**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Games
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are $n$ coins on the table forming a circle, and each coin is either facing up or facing down. Alice and Bob take turns to play the following game, and Alice goes first.

In each operation, the player chooses a facing-up coin, removes the coin, and flips the two coins that are adjacent to it. If (before the operation) there are only two coins left, then one will be removed and the other won't be flipped (as it would be flipped twice). If (before the operation) there is only one coin left, no coins will be flipped. If (before the operation) there are no facing-up coins, the player loses.

Decide who will win the game if they both play optimally. It can be proved that the game will end in a finite number of operations, and one of them will win.

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

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of the coins.

A string $s$ of length $n$ follows on the second line of each test case, containing only "U" and "D", representing that each coin is facing up or facing down.

Output: For each test case, print "YES" if Alice will win the game, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
3
5
UUDUD
5
UDDUD
2
UU
output:
YES
NO
NO

Note: In the first test case, the game may go as follows.

  * Alice chooses the first coin and $s$ becomes "DDUU".    * Bob chooses the last coin and $s$ becomes "UDD".    * Alice chooses the first coin and $s$ becomes "UU".    * Bob chooses the first coin and $s$ becomes "U".    * Alice chooses the only coin and $s$ becomes empty.    * Bob can't choose any coin now, and he loses the game.

It can be proved that Bob will always lose if they both play optimally.

Buggy Code:
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt(); // Number of test cases

        for (int i = 0; i < t; i++) {
            int n = scanner.nextInt(); // Number of coins
            scanner.nextLine(); // Consume newline
            String s = scanner.nextLine(); // Sequence of coins facing up or down

            // Count the number of "U"s in the sequence
            int countU = 0;
            for (char c : s.toCharArray()) {
                if (c == 'U') {
                    countU++;
                }
            }

            // Check the parity of the count of "U"s
            if (countU % 2 == 1) {
                // If there's an odd number of "U"s, check if any "U" is at an odd position
                // If yes, Alice wins, otherwise Bob wins
                boolean aliceWins = false;
                for (int j = 0; j < n; j++) {
                    if (s.charAt(j) == 'U' && j % 2 == 0) {
                        aliceWins = true;
                        break;
                    }
                }
                System.out.println(aliceWins ? "YES" : "NO");
            } else {
                // If there's an even number of "U"s, Bob wins
                System.out.println("NO");
            }
        }
        scanner.close();
    }
}

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

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

Think step by step.

Response:
```plaintext
3
5
UDDDU
4
UUUU
1
U
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Games
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are $n$ coins on the table forming a circle, and each coin is either facing up or facing down. Alice and Bob take turns to play the following game, and Alice goes first.

In each operation, the player chooses a facing-up coin, removes the coin, and flips the two coins that are adjacent to it. If (before the operation) there are only two coins left, then one will be removed and the other won't be flipped (as it would be flipped twice). If (before the operation) there is only one coin left, no coins will be flipped. If (before the operation) there are no facing-up coins, the player loses.

Decide who will win the game if they both play optimally. It can be proved that the game will end in a finite number of operations, and one of them will win.

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

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of the coins.

A string $s$ of length $n$ follows on the second line of each test case, containing only "U" and "D", representing that each coin is facing up or facing down.

Output: For each test case, print "YES" if Alice will win the game, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
3
5
UUDUD
5
UDDUD
2
UU
output:
YES
NO
NO

Note: In the first test case, the game may go as follows.

  * Alice chooses the first coin and $s$ becomes "DDUU".    * Bob chooses the last coin and $s$ becomes "UDD".    * Alice chooses the first coin and $s$ becomes "UU".    * Bob chooses the first coin and $s$ becomes "U".    * Alice chooses the only coin and $s$ becomes empty.    * Bob can't choose any coin now, and he loses the game.

It can be proved that Bob will always lose if they both play optimally.

Buggy Code:
#include <bits/stdc++.h>
#include<string>
#define ll long long
#define rep(i,a,b) for(int i=a;i<b;i++)
using namespace std;
int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t;cin>>t;
    while(t--){
        int n;cin>>n;
        string s;cin>>s;
        int u=0,d=0;
        int l=s.length();
        rep(i,0,l){
            if(s[i]=='U')
            u++;
            else
            d++;
        }
        if(d<u && d!=0)
        cout<<"YES"<<endl;
        else
        cout<<"NO"<<endl;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on scenarios where the code's logic fails. The buggy code incorrectly determines the winner based on the counts of 'U' and 'D' coins, without considering the game's actual rules and the strategic removal of coins.

### Analyzing the Buggy Code
The buggy code has the following logic:
- It counts the number of 'U' (up) and 'D' (down) coins.
- It then checks if the number of 'D' coins is less than the number of 'U' coins and not zero. If this condition is true, it prints "YES"; otherwise, it prints "NO".

This logic is flawed because it does not consider the actual game rules, such as the removal of coins and the flipping of adjacent coins.

### Challenging Test Input
To expose the bug, we need a test case where the number of 'D' coins is less than the number of 'U' coins, but Alice should still lose if they both play optimally. This can happen in scenarios where the arrangement of 'U' and 'D' coins makes it impossible for Alice to win.

### Example Test Case
Consider the following test case:
- 3 coins, with the arrangement "UUD".

In this case:
- The number of 'D' coins is 1, which is less than the number of 'U' coins (2).
- However, if Alice removes the first 'U' coin, the arrangement becomes "DD". Bob then removes one of the 'D' coins, and Alice is left with the last 'D' coin, which she cannot remove (since there are no adjacent 'U' coins to flip).

Thus, Alice loses in this scenario, even though the number of 'D' coins is less than the number of 'U' coins.

### Test Input
```plaintext
1
3
UUD
```

This test input will expose the bug in the buggy code, as it will incorrectly output "YES" instead of the correct output "NO".


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Games
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are $n$ coins on the table forming a circle, and each coin is either facing up or facing down. Alice and Bob take turns to play the following game, and Alice goes first.

In each operation, the player chooses a facing-up coin, removes the coin, and flips the two coins that are adjacent to it. If (before the operation) there are only two coins left, then one will be removed and the other won't be flipped (as it would be flipped twice). If (before the operation) there is only one coin left, no coins will be flipped. If (before the operation) there are no facing-up coins, the player loses.

Decide who will win the game if they both play optimally. It can be proved that the game will end in a finite number of operations, and one of them will win.

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

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of the coins.

A string $s$ of length $n$ follows on the second line of each test case, containing only "U" and "D", representing that each coin is facing up or facing down.

Output: For each test case, print "YES" if Alice will win the game, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
3
5
UUDUD
5
UDDUD
2
UU
output:
YES
NO
NO

Note: In the first test case, the game may go as follows.

  * Alice chooses the first coin and $s$ becomes "DDUU".    * Bob chooses the last coin and $s$ becomes "UDD".    * Alice chooses the first coin and $s$ becomes "UU".    * Bob chooses the first coin and $s$ becomes "U".    * Alice chooses the only coin and $s$ becomes empty.    * Bob can't choose any coin now, and he loses the game.

It can be proved that Bob will always lose if they both play optimally.

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

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

#define pb push_back
#define ff first
#define ss second
#define endl '\n'
#define all(a) (a).begin(), (a).end()
#define max3(a,b,c) max(max(a,b),c)
#define max4(a,b,c,d) max(max(a,b),max(c,d))
#define min3(a,b,c) min(min(a,b),c)
#define min4(a,b,c,d) min(min(a,b),min(c,d))
#define gap " "
#define vi vector<int>
#define vll vector<ll>
#define vpll vector<pair<ll,ll>>
#define rev_sort(v) sort(all(v),greater<int>())
const double PI = acos(-1);
const long long INF = 1e18;
const long long MOD = 1000000007;

using namespace std;
using namespace __gnu_pbds;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> ordered_set; // find_by_order, order_of_key

#ifdef ONLINE_JUDGE
#define debug(...) 42
#else
#include "debug.h"
#endif

/*-----------  END -----------*/

void solve()
{
  ll n; cin >> n;
  string s; cin >> s;
  ll cntu = 0, cntd = 0;
  ll cnt = 0;
  for (int i = 1; i < s.size()-1; ++i)
  {
    if (s[i] == 'U')
    {
      if (s[i-1] == 'D')
      {
        s[i-1] = 'U';
      }
      else s[i-1] = 'D';
      if (s[i+1] == 'D')
      {
        s[i+1] = 'U';
      }
      else s[i+1] = 'D';
      s.erase(i, 1);
      i -= 2;
      if (i==-1)
      {
        i++;
      }
      cnt++;
    }
  }
  assert(s.size() <= 3);
  debug(s, cnt);
  for (int i = 0; i < s.size(); ++i)
  {
    if (s[i] == 'U')
    {
      cnt++;
      s.erase(i, 1);
      i--;
    }
  }
  debug(s);
  if (cnt&1)
  {
    cout << "YES\n";
  }
  else cout << "NO\n";
}

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

  #ifndef ONLINE_JUDGE
    freopen("input.in", "r", stdin);
    freopen("output.out", "w", stdout);
    freopen("error.out", "w", stderr);
  #endif

  int ts;
  ts=1;
  cin>>ts;
  while(ts--){
    solve();
  }
}

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

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

Think step by step.

Response:
```plaintext
3
3
UUU
3
UUD
3
UDU
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Games
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are $n$ coins on the table forming a circle, and each coin is either facing up or facing down. Alice and Bob take turns to play the following game, and Alice goes first.

