**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $$$1$$$ and $$$n$$$ is written on each card: specifically, for each $$$i$$$ from $$$1$$$ to $$$n$$$, you have $$$a_i$$$ cards which have the number $$$i$$$ written on them.

There is also a shop which contains unlimited cards of each type. You have $$$k$$$ coins, so you can buy $$$k$$$ new cards in total, and the cards you buy can contain any integer between $$$1$$$ and $$$n$$$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $$$n$$$ which are a permutation of $$$[1, 2, \ldots, n]$$$. What's the maximum score you can get?

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t\ (1\le t\le 100)$$$. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$, $$$k$$$ ($$$1\le n \le 2 \cdot 10^5$$$, $$$0\le k \le 10^{12}$$$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^{12}$$$) — the number of cards of type $$$i$$$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $$$[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$$$ (including $$$11$$$ single $$$1$$$s), which contains $$$11$$$ subarrays consisting of a permutation of $$$[1]$$$.

In the second test case, we can buy $$$0$$$ cards of type $$$1$$$ and $$$4$$$ cards of type $$$2$$$, and then we rearrange the cards as following: $$$[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$$$. There are $$$8$$$ subarrays equal to $$$[1, 2]$$$ and $$$7$$$ subarrays equal to $$$[2, 1]$$$, which make a total of $$$15$$$ subarrays which are a permutation of $$$[1, 2]$$$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $$$[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$$$.

Buggy Code:
// Problem: C. Ski Resort
// Contest: Codeforces - Codeforces Round 878 (Div. 3)
// URL: https://codeforces.com/problemset/problem/1840/C
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)

/*
@author
youranonfriend
*/
#include <bits/stdc++.h>
#define int long long
using namespace std;
using vi = vector<int>;
#define pb push_back
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()

using pi = pair<int,int>;
#define f first
#define ss second
#define mp map<int,int>

void setIO(string name = "") {
	cin.tie(0)->sync_with_stdio(0);
	if (sz(name)) {
		freopen((name + ".in").c_str(), "r", stdin);
		freopen((name + ".out").c_str(), "w", stdout);
	}
}
bool ok(int m, vi a, int k){
	int s = 0;
	for(auto i : a){
		k-= max(0ll,m-i);
		if(k < 0) return 0;
	}
	return k >= 0;
}
void solve(){
	int n,k; cin >> n >> k;
	vi a(n);
	int mx = 0;
	for(auto &i : a) cin >> i, mx = max(mx,i);
	int l = 0, r = LLONG_MAX;
	for(int i=0; i < 200; i ++){
		int m = (l + r ) / 2;
		if(ok(m,a,k))l = m;
		else r = m;
	}
	int c = 0;
	for(auto &i : a){
		c += max(0ll,l-i);
		i = max(i,l);
		// cout << i << " ";
	}
	k-= c;
	// cout <<"--" << k << "--";
	sort(all(a));
	for(auto &i : a){
		if(!k) break;
		i ++;
		k--;
	}
	// for(auto i : a) cout << i << " ";
	int mn = INT_MAX;
	for(auto i : a) mn = min(mn,i);
	int cn = 0;
	// for(int i=1; i < n; i ++){
		// if(a[i] == a[i-1]) cn += a[i]-1;
		// else cn += min(a[i],a[i-1]);
	// }
	// cout << cn;
	bool o = 0;
	for(int i=0; i < n; i ++){
		if(a[i] == mn && !o) cn += a[i], o = 1;
		else cn += min(mn,a[i]-1);
	}
	cout << cn;
}
/*

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

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

5 6 3 3
1 4 3 2 1 4 3 2 1 4 3 2 1
*/

signed main() {
	setIO();
	int t = 1;
	cin >> t;
	while(t--){
		solve();
		cout << '\n';
	}
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $$$1$$$ and $$$n$$$ is written on each card: specifically, for each $$$i$$$ from $$$1$$$ to $$$n$$$, you have $$$a_i$$$ cards which have the number $$$i$$$ written on them.

There is also a shop which contains unlimited cards of each type. You have $$$k$$$ coins, so you can buy $$$k$$$ new cards in total, and the cards you buy can contain any integer between $$$1$$$ and $$$n$$$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $$$n$$$ which are a permutation of $$$[1, 2, \ldots, n]$$$. What's the maximum score you can get?

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t\ (1\le t\le 100)$$$. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$, $$$k$$$ ($$$1\le n \le 2 \cdot 10^5$$$, $$$0\le k \le 10^{12}$$$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^{12}$$$) — the number of cards of type $$$i$$$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $$$[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$$$ (including $$$11$$$ single $$$1$$$s), which contains $$$11$$$ subarrays consisting of a permutation of $$$[1]$$$.

In the second test case, we can buy $$$0$$$ cards of type $$$1$$$ and $$$4$$$ cards of type $$$2$$$, and then we rearrange the cards as following: $$$[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$$$. There are $$$8$$$ subarrays equal to $$$[1, 2]$$$ and $$$7$$$ subarrays equal to $$$[2, 1]$$$, which make a total of $$$15$$$ subarrays which are a permutation of $$$[1, 2]$$$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $$$[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$$$.

Buggy Code:
# In the name of ALLAH
import sys

# Function to compute power
def power(base, n):
    re = 1
    while n:
        if n & 1:
            re *= base
            n -= 1
        else:
            base *= base
            n //= 2
    return re

# Main function
def main():
    # Fast I/O in Python
    input = sys.stdin.readline
    print = sys.stdout.write

    t = int(input())
    for _ in range(t):
        n, k = map(int, input().split())
        v = list(map(int, input().split()))
        p = k
        if n == 1:
            print(str(v[0] + k) + '\n')
        else:
            l = 0
            h = k + max(v)
            ans = 0
            while l <= h:
                m = (l + h) // 2
                k = p
                for i in range(n):
                    if m > v[i]:
                        k -= (m - v[i])
                if k == 0:
                    ans = m
                    break
                elif k > 0:
                    ans = m
                    l = m + 1
                else:
                    h = m - 1
            m = ans
            k = p
            for i in range(n):
                if m > v[i]:
                    k -= m - v[i]
                    v[i] = m
            v.sort()
            for i in range(k):
                v[i] += 1
            q = sum(1 for i in range(n) if v[i] > m)
            result = (ans - 1) * n + 1 + q
            print(str(result) + '\n')

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider scenarios where the code might fail. The main points to focus on are:

1. **Edge Cases**: Test with `n = 1` and `k` very large.
2. **Large `k` and Small `n`**: Test with a small `n` and a large `k` to see how the code handles buying a lot of cards.
3. **Balanced Distribution**: Test with a balanced distribution of card counts to see if the code can handle equal or nearly equal counts.
4. **Highly Skewed Distribution**: Test with a highly skewed distribution of card counts to see if the code can handle very different counts.
5. **Special Values**: Test with specific values that might cause issues, such as `k` being just enough to make a difference.

Let's create a test input that includes these scenarios:

```plaintext
5
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
10 10
1 3 1 2 1 9 3 5 7 5
```

Explanation:
1. **Case 1**: `n = 1, k = 10, a = [1]`
   - This tests the edge case where `n` is 1, and `k` is large. The expected output is `11`.
2. **Case 2**: `n = 2, k = 4, a = [8, 4]`
   - This tests a small `n` with a large `k` and a skewed distribution. The expected output is `15`.
3. **Case 3**: `n = 3, k = 4, a = [6, 1, 8]`
   - This tests a small `n` with a large `k` and a highly skewed distribution. The expected output is `15`.
4. **Case 4**: `n = 3, k = 9, a = [7, 6, 2]`
   - This tests a small `n` with a large `k` and a balanced distribution. The expected output is `22`.
5. **Case 5**: `n = 10, k = 10, a = [1, 3, 1, 2, 1, 9, 3, 5, 7, 5]`
   - This tests a larger `n` with a large `k` and a highly skewed distribution. The expected output is `28`.

This test input should help expose any bugs 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: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $$$1$$$ and $$$n$$$ is written on each card: specifically, for each $$$i$$$ from $$$1$$$ to $$$n$$$, you have $$$a_i$$$ cards which have the number $$$i$$$ written on them.

There is also a shop which contains unlimited cards of each type. You have $$$k$$$ coins, so you can buy $$$k$$$ new cards in total, and the cards you buy can contain any integer between $$$1$$$ and $$$n$$$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $$$n$$$ which are a permutation of $$$[1, 2, \ldots, n]$$$. What's the maximum score you can get?

