**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 <iostream>
#include <vector>
#include <algorithm>
#include <random>
#include <ctime>
using namespace std;
#define int long long
// 图的邻接表表示
class Graph {
public:
    vector<vector<int>> edges;//有向边
    vector<int> type;//节点类型（1：出点  0：入点）
    vector<int> vis;//标记节点是否被访问
    vector<int> inColor, outColor;//inColor[i]:颜色为i的传入点数量  outColor[i]:颜色为i的传出点数量
    int n; // 图的节点数
    int k;//颜色数

    Graph(int nodes,int colorNum) : n(nodes),k(colorNum) {
        edges.resize(n+1);
        type.resize(n+1);
        vis.resize(n+1, 0);
        inColor.resize(k, 0);
        outColor.resize(k, 0);
    }

    // 添加边
    void addEdge(int u, int v) {
        edges[u].push_back(v);
    }

    // 使用DFS对第一个图进行染色
    void dfs1(int node, int color ) {
        vis[node] = 1;  // 标记当前节点u为已访问
        if (type[node] == 1) outColor[color]++;
        else inColor[color]++;

        for (int newnode : edges[node])
            if (vis[newnode] == 0)
                dfs1(newnode, (color + 1) % k);
    }

    // 使用DFS对第二个图进行染色
    void dfs2(int node, int color) {
        vis[node] = 1;  // 标记当前节点u为已访问
        if (type[node] == 1) inColor[(color +  1) % k]++;
        else outColor[(color + k - 1) % k]++;

        for (int newnode : edges[node])
            if (vis[newnode] == 0)
                dfs2(newnode, (color + 1) % k);
    }

    //计算传出点数量
    int outNodeNum() {
        int sum = 0;
        for (int i = 1; i <= n; i++)
            sum += type[i];
        return sum;
    }
};
// 生成KMP算法的next数组
vector<int> computeKMPNext(const vector<int>& pattern) {
    int m = pattern.size();
    vector<int> next(m, 0);
    int j = 0;

    for (int i = 1; i < m; i++) {
        while (j > 0 && pattern[i] != pattern[j]) {
            j = next[j - 1];
        }
        if (pattern[i] == pattern[j]) {
            j++;
        }
        next[i] = j;
    }
    return next;
}

// KMP字符串匹配算法
bool KMP(const vector<int>& text, const vector<int>& pattern) {
    int n = text.size(), m = pattern.size();
    vector<int> next = computeKMPNext(pattern);
    int j = 0;

    for (int i = 0; i < n; i++) {
        while (j > 0 && text[i] != pattern[j]) {
            j = next[j - 1];
        }
        if (text[i] == pattern[j]) {
            j++;
        }
        if (j == m) {
            return true; // 找到匹配
        }
    }
    return false;
}

// 判断两个数组是否循环同构
bool isCyclicallyIsomorphic(const vector<int>& A, const vector<int>& B) {
    if (A.size() != B.size()) {
        return false;
    }
    vector<int> extendedA = A;
    extendedA.insert(extendedA.end(), A.begin(), A.end()); // A+A
    return KMP(extendedA, B);
}

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

    while (t--) {
        int n, k;
        cin >> n >> k;
        Graph g1(n,k), g2(n,k);//创建两个图
        for (int i = 1; i <= n; i++) cin >> g1.type[i];//图一每个节点的类型
        int m1;//边数量
        cin >> m1;
        for (int i = 1; i <= m1; i++) {//图一的边
            int u, v;
            cin >> u >> v;
            g1.addEdge(u, v);
        }
        g1.dfs1(1, 0);//为图一染色

        for (int i = 1; i <= n; i++) cin >> g2.type[i];//图二每个节点的类型
        int m2;//边数量
        cin >> m2;
        for (int i = 1; i <= m2; i++) {//图二的边
            int u, v;
            cin >> u >> v;
            g2.addEdge(u, v);
        }
        g2.dfs2(1, 0);//为图二染色

        bool flag = false;

        //判断是否循环同构（这里采用KMP算法，也可使用哈希法）
		if (isCyclicallyIsomorphic(g1.inColor, g2.inColor) && isCyclicallyIsomorphic(g1.outColor, g2.outColor))
			flag = true;

        //一图全为传出点，另一图全为传入点，则两张图不会形成新的环
        if ((g1.outNodeNum() == n && g2.outNodeNum() == 0) || (g2.outNodeNum() == n && g1.outNodeNum() == 0))
            flag = true;

        // 输出结果
        cout << (flag == true ? "YES\n" : "NO\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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider scenarios that are likely to cause the code to fail. Here are the key points to consider:

1. **Cycle Length Divisibility**: Ensure that the cycles in the graphs are divisible by \( k \).
2. **Strongly Connected Graphs**: Both graphs must be strongly connected.
3. **Vertex Types**: The vertices must be correctly classified as incoming or outgoing.
4. **Edge Constraints**: The number of edges should be within the specified limits.

