**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Selection Sort
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Every student enrolled in the algorithms course is required to submit an assignment this week. The task is to implement an $$$O(n^2)$$$-time algorithm to sort $$$n$$$ given integers in non-decreasing order. Alice has already completed her assignment, and her implementation is shown below.

While you have access to Alice's code, you prefer not to simply copy it. Instead, you want to use Alice's sorting function as a building block for your own solution. There are two ways as listed below you can utilize her function, but each of them can be applied at most once. The order in which these two operations are invoked can be arbitrary.

- Prefix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s, i\texttt{)}$$$. This sorts the first $$$i$$$ elements in the array $$$s$$$.
- Suffix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s+n-i, i\texttt{)}$$$. This sorts the last $$$i$$$ elements in the array $$$s$$$.

Due to the time complexity of the sorting algorithm, the cost of performing either a prefix or suffix sort is $$$i^2$$$, where $$$i$$$ is the length of the chosen subarray. Your goal is to determine the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

For example, Let $$$s=[3,2,5,5,4,1]$$$. We can first perform a suffix sort of length $$$4$$$, and the array becomes $$$[3,2,1,4,5,5]$$$. Then, we perform a prefix sort of length $$$3$$$, and the array becomes $$$[1,2,3,4,5,5]$$$, which is a sorted array. The cost is $$$4^2+3^2=25$$$. Here is another example, let $$$s=[4,3,2,1]$$$. We can complete the sorting by performing only a prefix sort of length $$$4$$$, and the cost is $$$4^2=16$$$.

Input: The first line contains exactly one integer $$$n$$$ which indicates the number of integers in the array $$$s$$$. The second line contains the $$$n$$$ integers in $$$s=[s_0, s_1, \ldots, s_{n-1}]$$$.

- $$$1 \le n \le 10^6$$$
- For all $$$i$$$ ($$$0\le i < n$$$), $$$0\le s_i < 2^{31}-1$$$.

Output: Output an integer on a line, indicating the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

Examples:
input:
6
3 2 5 5 4 1
output:
25

input:
4
4 3 2 1
output:
16

Note:

Buggy Code:
//   _____     __          __          __
//  |  _  \    \ \        /  \        / /
//  | |_| |     \ \      / /\ \      / /
//  |  _  /      \ \    / /  \ \    / /
//  | | \ \       \ \  / /    \ \  / /
//  |_|  \_\       \_\/_/      \_\/_/
#include <bits/stdc++.h>
using namespace std;
// #define int long long
#define ll long long
#define endl '\n'
#define all(a) a.begin(), a.end()
#define pb push_back
#define x first
#define y second
#define debug(_x) cout << #_x << '=' << _x << endl
typedef pair<int, int> PII;
using i64 = long long;
const int mod = 1e9 + 7;
const int N = 1e6 + 10;

i64 a[N], dp[N], f[N], p[N];

void solve()
{
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        p[i] = i;
        dp[i] = 0;
    }
    sort(p + 1, p + 1 + n, [&](int x, int y)
         { return a[x] == a[y] ? x < y : a[x] < a[y]; });
    for (int i = n; i >= 1; i--)
    {
        f[i] = p[i] == i ? f[i + 1] + 1 : 0;
    }
    i64 mi = (n - f[1]) * (n - f[1]), cnt = 0;
    for (int i = 1; i <= n; i++)
    {
        dp[p[i]] = 1;
        while (dp[cnt + 1])
            cnt++;
        if (cnt == i)
        {
            mi = min(mi, i * i + (n - cnt - f[i + 1]) * (n - cnt - f[i + 1]));
        }
        else
            mi = min(mi, i * i + (n - cnt) * (n - cnt));
    }
    i64 ans = mi;
    reverse(a + 1, a + 1 + n);
    sort(p + 1, p + 1 + n, [&](int x, int y)
         { return a[x] == a[y] ? x < y : a[x] > a[y]; });
    for (int i = n; i >= 1; i--)
    {
        f[i] = p[i] == i ? f[i + 1] + 1 : 0;
        dp[i] = 0;
    }
    mi = (n - f[1]) * (n - f[1]), cnt = 0;
    for (int i = 1; i <= n; i++)
    {
        dp[p[i]] = 1;
        while (dp[cnt + 1])
            cnt++;
        if (cnt == i)
        {
            mi = min(mi, i * i + (n - cnt - f[i + 1]) * (n - cnt - f[i + 1]));
        }
        else
            mi = min(mi, i * i + (n - cnt) * (n - cnt));
    }
    ans = min(ans, mi);
    cout << ans;
}

signed main()
{
    ios::sync_with_stdio(0);
    // ios::sync_with_stdio(false);
    // cin.tie(nullptr);
    int _ = 1;
    // cin >> _;

    while (_--)
        solve();

    return 0;
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided buggy code, we need to consider edge cases that could cause the algorithm to fail. Here are some potential edge cases to consider:

1. **All elements are the same**: This can simplify the sorting process but might expose issues in the logic that handles identical elements.
2. **Already sorted array**: This should have a minimal cost, but the code might not handle it correctly.
3. **Reverse sorted array**: This should have a cost of \(n^2\), but the code might not handle it correctly.
4. **Array with a single element**: This is a trivial case but important to ensure the code handles it correctly.
5. **Array with a few elements**: This can help in understanding how the code handles small inputs.

Given the constraints and the need for a small-scale test input, let's consider an array that is already sorted but has a few elements to ensure it's non-trivial. This will help us check if the code correctly identifies the minimal cost for such an input.

Here is the test input:

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

This input is already sorted, and the expected output should be the minimum cost to sort it, which is 0 (since no sorting is needed). If the code correctly handles this case, it should output 0. If it fails, it will expose a bug in the logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Selection Sort
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Every student enrolled in the algorithms course is required to submit an assignment this week. The task is to implement an $$$O(n^2)$$$-time algorithm to sort $$$n$$$ given integers in non-decreasing order. Alice has already completed her assignment, and her implementation is shown below.

While you have access to Alice's code, you prefer not to simply copy it. Instead, you want to use Alice's sorting function as a building block for your own solution. There are two ways as listed below you can utilize her function, but each of them can be applied at most once. The order in which these two operations are invoked can be arbitrary.

- Prefix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s, i\texttt{)}$$$. This sorts the first $$$i$$$ elements in the array $$$s$$$.
- Suffix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s+n-i, i\texttt{)}$$$. This sorts the last $$$i$$$ elements in the array $$$s$$$.

Due to the time complexity of the sorting algorithm, the cost of performing either a prefix or suffix sort is $$$i^2$$$, where $$$i$$$ is the length of the chosen subarray. Your goal is to determine the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

For example, Let $$$s=[3,2,5,5,4,1]$$$. We can first perform a suffix sort of length $$$4$$$, and the array becomes $$$[3,2,1,4,5,5]$$$. Then, we perform a prefix sort of length $$$3$$$, and the array becomes $$$[1,2,3,4,5,5]$$$, which is a sorted array. The cost is $$$4^2+3^2=25$$$. Here is another example, let $$$s=[4,3,2,1]$$$. We can complete the sorting by performing only a prefix sort of length $$$4$$$, and the cost is $$$4^2=16$$$.

Input: The first line contains exactly one integer $$$n$$$ which indicates the number of integers in the array $$$s$$$. The second line contains the $$$n$$$ integers in $$$s=[s_0, s_1, \ldots, s_{n-1}]$$$.

- $$$1 \le n \le 10^6$$$
- For all $$$i$$$ ($$$0\le i < n$$$), $$$0\le s_i < 2^{31}-1$$$.

Output: Output an integer on a line, indicating the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

Examples:
input:
6
3 2 5 5 4 1
output:
25

input:
4
4 3 2 1
output:
16

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
signed main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);

	int n;
	cin >> n;
	vector<int> a(n),b(n);
	for(int i=0;i<n;i++) cin >> a[i], b[i] = a[i];
	sort(b.begin(),b.end());
	vector<int> cnt(n+1,0),pos1(n+1,0),pos2(n+1,0);
	for(int i=0;i<n;i++) {
		a[i] = lower_bound(b.begin(),b.end(),a[i]) - b.begin() + 1;
		cnt[a[i]]++;
		a[i] += cnt[a[i]] - 1;
	}
	set<int> s2;
	long long ans = n * n;
	for(int i=0;i<n;i++) s2.insert(a[i]);
	for(int i=0;i<n-1;i++) {
		s2.erase(a[i]);
		int num = *s2.begin();
		if( num <= i + 1 ) {
			long long val = (long long)( i + 1 ) * ( i + 1 ) + ( n - num + 1 ) * ( n - num + 1 );
			ans = min(ans, val);
		}
		else {
			ans = min(ans,(long long)(i+1)*(i+1)+(long long)(n-i-1)*(n-i-1));
			pos1[i] = 1;
		}
	}

	set<int> s3;
	for(int i=0;i<n;i++) s3.insert(a[i]);
	for(int i=n-1;i>=1;i--) {
		s3.erase(a[i]);
		int num = *s3.rbegin();
		if( num >= i + 1 ) {
			long long val = (long long)( n - i ) * ( n - i ) + num * num;
			ans = min(ans, val);
		}
		else {
			ans = min(ans,(long long)(n-i)*(n-i)+(long long)i*i);
			pos2[i] = 1;
		}
	}

	for(int i=0;i<n;i++) {
		if( a[i] == i + 1 ) ans = min(ans,(long long)(n-i-1)*(n-i-1));
		else break;
	}
	for(int i=n-1;i>=0;i--) {
		if( a[i] == i + 1 ) ans = min(ans,(long long)i*i);
		else break;
	}

	int idx1 = 0, idx2 = 0;
	while( idx1 < n ) {
		while( a[idx1] == idx1 + 1 && idx2 + 1 < n && ( a[idx2] == idx2 + 1 ) ) {
			if( ( idx1 == 0 || pos1[idx1-1] == 1 ) && ( idx2 == n - 1 || pos2[idx2+1] == 1 ) ) {
				ans = min(ans,(long long)idx1*idx1+(long long)(n-idx2-1)*(n-idx2-1));
			}
			idx2++;
		}
		idx1 = idx2 + 1, idx2 = idx1;
	}