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t\ (1\le t\le 100)$$$. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$, $$$k$$$ ($$$1\le n \le 2 \cdot 10^5$$$, $$$0\le k \le 10^{12}$$$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^{12}$$$) — the number of cards of type $$$i$$$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $$$[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$$$ (including $$$11$$$ single $$$1$$$s), which contains $$$11$$$ subarrays consisting of a permutation of $$$[1]$$$.

In the second test case, we can buy $$$0$$$ cards of type $$$1$$$ and $$$4$$$ cards of type $$$2$$$, and then we rearrange the cards as following: $$$[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$$$. There are $$$8$$$ subarrays equal to $$$[1, 2]$$$ and $$$7$$$ subarrays equal to $$$[2, 1]$$$, which make a total of $$$15$$$ subarrays which are a permutation of $$$[1, 2]$$$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $$$[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$$$.

Buggy Code:
def solve():
    n, k = map(int, input().split())
    a = list(map(int, input().split()))
    a.sort()
    ans = 0
    l = 1
    r = 2 * 10 ** 12
    while l <= r:
    	mid = (l + r) // 2
    	w = (n - (mid % n)) % n
    	for i in range(w):
    		a[i] += 1
    	sz = ((mid + n - 1) // n)
    	cur = 0
    	for i in range(n):
    		if a[i] < sz:
    			cur += sz - a[i]
    	if cur <= k:
    		ans = max(ans, mid - n + 1)
    		l = mid + 1
    	else:
    		r = mid - 1
    	for i in range(w):
    		a[i] -= 1
    print(ans)
t = int(input())
while t > 0:
    solve()
    t -= 1

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

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

Think step by step.

Response:
```plaintext
1
3 5
1 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $$$1$$$ and $$$n$$$ is written on each card: specifically, for each $$$i$$$ from $$$1$$$ to $$$n$$$, you have $$$a_i$$$ cards which have the number $$$i$$$ written on them.

There is also a shop which contains unlimited cards of each type. You have $$$k$$$ coins, so you can buy $$$k$$$ new cards in total, and the cards you buy can contain any integer between $$$1$$$ and $$$n$$$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $$$n$$$ which are a permutation of $$$[1, 2, \ldots, n]$$$. What's the maximum score you can get?

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t\ (1\le t\le 100)$$$. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$, $$$k$$$ ($$$1\le n \le 2 \cdot 10^5$$$, $$$0\le k \le 10^{12}$$$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^{12}$$$) — the number of cards of type $$$i$$$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $$$[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$$$ (including $$$11$$$ single $$$1$$$s), which contains $$$11$$$ subarrays consisting of a permutation of $$$[1]$$$.

In the second test case, we can buy $$$0$$$ cards of type $$$1$$$ and $$$4$$$ cards of type $$$2$$$, and then we rearrange the cards as following: $$$[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$$$. There are $$$8$$$ subarrays equal to $$$[1, 2]$$$ and $$$7$$$ subarrays equal to $$$[2, 1]$$$, which make a total of $$$15$$$ subarrays which are a permutation of $$$[1, 2]$$$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $$$[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$$$.

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

public class div1_942_a {

	static StringBuilder sb = new StringBuilder("");
	static int n;
	static long k;
	static long[] A;
	static long answer = 0;

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

		int t = Integer.parseInt(st.nextToken());

		for (int i = 0; i < t; i++) {
			st = new StringTokenizer(br.readLine());
			n = Integer.parseInt(st.nextToken());
			k = Long.parseLong(st.nextToken());
			answer = 0;

			st = new StringTokenizer(br.readLine());
			A = new long[n];

			for (int j = 0; j < n; j++) {
				A[j] = Long.parseLong(st.nextToken());
			}

			solve();
		}

		br.close();

		System.out.print(sb.toString());

	}

	static void solve() {
		long lo = A[0];

		for (int i = 1; i < n; i++) {
			lo = Math.min(lo, A[i]);
		}

		long hi = lo + k;

		long[] C = new long[n];
		for (int i = 0; i < n; i++) C[i] = A[i];
		while (lo != hi) {
			long mid = (lo + hi + 1L)/2L;

			if (works(mid)) {
				lo = mid;
			} else {
				hi = mid-1;
			}
			for (int i = 0; i < n; i++) A[i] = C[i];
		}

		sb.append(answer + "\n");
	}

	static boolean works(long mid) {
		boolean res = true;
		long greater = 0;
		long LN = n;

		long operations = 0;
		for (int i = 0; i < n; i++) {
			if (A[i] > mid) {
				greater++;
			} else {
				operations += mid-A[i];
				A[i] = mid;
			}
		}

		for (int i = 0; i < n; i++) {
			if (A[i] == mid && operations < k) {
				operations++;
				greater++;
				A[i]++;
			}
		}

		if (operations > k) res = false;

		if (res) {
			long local = LN * mid - LN + 1L + greater;
			if (local > answer) answer = local;
		}

		return res;
	}

}

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

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

Think step by step.

Response:
```plaintext
5
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Permutation Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: You have some cards. An integer between $$$1$$$ and $$$n$$$ is written on each card: specifically, for each $$$i$$$ from $$$1$$$ to $$$n$$$, you have $$$a_i$$$ cards which have the number $$$i$$$ written on them.

There is also a shop which contains unlimited cards of each type. You have $$$k$$$ coins, so you can buy $$$k$$$ new cards in total, and the cards you buy can contain any integer between $$$1$$$ and $$$n$$$.

After buying the new cards, you rearrange all your cards in a line. The score of a rearrangement is the number of (contiguous) subarrays of length $$$n$$$ which are a permutation of $$$[1, 2, \ldots, n]$$$. What's the maximum score you can get?

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t\ (1\le t\le 100)$$$. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$, $$$k$$$ ($$$1\le n \le 2 \cdot 10^5$$$, $$$0\le k \le 10^{12}$$$) — the number of distinct types of cards and the number of coins.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^{12}$$$) — the number of cards of type $$$i$$$ you have at the beginning.

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

Output: For each test case, output a single line containing an integer: the maximum score you can get.

Examples:
input:
8
1 10
1
2 4
8 4
3 4
6 1 8
3 9
7 6 2
5 3
6 6 7 4 6
9 7
7 6 1 7 6 2 4 3 3
10 10
1 3 1 2 1 9 3 5 7 5
9 8
5 8 7 5 1 3 2 9 8
output:
11
15
15
22
28
32
28
36

Note: In the first test case, the final (and only) array we can get is $$$[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]$$$ (including $$$11$$$ single $$$1$$$s), which contains $$$11$$$ subarrays consisting of a permutation of $$$[1]$$$.

In the second test case, we can buy $$$0$$$ cards of type $$$1$$$ and $$$4$$$ cards of type $$$2$$$, and then we rearrange the cards as following: $$$[1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2]$$$. There are $$$8$$$ subarrays equal to $$$[1, 2]$$$ and $$$7$$$ subarrays equal to $$$[2, 1]$$$, which make a total of $$$15$$$ subarrays which are a permutation of $$$[1, 2]$$$. It can also be proved that this is the maximum score we can get.

In the third test case, one of the possible optimal rearrangements is $$$[3, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 3]$$$.

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

public class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		Scanner scn = new Scanner(System.in);
		int t = scn.nextInt();
		while(t-->0){
		    int n = scn.nextInt();
		    long k = scn.nextLong();
		    long arr[] = new long[n];
		    for(int i = 0; i < n; i++){
		        arr[i] = scn.nextLong();
		    }
		  //  Arrays.sort(arr);
		    long ans = 0L;
		    long left = 1L, right = (long)1e18;
		    while(left<=right){
		        long mid = (right-left)/2+left;
		        long var_k = k;
		        boolean check = true;
		        long count = 0L;
		        for(int i = 0; i < n; i++){
		            if(arr[i]<mid){
		                if(var_k<mid-arr[i])check = false;
		                else var_k -= (mid-arr[i]);
		            }
		        }
		        for(int i = 0; i < n; i++){
		            if(arr[i]>mid)count++;
		        }
		        if(check){
		          //  System.out.println(mid+" if ");
		            left = mid+1L;
		            long num1 = 0L;
		            if(mid==1){
		                num1 = n+var_k+count;
		            }else{
		                num1 = (mid-1L)*n+1+var_k+count;
		            }
		            if(num1>ans){
		                ans = num1;
		            }
		        }else{
		          //  System.out.println(mid+" else ");
		            right = mid-1L;
		        }
		    }
		    if(k!=0)
		    System.out.println(ans);
		    else{
		        Arrays.sort(arr);
		        int idx = -1;
		        for(int i = 0; i < n-1; i++){
		            if(arr[i]!=arr[i+1]){
		                idx = n-(i+1);
		            }
		        }
		        if(idx==-1){
		            System.out.println((arr[0]-1)*n+1L);
		        }else{
		            System.out.println((arr[0]-1)*n+1L+idx);
		        }
		    }
		}

	}
}

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

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

Think step by step.