In each operation, the player chooses a facing-up coin, removes the coin, and flips the two coins that are adjacent to it. If (before the operation) there are only two coins left, then one will be removed and the other won't be flipped (as it would be flipped twice). If (before the operation) there is only one coin left, no coins will be flipped. If (before the operation) there are no facing-up coins, the player loses.

Decide who will win the game if they both play optimally. It can be proved that the game will end in a finite number of operations, and one of them will win.

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

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of the coins.

A string $s$ of length $n$ follows on the second line of each test case, containing only "U" and "D", representing that each coin is facing up or facing down.

Output: For each test case, print "YES" if Alice will win the game, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
3
5
UUDUD
5
UDDUD
2
UU
output:
YES
NO
NO

Note: In the first test case, the game may go as follows.

  * Alice chooses the first coin and $s$ becomes "DDUU".    * Bob chooses the last coin and $s$ becomes "UDD".    * Alice chooses the first coin and $s$ becomes "UU".    * Bob chooses the first coin and $s$ becomes "U".    * Alice chooses the only coin and $s$ becomes empty.    * Bob can't choose any coin now, and he loses the game.

It can be proved that Bob will always lose if they both play optimally.

Buggy Code:
def alice_wins(coins):
    # Count the number of 'U's at even and odd positions
    count_even = sum(coins[i] == 'U' for i in range(0, len(coins), 2))
    count_odd = sum(coins[i] == 'U' for i in range(1, len(coins), 2))

    # If the total number of 'U's is even, Alice wins if she starts with more 'U's on even positions
    # Otherwise, she wins if she starts with fewer 'U's on even positions
    return (count_even + count_odd) % 2 == 1 or count_even > count_odd

# Read the number of test cases
t = int(input())
for _ in range(t):
    # Read the number of coins and their initial configuration
    n = int(input())
    s = input()

    # Determine if Alice wins
    if alice_wins(s):
        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
4
1
U
2
UU
3
UUD
4
UUDD
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Games
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are $n$ coins on the table forming a circle, and each coin is either facing up or facing down. Alice and Bob take turns to play the following game, and Alice goes first.

In each operation, the player chooses a facing-up coin, removes the coin, and flips the two coins that are adjacent to it. If (before the operation) there are only two coins left, then one will be removed and the other won't be flipped (as it would be flipped twice). If (before the operation) there is only one coin left, no coins will be flipped. If (before the operation) there are no facing-up coins, the player loses.

Decide who will win the game if they both play optimally. It can be proved that the game will end in a finite number of operations, and one of them will win.

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

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of the coins.

A string $s$ of length $n$ follows on the second line of each test case, containing only "U" and "D", representing that each coin is facing up or facing down.

Output: For each test case, print "YES" if Alice will win the game, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
3
5
UUDUD
5
UDDUD
2
UU
output:
YES
NO
NO

Note: In the first test case, the game may go as follows.

  * Alice chooses the first coin and $s$ becomes "DDUU".    * Bob chooses the last coin and $s$ becomes "UDD".    * Alice chooses the first coin and $s$ becomes "UU".    * Bob chooses the first coin and $s$ becomes "U".    * Alice chooses the only coin and $s$ becomes empty.    * Bob can't choose any coin now, and he loses the game.

It can be proved that Bob will always lose if they both play optimally.

Buggy Code:
# in python memory limit exceeded can be beoz of storing more than the memory of int, hence use modulo if given in question
#in dp, precomputation before all test cases works sometimes
# t2 = perf_counter()
#ds=[input().strip().decode() for j in range(n)]   Array of strings (2d array of chars)
# A = input().split()                     # returns an array of strings
# A = [int(i) for i in input().split()]   # returns an array of integers
# C,D = [int(i) for i in input().split()] # returns two integer variables C & D

# A=[[0 for i in range(6)] for j in range(2)]       proper way to initialise 2d arrays. dont do like [0]*6 if you wanna change array in future
# A[0][0]=4

# for I/O for local system
import sys
from os import path
if(path.exists('Input.txt')):
    sys.stdin = open('Input.txt','r')
    sys.stdout = open('Output.txt','w')

# For fast I/O
input = sys.stdin.buffer.readline
# input = sys.stdin.readline
print = sys.stdout.write

# Import libraries here whenever required
from random import randint

# Use this because normal dict can sometimes give TLE
class mydict:
    def __init__(self, func=lambda: -1):
        self.random = randint(0, 1 << 32)
        self.default = func
        self.dict = {}
    def __getitem__(self, key):
        mykey = self.random ^ key
        if mykey not in self.dict:
            self.dict[mykey] = self.default()
        return self.dict[mykey]
    def get(self, key, default):
        mykey = self.random ^ key
        if mykey not in self.dict:
            return default
        return self.dict[mykey]
    def __setitem__(self, key, item):
        mykey = self.random ^ key
        self.dict[mykey] = item
    def getkeys(self):
        return [self.random ^ i for i in self.dict]
    def __str__(self):
        return f'{[(self.random ^ i, self.dict[i]) for i in self.dict]}'
    def sort(self):
        return sorted([self.random ^ i for i in self.dict])

# def manan_dict_ques():
#     d = mydict()
#     for i in b:
#         d[i] += 1
#     nd = mydict()
#     cnt = 0
#     ans = 0
#     for i in range(n):
#         val = a[i]
#         if(i >= m):
#             nd[a[i - m]] -= 1
#             if(nd[a[i - m]] < d[a[i - m]]):
#                 cnt -= 1
#         nd[val] += 1
#         if(nd[val] <= d[val]):
#             cnt += 1
#         if(i >= m - 1 and cnt >= k):
#             ans += 1
#     print(str(ans) + "\n")

def bs(arr,t,l,h):
    while h>=l:
        m=l+(h-l)//2
        if arr[m]==t:

            return True
        if arr[m]>t:
            h=m-1
        elif arr[m]<t:
            l=m+1
    return False

# Solver function
def issame(a,b):
    t=a[0]
    y=b[0]
    for i in range(1,len(a)):

        if i!=j:
            return False
    return True
def count(a):
    u=0
    d=0
    for i in a:
        if i=='U':
            u+=1
        else:
            d+=1
    return u,d

def cg(a):
    for i in range(1,len(a)):
        if i==n-1:
            if a[i]=='U' and a[i-1]=='U' and a[0]=="U":
                return i

        if a[i]=='U' and a[i-1]=='U' and a[i+1]=="U":
            return i

def rec(a):
    if len(a)==3:
        u,d=count(a)
        if u>d:
            return False
        else:
            return True

def sort_with_indices(arr):
    # Create a list of tuples where each tuple contains the original index and value
    indexed_arr = [(i, val) for i, val in enumerate(arr)]