	cout << ans << "\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
4
1 2 3 4
```

This test input is designed to expose the bug in the provided buggy code. The input array is already sorted, which is an edge case that can often reveal issues in sorting algorithms. The expected output should be 0, as no sorting is needed. However, the buggy code might not handle this case correctly, especially in the logic that calculates the minimum cost.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Selection Sort
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Every student enrolled in the algorithms course is required to submit an assignment this week. The task is to implement an $$$O(n^2)$$$-time algorithm to sort $$$n$$$ given integers in non-decreasing order. Alice has already completed her assignment, and her implementation is shown below.

While you have access to Alice's code, you prefer not to simply copy it. Instead, you want to use Alice's sorting function as a building block for your own solution. There are two ways as listed below you can utilize her function, but each of them can be applied at most once. The order in which these two operations are invoked can be arbitrary.

- Prefix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s, i\texttt{)}$$$. This sorts the first $$$i$$$ elements in the array $$$s$$$.
- Suffix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s+n-i, i\texttt{)}$$$. This sorts the last $$$i$$$ elements in the array $$$s$$$.

Due to the time complexity of the sorting algorithm, the cost of performing either a prefix or suffix sort is $$$i^2$$$, where $$$i$$$ is the length of the chosen subarray. Your goal is to determine the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

For example, Let $$$s=[3,2,5,5,4,1]$$$. We can first perform a suffix sort of length $$$4$$$, and the array becomes $$$[3,2,1,4,5,5]$$$. Then, we perform a prefix sort of length $$$3$$$, and the array becomes $$$[1,2,3,4,5,5]$$$, which is a sorted array. The cost is $$$4^2+3^2=25$$$. Here is another example, let $$$s=[4,3,2,1]$$$. We can complete the sorting by performing only a prefix sort of length $$$4$$$, and the cost is $$$4^2=16$$$.

Input: The first line contains exactly one integer $$$n$$$ which indicates the number of integers in the array $$$s$$$. The second line contains the $$$n$$$ integers in $$$s=[s_0, s_1, \ldots, s_{n-1}]$$$.

- $$$1 \le n \le 10^6$$$
- For all $$$i$$$ ($$$0\le i < n$$$), $$$0\le s_i < 2^{31}-1$$$.

Output: Output an integer on a line, indicating the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

Examples:
input:
6
3 2 5 5 4 1
output:
25

input:
4
4 3 2 1
output:
16

Note:

Buggy Code:
/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

public class Main {

    //------------------------------------------main--------------------------------------------------//
    public static void main(String[] args) throws java.lang.Exception {
        OutputStream outputStream = System.out;
        PrintWriter out = new PrintWriter(outputStream);
        FastReader sc = new FastReader();
        Task s = new Task(sc, out);
        s.solve();
        out.close();
        // your code goes here"
    }
}

class FastReader {

    BufferedReader br;
    StringTokenizer st;

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

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

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

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

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

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

    public int[] readIntArray(int n) throws IOException {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = nextInt();
        }
        return arr;
    }

    public long[] readLongArray(int n) throws IOException {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++) {
            arr[i] = nextLong();
        }
        return arr;
    }
}

class Task {

    PrintWriter out;
    FastReader sc;

    Task(FastReader sc, PrintWriter out) {
        this.sc = sc;
        this.out = out;
    }
    //----------------------------------------------inputfunction------------------------------------------//

    String st() {
        return sc.nextLine();
    }

    int i() {
        return sc.nextInt();
    }

    long l() {
        return sc.nextLong();
    }

    double d() {
        return sc.nextDouble();
    }

    //---------------------------------------------absolute----------------------------------------------//
    public long abs(long a) {
        return a < 0 ? -a : a;
    }

    public int abs(int a) {
        return a < 0 ? -a : a;
    }

    //-------------------------------------------max--------------------------------------------------//
    public int max(int a, int b) {
        return a >= b ? a : b;
    }

    public long max(long a, long b) {
        return a >= b ? a : b;
    }

    public int max(int a, int b, int c) {
        return Math.max(a, Math.max(b, c));
    }

    public long max(long a, long b, long c) {
        return Math.max(a, Math.max(b, c));
    }

    //------------------------------------------min---------------------------------------------------//
    public int min(int a, int b) {
        return a <= b ? a : b;
    }

    public long min(long a, long b) {
        return a <= b ? a : b;
    }

    public int min(int a, int b, int c) {
        return Math.min(a, Math.min(b, c));
    }

    public long min(long a, long b, long c) {
        return Math.min(a, Math.min(b, c));
    }

    //----------------------------------------sortingfunction---------------------------------------------//
    private void sort(int[] arr) {
        int n = arr.length;
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void sort(char[] arr) {
        int n = arr.length;
        List<Character> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(char[] arr) {
        int n = arr.length;
        List<Character> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void rsort(int[] arr) {
        int n = arr.length;
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }

    private void sort(long[] arr, int n) {
        List<Long> list = new ArrayList<>();
        for (int i = 1; i <= n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list);
        for (int i = 0; i < list.size(); i++) {
            arr[i+1] = list.get(i);
        }
    }

    private void rsort(long[] arr) {
        int n = arr.length;
        List<Long> list = new ArrayList<>();
        for (int i = 1; i < n; i++) {
            list.add(arr[i]);
        }
        Collections.sort(list, Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            arr[i] = list.get(i);
        }
    }
    //---------------------------------------power----------------------------------------------------//

    public long pow(long a, int b) {
        long ans = 1L;
        while (b > 0) {
            if ((b & 1) > 0) {
                ans *= a;
            }
            a *= a;
            b >>= 1;
        }
        return b;
    }
    //----------------------------------------binary_exponentiation-----------------------------------//

    public long exp(long a, long b, long mod) {
        long ans = 1L;
        while (b > 0) {
            if ((b & 1) > 0) {
                ans = (ans * a) % mod;
            }
            a = (a * a) % mod;
            b >>= 1;
        }
        return ans;
    }

    //---------------------------------------helpingmethods-------------------------------------------//
    //---------------------------------------main-----------------------------------------------------//
    int n;
    void remove(HashMap<Long,Integer> m, long val){
        m.put(val, m.get(val)-1);
    }

    long help(long[] a, int n){
        long[] sorted = new long[n+2], pre = new long[n+2], suf = new long[n+2];
        long[] preMax = new long[n+2], sufMin = new long[n+2];
        HashMap<Long, Integer> m = new HashMap<>(), l = new HashMap<>(), r  = new HashMap<>();
        sufMin[n+1] = n+1;
        long min = a[n],  max = a[1], ans = n;
        ans *= ans;
        for(int i = 1;i<=n;i++){
            sorted[i] = a[i];
            m.put(a[i], m.getOrDefault(a[i], 0)+1);
            preMax[i] = max(a[i], preMax[i-1]);
            sufMin[n-i+1] = min(a[n-i+1], sufMin[n-i+2]);
        }
        sort(sorted, n);
        for(int i = 1;i<=n;i++){
            if(!l.containsKey(sorted[i])){
                l.put(sorted[i], i);
            }
            r.put(sorted[i], i);
        }
        for(int i =2 ;i<=n;i++){
            if(a[i]>=max){
                max = a[i];
                pre[i] = pre[i-1];
            } else{
                pre[i] = i;
                pre[i] *= pre[i];
            }
        }
        for(int i = n-1; i>=1; i--){
            if(a[i]<=min){
                min = a[i];
                suf[i] = suf[i+1];
            } else{
                suf[i] = n-i+1;
                suf[i] *= suf[i];
            }
        }

        for(int i = n; i>=1; i--){
            remove(m, a[i]);
            if(preMax[i-1]>=sorted[i]){
                long last = preMax[i-1], count = 1l*m.get(last);
                long p = l.get(last) + count - 1;
                ans = min(ans, suf[i] + p*p);
            } else ans = min(ans, pre[i-1] + suf[i]);
        }

        for(int i = 1;i<=n;i++){
            m.put(a[i], m.getOrDefault(a[i], 0)+1);
        }
        for(int i = 1;i<=n;i++){
            remove(m, a[i]);
            if(sufMin[i+1]<=sorted[i]){
                long first = sufMin[i+1], count = 1l*m.get(first);
                long p = r.get(first) - count + 1;
                p = n - p + 1;
                ans = min(ans, pre[i] + p*p);
            } else ans = min(ans, pre[i] + suf[i+1]);
        }
        return ans;
    }
    public void solve() throws IOException {
        int n = i();
        long[] a = new long[n+2];
        for(int i =1 ;i<=n;i++){
            a[i] = l();
        }
        out.print(help(a, n));
    }
}

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

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

Think step by step.

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

1. **Edge Cases**:
   - **Single Element Array**: This should be trivially sorted with zero cost.
   - **Already Sorted Array**: This should also be trivially sorted with zero cost.
   - **Reverse Sorted Array**: This should require a prefix sort of the entire array, with a cost of \( n^2 \).
   - **Array with Duplicates**: This can sometimes cause issues if the algorithm doesn't handle duplicates correctly.

2. **Complex Scenarios**:
   - **Array with a Pivot Point**: An array where a small segment is already sorted and the rest is not.
   - **Array with a Single Outlier**: An array where all elements are almost sorted except for one element that is out of place.

3. **Small-Scale Test Input**:
   - We will focus on a small array to ensure it is valid and complete within the output length limit.

Let's create a test input that combines some of these elements:

- An array that is almost sorted but has one element out of place, which will test the algorithm's ability to handle such cases.

Here is the test input:

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

Explanation:
- The array is almost sorted except for the last two elements (6 and 5), which are out of order.
- This will test the algorithm's ability to handle a small unsorted segment and determine the optimal sorting strategy.

This input should be challenging enough to expose any issues 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: Selection Sort
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Every student enrolled in the algorithms course is required to submit an assignment this week. The task is to implement an $$$O(n^2)$$$-time algorithm to sort $$$n$$$ given integers in non-decreasing order. Alice has already completed her assignment, and her implementation is shown below.

While you have access to Alice's code, you prefer not to simply copy it. Instead, you want to use Alice's sorting function as a building block for your own solution. There are two ways as listed below you can utilize her function, but each of them can be applied at most once. The order in which these two operations are invoked can be arbitrary.

- Prefix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s, i\texttt{)}$$$. This sorts the first $$$i$$$ elements in the array $$$s$$$.
- Suffix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s+n-i, i\texttt{)}$$$. This sorts the last $$$i$$$ elements in the array $$$s$$$.

Due to the time complexity of the sorting algorithm, the cost of performing either a prefix or suffix sort is $$$i^2$$$, where $$$i$$$ is the length of the chosen subarray. Your goal is to determine the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

For example, Let $$$s=[3,2,5,5,4,1]$$$. We can first perform a suffix sort of length $$$4$$$, and the array becomes $$$[3,2,1,4,5,5]$$$. Then, we perform a prefix sort of length $$$3$$$, and the array becomes $$$[1,2,3,4,5,5]$$$, which is a sorted array. The cost is $$$4^2+3^2=25$$$. Here is another example, let $$$s=[4,3,2,1]$$$. We can complete the sorting by performing only a prefix sort of length $$$4$$$, and the cost is $$$4^2=16$$$.

Input: The first line contains exactly one integer $$$n$$$ which indicates the number of integers in the array $$$s$$$. The second line contains the $$$n$$$ integers in $$$s=[s_0, s_1, \ldots, s_{n-1}]$$$.

- $$$1 \le n \le 10^6$$$
- For all $$$i$$$ ($$$0\le i < n$$$), $$$0\le s_i < 2^{31}-1$$$.

Output: Output an integer on a line, indicating the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

Examples:
input:
6
3 2 5 5 4 1
output:
25

input:
4
4 3 2 1
output:
16

Note:

Buggy Code:
#author: sushmanth
from sys import stdin , stdout
input = stdin.readline

inp = lambda : list(map(int , input().split()))

"""
For a fixed prefix what is the largest suffix
For a fixed suffix what is the largest prefix
"""

def answer():

    count = dict()
    for x in a:
        count[x] = count.get(x , 0) + 1
    b = sorted(list(set(a)))
    c = sorted(a)

    pick = [0]
    for i in range(1 , n):
        if(a[i] != c[i]):
            pick.append(i)

    pick = pick[::-1]

    #left first and right intersecting
    left_ok = [False for i in range(n + 1)]
    left_ok[0] = True
    freq , ans = dict() , n * n
    pointer , s = 0 , 0
    for i in range(n):
        freq[a[i]] = freq.get(a[i] , 0) + 1
        while(pointer < len(b) and count.get(b[pointer] , 0) == freq.get(b[pointer] , 0)):
            s += freq[b[pointer]]
            pointer += 1

        if(pointer < len(b)):
            v = s + freq.get(b[pointer] , 0)

        ans = min(ans , (i+1)*(i+1) + (n-v)*(n-v))
        if((i + 1) == v):
            left_ok[i+1] = True

    #right first and left intersecting
    freq = dict()
    pointer , s = len(b) - 1 , 0
    right_ok = [False for i in range(n + 1)]
    right_ok[n] = True
    for i in range(n - 1 , -1 , -1):
        freq[a[i]] = freq.get(a[i] , 0) + 1
        while(pointer >= 0 and count.get(b[pointer] , 0) == freq.get(b[pointer] , 0)):
            s += freq[b[pointer]]
            pointer -= 1

        if(pointer >= 0):
            v = s + freq.get(b[pointer] , 0)

        #print(n-i , s)
        ans = min(ans , (n-v)*(n-v) + (n-i)*(n-i))
        if((n-i) == v):
            right_ok[i] = True

    #left and right non intersecting
    for i in range(n + 1):
        while(len(pick) and pick[-1] < i):
            pick.pop()

        if(len(pick)):
            if(left_ok[i] and right_ok[pick[-1]]):
                ans = min(ans , i*i + (n-pick[-1])*(n-pick[-1]))
        else:
            if(left_ok[i]):
                ans = min(ans , i*i)

    return ans

for T in range(1):

    n = int(input())
    a = inp()

    print(answer())

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

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

Think step by step.

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

This test case is designed to challenge the algorithm's ability to handle duplicate elements and to ensure that the prefix and suffix sorting operations are applied correctly. The array has two identical elements in the middle, which can confuse the algorithm if it does not properly account for the intersections between the prefix and suffix sorts. The expected output for this test case is 8, as the optimal solution involves sorting the prefix of length 2 and the suffix of length 2, each with a cost of 4.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Selection Sort
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Every student enrolled in the algorithms course is required to submit an assignment this week. The task is to implement an $$$O(n^2)$$$-time algorithm to sort $$$n$$$ given integers in non-decreasing order. Alice has already completed her assignment, and her implementation is shown below.

While you have access to Alice's code, you prefer not to simply copy it. Instead, you want to use Alice's sorting function as a building block for your own solution. There are two ways as listed below you can utilize her function, but each of them can be applied at most once. The order in which these two operations are invoked can be arbitrary.

- Prefix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s, i\texttt{)}$$$. This sorts the first $$$i$$$ elements in the array $$$s$$$.
- Suffix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s+n-i, i\texttt{)}$$$. This sorts the last $$$i$$$ elements in the array $$$s$$$.

Due to the time complexity of the sorting algorithm, the cost of performing either a prefix or suffix sort is $$$i^2$$$, where $$$i$$$ is the length of the chosen subarray. Your goal is to determine the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

For example, Let $$$s=[3,2,5,5,4,1]$$$. We can first perform a suffix sort of length $$$4$$$, and the array becomes $$$[3,2,1,4,5,5]$$$. Then, we perform a prefix sort of length $$$3$$$, and the array becomes $$$[1,2,3,4,5,5]$$$, which is a sorted array. The cost is $$$4^2+3^2=25$$$. Here is another example, let $$$s=[4,3,2,1]$$$. We can complete the sorting by performing only a prefix sort of length $$$4$$$, and the cost is $$$4^2=16$$$.

Input: The first line contains exactly one integer $$$n$$$ which indicates the number of integers in the array $$$s$$$. The second line contains the $$$n$$$ integers in $$$s=[s_0, s_1, \ldots, s_{n-1}]$$$.

- $$$1 \le n \le 10^6$$$
- For all $$$i$$$ ($$$0\le i < n$$$), $$$0\le s_i < 2^{31}-1$$$.

Output: Output an integer on a line, indicating the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

Examples:
input:
6
3 2 5 5 4 1
output:
25

input:
4
4 3 2 1
output:
16

Note:

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

public class M2041{
    public static void main(String[] args) throws IOException{
        BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);

        int n = Integer.parseInt(r.readLine());

        int[] nums = new int[n];
            int idx = 0;
            HashMap<Integer, Integer> cur = new HashMap<>();
        int[] sorted = new int[n];
        int[] reverse = new int[n];
        long ans = Long.MAX_VALUE;

        StringTokenizer st = new StringTokenizer(r.readLine());
        for (int i = 0; i < n; i++) {
            nums[i] = Integer.parseInt(st.nextToken());
            sorted[i] = nums[i];
        }
        Arrays.sort(sorted);

        //DISJOINT
        for (int i = 0; i < n; i++) {
            cur.put(nums[i], cur.getOrDefault(nums[i], 0) + 1);
            while(idx < n && cur.containsKey(sorted[idx])){
                cur.put(sorted[idx], cur.get(sorted[idx]) - 1);
                if(cur.get(sorted[idx]) == 0){
                    cur.remove(sorted[idx]);
                }

                idx++;
            }
            if(cur.size() == 0){
                int right = i + 1;
                while(right < n && nums[right] == sorted[right]){
                    right++;
                }

                ans = Math.min(ans, (long)(i + 1) * (i + 1) + (long)(n - right) * (n - right));
                // i = right - 1;
            }
        }

        //INTERSECT WITH LEFT FIRST
        cur = new HashMap<>();
        idx = 0;

        for (int i = 0; i < n; i++) {
            cur.put(nums[i], cur.getOrDefault(nums[i], 0) + 1);
            while(idx < n && cur.containsKey(sorted[idx])){
                cur.put(sorted[idx], cur.get(sorted[idx]) - 1);
                if(cur.get(sorted[idx]) == 0){
                    cur.remove(sorted[idx]);
                }
                idx++;
            }
            ans = Math.min(ans, (long)(i + 1) * (i + 1) + (long)(n - idx) * (n - idx));
        }

        //INTERSECT WITH RIGHT FIRST
        cur = new HashMap<>();
        idx = n - 1;
        for (int i = n - 1; i >= 0; i--) {
            cur.put(nums[i], cur.getOrDefault(nums[i], 0) + 1);
            while(idx >= 0 && cur.containsKey(sorted[idx])){
                cur.put(sorted[idx], cur.get(sorted[idx]) - 1);
                if(cur.get(sorted[idx]) == 0){
                    cur.remove(sorted[idx]);
                }
                idx--;
            }
            ans = Math.min(ans, (long)(idx + 1) * (idx + 1) + (long)(n - i) * (n - i));
        }

        pw.println(ans);
        pw.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
4
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: Selection Sort
time_limit_ms: 2000
memory_limit_mb: 1024
Description: Every student enrolled in the algorithms course is required to submit an assignment this week. The task is to implement an $$$O(n^2)$$$-time algorithm to sort $$$n$$$ given integers in non-decreasing order. Alice has already completed her assignment, and her implementation is shown below.

While you have access to Alice's code, you prefer not to simply copy it. Instead, you want to use Alice's sorting function as a building block for your own solution. There are two ways as listed below you can utilize her function, but each of them can be applied at most once. The order in which these two operations are invoked can be arbitrary.

- Prefix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s, i\texttt{)}$$$. This sorts the first $$$i$$$ elements in the array $$$s$$$.
- Suffix sort: choose a length $$$i \in \{1, 2, \ldots, n\}$$$ and call $$$\texttt{alicesort(}s+n-i, i\texttt{)}$$$. This sorts the last $$$i$$$ elements in the array $$$s$$$.