Response:
```plaintext
1
3 1
1 2 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 (Hard)
time_limit_ms: 1500
memory_limit_mb: 256
Description: In the hard version, $$$2 \leq \sum n \leq 10^5$$$ and $$$1 \leq \sum m \leq 5 \times 10^{5}$$$

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 $$$c$$$ ($$$2 \leq n \leq 10^5$$$, $$$1 \leq m \leq 5 \times 10^{5}$$$, $$$1 \leq c \leq 10^9$$$)

$$$m$$$ 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:
4
4 6 5
4 3
2 3
2 4
1 2
4 1
3 1
6 6 2
1 4
2 5
3 6
1 5
3 5
6 5
6 5 7
1 4
2 5
3 6
3 5
6 5
7 5 4
1 4
3 6
3 5
6 5
2 7
output:
-1
20
25
33

Note: In the first test case of the sample input, there is no way to divide the prison according to the Ministry's requirements.

In the second test case, consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{2, 3, 5, 6\}$$$ and $$$\{1, 4\}$$$ cells respectively. There are no new corridors built. The total funding is $$$4^2 + 2^2 = 20$$$. You can verify this is the minimum funding required.

In the third test case, build a corridor between $$$2$$$ and $$$4$$$. Consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{3, 5, 6\}$$$ and $$$\{1, 2, 4\}$$$ cells respectively. The total funding is $$$3^2 + 3^2 + 7 \times 1 = 25$$$. You can verify this is the minimum funding required.

In the fourth test case, build a corridor between $$$2$$$ and $$$4$$$ and between $$$5$$$ and $$$7$$$. Consider the corridor between cells $$$5$$$ and $$$7$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{1, 2, 4, 7\}$$$ and $$$\{3, 5, 6\}$$$ cells respectively. The total funding is $$$4^2 + 3^2 + 4 \times 2 = 33$$$. You can verify this is the minimum funding required.

Note for all test cases that there may be multiple ways to get the same funding but there is no other division which will have a more optimal minimum funding.

Buggy Code:
#include <algorithm>
#include <cmath>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <vector>
#include <cstring>
#include <numeric>

using namespace std;
const int N = 310;
int fir[N], nxt[N<<1], to[N<<1], cnt;
int n,m,c,t;
int color[N], dep[N], top[N], siz[N];
vector<vector<int>> split;
void addedge(int x,int y){
    to[++cnt] = y;
    nxt[cnt] = fir[x];
    fir[x] = cnt;
    to[++cnt] = x;
    nxt[cnt] = fir[y];
    fir[y] = cnt;
}
void dfs(int p, int fa, int cr){
    color[p] = cr;
    siz[p] = 1;
    dep[p] = dep[fa] + 1;
    top[p] = dep[p];
    for (int h = fir[p];h;h = nxt[h]){
        int u = to[h];
        if (color[u] != 0) {
            if (u != fa && top[u] < top[p]){
                top[p] = top[u];
            }
            continue;
        }
        dfs(u, p, cr);
        siz[p] += siz[u];
        top[p] = min(top[p], top[u]);
    }
}
void solve(vector<int> &choice, vector<vector<int>> &split){
    int cnt = std::reduce(choice.begin(), choice.end());

    vector<vector<int>> dp(n+1, vector<int>(2, 0));
    dp[0][0] = 1;
    for (int i = 1;i < choice.size();i++){
        for (int j = 0;j < choice[i];j++){
            for (int k = n;k >= i;k--){
                dp[k][1] |= dp[k-i][1];
            }
        }
        for (auto v: split[i]){
            for (int j = n;j >= 0;j--){
                if (j-v >= 0)
                    dp[j][1] |= dp[j-v][0];
                if (j-i+v >= 0)
                    dp[j][1] |= dp[j-i+v][0];
            }
        }
        for (int j = 0;j < choice[i];j++){
            for (int k = n;k >= i;k--){
                dp[k][0] |= dp[k-i][0];
            }
        }
    }
    auto ans = 1ll*n*n*(c+1);
    auto base = 1ll*c*(cnt-1);

    for (int i = 1;i < n;i++){
        if (dp[i][1]){
            ans = min(ans, base + 1ll*i*i + 1ll*(n-i)*(n-i));
        }
        if (dp[i][0]){
            ans = min(ans, base + 1ll*i*i + 1ll*(n-i)*(n-i));
        }
    }
    if (ans == 1ll*n*n*(c+1)){
        printf("-1\n");
    } else {
        printf("%lld\n", ans);
    }
}
int main() {
    scanf("%d",&t);
    while (t--){
        scanf("%d%d%d",&n,&m,&c);
        for (int i = 1;i <= n;i++) fir[i] = 0;
        cnt = 0;
        for (int i = 1;i <= m;i++){
            int x,y;
            scanf("%d%d",&x,&y);
            addedge(x,y);
        }
        for (int i = 0;i <= n;i++){
            siz[i] = 0;
            color[i] = 0;
            top[i] = 0;
            dep[i] = 0;
        }
        for (int i = 1;i <= n;i++){
            if (color[i] == 0){
                dfs(i, 0, i);
            }
        }
        split.clear();
        split.resize(n+1);
        vector<int> choice(n+1);
        for (int i = 1;i <= n;i++){
            if (top[i] >= dep[i]){
                split[siz[color[i]]].push_back(siz[i]);
            }
            if (color[i] == i){
                choice[siz[i]]++;
            }
        }
        for (int i = 1;i <= n;i++){
            while (choice[i] >= 4){
                choice[i] -= 2;
                choice[i*2]++;
            }
        }
        solve(choice, split);
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
5 6 3
1 2
2 3
3 4
4 5
1 3
2 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 (Hard)
time_limit_ms: 1500
memory_limit_mb: 256
Description: In the hard version, $$$2 \leq \sum n \leq 10^5$$$ and $$$1 \leq \sum m \leq 5 \times 10^{5}$$$

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 $$$c$$$ ($$$2 \leq n \leq 10^5$$$, $$$1 \leq m \leq 5 \times 10^{5}$$$, $$$1 \leq c \leq 10^9$$$)

$$$m$$$ 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:
4
4 6 5
4 3
2 3
2 4
1 2
4 1
3 1
6 6 2
1 4
2 5
3 6
1 5
3 5
6 5
6 5 7
1 4
2 5
3 6
3 5
6 5
7 5 4
1 4
3 6
3 5
6 5
2 7
output:
-1
20
25
33

Note: In the first test case of the sample input, there is no way to divide the prison according to the Ministry's requirements.

In the second test case, consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{2, 3, 5, 6\}$$$ and $$$\{1, 4\}$$$ cells respectively. There are no new corridors built. The total funding is $$$4^2 + 2^2 = 20$$$. You can verify this is the minimum funding required.

In the third test case, build a corridor between $$$2$$$ and $$$4$$$. Consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{3, 5, 6\}$$$ and $$$\{1, 2, 4\}$$$ cells respectively. The total funding is $$$3^2 + 3^2 + 7 \times 1 = 25$$$. You can verify this is the minimum funding required.

In the fourth test case, build a corridor between $$$2$$$ and $$$4$$$ and between $$$5$$$ and $$$7$$$. Consider the corridor between cells $$$5$$$ and $$$7$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{1, 2, 4, 7\}$$$ and $$$\{3, 5, 6\}$$$ cells respectively. The total funding is $$$4^2 + 3^2 + 4 \times 2 = 33$$$. You can verify this is the minimum funding required.

Note for all test cases that there may be multiple ways to get the same funding but there is no other division which will have a more optimal minimum funding.

Buggy Code:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<bitset>
using namespace std;
#define ll long long
#define il inline
#define N 100005
il ll read() {
	ll x=0,f=1;char c=getchar();
	while(c<'0'||c>'9') {if(c=='-') {f=-1;} c=getchar();}
	while(c>='0'&&c<='9') {x=(x<<3)+(x<<1)+(c^48);c=getchar();}
	return x*f;
}
struct Edge {
	int nxt,v;
};
Edge edge[N*10];
int head[N],num_edge;
il void add_edge(int u,int v) {
	edge[++num_edge].nxt=head[u],edge[num_edge].v=v;
	head[u]=num_edge;
}
vector<int> g[N];
ll n,m,c,k;
int dfn[N],low[N],tim,ecc,belong[N],siz[N];
bool evis[N*10],vvis[N],cvis[N];
vector<int> bin[N],sizbin[N];
bitset<N> dp[2][2];
il void tarjan(int u,int fa) {
	dfn[u]=low[u]=++tim;
	for(int i=head[u];i;i=edge[i].nxt) {
		int v=edge[i].v;
		if(v==fa) continue;
		if(!dfn[v]) {
			tarjan(v,u);
			low[u]=min(low[u],low[v]);
			if(dfn[u]<low[v]) evis[i]=evis[i^1]=1;
		}
		else low[u]=min(low[u],dfn[v]);
	}
}
il void dfs(int u,int id) {
	vvis[u]=1,bin[id].push_back(u),belong[u]=id;
	for(int i=head[u];i;i=edge[i].nxt) {
		if(evis[i]) continue;
		int v=edge[i].v;
		if(!vvis[v]) dfs(v,id);
	}
}
il void dfs2(int u,int fa) {
	siz[u]=bin[u].size();
	for(auto v:g[u]) {
		if(v==fa) continue;
		dfs2(v,u);
		siz[u]+=siz[v];
	}
}
il void dfs3(int u,int fa,int rot) {
	sizbin[rot].push_back(siz[u]);
	for(auto v:g[u]) if(v!=fa) dfs3(v,u,rot);
}
il void mian() {
