**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Palindrome Everywhere
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a cycle with $n$ vertices numbered from $0$ to $n-1$. For each $0\le i\le n-1$, there is an undirected edge between vertex $i$ and vertex $((i+1)\bmod n)$ with the color $c_i$ ($c_i=\texttt{R}$ or $\texttt{B}$).

Determine whether the following condition holds for every pair of vertices $(i,j)$ ($0\le i<j\le n-1$):

  * There exists a palindrome route between vertex $i$ and vertex $j$. Note that the route may not be simple. Formally, there must exist a sequence $p=[p_0,p_1,p_2,\ldots,p_m]$ such that:     * $p_0=i$, $p_m=j$;      * For each $0\leq x\le m-1$, either $p_{x+1}=(p_x+1)\bmod n$ or $p_{x+1}=(p_{x}-1)\bmod n$;      * For each $0\le x\le y\le m-1$ satisfying $x+y=m-1$, the edge between $p_x$ and $p_{x+1}$ has the same color as the edge between $p_y$ and $p_{y+1}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer $n$ ($3\leq n\leq10^6$) — the number of vertices in the cycle.

The second line contains a string $c$ of length $n$ ($c_i=\texttt{R}$ or $\texttt{B}$) — the color of each edge.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print "YES" (without quotes) if there is a palindrome route between any pair of nodes, and "NO" (without quotes) 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:
7
5
RRRRR
5
RRRRB
5
RBBRB
6
RBRBRB
6
RRBBRB
5
RBRBR
12
RRBRRBRRBRRB
output:
YES
YES
YES
NO
NO
YES
NO

Note: In the first test case, it is easy to show that there is a palindrome route between any two vertices.

In the second test case, for any two vertices, there exists a palindrome route with only red edges.

In the third test case, the cycle is as follows: $0\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}3\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}0$. Take $(i,j)=(0,3)$ as an example, then $0\color{red}{\overset{\texttt{R}}{\longrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longrightarrow}}3\color{red}{\overset{\texttt{R}}{\longrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longrightarrow}}0\color{blue}{\overset{\texttt{B}}{\longrightarrow}}4\color{red}{\overset{\texttt{R}}{\longrightarrow}}3$ is a palindrome route. Thus, the condition holds for $(i,j)=(0,3)$.

In the fourth test case, when $(i,j)=(0,2)$, there does not exist a palindrome route.

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

int n;
char s[1000005];

int main(void) {
    int tc;
    scanf("%d",&tc);
    for(int t=0;t<tc;t++) {
        scanf("%d",&n);
        scanf("%s",s);
        int r=0;
        int b=0;
        int r2=0;
        int b2=0;
        vector<int> vr;
        vector<int> vb;
        for(int i=0;i<n;i++) {
            if (s[i]=='R') {
                r++;
            }
            if (s[i]=='B') {
                b++;
            }
            if (s[i]==s[(i+1)%n]) {
                if (s[i]=='R') {
                    r2++;
                    vr.push_back(i);
                }
                else {
                    b2++;
                    vb.push_back(i);
                }
            }
        }
        if (b==0||r==0) {
            printf("YES\n");
            continue;
        }
        if (r2==0&&b2==0) {
            printf("NO\n");
            continue;
        }
        if (r2>0&&b2>0) {
            printf("NO\n");
            continue;
        }
        if (r2>0) {
            sort(vr.begin(),vr.end());
            int cnt=0;
            for(int i=0;i<vr.size();i++) {
                if (i+1==vr.size()) {
                    if (vr[i]==n-1&&vr[0]==0) {
                        cnt++;
                    }
                }
                else {
                    if (vr[i+1]==vr[i]+1) {
                        cnt++;
                    }
                }
            }
            assert(cnt!=vr.size());
            if (cnt+1==vr.size()) {
                printf("YES\n");
            }
            else {
                printf("NO\n");
            }
        }
        else {
            sort(vb.begin(),vb.end());
            int cnt=0;
            for(int i=0;i<vb.size();i++) {
                if (i+1==vb.size()) {
                    if (vb[i]==n-1&&vb[0]==0) {
                        cnt++;
                    }
                }
                else {
                    if (vb[i+1]==vb[i]+1) {
                        cnt++;
                    }
                }
            }
            assert(cnt!=vb.size());
            if (cnt+1==vb.size()) {
                printf("YES\n");
            }
            else {
                printf("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
3
4
RRBB
4
RBRB
4
BRBR
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Palindrome Everywhere
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a cycle with $n$ vertices numbered from $0$ to $n-1$. For each $0\le i\le n-1$, there is an undirected edge between vertex $i$ and vertex $((i+1)\bmod n)$ with the color $c_i$ ($c_i=\texttt{R}$ or $\texttt{B}$).