Given the code's logic, particularly the way it handles the coloring and the KMP-based cyclic isomorphism check, we can create a test case that challenges the following aspects:
- **Cyclic Isomorphism**: The graphs should have similar cyclic structures but not necessarily identical.
- **Vertex Types**: The vertex types should be such that the KMP-based check fails to detect a valid configuration.

Here's a small-scale test input that meets these criteria:

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

### Explanation:
1. **Graph 1**:
   - Vertices: 1, 2, 3, 4
   - Edges: 1 -> 2, 2 -> 3, 3 -> 4, 4 -> 1
   - Types: 1 (outgoing), 0 (incoming), 0 (incoming), 1 (outgoing)
   - This graph is strongly connected, and the cycle length is 4, which is divisible by \( k = 2 \).

2. **Graph 2**:
   - Vertices: 1, 2, 3, 4
   - Edges: 1 -> 3, 3 -> 2, 2 -> 4, 4 -> 1
   - Types: 0 (incoming), 1 (outgoing), 1 (outgoing), 0 (incoming)
   - This graph is also strongly connected, and the cycle length is 4, which is divisible by \( k = 2 \).

### Why This Test Case is Challenging:
- **Cyclic Isomorphism**: Both graphs have the same cycle length (4) and are strongly connected, but the vertex types are different. The KMP-based check might fail to correctly identify the cyclic isomorphism due to the different vertex types.
- **Vertex Types**: The vertex types are such that the KMP-based check might not correctly match the inColor and outColor arrays, leading to a false negative.

This test case should expose any issues in the KMP-based cyclic isomorphism check and the way the code handles vertex types.


**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:
from bisect import bisect_left, bisect_right
from itertools import combinations, permutations, product
from collections import defaultdict, Counter, deque
from heapq import heappush, heappop
from math import gcd, isqrt, lcm
from random import randint, shuffle
import sys

lol = lambda func=int: list(map(func, sys.stdin.readline().split()))
t = lol()[0]

def compute_lps(pattern):
    """Builds the LPS (Longest Prefix Suffix) array."""
    lps = [0] * len(pattern)
    j = 0  # Length of the previous longest prefix suffix

    for i in range(1, len(pattern)):
        if pattern[i] == pattern[j]:
            j += 1
            lps[i] = j
        else:
            if j != 0:
                j = lps[j - 1]
                i -= 1  # Stay at the same i
            else:
                lps[i] = 0
    return lps

def kmp_search(text, pattern):
    """KMP search to find the pattern in the text."""
    lps = compute_lps(pattern)
    i = j = 0  # i -> text, j -> pattern

    while i < len(text):
        if text[i] == pattern[j]:
            i += 1
            j += 1

        if j == len(pattern):
            return True  # Found pattern in text
        elif i < len(text) and text[i] != pattern[j]:
            if j != 0:
                j = lps[j - 1]
            else:
                i += 1
    return False

def are_cyclic_rotations(str1, str2):
    """Checks if str2 is a cyclic rotation of str1."""
    if len(str1) != len(str2):
        return False
    return kmp_search(str1 + str1, str2)

for case in range(t):
    n, k = lol()

    a1 = lol()
    g1 = [[] for _ in range(n)]
    m1 = lol()[0]
    for _ in range(m1):
        x, y = lol()
        x -= 1
        y -= 1
        g1[x].append(y)
    a2 = lol()
    g2 = [[] for _ in range(n)]
    m2 = lol()[0]
    for _ in range(m2):
        x, y = lol()
        x -= 1
        y -= 1
        g2[x].append(y)
    if case == 167:
        print(n, k)
        print(*a1)
        print(m1)
        for i in range(m1):
            for j in g1[i]:
                print(i + 1, j + 1)
        print(*a2)
        print(m2)
        for i in range(m2):
            for j in g2[i]:
                print(i + 1, j + 1)

    if 0 not in a1 and 1 not in a2:
        print("YES")
        continue
    if 0 not in a2 and 1 not in a1:
        print("YES")
        continue

    pos1 = [-1] * n
    pos2 = [-1] * n
    pos1[0] = 0
    stack = [0]
    while stack:
        v = stack.pop()
        for u in g1[v]:
            if pos1[u] == -1:
                pos1[u] = (pos1[v] + 1) % k
                stack.append(u)
    pos2[0] = 0
    stack = [0]
    while stack:
        v = stack.pop()
        for u in g2[v]:
            if pos2[u] == -1:
                pos2[u] = (pos2[v] + 1) % k
                stack.append(u)
    cnt1 = [[0, 0] for _ in range(k)]
    cnt2 = [[0, 0] for _ in range(k)]
    cnt3 = [[0, 0] for _ in range(k)]

    for i in range(n):
        cnt1[pos1[i]][a1[i]] += 1
        if a2[i] == 1:
            cnt2[pos2[i] - 1][0] += 1
        else:
            cnt2[pos2[i] + 1 - k][1] += 1
        if a2[i] == 1:
            cnt3[pos2[i] + 1 - k][0] += 1
        else:
            cnt3[pos2[i] - 1][1] += 1
    # print(cnt1, cnt2)

    cnt1 = [tuple(x) for x in cnt1]
    cnt2 = [tuple(x) for x in cnt2]
    # print(cnt1, cnt2)
    ans = are_cyclic_rotations(cnt1, cnt2) or are_cyclic_rotations(cnt1, cnt3)
    if t == 3 or t == 1 or True:
        print("YES" if ans else "NO")

**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 code, we need to focus on edge cases and scenarios that maximize the failure probability. Here are the steps to think through this:

1. **Understanding the Code's Logic**:
   - The code checks if the graphs can be connected such that the resulting graph's cycles are divisible by `k`.
   - It uses the KMP algorithm to check if two arrays (representing the positions of incoming and outgoing vertices) are cyclic rotations of each other.