	n=read(),m=read(),c=read(),k=-1,num_edge=1;
	for(int i=1,u,v;i<=m;i++) {
		u=read(),v=read();
		add_edge(u,v);
		add_edge(v,u);
	}
	for(int i=1;i<=n;i++) if(!dfn[i]) ++k,tarjan(i,0);
	ll ans=k*c;
	for(int i=1;i<=n;i++) if(!vvis[i]) dfs(i,++ecc);
	if(ecc==1) {
		printf("-1\n");
		return;
	}
	for(int u=1;u<=n;u++) {
		for(int i=head[u];i;i=edge[i].nxt) {
			if(!evis[i]) continue;
			int uu=belong[u],vv=belong[edge[i].v];
			g[uu].push_back(vv);
		}
	}
	dp[0][0][0]=dp[1][0][0]=1;
	int chk=1;
	for(int i=1;i<=ecc;i++) if(!cvis[i]) {
		chk^=1;
		dfs2(i,0);
		int sz=siz[i];
		dfs3(i,0,i);
		dp[0][chk]=dp[0][chk^1]|(dp[0][chk^1]<<sz);
		dp[1][chk]=dp[1][chk^1]|(dp[1][chk^1]<<sz);
		for(auto s:sizbin[i]) {
			dp[0][chk]|=(dp[1][chk^1]<<s);
			dp[0][chk]|=(dp[1][chk^1]<<(sz-s));
		}
	}
	ll minn=1e18;
	for(ll i=n/2;i>=0;i--) {
		if(dp[0][chk][i]||dp[1][chk][i]) minn=min(minn,i*i+(n-i)*(n-i));
	}
	printf("%lld\n",ans+minn);
}
il void clean() {
	for(int i=0;i<=1;i++) for(int j=0;j<=1;j++) dp[i][j].set(0);
	for(int i=1;i<=n;i++) {
		head[i]=cvis[i]=dfn[i]=low[i]=siz[i]=vvis[i]=belong[i]=0;
		bin[i].clear(),g[i].clear(),sizbin[i].clear();
	}
	for(int i=2;i<=num_edge;i++) evis[i]=0;
	num_edge=tim=ecc=c=k=m=n=0;
}
int main() {
	int t=read();
	while(t--) {
		mian();
		clean();
	}
	return 0;
}

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

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

Think step by step.

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

1. **Edge Cases**:
   - A fully connected graph (complete graph) where every cell is connected to every other cell.
   - A graph with a single bridge (a critical edge whose removal disconnects the graph).
   - A graph with multiple components (disconnected subgraphs).
   - A graph where the only way to divide it into two complexes is by building new corridors.

2. **Scenarios to Maximize Failure Probability**:
   - A graph with a single bridge, where the bridge is the only way to connect two large complexes.
   - A graph where the only way to divide it is by building new corridors, and the cost of building these corridors is high.

Let's focus on a small-scale test input that fits these criteria:

### Test Input

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

### Explanation

1. **First Test Case**:
   - **n = 5, m = 6, c = 10**
   - **Edges**: (1, 2), (2, 3), (3, 4), (4, 5), (1, 5), (2, 4)
   - This graph is fully connected, and it is impossible to divide it into two complexes with exactly one corridor connecting them. The expected output is `-1`.

2. **Second Test Case**:
   - **n = 6, m = 5, c = 1**
   - **Edges**: (1, 2), (2, 3), (3, 4), (4, 5), (5, 1)
   - This graph forms a cycle, and it can be divided into two complexes with one corridor connecting them. The expected output is the minimum funding required to divide the graph.

3. **Third Test Case**:
   - **n = 6, m = 5, c = 10**
   - **Edges**: (1, 2), (2, 3), (3, 4), (4, 5), (5, 1)
   - This graph is similar to the second test case but with a higher cost for building new corridors. The expected output is the minimum funding required to divide the graph, considering the higher cost of building corridors.

This test input should expose the bug in the provided buggy code by testing various edge cases and scenarios where the division of the prison into two complexes is either impossible or requires careful consideration of the cost of building new corridors.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Min-Fund Prison (Hard)
time_limit_ms: 1500
memory_limit_mb: 256
Description: In the hard version, $$$2 \leq \sum n \leq 10^5$$$ and $$$1 \leq \sum m \leq 5 \times 10^{5}$$$

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 $$$c$$$ ($$$2 \leq n \leq 10^5$$$, $$$1 \leq m \leq 5 \times 10^{5}$$$, $$$1 \leq c \leq 10^9$$$)

$$$m$$$ 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:
4
4 6 5
4 3
2 3
2 4
1 2
4 1
3 1
6 6 2
1 4
2 5
3 6
1 5
3 5
6 5
6 5 7
1 4
2 5
3 6
3 5
6 5
7 5 4
1 4
3 6
3 5
6 5
2 7
output:
-1
20
25
33

Note: In the first test case of the sample input, there is no way to divide the prison according to the Ministry's requirements.

In the second test case, consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{2, 3, 5, 6\}$$$ and $$$\{1, 4\}$$$ cells respectively. There are no new corridors built. The total funding is $$$4^2 + 2^2 = 20$$$. You can verify this is the minimum funding required.

In the third test case, build a corridor between $$$2$$$ and $$$4$$$. Consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{3, 5, 6\}$$$ and $$$\{1, 2, 4\}$$$ cells respectively. The total funding is $$$3^2 + 3^2 + 7 \times 1 = 25$$$. You can verify this is the minimum funding required.

In the fourth test case, build a corridor between $$$2$$$ and $$$4$$$ and between $$$5$$$ and $$$7$$$. Consider the corridor between cells $$$5$$$ and $$$7$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{1, 2, 4, 7\}$$$ and $$$\{3, 5, 6\}$$$ cells respectively. The total funding is $$$4^2 + 3^2 + 4 \times 2 = 33$$$. You can verify this is the minimum funding required.

Note for all test cases that there may be multiple ways to get the same funding but there is no other division which will have a more optimal minimum funding.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pb emplace_back
int n, m, cost, num;
int head[100010], cnt_e = 1;
struct edge {
	int v, nxt;
	edge(int _v = 0, int _nxt = 0) {
		v = _v, nxt = _nxt;
	}
} e[1000010];
void adde(int u, int v) {
	e[++cnt_e] = edge(v, head[u]), head[u] = cnt_e;
}
int dfc, tp, stk[100010], dfn[100010], low[100010], id[100010];
vector < int > vec[100010], edcc[100010];
int sz[100010];
void tarjan(int u, int pre) {
	dfn[u] = low[u] = ++dfc;
	stk[++tp] = u;
	for(int i = head[u]; i; i = e[i].nxt)
		if(pre != (i ^ 1)) {
			int v = e[i].v;
			if(!dfn[v]) tarjan(v, i), low[u] = min(low[u], low[v]);
			else if(low[v]) low[u] = min(low[u], dfn[v]);
		}
	if(dfn[u] == low[u]) {
		int v;
		++num;
		do {
			edcc[num].pb(v = stk[tp--]);
			id[v] = num;
		} while(v ^ u);
	}
}
int c, cnt[100010];
vector < int > as[100010];
void dfs(int u, int ff) {
	sz[u] = edcc[u].size();
	for(int v : vec[u])
		if(v != ff)
			dfs(v, u), sz[u] += sz[v];
	as[c].pb(sz[u]);
}
int fa[100010], ax[500010], ay[500010], bel[500010];
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
vector < int > tmp[100010];
bitset < 1000010 > b1, b2;
void solve() {
	cin >> n >> m >> cost;
	for(int i = 1; i <= n; ++i) fa[i] = i;
	for(int i = 1; i <= m; ++i)
		cin >> ax[i] >> ay[i], adde(ax[i], ay[i]), adde(ay[i], ax[i]), fa[find(ax[i])] = find(ay[i]);
	for(int i = 1; i <= m; ++i) bel[i] = find(ax[i]), tmp[bel[i]].pb(i);
	c = 0;
	for(int i = 1; i <= n; ++i)
		if(!dfn[i]) {
			for(int j = 1; j <= num; ++j) sz[j] = 0;
			num = tp = dfc = 0;
			tarjan(i, 0);
			cnt[++c] = dfc;
			for(int j : tmp[find(i)])
				if(id[ax[j]] != id[ay[j]])
//					cout << " " << id[ax[j]] << " " << id[ay[j]] << "  " << ax[j] << " " << ay[j] << "\n",
					vec[id[ax[j]]].pb(id[ay[j]]), vec[id[ay[j]]].pb(id[ax[j]]);
			dfs(1, 0);
			as[c].pop_back();
			for(int j = 1; j <= num; ++j) edcc[j].clear();
		}
	b1.reset(), b2.reset();
//	for(int i : as[1])
//		cout << "? " << i << "\n";
	b1[0] = 1;
	for(int i = 1; i <= c; ++i) {
		b2 |= b2 << cnt[i];
		for(int j : as[i]) b2 |= b1 << j;
		b1 |= b1 << cnt[i];
	}
	ll ans = 1e18;
	for(int i = 1; i < n; ++i)
		if(b1[i] || b2[i])
			ans = min(ans, i * 1ll * i + (n - i) * 1ll * (n - i));
	if(ans == 1e18)
		return cout << "-1\n", void();
	ans += (c - 1) * 1ll * cost;
	cout << ans << "\n";
	for(int i = 1; i <= n; ++i) dfn[i] = low[i] = head[i] = 0, tmp[i].clear();
	for(int i = 1; i <= c; ++i) as[i].clear();
	cnt_e = 1;
}
int main() {
	ios :: sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	int T;
	cin >> T;
	while(T--)
		solve();
	return 0;
}
/*
1
6 6 2
1 4
2 5
3 6
1 5
3 5
6 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:
```plaintext
1
4 4 10
1 2
2 3
3 4
4 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Min-Fund Prison (Hard)
time_limit_ms: 1500
memory_limit_mb: 256
Description: In the hard version, $$$2 \leq \sum n \leq 10^5$$$ and $$$1 \leq \sum m \leq 5 \times 10^{5}$$$