Due to the time complexity of the sorting algorithm, the cost of performing either a prefix or suffix sort is $$$i^2$$$, where $$$i$$$ is the length of the chosen subarray. Your goal is to determine the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

For example, Let $$$s=[3,2,5,5,4,1]$$$. We can first perform a suffix sort of length $$$4$$$, and the array becomes $$$[3,2,1,4,5,5]$$$. Then, we perform a prefix sort of length $$$3$$$, and the array becomes $$$[1,2,3,4,5,5]$$$, which is a sorted array. The cost is $$$4^2+3^2=25$$$. Here is another example, let $$$s=[4,3,2,1]$$$. We can complete the sorting by performing only a prefix sort of length $$$4$$$, and the cost is $$$4^2=16$$$.

Input: The first line contains exactly one integer $$$n$$$ which indicates the number of integers in the array $$$s$$$. The second line contains the $$$n$$$ integers in $$$s=[s_0, s_1, \ldots, s_{n-1}]$$$.

- $$$1 \le n \le 10^6$$$
- For all $$$i$$$ ($$$0\le i < n$$$), $$$0\le s_i < 2^{31}-1$$$.

Output: Output an integer on a line, indicating the minimum cost to sort the input array $$$s$$$ of $$$n$$$ integers in non-decreasing order using Alice's function, following the rules mentioned above.

Examples:
input:
6
3 2 5 5 4 1
output:
25

input:
4
4 3 2 1
output:
16

Note:

Buggy Code:
from sys import stdin
input = lambda: stdin.readline().rstrip('\r\n')

n = int(input())
a = list(map(int, input().split()))
b = sorted(range(n), key=lambda x: a[x])

def solve(b):
    bad = []
    for i, j in enumerate(b):
        if i != j: bad += [i]

    mex = 0
    vis = [0]*n
    bad = [n] + bad[::-1]
    ans = (n - bad[-1]) ** 2

    for i, j in enumerate(b, 1):
        vis[j] = 1
        while mex < n and vis[mex]:
            mex += 1
        while bad[-1] < mex:
            bad.pop()

        ans = min(ans, (n - max(mex, bad[-1]))**2 + i**2)

    return ans

ans = solve([n - i - 1 for i in b[::-1]])
ans = min(ans, solve(b))
print(ans)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that are likely to cause issues. The code's main logic involves identifying "bad" indices (where the element is not in its correct position) and then trying to minimize the cost of sorting using prefix and suffix sorts. Here are the key points to consider:

1. **Arrays with all elements in the correct position**: This should ideally have a cost of 0, but we need to ensure the code handles this correctly.
2. **Arrays with a few elements out of order**: This can help test the logic for identifying and handling "bad" indices.
3. **Arrays with elements in reverse order**: This can test the worst-case scenario for sorting.
4. **Arrays with repeated elements**: This can test how the code handles duplicate values.
5. **Arrays with a single element out of order**: This can test the edge case where only one element needs to be moved.

Given these considerations, let's create a small-scale test input that is likely to expose any bugs:

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

### Explanation:
- **Array**: `[1, 3, 2, 4, 5]`
- **Expected Output**: The array is almost sorted, with only one element out of order. The optimal solution would be to perform a prefix sort of length 3, which sorts the first 3 elements. The cost would be \(3^2 = 9\).

This test input is small enough to be manageable but complex enough to challenge the code's logic for identifying and handling "bad" indices. It should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: C+K+S
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given two strongly connected$^{\dagger}$ directed graphs, each with exactly $n$ vertices, but possibly different numbers of edges. Upon closer inspection, you noticed an important feature — the length of any cycle in these graphs is divisible by $k$.

Each of the $2n$ vertices belongs to exactly one of two types: incoming or outgoing. For each vertex, its type is known to you.

You need to determine whether it is possible to draw exactly $n$ directed edges between the source graphs such that the following four conditions are met:

  * The ends of any added edge lie in different graphs.    * From each outgoing vertex, exactly one added edge originates.    * Into each incoming vertex, exactly one added edge enters.    * In the resulting graph, the length of any cycle is divisible by $k$.

$^{\dagger}$A strongly connected graph is a graph in which there is a path from every vertex to every other vertex.

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

The first line of each test case contains two integers $n$ and $k$ ($2 \le k \le n \le 2 \cdot 10^5$) — the number of vertices in each graph and the value by which the length of each cycle is divisible.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($a_i \in \\{0, 1\\}$). If $a_i = 0$, then vertex $i$ of the first graph is incoming. If $a_i = 1$, then vertex $i$ of the first graph is outgoing.

The third line of each test case contains a single integer $m_1$ ($1 \le m_1 \le 5 \cdot 10^5$) — the number of edges in the first graph.

The next $m_1$ lines contain descriptions of the edges of the first graph. The $i$-th of them contains two integers $v_i$ and $u_i$ ($1 \le v_i, u_i \le n$) — an edge in the first graph leading from vertex $v_i$ to vertex $u_i$.

Next, in the same format, follows the description of the second graph.

The next line contains $n$ integers $b_1, b_2, \ldots, b_n$ ($b_i \in \\{0, 1\\}$). If $b_i = 0$, then vertex $i$ of the second graph is incoming. If $b_i = 1$, then vertex $i$ of the second graph is outgoing.

The next line contains a single integer $m_2$ ($1 \le m_2 \le 5 \cdot 10^5$) — the number of edges in the second graph.

The next $m_2$ lines contain descriptions of the edges of the second graph. The $i$-th of them contains two integers $v_i$ and $u_i$ ($1 \le v_i, u_i \le n$) — an edge in the second graph leading from vertex $v_i$ to vertex $u_i$.

It is guaranteed that both graphs are strongly connected, and the lengths of all cycles are divisible by $k$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$. It is guaranteed that the sum of $m_1$ and the sum of $m_2$ over all test cases does not exceed $5 \cdot 10^5$.

Output: For each test case, output "YES" (without quotes) if it is possible to draw $n$ new edges such that all conditions are met, and "NO" (without quotes) otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
3
4 2
1 0 0 1
4
1 2
2 3
3 4
4 1
1 0 0 1
4
1 3
3 2
2 4
4 1
3 3
0 0 0
3
1 2
2 3
3 1
1 1 0
3
1 2
2 3
3 1
4 2
1 1 1 1
4
1 2
2 3
3 4
4 1
0 0 0 0
6
1 2
2 1
1 3
3 1
1 4
4 1
output:
YES
NO
YES

Note: In the first test case, it is possible to draw edges from the first graph to the second graph as $(1, 3)$ and $(4, 2)$ (the first number in the pair is the vertex number in the first graph, and the second number in the pair is the vertex number in the second graph), and from the second graph to the first graph as $(1, 2)$, $(4, 3)$ (the first number in the pair is the vertex number in the second graph, and the second number in the pair is the vertex number in the first graph).

In the second test case, there are a total of $4$ incoming vertices and $2$ outgoing vertices, so it is not possible to draw $3$ edges.

Buggy Code:
#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define pl pair<ll, ll>
#define pi pair<int, int>
#define minpq priority_queue<ll, vector<ll>, greater<ll>>
using namespace std;

#if 1
void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '['; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "]";}
void _print() {cerr << endl << flush;}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#define debug(x...) cerr << "*["<<__LINE__<<"]\t"<< #x << " = "; _print(x)
#endif

const ll mod = 1e9 + 7;

template<typename T> bool chkmin(T &a, T b){return (b < a) ? a = b, 1 : 0;}
template<typename T> bool chkmax(T &a, T b){return (b > a) ? a = b, 1 : 0;}
ll gcd(ll a, ll b) {if(b == 0){return a;} return gcd(b, a % b);}

vector<int> prefix_function(vector<int> s) {
    int n = s.size();
    vector<int> pf(n);
    for (int i = 1; i < n; i++) {
        int j = pf[i - 1];
        while (j > 0 && s[i] != s[j]) j = pf[j - 1];
        if (s[i] == s[j]) j++;
        pf[i] = j;
    }
    return pf;
}

int is_cyclic_shift(vector<int> a, vector<int> b, int k){
	if (a.size() != b.size()) return -1;
	if (a.size() == 0 && b.size() == 0) return 0;
	vector<int> na, nb;
	for (int i = 0; i < a.size(); i ++) na.push_back((a[(i + 1) % a.size()] - a[i] + k) % k);
	for (int i = 0; i < b.size(); i ++) nb.push_back((b[(i + 1) % b.size()] - b[i] + k) % k);
	vector<int> c = na;
	c.push_back(1 << 30);
	for (int _ = 0; _ < 2; _ ++){
		for (int i = 0; i < nb.size(); i ++) c.push_back(nb[i]);
	}
	vector<int> pf = prefix_function(c);
	for (int i = 0; i < pf.size(); i ++) if (pf[i] == na.size()) {
		return (b[i - 2 * a.size()] - a[0] + k) % k;
	}
	return -1;
}

void solve(){
	int n, k; cin >> n >> k;
	vector<vector<int>> ga(n, vector<int>()), gb(n, vector<int>());
	vector<int> ina(n), inb(n);
	for (int i = 0; i < n; i ++) cin >> ina[i];
	int m1; cin >> m1;
	for (int i = 0; i < m1; i ++){
		int x, y; cin >> x >> y; x --; y --;
		ga[x].push_back(y);
	}
	for (int i = 0; i < n; i ++) cin >> inb[i];
	int m2; cin >> m2;
	for (int i = 0; i < m2; i ++){
		int x, y; cin >> x >> y; x --; y --;
		gb[x].push_back(y);
	}
	vector<int> ca(n, -1), cb(n, -1);
	function<void(int, int, vector<vector<int>>&, vector<int>&)> dfs = [&](int x, int curc, vector<vector<int>> &g, vector<int> &c){
		c[x] = curc;
		for (int to : g[x]) if (c[to] == -1) dfs(to, (curc + 1) % k, g, c);
	};
	dfs(0, 0, ga, ca); dfs(0, 0, gb, cb);
	//debug(ca); debug(cb);
	vector<int> inca, outca, incb, outcb;
	for (int i = 0; i < n; i ++){
		if (ina[i]) inca.push_back(ca[i]);
		else outca.push_back((ca[i] + 1) % k);
		if (inb[i]) incb.push_back(cb[i]);
		else outcb.push_back((cb[i] + 1) % k);
	}
	sort(inca.begin(), inca.end());
	sort(incb.begin(), incb.end());
	sort(outca.begin(), outca.end());
	sort(outcb.begin(), outcb.end());
	int ans1 = is_cyclic_shift(inca, outcb, k), ans2 = is_cyclic_shift(outca, incb, k);
	bool good = ans1 >= 0 && ans2 >= 0 && ans1 == ans2;
	if (inca.size() == 0 && outcb.size() == 0 && incb.size() == outca.size()) good = true;
	if (incb.size() == 0 && outca.size() == 0 && inca.size() == outcb.size()) good = true;
	cout << (good ? "YES\n" : "NO\n");
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	int tc = 1;
	cin >> tc;
	while(tc --){
		solve();
	}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios that might cause the code to fail. The code is designed to determine if it is possible to draw exactly \( n \) directed edges between two strongly connected graphs such that the resulting graph maintains certain properties. Here are the steps to create a challenging test input:

1. **Balance of Incoming and Outgoing Vertices**: Ensure that the number of incoming and outgoing vertices is balanced in a way that the code might have difficulty handling.

2. **Cycle Lengths**: Ensure that the cycle lengths in both graphs are divisible by \( k \), but in a way that might expose issues in the cycle detection or matching logic.

3. **Strongly Connected Graphs**: Both graphs must be strongly connected, meaning there is a path from every vertex to every other vertex.

4. **Edge Cases**: Consider edge cases where the number of vertices or edges is at the boundary of the constraints.

Based on these considerations, let's construct a test input:

```plaintext
1
5 3
0 1 0 1 0
5
1 2
2 3
3 4
4 5
5 1
0 1 0 1 0
5
1 3
3 5
5 2
2 4
4 1
```

### Explanation:
- **Number of Test Cases**: 1
- **Number of Vertices and \( k \)**: 5 vertices and \( k = 3 \)
- **Types of Vertices in the First Graph**:
  - Incoming: 1, 3, 5
  - Outgoing: 2, 4
- **Edges in the First Graph**: A cycle of length 5, which is divisible by 3.
- **Types of Vertices in the Second Graph**:
  - Incoming: 1, 3, 5
  - Outgoing: 2, 4
- **Edges in the Second Graph**: A cycle of length 5, which is divisible by 3.

This test case is designed to be small and valid while challenging the code to handle balanced vertex types and cycles of a specific length. The strongly connected nature of both graphs and the divisibility of cycle lengths by \( k \) should make it a good candidate to expose any bugs in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: C+K+S
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given two strongly connected$^{\dagger}$ directed graphs, each with exactly $n$ vertices, but possibly different numbers of edges. Upon closer inspection, you noticed an important feature — the length of any cycle in these graphs is divisible by $k$.

Each of the $2n$ vertices belongs to exactly one of two types: incoming or outgoing. For each vertex, its type is known to you.

You need to determine whether it is possible to draw exactly $n$ directed edges between the source graphs such that the following four conditions are met:

  * The ends of any added edge lie in different graphs.    * From each outgoing vertex, exactly one added edge originates.    * Into each incoming vertex, exactly one added edge enters.    * In the resulting graph, the length of any cycle is divisible by $k$.

$^{\dagger}$A strongly connected graph is a graph in which there is a path from every vertex to every other vertex.

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

The first line of each test case contains two integers $n$ and $k$ ($2 \le k \le n \le 2 \cdot 10^5$) — the number of vertices in each graph and the value by which the length of each cycle is divisible.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($a_i \in \\{0, 1\\}$). If $a_i = 0$, then vertex $i$ of the first graph is incoming. If $a_i = 1$, then vertex $i$ of the first graph is outgoing.

The third line of each test case contains a single integer $m_1$ ($1 \le m_1 \le 5 \cdot 10^5$) — the number of edges in the first graph.

The next $m_1$ lines contain descriptions of the edges of the first graph. The $i$-th of them contains two integers $v_i$ and $u_i$ ($1 \le v_i, u_i \le n$) — an edge in the first graph leading from vertex $v_i$ to vertex $u_i$.

Next, in the same format, follows the description of the second graph.

The next line contains $n$ integers $b_1, b_2, \ldots, b_n$ ($b_i \in \\{0, 1\\}$). If $b_i = 0$, then vertex $i$ of the second graph is incoming. If $b_i = 1$, then vertex $i$ of the second graph is outgoing.

The next line contains a single integer $m_2$ ($1 \le m_2 \le 5 \cdot 10^5$) — the number of edges in the second graph.

The next $m_2$ lines contain descriptions of the edges of the second graph. The $i$-th of them contains two integers $v_i$ and $u_i$ ($1 \le v_i, u_i \le n$) — an edge in the second graph leading from vertex $v_i$ to vertex $u_i$.

It is guaranteed that both graphs are strongly connected, and the lengths of all cycles are divisible by $k$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$. It is guaranteed that the sum of $m_1$ and the sum of $m_2$ over all test cases does not exceed $5 \cdot 10^5$.

Output: For each test case, output "YES" (without quotes) if it is possible to draw $n$ new edges such that all conditions are met, and "NO" (without quotes) otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
3
4 2
1 0 0 1
4
1 2
2 3
3 4
4 1
1 0 0 1
4
1 3
3 2
2 4
4 1
3 3
0 0 0
3
1 2
2 3
3 1
1 1 0
3
1 2
2 3
3 1
4 2
1 1 1 1
4
1 2
2 3
3 4
4 1
0 0 0 0
6
1 2
2 1
1 3
3 1
1 4
4 1
output:
YES
NO
YES

Note: In the first test case, it is possible to draw edges from the first graph to the second graph as $(1, 3)$ and $(4, 2)$ (the first number in the pair is the vertex number in the first graph, and the second number in the pair is the vertex number in the second graph), and from the second graph to the first graph as $(1, 2)$, $(4, 3)$ (the first number in the pair is the vertex number in the second graph, and the second number in the pair is the vertex number in the first graph).

In the second test case, there are a total of $4$ incoming vertices and $2$ outgoing vertices, so it is not possible to draw $3$ edges.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int N=5e5+5,p1=19491001,mod1=1e9+7;
int T,n,a[N],k,dis[N],pw1[N],m1,m2;
vector<int>son[N];
bool vis[N];
//
void dfs1(int x)
{
	vis[x]=1;
	for(auto y:son[x])if(!vis[y])dis[y]=(dis[x]+1)%k,dfs1(y);
}
int h1[N];
bool cmp(int x,int y){return dis[x]<dis[y];}
bool iscor[2][N];
int hs1(int l,int r){return (h1[r]-1ll*h1[l-1]*pw1[r-l+1]%mod1+mod1)%mod1;}
int d[N],c[N];
void slv(vector<int>a,vector<int>b,int tp)
{
	sort(a.begin(),a.end()),sort(b.begin(),b.end());
	int len=a.size();
	for(int i=0;i<len;i++)c[i+1]=(a[(i+1)%len]-a[i]+k)%k,d[i+1]=(b[(i+1)%len]-b[i]+k)%k;
	for(int i=1;i<=len;i++)h1[i]=(1ll*h1[i-1]*p1+c[i])%mod1;
	int hh=0;
	for(int i=1;i<=len;i++)hh=(1ll*hh*p1+d[i])%mod1;
	for(int i=0;i<len;i++)
	{
		int hh0=hs1(i+1,len),hh1=hs1(1,i);
		hh0=(1ll*hh0*pw1[i]+hh1)%mod1;
		if(hh0==hh)
		{
			iscor[tp][(a[0]-b[i]+k)%k]=1;
		}
	}
}
void solve()
{
	scanf("%d %d",&n,&k);
	for(int i=1;i<=n*2;i++)vis[i]=0,dis[i]=0;
	for(int i=1;i<=n;i++)scanf("%d",&a[i]),son[i].clear();
	scanf("%d",&m1);
	for(int i=1,x,y;i<=m1;i++)scanf("%d %d",&x,&y),son[x].push_back(y);
	for(int i=n+1;i<=n*2;i++)scanf("%d",&a[i]),son[i].clear();
	scanf("%d",&m2);
	for(int i=1,x,y;i<=m2;i++)scanf("%d %d",&x,&y),son[x+n].push_back(y+n);
	dfs1(1),dfs1(n+1);
	vector<int>tempa0,tempa1,tempb0,tempb1;
	for(int i=1;i<=n;i++)if(a[i])tempa1.push_back(dis[i]);else tempa0.push_back(dis[i]);
	for(int i=n+1;i<=n*2;i++)if(a[i])tempb1.push_back(dis[i]);else tempb0.push_back(dis[i]);
	if(tempa1.size()!=tempb0.size()){printf("NO\n");return ;}
	if(!tempa0.size() || !tempb0.size()){printf("YES\n");return ;}
	for(int i=0;i<k;i++)iscor[0][i]=iscor[1][i]=0;
	slv(tempa1,tempb0,0);
	slv(tempb1,tempa0,1);
	bool fl=0;
	for(int i=0;i<k;i++)if(iscor[0][i])
	{
		if(iscor[1][(-2%k-i+k*3)%k])fl=1;
	}
	if(fl)printf("YES\n");
	else printf("NO\n");
}
int main()
{
	pw1[0]=1;for(int i=1;i<N;i++)pw1[i]=1ll*pw1[i-1]*p1%mod1;
	scanf("%d",&T);
	while(T--)solve();
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
4 2
1 0 0 1
4
1 2
2 3
3 4
4 1
1 0 0 1
4
1 3
3 2
2 4
4 1
3 3
0 0 0
3
1 2
2 3
3 1
1 1 0
3
1 2
2 3
3 1
4 2
1 1 1 1
4
1 2
2 3
3 4
4 1
0 0 0 0
6
1 2
2 1
1 3
3 1
1 4
4 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: C+K+S
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given two strongly connected$^{\dagger}$ directed graphs, each with exactly $n$ vertices, but possibly different numbers of edges. Upon closer inspection, you noticed an important feature — the length of any cycle in these graphs is divisible by $k$.

Each of the $2n$ vertices belongs to exactly one of two types: incoming or outgoing. For each vertex, its type is known to you.

You need to determine whether it is possible to draw exactly $n$ directed edges between the source graphs such that the following four conditions are met:

  * The ends of any added edge lie in different graphs.    * From each outgoing vertex, exactly one added edge originates.    * Into each incoming vertex, exactly one added edge enters.    * In the resulting graph, the length of any cycle is divisible by $k$.

$^{\dagger}$A strongly connected graph is a graph in which there is a path from every vertex to every other vertex.

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

The first line of each test case contains two integers $n$ and $k$ ($2 \le k \le n \le 2 \cdot 10^5$) — the number of vertices in each graph and the value by which the length of each cycle is divisible.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($a_i \in \\{0, 1\\}$). If $a_i = 0$, then vertex $i$ of the first graph is incoming. If $a_i = 1$, then vertex $i$ of the first graph is outgoing.

The third line of each test case contains a single integer $m_1$ ($1 \le m_1 \le 5 \cdot 10^5$) — the number of edges in the first graph.

The next $m_1$ lines contain descriptions of the edges of the first graph. The $i$-th of them contains two integers $v_i$ and $u_i$ ($1 \le v_i, u_i \le n$) — an edge in the first graph leading from vertex $v_i$ to vertex $u_i$.

Next, in the same format, follows the description of the second graph.

The next line contains $n$ integers $b_1, b_2, \ldots, b_n$ ($b_i \in \\{0, 1\\}$). If $b_i = 0$, then vertex $i$ of the second graph is incoming. If $b_i = 1$, then vertex $i$ of the second graph is outgoing.

The next line contains a single integer $m_2$ ($1 \le m_2 \le 5 \cdot 10^5$) — the number of edges in the second graph.

The next $m_2$ lines contain descriptions of the edges of the second graph. The $i$-th of them contains two integers $v_i$ and $u_i$ ($1 \le v_i, u_i \le n$) — an edge in the second graph leading from vertex $v_i$ to vertex $u_i$.

It is guaranteed that both graphs are strongly connected, and the lengths of all cycles are divisible by $k$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$. It is guaranteed that the sum of $m_1$ and the sum of $m_2$ over all test cases does not exceed $5 \cdot 10^5$.

Output: For each test case, output "YES" (without quotes) if it is possible to draw $n$ new edges such that all conditions are met, and "NO" (without quotes) otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
3
4 2
1 0 0 1
4
1 2
2 3
3 4
4 1
1 0 0 1
4
1 3
3 2
2 4
4 1
3 3
0 0 0
3
1 2
2 3
3 1
1 1 0
3
1 2
2 3
3 1
4 2
1 1 1 1
4
1 2
2 3
3 4
4 1
0 0 0 0
6
1 2
2 1
1 3
3 1
1 4
4 1
output:
YES
NO
YES

Note: In the first test case, it is possible to draw edges from the first graph to the second graph as $(1, 3)$ and $(4, 2)$ (the first number in the pair is the vertex number in the first graph, and the second number in the pair is the vertex number in the second graph), and from the second graph to the first graph as $(1, 2)$, $(4, 3)$ (the first number in the pair is the vertex number in the second graph, and the second number in the pair is the vertex number in the first graph).

In the second test case, there are a total of $4$ incoming vertices and $2$ outgoing vertices, so it is not possible to draw $3$ edges.

Buggy Code:
#include <algorithm>
#include <cassert>
#include <cmath>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <unordered_map>
#define lson (p << 1)
#define rson (p << 1 | 1)
using namespace std;
using ll = long long;
using ull = unsigned long long;
const int maxn = 2e6 + 10;
const int mod = 1e9 + 7;
int n, k;
int a[2][maxn];
int deg[maxn];

ll ksm(ll x, ll y) {
  ll ret = 1;
  do {
    if (y % 2 == 1) {
      ret = ret * x % mod;
    }
    x = x * x % mod;
  } while (y >>= 1);
  return ret;
}

void solve() {
  cin >> n >> k;
  int cnt00 = 0, cnt01 = 0, cnt10 = 0, cnt11 = 0;
  for (int i = 1; i <= n; i++) {
    deg[i] = 0;
    cin >> a[0][i];
    if (a[0][i]) {
      cnt01++;
    } else {
      cnt00++;
    }
  }
  int flag = 0;

  int m;
  cin >> m;
  vector<pair<int, int>> vec;
  for (int i = 1; i <= n; i++) {
    int u, v;
    cin >> u >> v;
    vec.push_back({u, v});
    deg[u]++;
    deg[v]++;
  }
  for (auto p : vec) {
    if (deg[p.second] != 2 && deg[p.first] != 2) {
      if (a[0][p.first] == 0 && a[1][p.second] == 1) {
        continue;
      } else {
        flag = 1;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    cin >> a[1][i];
    if (a[1][i]) {
      cnt11++;
    } else {
      cnt10++;
    }
  }
  if (cnt00 != cnt11) {
    flag = 1;
  }
  cin >> m;
  vec.clear();
  for (int i = 1; i <= n; i++) {
    deg[i] = 0;
  }
  for (int i = 1; i <= n; i++) {
    int u, v;
    cin >> u >> v;
    vec.push_back({u, v});
    deg[u]++;
    deg[v]++;
  }
  for (auto p : vec) {
    if (deg[p.second] != 2 && deg[p.first] != 2) {
      if (a[0][p.first] == 0 && a[1][p.second] == 1) {
        continue;
      } else {
        flag = 1;
      }
    }
  }
  if (flag) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
  }
}

int main() {
  int _ = 1;
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> _;