2. **Identifying Potential Bugs**:
   - The code might fail if the graphs have a specific structure that makes it impossible to satisfy the conditions.
   - The code might not handle edge cases correctly, such as when the graphs are very small or have a specific configuration of edges and vertex types.

3. **Edge Cases to Consider**:
   - **Small Graphs**: Test with the smallest possible graphs.
   - **Graphs with Few Edges**: Test with graphs that have the minimum number of edges required to be strongly connected.
   - **Graphs with Specific Vertex Types**: Test with graphs where the number of incoming and outgoing vertices is not balanced.
   - **Graphs with Cycles of Different Lengths**: Test with graphs where the cycle lengths are divisible by `k` but in a non-trivial way.

4. **Generating the Test Input**:
   - We will create a small graph with `n = 4` and `k = 2`.
   - We will ensure that the graphs are strongly connected and the cycle lengths are divisible by `k`.
   - We will configure the vertex types in a way that makes it difficult to satisfy the conditions.

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

```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:
- **Test Case 1**:
  - `n = 4` and `k = 2`.
  - The first graph has vertices with types: `1 0 0 1` (outgoing, incoming, incoming, outgoing).
  - The first graph has edges: `1 -> 2`, `2 -> 3`, `3 -> 4`, `4 -> 1`.
  - The second graph has vertices with types: `1 0 0 1` (outgoing, incoming, incoming, outgoing).
  - The second graph has edges: `1 -> 3`, `3 -> 2`, `2 -> 4`, `4 -> 1`.

This test case ensures that the graphs are strongly connected and the cycle lengths are divisible by `k`. The vertex types are configured to make it difficult to satisfy the conditions, especially the requirement that the length of any cycle in the resulting graph is divisible by `k`. This should help in exposing 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:
def check_feasibility(n, k, a_vertices, b_vertices):
    # Count incoming and outgoing vertices in each graph
    a_incoming_count = sum(1 for x in a_vertices if x == 0)
    a_outgoing_count = sum(1 for x in a_vertices if x == 1)

    b_incoming_count = sum(1 for x in b_vertices if x == 0)
    b_outgoing_count = sum(1 for x in b_vertices if x == 1)

    # Feasibility check
    return a_outgoing_count == b_incoming_count and a_incoming_count == b_outgoing_count

# Read the number of test cases
T = int(input())
results = []

for _ in range(T):
    n, k = map(int, input().split())
    a_vertices = list(map(int, input().split()))
    m1 = int(input())  # Number of edges in the first graph; unused in solution
    for _ in range(m1):  # Read edges; unused in solution
        input()

    b_vertices = list(map(int, input().split()))
    m2 = int(input())  # Number of edges in the second graph; unused in solution
    for _ in range(m2):  # Read edges; unused in solution
        input()

    if check_feasibility(n, k, a_vertices, b_vertices):
        results.append("YES")
    else:
        results.append("NO")

print("\n".join(results))

**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:
import java.util.Scanner;

public class ColorThePicture {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt(); // عدد الحالات
        StringBuilder result = new StringBuilder();

        while (t-- > 0) {
            int n = scanner.nextInt(); // عدد القمم
            int k = scanner.nextInt(); // القيمة التي يجب أن يكون طول كل حلقة قابلاً للقسمة عليها

            int[] a = new int[n]; // نوع القمم في الرسم البياني الأول
            for (int i = 0; i < n; i++) {
                a[i] = scanner.nextInt();
            }

            int m1 = scanner.nextInt(); // عدد الحواف في الرسم البياني الأول
            for (int i = 0; i < m1; i++) {
                scanner.nextInt(); // تجاهل إدخال الحواف
                scanner.nextInt();
            }

            int[] b = new int[n]; // نوع القمم في الرسم البياني الثاني
            for (int i = 0; i < n; i++) {
                b[i] = scanner.nextInt();
            }

            int m2 = scanner.nextInt(); // عدد الحواف في الرسم البياني الثاني
            for (int i = 0; i < m2; i++) {
                scanner.nextInt(); // تجاهل إدخال الحواف
                scanner.nextInt();
            }

            // احصاء الـ incoming و outgoing
            int outgoingA = 0, incomingA = 0;
            for (int i = 0; i < n; i++) {
                if (a[i] == 1) outgoingA++;
                else incomingA++;
            }

            int outgoingB = 0, incomingB = 0;
            for (int i = 0; i < n; i++) {
                if (b[i] == 1) outgoingB++;
                else incomingB++;
            }

            // التحقق من إمكانية رسم الحواف
            if (outgoingA == incomingB && outgoingB == incomingA) {
                result.append("YES\n");
            } else {
                result.append("NO\n");
            }
        }

        System.out.print(result);
        scanner.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
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
```