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 $$$c$$$ ($$$2 \leq n \leq 10^5$$$, $$$1 \leq m \leq 5 \times 10^{5}$$$, $$$1 \leq c \leq 10^9$$$)

$$$m$$$ 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:
4
4 6 5
4 3
2 3
2 4
1 2
4 1
3 1
6 6 2
1 4
2 5
3 6
1 5
3 5
6 5
6 5 7
1 4
2 5
3 6
3 5
6 5
7 5 4
1 4
3 6
3 5
6 5
2 7
output:
-1
20
25
33

Note: In the first test case of the sample input, there is no way to divide the prison according to the Ministry's requirements.

In the second test case, consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{2, 3, 5, 6\}$$$ and $$$\{1, 4\}$$$ cells respectively. There are no new corridors built. The total funding is $$$4^2 + 2^2 = 20$$$. You can verify this is the minimum funding required.

In the third test case, build a corridor between $$$2$$$ and $$$4$$$. Consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{3, 5, 6\}$$$ and $$$\{1, 2, 4\}$$$ cells respectively. The total funding is $$$3^2 + 3^2 + 7 \times 1 = 25$$$. You can verify this is the minimum funding required.

In the fourth test case, build a corridor between $$$2$$$ and $$$4$$$ and between $$$5$$$ and $$$7$$$. Consider the corridor between cells $$$5$$$ and $$$7$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{1, 2, 4, 7\}$$$ and $$$\{3, 5, 6\}$$$ cells respectively. The total funding is $$$4^2 + 3^2 + 4 \times 2 = 33$$$. You can verify this is the minimum funding required.

Note for all test cases that there may be multiple ways to get the same funding but there is no other division which will have a more optimal minimum funding.

Buggy Code:
from collections import defaultdict,deque
T = int(input())
class ConnectedComponents:
  def merge(self, parent, x):
    if parent[x] == x:
      return x
    return self.merge(parent, parent[x])

  def connectedComponents(self, n, edges):
    parent = [i for i in range(n)]
    for x in edges:
      parent[self.merge(parent, x[0])] = self.merge(parent, x[1])

    for i in range(n):
      parent[i] = self.merge(parent, parent[i])

    m = defaultdict(list)
    for i in range(n):
      m[parent[i]].append(i)

    freq = []
    for it in m.items():
      l = it[1]
      freq.append(len(l))

    return freq

while(T):
    [N,M,cost] = list(map(int,input().strip().split(" ")))
    adj = []
    degree = [0]*(N+1)
    for _ in range(M):
        [u,v] = list(map(int,input().strip().split(" ")))
        adj.append([u-1,v-1])
        degree[u-1] += 1
        degree[v-1] += 1

    visited = set()
    obj = ConnectedComponents()
    partitions = obj.connectedComponents(N,adj)
    ans = (len(partitions)-1)*cost
    if ans == 0:
        part1 = max(degree)
        part2 = N-part1
        ans += pow(part1,2) + pow(part2,2)
        if part2 == 1:
            ans = -1
    else:
        part1 = (N//2)
        part2 = N-part1
        ans += pow(part1,2) + pow(part2,2)
    print(ans)
    T -= 1

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

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

Think step by step.

Response:
```plaintext
1
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 (Hard)
time_limit_ms: 1500
memory_limit_mb: 256
Description: In the hard version, $$$2 \leq \sum n \leq 10^5$$$ and $$$1 \leq \sum m \leq 5 \times 10^{5}$$$

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 $$$c$$$ ($$$2 \leq n \leq 10^5$$$, $$$1 \leq m \leq 5 \times 10^{5}$$$, $$$1 \leq c \leq 10^9$$$)

$$$m$$$ 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:
4
4 6 5
4 3
2 3
2 4
1 2
4 1
3 1
6 6 2
1 4
2 5
3 6
1 5
3 5
6 5
6 5 7
1 4
2 5
3 6
3 5
6 5
7 5 4
1 4
3 6
3 5
6 5
2 7
output:
-1
20
25
33

Note: In the first test case of the sample input, there is no way to divide the prison according to the Ministry's requirements.

In the second test case, consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{2, 3, 5, 6\}$$$ and $$$\{1, 4\}$$$ cells respectively. There are no new corridors built. The total funding is $$$4^2 + 2^2 = 20$$$. You can verify this is the minimum funding required.

In the third test case, build a corridor between $$$2$$$ and $$$4$$$. Consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{3, 5, 6\}$$$ and $$$\{1, 2, 4\}$$$ cells respectively. The total funding is $$$3^2 + 3^2 + 7 \times 1 = 25$$$. You can verify this is the minimum funding required.

In the fourth test case, build a corridor between $$$2$$$ and $$$4$$$ and between $$$5$$$ and $$$7$$$. Consider the corridor between cells $$$5$$$ and $$$7$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{1, 2, 4, 7\}$$$ and $$$\{3, 5, 6\}$$$ cells respectively. The total funding is $$$4^2 + 3^2 + 4 \times 2 = 33$$$. You can verify this is the minimum funding required.

Note for all test cases that there may be multiple ways to get the same funding but there is no other division which will have a more optimal minimum funding.

Buggy Code:
from collections import defaultdict,deque
T = int(input())
def findComponents(node,adj,visited):
    q = deque([node])
    visited.add(node)
    count = 0
    while(q):
        v = q.popleft()
        count += 1
        for nn in adj[v]:
            if nn not in visited:
                visited.add(nn)
                q.append(nn)
    return count

while(T):
    [N,M,cost] = list(map(int,input().strip().split(" ")))
    adj = defaultdict(list)
    degree = [0]*(N+1)
    for _ in range(M):
        [u,v] = list(map(int,input().strip().split(" ")))
        adj[u].append(v)
        adj[v].append(u)
        degree[u] += 1
        degree[v] += 1

    visited = set()
    partitions = []
    for i in range(1,N):
        if i not in visited:
            val = findComponents(i,adj,visited)
            partitions.append(val)

    ans = (len(partitions)-1)*cost
    part1 = part2 = 0
    if ans == 0:
        part1 = max(degree)
        part2 = N-part1
        ans += pow(part1,2) + pow(part2,2)
        if part2 == 1:
            ans = -1
    else:
        part1 = (N//2)
        part2 = N-part1
        ans += pow(part1,2) + pow(part2,2)
    print(ans)
    T -= 1

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

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

Think step by step.

Response:
```plaintext
1
5 4 3
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 (Hard)
time_limit_ms: 1500
memory_limit_mb: 256
Description: In the hard version, $$$2 \leq \sum n \leq 10^5$$$ and $$$1 \leq \sum m \leq 5 \times 10^{5}$$$