    # Sort the list of tuples based on the values
    sorted_arr = sorted(indexed_arr, key=lambda x: x[1])

    # Extract the sorted indices
    sorted_indices = [index for index, _ in sorted_arr]
    oa=[_ for index, _ in sorted_arr]
    return sorted_indices, oa
def solve():
    n=int(input())
    s=input().strip().decode()
    if n>=3:
        u=0
        for i in s:
            if i=="U":
                u+=1
        if u==0:
            print("No\n")
        elif (n%2!=0 and u%2!=0) or (n%2==0 and u%2==0):
            print("YES\n")
        else:
            print("NO\n")
    elif n==2:
        u=0
        for i in s:
            if i=="U":
                u+=1
        if u==1:
            print("YES\n")
        else:
            print("No\n")
    else:
        if s[0]=="U":
            print("yes\n")
        else:
            print("No\n")

#Main
for _ in range(int(input())):
    solve()

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

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

Think step by step.

Response:
```plaintext
3
3
UUU
3
UDD
1
D
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Games
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are $n$ coins on the table forming a circle, and each coin is either facing up or facing down. Alice and Bob take turns to play the following game, and Alice goes first.

In each operation, the player chooses a facing-up coin, removes the coin, and flips the two coins that are adjacent to it. If (before the operation) there are only two coins left, then one will be removed and the other won't be flipped (as it would be flipped twice). If (before the operation) there is only one coin left, no coins will be flipped. If (before the operation) there are no facing-up coins, the player loses.

Decide who will win the game if they both play optimally. It can be proved that the game will end in a finite number of operations, and one of them will win.

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

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of the coins.

A string $s$ of length $n$ follows on the second line of each test case, containing only "U" and "D", representing that each coin is facing up or facing down.

Output: For each test case, print "YES" if Alice will win the game, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
3
5
UUDUD
5
UDDUD
2
UU
output:
YES
NO
NO

Note: In the first test case, the game may go as follows.

  * Alice chooses the first coin and $s$ becomes "DDUU".    * Bob chooses the last coin and $s$ becomes "UDD".    * Alice chooses the first coin and $s$ becomes "UU".    * Bob chooses the first coin and $s$ becomes "U".    * Alice chooses the only coin and $s$ becomes empty.    * Bob can't choose any coin now, and he loses the game.

It can be proved that Bob will always lose if they both play optimally.

Buggy Code:
import java.util.Scanner;

public class cf10 {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while (t>0) {
            int n=sc.nextInt();
            String str=sc.next();
            int alice=0,bob=0;
            for(int i=1;i<=n;i++){

                if(i<=n/2){
                    if(str.charAt(i-1)=='U'){
                        alice++;
                    }
                }

                else{
                        if(str.charAt(i-1)=='U'){
                            bob++;
                        }

                }

            }

            if(alice>bob){
                System.out.println("YES");
            }
            else{
                System.out.println("NO");
            }

            t--;
        }

    }

}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to consider scenarios where the code's logic fails to correctly determine the winner of the game. The buggy code simply counts the number of 'U' (facing-up) coins in the first half and the second half of the string and compares these counts to decide the winner. This approach is flawed because it does not account for the game's rules and the dynamic changes in the coin configuration during the game.

### Key Points to Consider:
1. **Initial Configuration**: The initial configuration of the coins can significantly affect the outcome of the game.
2. **Game Dynamics**: The game involves removing a facing-up coin and flipping the adjacent coins, which can change the configuration dynamically.
3. **Optimal Play**: Both players play optimally, so the initial configuration and the sequence of moves are crucial.

### Edge Cases:
1. **Single Coin**: The game ends immediately if there is only one coin.
2. **Two Coins**: The game ends with one coin left if there are two coins.
3. **No Facing-Up Coins**: The player whose turn it is loses if there are no facing-up coins.
4. **Symmetric Configurations**: Configurations where the number of facing-up coins is symmetric but the game dynamics are not considered.

### Challenging Test Input:
Let's create a test input that exposes the bug by focusing on a symmetric configuration where the number of facing-up coins is the same in both halves, but the game dynamics should lead to a different outcome.