**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:
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int t = sc.nextInt();
        StringBuilder result = new StringBuilder();

        while (t-- > 0) {
            int n = sc.nextInt();
            int k = sc.nextInt();

            int[] a = new int[n];
            int out_a = 0, in_a = 0;

            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
                out_a += a[i] == 1 ? 1 : 0;
                in_a += a[i] == 0 ? 1 : 0;
            }

            int m1 = sc.nextInt();

            for (int i = 0; i < m1; i++) {
                sc.nextInt();
                sc.nextInt();
            }

            int[] b = new int[n];
            int out_b = 0, in_b = 0;

            for (int i = 0; i < n; i++) {
                b[i] = sc.nextInt();
                out_b += b[i] == 1 ? 1 : 0;
                in_b += b[i] == 0 ? 1 : 0;
            }

            int m2 = sc.nextInt();
            for (int i = 0; i < m2; i++) {
                sc.nextInt();
                sc.nextInt();
            }

          	String res = (out_a == in_b && out_b == in_a) ? "YES\n" : "NO\n";
          	result.append(res);
        }

        System.out.print(result);
        sc.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
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
```


**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:
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int t = sc.nextInt();
        StringBuilder result = new StringBuilder();

        while (t-- > 0) {
            int n = sc.nextInt();
            int k = sc.nextInt();

            int[] a = new int[n];
            int out_a = 0, in_a = 0;

            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
                out_a += a[i] == 1 ? 1 : 0;
                in_a += a[i] == 0 ? 1 : 0;
            }

            int m1 = sc.nextInt();

            for (int i = 0; i < m1; i++) {
                sc.nextInt();
                sc.nextInt();
            }

            int[] b = new int[n];
            int out_b = 0, in_b = 0;

            for (int i = 0; i < n; i++) {
                b[i] = sc.nextInt();
                out_b += b[i] == 1 ? 1 : 0;
                in_b += b[i] == 0 ? 1 : 0;
            }

            int m2 = sc.nextInt();
            for (int i = 0; i < m2; i++) {
                sc.nextInt();
                sc.nextInt();
            }

            result.append((out_a == in_b && out_b == in_a) ? "YES\n" : "NO\n");
        }

        System.out.print(result);
        sc.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
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
```

### Explanation:

1. **Test Case 1:**
   - **n = 4, k = 2**
   - **First Graph:**
     - Vertices: 1 (outgoing), 2 (incoming), 3 (incoming), 4 (outgoing)
     - Edges: (1, 2), (2, 3), (3, 4), (4, 1)
   - **Second Graph:**
     - Vertices: 1 (incoming), 2 (outgoing), 3 (outgoing), 4 (incoming)
     - Edges: (1, 3), (3, 2), (2, 4), (4, 1)
   - **Expected Output:** YES
   - **Explanation:** The vertices can be matched as follows: (1, 3), (4, 2), (2, 1), (3, 4).

2. **Test Case 2:**
   - **n = 3, k = 3**
   - **First Graph:**
     - Vertices: 1 (incoming), 2 (incoming), 3 (incoming)
     - Edges: (1, 2), (2, 3), (3, 1)
   - **Second Graph:**
     - Vertices: 1 (outgoing), 2 (outgoing), 3 (incoming)
     - Edges: (1, 2), (2, 3), (3, 1)
   - **Expected Output:** NO
   - **Explanation:** There are 3 incoming vertices in the first graph and only 1 outgoing vertex in the second graph, making it impossible to draw 3 edges.

3. **Test Case 3:**
   - **n = 4, k = 2**
   - **First Graph:**
     - Vertices: 1 (outgoing), 2 (outgoing), 3 (outgoing), 4 (outgoing)
     - Edges: (1, 2), (2, 3), (3, 4), (4, 1)
   - **Second Graph:**
     - Vertices: 1 (incoming), 2 (incoming), 3 (incoming), 4 (incoming)
     - Edges: (1, 2), (2, 1), (1, 3), (3, 1), (1, 4), (4, 1)
   - **Expected Output:** YES
   - **Explanation:** The vertices can be matched as follows: (1, 1), (2, 2), (3, 3), (4, 4).

These test cases are designed to expose the bug in the provided buggy code by checking different scenarios where the number of incoming and outgoing vertices are balanced and unbalanced.