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 $$$c$$$ ($$$2 \leq n \leq 10^5$$$, $$$1 \leq m \leq 5 \times 10^{5}$$$, $$$1 \leq c \leq 10^9$$$)

$$$m$$$ 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:
4
4 6 5
4 3
2 3
2 4
1 2
4 1
3 1
6 6 2
1 4
2 5
3 6
1 5
3 5
6 5
6 5 7
1 4
2 5
3 6
3 5
6 5
7 5 4
1 4
3 6
3 5
6 5
2 7
output:
-1
20
25
33

Note: In the first test case of the sample input, there is no way to divide the prison according to the Ministry's requirements.

In the second test case, consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{2, 3, 5, 6\}$$$ and $$$\{1, 4\}$$$ cells respectively. There are no new corridors built. The total funding is $$$4^2 + 2^2 = 20$$$. You can verify this is the minimum funding required.

In the third test case, build a corridor between $$$2$$$ and $$$4$$$. Consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{3, 5, 6\}$$$ and $$$\{1, 2, 4\}$$$ cells respectively. The total funding is $$$3^2 + 3^2 + 7 \times 1 = 25$$$. You can verify this is the minimum funding required.

In the fourth test case, build a corridor between $$$2$$$ and $$$4$$$ and between $$$5$$$ and $$$7$$$. Consider the corridor between cells $$$5$$$ and $$$7$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{1, 2, 4, 7\}$$$ and $$$\{3, 5, 6\}$$$ cells respectively. The total funding is $$$4^2 + 3^2 + 4 \times 2 = 33$$$. You can verify this is the minimum funding required.

Note for all test cases that there may be multiple ways to get the same funding but there is no other division which will have a more optimal minimum funding.

Buggy Code:
from collections import defaultdict,deque
T = int(input())
def findComponents(node,adj,visited):
    q = deque([node])
    visited.add(node)
    count = 0
    while(q):
        v = q.popleft()
        count += 1
        for nn in adj[v]:
            if nn not in visited:
                visited.add(nn)
                q.append(nn)
    return count

def subarray_closest_sum(arr, n, k):
    start = end = 0
    curr_sum = arr[0]
    min_diff = abs(curr_sum - k)
    val = k-curr_sum

    while end < n - 1:
        if curr_sum < k:
            end += 1
            curr_sum += arr[end]
        else:
            curr_sum -= arr[start]
            start += 1
        if abs(curr_sum - k) < min_diff:
            min_diff = abs(curr_sum - k)
            val = k-curr_sum
    return val

while(T):
    [N,M,cost] = list(map(int,input().strip().split(" ")))
    adj = defaultdict(list)
    degree = [0]*(N+1)
    for _ in range(M):
        [u,v] = list(map(int,input().strip().split(" ")))
        adj[u].append(v)
        adj[v].append(u)
        degree[u] += 1
        degree[v] += 1

    visited = set()
    partitions = []
    for i in range(1,N):
        if i not in visited:
            val = findComponents(i,adj,visited)
            partitions.append(val)

    ans = (len(partitions)-1)*cost
    part1 = part2 = 0
    if ans == 0:
        part1 = max(degree)
        part2 = N-part1
        ans += pow(part1,2) + pow(part2,2)
        if part2 == 1:
            ans = -1
    else:
        #min_diff_bw_parts = subarray_closest_sum(partitions,len(partitions),N//2)
        part1 = (N//2)
        part2 = N-part1
        ans += pow(part1,2) + pow(part2,2)
    print(ans)
    T -= 1

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Min-Fund Prison (Hard)
time_limit_ms: 1500
memory_limit_mb: 256
Description: In the hard version, $$$2 \leq \sum n \leq 10^5$$$ and $$$1 \leq \sum m \leq 5 \times 10^{5}$$$

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 $$$c$$$ ($$$2 \leq n \leq 10^5$$$, $$$1 \leq m \leq 5 \times 10^{5}$$$, $$$1 \leq c \leq 10^9$$$)

$$$m$$$ 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:
4
4 6 5
4 3
2 3
2 4
1 2
4 1
3 1
6 6 2
1 4
2 5
3 6
1 5
3 5
6 5
6 5 7
1 4
2 5
3 6
3 5
6 5
7 5 4
1 4
3 6
3 5
6 5
2 7
output:
-1
20
25
33

Note: In the first test case of the sample input, there is no way to divide the prison according to the Ministry's requirements.

In the second test case, consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{2, 3, 5, 6\}$$$ and $$$\{1, 4\}$$$ cells respectively. There are no new corridors built. The total funding is $$$4^2 + 2^2 = 20$$$. You can verify this is the minimum funding required.

In the third test case, build a corridor between $$$2$$$ and $$$4$$$. Consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{3, 5, 6\}$$$ and $$$\{1, 2, 4\}$$$ cells respectively. The total funding is $$$3^2 + 3^2 + 7 \times 1 = 25$$$. You can verify this is the minimum funding required.

In the fourth test case, build a corridor between $$$2$$$ and $$$4$$$ and between $$$5$$$ and $$$7$$$. Consider the corridor between cells $$$5$$$ and $$$7$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{1, 2, 4, 7\}$$$ and $$$\{3, 5, 6\}$$$ cells respectively. The total funding is $$$4^2 + 3^2 + 4 \times 2 = 33$$$. You can verify this is the minimum funding required.

Note for all test cases that there may be multiple ways to get the same funding but there is no other division which will have a more optimal minimum funding.

Buggy Code:
#include<bits/stdc++.h>
#define maxn 200010
#define int long long
inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();return x*f;}
void write(int x){if(x<0)putchar('-'),x=-x;if(x>9)write(x/10);putchar(x%10+'0');return;}
using namespace std;
int n,m,c,anss=1e18,cnt,fa[maxn],siz[maxn],vis[maxn],sz[maxn];
int dfn[maxn],low[maxn],fl=0;
vector<int>e[maxn];
bitset<maxn>b,ans,f,g;
int find(int x){
	return x==fa[x]?x:fa[x]=find(fa[x]);
}
void dfs(int x,int fa){
	dfn[x]=low[x]=++cnt;
	sz[x]=1;
	for(int i:e[x])
	 if(!dfn[i]){
	 	dfs(i,x);
	 	sz[x]+=sz[i];
	 	low[x]=min(low[x],low[i]);
	 }
	 else if(i!=fa) low[x]=min(low[x],dfn[i]);
	if(low[x]==dfn[x]){
//		cout<<x;
		ans|=(f<<sz[x]);
		ans|=(f<<(siz[find(x)]-sz[x]));
	}
	else fl=1;
}
void solve(){
	vector<int>rt;
	b=ans=fl=0;
	cnt=0;
	anss=1e18;
	scanf("%lld%lld%lld",&n,&m,&c);
	for(int i=1;i<=n;i++) fa[i]=i;
	for(int i=1;i<=m;i++){
		int u,v;
		scanf("%lld%lld",&u,&v);
		e[u].push_back(v);
		e[v].push_back(u);
		fa[find(u)]=find(v);
	}
	for(int i=1;i<=n;i++) siz[find(i)]++;
	for(int i=1;i<=n;i++)
	 if(!vis[find(i)]){
	 	int len=rt.size(),u=find(i);
		rt.push_back(u),vis[u]=1;
		len++;
	}
	int len=rt.size();
	f[0]=1;
	for(int i=0;i<len;i++){
		dfs(rt[i],0);
		f=ans;
	}
	for(int i=0;i<=n;i++)
	 if(ans[i]){
	 	anss=min(anss,i*i+(n-i)*(n-i)+(len-1)*c);
	 }
	if(anss==n*n) anss=-1;
	printf("%lld\n",anss);
	for(int i=1;i<=n;i++) e[i].clear(),siz[i]=vis[i]=dfn[i]=low[i]=0;
}
signed main(){
	int T;
	scanf("%lld",&T);
	while(T--) solve();
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
5 10 1
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
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 (Hard)
time_limit_ms: 1500
memory_limit_mb: 256
Description: In the hard version, $$$2 \leq \sum n \leq 10^5$$$ and $$$1 \leq \sum m \leq 5 \times 10^{5}$$$

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 $$$c$$$ ($$$2 \leq n \leq 10^5$$$, $$$1 \leq m \leq 5 \times 10^{5}$$$, $$$1 \leq c \leq 10^9$$$)

$$$m$$$ 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:
4
4 6 5
4 3
2 3
2 4
1 2
4 1
3 1
6 6 2
1 4
2 5
3 6
1 5
3 5
6 5
6 5 7
1 4
2 5
3 6
3 5
6 5
7 5 4
1 4
3 6
3 5
6 5
2 7
output:
-1
20
25
33

Note: In the first test case of the sample input, there is no way to divide the prison according to the Ministry's requirements.