  while (_--) {
    solve();
  }
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the given buggy code, we need to focus on the key issues in the algorithm:

1. **Counting and Matching In/Out Degrees**: The code checks if the number of incoming vertices in one graph matches the number of outgoing vertices in the other graph. If these counts do not match, the code sets a flag and outputs "NO".

2. **Cycle Length Divisibility**: The code does not explicitly check if the cycles in the resulting graph are divisible by \( k \). However, it does check some conditions that might indirectly affect this.

3. **Edge Degree Check**: The code checks if the degree of the vertices in the edges is not equal to 2. This is a bit odd and might not be the correct approach to ensure the cycles are divisible by \( k \).

Given these points, let's create a test case that maximizes the probability of exposing these issues:

### Test Case:
```plaintext
1
4 2
1 0 0 1
4
1 2
2 3
3 4
4 1
1 0 0 1
4
1 2
2 3
3 4
4 1
```

### Explanation:
1. **Graph Size and \( k \)**: We use \( n = 4 \) and \( k = 2 \). This ensures that any cycle in the graph must have an even length.
2. **Vertex Types**: Both graphs have the same vertex types: one outgoing (1) and three incoming (0). This ensures that the number of incoming and outgoing vertices match between the two graphs.
3. **Graph Edges**: Both graphs are simple cycles of length 4, which are strongly connected and have cycles divisible by 2.

### Expected Behavior:
- The code should output "YES" because:
  - The number of incoming vertices in the first graph (3) matches the number of outgoing vertices in the second graph (1).
  - The number of outgoing vertices in the first graph (1) matches the number of incoming vertices in the second graph (3).
  - Both graphs are strongly connected and have cycles of length 4, which is divisible by 2.

### Why This Test Case is Challenging:
- It tests the basic matching of in/out degrees.
- It ensures the graphs are strongly connected and have cycles divisible by \( k \).
- It does not trigger the degree check (since all vertices in the cycles have degree 2), which might be a bug in the code.

This test case should help expose any issues in the code's logic for matching in/out degrees and ensuring cycle length divisibility.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: C+K+S
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given two strongly connected$^{\dagger}$ directed graphs, each with exactly $n$ vertices, but possibly different numbers of edges. Upon closer inspection, you noticed an important feature — the length of any cycle in these graphs is divisible by $k$.

Each of the $2n$ vertices belongs to exactly one of two types: incoming or outgoing. For each vertex, its type is known to you.

You need to determine whether it is possible to draw exactly $n$ directed edges between the source graphs such that the following four conditions are met:

  * The ends of any added edge lie in different graphs.    * From each outgoing vertex, exactly one added edge originates.    * Into each incoming vertex, exactly one added edge enters.    * In the resulting graph, the length of any cycle is divisible by $k$.

$^{\dagger}$A strongly connected graph is a graph in which there is a path from every vertex to every other vertex.

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

The first line of each test case contains two integers $n$ and $k$ ($2 \le k \le n \le 2 \cdot 10^5$) — the number of vertices in each graph and the value by which the length of each cycle is divisible.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($a_i \in \\{0, 1\\}$). If $a_i = 0$, then vertex $i$ of the first graph is incoming. If $a_i = 1$, then vertex $i$ of the first graph is outgoing.

The third line of each test case contains a single integer $m_1$ ($1 \le m_1 \le 5 \cdot 10^5$) — the number of edges in the first graph.

The next $m_1$ lines contain descriptions of the edges of the first graph. The $i$-th of them contains two integers $v_i$ and $u_i$ ($1 \le v_i, u_i \le n$) — an edge in the first graph leading from vertex $v_i$ to vertex $u_i$.

Next, in the same format, follows the description of the second graph.

The next line contains $n$ integers $b_1, b_2, \ldots, b_n$ ($b_i \in \\{0, 1\\}$). If $b_i = 0$, then vertex $i$ of the second graph is incoming. If $b_i = 1$, then vertex $i$ of the second graph is outgoing.

The next line contains a single integer $m_2$ ($1 \le m_2 \le 5 \cdot 10^5$) — the number of edges in the second graph.

The next $m_2$ lines contain descriptions of the edges of the second graph. The $i$-th of them contains two integers $v_i$ and $u_i$ ($1 \le v_i, u_i \le n$) — an edge in the second graph leading from vertex $v_i$ to vertex $u_i$.

It is guaranteed that both graphs are strongly connected, and the lengths of all cycles are divisible by $k$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$. It is guaranteed that the sum of $m_1$ and the sum of $m_2$ over all test cases does not exceed $5 \cdot 10^5$.

Output: For each test case, output "YES" (without quotes) if it is possible to draw $n$ new edges such that all conditions are met, and "NO" (without quotes) otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
3
4 2
1 0 0 1
4
1 2
2 3
3 4
4 1
1 0 0 1
4
1 3
3 2
2 4
4 1
3 3
0 0 0
3
1 2
2 3
3 1
1 1 0
3
1 2
2 3
3 1
4 2
1 1 1 1
4
1 2
2 3
3 4
4 1
0 0 0 0
6
1 2
2 1
1 3
3 1
1 4
4 1
output:
YES
NO
YES

Note: In the first test case, it is possible to draw edges from the first graph to the second graph as $(1, 3)$ and $(4, 2)$ (the first number in the pair is the vertex number in the first graph, and the second number in the pair is the vertex number in the second graph), and from the second graph to the first graph as $(1, 2)$, $(4, 3)$ (the first number in the pair is the vertex number in the second graph, and the second number in the pair is the vertex number in the first graph).

In the second test case, there are a total of $4$ incoming vertices and $2$ outgoing vertices, so it is not possible to draw $3$ edges.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
template<typename T>
using vec = vector<T>;
using vi = vec<int>;
using ii = pair<int, int>;
using vii = vec<ii>;
using i64 = long long;
using vi64 = vec<i64>;
using iii = tuple<int, int, int>;
using str = string;
template<typename Key, typename Value>
using hash_map = unordered_map<Key, Value>;
template<typename T>
using hash_set = unordered_set<T>;
template<typename T>
using min_heap = priority_queue<T, vec<T>, greater<T>>;
template<typename T>
using max_heap = priority_queue<T>;

// Doing this way so it works for both string and vi
template<typename T>
vi prefixFunction(const T& s) {
    int n = (int)s.size();
    vi pi(n);
    pi[0] = 0;
    for (int i = 1; i < n; i++) {
        int j = pi[i-1];
        while (j > 0 && s[i] != s[j])
            j = pi[j-1];
        if (s[i] == s[j])
            j++;
        pi[i] = j;
    }
    return pi;
}

template<typename T>
vi zFunction(const T& s) {
    int n = (int) s.length();
    vi z(n);
    for (int i = 1, l = 0, r = 0; i < n; ++i) {
        if (i <= r)
            z[i] = min (r - i + 1, z[i - l]);
        while (i + z[i] < n && s[z[i]] == s[i + z[i]])
            ++z[i];
        if (i + z[i] - 1 > r)
            l = i, r = i + z[i] - 1;
    }
    return z;
}

template<typename T, typename U>
vi matching_positions(const T& text, const T& pattern, U separator){
    T aux = pattern;
    aux.push_back(separator);
    aux.insert(aux.end(), text.begin(), text.end());
    vi pi = prefixFunction(aux);
    vi ans;
    for(int i = pattern.size(); i < aux.size(); i++)
        if(pi[i] == pattern.size())
            ans.push_back(i - 2 * pattern.size());
    return ans;
}

// Code below

int k, n;
vi a, b;
int m1, m2;
vec<vi> al_a, al_b;

vi color_a, color_b;

void color_graph_a(int u, int color){
    if(color_a[u] != -1)
        return;
    color_a[u] = color;
    for(int v : al_a[u])
        color_graph_a(v, (color+1)%k);
}

void color_graph_b(int u, int color){
    if(color_b[u] != -1)
        return;
    color_b[u] = color;
    for(int v : al_b[u])
        color_graph_b(v, (color+1)%k);
}

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

    a = vi(n+1);
    for(int i = 1; i <= n; i++)
        cin >> a[i];
    al_a = vec<vi>(n+1);
    cin >> m1;
    for(int i = 0; i < m1; i++){
        int u, v; cin >> u >> v;
        al_a[u].push_back(v);
        al_a[v].push_back(u);
    }

    b = vi(n+1);
    for(int i = 1; i <= n; i++)
        cin >> b[i];
    al_b = vec<vi>(n+1);
    cin >> m2;
    for(int i = 0; i < m2; i++){
        int u, v; cin >> u >> v;
        al_b[u].push_back(v);
        al_b[v].push_back(u);
    }

    int a_incoming_size = 0;
    int b_outgoing_size = 0;
    for(int i = 1; i <= n; i++)
        if(a[i] == 0)
            a_incoming_size++;
    for(int i = 1; i <= n; i++)
        if(b[i])
            b_outgoing_size++;

    if(a_incoming_size != b_outgoing_size){
        cout << "NO" << endl;
        return;
    }

    if(all_of(a.begin() + 1, a.end(), [](int x){return x == a[1];})){
        cout << "YES" << endl;
        return;
    }

    color_a = vi(n+1, -1);
    color_b = vi(n+1, -1);

    color_graph_a(1, 0);
    color_graph_b(1, 0);

    vi freq_a_outgoing(k, 0);
    vi freq_a_incoming(k, 0);
    vi freq_b_outgoing(k, 0);
    vi freq_b_incoming(k, 0);

    for(int u = 1; u <= n; u++){
        if(a[u])
            freq_a_outgoing[color_a[u]]++;
        else
            freq_a_incoming[color_a[u]]++;
    }

    for(int u = 1; u <= n; u++){
        if(b[u])
            freq_b_outgoing[color_b[u]]++;
        else
            freq_b_incoming[color_b[u]]++;
    }

    vi twice_freq_b_incoming = freq_b_incoming;
    twice_freq_b_incoming.insert(twice_freq_b_incoming.end(), freq_b_incoming.begin(), freq_b_incoming.end());
    twice_freq_b_incoming.pop_back();
    vi shifts1 = matching_positions(twice_freq_b_incoming, freq_a_outgoing, (int)1e9);

    vi twice_freq_b_outgoing = freq_b_outgoing;
    twice_freq_b_outgoing.insert(twice_freq_b_outgoing.end(), freq_b_outgoing.begin(), freq_b_outgoing.end());
    twice_freq_b_outgoing.pop_back();
    vi shifts2 = matching_positions(twice_freq_b_outgoing, freq_a_incoming, (int)1e9);

    vi working_shifts(k, 0);
    for(int shift : shifts1)
        working_shifts[shift]++;
    for(int shift : shifts2)
        working_shifts[(shift+2)%k]++;
    for(int shift : working_shifts)
        if(shift == 2){
            cout << "YES" << endl;
            return;
        }

    cout << "NO" << endl;
    return;
}

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

    int testCases = 1;
    cin >> testCases;

    while(testCases--)
        solve();

    return 0;
}

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

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

Think step by step.

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

1. **Understand the Bug**: The code seems to have issues with the way it processes the colorings and the matching of the frequency of incoming and outgoing vertices. It also has a specific check for all vertices being of the same type in one graph, which might be a potential edge case.

2. **Identify Edge Cases**:
   - **All vertices of the same type**: This is already handled in the code, so we should ensure our test case does not fall into this category.
   - **Different numbers of incoming and outgoing vertices**: This is also checked, so we need to ensure the counts are balanced.
   - **Cycles and colorings**: The code uses colorings to ensure that the cycles are of length divisible by \( k \). We need to create a graph where the colorings might not align properly.

3. **Create a Small-Scale Test Case**:
   - **Graphs with the same number of vertices and edges**.
   - **Balanced incoming and outgoing vertices**.
   - **Graphs that are strongly connected**.
   - **Cycles that are divisible by \( k \)**.
   - **Colorings that do not align properly**.

Let's create a test case with \( n = 4 \) and \( k = 2 \):

- **First Graph**:
  - Vertices: 4
  - Edges: 4 (forming a cycle)
  - Types: 2 incoming, 2 outgoing

- **Second Graph**:
  - Vertices: 4
  - Edges: 4 (forming a cycle)
  - Types: 2 incoming, 2 outgoing

We will ensure that the colorings do not align properly.

