**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DFS Checker (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, the given tree is a perfect binary tree and the constraints on $n$ and $q$ are lower. You can make hacks only if both versions of the problem are solved.

You are given a perfect binary tree$^\dagger$ consisting of $n$ vertices. The vertices are numbered from $1$ to $n$, and the root is the vertex $1$. You are also given a permutation $p_1, p_2, \ldots, p_n$ of $[1,2,\ldots,n]$.

You need to answer $q$ queries. For each query, you are given two integers $x$, $y$; you need to swap $p_x$ and $p_y$ and determine if $p_1, p_2, \ldots, p_n$ is a valid DFS order$^\ddagger$ of the given tree.

Please note that the swaps are persistent through queries.

$^\dagger$ A perfect binary tree is a tree with vertex $1$ as its root, with size $n=2^k-1$ for a positive integer $k$, and where the parent of each vertex $i$ ($1<i\le n$) is $\left\lfloor\frac{i}{2}\right\rfloor$. Thus, all leaves of this tree are at a distance $k - 1$ from the root.

$^\ddagger$ A DFS order is found by calling the following $\texttt{dfs}$ function on the given tree.

              dfs_order = []          function dfs(v):         append v to the back of dfs_order         pick an arbitrary permutation s of children of v         for child in s:             dfs(child)     dfs(1)

Note that the DFS order is not unique.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $q$ ($3\le n\le 65\,535$, $2\le q\le 5 \cdot 10^4$) — the number of vertices in the tree and the number of queries. It is guaranteed that $n=2^k-1$ for a positive integer $k$.

The next line contains $n-1$ integers $a_2,a_3,\ldots,a_n$ ($1\le a_i<i$) — the parent of each vertex in the given tree. It is guaranteed that $a_i=\left\lfloor\frac{i}{2}\right\rfloor$.

The next line contains $n$ integers $p_1,p_2,\ldots,p_n$ ($1\le p_i\le n$, all $p_i$ are distinct) — the initial permutation $p$.

The next $q$ lines each contain two integers $x$, $y$ ($1\le x,y\le n,x\neq y$) — the positions of the elements to swap in the permutation.

It is guaranteed that the sum of all $n$ does not exceed $65\,535$, and the sum of all $q$ does not exceed $5 \cdot 10^4$.

Output: For each test case, print $q$ lines corresponding to the $q$ queries. For each query, output $\texttt{YES}$ if there is a DFS order that exactly equals the current permutation, and output $\texttt{NO}$ otherwise.

You can output $\texttt{Yes}$ and $\texttt{No}$ in any case (for example, strings $\texttt{yEs}$, $\texttt{yes}$, $\texttt{Yes}$ and $\texttt{YES}$ will be recognized as a positive response).

Examples:
input:
2
3 3
1 1
1 2 3
2 3
3 2
1 3
7 4
1 1 2 2 3 3
1 2 3 4 5 6 7
3 5
2 5
3 7
4 6
output:
YES
YES
NO
YES
NO
NO
YES

Note: In the first test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,3,2],[1,2,3],[3,2,1]$, respectively. The first two permutations are valid DFS orders; the third is not a DFS order.

In the second test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,2,5,4,3,6,7],[1,3,5,4,2,6,7],[1,3,7,4,2,6,5],[1,3,7,6,2,4,5]$, respectively.

Buggy Code:
// package CD_August_round;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class CDB {

    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(reader.readLine().trim());

        while (t-- > 0) {
            StringTokenizer st = new StringTokenizer(reader.readLine());
            int n = Integer.parseInt(st.nextToken());
            int q = Integer.parseInt(st.nextToken());

            int[] a = new int[n];
            Arrays.fill(a, -1);
            st = new StringTokenizer(reader.readLine());
            for (int i = 1; i < n; i++) a[i] = Integer.parseInt(st.nextToken()) - 1;

            int[] p = new int[n];
            st = new StringTokenizer(reader.readLine());
            for (int i = 0; i < n; i++) p[i] = Integer.parseInt(st.nextToken()) - 1;

            int[] pos = new int[n];
            for (int i = 0; i < n; i++) pos[p[i]] = i;

            int[] f = new int[n];
            List<Integer>[] adj = new ArrayList[n];
            for (int i = 0; i < n; i++) adj[i] = new ArrayList<>();
            for (int i = n - 1; i >= 0; i--) {
                if (i > 0) adj[a[i]].add(i);
                for (int j : adj[i]) f[i] += f[j];
                f[i]++;
            }

            Set<Integer>[] s = new HashSet[n];
            for (int i = 0; i < n; i++) s[i] = new HashSet<>();
            int bad = 0;

            for (int i = 0; i < n; i++) {
                if (a[p[i]] != -1) add(s, a, f, p, bad, i);
            }

            for (int i = 0; i < n; i++) {
                bad += calc(s, pos, i);
            }

            while (q-- > 0) {
                st = new StringTokenizer(reader.readLine());
                int x = Integer.parseInt(st.nextToken()) - 1;
                int y = Integer.parseInt(st.nextToken()) - 1;

                Set<Integer> change = new HashSet<>(Arrays.asList(p[x], p[y], a[p[x]], a[p[y]]));

                for (int z : change) {
                    if (z != -1) bad -= calc(s, pos, z);
                }

                if (a[p[x]] != -1) del(s, a, f, p, bad, x);
                if (a[p[y]] != -1) del(s, a, f, p, bad, y);

                int temp = p[x];
                p[x] = p[y];
                p[y] = temp;

                temp = pos[p[x]];
                pos[p[x]] = pos[p[y]];
                pos[p[y]] = temp;

                if (a[p[x]] != -1) add(s, a, f, p, bad, x);
                if (a[p[y]] != -1) add(s, a, f, p, bad, y);

                for (int z : change) {
                    if (z != -1) bad += calc(s, pos, z);
                }

                System.out.println(bad == 0 ? "YES" : "NO");
            }
        }
    }

    private static void add(Set<Integer>[] s, int[] a, int[] f, int[] p, int bad, int x) {
        int i = a[p[x]];
        s[i].add(x);
        Integer[] arr = s[i].toArray(new Integer[0]);
        Arrays.sort(arr);
        updateBad(arr, f, p, bad, x, true);
    }

    private static void del(Set<Integer>[] s, int[] a, int[] f, int[] p, int bad, int x) {
        int i = a[p[x]];
        s[i].remove(x);
        Integer[] arr = s[i].toArray(new Integer[0]);
        Arrays.sort(arr);
        updateBad(arr, f, p, bad, x, false);
    }

    private static void updateBad(Integer[] arr, int[] f, int[] p, int bad, int x, boolean add) {
        int l = -1, r = -1;
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] == x) {
                if (i > 0) l = arr[i - 1];
                if (i < arr.length - 1) r = arr[i + 1];
            }
        }
        if (l != -1 && r != -1) {
            if (add) {
                if (l + f[p[l]] != r) bad++;
            } else {
                if (l + f[p[l]] != r) bad--;
            }
        }
    }

    private static int calc(Set<Integer>[] s, int[] pos, int i) {
        if (s[i].isEmpty()) return 0;
        int x = pos[i];
        int y = s[i].stream().min(Integer::compareTo).orElse(-1);
        return y < x ? 1 : 0;
    }
}

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

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

Think step by step.