Determine whether the following condition holds for every pair of vertices $(i,j)$ ($0\le i<j\le n-1$):

  * There exists a palindrome route between vertex $i$ and vertex $j$. Note that the route may not be simple. Formally, there must exist a sequence $p=[p_0,p_1,p_2,\ldots,p_m]$ such that:     * $p_0=i$, $p_m=j$;      * For each $0\leq x\le m-1$, either $p_{x+1}=(p_x+1)\bmod n$ or $p_{x+1}=(p_{x}-1)\bmod n$;      * For each $0\le x\le y\le m-1$ satisfying $x+y=m-1$, the edge between $p_x$ and $p_{x+1}$ has the same color as the edge between $p_y$ and $p_{y+1}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer $n$ ($3\leq n\leq10^6$) — the number of vertices in the cycle.

The second line contains a string $c$ of length $n$ ($c_i=\texttt{R}$ or $\texttt{B}$) — the color of each edge.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print "YES" (without quotes) if there is a palindrome route between any pair of nodes, and "NO" (without quotes) 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:
7
5
RRRRR
5
RRRRB
5
RBBRB
6
RBRBRB
6
RRBBRB
5
RBRBR
12
RRBRRBRRBRRB
output:
YES
YES
YES
NO
NO
YES
NO

Note: In the first test case, it is easy to show that there is a palindrome route between any two vertices.

In the second test case, for any two vertices, there exists a palindrome route with only red edges.

In the third test case, the cycle is as follows: $0\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}3\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}0$. Take $(i,j)=(0,3)$ as an example, then $0\color{red}{\overset{\texttt{R}}{\longrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longrightarrow}}3\color{red}{\overset{\texttt{R}}{\longrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longrightarrow}}0\color{blue}{\overset{\texttt{B}}{\longrightarrow}}4\color{red}{\overset{\texttt{R}}{\longrightarrow}}3$ is a palindrome route. Thus, the condition holds for $(i,j)=(0,3)$.

In the fourth test case, when $(i,j)=(0,2)$, there does not exist a palindrome route.