In the second test case, consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{2, 3, 5, 6\}$$$ and $$$\{1, 4\}$$$ cells respectively. There are no new corridors built. The total funding is $$$4^2 + 2^2 = 20$$$. You can verify this is the minimum funding required.

In the third test case, build a corridor between $$$2$$$ and $$$4$$$. Consider the corridor between cells $$$1$$$ and $$$5$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{3, 5, 6\}$$$ and $$$\{1, 2, 4\}$$$ cells respectively. The total funding is $$$3^2 + 3^2 + 7 \times 1 = 25$$$. You can verify this is the minimum funding required.

In the fourth test case, build a corridor between $$$2$$$ and $$$4$$$ and between $$$5$$$ and $$$7$$$. Consider the corridor between cells $$$5$$$ and $$$7$$$ as the connection between the $$$2$$$ complexes consisting of $$$\{1, 2, 4, 7\}$$$ and $$$\{3, 5, 6\}$$$ cells respectively. The total funding is $$$4^2 + 3^2 + 4 \times 2 = 33$$$. You can verify this is the minimum funding required.

Note for all test cases that there may be multiple ways to get the same funding but there is no other division which will have a more optimal minimum funding.

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

using ll = long long;
using ull = unsigned long long;
using fl = long double;
template<typename T>
using min_heap = priority_queue<T, vector<T>, greater<T>>;
template<typename S, typename T>
void xmin(S&a, T const&b){if(b<a) a=b;}
template<typename S, typename T>
void xmax(S&a, T const&b){if(b>a) a=b;}

namespace debug_{

template<size_t N> struct priority : priority<N-1> {};
template<> struct priority<0> {};

template<bool enabled> struct Debug;
template<typename T, typename SFINAE = void> struct Printer : enable_if<false> {};
// pair and tuple
template<typename S, typename T>
struct Printer<pair<S, T> > : enable_if<true> {
    template<typename D>
    static void print(D const&d, pair<S, T> const&val){
        d << "(" << val.first << ", " << val.second << ")";
    }
};
template<typename... S>
struct Printer<tuple<S ...> > : enable_if<true> {
    template<size_t i, size_t n, typename D>
    static void print_rec(D const&d, tuple<S ...> const&val){
        d << get<i>(val);
        if(i+1 < n) {
            d << ", ";
            print_rec<i+(i+1<n), n>(d, val);
        }
    }
    template<typename D>
    static void print(D const&d, tuple<S ...> const&val){
        d << "(";
        print_rec<0, tuple_size<tuple<S...> >::value >(d, val);
        d << ")";
    }
};
template<>
struct Printer<tuple<> > : enable_if<true> {
    template<typename D>
    static void print(D const&d, tuple<> const&){
        d << "()";
    }
};
// non-iterable container
template<typename T>
struct Hacked_Container : T{
    static auto get(T const& val){
        return val.*&Hacked_Container::c;
    }
};
template<typename T> struct is_hackable_container : false_type{};
template<typename... T> struct is_hackable_container<stack<T...> > : true_type{};
template<typename... T> struct is_hackable_container<queue<T...> > : true_type{};
template<typename... T> struct is_hackable_container<priority_queue<T...> > : true_type{};
template<typename T>
struct Printer<T, enable_if_t<is_hackable_container<T>::value> >  : enable_if<true> {
    template<typename D>
    static void print(D const&d, T const&s){
        d << Hacked_Container<T>::get(s);
    }
};
// custom printable struct
template<typename T>
struct Printer<T,enable_if_t<sizeof((declval<T>().debug_print_(2), 1)) != -1>> : enable_if<true>{
    template<typename D>
    static void print(D const&d, T const&val){
        val.debug_print_(d);
    }
};

template <>
struct Debug<false>{
    template<typename T, typename = enable_if_t<is_same<int, decltype((declval<Debug<true>>() << declval<T>()), (int)1)>::value > >
    friend const Debug& operator<<(Debug const&o, T const&){
        return o;
    }
    template<typename... Args>
    void pprint_all(string const&names, Args const& ... vals){}
};
template <>
struct Debug<true>{
    template<typename T, typename = decltype(cerr << declval<T>())>
    void print(T const&val, priority<30>) const {
        cerr << val;
    }
    template<typename T, typename = decltype((begin(declval<T>()), end(declval<T>())))>
    void print(T const&val, priority<20>) const {
        *this << "[";
        bool first = true;
        for(auto const&e:val){
            if(!first) *this << ", ";
            *this << e;
            first = false;
        }
        *this << "]";
    }
    template<typename T, typename = typename Printer<T>::type>
    void print(T const&val, priority<50>) const {
        Printer<T>::print(*this, val);
    }
    void pprint_all_rec(string const&, size_t) const {}
    template<typename T, typename... Args>
    void pprint_all_rec(string const&names, size_t pos, T const&val, Args const& ... rest) const {
        if(pos != 0) *this << ", ";
        assert(pos != string::npos);
        size_t next = names.find(",", pos);
        if(next == string::npos){
                *this << names.substr(pos);
        } else {
            *this << names.substr(pos, next-pos);
            ++next;
        }
        *this << "=" << val;
        pprint_all_rec(names, next, rest...);
    }

    template<typename... Args>
    void pprint_all(string const&names, Args const& ... vals) const {
        *this << "(";
        pprint_all_rec(names, 0, vals...);
        *this << ")";
    }
    template<typename T, typename = decltype(declval<Debug>().print(declval<T>(), priority<99>{}))>
    friend const Debug& operator<<(Debug const &o, T const& val){
        o.print(val, priority<99>{});
        return o;
    }
};
}

#define make_printable(...)\
template<typename T>\
void debug_print_(T const& d)const{\
    d.pprint_all(#__VA_ARGS__, __VA_ARGS__);\
}
#ifndef LOCAL_RUN
debug_::Debug<false> debug;
#else

 debug_::Debug<true> debug;
// debug_::Debug<false> debug; // disable debug printing

#endif

#define named(x) string(#x) << " : " <<  x

struct Artpoints{
    typedef pair<int, int> PII;

    int N;
    vector<vector<int>> l;
    vector <PII> brid, non;
    vector<int> vis, seen, par, ap;
    int cnt, root;

    Artpoints(int n) : l(n), vis(n), seen(n), par(n), ap(n){
        N = n;