Response:
```plaintext
2
3 2
1 1
1 2 3
2 3
3 2
7 3
1 1 2 2 3 3
1 2 3 4 5 6 7
3 4
4 6
6 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DFS Checker (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, the given tree is a perfect binary tree and the constraints on $n$ and $q$ are lower. You can make hacks only if both versions of the problem are solved.

You are given a perfect binary tree$^\dagger$ consisting of $n$ vertices. The vertices are numbered from $1$ to $n$, and the root is the vertex $1$. You are also given a permutation $p_1, p_2, \ldots, p_n$ of $[1,2,\ldots,n]$.

You need to answer $q$ queries. For each query, you are given two integers $x$, $y$; you need to swap $p_x$ and $p_y$ and determine if $p_1, p_2, \ldots, p_n$ is a valid DFS order$^\ddagger$ of the given tree.

Please note that the swaps are persistent through queries.

$^\dagger$ A perfect binary tree is a tree with vertex $1$ as its root, with size $n=2^k-1$ for a positive integer $k$, and where the parent of each vertex $i$ ($1<i\le n$) is $\left\lfloor\frac{i}{2}\right\rfloor$. Thus, all leaves of this tree are at a distance $k - 1$ from the root.

$^\ddagger$ A DFS order is found by calling the following $\texttt{dfs}$ function on the given tree.

              dfs_order = []          function dfs(v):         append v to the back of dfs_order         pick an arbitrary permutation s of children of v         for child in s:             dfs(child)     dfs(1)

Note that the DFS order is not unique.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $q$ ($3\le n\le 65\,535$, $2\le q\le 5 \cdot 10^4$) — the number of vertices in the tree and the number of queries. It is guaranteed that $n=2^k-1$ for a positive integer $k$.

The next line contains $n-1$ integers $a_2,a_3,\ldots,a_n$ ($1\le a_i<i$) — the parent of each vertex in the given tree. It is guaranteed that $a_i=\left\lfloor\frac{i}{2}\right\rfloor$.

The next line contains $n$ integers $p_1,p_2,\ldots,p_n$ ($1\le p_i\le n$, all $p_i$ are distinct) — the initial permutation $p$.

The next $q$ lines each contain two integers $x$, $y$ ($1\le x,y\le n,x\neq y$) — the positions of the elements to swap in the permutation.

It is guaranteed that the sum of all $n$ does not exceed $65\,535$, and the sum of all $q$ does not exceed $5 \cdot 10^4$.

Output: For each test case, print $q$ lines corresponding to the $q$ queries. For each query, output $\texttt{YES}$ if there is a DFS order that exactly equals the current permutation, and output $\texttt{NO}$ otherwise.

You can output $\texttt{Yes}$ and $\texttt{No}$ in any case (for example, strings $\texttt{yEs}$, $\texttt{yes}$, $\texttt{Yes}$ and $\texttt{YES}$ will be recognized as a positive response).

Examples:
input:
2
3 3
1 1
1 2 3
2 3
3 2
1 3
7 4
1 1 2 2 3 3
1 2 3 4 5 6 7
3 5
2 5
3 7
4 6
output:
YES
YES
NO
YES
NO
NO
YES

Note: In the first test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,3,2],[1,2,3],[3,2,1]$, respectively. The first two permutations are valid DFS orders; the third is not a DFS order.

In the second test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,2,5,4,3,6,7],[1,3,5,4,2,6,7],[1,3,7,4,2,6,5],[1,3,7,6,2,4,5]$, respectively.

Buggy Code:
import java.util.*;
import java.io.*;
public class D1 {
  static boolean[] good;
  static int bads;
  static int[] parent;
  static int[] p;
  static int n;
  public static void main(String[] args) throws IOException {
    BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);

    StringTokenizer st = new StringTokenizer(r.readLine());
    int T = Integer.parseInt(st.nextToken());
    while(T-- > 0) {
      st = new StringTokenizer(r.readLine());
      n = Integer.parseInt(st.nextToken());
      int q = Integer.parseInt(st.nextToken());
      parent = new int[n + 1];
      st = new StringTokenizer(r.readLine());
      for(int i = 2; i <= n; i++) {
        parent[i] = Integer.parseInt(st.nextToken());
      }
      st = new StringTokenizer(r.readLine());
      p = new int[n + 1];
      p[0] = -1;
      for(int i = 1; i <= n; i++) {
        p[i] = Integer.parseInt(st.nextToken());
      }
      good = new boolean[n];
      good[0] = true;
      bads = 0;
      for(int i = 1; i < n; i++) {
        good[i] = true;
        process(i);
      }

      for(int i = 0; i < q; i++) {
        st = new StringTokenizer(r.readLine());
        int x = Integer.parseInt(st.nextToken());
        int y = Integer.parseInt(st.nextToken());
        int temp = p[x];
        p[x] = p[y];
        p[y] = temp;

        if(x < n) {
          process(x);
        }
        if(x > 1) {
          process(x - 1);
        }
        if(y < n) {
          process(y);
        }
        if(y > 1) {
          process(y - 1);
        }
        if(bads == 0) {
          pw.println("Yes");
        } else {
          pw.println("No");
        }
      }
    }

    pw.close();
  }

  public static void process(int i) {
    int v = p[i];
    boolean temp = good[i];
    if(v <= n/2) {
      //parent-child case
      good[i] = v == parent[p[i + 1]];
      if(temp && !good[i]) {bads++;}
      if(!temp && good[i]) {bads--;}
    } else {
      //leaf case
      int back = 1;
      while(true) {
        if(2 * back > i) {
          int next;
          if(p[i + 1 - back] % 2 == 0) {
            next = p[i + 1 - back] + 1;
          } else {
            next = p[i + 1 - back] - 1;
          }
          good[i] = p[i + 1] == next;
          break;
        } else {
          int c1 = p[i + 1 - back];
          int c2 = p[i + 1 - 2*back];
          if(parent[c1] == parent[c2]) {
            back*=2;
            back++;
          } else {
            int next;
            if(p[i + 1 - back] % 2 == 0) {
              next = p[i + 1 - back] + 1;
            } else {
              next = p[i + 1 - back] - 1;
            }
            good[i] = p[i + 1] == next;
            break;
          }
        }
      }
      if(temp && !good[i]) {bads++;}
      if(!temp && good[i]) {bads--;}
    }
  }
}

**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 algorithm, let's think through the possible edge cases and scenarios where the code might fail:

1. **Leaf Nodes and Parent-Child Relationships**: The code seems to have a specific check for leaf nodes and parent-child relationships. A scenario where these relationships are violated might expose a bug.
2. **Swapping Elements at the Boundary**: Swapping elements that are at the boundary of being a leaf node or a parent node can be tricky.
3. **Small Tree with Specific Structure**: A small tree with a specific structure that can be easily visualized and understood can help in identifying the bug.

Given these considerations, let's create a test case with a small perfect binary tree and a permutation that will expose the bug when elements are swapped.

### Test Case

- **Tree Structure**: A perfect binary tree with 7 nodes.
- **Initial Permutation**: A permutation that is a valid DFS order.
- **Queries**: Queries that swap elements to create invalid DFS orders.

### Test Input

```plaintext
1
7 4
1 1 2 2 3 3
1 2 3 4 5 6 7
3 5
2 5
3 7
4 6
```

### Explanation

- **Tree Structure**: The tree is a perfect binary tree with 7 nodes. The parent array is `[1, 1, 2, 2, 3, 3]`.
- **Initial Permutation**: The initial permutation `[1, 2, 3, 4, 5, 6, 7]` is a valid DFS order.
- **Queries**:
  - **Query 1**: Swap `p[3]` and `p[5]`. The permutation becomes `[1, 2, 5, 4, 3, 6, 7]`. This is a valid DFS order.
  - **Query 2**: Swap `p[2]` and `p[5]`. The permutation becomes `[1, 5, 3, 4, 2, 6, 7]`. This is not a valid DFS order.
  - **Query 3**: Swap `p[3]` and `p[7]`. The permutation becomes `[1, 5, 7, 4, 2, 6, 3]`. This is not a valid DFS order.
  - **Query 4**: Swap `p[4]` and `p[6]`. The permutation becomes `[1, 5, 7, 6, 2, 4, 3]`. This is not a valid DFS order.

This test case is designed to expose the bug in the code by creating scenarios where the parent-child relationships and leaf node checks are violated.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DFS Checker (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, the given tree is a perfect binary tree and the constraints on $n$ and $q$ are lower. You can make hacks only if both versions of the problem are solved.

You are given a perfect binary tree$^\dagger$ consisting of $n$ vertices. The vertices are numbered from $1$ to $n$, and the root is the vertex $1$. You are also given a permutation $p_1, p_2, \ldots, p_n$ of $[1,2,\ldots,n]$.

You need to answer $q$ queries. For each query, you are given two integers $x$, $y$; you need to swap $p_x$ and $p_y$ and determine if $p_1, p_2, \ldots, p_n$ is a valid DFS order$^\ddagger$ of the given tree.

Please note that the swaps are persistent through queries.

$^\dagger$ A perfect binary tree is a tree with vertex $1$ as its root, with size $n=2^k-1$ for a positive integer $k$, and where the parent of each vertex $i$ ($1<i\le n$) is $\left\lfloor\frac{i}{2}\right\rfloor$. Thus, all leaves of this tree are at a distance $k - 1$ from the root.

$^\ddagger$ A DFS order is found by calling the following $\texttt{dfs}$ function on the given tree.

              dfs_order = []          function dfs(v):         append v to the back of dfs_order         pick an arbitrary permutation s of children of v         for child in s:             dfs(child)     dfs(1)

Note that the DFS order is not unique.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $q$ ($3\le n\le 65\,535$, $2\le q\le 5 \cdot 10^4$) — the number of vertices in the tree and the number of queries. It is guaranteed that $n=2^k-1$ for a positive integer $k$.

The next line contains $n-1$ integers $a_2,a_3,\ldots,a_n$ ($1\le a_i<i$) — the parent of each vertex in the given tree. It is guaranteed that $a_i=\left\lfloor\frac{i}{2}\right\rfloor$.

The next line contains $n$ integers $p_1,p_2,\ldots,p_n$ ($1\le p_i\le n$, all $p_i$ are distinct) — the initial permutation $p$.

The next $q$ lines each contain two integers $x$, $y$ ($1\le x,y\le n,x\neq y$) — the positions of the elements to swap in the permutation.

It is guaranteed that the sum of all $n$ does not exceed $65\,535$, and the sum of all $q$ does not exceed $5 \cdot 10^4$.

Output: For each test case, print $q$ lines corresponding to the $q$ queries. For each query, output $\texttt{YES}$ if there is a DFS order that exactly equals the current permutation, and output $\texttt{NO}$ otherwise.

You can output $\texttt{Yes}$ and $\texttt{No}$ in any case (for example, strings $\texttt{yEs}$, $\texttt{yes}$, $\texttt{Yes}$ and $\texttt{YES}$ will be recognized as a positive response).

Examples:
input:
2
3 3
1 1
1 2 3
2 3
3 2
1 3
7 4
1 1 2 2 3 3
1 2 3 4 5 6 7
3 5
2 5
3 7
4 6
output:
YES
YES
NO
YES
NO
NO
YES

Note: In the first test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,3,2],[1,2,3],[3,2,1]$, respectively. The first two permutations are valid DFS orders; the third is not a DFS order.

In the second test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,2,5,4,3,6,7],[1,3,5,4,2,6,7],[1,3,7,4,2,6,5],[1,3,7,6,2,4,5]$, respectively.

Buggy Code:
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
#define pb push_back
const int mod =  1000000007 ;
template<class T> using ordered_set =tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update> ;
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
    int t;
    cin>>t;
    while(t--){
        int n,q;
        cin>>n>>q;
        vector<set<int>>gra(n+1);
        for(int i=2;i<=n;i++){
            int x;
            cin>>x;
            gra[x].insert(i);
        }
        vector<int>a(n+1);
        for(int i=1;i<=n;i++){
            cin>>a[i];
        }

        while(q--){
                    vector<int>ch(n+1);
        for(int i=1;i<=n;i++){
            ch[i]=gra[i].size();
        }
int l,r;
cin>>l>>r;
swap(a[l],a[r]);
bool flag=true;
stack<int>s;
s.push(a[1]);
for(int i=2;i<=n;i++){
    if(s.empty()){
        flag=false;
        break;
    }
   int x=s.top();
   if(gra[x].find(a[i])==gra[x].end()){
    flag=false;
    break;
   }
   else{
    ch[x]--;
   }
if(ch[a[i]]>0){
    s.push(a[i]);
}
if(ch[s.top()]==0){
    s.pop();
}
}

    if(flag){
            cout<<"YES"<<endl;
        }
        else cout<<"NO"<<endl;
        }

    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
7 4
1 1 2 2 3 3
1 2 3 4 5 6 7
3 5
2 5
3 7
4 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DFS Checker (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, the given tree is a perfect binary tree and the constraints on $n$ and $q$ are lower. You can make hacks only if both versions of the problem are solved.

You are given a perfect binary tree$^\dagger$ consisting of $n$ vertices. The vertices are numbered from $1$ to $n$, and the root is the vertex $1$. You are also given a permutation $p_1, p_2, \ldots, p_n$ of $[1,2,\ldots,n]$.

You need to answer $q$ queries. For each query, you are given two integers $x$, $y$; you need to swap $p_x$ and $p_y$ and determine if $p_1, p_2, \ldots, p_n$ is a valid DFS order$^\ddagger$ of the given tree.

Please note that the swaps are persistent through queries.

$^\dagger$ A perfect binary tree is a tree with vertex $1$ as its root, with size $n=2^k-1$ for a positive integer $k$, and where the parent of each vertex $i$ ($1<i\le n$) is $\left\lfloor\frac{i}{2}\right\rfloor$. Thus, all leaves of this tree are at a distance $k - 1$ from the root.

$^\ddagger$ A DFS order is found by calling the following $\texttt{dfs}$ function on the given tree.

              dfs_order = []          function dfs(v):         append v to the back of dfs_order         pick an arbitrary permutation s of children of v         for child in s:             dfs(child)     dfs(1)

Note that the DFS order is not unique.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $q$ ($3\le n\le 65\,535$, $2\le q\le 5 \cdot 10^4$) — the number of vertices in the tree and the number of queries. It is guaranteed that $n=2^k-1$ for a positive integer $k$.

The next line contains $n-1$ integers $a_2,a_3,\ldots,a_n$ ($1\le a_i<i$) — the parent of each vertex in the given tree. It is guaranteed that $a_i=\left\lfloor\frac{i}{2}\right\rfloor$.

The next line contains $n$ integers $p_1,p_2,\ldots,p_n$ ($1\le p_i\le n$, all $p_i$ are distinct) — the initial permutation $p$.

The next $q$ lines each contain two integers $x$, $y$ ($1\le x,y\le n,x\neq y$) — the positions of the elements to swap in the permutation.

It is guaranteed that the sum of all $n$ does not exceed $65\,535$, and the sum of all $q$ does not exceed $5 \cdot 10^4$.

Output: For each test case, print $q$ lines corresponding to the $q$ queries. For each query, output $\texttt{YES}$ if there is a DFS order that exactly equals the current permutation, and output $\texttt{NO}$ otherwise.

You can output $\texttt{Yes}$ and $\texttt{No}$ in any case (for example, strings $\texttt{yEs}$, $\texttt{yes}$, $\texttt{Yes}$ and $\texttt{YES}$ will be recognized as a positive response).

Examples:
input:
2
3 3
1 1
1 2 3
2 3
3 2
1 3
7 4
1 1 2 2 3 3
1 2 3 4 5 6 7
3 5
2 5
3 7
4 6
output:
YES
YES
NO
YES
NO
NO
YES

Note: In the first test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,3,2],[1,2,3],[3,2,1]$, respectively. The first two permutations are valid DFS orders; the third is not a DFS order.

In the second test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,2,5,4,3,6,7],[1,3,5,4,2,6,7],[1,3,7,4,2,6,5],[1,3,7,6,2,4,5]$, respectively.

Buggy Code:
import sys
from collections import defaultdict, deque, Counter
import math
import heapq
import bisect
import random

dfsss,hashingggg = 1,0
def cinput():
    return sys.stdin.readline().rstrip()
def cprint(output):
    sys.stdout.write(output + '\n')
    sys.stdout.flush()
def rint():
    return int(cinput())
def rlist():
    return list(map(int, cinput().split()))
def rstring():
    return cinput()
def ristring():
    return list(map(int,rstring()))
def rcstring():
    s = cinput()
    return(list(s))
def rslist():
    return list(cinput().split())
def rints():
    return list(map(int, cinput().split()))
def plist(lst):
    cprint(" ".join(map(str, lst)))
def print_2d_list(matrix):
    for row in matrix:
        cprint(" ".join(map(str, row)))
if dfsss:
    from types import GeneratorType

    def bootstrap(f, stack=[]):
        def wrappedfunc(*args, **kwargs):
            if stack:
                return f(*args, **kwargs)
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to

        return wrappedfunc
#rnd = random.randint(x,ys)
if hashingggg:
    RANDOM = random.randrange(2**62)
    def w(x):
        return x ^ RANDOM
inf = 10**18 + 10
testcases = rint()
while testcases:
    testcases -= 1
    n,q = rints()
    child = [[] for i in range(n)]
    par = [0] + rlist()

    for i in range(1,n):
        child[par[i]-1].append(i)
    for i in range(1,n):
        par[i] -= 1
    sibling = [-1]*n
    #print(child)

    @bootstrap
    def dfs(node):
        if not child[node]:
            yield None
        #print(child[node])
        u,v = child[node]
        sibling[u] = v
        sibling[v] = u
        yield dfs(u)
        yield dfs(v)
        yield None

    dfs(0)

    per = rlist()
    for i in range(n):
        per[i] -= 1

    def check():

        if per[0] != 0:
            return False
        for i in range(n-1):
            p = per[i]
            if child[p]:
                u,v = child[p]
                #print(u,v)
                if per[i+1] != u and per[i+1] != v:
                    return False
            else:
                if per[i+1] != sibling[p] and per[i+1] != sibling[par[p]]:
                    return False
        return True

    for j in range(q):
        x,y = rints()
        x -= 1
        y -= 1
        per[x],per[y] = per[y],per[x]
        if check():
            print("YES")
        else:
            print("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:
```plaintext
2
3 2
1 1
1 2 3
2 1
1 3
7 3
1 1 2 2 3 3
1 2 3 4 5 6 7
3 1
2 6
4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DFS Checker (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, the given tree is a perfect binary tree and the constraints on $n$ and $q$ are lower. You can make hacks only if both versions of the problem are solved.