Buggy Code:
//这回只花了114514min就打完了。
//真好。记得多手造几组。ACM拍什么拍。
#include "bits/stdc++.h"
using namespace std;
using ui = unsigned; using db = long double; using ll = long long; using ull = unsigned long long; using lll = __int128;
template<class T1, class T2> istream &operator>>(istream &cin, pair<T1, T2> &a) { return cin >> a.first >> a.second; }
template <std::size_t Index = 0, typename... Ts> typename std::enable_if<Index == sizeof...(Ts), void>::type tuple_read(std::istream &is, std::tuple<Ts...> &t) { }
template <std::size_t Index = 0, typename... Ts> typename std::enable_if < Index < sizeof...(Ts), void>::type tuple_read(std::istream &is, std::tuple<Ts...> &t) { is >> std::get<Index>(t); tuple_read<Index + 1>(is, t); }
template <typename... Ts>std::istream &operator>>(std::istream &is, std::tuple<Ts...> &t) { tuple_read(is, t); return is; }
template<class T1> istream &operator>>(istream &cin, valarray<T1> &a);
template<class T1> istream &operator>>(istream &cin, vector<T1> &a) { for (auto &x : a) cin >> x; return cin; }
template<class T1> istream &operator>>(istream &cin, valarray<T1> &a) { for (auto &x : a) cin >> x; return cin; }
template<class T1, class T2> bool cmin(T1 &x, const T2 &y) { if (y < x) { x = y; return 1; } return 0; }
template<class T1, class T2> bool cmax(T1 &x, const T2 &y) { if (x < y) { x = y; return 1; } return 0; }
template<class T1> vector<T1> range(T1 l, T1 r, T1 step = 1) { assert(step > 0); int n = (r - l + step - 1) / step, i; vector<T1> res(n); for (i = 0; i < n; i++) res[i] = l + step * i; return res; }
template<class T1> basic_string<T1> operator*(const basic_string<T1> &s, int m) { auto r = s; m *= s.size(); r.resize(m); for (int i = s.size(); i < m; i++) r[i] = r[i - s.size()]; return r; }
istream &operator>>(istream &cin, lll &x) { bool flg = 0; x = 0; static string s; cin >> s; if (s[0] == '-') flg = 1, s = s.substr(1); for (char c : s) x = x * 10 + (c - '0'); if (flg) x = -x; return cin; }
ostream &operator<<(ostream &cout, lll x) { static char s[60]; if (x < 0) cout << '-', x = -x; int tp = 1; s[0] = '0' + (x % 10); while (x /= 10) s[tp++] = '0' + (x % 10); while (tp--) cout << s[tp]; return cout; }
#if !defined(ONLINE_JUDGE)&&defined(LOCAL)
#include "my_header/debug.h"
#else
#define dbg(...) ;
#endif
template<class T1, class T2> ostream &operator<<(ostream &cout, const pair<T1, T2> &a) { return cout << a.first << ' ' << a.second; }
template<class T1, class T2> ostream &operator<<(ostream &cout, const vector<pair<T1, T2>> &a) { for (auto &x : a) cout << x << '\n'; return cout; }
template<class T1> ostream &operator<<(ostream &cout, vector<T1> a) { int n = a.size(); if (!n) return cout; cout << a[0]; for (int i = 1; i < n; i++) cout << ' ' << a[i]; return cout; }
template<class T1> ostream &operator<<(ostream &cout, const valarray<T1> &a) { int n = a.size(); if (!n) return cout; cout << a[0]; for (int i = 1; i < n; i++) cout << ' ' << a[i]; return cout; }
template<class T1> ostream &operator<<(ostream &cout, const vector<valarray<T1>> &a) { int n = a.size(); if (!n) return cout; cout << a[0]; for (int i = 1; i < n; i++) cout << '\n' << a[i]; return cout; }
template<class T1> ostream &operator<<(ostream &cout, const vector<vector<T1>> &a) { int n = a.size(); if (!n) return cout; cout << a[0]; for (int i = 1; i < n; i++) cout << '\n' << a[i]; return cout; }
#define all(x) (x).begin(),(x).end()
#define print(...) cout<<format(__VA_ARGS__)
#define println(...) cout<<format(__VA_ARGS__)<<'\n'
#define err(...) cerr<<format(__VA_ARGS__)
#define errln(...) cerr<<format(__VA_ARGS__)<<'\n'
int main()
{
	ios::sync_with_stdio(0); cin.tie(0);
	cout << fixed << setprecision(15);
	int T; cin >> T;
	while (T--)
	{
		auto solve = [&]() {
			int n, m, i, j;
			string s;
			cin >> n >> s;
			{
				auto x = s.find('R' ^ 'B' ^ s[0]);
				if (x == s.npos) return 1;
				rotate(s.begin(), x + all(s));
			}
			assert(s[0] != s[n - 1]);
			vector<int> a(n);
			for (i = 0; i < n; i++) a[i] = s[i] == 'R';
			int flg = 0;
			for (i = j = 0; i < n; i = j)
			{
				while (j < n && a[i] == a[j]) ++j;
				flg |= (j - i > 1) << a[i];
			}
			if (flg == 3 || flg == 0) return 0;
			if (flg == 1) for (int &x : a) x ^= 1;
			int c = count(all(a), 0);
			if (c >= 3) return 0;
			if (c <= 1) return 1;
			return n & 1;
		};
		if (solve()) cout << "YES\n";
		else cout << "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
5
RBRBR
```

### Explanation:
This test case is designed to challenge the buggy code by providing a specific pattern of colors that may not be handled correctly.

- The string "RBRBR" is a cycle of 5 vertices with alternating red (R) and blue (B) edges.
- The expected output is "YES" because there exists a palindrome route between any pair of vertices in this cycle. For example:
  - Between vertices 0 and 2: `0 -> 1 -> 2` (RBR).
  - Between vertices 1 and 3: `1 -> 2 -> 3` (BRB).

This test case is small and valid, ensuring that it fits within the constraints and can be used to test the algorithm effectively. The alternating pattern of colors is a common edge case that can expose bugs in algorithms that do not correctly handle symmetry or palindrome conditions.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Palindrome Everywhere