```plaintext
3
4
UUUU
4
UDDU
4
UDUD
```

### Explanation:
1. **Test Case 1: `UUUU`**
   - Initial configuration: `UUUU`
   - Alice can remove any 'U' and flip the adjacent coins, resulting in `DDUU`.
   - Bob can remove any 'U' and flip the adjacent coins, resulting in `UDD`.
   - Alice can remove the first 'U' and flip the adjacent coins, resulting in `UU`.
   - Bob can remove the first 'U' and flip the adjacent coins, resulting in `U`.
   - Alice can remove the last 'U', and the game ends.
   - **Expected Output: YES**

2. **Test Case 2: `UDDU`**
   - Initial configuration: `UDDU`
   - Alice can remove the first 'U' and flip the adjacent coins, resulting in `DUD`.
   - Bob can remove the last 'U' and flip the adjacent coins, resulting in `DU`.
   - Alice can remove the first 'U', and the game ends.
   - **Expected Output: YES**

3. **Test Case 3: `UDUD`**
   - Initial configuration: `UDUD`
   - Alice can remove the first 'U' and flip the adjacent coins, resulting in `DUD`.
   - Bob can remove the last 'U' and flip the adjacent coins, resulting in `DU`.
   - Alice can remove the first 'U', and the game ends.
   - **Expected Output: YES**

However, the buggy code will incorrectly output `NO` for all these cases because it only counts the number of 'U' coins in the first and second halves without considering the game dynamics.

This test input should effectively expose the bug in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Games
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are $n$ coins on the table forming a circle, and each coin is either facing up or facing down. Alice and Bob take turns to play the following game, and Alice goes first.

In each operation, the player chooses a facing-up coin, removes the coin, and flips the two coins that are adjacent to it. If (before the operation) there are only two coins left, then one will be removed and the other won't be flipped (as it would be flipped twice). If (before the operation) there is only one coin left, no coins will be flipped. If (before the operation) there are no facing-up coins, the player loses.

Decide who will win the game if they both play optimally. It can be proved that the game will end in a finite number of operations, and one of them will win.

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

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of the coins.

A string $s$ of length $n$ follows on the second line of each test case, containing only "U" and "D", representing that each coin is facing up or facing down.

Output: For each test case, print "YES" if Alice will win the game, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
3
5
UUDUD
5
UDDUD
2
UU
output:
YES
NO
NO

Note: In the first test case, the game may go as follows.

  * Alice chooses the first coin and $s$ becomes "DDUU".    * Bob chooses the last coin and $s$ becomes "UDD".    * Alice chooses the first coin and $s$ becomes "UU".    * Bob chooses the first coin and $s$ becomes "U".    * Alice chooses the only coin and $s$ becomes empty.    * Bob can't choose any coin now, and he loses the game.

It can be proved that Bob will always lose if they both play optimally.

Buggy Code:
import java.io.*;
// import java.lang.reflect.Array;
import java.util.*;

public class Solution {
    public static void main(String[] args) throws IOException {
        Scanner scan=new Scanner(System.in);
        int t = scan.nextInt();
        for(int i=0;i<t;i++){
            ArrayList<Character> c = new ArrayList<>();
            int n = scan.nextInt();
            String s=scan.next();
            for(int j=0 ; j<n ; j++){
                c.add(s.charAt(j));
            }
            int counter=0;
            while(true){
                int index=findFlip(c);
                if(index>=0 && index<c.size()-1){
                    c=flipCoins(index,c);
                }else{
                    end(counter);
                    break;
                }
                counter+=1;
            }
        }
        scan.close();
    }
    public static void end(int t){
        if(t%2==0){
            System.out.println("Yes");
        } else{
            System.out.println("No");
        }
    }
    public static char flip(Character c) {
        if(c.equals('D')){
            return 'U';
        } else{
            return 'D';
        }
    }
    public static int findFlip(ArrayList<Character> c) {
        for(int i=0;i<c.size();i++){
            if(c.get(i)=='U'){
              return i;
            }
        }
        return c.size()+1;
    }
    public static ArrayList<Character> flipByIndex(int index, ArrayList<Character> c){
        if(index==0){
            char after=flip(c.get(1));
            char before=flip(c.get(c.size()-1));
            c.set(index+1,after);
            c.set(c.size()-1,before);
        } else if(index==c.size()-1){
            char after=flip(c.get(0));
            char before=flip(c.get(index-1));
            c.set(index-1,before);
            c.set(0,after);
        } else{
            char after=flip(c.get(index+1));
            char before=flip(c.get(index-1));
            c.set(index+1,after);
            c.set(index-1,before);
        }
        return c;
    }
    public static ArrayList<Character> flipCoins(int index, ArrayList<Character> c){
        int size=c.size();
        if(size==2 && index>=0){
           c.remove(index);
           return c;
        } else{
            c=flipByIndex(index,c);
            c.remove(index);
            return c;
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
3
1
U
2
UU
3
UDD
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Games
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are $n$ coins on the table forming a circle, and each coin is either facing up or facing down. Alice and Bob take turns to play the following game, and Alice goes first.

In each operation, the player chooses a facing-up coin, removes the coin, and flips the two coins that are adjacent to it. If (before the operation) there are only two coins left, then one will be removed and the other won't be flipped (as it would be flipped twice). If (before the operation) there is only one coin left, no coins will be flipped. If (before the operation) there are no facing-up coins, the player loses.

Decide who will win the game if they both play optimally. It can be proved that the game will end in a finite number of operations, and one of them will win.

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

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of the coins.

A string $s$ of length $n$ follows on the second line of each test case, containing only "U" and "D", representing that each coin is facing up or facing down.

Output: For each test case, print "YES" if Alice will win the game, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
3
5
UUDUD
5
UDDUD
2
UU
output:
YES
NO
NO

Note: In the first test case, the game may go as follows.

  * Alice chooses the first coin and $s$ becomes "DDUU".    * Bob chooses the last coin and $s$ becomes "UDD".    * Alice chooses the first coin and $s$ becomes "UU".    * Bob chooses the first coin and $s$ becomes "U".    * Alice chooses the only coin and $s$ becomes empty.    * Bob can't choose any coin now, and he loses the game.

It can be proved that Bob will always lose if they both play optimally.

Buggy Code:
no_of_inputs = int(input())
test_length = []
tests = []

for number in range(no_of_inputs):
    test_length.append(input())
    tests.append(input())

for problem in tests:
    length_of_test_case = len(problem)

    index = 0
    letter_array = []
    turns = 0

    for letter in problem:
        letter_array.append(letter)

    while len(letter_array) > 1:

        if letter_array[index] == "U":
            turns += 1
            try:
               if letter_array[index - 1] == "D":
                   letter_array[index - 1] = "U"
               elif letter_array[index - 1] == "U":
                   letter_array[index - 1] = "D"
            except:
               pass
            try:
               if letter_array[index + 1] == "D":
                   letter_array[index + 1] = "U"
               elif letter_array[index + 1] == "U":
                   letter_array[index + 1] = "D"
            except:
               pass
            letter_array.pop(index)
            index = 0

        index += 1

    if "D" in letter_array:
       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
1
3
UUU
```

### Explanation:
- **Test Case: 3 coins, all facing up ("UUU")**
  - **Initial State:** `UUU`
  - **Alice's Turn:**
    - Alice can choose any of the three "U" coins. Let's say she chooses the first "U". The state becomes `DUU`.
  - **Bob's Turn:**
    - Bob can choose the second "U". The state becomes `DDU`.
  - **Alice's Turn:**
    - Alice can choose the first "D". The state becomes `DU`.
  - **Bob's Turn:**
    - Bob can choose the second "D". The state becomes `D`.
  - **Alice's Turn:**
    - Alice can choose the only remaining "D". The state becomes empty.
  - **Result:**
    - Bob can't make a move, so Alice wins.

This test case is designed to expose the bug in the buggy code where it incorrectly handles the flipping and removal of coins, especially when the coins are all facing up. The expected output is "YES" because Alice wins, but the buggy code might incorrectly handle the sequence of operations and produce the wrong result.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Games
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are $n$ coins on the table forming a circle, and each coin is either facing up or facing down. Alice and Bob take turns to play the following game, and Alice goes first.

In each operation, the player chooses a facing-up coin, removes the coin, and flips the two coins that are adjacent to it. If (before the operation) there are only two coins left, then one will be removed and the other won't be flipped (as it would be flipped twice). If (before the operation) there is only one coin left, no coins will be flipped. If (before the operation) there are no facing-up coins, the player loses.

Decide who will win the game if they both play optimally. It can be proved that the game will end in a finite number of operations, and one of them will win.

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

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of the coins.

A string $s$ of length $n$ follows on the second line of each test case, containing only "U" and "D", representing that each coin is facing up or facing down.

Output: For each test case, print "YES" if Alice will win the game, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
3
5
UUDUD
5
UDDUD
2
UU
output:
YES
NO
NO

Note: In the first test case, the game may go as follows.

  * Alice chooses the first coin and $s$ becomes "DDUU".    * Bob chooses the last coin and $s$ becomes "UDD".    * Alice chooses the first coin and $s$ becomes "UU".    * Bob chooses the first coin and $s$ becomes "U".    * Alice chooses the only coin and $s$ becomes empty.    * Bob can't choose any coin now, and he loses the game.

It can be proved that Bob will always lose if they both play optimally.

Buggy Code:
def who_wins_game(n, s):
    # If there is only one coin, Alice wins
    if n == 1:
        return "YES"

    # If there are two coins, Bob wins if they are facing down
    if n == 2:
        if s == "DD" or s=='UU':
            return "NO"
        else:
            return "YES"

    # If there are more than two coins
    # Alice wins if there are an odd number of facing up coins initially
    if s.count('U') % 2 == 1:
        return "YES"
    else:
        return "NO"

def answer(len_str,test_str):
    for i in range(len(len_str)):
        print(who_wins_game(len_str[i], test_str[i]))
t = int(input())
len_str = []
test_str = []
for i in range(t):
    n = int(input())
    len_str.append(n)
    s = input()
    test_str.append(s)
answer(len_str,test_str)
# def presents(n,l):
#   return
# n = int(input())
# l = input().strip().split()
# print(presents(n,l))

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that the code might handle incorrectly. The buggy code has the following issues:

1. **Incorrect handling of the case with two coins**:
   - The code incorrectly states that if there are two coins and they are both facing up or both facing down, Bob wins. However, if both coins are facing up, Alice can win by removing one of them and flipping the other, leaving Bob with a single coin that he cannot flip.

2. **Incorrect handling of the case with more than two coins**:
   - The code assumes that Alice wins if there are an odd number of facing-up coins initially. This is not always correct, especially when the arrangement of the coins and the specific rules of the game are considered.

To create a test input that exposes these issues, we can include the following scenarios:

1. **Two coins, both facing up**:
   - This should result in Alice winning, but the buggy code will incorrectly state that Bob wins.

2. **Three coins, all facing up**:
   - This should result in Alice winning, and the buggy code should correctly state that Alice wins.

3. **Four coins, two facing up and two facing down**:
   - This is a more complex scenario where the arrangement of the coins matters. The buggy code might incorrectly state that Bob wins.