    }

    void dfs(int x){
      if(vis[x] != -1)
        return;
      vis[x] = seen[x] = cnt++;

      int adj = 0;
      for(int i = 0; i < (int)l[x].size(); i++){
        int v = l[x][i];
        if(par[x] == v)
          continue;
        if(vis[v] == -1){
          adj++;
          par[v] = x;
          dfs(v);
          seen[x] = min(seen[x], seen[v]);
          if(seen[v] >= vis[x] && x != root)
            ap[x] = 1;
          if(seen[v] == vis[v])
            brid.push_back(make_pair(v, x));
          else non.emplace_back(v, x);
        }
        else{
          seen[x] = min(seen[x], vis[v]);
          seen[v] = min(seen[x], seen[v]);
        }
      }
      if(x == root) ap[x] = (adj>1);
    }

    void bridges(){
      brid.clear();
      for(int i = 0; i < N; i++){
        vis[i] = seen[i] = par[i] = -1;
        ap[i] = 0;
      }
      cnt = 0;
      for(int i = 0; i < N; i++)
        if(vis[i] == -1){
          root = i;
          dfs(i);
        }
    }
};

struct DSU{
    DSU(int n) : p(n, -1) {}
    int f(int x){
        return p[x] < 0 ? x : p[x] = f(p[x]);
    }
    void u(int a, int b){
        a=f(a);
        b=f(b);
        if(a==b) return;
        p[a] += p[b];
        p[b] = a;
    }
    vector<int> p;
};

struct Component{
    int total = 0;
    vector<int> sub;
};

const ll inf = 1.01e18;
constexpr uint64_t mod = (1ull<<61) - 1;
/*
static constexpr uint64_t add(uint64_t a, uint64_t b){
    a+=b+1;
    a = (a&mod) + (a>>61);
    return a-1;
}
static constexpr uint64_t sub(uint64_t a, uint64_t b){
    return add(a, mod-b);
}
static constexpr uint64_t modmul(uint64_t a, uint64_t b){
    uint64_t l1 = (uint32_t)a, h1 = a>>32, l2 = (uint32_t)b, h2 = b>>32;
    uint64_t l = l1*l2, m = l1*h2 + l2*h1, h = h1*h2;
    uint64_t ret = (l&mod) + (l>>61) + (h << 3) + (m >> 29) + (m << 35 >> 3) + 1;
    ret = (ret & mod) + (ret>>61);
    ret = (ret & mod) + (ret>>61);
    return ret-1;
}*/
static constexpr uint64_t add(uint64_t a, uint64_t b){
    return a+b;
}
static constexpr uint64_t sub(uint64_t a, uint64_t b){
    return a-b;
}
static constexpr uint64_t modmul(uint64_t a, uint64_t b){
    return a*b;
}

void solve(){
    /// SOLVE HERE
    int n, m;
    ll c;
    cin >> n >> m >> c;
    vector<vector<int> > g(n);
    DSU c1(n), c2(n);
    Artpoints art(n);
    for(int i=0; i<m; ++i){
        int a, b;
        cin >> a >> b;
        --a; --b;
        c1.u(a, b);
        g[a].push_back(b);
        g[b].push_back(a);
    }
    art.l = std::move(g);
    art.bridges();
    for(auto &[u, v] : art.non){
        c2.u(u, v);
    };
    vector<vector<int> > g2(n);
    for(auto &e : art.brid){
        const int u = c2.f(e.first), v = c2.f(e.second);
        g2[u].push_back(v);
        g2[v].push_back(u);
    }
    auto dfs = [&](auto dfs, int u, int p, Component &c) -> int{
        const int me = -c2.p[u];
        int total = me;
        for(int v : g2[u]) if(v != p){
            total += dfs(dfs, v, u, c);
        }
        c.sub.push_back(total);
        return total;
    };
    vector<Component> components;
    vector<vector<int> > sides(n+1);
    vector<int> cnt(n+1);
    for(int i=0; i<n; ++i) if(c1.p[i] < 0){
        components.emplace_back();
        auto &c = components.back();
        dfs(dfs, c2.f(i), -1, c);
        c.total = c.sub.back();
        c.sub.pop_back();
        ++cnt[c.total];
        auto &s = sides[c.total];
        s.insert(s.end(), c.sub.begin(), c.sub.end());
    }
    // debug << named(cnt) << "\n" << named(sides) << "\n";
    const int U = n/2+5;
    vector<uint64_t> dp(U+1);
    vector<char> can(U+1);
    dp[0] = 1;
    mt19937 rng(5152819052190ull);
    auto r = [&](){
        return 1ull|uniform_int_distribution<uint64_t>(1, 1e18)(rng);
    };
    vector<uint64_t> ha(U+1);
    for(auto &e : ha) e = r();
    auto A = [&](int x){
        auto const y = ha[x];
        for(int i=U-x; i>=0; --i){
            dp[i+x] = add(dp[i+x], modmul(dp[i], y));
        }
    };
    auto S = [&](int x){
        auto const y = ha[x];
        for(int i=0; i<=U-x; ++i){
            dp[i+x] = sub(dp[i+x], modmul(dp[i], y));
        }
    };
    auto P = [&](){
        for(int i=0; i<=U; ++i){
            can[i] |= !!dp[i];
        }
    };

    for(int i=0; i<=n; ++i){
        while(cnt[i] > 3){
            cnt[i] -= 2;
            cnt[2*i] += 1;
        }
        auto &s = sides[i];
        sort(s.begin(), s.end());
        s.erase(unique(s.begin(), s.end()), s.end());
        for(int j=0; j<cnt[i]; ++j) A(i);
    }

    ll ret = inf;
    // bridge is a new edge
    if(components.size() > 1){
        P();
        for(int i=0; i<=U; ++i) if(can[i]){
            const ll A = i;
            const ll B = n - i;
            xmin(ret, A*A + B*B);
        }
    }

    for(int i=0; i<=n; ++i) if(cnt[i]){
        S(i);
        for(int e : sides[i]){
            A(e);
            P();
            S(e);
        }
        A(i);
    }
    for(int i=0; i<=U; ++i) if(can[i]){
        const ll A = i;
        const ll B = n - i;
        xmin(ret, A*A + B*B);
    }
    ret += ((ll)components.size() - 1) * c;
    if(ret >= inf) ret = -1;
    cout << ret << "\n";

}

signed gen(int T){
    mt19937 rng(43151);
    auto get_rand = [&](int64_t l, int64_t r){
        return uniform_int_distribution<int64_t>(l, r)(rng);
    }; (void) get_rand;
    auto get_double = [&](double l, double r){
        return uniform_real_distribution<double>(l, r)(rng);
    };  (void) get_double;
    ofstream o("gen.txt");
    o << T << "\n";
    for(int cas=0;cas<T;++cas){
        /// GEN HERE
        const int n = 1e1;
        vector<int> a(n);
        iota(a.begin(), a.end(), 1);
        shuffle(a.begin(), a.end(), rng);
        o << n << " " << n-1 << " " << (int)1e9 << "\n";
        for(int i=1; i<n; ++i){
            o << a[i] << " " << a[i-1] << "\n";
        }

        o << "\n";
    }
    o << endl;
    o.close();
    return 0;
}

signed main()
{
    #ifdef LOCAL_RUN
    freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    cin.exceptions(ios::badbit | ios::eofbit | ios::failbit);
    cin.tie(nullptr); ios_base::sync_with_stdio(false);
    int TTT; cin >> TTT;
    if(TTT < 0) return gen(-TTT);
    while(TTT--){
    #else
    cin.tie(nullptr); ios_base::sync_with_stdio(false);
    int TTT; cin >> TTT;
    while(TTT--){
    #endif // LOCAL_RUN

    solve();

    #ifdef LOCAL_RUN
    cout << flush;
    #endif // LOCAL_RUN
    }
    return 0;
}

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

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

Think step by step.

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

1. **Understand the Problem**: The problem requires dividing the prison into two complexes with exactly one corridor connecting them, and the goal is to minimize the total funding required.

2. **Identify Potential Bugs**:
   - The code might not handle cases where the graph is already connected and cannot be divided into two complexes.
   - The code might not handle cases where the graph is disconnected and cannot be divided as required.
   - The code might have issues with the dynamic programming (DP) part, especially in handling large numbers and edge cases.

3. **Create Test Cases**:
   - **Case 1**: A fully connected graph that cannot be divided into two complexes.
   - **Case 2**: A graph with a single bridge that can be divided into two complexes.
   - **Case 3**: A graph with multiple components that cannot be divided as required.
   - **Case 4**: A graph with a large number of vertices and edges to test performance and correctness.

4. **Generate the Test Input**:
   - For simplicity and to ensure the test input is small and complete, we will focus on the first two cases.

Here is the test input:

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

**Explanation**:
- **First Test Case**:
  - **n = 4, m = 6, c = 5**: A fully connected graph with 4 vertices and 6 edges.
  - **Edges**: (4, 3), (2, 3), (2, 4), (1, 2), (4, 1), (3, 1)
  - **Expected Output**: -1 (The graph cannot be divided into two complexes with exactly one corridor connecting them.)

- **Second Test Case**:
  - **n = 5, m = 4, c = 2**: A graph with 5 vertices and 4 edges, forming a linear structure.
  - **Edges**: (1, 2), (2, 3), (3, 4), (4, 5)
  - **Expected Output**: 13 (The graph can be divided into two complexes with exactly one corridor connecting them. The complexes will have 2 and 3 vertices, and the total funding will be \(2^2 + 3^2 = 13\).)

This test input should help in identifying and exposing potential bugs in the provided code.