You are given a perfect binary tree$^\dagger$ consisting of $n$ vertices. The vertices are numbered from $1$ to $n$, and the root is the vertex $1$. You are also given a permutation $p_1, p_2, \ldots, p_n$ of $[1,2,\ldots,n]$.

You need to answer $q$ queries. For each query, you are given two integers $x$, $y$; you need to swap $p_x$ and $p_y$ and determine if $p_1, p_2, \ldots, p_n$ is a valid DFS order$^\ddagger$ of the given tree.

Please note that the swaps are persistent through queries.

$^\dagger$ A perfect binary tree is a tree with vertex $1$ as its root, with size $n=2^k-1$ for a positive integer $k$, and where the parent of each vertex $i$ ($1<i\le n$) is $\left\lfloor\frac{i}{2}\right\rfloor$. Thus, all leaves of this tree are at a distance $k - 1$ from the root.

$^\ddagger$ A DFS order is found by calling the following $\texttt{dfs}$ function on the given tree.

              dfs_order = []          function dfs(v):         append v to the back of dfs_order         pick an arbitrary permutation s of children of v         for child in s:             dfs(child)     dfs(1)

Note that the DFS order is not unique.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $q$ ($3\le n\le 65\,535$, $2\le q\le 5 \cdot 10^4$) — the number of vertices in the tree and the number of queries. It is guaranteed that $n=2^k-1$ for a positive integer $k$.

The next line contains $n-1$ integers $a_2,a_3,\ldots,a_n$ ($1\le a_i<i$) — the parent of each vertex in the given tree. It is guaranteed that $a_i=\left\lfloor\frac{i}{2}\right\rfloor$.

The next line contains $n$ integers $p_1,p_2,\ldots,p_n$ ($1\le p_i\le n$, all $p_i$ are distinct) — the initial permutation $p$.

The next $q$ lines each contain two integers $x$, $y$ ($1\le x,y\le n,x\neq y$) — the positions of the elements to swap in the permutation.

It is guaranteed that the sum of all $n$ does not exceed $65\,535$, and the sum of all $q$ does not exceed $5 \cdot 10^4$.

Output: For each test case, print $q$ lines corresponding to the $q$ queries. For each query, output $\texttt{YES}$ if there is a DFS order that exactly equals the current permutation, and output $\texttt{NO}$ otherwise.

You can output $\texttt{Yes}$ and $\texttt{No}$ in any case (for example, strings $\texttt{yEs}$, $\texttt{yes}$, $\texttt{Yes}$ and $\texttt{YES}$ will be recognized as a positive response).

Examples:
input:
2
3 3
1 1
1 2 3
2 3
3 2
1 3
7 4
1 1 2 2 3 3
1 2 3 4 5 6 7
3 5
2 5
3 7
4 6
output:
YES
YES
NO
YES
NO
NO
YES

Note: In the first test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,3,2],[1,2,3],[3,2,1]$, respectively. The first two permutations are valid DFS orders; the third is not a DFS order.

In the second test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,2,5,4,3,6,7],[1,3,5,4,2,6,7],[1,3,7,4,2,6,5],[1,3,7,6,2,4,5]$, respectively.

Buggy Code:
from collections import deque
def get_k(n):
    m = n+1
    k = 0
    while m > 1:
        m //= 2
        k += 1
    return k

def solve():
    n,q  = [int(e) for e in input().split(' ')]
    tmp = input()
    perm = [int(e) for e in input().split(' ')]
    k = get_k(n)
    idx = 0
    mp = {e:[] for e in range(n)}
    def dfs(root, parent, height):
        nonlocal idx,k
        mp[root].append(parent)
        if height == k:
            mp[root].append(-1)
            mp[root].append(-1)
            return
        idx += 1
        mp[root].append(idx)
        dfs(idx,root,height+1)
        idx += 1
        mp[root].append(idx)
        dfs(idx,root,height+1)
        return

    def check(x):
        if x == -1:
            return True
        pidx, lidx, ridx = mp[x]
        if pidx != -1 and perm[pidx] != perm[x]//2:
            return False
        if lidx != -1 and perm[x] != perm[lidx]//2:
            return False
        if ridx != -1 and perm[x] != perm[ridx]//2:
            return False
        return check(pidx)

    dfs(0,-1,1)

    queue = deque()
    for _ in range(q):
        x,y = [int(e) for e in input().split(' ')]
        x -= 1
        y -= 1
        queue.append((x,y))
        # print("before",x,y,perm, queue)
        perm[x], perm[y] = perm[y], perm[x]
        valid = True
        # if check(x) and check(y):
        #     valid = True
        for _ in range(len(queue)):
            tx,ty = queue[0]
            if not (check(tx) and check(ty)):
                valid = False
                break
            queue.popleft()
        # print("after",perm, queue)
        if valid:
            print("YES")
        else:
            print("NO")

t = int(input())
while t != 0:
    t -= 1
    solve()

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

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

Think step by step.