```plaintext
1
4 2
1 0 0 1
4
1 2
2 3
3 4
4 1
1 0 0 1
4
1 3
3 2
2 4
4 1
```

Explanation:
- **First Graph**:
  - Vertices: 1 (outgoing), 2 (incoming), 3 (incoming), 4 (outgoing)
  - Edges: (1, 2), (2, 3), (3, 4), (4, 1)
  - Cycle: 1 -> 2 -> 3 -> 4 -> 1 (length 4, divisible by 2)
  - Colorings: 1 -> 0, 2 -> 1, 3 -> 0, 4 -> 1

- **Second Graph**:
  - Vertices: 1 (outgoing), 2 (incoming), 3 (incoming), 4 (outgoing)
  - Edges: (1, 3), (3, 2), (2, 4), (4, 1)
  - Cycle: 1 -> 3 -> 2 -> 4 -> 1 (length 4, divisible by 2)
  - Colorings: 1 -> 0, 2 -> 0, 3 -> 1, 4 -> 1

In this test case, the colorings of the vertices do not align properly, which should expose the bug in the code. The code will have to handle the mismatch in colorings to determine if it can still form the required edges.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: C+K+S
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given two strongly connected$^{\dagger}$ directed graphs, each with exactly $n$ vertices, but possibly different numbers of edges. Upon closer inspection, you noticed an important feature — the length of any cycle in these graphs is divisible by $k$.

Each of the $2n$ vertices belongs to exactly one of two types: incoming or outgoing. For each vertex, its type is known to you.

You need to determine whether it is possible to draw exactly $n$ directed edges between the source graphs such that the following four conditions are met:

  * The ends of any added edge lie in different graphs.    * From each outgoing vertex, exactly one added edge originates.    * Into each incoming vertex, exactly one added edge enters.    * In the resulting graph, the length of any cycle is divisible by $k$.

$^{\dagger}$A strongly connected graph is a graph in which there is a path from every vertex to every other vertex.

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

The first line of each test case contains two integers $n$ and $k$ ($2 \le k \le n \le 2 \cdot 10^5$) — the number of vertices in each graph and the value by which the length of each cycle is divisible.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($a_i \in \\{0, 1\\}$). If $a_i = 0$, then vertex $i$ of the first graph is incoming. If $a_i = 1$, then vertex $i$ of the first graph is outgoing.

The third line of each test case contains a single integer $m_1$ ($1 \le m_1 \le 5 \cdot 10^5$) — the number of edges in the first graph.

The next $m_1$ lines contain descriptions of the edges of the first graph. The $i$-th of them contains two integers $v_i$ and $u_i$ ($1 \le v_i, u_i \le n$) — an edge in the first graph leading from vertex $v_i$ to vertex $u_i$.

Next, in the same format, follows the description of the second graph.

The next line contains $n$ integers $b_1, b_2, \ldots, b_n$ ($b_i \in \\{0, 1\\}$). If $b_i = 0$, then vertex $i$ of the second graph is incoming. If $b_i = 1$, then vertex $i$ of the second graph is outgoing.

The next line contains a single integer $m_2$ ($1 \le m_2 \le 5 \cdot 10^5$) — the number of edges in the second graph.

The next $m_2$ lines contain descriptions of the edges of the second graph. The $i$-th of them contains two integers $v_i$ and $u_i$ ($1 \le v_i, u_i \le n$) — an edge in the second graph leading from vertex $v_i$ to vertex $u_i$.

It is guaranteed that both graphs are strongly connected, and the lengths of all cycles are divisible by $k$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$. It is guaranteed that the sum of $m_1$ and the sum of $m_2$ over all test cases does not exceed $5 \cdot 10^5$.

Output: For each test case, output "YES" (without quotes) if it is possible to draw $n$ new edges such that all conditions are met, and "NO" (without quotes) otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
3
4 2
1 0 0 1
4
1 2
2 3
3 4
4 1
1 0 0 1
4
1 3
3 2
2 4
4 1
3 3
0 0 0
3
1 2
2 3
3 1
1 1 0
3
1 2
2 3
3 1
4 2
1 1 1 1
4
1 2
2 3
3 4
4 1
0 0 0 0
6
1 2
2 1
1 3
3 1
1 4
4 1
output:
YES
NO
YES

Note: In the first test case, it is possible to draw edges from the first graph to the second graph as $(1, 3)$ and $(4, 2)$ (the first number in the pair is the vertex number in the first graph, and the second number in the pair is the vertex number in the second graph), and from the second graph to the first graph as $(1, 2)$, $(4, 3)$ (the first number in the pair is the vertex number in the second graph, and the second number in the pair is the vertex number in the first graph).

In the second test case, there are a total of $4$ incoming vertices and $2$ outgoing vertices, so it is not possible to draw $3$ edges.

Buggy Code:
// IN THE NAME OF GOD
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define endl '\n'
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define fuck(x) cout<<"("<<#x<<" : "<<x<<")\n"
#define pb push_back
#define pf push_front
#define Mp make_pair
#define size(x) (ll)x.size()

const int N = 6e5 + 23, lg = 19;
ll Mod = 1e9 + 7;

ll MOD(ll a, ll mod=Mod) {a%=mod; (a<0)&&(a+=mod); return a;}
ll poww(ll a, ll b, ll mod=Mod) {
    ll res = 1;
    while(b > 0) {
        if(b%2 == 1) res = MOD(res * a, mod);
        b /= 2;
        a = MOD(a * a, mod);
    }
    return res;
}

int t, n, k, m1, m2, a[N], b[N], mark[2][N], cnt[2][2][N];
vector<int> adj[2][N];

void dfs1(int v, int now=0) {
    mark[0][v] = 1;
    cnt[0][a[v]][now] ++;
    for(int u : adj[0][v]) {
        if(mark[0][u] == 1) continue;
        dfs1(u, MOD(now + 1, k));
    }
}

void dfs2(int v, int now=0) {
    mark[1][v] = 1;
    cnt[1][b[v]][now] ++;
    for(int u : adj[1][v]) {
        if(mark[1][u] == 1) continue;
        dfs2(u, MOD(now + 1, k));
    }
}

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

    cin>>t;
    while(t--) {
        cin>>n>>k;
        int sum = 0, sum2 = 0;
        for(int i=1; i<=n; i++) {
            cin>>a[i];
            a[i] ^= 1;
            sum += a[i];
            sum2 += a[i];
        }
        cin>>m1;
        for(int v,u,i=1; i<=m1; i++) {
            cin>>v>>u; adj[0][v].pb(u);
        }

        for(int i=1; i<=n; i++) {
            cin>>b[i];
            b[i] ^= 1;
            sum2 += b[i];
        }
        cin>>m2;
        for(int v,u,i=1; i<=m2; i++) {
            cin>>v>>u; adj[1][v].pb(u);
        }

        dfs1(1);
        dfs2(1);

        int ok = 1;
        for(int i=0; i<k; i++) {
            int nxt = MOD(i + 1, k);
            for(int j=0; j<1; j++) {
                if(cnt[0][j][i] != cnt[1][1-j][nxt]) ok = 0;
                if(cnt[1][j][i] != cnt[0][1-j][nxt]) ok = 0;
            }
        }

        // fuck(ok);
        if((ok == 1 || sum == n || sum == 0) && sum2 == n){
            cout<<"YES\n";
        } else {
            cout<<"NO\n";
        }

        for(int i=0; i<=n; i++) {
            a[i] = b[i] = 0;
            for(int j=0; j<2; j++) {
                adj[j][i].clear();
                mark[j][i] = cnt[j][0][i] = cnt[j][1][i] = 0;
            }
        }
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
4 3
1 0 0 1
4
1 2
2 3
3 4
4 1
0 1 1 0
4
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: C+K+S
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given two strongly connected$^{\dagger}$ directed graphs, each with exactly $n$ vertices, but possibly different numbers of edges. Upon closer inspection, you noticed an important feature — the length of any cycle in these graphs is divisible by $k$.

Each of the $2n$ vertices belongs to exactly one of two types: incoming or outgoing. For each vertex, its type is known to you.

You need to determine whether it is possible to draw exactly $n$ directed edges between the source graphs such that the following four conditions are met:

  * The ends of any added edge lie in different graphs.    * From each outgoing vertex, exactly one added edge originates.    * Into each incoming vertex, exactly one added edge enters.    * In the resulting graph, the length of any cycle is divisible by $k$.

$^{\dagger}$A strongly connected graph is a graph in which there is a path from every vertex to every other vertex.

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

The first line of each test case contains two integers $n$ and $k$ ($2 \le k \le n \le 2 \cdot 10^5$) — the number of vertices in each graph and the value by which the length of each cycle is divisible.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($a_i \in \\{0, 1\\}$). If $a_i = 0$, then vertex $i$ of the first graph is incoming. If $a_i = 1$, then vertex $i$ of the first graph is outgoing.

The third line of each test case contains a single integer $m_1$ ($1 \le m_1 \le 5 \cdot 10^5$) — the number of edges in the first graph.

The next $m_1$ lines contain descriptions of the edges of the first graph. The $i$-th of them contains two integers $v_i$ and $u_i$ ($1 \le v_i, u_i \le n$) — an edge in the first graph leading from vertex $v_i$ to vertex $u_i$.

Next, in the same format, follows the description of the second graph.

The next line contains $n$ integers $b_1, b_2, \ldots, b_n$ ($b_i \in \\{0, 1\\}$). If $b_i = 0$, then vertex $i$ of the second graph is incoming. If $b_i = 1$, then vertex $i$ of the second graph is outgoing.

The next line contains a single integer $m_2$ ($1 \le m_2 \le 5 \cdot 10^5$) — the number of edges in the second graph.

The next $m_2$ lines contain descriptions of the edges of the second graph. The $i$-th of them contains two integers $v_i$ and $u_i$ ($1 \le v_i, u_i \le n$) — an edge in the second graph leading from vertex $v_i$ to vertex $u_i$.