Here is the test input:

```plaintext
3
2
UU
3
UUU
4
UDDU
```

This test input includes:
- A case with two coins both facing up, which should result in Alice winning.
- A case with three coins all facing up, which should result in Alice winning.
- A case with four coins, two facing up and two facing down, which should result in Alice winning if she plays optimally.

This will help expose the bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Games
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are $n$ coins on the table forming a circle, and each coin is either facing up or facing down. Alice and Bob take turns to play the following game, and Alice goes first.

In each operation, the player chooses a facing-up coin, removes the coin, and flips the two coins that are adjacent to it. If (before the operation) there are only two coins left, then one will be removed and the other won't be flipped (as it would be flipped twice). If (before the operation) there is only one coin left, no coins will be flipped. If (before the operation) there are no facing-up coins, the player loses.

Decide who will win the game if they both play optimally. It can be proved that the game will end in a finite number of operations, and one of them will win.

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

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of the coins.

A string $s$ of length $n$ follows on the second line of each test case, containing only "U" and "D", representing that each coin is facing up or facing down.

Output: For each test case, print "YES" if Alice will win the game, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
3
5
UUDUD
5
UDDUD
2
UU
output:
YES
NO
NO

Note: In the first test case, the game may go as follows.

  * Alice chooses the first coin and $s$ becomes "DDUU".    * Bob chooses the last coin and $s$ becomes "UDD".    * Alice chooses the first coin and $s$ becomes "UU".    * Bob chooses the first coin and $s$ becomes "U".    * Alice chooses the only coin and $s$ becomes empty.    * Bob can't choose any coin now, and he loses the game.

It can be proved that Bob will always lose if they both play optimally.

Buggy Code:
import java.util.Scanner;

public class Coin_Games_1 {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        sc.nextLine();
        while(t-- > 0) {
            int n = sc.nextInt();
            sc.nextLine();
            StringBuilder sb = new StringBuilder(sc.nextLine());

            int c = 0;
            if(n <= 2) {
                for(int i = 0; i < n; i++) {
                    if(sb.charAt(i) == 'D') {
                        c++;
                    }
                }
            } else {
                for(int i = 0; i < sb.length(); i++) {
                    if(sb.charAt(i) == 'D' && sb.charAt((i + 1) % sb.length()) == 'U' && sb.charAt((i + 2) % sb.length()) == 'D') {
                        c++;
                        int m = (i + 1) % sb.length();
                        int p = (i + 2) % sb.length();
                        sb.setCharAt(i, 'U');
                        sb.setCharAt(p, 'U');
                        i+=2;
                    }
                }
                for(int i = 0; i < sb.length(); i++) {
                    if(sb.charAt(i) == 'D') {
                        c++;
                    }
                }
            }
            if(c % 2 != 0) {
                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
3
5
UDDDU
5
UUDDU
4
UUDD
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Games
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are $n$ coins on the table forming a circle, and each coin is either facing up or facing down. Alice and Bob take turns to play the following game, and Alice goes first.

In each operation, the player chooses a facing-up coin, removes the coin, and flips the two coins that are adjacent to it. If (before the operation) there are only two coins left, then one will be removed and the other won't be flipped (as it would be flipped twice). If (before the operation) there is only one coin left, no coins will be flipped. If (before the operation) there are no facing-up coins, the player loses.

Decide who will win the game if they both play optimally. It can be proved that the game will end in a finite number of operations, and one of them will win.

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

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of the coins.

A string $s$ of length $n$ follows on the second line of each test case, containing only "U" and "D", representing that each coin is facing up or facing down.

Output: For each test case, print "YES" if Alice will win the game, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
3
5
UUDUD
5
UDDUD
2
UU
output:
YES
NO
NO

Note: In the first test case, the game may go as follows.

  * Alice chooses the first coin and $s$ becomes "DDUU".    * Bob chooses the last coin and $s$ becomes "UDD".    * Alice chooses the first coin and $s$ becomes "UU".    * Bob chooses the first coin and $s$ becomes "U".    * Alice chooses the only coin and $s$ becomes empty.    * Bob can't choose any coin now, and he loses the game.

It can be proved that Bob will always lose if they both play optimally.

Buggy Code:
import java.util.Scanner;

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

        int t = sc.nextInt();

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

            int count_U = 0;
            int count_D = 0;

            String s = sc.next();

            for(int i=0; i<length; i++)
            {
                if(s.charAt(i) == 'U')
                {
                    count_U++;
                }
                else
                {
                    count_D++;
                }
            }