Response:
```plaintext
2
3 2
1 1
1 2 3
1 2
2 3
7 3
1 1 2 2 3 3
1 2 3 4 5 6 7
1 4
4 7
7 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DFS Checker (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, the given tree is a perfect binary tree and the constraints on $n$ and $q$ are lower. You can make hacks only if both versions of the problem are solved.

You are given a perfect binary tree$^\dagger$ consisting of $n$ vertices. The vertices are numbered from $1$ to $n$, and the root is the vertex $1$. You are also given a permutation $p_1, p_2, \ldots, p_n$ of $[1,2,\ldots,n]$.

You need to answer $q$ queries. For each query, you are given two integers $x$, $y$; you need to swap $p_x$ and $p_y$ and determine if $p_1, p_2, \ldots, p_n$ is a valid DFS order$^\ddagger$ of the given tree.

Please note that the swaps are persistent through queries.

$^\dagger$ A perfect binary tree is a tree with vertex $1$ as its root, with size $n=2^k-1$ for a positive integer $k$, and where the parent of each vertex $i$ ($1<i\le n$) is $\left\lfloor\frac{i}{2}\right\rfloor$. Thus, all leaves of this tree are at a distance $k - 1$ from the root.

$^\ddagger$ A DFS order is found by calling the following $\texttt{dfs}$ function on the given tree.

              dfs_order = []          function dfs(v):         append v to the back of dfs_order         pick an arbitrary permutation s of children of v         for child in s:             dfs(child)     dfs(1)

Note that the DFS order is not unique.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $q$ ($3\le n\le 65\,535$, $2\le q\le 5 \cdot 10^4$) — the number of vertices in the tree and the number of queries. It is guaranteed that $n=2^k-1$ for a positive integer $k$.

The next line contains $n-1$ integers $a_2,a_3,\ldots,a_n$ ($1\le a_i<i$) — the parent of each vertex in the given tree. It is guaranteed that $a_i=\left\lfloor\frac{i}{2}\right\rfloor$.

The next line contains $n$ integers $p_1,p_2,\ldots,p_n$ ($1\le p_i\le n$, all $p_i$ are distinct) — the initial permutation $p$.

The next $q$ lines each contain two integers $x$, $y$ ($1\le x,y\le n,x\neq y$) — the positions of the elements to swap in the permutation.

It is guaranteed that the sum of all $n$ does not exceed $65\,535$, and the sum of all $q$ does not exceed $5 \cdot 10^4$.

Output: For each test case, print $q$ lines corresponding to the $q$ queries. For each query, output $\texttt{YES}$ if there is a DFS order that exactly equals the current permutation, and output $\texttt{NO}$ otherwise.

You can output $\texttt{Yes}$ and $\texttt{No}$ in any case (for example, strings $\texttt{yEs}$, $\texttt{yes}$, $\texttt{Yes}$ and $\texttt{YES}$ will be recognized as a positive response).

Examples:
input:
2
3 3
1 1
1 2 3
2 3
3 2
1 3
7 4
1 1 2 2 3 3
1 2 3 4 5 6 7
3 5
2 5
3 7
4 6
output:
YES
YES
NO
YES
NO
NO
YES

Note: In the first test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,3,2],[1,2,3],[3,2,1]$, respectively. The first two permutations are valid DFS orders; the third is not a DFS order.

In the second test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,2,5,4,3,6,7],[1,3,5,4,2,6,7],[1,3,7,4,2,6,5],[1,3,7,6,2,4,5]$, respectively.

Buggy Code:
import java.io.*;
import java.util.*;
import java.util.stream.Collectors;
import java.util.stream.IntStream;
import java.util.stream.Stream;

public class Main implements Runnable {

  BufferedReader in;
  PrintWriter out;
  StringTokenizer tok = new StringTokenizer("");

  int[] numbers = new int[52];

  int[] permutation = new int[65540];
  Set<Integer> completedSubtree = new HashSet<>();
  int n;
  Stack<Integer> incompleteSubtrees;

  public static void main(String[] args) {
    new Thread(null, new Main(), "", 256 * (1L << 20)).start();
  }

  public void run() {
    try {
      long t1 = System.currentTimeMillis();
      if (System.getProperty("ONLINE_JUDGE") != null) {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
      } else {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
      }
      Locale.setDefault(Locale.US);
      solve();
      in.close();
      out.close();
      long t2 = System.currentTimeMillis();
      System.err.println("Time = " + (t2 - t1));
    } catch (Throwable t) {
      t.printStackTrace(System.err);
      System.exit(-1);
    }
  }

  //Util Functions
  String readString() throws IOException {
    while (!tok.hasMoreTokens()) {
      tok = new StringTokenizer(in.readLine());
    }
    return tok.nextToken();
  }

  int readInt() throws IOException {
    return Integer.parseInt(readString());
  }

  long readLong() throws IOException {
    return Long.parseLong(readString());
  }

  double readDouble() throws IOException {
    return Double.parseDouble(readString());
  }

  ArrayList<Integer> readIntArrayList(int length) throws IOException {
    ArrayList array = new ArrayList();
    for(int input = 0; input < length; input++){
      array.add(readInt());
    }
    return array;
  }

  int[] readIntList(int length) throws IOException {
    int[] array = new int[length];
    for(int input = 0; input < length; input++){
      array[input] = readInt();
    }
    return array;
  }

  void readIntList(int[] array,int length) throws IOException {
    for(int input = 0; input < length; input++){
      array[input] = readInt();
    }
  }

  // Helper functions

  int[] bitsRep(int n){
    int[] bits = new int[22];
    int i = 0;
    while(n > 0){
      bits[i++] = n % 2;
      n = n / 2;
    }
    return bits;
  }

  boolean[] bitsBool(int[] intArray){
    boolean[] bits = new boolean[intArray.length];
    for(int i = 0; i < intArray.length; i++)
      bits[i] = intArray[i] > 0;

    return bits;
  }

  int[] arrayTopla(int[] a, int[] b){
    int[] sum = new int[Math.max(a.length,b.length)];
    for(int i = 0; i < sum.length; i++){
      if(i < a.length){
        sum[i] += a[i];
      }
      if(i < b.length){
        sum[i] += b[i];
      }
    }
    return sum;
  }

  int[] arrayCikar(int[] a, int[] b){
    int[] fark = new int[Math.max(a.length,b.length)];
    for(int i = 0; i < fark.length; i++){
      if(i < a.length){
        fark[i] += a[i];
      }
      if(i < b.length){
        fark[i] -= b[i];
      }
    }
    return fark;
  }
  // Helper Class

  int binarySearch(List<Integer> list, int eleman) {
    int l = 0;
    int r = list.size();
    while (r - l > 1) {
      int m = (l + r) / 2;
      if (list.get(m) > eleman) {
        r = m;
      } else {
        l = m;
      }
    }
    if(list.size() <= l){
      return -1;
    }
    if(list.get(l) <= eleman){
      if(list.size() == l + 1)
        return -1;
      else{
        return l + 1;
      }
    }
    return l;
  }

  void swap(int first,int second){
    int temp = permutation[first];
    permutation[first] = permutation[second];
    permutation[second] = temp;
  }
  boolean isLeaf(int node){
    return node * 2 > n;
  }

  void solve() throws IOException {
    int t = readInt();
    for (int problem = 0; problem < t; problem++){
      n = readInt();
      int q = readInt();

      for(int i = 0; i < n - 1; i++){
        readInt();
      }
      for(int i = 0; i < n; i++){
        permutation[i] = readInt();
      }

      for(int i = 0; i < q; i++){
        int first = readInt() - 1;
        int second = readInt() - 1;
        swap(first,second);
        completedSubtree = new HashSet<>();
        incompleteSubtrees = new Stack<>();
        if(permutation[0] != 1){
          out.println("NO");
          continue;
        }
        else{
          incompleteSubtrees.push(1);
        }
        boolean flag = true;
        for(int j = 1; j < n; j++){
          int current = permutation[j];
          int incomplete = incompleteSubtrees.peek();
          if(!isParentOf(current,incomplete)){
            flag = false;
            break;
          }
          if(isLeaf(current)){
            completedSubtree.add(current);
            if(allChildrenAreComplete(incomplete)) {
              completedSubtree.add(incomplete);
              incompleteSubtrees.pop();
            }
          }
          else{
            incompleteSubtrees.push(current);
          }
        }
        if(flag){
          out.println("YES");
        }
        else{
          out.println("NO");
        }
      }

    }
  }

  private boolean allChildrenAreComplete(int incomplete) {
    return completedSubtree.contains(incomplete * 2) && completedSubtree.contains(incomplete * 2 + 1);
  }

  private boolean isParentOf(int first, int second) {
    return first / 2 == second;
  }
}

**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 code, we need to consider edge cases and scenarios that can maximize the failure probability. The buggy code seems to have issues with correctly handling the DFS order validation, especially when the tree structure and the permutation are manipulated.

### Key Points to Consider:
1. **Tree Structure**: Ensure the tree is a perfect binary tree.
2. **Permutation**: The initial permutation should be a valid DFS order.
3. **Queries**: Design queries that swap elements in such a way that the resulting permutation might not be a valid DFS order.

### Test Input:
Let's create a test case with a small tree and a few queries to test the edge cases.

#### Tree Structure:
- \( n = 7 \) (a perfect binary tree with 3 levels)
- The tree structure is as follows:
  - Root: 1
  - Children of 1: 2, 3
  - Children of 2: 4, 5
  - Children of 3: 6, 7

#### Initial Permutation:
- A valid DFS order: \( p = [1, 2, 4, 5, 3, 6, 7] \)

#### Queries:
1. Swap \( p_2 \) and \( p_3 \) (swapping 2 and 4)
2. Swap \( p_5 \) and \( p_6 \) (swapping 3 and 6)
3. Swap \( p_1 \) and \( p_7 \) (swapping 1 and 7)

### Expected Output:
1. After the first swap, the permutation becomes \( [1, 4, 2, 5, 3, 6, 7] \), which is not a valid DFS order.
2. After the second swap, the permutation becomes \( [1, 4, 2, 5, 6, 3, 7] \), which is not a valid DFS order.
3. After the third swap, the permutation becomes \( [7, 4, 2, 5, 6, 3, 1] \), which is not a valid DFS order.

### Test Input:
```plaintext
1
7 3
1 1 2 2 3 3
1 2 4 5 3 6 7
2 3
5 6
1 7
```

This test input is designed to challenge the code by creating scenarios where the resulting permutation is not a valid DFS order, thus exposing the bug in the DFS order validation logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DFS Checker (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, the given tree is a perfect binary tree and the constraints on $n$ and $q$ are lower. You can make hacks only if both versions of the problem are solved.

You are given a perfect binary tree$^\dagger$ consisting of $n$ vertices. The vertices are numbered from $1$ to $n$, and the root is the vertex $1$. You are also given a permutation $p_1, p_2, \ldots, p_n$ of $[1,2,\ldots,n]$.

You need to answer $q$ queries. For each query, you are given two integers $x$, $y$; you need to swap $p_x$ and $p_y$ and determine if $p_1, p_2, \ldots, p_n$ is a valid DFS order$^\ddagger$ of the given tree.

Please note that the swaps are persistent through queries.

$^\dagger$ A perfect binary tree is a tree with vertex $1$ as its root, with size $n=2^k-1$ for a positive integer $k$, and where the parent of each vertex $i$ ($1<i\le n$) is $\left\lfloor\frac{i}{2}\right\rfloor$. Thus, all leaves of this tree are at a distance $k - 1$ from the root.

$^\ddagger$ A DFS order is found by calling the following $\texttt{dfs}$ function on the given tree.

              dfs_order = []          function dfs(v):         append v to the back of dfs_order         pick an arbitrary permutation s of children of v         for child in s:             dfs(child)     dfs(1)

Note that the DFS order is not unique.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $q$ ($3\le n\le 65\,535$, $2\le q\le 5 \cdot 10^4$) — the number of vertices in the tree and the number of queries. It is guaranteed that $n=2^k-1$ for a positive integer $k$.

The next line contains $n-1$ integers $a_2,a_3,\ldots,a_n$ ($1\le a_i<i$) — the parent of each vertex in the given tree. It is guaranteed that $a_i=\left\lfloor\frac{i}{2}\right\rfloor$.

The next line contains $n$ integers $p_1,p_2,\ldots,p_n$ ($1\le p_i\le n$, all $p_i$ are distinct) — the initial permutation $p$.

The next $q$ lines each contain two integers $x$, $y$ ($1\le x,y\le n,x\neq y$) — the positions of the elements to swap in the permutation.

It is guaranteed that the sum of all $n$ does not exceed $65\,535$, and the sum of all $q$ does not exceed $5 \cdot 10^4$.

Output: For each test case, print $q$ lines corresponding to the $q$ queries. For each query, output $\texttt{YES}$ if there is a DFS order that exactly equals the current permutation, and output $\texttt{NO}$ otherwise.

You can output $\texttt{Yes}$ and $\texttt{No}$ in any case (for example, strings $\texttt{yEs}$, $\texttt{yes}$, $\texttt{Yes}$ and $\texttt{YES}$ will be recognized as a positive response).

Examples:
input:
2
3 3
1 1
1 2 3
2 3
3 2
1 3
7 4
1 1 2 2 3 3
1 2 3 4 5 6 7
3 5
2 5
3 7
4 6
output:
YES
YES
NO
YES
NO
NO
YES

Note: In the first test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,3,2],[1,2,3],[3,2,1]$, respectively. The first two permutations are valid DFS orders; the third is not a DFS order.

In the second test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,2,5,4,3,6,7],[1,3,5,4,2,6,7],[1,3,7,4,2,6,5],[1,3,7,6,2,4,5]$, respectively.

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

#define IOS                  \
    ios::sync_with_stdio(0); \
    cin.tie(0);              \
    cout.tie(0);
#define endl "\n"
#define int long long
#define F first
#define S second
#define all(x) (x).begin(), (x).end()

const int inf = 1e18;

// int dx[4]={-1,0,1,0};
// int dy[4]={0,1,0,-1};

// int dx[8]={0,-1,-1,-1,0,1,1,1};
// int dy[8]={1,1,0,-1,-1,-1,0,1};

const int N = 200200;
const int mod = 1e9 + 7;
using ii = pair<int, int>;

bool ans = 1;
vector<int> sz;

int dfs(int node, int par, vector<vector<int>> &g)
{
    sz[node] = 1;
    for (auto c : g[node])
    {
        if (c != par)
        {
            sz[node] += dfs(c, node, g);
        }
    }

    return sz[node];
}

void mfs(int node, int par, vector<vector<int>> &g, vector<int> &p)
{
    for (auto c : g[node])
    {
        if (c != par)
        {
            int diff = -p[node] + p[c];
            if (diff < 0)
            {
                ans = 0;
                return;
            }

            if (diff > sz[node]-1)
            {
                ans = 0;
                return;
            }

            mfs(c, node, g, p);
        }
    }
}

void solver()
{
    int n, q;
    cin >> n >> q;

    vector<vector<int>> g(n + 1);
    sz.clear();
    sz.resize(n + 1);

    for (int i = 2; i <= n; i++)
    {
        int p;
        cin >> p;
        g[i].push_back(p);
        g[p].push_back(i);
    }

    vector<int> p(n + 1);

    for (int i = 1; i <= n; i++)
    {
        cin >> p[i];
    }

    dfs(1, 0, g);
    // for(int i=1;i<=n;i++)
    // {
    //     cout<<sz[i]<<" ";
    // }

    // cout<<endl;

    while (q--)
    {
        int l, r;
        cin >> l >> r;

        ans = 1;

        swap(p[l], p[r]);

        vector<int> cc(n + 1);

        for (int i = 1; i <= n; i++)
        {
            cc[p[i]] = i;
        }

        // for(int i=1;i<=n;i++)
        // {
        //     cout<<pp[i]<<" ";
        // }

        // cout<<endl;

        // for (int i = 1; i <= n; i++)
        // {
        //     cout << cc[i] << " ";
        // }

        // cout << endl;

        mfs(1, 0, g, cc);
        if (ans)
        {
            cout << "YES" << endl;
        }
        else
        {
            cout << "NO" << endl;
        }
    }
}

signed main()
{
    IOS int _t;
    cin >> _t;
    while (_t--)
        solver();
    return 0;
}

#undef int

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

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

Think step by step.

Response:
```plaintext
2
3 3
1 1
1 2 3
2 3
3 2
1 3
7 4
1 1 2 2 3 3
1 2 3 4 5 6 7
3 5
2 5
3 7
4 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: DFS Checker (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 512
Description: This is the easy version of the problem. In this version, the given tree is a perfect binary tree and the constraints on $n$ and $q$ are lower. You can make hacks only if both versions of the problem are solved.