**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:
import java.util.*;

public class Main {
	public static void main(String[] args) {
      	Scanner sc = new Scanner(System.in);

      	int t = sc.nextInt();

      	while(t-- > 0) {
          	int n = sc.nextInt();
          	int k = sc.nextInt();

			int[] a = new int[n];
			int out_a = 0, in_a = 0;

          	for(int i = 0; i < n; i++) {
              	a[i] = sc.nextInt();

              	if(a[i] == 1) {
                  	out_a++;
                }

              	else {
                  	in_a++;
                }
            }

          	int m1 = sc.nextInt();

          	for(int i = 0; i < m1; i++) {
              	int u = sc.nextInt();
              	int v = sc.nextInt();
            }

          	int[] b = new int[n];
          	int out_b = 0, in_b = 0;

          	for(int i = 0; i < n; i++) {
              	b[i] = sc.nextInt();

              	if(b[i] == 1) {
                  	out_b++;
                }

              	else {
                  	in_b++;
                }
            }

          	int m2 = sc.nextInt();

          	for(int i = 0; i < m2; i++) {
              	int u = sc.nextInt();
              	int v = sc.nextInt();
            }

          	if(out_a == in_b && out_b == in_a) {
              	System.out.println("YES");
            }

          	else {
              	System.out.println("NO");
            }
        }

      	sc.close();
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on the logic that checks whether the number of outgoing vertices in the first graph matches the number of incoming vertices in the second graph, and vice versa. The code assumes that if these counts match, the solution is always possible, which is not necessarily true. We need a test case where these counts match but the solution is still impossible due to the cycle length constraint.

Let's consider a small-scale test input where the counts match but the cycle length constraint is violated.

### Test Input
```plaintext
1
4 3
1 0 1 0
4
1 2
2 3
3 4
4 1
0 1 0 1
4
1 2
2 3
3 4
4 1
```

### Explanation
- **n = 4, k = 3**: We have 4 vertices in each graph, and the length of any cycle must be divisible by 3.
- **a = [1, 0, 1, 0]**: The first graph has 2 outgoing vertices (1 and 3) and 2 incoming vertices (2 and 4).
- **b = [0, 1, 0, 1]**: The second graph has 2 outgoing vertices (2 and 4) and 2 incoming vertices (1 and 3).

The code will check:
- `out_a = 2` (vertices 1 and 3)
- `in_a = 2` (vertices 2 and 4)
- `out_b = 2` (vertices 2 and 4)
- `in_b = 2` (vertices 1 and 3)

Since `out_a == in_b` and `out_b == in_a`, the code will output "YES". However, this is incorrect because the cycle length constraint is violated.

### Why This Test Input Exposes the Bug
- The cycle length in both graphs is 4, which is not divisible by 3.
- Even though the counts match, the cycle length constraint makes it impossible to draw the edges such that all cycles in the resulting graph have lengths divisible by 3.

This test input should help expose the bug in the given 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:
def solve():
    import sys
    input = sys.stdin.read().split()

    index = 0
    t = int(input[index])
    index += 1

    results = []

    for _ in range(t):
        n = int(input[index])
        k = int(input[index + 1])
        index += 2

        a = list(map(int, input[index:index + n]))
        index += n

        m1 = int(input[index])
        index += 1

        for _ in range(m1):
            v = int(input[index])
            u = int(input[index + 1])
            index += 2

        b = list(map(int, input[index:index + n]))
        index += n

        m2 = int(input[index])
        index += 1

        for _ in range(m2):
            v = int(input[index])
            u = int(input[index + 1])
            index += 2

        # Count incoming and outgoing vertices in the first graph
        out1 = sum(a)
        inc1 = n - out1

        # Count incoming and outgoing vertices in the second graph
        out2 = sum(b)
        inc2 = n - out2

        # Check feasibility
        if out1 == inc2 and out2 == inc1:
            results.append("YES")
        else:
            results.append("NO")

    print("\n".join(results))

if __name__ == "__main__":
    solve()

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

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

Think step by step.

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


**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:
import collections,sys,math,functools,operator,itertools,bisect,heapq,decimal,string,time,random
#https://github.com/shakayami/ACL-for-python/blob/master/string.py
def z_algorithm(s):
    n=len(s)
    if n==0:
        return []
    z=[0]*n
    i=1;j=0
    while(i<n):
        z[i]=0 if (j+z[j]<=i) else min(j+z[j]-i,z[i-j])
        while((i+z[i]<n) and (s[z[i]]==s[i+z[i]])):
            z[i]+=1
        if (j+z[j]<i+z[i]):
            j=i
        i+=1
    z[0]=n
    return z
input = sys.stdin.readline
#sys.setrecursionlimit(10**9)
#n = int(input())
#alist = list(map(int,input().split()))
#alist = []
#s = input()
#n,m = map(int,input().split())
#for i in range(n):
#    alist.append(list(map(int,input().split())))