            if(count_U > count_D && count_D>0)
            {
                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
3
3
UUU
4
UUUU
5
UDDDU
```

**Explanation:**

1. **Test Case 1:**
   - **Input:** `3\nUUU`
   - **Expected Output:** `NO`
   - **Reason:** Alice starts and removes one 'U', leaving "UU". Bob then removes one 'U', leaving "U". Alice removes the last 'U', and Bob has no moves left, so Bob loses.

2. **Test Case 2:**
   - **Input:** `4\nUUUU`
   - **Expected Output:** `NO`
   - **Reason:** Alice starts and removes one 'U', leaving "UUU". Bob then removes one 'U', leaving "UU". Alice removes one 'U', leaving "U". Bob removes the last 'U', and Alice has no moves left, so Alice loses.

3. **Test Case 3:**
   - **Input:** `5\nUDDDU`
   - **Expected Output:** `YES`
   - **Reason:** Alice starts and removes the first 'U', leaving "DDD". Bob then removes the first 'D', leaving "DD". Alice removes the first 'D', leaving "D". Bob has no moves left, so Bob loses.

These test cases are designed to challenge the buggy code by focusing on edge cases where the number of 'U' and 'D' coins are critical to the outcome. The buggy code incorrectly assumes that the game can be won based on a simple count of 'U' and 'D' coins, which is not always correct.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Games
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are $n$ coins on the table forming a circle, and each coin is either facing up or facing down. Alice and Bob take turns to play the following game, and Alice goes first.

In each operation, the player chooses a facing-up coin, removes the coin, and flips the two coins that are adjacent to it. If (before the operation) there are only two coins left, then one will be removed and the other won't be flipped (as it would be flipped twice). If (before the operation) there is only one coin left, no coins will be flipped. If (before the operation) there are no facing-up coins, the player loses.

Decide who will win the game if they both play optimally. It can be proved that the game will end in a finite number of operations, and one of them will win.

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

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of the coins.

A string $s$ of length $n$ follows on the second line of each test case, containing only "U" and "D", representing that each coin is facing up or facing down.

Output: For each test case, print "YES" if Alice will win the game, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
3
5
UUDUD
5
UDDUD
2
UU
output:
YES
NO
NO

Note: In the first test case, the game may go as follows.

  * Alice chooses the first coin and $s$ becomes "DDUU".    * Bob chooses the last coin and $s$ becomes "UDD".    * Alice chooses the first coin and $s$ becomes "UU".    * Bob chooses the first coin and $s$ becomes "U".    * Alice chooses the only coin and $s$ becomes empty.    * Bob can't choose any coin now, and he loses the game.

It can be proved that Bob will always lose if they both play optimally.

Buggy Code:
#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <string>
#include <algorithm>
#include <numeric>
#include <climits>
#include <functional>
#include <queue>
#define ll long long int
#define all(x) x.begin(), x.end()

using namespace std;

void solve() {
  string s;
  int n;
  cin>>n;
  cin>>s;

  bool ok = false;

  map<string,bool> dp;

  dp["U"] = true;
  dp["UU"] = false;
  dp["UD"] = true;

  auto convert = [](string s, int i) {
    if(s.size()<3) {
      s.erase(s.begin()+i);
      return s;
    }

    s[(i+1)%s.size()] = s[(i+1)%s.size()] == 'U'?'D':'U';
    int t = i-1<0?(s.size()-1):i-1;

    s[t] = s[t]=='U'?'D':'U';

    s.erase(s.begin()+i);

    return s;

  };

  function<bool(string)> check = [&dp,&check,&convert](string s) {
    bool ok = false;

    if(dp.find(s)!=dp.end()) {
      return dp[s];
    }

    for(int i=0;i<s.size();i++) { if(s[i]!='U')
        continue;
      string t = convert(s, i);

      ok |= (!check(t));
    }

    dp[s] = ok;

    return ok;
  };

  if(check(s))
    cout<<"YES\n";
  else
   cout<<"NO\n";

}

int main()
{
    int t;
    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
3
UUU
4
UUUU
1
U
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Games
time_limit_ms: 1000
memory_limit_mb: 256
Description: There are $n$ coins on the table forming a circle, and each coin is either facing up or facing down. Alice and Bob take turns to play the following game, and Alice goes first.

In each operation, the player chooses a facing-up coin, removes the coin, and flips the two coins that are adjacent to it. If (before the operation) there are only two coins left, then one will be removed and the other won't be flipped (as it would be flipped twice). If (before the operation) there is only one coin left, no coins will be flipped. If (before the operation) there are no facing-up coins, the player loses.

Decide who will win the game if they both play optimally. It can be proved that the game will end in a finite number of operations, and one of them will win.

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

The first line of each test case contains only one positive integer $n$ ($1 \leq n \leq 100$), representing the number of the coins.

A string $s$ of length $n$ follows on the second line of each test case, containing only "U" and "D", representing that each coin is facing up or facing down.

Output: For each test case, print "YES" if Alice will win the game, and "NO" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
3
5
UUDUD
5
UDDUD
2
UU
output:
YES
NO
NO

Note: In the first test case, the game may go as follows.

  * Alice chooses the first coin and $s$ becomes "DDUU".    * Bob chooses the last coin and $s$ becomes "UDD".    * Alice chooses the first coin and $s$ becomes "UU".    * Bob chooses the first coin and $s$ becomes "U".    * Alice chooses the only coin and $s$ becomes empty.    * Bob can't choose any coin now, and he loses the game.

It can be proved that Bob will always lose if they both play optimally.

Buggy Code:
#include <iostream>
#include <iomanip>
#include <fstream>
#include <vector>
#include <set>
#include <queue>
#include <cmath>
#include <map>
#include <algorithm>
#include <numeric>
#include <stack>
#include <cstring>
#include <bitset>
#include <climits>
#include <valarray>
#include <list>
#include <unordered_map>
#include <unordered_set>
#include <complex>
#define int long long
#define double long double
#define endl '\n'
#define mod 1000000007
#define inf 2000000000000000000
using namespace std;
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;
// #define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
int count(string s){
    int cnt = 0;
    for (int i = 0; i < s.size(); i++) cnt += (s[i] == 'U');
    return cnt;
}
int n;
int prevv = -1;
map<pair<string,int>, int> mp;
int f(string s, int turn){
    if (mp[{s,turn}] != 0) return mp[{s,turn}];
    if (count(s) == 0) {
        return 1;
    }
    int ans = 1;
    for (int i = 0; i < s.size(); i++){
        if (s[i] == 'D') continue;
        string cur = s;
        int next = i+1;
        if (next == s.size()) next = 0;
        if (s.size() > 2){
            int prev = i-1;
            if (prev == -1) prev = s.size()-1;

            if (cur[prev] == 'U') cur[prev] = 'D';
            else cur[prev] = 'U';

            if (cur[next] == 'U') cur[next] = 'D';
            else cur[next] = 'U';
        }

        cur.erase(cur.begin()+i);

        int aa = f(cur, !turn);

        if (aa == 1) ans = 2;
    }
    prevv = ans;
    return mp[{s,turn}] = ans;
}
void solve(){
    cin >> n;
    string s; cin >> s;
    mp.clear();
    int ans = f(s,0);
    if (ans == 1){
        cout << "NO\n";
    } else {
        cout <<"YES\n";
    }

}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--) {
        solve();
    }
}

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

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

Think step by step.

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

1. **Edge Cases**:
   - Single coin: This is a simple edge case where the game ends immediately if the coin is facing up.
   - Two coins: This is another edge case where the rules change when there are only two coins left.
   - No facing-up coins: This is a critical edge case where the player loses immediately.

2. **Complex Cases**:
   - Multiple facing-up coins: This can help in testing the logic for multiple moves and the correctness of the game state transitions.
   - Cyclic nature: Since the coins form a circle, the transitions should be handled correctly, especially for the first and last coins.

3. **Optimal Play**:
   - Ensure that the test cases are designed to test the optimal play strategy, which means both players will make the best possible moves to win.

Given these considerations, let's create a small-scale test input that includes these edge cases and complex scenarios:

```plaintext
4
1
U
2
UU
3
UUD
5
UUDUD
```

### Explanation of the Test Input:
1. **Test Case 1**:
   - **n = 1, s = "U"**
   - Alice removes the only coin and wins immediately.
   - Expected output: "YES"

2. **Test Case 2**:
   - **n = 2, s = "UU"**
   - Alice removes one coin, and Bob removes the other. Bob wins.
   - Expected output: "NO"

3. **Test Case 3**:
   - **n = 3, s = "UUD"**
   - Alice removes the first coin, and the string becomes "DU".
   - Bob removes the only facing-up coin, and Alice loses.
   - Expected output: "NO"

4. **Test Case 4**:
   - **n = 5, s = "UUDUD"**
   - This is a more complex case where multiple moves are involved. Alice can make a move that forces Bob into a losing position.
   - Expected output: "YES"

This test input should help in exposing any bugs in the given buggy code, especially in handling edge cases and ensuring the correctness of the game state transitions.


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

You are given a rooted tree 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$^\dagger$ of the given tree.

Please note that the swaps are persistent through queries.

$^\dagger$ 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 $n$, $q$ ($2\le n\le 3\cdot 10^5$, $2\le q\le 10^5$) — the number of vertices in the tree and the number of queries.

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.

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 $3\cdot 10^5$, and the sum of all $q$ does not exceed $10^5$.

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:
3
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
5 4
1 1 3 4
2 3 4 5 1
5 1
4 5
3 4
2 3
output:
YES
YES
NO
YES
NO
NO
YES
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>
#define Fst first
#define Snd second
#define MP make_pair
#define ll long long
#define PII pair<int,int>
#define PIL pair<int,ll>
#define PLI pair<ll,int>
#define PLL pair<ll,ll>
#define PDD pair<double,double>
using namespace std;
int tac,n,m;
int fa[300010];
int a[300010],dep[300010],lca[300010][25];
inline int goup(int x,int y)
{
    for (int i=0;i<20;i++) if (y&(1<<i)) x=(x==-1 ? -1 : lca[x][i]);
    return x;
}
inline int LCA(int x,int y)
{
    if (dep[x]<dep[y]) swap(x,y);
    x=goup(x,dep[x]-dep[y]); if (x==y) return x;
    int r=0;
    for (int i=19;i>=0;i--)
    {
        if (lca[x][i]!=lca[y][i]) x=lca[x][i],y=lca[y][i];
        else if (lca[x][i]!=-1 && dep[lca[x][i]]>dep[r]) r=lca[x][i];
    }
    return r;
}
inline int DIST(int x,int y)
{
    return dep[x]+dep[y]-dep[LCA(x,y)]*2;
}
inline void Main()
{
    scanf("%d%d",&n,&m); fa[0]=-1;
    for (int i=1;i<n;i++) scanf("%d",&fa[i]),fa[i]--;
    for (int i=0;i<n;i++) scanf("%d",&a[i]),a[i]--;
    for (int i=0;i<n;i++) lca[i][0]=fa[i];
    for (int i=0;i<20;i++)
        for (int j=0;j<n;j++)
            lca[j][i+1]=(lca[j][i]==-1 ? -1 : lca[lca[j][i]][i]);
    dep[0]=0; for (int i=1;i<n;i++) dep[i]=dep[fa[i]]+1;
    ll al=0; for (int i=0;i<n;i++) al+=DIST(a[i],a[(i+1)%n]);
    while (m--)
    {
        int x,y; scanf("%d%d",&x,&y),x--,y--;
        set<PII> st;
        st.insert(MP((x-1+n)%n,x)); st.insert(MP(x,(x+1)%n));
        st.insert(MP((y-1+n)%n,y)); st.insert(MP(y,(y+1)%n));
        for (set<PII>::iterator it=st.begin();it!=st.end();it++) al-=DIST(a[it->Fst],a[it->Snd]);
        swap(a[x],a[y]);
        for (set<PII>::iterator it=st.begin();it!=st.end();it++) al+=DIST(a[it->Fst],a[it->Snd]);
        if (!a[0] && al==(n-1)*2) puts("YES");
        else puts("NO");
    }
}
int main()
{
    scanf("%d",&tac);
    while (tac--) Main();
    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
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
5 4
5 4
1 1 3 4
2 3 4 5 1
5 1
4 5
3 4
2 3
```


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

You are given a rooted tree 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$^\dagger$ of the given tree.