You are given a perfect binary tree$^\dagger$ consisting of $n$ vertices. The vertices are numbered from $1$ to $n$, and the root is the vertex $1$. You are also given a permutation $p_1, p_2, \ldots, p_n$ of $[1,2,\ldots,n]$.

You need to answer $q$ queries. For each query, you are given two integers $x$, $y$; you need to swap $p_x$ and $p_y$ and determine if $p_1, p_2, \ldots, p_n$ is a valid DFS order$^\ddagger$ of the given tree.

Please note that the swaps are persistent through queries.

$^\dagger$ A perfect binary tree is a tree with vertex $1$ as its root, with size $n=2^k-1$ for a positive integer $k$, and where the parent of each vertex $i$ ($1<i\le n$) is $\left\lfloor\frac{i}{2}\right\rfloor$. Thus, all leaves of this tree are at a distance $k - 1$ from the root.

$^\ddagger$ A DFS order is found by calling the following $\texttt{dfs}$ function on the given tree.

              dfs_order = []          function dfs(v):         append v to the back of dfs_order         pick an arbitrary permutation s of children of v         for child in s:             dfs(child)     dfs(1)

Note that the DFS order is not unique.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $q$ ($3\le n\le 65\,535$, $2\le q\le 5 \cdot 10^4$) — the number of vertices in the tree and the number of queries. It is guaranteed that $n=2^k-1$ for a positive integer $k$.

The next line contains $n-1$ integers $a_2,a_3,\ldots,a_n$ ($1\le a_i<i$) — the parent of each vertex in the given tree. It is guaranteed that $a_i=\left\lfloor\frac{i}{2}\right\rfloor$.

The next line contains $n$ integers $p_1,p_2,\ldots,p_n$ ($1\le p_i\le n$, all $p_i$ are distinct) — the initial permutation $p$.

The next $q$ lines each contain two integers $x$, $y$ ($1\le x,y\le n,x\neq y$) — the positions of the elements to swap in the permutation.

It is guaranteed that the sum of all $n$ does not exceed $65\,535$, and the sum of all $q$ does not exceed $5 \cdot 10^4$.

Output: For each test case, print $q$ lines corresponding to the $q$ queries. For each query, output $\texttt{YES}$ if there is a DFS order that exactly equals the current permutation, and output $\texttt{NO}$ otherwise.

You can output $\texttt{Yes}$ and $\texttt{No}$ in any case (for example, strings $\texttt{yEs}$, $\texttt{yes}$, $\texttt{Yes}$ and $\texttt{YES}$ will be recognized as a positive response).

Examples:
input:
2
3 3
1 1
1 2 3
2 3
3 2
1 3
7 4
1 1 2 2 3 3
1 2 3 4 5 6 7
3 5
2 5
3 7
4 6
output:
YES
YES
NO
YES
NO
NO
YES

Note: In the first test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,3,2],[1,2,3],[3,2,1]$, respectively. The first two permutations are valid DFS orders; the third is not a DFS order.

In the second test case, the permutation $p_1, p_2, \ldots, p_n$ after each modification is $[1,2,5,4,3,6,7],[1,3,5,4,2,6,7],[1,3,7,4,2,6,5],[1,3,7,6,2,4,5]$, respectively.

Buggy Code:
import java.util.*;

public class Main {

    // Function to perform DFS on the tree and validate if it matches the current permutation
    static boolean validateDFSOrder(int[] parent, int[] p, int n) {
        // Stack to simulate DFS traversal
        Stack<Integer> stack = new Stack<>();
        int[] position = new int[n + 1];

        // Store positions of nodes in the current permutation
        for (int i = 0; i < n; i++) {
            position[p[i]] = i;
        }

        // Root is always the first node in DFS traversal (node 1)
        stack.push(1);
        int currentIndex = 0;

        while (!stack.isEmpty()) {
            int node = stack.pop();

            // The current node in DFS should match the current node in the permutation
            if (p[currentIndex] != node) {
                return false;
            }
            currentIndex++;

            // Get the children of the current node
            List<Integer> children = new ArrayList<>();
            for (int i = 2; i <= n; i++) {
                if (parent[i] == node) {
                    children.add(i);
                }
            }

            // Sort the children based on their position in the permutation
            Collections.sort(children, (a, b) -> position[b] - position[a]);

            // Push children onto the stack in reverse order to ensure DFS
            for (int child : children) {
                stack.push(child);
            }
        }
        return true;
    }

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

        // Read number of test cases
        int t = sc.nextInt();

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

            // Read parent array
            int[] parent = new int[n + 1];
            for (int i = 2; i <= n; i++) {
                parent[i] = sc.nextInt();
            }

            // Read the initial permutation p
            int[] p = new int[n];
            for (int i = 0; i < n; i++) {
                p[i] = sc.nextInt();
            }

            // Process each query
            for (int i = 0; i < q; i++) {
                int x = sc.nextInt() - 1; // Converting 1-based to 0-based index
                int y = sc.nextInt() - 1; // Converting 1-based to 0-based index

                // Swap the elements at position x and y
                int temp = p[x];
                p[x] = p[y];
                p[y] = temp;