It is guaranteed that both graphs are strongly connected, and the lengths of all cycles are divisible by $k$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$. It is guaranteed that the sum of $m_1$ and the sum of $m_2$ over all test cases does not exceed $5 \cdot 10^5$.

Output: For each test case, output "YES" (without quotes) if it is possible to draw $n$ new edges such that all conditions are met, and "NO" (without quotes) otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
3
4 2
1 0 0 1
4
1 2
2 3
3 4
4 1
1 0 0 1
4
1 3
3 2
2 4
4 1
3 3
0 0 0
3
1 2
2 3
3 1
1 1 0
3
1 2
2 3
3 1
4 2
1 1 1 1
4
1 2
2 3
3 4
4 1
0 0 0 0
6
1 2
2 1
1 3
3 1
1 4
4 1
output:
YES
NO
YES

Note: In the first test case, it is possible to draw edges from the first graph to the second graph as $(1, 3)$ and $(4, 2)$ (the first number in the pair is the vertex number in the first graph, and the second number in the pair is the vertex number in the second graph), and from the second graph to the first graph as $(1, 2)$, $(4, 3)$ (the first number in the pair is the vertex number in the second graph, and the second number in the pair is the vertex number in the first graph).

In the second test case, there are a total of $4$ incoming vertices and $2$ outgoing vertices, so it is not possible to draw $3$ edges.

Buggy Code:
#include<bits/stdc++.h>
#define pii pair<int,int>
#define fi first
#define se second
#define int long long
#define endl '\n'
#define ull unsigned long long
using namespace std;
const int N=4e5+5;
const ull B=379;
vector<int>G[N];
int n,k,m;
int p[N],d[N];
bool vis[N];
ull s[N],pw[N],c[2][2][N];
int ct[2][2];
void dfs(int u){
    vis[u]=1;
    for(auto v:G[u])
        if(!vis[v]){d[v]=(d[u]+1)%k;dfs(v);}
}
void get(int op){
    for(int i=1;i<=n;i++){
        cin>>p[i];
        ct[op][p[i]]++;
    }
    cin>>m;
    for(int i=1;i<=n;i++){
        G[i].clear();
        vis[i]=0;
    }
    d[1]=0;
    for(int i=1;i<=m;i++){
        int u,v;
        cin>>u>>v;
        G[u].push_back(v);
    }
    dfs(1);
    for(int i=0;i<=k;i++)
        c[op][0][i]=c[op][1][i]=0;
    for(int i=1;i<=n;i++)c[op][p[i]][d[i]+1]++;
}
ull ask(int l,int r){
    return s[r]-s[l-1]*pw[r-l+1];
}
set<int>st[2];
void check(int op){
    s[0]=0;pw[0]=1;
    for(int i=1;i<=k;i++){
        pw[i]=pw[i-1]*B;
        s[i]=s[i-1]*B+(c[op][0][i]+1);
    }
    for(int i=k+1;i<=2*k;i++)
        s[i]=s[i-1]*B+(c[op][0][i-k]+1);
    ull hs=0;
    for(int i=1;i<=k;i++)hs=hs*B+(c[op^1][1][i]+1);
    st[op].clear();
    for(int i=1;i<=k;i++)
        if(ask(i,i+k-1)==hs)st[op].insert(i);
    return;
}
void work(){
    memset(ct,0,sizeof(ct));
    cin>>n>>k;
    get(0);get(1);
    check(0);check(1);
    if(ct[0][0]!=ct[1][1]){
        cout<<"NO"<<endl;
        return;
    }
    if(ct[0][0]==n||ct[1][0]==n){
        cout<<"YES"<<endl;
        return;
    }
    if(st[0].size()==0||st[1].size()==0){
        cout<<"NO"<<endl;
        return;
    }
    for(auto v:st[1])
        if(st[0].find(v)!=st[0].end()){
            cout<<"YES"<<endl;
            return;
        }
    cout<<"NO"<<endl;
    return;
}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int T=1;
    cin>>T;
    while(T--)work();
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
4 2
1 0 0 1
4
1 2
2 3
3 4
4 1
1 0 0 1
4
1 3
3 2
2 4
4 1
3 3
0 0 0
3
1 2
2 3
3 1
1 1 0
3
1 2
2 3
3 1
4 2
1 1 1 1
4
1 2
2 3
3 4
4 1
0 0 0 0
6
1 2
2 1
1 3
3 1
1 4
4 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: C+K+S
time_limit_ms: 3000
memory_limit_mb: 256
Description: You are given two strongly connected$^{\dagger}$ directed graphs, each with exactly $n$ vertices, but possibly different numbers of edges. Upon closer inspection, you noticed an important feature — the length of any cycle in these graphs is divisible by $k$.

Each of the $2n$ vertices belongs to exactly one of two types: incoming or outgoing. For each vertex, its type is known to you.

You need to determine whether it is possible to draw exactly $n$ directed edges between the source graphs such that the following four conditions are met:

  * The ends of any added edge lie in different graphs.    * From each outgoing vertex, exactly one added edge originates.    * Into each incoming vertex, exactly one added edge enters.    * In the resulting graph, the length of any cycle is divisible by $k$.

$^{\dagger}$A strongly connected graph is a graph in which there is a path from every vertex to every other vertex.

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

The first line of each test case contains two integers $n$ and $k$ ($2 \le k \le n \le 2 \cdot 10^5$) — the number of vertices in each graph and the value by which the length of each cycle is divisible.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($a_i \in \\{0, 1\\}$). If $a_i = 0$, then vertex $i$ of the first graph is incoming. If $a_i = 1$, then vertex $i$ of the first graph is outgoing.

The third line of each test case contains a single integer $m_1$ ($1 \le m_1 \le 5 \cdot 10^5$) — the number of edges in the first graph.

The next $m_1$ lines contain descriptions of the edges of the first graph. The $i$-th of them contains two integers $v_i$ and $u_i$ ($1 \le v_i, u_i \le n$) — an edge in the first graph leading from vertex $v_i$ to vertex $u_i$.

Next, in the same format, follows the description of the second graph.

The next line contains $n$ integers $b_1, b_2, \ldots, b_n$ ($b_i \in \\{0, 1\\}$). If $b_i = 0$, then vertex $i$ of the second graph is incoming. If $b_i = 1$, then vertex $i$ of the second graph is outgoing.

The next line contains a single integer $m_2$ ($1 \le m_2 \le 5 \cdot 10^5$) — the number of edges in the second graph.

The next $m_2$ lines contain descriptions of the edges of the second graph. The $i$-th of them contains two integers $v_i$ and $u_i$ ($1 \le v_i, u_i \le n$) — an edge in the second graph leading from vertex $v_i$ to vertex $u_i$.

It is guaranteed that both graphs are strongly connected, and the lengths of all cycles are divisible by $k$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$. It is guaranteed that the sum of $m_1$ and the sum of $m_2$ over all test cases does not exceed $5 \cdot 10^5$.

Output: For each test case, output "YES" (without quotes) if it is possible to draw $n$ new edges such that all conditions are met, and "NO" (without quotes) otherwise.

You may output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
3
4 2
1 0 0 1
4
1 2
2 3
3 4
4 1
1 0 0 1
4
1 3
3 2
2 4
4 1
3 3
0 0 0
3
1 2
2 3
3 1
1 1 0
3
1 2
2 3
3 1
4 2
1 1 1 1
4
1 2
2 3
3 4
4 1
0 0 0 0
6
1 2
2 1
1 3
3 1
1 4
4 1
output:
YES
NO
YES

Note: In the first test case, it is possible to draw edges from the first graph to the second graph as $(1, 3)$ and $(4, 2)$ (the first number in the pair is the vertex number in the first graph, and the second number in the pair is the vertex number in the second graph), and from the second graph to the first graph as $(1, 2)$, $(4, 3)$ (the first number in the pair is the vertex number in the second graph, and the second number in the pair is the vertex number in the first graph).

In the second test case, there are a total of $4$ incoming vertices and $2$ outgoing vertices, so it is not possible to draw $3$ edges.

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

#define fr(i, a, b) for (int i = (a); i <= (b); ++i)
#define rf(i, a, b) for (int i = (a); i >= (b); --i)
#define fe(x, y) for (auto& x : y)

#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mt make_tuple

#define all(x) (x).begin(), (x).end()
#define sz(x) (int)(x).size()
#define pw(x) (1LL << (x))

using namespace std;

#ifdef LOCAL
mt19937_64 rng(228);
#else
mt19937_64 rng(chrono::system_clock::now().time_since_epoch().count());
#endif

#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template <typename T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define fbo find_by_order
#define ook order_of_key

template <typename T>
bool umn(T& a, T b) {
    return a > b ? a = b, 1 : 0;
}
template <typename T>
bool umx(T& a, T b) {
    return a < b ? a = b, 1 : 0;
}

using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template <typename T>
using ve = vector<T>;

ll rnd(ll l, ll r) {
    return uniform_int_distribution<ll>(l, r)(rng);
}

const int N = 5e5 + 5;

int n, k;
int a[N], b[N];
int m1, m2;

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

    int cnt1[2] = {0, 0};
    int cnt2[2] = {0, 0};

    fr (i, 1, n) {
        cin >> a[i];
        cnt1[a[i]]++;
    }

    cin >> m1;
    fr (i, 1, m1) {
        int a, b;
        cin >> a >> b;
    }

    fr (i, 1, n) {
        cin >> b[i];
        cnt2[b[i]]++;
    }

    cin >> m2;
    fr (i, 1, m2) {
        int a, b;
        cin >> a >> b;
    }

    if (cnt1[0] != cnt2[1]) {
        cout << "NO\n";
        return;
    }

    if (k == 2 || k == 4) {
        cout << "YES\n";
        return;
    }

    cout << "NO\n";
}

int main() {
#ifdef LOCAL
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
#else
    ios::sync_with_stdio(0);
    cin.tie(0);
#endif

    int test;
    cin >> test;

    while (test--) {
        solve();
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
4 3
1 0 0 1
4
1 2
2 3
3 4
4 1
1 0 0 1
4
1 3
3 2
2 4
4 1
3 3
0 0 0
3
1 2
2 3
3 1
1 1 0
3
1 2
2 3
3 1
4 3
1 1 1 1
4
1 2
2 3
3 4
4 1
0 0 0 0
6
1 2
2 1
1 3
3 1
1 4
4 1
```