def solve():
    n,k = map(int,input().split())
    a1 = list(map(int,input().split()))
    m1 = int(input())
    edge1 = [[] for i in range(n)]
    for i in range(m1):
        u,v = map(int,input().split())
        u-=1
        v-=1
        edge1[u].append(v)
        edge1[v].append(u)
    a2 = list(map(int,input().split()))
    m2 = int(input())
    edge2 = [[] for i in range(n)]
    for i in range(m2):
        u,v = map(int,input().split())
        u-=1
        v-=1
        edge2[u].append(v)
        edge2[v].append(u)
    start = 0
    d = collections.deque()
    dist1 = [10**18 for i in range(n)]
    d.append(start)
    dist1[start] = 0
    d1 = [0 for i in range(n)]
    while d:
        now = d.popleft()
        for i in edge1[now]:
            if dist1[i] > dist1[now] + 1:
                dist1[i] = dist1[now] + 1
                d.append(i)
                d1[i] = (d1[now]+1)%k
    start = 0
    d = collections.deque()
    dist1 = [10**18 for i in range(n)]
    d.append(start)
    dist1[start] = 0
    d2 = [0 for i in range(n)]
    while d:
        now = d.popleft()
        for i in edge2[now]:
            if dist1[i] > dist1[now] + 1:
                dist1[i] = dist1[now] + 1
                d.append(i)
                d2[i] = (d2[now]+1)%k
    if (a1.count(0) == n and a2.count(1)) == n or (a1.count(1) == n and a2.count(0)):
        print('YES')
        return
    if a1.count(0) != a2.count(1) or a1.count(1) != a2.count(0):
        print("NO")
        return
    b1 = [0 for i in range(k)]
    for i in d1:
        b1[i] += 1
    s1 = set()
    x = b1[:] + [-1] + b1[:] + b1[:]
    y = z_algorithm(x)
    for i in range(k):
        if y[k+1+i] == k:
            s1.add(i)
    b2 = [0 for i in range(k)]
    for i in d2:
        b2[i] += 1
    s2 = set()
    x = b2[:] + [-1] + b2[:] + b2[:]
    y = z_algorithm(x)
    for i in range(k):
        if y[k+1+i] == k:
            s2.add(i)
    for i in s1:
        x = (2 - i)%k
        if x in s2:
            print('YES')
            return
    print('NO')

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

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that are likely to cause the algorithm to fail. Here are the steps to think through this:

1. **Understanding the Bug**:
   - The buggy code incorrectly calculates the distances and distances modulo `k` for both graphs.
   - It also has a flawed logic for checking the cycle lengths and the conditions for drawing edges.

2. **Edge Cases**:
   - **Graphs with Few Vertices**: Small graphs can quickly reveal issues in the algorithm.
   - **Graphs with Different Cycle Lengths**: Ensure that the cycle lengths are divisible by `k` but test different configurations.
   - **Graphs with Different Edge Counts**: Test graphs with different numbers of edges to see if the algorithm handles them correctly.
   - **Graphs with All Incoming or All Outgoing Vertices**: These cases can expose issues in the edge drawing logic.

3. **Challenging Test Input**:
   - **Test Case 1**: Two graphs with 3 vertices, where one graph has all incoming vertices and the other has all outgoing vertices.
   - **Test Case 2**: Two graphs with 4 vertices, where the cycle lengths are divisible by `k` but the number of edges is different.
   - **Test Case 3**: Two graphs with 5 vertices, where the cycle lengths are divisible by `k` but the vertices have mixed types.

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

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

This test input includes:
- **Test Case 1**: Two graphs with 3 vertices, where one graph has all incoming vertices and the other has all outgoing vertices.
- **Test Case 2**: Two graphs with 4 vertices, where the cycle lengths are divisible by 2, and the vertices have mixed types.
- **Test Case 3**: Two graphs with 5 vertices, where the cycle lengths are divisible by 3, and the vertices have mixed types.

This should help in exposing the bugs in the given algorithm.


**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:
def solve():
    import sys
    input = sys.stdin.read
    data = input().split()

    index = 0
    T = int(data[index])
    index += 1

    results = []

    for _ in range(T):
        n, k = map(int, [data[index], data[index + 1]])
        index += 2

        # Read the first graph
        graph1_incoming_outgoing = list(map(int, data[index:index + n]))
        index += n

        m1 = int(data[index])
        index += 1

        for _ in range(m1):
            index += 2

        # Read the second graph
        graph2_incoming_outgoing = list(map(int, data[index:index + n]))
        index += n

        m2 = int(data[index])
        index += 1

        for _ in range(m2):
            index += 2

        # Calculate the number of incoming and outgoing vertices
        incoming_count = sum(1 - x for x in graph1_incoming_outgoing) + sum(1 - y for y in graph2_incoming_outgoing)
        outgoing_count = sum(graph1_incoming_outgoing) + sum(graph2_incoming_outgoing)

        # Check if the counts match
        if incoming_count == outgoing_count:
            results.append("YES")
        else:
            results.append("NO")

    print("\n".join(results))

solve()

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

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

Think step by step.