                // Check if the current permutation is a valid DFS order
                if (validateDFSOrder(parent, p, n)) {
                    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:
```plaintext
2
3 3
1 1
1 2 3
2 3
3 2
1 3
7 4
1 1 2 2 3 3
1 2 3 4 5 6 7
3 5
2 5
3 7
4 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Arithmancy (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: Professor Vector is preparing to teach her Arithmancy class. She needs to prepare $$$n$$$ distinct magic words for the class. Each magic word is a string consisting of characters X and O. A spell is a string created by concatenating two magic words together. The power of a spell is equal to the number of its different non-empty substrings. For example, the power of the spell XOXO is equal to 7, because it has 7 different substrings: X, O, XO, OX, XOX, OXO and XOXO.

Each student will create their own spell by concatenating two magic words. Since the students are not very good at magic yet, they will choose each of the two words independently and uniformly at random from the $$$n$$$ words provided by Professor Vector. It is therefore also possible that the two words a student chooses are the same. Each student will then compute the power of their spell, and tell it to Professor Vector. In order to check their work, and of course to impress the students, Professor Vector needs to find out which two magic words and in which order were concatenated by each student.

Your program needs to perform the role of Professor Vector: first, create $$$n$$$ distinct magic words, and then handle multiple requests where it is given the spell power and needs to determine the indices of the two magic words, in the correct order, that were used to create the corresponding spell.

Input: None

Output: None

Examples:
input:
2

2
15

11
output:
XOXO
X

1 1

2 1

Note:

Buggy Code:
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <algorithm>
#include <string>
#include <stack>
#include <vector>
#include <cstdlib>
#include <cassert>
#include <climits>
#include <iomanip>
#include <array>
#include <complex>
#include <set>
#include <unordered_set>
#include <functional>
#include <chrono>

using namespace std;

#define abs(x) (x < 0 ? (-x) : x)
#define bit(x, i) (x & (1 << i))
#define max(a, b) (a < b ? b : a)
#define min(a, b) (a < b ? a : b)
#define char2Int(c) (c - '0')
#define setAll(a, x) memset(a, x, sizeof(a))

#define MOD 1000000007
#define PI 3.1415926535897932384626
#define YES "YES"
#define NO "NO"
#define NEWLINE '\n'

#define ll long long
#define ull unsigned long long
#define db long double
#define ui unsigned int
#define pi pair<int, int>
#define pl pair<ll, ll>
#define pd pair<db, db>

#define mp make_pair
#define FOR(i, n) for (i = 0; i < n; i++)
#define FOR_T(i, j, n) for (i = j; i < n; i++)
#define FOR_REV(i, n) for (i = n; i >= 0; i--)
#define println(s) cout << s << endl
#define print(a, b) cout << a << ' ' << b << endl
#define prints(a) cout << a

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<char> vc;

typedef struct node
{
    int data;
    struct node *left;
    struct node *right;
} Node_bi;

// Print an iterable of length n
template <typename LIST>
void printlist(LIST l, ll n)
{
    ll i;
    FOR(i, n)
    {
        prints(l[i]);
        prints(" ");
    }
    prints('\n');
}

// Print the Matrix
template <typename Mat>
void printMatrix(Mat &M, int m, int n)
{
    int i, j;
    FOR(i, m)
    {
        FOR(j, n)
        {
            prints(M[i][j]);
            prints(" ");
        }

        prints(NEWLINE);
    }
}

template <typename A, typename B>
void printPair(const pair<A, B> &p)
{
    print(p.first, p.second);
}

// Graph Functions
void add_edge(vi adj[], int u, int v)
{
    adj[u].push_back(v);
    adj[v].push_back(u);
}

unordered_set<string> findAllSubstrings(const string &str)
{
    unordered_set<std::string> substrings;

    for (size_t start = 0; start < str.length(); ++start)
    {
        for (size_t length = 1; start + length - 1 < str.length(); ++length)
        {
            std::string substring = str.substr(start, length);
            substrings.insert(substring);
        }
    }

    return substrings;
}

void solve()
{
    // Enter the solution here
    int n;
    scanf("%d", &n);

    string s1[1] = {"X"};
    string s2[2] = {"XOXO", "X"};
    string s3[3] = {"XXOXO", "XXOX", "XOX"};

    int i;
    FOR(i, n)
    {
        if (n == 1)
        {
            println(s1[i]);
        }
        else if (n == 2)
        {
            println(s2[i]);
        }
        else if (n == 3)
        {
            println(s3[i]);
        }
    }

    // power: pair
    map<int, pair<int, int>> result;

    if (n == 1)
    {
        result[1] = mp(1, 1);
    }
    if (n == 2)
    {
        int i, j;
        FOR(i, n)
        {
            FOR_T(j, i, n)
            {
                string A = s2[i] + s2[j];
                int powerA = findAllSubstrings(A).size();
                result[powerA] = mp(i + 1, j + 1);
                if (i != j)
                {
                    string B = s2[j] + s2[i];
                    int powerB = findAllSubstrings(B).size();
                    result[powerB] = mp(j + 1, i + 1);
                }
            }
        }
    }
    if (n == 3)
    {
        int i, j;
        FOR(i, n)
        {
            FOR_T(j, i, n)
            {
                string A = s3[i] + s3[j];
                int powerA = findAllSubstrings(A).size();
                result[powerA] = mp(i + 1, j + 1);

                if (i != j)
                {
                    string B = s3[j] + s3[i];
                    int powerB = findAllSubstrings(B).size();
                    result[powerB] = mp(j + 1, i + 1);
                }
            }
        }
    }

    fflush(stdout);
    // prepare n magic words;

    int q;
    scanf("%d", &q);

    int j;
    FOR(j, q)
    {
        int p;
        scanf("%d", &p);

        // Print the two magic words in order;
        printPair(result[p]);
    }
}

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

    solve();

    return 0;
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability. The code has predefined magic words for \( n = 1 \), \( n = 2 \), and \( n = 3 \). The bug might be exposed when the input \( n \) is different from these predefined values, or when the power of the spell is not correctly matched to the indices of the magic words.

### Step-by-Step Analysis:

1. **Predefined Values**: The code only handles \( n = 1 \), \( n = 2 \), and \( n = 3 \) with specific magic words. Any other value of \( n \) will not be handled correctly.
2. **Power Calculation**: The power of a spell is calculated using the number of distinct substrings. The code uses a fixed set of magic words for \( n = 2 \) and \( n = 3 \), which might not cover all possible cases.
3. **Query Handling**: The code uses a map to store the power of the spells and their corresponding indices. If the power is not correctly calculated or if the indices are not correctly stored, the queries will fail.

### Test Input:
To expose the bug, we can use an input where \( n \) is not 1, 2, or 3, and the power of the spell is a value that should not be possible with the predefined magic words. For example, if \( n = 4 \), the code will not handle it correctly.

Let's choose \( n = 4 \) and provide a query for a power that is not possible with the predefined magic words.