Please note that the swaps are persistent through queries.

$^\dagger$ 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 $n$, $q$ ($2\le n\le 3\cdot 10^5$, $2\le q\le 10^5$) — the number of vertices in the tree and the number of queries.

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.

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 $3\cdot 10^5$, and the sum of all $q$ does not exceed $10^5$.

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:
3
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
5 4
1 1 3 4
2 3 4 5 1
5 1
4 5
3 4
2 3
output:
YES
YES
NO
YES
NO
NO
YES
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 <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <set>

#define MAXNUM __INT_MAX__

int main() {
    int t;
    std::cin >> t;
    for (int tt = 0; tt < t; ++tt) {
        int n, q;
        std::cin >> n >> q;
        std::vector<int> parent(n + 1, 0);
        std::vector<int> perm(n);
        std::vector<int> perm_idx(n + 1);
        std::vector<std::map<int, int>> childs(n + 1); // order of childs in permutation
        for (int i = 2; i <= n; ++i) {
            std::cin >> parent[i];
        }

        for (int i = 0; i < n; ++i) {
            int curr;
            std::cin >> curr;
            perm[i] = curr;
            perm_idx[curr] = i;
            if (parent[curr]) {
                childs[parent[curr]][i] = curr;
            }
        }
        std::vector<int> indent(n + 1, 0); // number of childs and childs of childs in total
        for (int i = n; i >= 1; --i) {
            for (auto ch : childs[i]) {
                indent[i] += indent[ch.second] + 1;
            }
        }

        int wrong_vs = 0;   // if 0 - permutation is correct
        std::vector<bool> right_vs(n + 1, true);

        auto set_rightness = [&](int i){
            bool was_wrong = !right_vs[i];
            if (childs[i].size() > 0 && perm_idx[i] + 1 != childs[i].begin()->first) {
                right_vs[i] = false;
            } else if (
                parent[i] && std::next(childs[parent[i]].find(perm_idx[i])) != childs[parent[i]].end()
                && std::next(childs[parent[i]].find(perm_idx[i]))->first != perm_idx[i] + indent[i] + 1
            ) {
                right_vs[i] = false;
            } else {
                right_vs[i] = true;
            }
            if (was_wrong && right_vs[i]) {
                wrong_vs--;
            } else if (!was_wrong && !right_vs[i]) {
                wrong_vs++;
            }
        };
        for (int i = 1; i <= n; ++i) {
            set_rightness(i);
        }

        for (int qq = 0; qq < q; ++qq) {
            // query
            int v1idx, v2idx;
            std::cin >> v1idx >> v2idx;
            v1idx--;
            v2idx--;
            int v1 = perm[v1idx], v2 = perm[v2idx];

            if (parent[v1]) {
                childs[parent[v1]].erase(v1idx);
            }
            if (parent[v2]) {
                childs[parent[v2]].erase(v2idx);
            }

            int prev_idx1_1 = -1;
            if (parent[v1]) {
                auto lb = childs[parent[v1]].lower_bound(v1idx);
                if (lb != childs[parent[v1]].end() && lb != childs[parent[v1]].begin()) {
                    --lb;
                    prev_idx1_1 = lb->first;
                } else {
                    prev_idx1_1 = perm_idx[parent[v1]];
                }
            }
            int prev_idx2_1 = -1;
            if (parent[v2]) {
                auto lb = childs[parent[v2]].lower_bound(v2idx);
                if (lb != childs[parent[v2]].end() && lb != childs[parent[v2]].begin()) {
                    --lb;
                    prev_idx2_1 = lb->first;
                } else {
                    prev_idx2_1 = perm_idx[parent[v2]];
                }
            }

            // change idxs
            std::swap(v1idx, v2idx);
            std::swap(perm[perm_idx[v1]], perm[perm_idx[v2]]);
            std::swap(perm_idx[v1], perm_idx[v2]);
            if (parent[v1]) {
                childs[parent[v1]][v1idx] = v1;
            }
            if (parent[v2]) {
                childs[parent[v2]][v2idx] = v2;
            }

            int prev_idx1 = -1;
            if (parent[v1]) {
                auto lb = childs[parent[v1]].lower_bound(v1idx);
                if (lb != childs[parent[v1]].end() && lb != childs[parent[v1]].begin()) {
                    --lb;
                    prev_idx1 = lb->first;
                } else {
                    prev_idx1 = perm_idx[parent[v1]];
                }
            }
            int prev_idx2 = -1;
            if (parent[v2]) {
                auto lb = childs[parent[v2]].lower_bound(v2idx);
                if (lb != childs[parent[v2]].end() && lb != childs[parent[v2]].begin()) {
                    --lb;
                    prev_idx2 = lb->first;
                } else {
                    prev_idx2 = perm_idx[parent[v2]];
                }
            }

            // check previous and vertices
            if (parent[v1]) {
                set_rightness(perm[prev_idx1]);
                set_rightness(perm[prev_idx1_1]);
            }
            if (parent[v2]) {
                set_rightness(perm[prev_idx2]);
                set_rightness(perm[prev_idx2_1]);
            }
            set_rightness(v1);
            set_rightness(v2);