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

1. **不平衡的输入输出顶点数量**：确保输入和输出顶点的数量不匹配，这样可以测试代码是否正确处理这种情况。
2. **强连通图**：确保生成的图是强连通的，并且所有循环的长度都能被 `k` 整除。
3. **小规模测试用例**：保持测试用例的小规模，以确保输出符合要求。

### 生成测试用例

考虑以下测试用例：

1. **测试用例1**：输入和输出顶点数量不匹配。
2. **测试用例2**：输入和输出顶点数量匹配，但图的结构使得无法满足条件。

### 具体生成

```plaintext
2
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
```

### 解释

1. **测试用例1**：
   - `n = 3`, `k = 3`
   - 第一个图的顶点类型：`0 0 0`（全部为输入）
   - 第一个图的边：`1 -> 2`, `2 -> 3`, `3 -> 1`（强连通，循环长度为3）
   - 第二个图的顶点类型：`1 1 0`（1个输入，2个输出）
   - 第二个图的边：`1 -> 2`, `2 -> 3`, `3 -> 1`（强连通，循环长度为3）

   由于第一个图没有输出顶点，而第二个图有2个输出顶点，因此无法满足条件。

2. **测试用例2**：
   - `n = 4`, `k = 2`
   - 第一个图的顶点类型：`1 1 1 1`（全部为输出）
   - 第一个图的边：`1 -> 2`, `2 -> 3`, `3 -> 4`, `4 -> 1`（强连通，循环长度为4）
   - 第二个图的顶点类型：`0 0 0 0`（全部为输入）
   - 第二个图的边：`1 -> 2`, `2 -> 1`, `1 -> 3`, `3 -> 1`, `1 -> 4`, `4 -> 1`（强连通，循环长度为2）

   由于第一个图没有输入顶点，而第二个图没有输出顶点，因此无法满足条件。

这两个测试用例应该能够有效暴露代码中的问题。


**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:
from collections import defaultdict
import sys

def read_int():
    return int(sys.stdin.readline())

def read_ints():
    return list(map(int, sys.stdin.readline().split()))

def dfs(graph, node, visited):
    stack = [node]
    while stack:
        v = stack.pop()
        if v not in visited:
            visited.add(v)
            for neighbor in graph[v]:
                if neighbor not in visited:
                    stack.append(neighbor)

def can_connect(n, k, a, edges1, b, edges2):
    outgoing_count_first = sum(a)
    incoming_count_first = n - outgoing_count_first
    outgoing_count_second = sum(b)
    incoming_count_second = n - outgoing_count_second

    # Check if the counts match
    if outgoing_count_first != incoming_count_second or incoming_count_first != outgoing_count_second:
        return "NO"

    # Build adjacency lists for both graphs
    graph1 = defaultdict(list)
    graph2 = defaultdict(list)

    for u, v in edges1:
        graph1[u].append(v)

    for u, v in edges2:
        graph2[u].append(v)

    # Perform DFS on both graphs to check connectivity and cycle properties
    visited1 = set()
    dfs(graph1, 1, visited1)
    visited2 = set()
    dfs(graph2, 1, visited2)

    if len(visited1) == n and len(visited2) == n:
        return "YES"
    else:
        return "NO"

# Main function to handle multiple test cases
def main():
    t = read_int()
    results = []
    for _ in range(t):
        n, k = read_ints()
        a = read_ints()
        m1 = read_int()
        edges1 = [read_ints() for _ in range(m1)]
        b = read_ints()
        m2 = read_int()
        edges2 = [read_ints() for _ in range(m2)]

        result = can_connect(n, k, a, edges1, b, edges2)
        results.append(result)

    for result in results:
        print(result)

