$$.

Due to budget cuts and the ongoing fight against the Death Eaters, you must find the $$$\textbf{minimum total funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Note: The total funding is the sum of the funding required for the $$$2$$$ complexes and the corridors built. If after the division, the two complexes have $$$x$$$ and $$$y$$$ cells respectively and you have built a total of $$$a$$$ corridors, the total funding will be $$$x^2 + y^2 + c \times a$$$. Note that $$$x+y=n$$$.

Input: The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow.

The first line of each test case consists of three integers $$$n, m$$$ and $$$c$$$ ($$$2 \leq n \leq 10^5$$$, $$$m = n - 1$$$, $$$1 \leq c \leq 10^9$$$)

$$$m$$$ lines follow, each consisting of 2 integers — $$$u_i, v_i$$$ indicating a corridor is present between cells $$$u_i$$$ and $$$v_i$$$ ($$$1 \leq u_i, v_i \leq n$$$, $$$u_i \neq v_i$$$)

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

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

It is guaranteed that there exists at most one corridor between any two cells.

Output: Print the $$$\textbf{minimum funding}$$$ required to divide the prison as per the Ministry's requirements or $$$-1$$$ if no division is possible.

Examples:
input:
2
2 1 3
1 2
8 7 76
3 1
3 2
2 4
2 5
4 6
4 7
7 8
output:
2
32

Note:

Buggy Code:
/**
 * G_1_Min_Fund_Prison_Easy
 */
import java.io.*;
import java.util.*;

public class G_1_Min_Fund_Prison_Easy {
    static int n, m, c;
    static ArrayList<Integer>adj[];
    static int deg[];
    static long ans;
    static long inf = 1000000000;
    static int ch[];
    static void dfs(int x, int par)
    {
        for(int j=0; j<adj[x].size(); j++)
        {
            int y = adj[x].get(j);
            if(y!=par){
                dfs(y, x);
                ch[x] += ch[y];
            }
        }
        ch[x]++;
        long val1 = ch[x];
        val1 *= ch[x];
        long val2 = (n-ch[x]);
        val2 *= (n-ch[x]);
        //System.out.println(x + " " + val1+val2);
        ans = Math.min(ans, val1+val2);
    }
    static void solve(int tc, Scanner sc)
    {
        n = sc.nextInt();
        m = sc.nextInt();
        c = sc.nextInt();
        adj = new ArrayList[n+1];
        deg = new int[n+1];
        Arrays.setAll(adj, key->new ArrayList<>());
        for(int i=1; i<=m; i++)
        {
            int u = sc.nextInt();
            int v = sc.nextInt();
            adj[u].add(v);
            deg[v]++;
        }
        int x=-1;
        for(int i=1; i<=n; i++){
            if(deg[i]==0){
                x = i;
                break;
            }
        }
        //System.out.println(x);
        ans = n;
        ans *= n;
        ch  = new int[n+1];
        for(int i=0; i<=n; i++){
            ch[i] = 0;
        }
        dfs(x, -1);
        // for(int i=1; i<=n; i++){
        //     System.out.print(ch[i]+ " ");
        // }
        // System.out.println();
        System.out.println(ans);
    }
    public static void main(String[] args) throws Exception{
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for(int i=1; i<=t; i++){
            solve(i, sc);
        }
        sc.close();
    }
}

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

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

Think step by step.

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