            if (wrong_vs == 0) {
                std::cout << "YES" << std::endl;
            } else {
                std::cout << "NO" << std::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
3
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
5 4
1 1 3 4
2 3 4 5 1
5 1
4 5
3 4
2 3
```


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

You are given a rooted tree 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$^\dagger$ of the given tree.

Please note that the swaps are persistent through queries.

$^\dagger$ 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 $n$, $q$ ($2\le n\le 3\cdot 10^5$, $2\le q\le 10^5$) — the number of vertices in the tree and the number of queries.

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.

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 $3\cdot 10^5$, and the sum of all $q$ does not exceed $10^5$.

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:
3
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
5 4
1 1 3 4
2 3 4 5 1
5 1
4 5
3 4
2 3
output:
YES
YES
NO
YES
NO
NO
YES
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:
#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")
#include <vector>
#pragma GCC target ("avx2")

#include<bits/stdc++.h>
#include<math.h>
using namespace std;

typedef long long int ll;
typedef long double ld;
typedef pair<ll, ll> pl;
typedef vector<ll> vl;
#define FD(i, r, l) for(ll i = r; i > (l); --i)

#define K first
#define V second
#define G(x) ll x; cin >> x;
#define GD(x) ld x; cin >> x;
#define GS(s) string s; cin >> s;
#define EX(x) { cout << x << '\n'; exit(0); }
#define A(a) (a).begin(), (a).end()
#define F(i, l, r) for (ll i = l; i < (r); ++i)

template<typename A, typename B>
A& chmax(A &a, B b) { return a < b ? (a=b): a; }

template<typename A, typename B>
A& chmin(A &a, B b) { return a > b ? (a=b): a; }

const ll N = 3e5 + 10;

ll sz[N];
vl adj[N];

#define NN N

namespace seg {
    typedef pl T;
    T id={0, 0};
    T f(T a, T b) {return {a.K ^ b.K, a.V ^ b.V};}

    T t[2 * NN];
    ll n=NN;  // array size

    void modify(ll p, T value) {  // set value at position p
      for (p+=n, t[p] = value; p /= 2;) t[p] = f(t[2*p], t[2*p+1]);
    }

    T query(ll l, ll r) { // fold f on interval [l, r)
      T resl=id, resr=id;
      for (l += n, r += n; l < r; l /= 2, r /= 2) {
        if (l&1) resl = f(resl, t[l++]);
        if (r&1) resr = f(t[--r], resr);
      }
      return f(resl, resr);
    }
}

// ll par[NN];

pl xhash[NN];

pl subhash[NN];

ll par[NN];

ll dfs(ll i, ll p = 0) {
    par[i] = p;
    subhash[i] = xhash[i];
    for (ll x: adj[i]) {
        sz[i] += dfs(x, i);
        subhash[i].K ^= subhash[x].K, subhash[i].V^=subhash[x].V;
    }
    return ++sz[i];
}

int main(){
//    freopen("a.in", "r", stdin);
//    freopen("a.out", "w", stdout);

    ios_base::sync_with_stdio(false); cin.tie(0);
    cout << fixed << setprecision(20);

    random_device rd;
    mt19937_64 rng(rd());
    G(t) while(t--) {
        G(n) G(q)

        F(i, 0, n+1) {
            sz[i] = 0, adj[i].clear(), seg::modify(i, seg::id);
            xhash[i] = {rng(), rng()};
        }
        F(i, 2, n + 1) {
            G(x) adj[x].push_back(i);
        }

        dfs(1);

        vl perm(n );
        vl pos(n + 1); // positions of i in permutation
        F(i, 0, n) {
            cin >> perm[i];
            pos[perm[i]] = i;
            seg::modify(i, xhash[perm[i]]);
        }

        auto good = [&](ll x) -> bool {
            if (pos[x] + sz[x] > n) return false;
            return seg::query(pos[x], pos[x] + sz[x]) == subhash[x];
        };

        set<ll> bad;
        F(i, 1, n+1) if (!good(i)) bad.insert(i);

        while (q--) {
            G(x) G(y)
            --x,--y;

            x = perm[x], y = perm[y];
            bad.insert(x), bad.insert(y);
            bad.insert(par[x]); bad.insert(par[y]);

            if (pos[x] > 0) bad.insert(perm[pos[x] - 1]);
            if (pos[y] > 0) bad.insert(perm[pos[y] - 1]);

            // F(i, 1, n+1) bad.insert(i);

            bad.erase(0);

            ll px = pos[x], py = pos[y];
            swap(perm[px], perm[py]);
            seg::modify(px, xhash[y]);
            seg::modify(py, xhash[x]);
            pos[x] = py, pos[y] = px;

            while (bad.size()) {
                auto v = bad.begin();
                if (good(*v)) {
                    bad.erase(v);
                    continue;
                } else break;
            }

            cout << (bad.empty() ? "Yes":"No") << '\n';
        }

    }
}

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

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

Think step by step.

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


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

You are given a rooted tree 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$^\dagger$ of the given tree.

Please note that the swaps are persistent through queries.

$^\dagger$ 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 $n$, $q$ ($2\le n\le 3\cdot 10^5$, $2\le q\le 10^5$) — the number of vertices in the tree and the number of queries.

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.

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 $3\cdot 10^5$, and the sum of all $q$ does not exceed $10^5$.

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:
3
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
5 4
1 1 3 4
2 3 4 5 1
5 1
4 5
3 4
2 3
output:
YES
YES
NO
YES
NO
NO
YES
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
from types import GeneratorType

def bootstrap(f, stack=None):
    if stack is None:
        stack = []

    def func(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            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 func

class HLD:
    def __init__(self, g: list):
        n = len(g)
        m = n.bit_length()
        sz = [1] * n
        depth = [-1] * n
        depth[0] = 0
        # pa[u][j]：u的第j级祖先,pa[u][0]代表u的父亲
        pa = [[-1] * m for _ in range(n)]

        q = deque([0])
        level_order = []
        while q:
            u = q.popleft()
            level_order.append(u)
            for v in g[u]:
                if depth[v] >= 0:
                    continue
                depth[v] = depth[u] + 1
                pa[v][0] = u
                q.append(v)

        for u in level_order[::-1]:
            if (fa := pa[u][0]) != -1:
                sz[fa] += sz[u]

        for i in range(m - 1):
            for x in range(n):
                if (p := pa[x][i]) != -1:
                    pa[x][i + 1] = pa[p][i]

        self.sz = sz
        self.depth = depth
        self.pa = pa

    def get_kth_ancestor(self, node: int, k: int) -> int:
        for i in range(k.bit_length()):
            if k >> i & 1:
                node = self.pa[node][i]
        return node

    def get_lca(self, x: int, y: int) -> int:
        if self.depth[x] > self.depth[y]:
            x, y = y, x
        y = self.get_kth_ancestor(y, self.depth[y] - self.depth[x])
        if y == x:
            return x
        for i in range(len(self.pa[x]) - 1, -1, -1):
            px, py = self.pa[x][i], self.pa[y][i]
            if px != py:
                x, y = px, py
        return self.pa[x][0]

def cf2002D():
    n, q = map(int, input().split())
    a = [-1] + list(map(int, input().split()))
    p = list(map(int, input().split()))
    g = [[] for _ in range(n)]
    for i in range(1, n):
        a[i] -= 1
        g[a[i]].append(i)

    hld = HLD(g)
    for i in range(n):
        p[i] -= 1

    cnt = 0
    good = [0] * n

    def check(i: int) -> None:
        nonlocal cnt
        if i < 1 or i >= n:
            return

        cnt -= good[i]
        good[i] = 1 if hld.get_lca(p[i], p[i - 1]) == a[p[i]] else 0
        cnt += good[i]

    for i in range(1, n):
        check(i)

    for _ in range(q):
        x, y = map(int, input().split())
        x -= 1
        y -= 1
        p[x], p[y] = p[y], p[x]
        check(x)
        check(x + 1)
        check(y)
        check(y + 1)
        if cnt == n - 1:
            print("YES")
        else:
            print("NO")
    return

t = int(input())
for i in range(t):
    cf2002D()

**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 focus on edge cases and scenarios that can cause the algorithm to fail. The code involves a combination of tree traversal and permutation checks, so we should consider the following:

1. **Tree Structure**: The tree should have a structure that can cause issues with the DFS order checks.
2. **Permutation**: The initial permutation should be such that it is a valid DFS order, but swaps can easily make it invalid.
3. **Queries**: The queries should be designed to make the permutation invalid or to test edge cases in the tree structure.

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

1. **Tree Structure**: A simple tree with a root and a few children.
2. **Permutation**: A permutation that is a valid DFS order.
3. **Queries**: Queries that swap elements to make the permutation invalid.

Here is a test input that meets these criteria:

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

### Explanation:
- **Tree Structure**: The tree has 5 nodes. Node 1 is the root, and nodes 2 and 3 are its children. Nodes 4 and 5 are children of nodes 2 and 3, respectively.
- **Permutation**: The initial permutation `[1, 2, 3, 4, 5]` is a valid DFS order.
- **Queries**:
  - Swap 2 and 4: The permutation becomes `[1, 4, 3, 2, 5]`, which is not a valid DFS order.
  - Swap 3 and 5: The permutation becomes `[1, 4, 5, 2, 3]`, which is not a valid DFS order.
  - Swap 1 and 3: The permutation becomes `[3, 4, 1, 2, 5]`, which is not a valid DFS order.
  - Swap 2 and 5: The permutation becomes `[3, 5, 1, 2, 4]`, which is not a valid DFS order.

This test input should expose the bug in the provided code by making the permutation invalid after each swap.