```plaintext
4

1 15
```

### Explanation:
- **n = 4**: The code does not have predefined magic words for \( n = 4 \), so it will not handle this case correctly.
- **1 15**: The first query is for \( n = 1 \), which is handled correctly by the code. The second query is for a power of 15, which is not possible with the predefined magic words for \( n = 4 \).

This input will likely expose the bug in the code, as it will either fail to handle \( n = 4 \) or return incorrect indices for the power of 15.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Arithmancy (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: Professor Vector is preparing to teach her Arithmancy class. She needs to prepare $$$n$$$ distinct magic words for the class. Each magic word is a string consisting of characters X and O. A spell is a string created by concatenating two magic words together. The power of a spell is equal to the number of its different non-empty substrings. For example, the power of the spell XOXO is equal to 7, because it has 7 different substrings: X, O, XO, OX, XOX, OXO and XOXO.

Each student will create their own spell by concatenating two magic words. Since the students are not very good at magic yet, they will choose each of the two words independently and uniformly at random from the $$$n$$$ words provided by Professor Vector. It is therefore also possible that the two words a student chooses are the same. Each student will then compute the power of their spell, and tell it to Professor Vector. In order to check their work, and of course to impress the students, Professor Vector needs to find out which two magic words and in which order were concatenated by each student.

Your program needs to perform the role of Professor Vector: first, create $$$n$$$ distinct magic words, and then handle multiple requests where it is given the spell power and needs to determine the indices of the two magic words, in the correct order, that were used to create the corresponding spell.

Input: None

Output: None

Examples:
input:
2

2
15

11
output:
XOXO
X

1 1

2 1

Note:

Buggy Code:
// sun's code

#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <cstring>
#include <iomanip>
#include <algorithm>
#include <stack>
#include <deque>
#include <unordered_map>
#include <unordered_set>
#include <set>
#include <bitset>
#include <math.h>
#include <cstdio>
#include <ctime>
#include <cstdlib>
#include <string>
#include <fstream>
#include <iterator>
#include <climits>
#include <chrono>
#include <random>

/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
*/

using namespace std;

/*
using namespace __gnu_pbds;

#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define ordered_multiset tree<int, null_type,less_equal<int>, rb_tree_tag,tree_order_statistics_node_update>
*/

#define int long long
#define mask(i) (1ll<<(i))
#define ii pair<int, int>
#define fi first
#define se second
#define iii pair<int, ii>
#define vii vector<ii>
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define mi map<int,int>
#define reset(x,val) memset((x),(val),sizeof(x))
#define ll long long
#define ull unsigned long long
#define ld long double
#define all(x) x.begin(), x.end()
#define turn_on(i,m) (m |= (1LL << i))
#define turn_off(i,m) (m &= ~(1LL << i))
#define getbit(i, m) ((m >> i) & 1)
#define qii queue<ii>
#define pdd pair<double, double>

const int maxn = 2e5 + 123;
const int INF = 1e18;
const int mod = 1e9 + 7;
const int base = 311;
const int N = 1e6 + 123;

template <class T> inline T sqr(T x) { return x * x; };
template <class T> inline bool maximize(T& a, T b) { return (a < b) ? a = b, true : false; }
template <class T> inline bool minimize(T& a, T b) { return (a > b) ? a = b, true : false; }
template <class T> inline int Power(T x, int y) { if (!y) return 1; if (y & 1) return sqr(Power(x, y / 2)) % mod * x % mod; return sqr(Power(x, y / 2)) % mod; }

inline int gcd(int x, int y) { int r = 0; while (y > 0) { r = x % y; x = y; y = r; } return x; }
inline int lcm(int x, int y) { return x * y / gcd(x, y); }

int dx[] = { -1,0,1,0 };
int dy[] = { 0,1,0,-1 };

template <class T> inline void print(T value1, T value2, bool ok)
{
	if (!ok) cout << value1 << endl;
	else cout << value1 << ' ' << value2 << endl;
}

int n, q;

int countSub(string value)
{
	map<string, int> f;
	for (int i = 0; i < value.size(); ++i)
		for (int len = 1; i + len - 1 < value.size(); ++len) f[value.substr(i, len)] ++;

	cout << f.size() << " " << value << endl;
	return f.size();
}

string check(string a, string b, string c)
{
	map<int, int> f;

	f[countSub(a + a)]++;
	f[countSub(a + b)]++;
	f[countSub(a + c)]++;
	f[countSub(b + a)]++;
	f[countSub(b + b)]++;
	f[countSub(b + c)]++;
	f[countSub(c + a)]++;
	f[countSub(c + b)]++;
	f[countSub(c + c)]++;

	cout << f.size() << endl;
	for (auto v : f) cout << v.fi << ' ';
	cout << endl;
	if (f.size() == 9) return "__miscitaofvh";
	return "__sun";
}

void solve()
{
	//cout << check("OXOXOXOXOX", "XOXXXXOOOO", "XXXX");

	//return;
	cin >> n;

	if (n == 1)
	{
		if (n == 1)
		{
			print("X", "", 0);
			cout << flush;
			cin >> q;

			while (q--)
			{
				cin >> n;
				print(1, 1, 1);
			}
		}
	}

	if (n == 2)
	{
		print("XOXO", "", 0);
		print("X", "", 0);
		cout << flush;
		cin >> q;

		while (q--)
		{
			cin >> n;
			if (n == 15) print(1, 1, 1);
			if (n == 11) print(2, 1, 1);
			if (n == 9)  print(1, 2, 1);
			if (n == 2)  print(1, 1, 1);
		}

	}

	if (n == 3)
	{
		print("OXOXOXOXOX", "", 0);
		print("XOXXXXOOOO", "", 0);
		print("XXXX", "", 0);
		cout << flush;
		cin >> q;
		while (q--)
		{
			cin >> n;
			if (n == 39) print(1, 1, 1);
			if (n == 150) print(1, 2, 1);
			if (n == 59) print(1, 3, 1);
			if (n == 151) print(2, 1, 1);
			if (n == 139) print(2, 2, 1);
			if (n == 76) print(2, 3, 1);
			if (n == 54) print(3, 1, 1);
			if (n == 74) print(3, 2, 1);
			if (n == 8) print(3, 3, 1);
		}
	}
	cout << flush;
}

int32_t main()
{
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);

	solve();

	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3

2
139
74
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Arithmancy (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: Professor Vector is preparing to teach her Arithmancy class. She needs to prepare $$$n$$$ distinct magic words for the class. Each magic word is a string consisting of characters X and O. A spell is a string created by concatenating two magic words together. The power of a spell is equal to the number of its different non-empty substrings. For example, the power of the spell XOXO is equal to 7, because it has 7 different substrings: X, O, XO, OX, XOX, OXO and XOXO.

Each student will create their own spell by concatenating two magic words. Since the students are not very good at magic yet, they will choose each of the two words independently and uniformly at random from the $$$n$$$ words provided by Professor Vector. It is therefore also possible that the two words a student chooses are the same. Each student will then compute the power of their spell, and tell it to Professor Vector. In order to check their work, and of course to impress the students, Professor Vector needs to find out which two magic words and in which order were concatenated by each student.

Your program needs to perform the role of Professor Vector: first, create $$$n$$$ distinct magic words, and then handle multiple requests where it is given the spell power and needs to determine the indices of the two magic words, in the correct order, that were used to create the corresponding spell.

Input: None

Output: None

Examples:
input:
2

2
15

11
output:
XOXO
X

1 1

2 1

Note:

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

using namespace std;

string str[32];
long long table[32][32];
bool check[1000020];
set<int> value;
const long long MOD=1000000007;
mt19937 rd(chrono::steady_clock::now().time_since_epoch().count());
#define rand rd
long long Rand(long long l, long long h) {
    assert(l <= h);
    return l + rd() * 1LL * rd() % (h - l + 1);
}
int sscount(int pos1,int pos2){
    set<long long> ss;
    string tmp=str[pos1]+str[pos2];
    long long k=0;
    for (int i=0;i<tmp.size();i++){
        for (int j=i+1;j<tmp.size();j++){
            k=0;
            for (int l=i;l<=j;l++){
                k=(k*31+(tmp[l]-'a'+1))%MOD;
            }
            ss.insert(k);
        }
    }
    return ss.size();
}
void solve(){
    //The code goes here
    int n,q;
    cin >> n;
    for (int i=1;i<=n;i++){
        cout << str[i] << endl;
        cout.flush();
    }
    cin >> q;
    for (int i=1;i<=q;i++){
        int num;
        cin >> num;
        bool found=false;
        for (int j=1;j<=n;j++){
            for (int k=1;k<=n;k++){
                if (table[j][k]==num){
                    cout << j << " " << k << endl;
                    cout.flush();
                    found=true;
                    break;
                }
            }
            if (found) break;
        }
    }
}
int main()
{

    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int t=1;
    for (int j=0;j<1;j++){
        str[1]+=((Rand(0,1)==0)? "X": "O");
    }
    table[1][1]=sscount(1,1);
    check[table[1][1]]=true;
    for (int i=2;i<=3;i++){
        bool ok=false;
        do{
            str[i]="";
            for (int j=0;j<((i<=7)? i: ((i<=20)? i*1.2: ((i<=25)? i*1.33: i*1.75)));j++){
                str[i]+=(((Rand(0,1)==0)? "X": "O"));
            }
            for (int j=1;j<=i;j++){
                if (check[sscount(i,j)] || check[sscount(j,i)]){
                    break;
                }
                if (j==i-1){
                    ok=true;
                    for (int k=1;k<=i;k++){
                        table[k][j]=sscount(k,i);
                        check[table[k][i]]=true;
                        if (k!=i){
                            table[i][k]=sscount(i,k);
                            check[table[i][k]]=true;
                        }
                    }
                }
            }
        }
        while (!ok);
    }
    //cin >> t;
    while (t--){
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3

2
15
2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Arithmancy (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: Professor Vector is preparing to teach her Arithmancy class. She needs to prepare $$$n$$$ distinct magic words for the class. Each magic word is a string consisting of characters X and O. A spell is a string created by concatenating two magic words together. The power of a spell is equal to the number of its different non-empty substrings. For example, the power of the spell XOXO is equal to 7, because it has 7 different substrings: X, O, XO, OX, XOX, OXO and XOXO.

Each student will create their own spell by concatenating two magic words. Since the students are not very good at magic yet, they will choose each of the two words independently and uniformly at random from the $$$n$$$ words provided by Professor Vector. It is therefore also possible that the two words a student chooses are the same. Each student will then compute the power of their spell, and tell it to Professor Vector. In order to check their work, and of course to impress the students, Professor Vector needs to find out which two magic words and in which order were concatenated by each student.

Your program needs to perform the role of Professor Vector: first, create $$$n$$$ distinct magic words, and then handle multiple requests where it is given the spell power and needs to determine the indices of the two magic words, in the correct order, that were used to create the corresponding spell.

Input: None

Output: None

Examples:
input:
2

2
15

11
output:
XOXO
X

1 1

2 1

Note:

Buggy Code:
// sun's code

#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <cstring>
#include <iomanip>
#include <algorithm>
#include <stack>
#include <deque>
#include <unordered_map>
#include <unordered_set>
#include <set>
#include <bitset>
#include <math.h>
#include <cstdio>
#include <ctime>
#include <cstdlib>
#include <string>
#include <fstream>
#include <iterator>
#include <climits>
#include <chrono>
#include <random>

/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
*/

using namespace std;

/*
using namespace __gnu_pbds;

#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define ordered_multiset tree<int, null_type,less_equal<int>, rb_tree_tag,tree_order_statistics_node_update>
*/

#define int long long
#define mask(i) (1ll<<(i))
#define ii pair<int, int>
#define fi first
#define se second
#define iii pair<int, ii>
#define vii vector<ii>
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define mi map<int,int>
#define reset(x,val) memset((x),(val),sizeof(x))
#define ll long long
#define ull unsigned long long
#define ld long double
#define all(x) x.begin(), x.end()
#define turn_on(i,m) (m |= (1LL << i))
#define turn_off(i,m) (m &= ~(1LL << i))
#define getbit(i, m) ((m >> i) & 1)
#define qii queue<ii>
#define pdd pair<double, double>

const int maxn = 2e5 + 123;
const int INF = 1e18;
const int mod = 1e9 + 7;
const int base = 311;
const int N = 1e6 + 123;

template <class T> inline T sqr(T x) { return x * x; };
template <class T> inline bool maximize(T& a, T b) { return (a < b) ? a = b, true : false; }
template <class T> inline bool minimize(T& a, T b) { return (a > b) ? a = b, true : false; }
template <class T> inline int Power(T x, int y) { if (!y) return 1; if (y & 1) return sqr(Power(x, y / 2)) % mod * x % mod; return sqr(Power(x, y / 2)) % mod; }

inline int gcd(int x, int y) { int r = 0; while (y > 0) { r = x % y; x = y; y = r; } return x; }
inline int lcm(int x, int y) { return x * y / gcd(x, y); }

int dx[] = { -1,0,1,0 };
int dy[] = { 0,1,0,-1 };

template <class T> inline void print(T value1, T value2, bool ok)
{
	if (!ok) cout << value1 << endl;
	else cout << value1 << ' ' << value2 << endl;
	cout << flush;
}

int n, q;

int countSub(string value)
{
	map<string, int> f;
	for (int i = 0; i < value.size(); ++i)
		for (int len = 1; i + len - 1 < value.size(); ++len) f[value.substr(i, len)] ++;

	cout << f.size() << " " << value << endl;
	return f.size();
}

string check(string a, string b, string c)
{
	map<int, int> f;

	f[countSub(a + a)]++;
	f[countSub(a + b)]++;
	f[countSub(a + c)]++;
	f[countSub(b + a)]++;
	f[countSub(b + b)]++;
	f[countSub(b + c)]++;
	f[countSub(c + a)]++;
	f[countSub(c + b)]++;
	f[countSub(c + c)]++;

	cout << f.size() << endl;
	for (auto v : f) cout << v.fi << ' ';
	cout << endl;
	if (f.size() == 9) return "__miscitaofvh";
	return "__sun";
}

void solve()
{
	//cout << check("OXOXOXOXOX", "XOXXXXOOOO", "XXXX");

	//return;
	cin >> n;

	if (n == 1)
	{
		if (n == 1)
		{
			print("X", "", 0);
			cin >> q;

			while (q--)
			{
				cin >> n;
				print(1, 1, 1);
			}

		}
	}

	if (n == 2)
	{
		print("XOXO", "", 0);
		print("X", "", 0);
		cin >> q;

		while (q--)
		{
			cin >> n;
			if (n == 15) print(1, 1, 1);
			if (n == 11) print(2, 1, 1);
			if (n == 9)  print(1, 2, 1);
			if (n == 2)  print(1, 1, 1);
		}

	}

	if (n == 3)
	{
		print("OXOXOXOXOX", "", 0);
		print("XOXXXXOOOO", "", 0);
		print("XXXX", "", 0);

		cin >> q;
		while (q--)
		{
			cin >> n;
			if (n == 39) print(1, 1, 1);
			if (n == 150) print(1, 2, 1);
			if (n == 59) print(1, 3, 1);
			if (n == 151) print(2, 1, 1);
			if (n == 139) print(2, 2, 1);
			if (n == 76) print(2, 3, 1);
			if (n == 54) print(3, 1, 1);
			if (n == 74) print(3, 2, 1);
			if (n == 8) print(3, 3, 1);
		}
	}
}

int32_t main()
{
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);

	solve();

	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. The code provided handles specific cases for \( n = 1 \), \( n = 2 \), and \( n = 3 \). We will create a test input for \( n = 3 \) because it has more complex interactions and a larger set of possible spell powers.

### Steps to Generate the Test Input

1. **Choose \( n = 3 \)**: This will ensure that the code generates three distinct magic words and handles multiple requests for spell powers.
2. **Generate Magic Words**: We need to ensure that the magic words are distinct and can create a variety of spell powers.
3. **Calculate Spell Powers**: For each pair of magic words, calculate the power of the spell formed by concatenating them.
4. **Create Test Cases**: Choose a few spell powers that should be handled correctly and a few that might expose the bug.

### Chosen Magic Words
Let's use the following magic words:
- "OXOXOXOXOX" (10 characters)
- "XOXXXXOOOO" (10 characters)
- "XXXX" (4 characters)

### Calculated Spell Powers
- "OXOXOXOXOX" + "OXOXOXOXOX" -> 39
- "OXOXOXOXOX" + "XOXXXXOOOO" -> 150
- "OXOXOXOXOX" + "XXXX" -> 59
- "XOXXXXOOOO" + "OXOXOXOXOX" -> 151
- "XOXXXXOOOO" + "XOXXXXOOOO" -> 139
- "XOXXXXOOOO" + "XXXX" -> 76
- "XXXX" + "OXOXOXOXOX" -> 54
- "XXXX" + "XOXXXXOOOO" -> 74
- "XXXX" + "XXXX" -> 8

### Test Case Input
We will create a test input with \( n = 3 \) and 4 queries. The queries will include:
- A power that should be correctly identified (e.g., 39)
- A power that should be correctly identified (e.g., 150)
- A power that might expose the bug (e.g., 59)
- A power that might expose the bug (e.g., 74)

### Final Test Input
```plaintext
3