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
```plaintext
1
4 3
0 1 0 1
4
1 2
2 3
3 4
4 1
1 0 1 0
4
1 3
3 2
2 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 LL long long
using namespace std;
const int N=4e5+10,md=1e9+9;
int cnt[2][2][N],tp[2][N],dis[2][N],n,k,vis[N],ins[2][N],ct[2][2];
LL pre[2][2][N],pw[N],ni[N],B=1331;
vector<int> e[2][N];
LL quick(LL a,int b){
	LL ans=1;
	while(b){
		if(b&1) ans=ans*a%md;
		a=a*a%md;
		b>>=1;
	}
	return ans;
}
void init(){
	pw[0]=ni[0]=1;
	int nn=200000;
	for(int i=1;i<=nn;i++) pw[i]=pw[i-1]*B%md;
	ni[nn]=quick(pw[nn],md-2);
	for(int i=nn-1;i>=1;i--) ni[i]=quick(pw[i],md-2);
}
void iiit(){
//	for(int r=0;r<=1;r++){
//		for(int l=0;l<=1;l++){
//			cout<<r<<" ?? "<<l<<endl;
//			for(int i=1;i<=k;i++) cout<<cnt[r][l][i]<<" ";
//			puts("");
//	    }
//	}
	for(int r=0;r<=1;r++){
		for(int l=0;l<=1;l++){
			for(int i=1;i<=k;i++) pre[r][l][i]=(pre[r][l][i-1]+cnt[r][l][i]*pw[i]);
		}
	}
}
LL Hash(int d1,int d2,int r){
//	cout<<d1<<" "<<d2<<" "<<k-r+1<<endl;
	return (pre[d1][d2][r]*pw[k-r]%md+(pre[d1][d2][k]-pre[d1][d2][r]+md)%md*ni[r]%md)%md;
}
void dfs(int d){
	queue<int> q;
	dis[d][1]=0;
	q.push(1);
	for(int i=1;i<=n;i++) ins[d][i]=0,dis[d][i]=0;
	while(q.size()){
		int x=q.front();
		q.pop();
		if(ins[d][x]) continue;
		cnt[d][tp[d][x]][dis[d][x]+1]++;
		ins[d][x]=1;
		for(int y:e[d][x]){
			if(ins[d][y]==1) continue;
			dis[d][y]=(dis[d][x]+1)%k;
			q.push(y);
		}
	}
}
int rd(){
	int ww=0,ff=1;
	char ch=getchar();
	while(ch<'0'||ch>'9') {
		if(ch=='-') ff=-1;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9'){
		ww=(ww<<3)+(ww<<1)+ch-'0';
		ch=getchar();
	}
	return ww*ff;
}
int main(){
	int T=rd(),tt=0,ttt=T,nw,kw;
	init();
	while(T--){
		tt++;
		n=rd(),k=rd();
		if(tt==1) nw=n,kw=k;
		ct[0][0]=ct[0][1]=ct[1][0]=ct[1][1]=0;
		if(tt==72&&ttt==10000&&nw!=2&&kw==2) cout<<n<<" "<<k<<endl;
		for(int d=0;d<=1;d++){
			for(int dd=0;dd<=1;dd++) for(int i=0;i<=k;i++) pre[d][dd][i]=cnt[d][dd][i]=0;
			for(int i=1;i<=n;i++) e[d][i].clear();
			for(int i=1;i<=n;i++) {
				tp[d][i]=rd();
				if(tt==72&&ttt==10000&&nw!=2&&kw==2) cout<<tp[d][i]<<endl;
				ct[d][tp[d][i]]++;
			}
			int m=rd();
			for(int i=1;i<=m;i++) {
				int x=rd(),y=rd();
				if(tt==72&&ttt==10000&&nw!=2&&kw==2) cout<<x<<" "<<y<<endl;
				e[d][x].push_back(y);
			}
		}
//		cout<<ct[0][1]<<" "<<ct[1][0]<<" "<<ct[0][0]<<" "<<ct[1][1]<<endl;
		if(ct[0][1]!=ct[1][0]||ct[0][0]!=ct[1][1]){
		 	puts("No");
		 	continue;
		}
		if(ct[0][1]==n||ct[0][0]==n) {
			puts("Yes");
			continue;
		}
//		puts("!!!");
		dfs(0),dfs(1);
//		for(int i=1;i<=k;i++) cout<<cnt[0][1][i]<<" ";puts("");
		iiit();
//		cout<<"hh"<<Hash(0,1,1)<<" "<<pre[1][0][2]<<endl;
		for(int i=0;i<=k;i++) vis[i]=0;
		for(int i=0;i<k;i++) {
			if(Hash(0,1,i)==pre[1][0][k]) {
				vis[i]=1;
//				cout<<i<<"?A?A?"<<endl;
			}
		}
		int flag=0;
		for(int i=0;i<=k;i++){
			if(Hash(1,1,i)==pre[0][0][k]){
				if(k-2-i>=0){
					if(vis[k-2-i]) {
						flag=1;
						break;
					}
				}else {
					if(vis[k+k-2-i]){
						flag=1;
						break;
					}
				}
			}
		}
		if(ttt==10000&&nw!=2&&kw==2) continue;
		puts(flag?"Yes":"No");
	}
	return 0;
}
/*
4
3 2
0 1 1
6
1 2
1 2
1 2
1 3
2 1
3 1
0 1 0
5
1 2
2 1
2 1
2 3
3 2
*/

**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 code, we need to consider the following:

1. **Edge Cases**: Test the boundaries of the input constraints, such as the minimum and maximum values for \( n \) and \( k \).
2. **Graph Properties**: Ensure that the graphs are strongly connected and that the lengths of all cycles are divisible by \( k \).
3. **Vertex Types**: Ensure that the number of incoming and outgoing vertices is balanced between the two graphs.
4. **Cycle Lengths**: Test scenarios where the cycle lengths are critical to the problem's constraints.

Given these considerations, let's create a test case that is small but challenging:

### Test Case
- **Number of Test Cases**: 1
- **Number of Vertices**: 4
- **Divisibility Factor**: 2
- **Graphs**:
  - **First Graph**:
    - Vertices: 4
    - Types: 1 (outgoing), 0 (incoming), 0 (incoming), 1 (outgoing)
    - Edges: 4
      - (1, 2)
      - (2, 3)
      - (3, 4)
      - (4, 1)
  - **Second Graph**:
    - Vertices: 4
    - Types: 0 (incoming), 1 (outgoing), 0 (incoming), 1 (outgoing)
    - Edges: 4
      - (1, 3)
      - (3, 2)
      - (2, 4)
      - (4, 1)

### Explanation
- The graphs are strongly connected and the lengths of all cycles are divisible by 2.
- The number of incoming and outgoing vertices is balanced between the two graphs.
- The cycle lengths in both graphs are 4, which is divisible by 2.

### Test Input
```plaintext
1
4 2
1 0 0 1
4
1 2
2 3
3 4
4 1
0 1 0 1
4
1 3
3 2
2 4
4 1
```

This test case should be challenging enough to expose any bugs in the code, particularly those related to the cycle length and vertex type constraints.