4
39
150
59
74
```

This test input will help in verifying whether the code correctly identifies the pairs of magic words for the given spell powers and whether it handles edge cases appropriately.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Arithmancy (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: Professor Vector is preparing to teach her Arithmancy class. She needs to prepare $$$n$$$ distinct magic words for the class. Each magic word is a string consisting of characters X and O. A spell is a string created by concatenating two magic words together. The power of a spell is equal to the number of its different non-empty substrings. For example, the power of the spell XOXO is equal to 7, because it has 7 different substrings: X, O, XO, OX, XOX, OXO and XOXO.

Each student will create their own spell by concatenating two magic words. Since the students are not very good at magic yet, they will choose each of the two words independently and uniformly at random from the $$$n$$$ words provided by Professor Vector. It is therefore also possible that the two words a student chooses are the same. Each student will then compute the power of their spell, and tell it to Professor Vector. In order to check their work, and of course to impress the students, Professor Vector needs to find out which two magic words and in which order were concatenated by each student.

Your program needs to perform the role of Professor Vector: first, create $$$n$$$ distinct magic words, and then handle multiple requests where it is given the spell power and needs to determine the indices of the two magic words, in the correct order, that were used to create the corresponding spell.

Input: None

Output: None

Examples:
input:
2

2
15

11
output:
XOXO
X

1 1

2 1

Note:

Buggy Code:
import random
import sys

def generate_magic_words(n):
    words = []
    for _ in range(n):
        word_length = random.randint(1, 30 * n)
        word = ''.join(random.choice('XO') for _ in range(word_length))
        words.append(word)
    return words

def calculate_spell_power(word1, word2):
    spell = word1 + word2
    substrings = set()
    for i in range(len(spell)):
        for j in range(i + 1, len(spell) + 1):
            substrings.add(spell[i:j])
    return len(substrings)

def find_magic_words(words, power):
    for i in range(len(words)):
        for j in range(len(words)):
            if calculate_spell_power(words[i], words[j]) == power:
                return i + 1, j + 1
    return None

def main():
    n = int(input())
    words = generate_magic_words(n)
    for word in words:
        print(word)
    sys.stdout.flush()

    q = int(input())
    for _ in range(q):
        power = int(input())
        result = find_magic_words(words, power)
        print(result[0], result[1])
        sys.stdout.flush()

if __name__ == "__main__":
    main()

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

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

Think step by step.

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

1. **Word Generation**: The words generated should be distinct and have varying lengths to test the substring calculation.
2. **Spell Power Calculation**: The spell power should be calculated correctly, and the test input should include cases where the power is unique and cases where it is not.
3. **Edge Cases**: Include edge cases such as very short and very long words, and cases where the words are identical.

Let's create a small-scale test input with the following specifications:

- **Number of Magic Words (n)**: 3
- **Magic Words**:
  - Word 1: "X"
  - Word 2: "O"
  - Word 3: "XO"

- **Queries**:
  - Query 1: Spell Power 3 (Result: 1 2)
  - Query 2: Spell Power 4 (Result: 3 1)
  - Query 3: Spell Power 4 (Result: 1 3)

This test input will help to expose any issues in the word generation and power calculation.

Here is the test input:

```plaintext
3

3
4
4
```

Explanation:
- **Word 1: "X"**
- **Word 2: "O"**
- **Word 3: "XO"**

- **Query 1**: The spell "XO" has a power of 3 (substrings: X, O, XO).
- **Query 2**: The spell "XOX" has a power of 4 (substrings: X, O, XOX, X, OX, X, XO).
- **Query 3**: The spell "XOX" has a power of 4 (substrings: X, O, XOX, X, OX, X, XO).

This test input should be sufficient to identify 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: Arithmancy (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: Professor Vector is preparing to teach her Arithmancy class. She needs to prepare $$$n$$$ distinct magic words for the class. Each magic word is a string consisting of characters X and O. A spell is a string created by concatenating two magic words together. The power of a spell is equal to the number of its different non-empty substrings. For example, the power of the spell XOXO is equal to 7, because it has 7 different substrings: X, O, XO, OX, XOX, OXO and XOXO.

Each student will create their own spell by concatenating two magic words. Since the students are not very good at magic yet, they will choose each of the two words independently and uniformly at random from the $$$n$$$ words provided by Professor Vector. It is therefore also possible that the two words a student chooses are the same. Each student will then compute the power of their spell, and tell it to Professor Vector. In order to check their work, and of course to impress the students, Professor Vector needs to find out which two magic words and in which order were concatenated by each student.

Your program needs to perform the role of Professor Vector: first, create $$$n$$$ distinct magic words, and then handle multiple requests where it is given the spell power and needs to determine the indices of the two magic words, in the correct order, that were used to create the corresponding spell.

Input: None

Output: None

Examples:
input:
2

2
15

11
output:
XOXO
X

1 1

2 1

Note:

Buggy Code:
def f(s):
    r = set()
    for i in range(0, len(s)):
        for g in range(i + 1, len(s) + 1):
            r.add(s[i:g])
    return len(r)

n = int(input())
words = []
if n == 1:
    words = ["X"]
if n == 2:
    words = ["X", "XOO"]
if n == 3:
    words = ["X", "XOO", "XOXO"]
print(*words, sep='\n')
ans = {}
for i in range(0, len(words)):
    for g in range(0, len(words)):
        ans[f(words[i] + words[g])] = [i + 1, g + 1]
q = int(input())
for v in range(0, q):
    x = int(input())
    print(*ans[x])

**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

15

10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Arithmancy (Easy)
time_limit_ms: 2000
memory_limit_mb: 256
Description: Professor Vector is preparing to teach her Arithmancy class. She needs to prepare $$$n$$$ distinct magic words for the class. Each magic word is a string consisting of characters X and O. A spell is a string created by concatenating two magic words together. The power of a spell is equal to the number of its different non-empty substrings. For example, the power of the spell XOXO is equal to 7, because it has 7 different substrings: X, O, XO, OX, XOX, OXO and XOXO.

Each student will create their own spell by concatenating two magic words. Since the students are not very good at magic yet, they will choose each of the two words independently and uniformly at random from the $$$n$$$ words provided by Professor Vector. It is therefore also possible that the two words a student chooses are the same. Each student will then compute the power of their spell, and tell it to Professor Vector. In order to check their work, and of course to impress the students, Professor Vector needs to find out which two magic words and in which order were concatenated by each student.

Your program needs to perform the role of Professor Vector: first, create $$$n$$$ distinct magic words, and then handle multiple requests where it is given the spell power and needs to determine the indices of the two magic words, in the correct order, that were used to create the corresponding spell.

Input: None

Output: None

Examples:
input:
2

2
15

11
output:
XOXO
X

1 1

2 1

Note:

Buggy Code:
# /**
#  *     author: brownfox2k6
#  *    created: 10/05/2024 15:00:39 Hanoi, Vietnam
# **/

s = ["XX", "OOO", "XOXOXO"]

n = int(input())
print('\n'.join(s[:n]), flush=True)

for _ in range(int(input())):
  x = int(input())
  ans = [-1, -1]
  for u in range(n):
    for v in range(n):
      t = s[u] + s[v]
      m = len(t)
      st = set()
      for i in range(m):
        for j in range(i+1, m+1):
          st.add(t[i:j])
      if len(st) == x:
        ans = [u+1, v+1]
  assert(ans[0] != -1)
  print(*ans, flush=True)

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and scenarios that can cause the code to fail. Here are the steps to think through this problem:

1. **Distinct Magic Words**: Ensure the magic words are distinct and cover a variety of lengths and patterns.
2. **Spell Power Calculation**: The spell power is the number of unique non-empty substrings. We need to ensure that the power calculation is correct and can be challenging.
3. **Edge Cases**:
   - Short magic words
   - Long magic words
   - Magic words with repetitive patterns
   - Magic words with alternating patterns

Given the constraints and the need for a small-scale test input, let's create a test input with a few magic words that cover these edge cases.

### Test Input
```plaintext
3

2
15
10
```

### Explanation
1. **Magic Words**:
   - `XX` (2 characters, repetitive pattern)
   - `OOO` (3 characters, repetitive pattern)
   - `XOXOXO` (6 characters, alternating pattern)

2. **Requests**:
   - **2**: The power of the spell "XX" + "XX" is 2 (substrings: X, XX).
   - **15**: The power of the spell "XOXOXO" + "XOXOXO" is 15 (many unique substrings).
   - **10**: The power of the spell "OOO" + "XOXOXO" is 10 (a mix of repetitive and alternating patterns).

This test input is designed to cover:
- Short and long magic words
- Repetitive and alternating patterns
- A variety of spell powers to ensure the power calculation is correct and the indices are determined accurately.

By using this test input, we can effectively expose any bugs in the provided code, especially in the power calculation and the determination of the correct indices.