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a cycle with $n$ vertices numbered from $0$ to $n-1$. For each $0\le i\le n-1$, there is an undirected edge between vertex $i$ and vertex $((i+1)\bmod n)$ with the color $c_i$ ($c_i=\texttt{R}$ or $\texttt{B}$).

Determine whether the following condition holds for every pair of vertices $(i,j)$ ($0\le i<j\le n-1$):

  * There exists a palindrome route between vertex $i$ and vertex $j$. Note that the route may not be simple. Formally, there must exist a sequence $p=[p_0,p_1,p_2,\ldots,p_m]$ such that:     * $p_0=i$, $p_m=j$;      * For each $0\leq x\le m-1$, either $p_{x+1}=(p_x+1)\bmod n$ or $p_{x+1}=(p_{x}-1)\bmod n$;      * For each $0\le x\le y\le m-1$ satisfying $x+y=m-1$, the edge between $p_x$ and $p_{x+1}$ has the same color as the edge between $p_y$ and $p_{y+1}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer $n$ ($3\leq n\leq10^6$) — the number of vertices in the cycle.

The second line contains a string $c$ of length $n$ ($c_i=\texttt{R}$ or $\texttt{B}$) — the color of each edge.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print "YES" (without quotes) if there is a palindrome route between any pair of nodes, and "NO" (without quotes) 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:
7
5
RRRRR
5
RRRRB
5
RBBRB
6
RBRBRB
6
RRBBRB
5
RBRBR
12
RRBRRBRRBRRB
output:
YES
YES
YES
NO
NO
YES
NO

Note: In the first test case, it is easy to show that there is a palindrome route between any two vertices.

In the second test case, for any two vertices, there exists a palindrome route with only red edges.

In the third test case, the cycle is as follows: $0\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}3\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}0$. Take $(i,j)=(0,3)$ as an example, then $0\color{red}{\overset{\texttt{R}}{\longrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longrightarrow}}3\color{red}{\overset{\texttt{R}}{\longrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longrightarrow}}0\color{blue}{\overset{\texttt{B}}{\longrightarrow}}4\color{red}{\overset{\texttt{R}}{\longrightarrow}}3$ is a palindrome route. Thus, the condition holds for $(i,j)=(0,3)$.

In the fourth test case, when $(i,j)=(0,2)$, there does not exist a palindrome route.

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

using namespace std;

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

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

const int N = 1e6 + 23, lg = 18;
ll Mod = 1e9 + 7;

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

int t, n, a[N], tmp, sz[N], par[N];

int getPar(int v) {
    return (par[v]==0?v:par[v]=getPar(par[v]));
}

void merge(int v, int u) {
    v=getPar(v), u=getPar(u);
    if(v == u) return;
    sz[v] += sz[u];
    par[u] = v;
    tmp++;
}

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

    cin>>t;
    while(t--) {
        tmp = 0;
        cin>>n;
        fill(sz+1, sz+n+1, 1);
        for(int i=1; i<=n; i++) {
            char ch; cin>>ch;
            if(ch == 'R') a[i] = 1;
            else a[i] = 0;
            if(i > 1 && a[i-1] == a[i]) merge(i, i-1);
        }
        if(a[n] == a[1]) merge(1, n);

        int cnt = 0, oddc = 0;
        for(int i=1; i<=n; i++) {
            if(a[i]==1 && sz[getPar(i)]>1) cnt|=2;
            if(a[i]==0 && sz[getPar(i)]>1) cnt|=1;
        }

        for(int i=1; i<=n; i++) {
            if(cnt == 2 && getPar(i)==i && a[i]==1 && sz[i]%2==0) oddc++;
            if(cnt == 1 && getPar(i)==i && a[i]==0 && sz[i]%2==0) oddc++;
        }

        if(sz[getPar(1)]==n-1 || sz[getPar(n)]==n-1) {
            cout<<"YES\n";
        } else if((cnt == 3 || cnt == 0)) {
            cout<<"NO\n";
            // fuck("cnt");
        } else if(oddc > 1) {
            cout<<"NO\n";
            // fuck("coddc");
        } else {
            cout<<"YES\n";
        }

        for(int i=0; i<=n; i++) {
            sz[i] = par[i] = 0;
        }
    }

    return 0;
}

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

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

Think step by step.

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

1. **Identify the Bug**: The bug in the code seems to be related to the logic for determining if there is a palindrome route between any two vertices. Specifically, the code might not handle certain edge cases correctly, such as cycles with alternating colors or cycles with specific lengths that could cause issues in the merge operations.

2. **Edge Cases to Consider**:
   - **Alternating Colors**: A cycle with alternating colors (e.g., "RBRBRB") should be a valid palindrome route, but the code might fail to recognize it.
   - **Cycles with Specific Lengths**: Cycles with lengths that are multiples of 2 or 3 might cause issues in the merge operations.
   - **Single Color Cycles**: Cycles with a single color (e.g., "RRRRR") should always be valid, but the code might have issues with the merge logic.
   - **Mixed Color Cycles**: Cycles with a mix of colors in specific patterns (e.g., "RRBBRR") might cause issues in the merge operations.

3. **Generate the Test Input**:
   - We will create a test input with a single test case that has a cycle with alternating colors and a specific length that might cause issues in the merge operations.

Here is the test input:

```plaintext
1
6
RBRBRB
```

This test case has a cycle with 6 vertices and alternating colors "RBRBRB". According to the problem description, there should be a palindrome route between any two vertices in this cycle. However, the buggy code might fail to recognize this correctly due to issues in the merge logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Palindrome Everywhere
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a cycle with $n$ vertices numbered from $0$ to $n-1$. For each $0\le i\le n-1$, there is an undirected edge between vertex $i$ and vertex $((i+1)\bmod n)$ with the color $c_i$ ($c_i=\texttt{R}$ or $\texttt{B}$).

Determine whether the following condition holds for every pair of vertices $(i,j)$ ($0\le i<j\le n-1$):

  * There exists a palindrome route between vertex $i$ and vertex $j$. Note that the route may not be simple. Formally, there must exist a sequence $p=[p_0,p_1,p_2,\ldots,p_m]$ such that:     * $p_0=i$, $p_m=j$;      * For each $0\leq x\le m-1$, either $p_{x+1}=(p_x+1)\bmod n$ or $p_{x+1}=(p_{x}-1)\bmod n$;      * For each $0\le x\le y\le m-1$ satisfying $x+y=m-1$, the edge between $p_x$ and $p_{x+1}$ has the same color as the edge between $p_y$ and $p_{y+1}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer $n$ ($3\leq n\leq10^6$) — the number of vertices in the cycle.

The second line contains a string $c$ of length $n$ ($c_i=\texttt{R}$ or $\texttt{B}$) — the color of each edge.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print "YES" (without quotes) if there is a palindrome route between any pair of nodes, and "NO" (without quotes) 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:
7
5
RRRRR
5
RRRRB
5
RBBRB
6
RBRBRB
6
RRBBRB
5
RBRBR
12
RRBRRBRRBRRB
output:
YES
YES
YES
NO
NO
YES
NO

Note: In the first test case, it is easy to show that there is a palindrome route between any two vertices.

In the second test case, for any two vertices, there exists a palindrome route with only red edges.

In the third test case, the cycle is as follows: $0\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}3\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}0$. Take $(i,j)=(0,3)$ as an example, then $0\color{red}{\overset{\texttt{R}}{\longrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longrightarrow}}3\color{red}{\overset{\texttt{R}}{\longrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longrightarrow}}0\color{blue}{\overset{\texttt{B}}{\longrightarrow}}4\color{red}{\overset{\texttt{R}}{\longrightarrow}}3$ is a palindrome route. Thus, the condition holds for $(i,j)=(0,3)$.

In the fourth test case, when $(i,j)=(0,2)$, there does not exist a palindrome route.

Buggy Code:
import sys
import os
from io import BytesIO, IOBase

# region fastio
BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

if sys.version_info[0] < 3:
    sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

file = sys.stdin
if os.environ.get('USER') == "loic":
    file = open("data.in")

line = lambda: file.readline().split()
ui = lambda: int(line()[0])
mi = lambda: map(int,line())
li = lambda: list(mi())
lsi = lambda: list(map(int,line()[0]))
#######################################################################

def solve():

    r = A.count('R')
    b = A.count('B')

    if r <= 1 or b <= 1:
        return "YES"

    if N % 2 == 1:
        c = 0
        for i in range(N):
            if A[i] == A[i-1]:
                c += 1
        if c == 1:
            return "YES"

        if r == 2 or b == 2:
            val = 'R' if r == 2 else 'B'
            f = A.index(val)
            s = A.index(val, f + 1, N)
            if s - f != 1 and s - f != N - 1:
                return "YES"

    return "NO"

for test in range(1,ui()+1):
    N = ui()
    A = list(line()[0])

    if test == 526:
        print("_".join(v for v in A))

    print(solve())

file.close()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. Let's analyze the code and identify potential issues:

1. **Edge Cases for Small Cycles:**
   - Cycles with 3 or 4 vertices.
2. **Symmetry and Palindromes:**
   - Cycles with a symmetric color pattern.
3. **Single Color Cycles:**
   - Cycles with all edges of the same color.
4. **Mixed Color Cycles:**
   - Cycles with a mix of colors that might still form palindromic routes.
5. **Odd and Even Length Cycles:**
   - Cycles with an odd number of vertices and cycles with an even number of vertices.

Given these points, let's construct a test input that can expose the bug:

```plaintext
7
3
RRR
4
RRRR
5
RRRRR
6
RBRBRB
7
RRBRRBR
8
RRBRRBRR
9
RRBRBRRBR
10
RRBRRBRRBR
```

### Explanation:
1. **3 Vertices (RRR):**
   - All edges are the same color, so there is a palindrome route between any pair of vertices.
2. **4 Vertices (RRRR):**
   - All edges are the same color, so there is a palindrome route between any pair of vertices.
3. **5 Vertices (RRRRR):**
   - All edges are the same color, so there is a palindrome route between any pair of vertices.
4. **6 Vertices (RBRBRB):**
   - The cycle is symmetric, so there is a palindrome route between any pair of vertices.
5. **7 Vertices (RRBRRBR):**
   - The cycle is symmetric, so there is a palindrome route between any pair of vertices.
6. **8 Vertices (RRBRRBRR):**
   - The cycle is symmetric, so there is a palindrome route between any pair of vertices.
7. **9 Vertices (RRBRBRRBR):**
   - The cycle is symmetric, so there is a palindrome route between any pair of vertices.
8. **10 Vertices (RRBRRBRRBR):**
   - The cycle is symmetric, so there is a palindrome route between any pair of vertices.

This test input covers various cycle lengths and color patterns, including symmetric and single-color cycles, which should help expose any issues in the buggy code.

7
3
RRR
4
RRRR
5
RRRRR
6
RBRBRB
7
RRBRRBR
8
RRBRRBRR
9
RRBRBRRBR
10
RRBRRBRRBR


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Palindrome Everywhere
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a cycle with $n$ vertices numbered from $0$ to $n-1$. For each $0\le i\le n-1$, there is an undirected edge between vertex $i$ and vertex $((i+1)\bmod n)$ with the color $c_i$ ($c_i=\texttt{R}$ or $\texttt{B}$).

Determine whether the following condition holds for every pair of vertices $(i,j)$ ($0\le i<j\le n-1$):

  * There exists a palindrome route between vertex $i$ and vertex $j$. Note that the route may not be simple. Formally, there must exist a sequence $p=[p_0,p_1,p_2,\ldots,p_m]$ such that:     * $p_0=i$, $p_m=j$;      * For each $0\leq x\le m-1$, either $p_{x+1}=(p_x+1)\bmod n$ or $p_{x+1}=(p_{x}-1)\bmod n$;      * For each $0\le x\le y\le m-1$ satisfying $x+y=m-1$, the edge between $p_x$ and $p_{x+1}$ has the same color as the edge between $p_y$ and $p_{y+1}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer $n$ ($3\leq n\leq10^6$) — the number of vertices in the cycle.

The second line contains a string $c$ of length $n$ ($c_i=\texttt{R}$ or $\texttt{B}$) — the color of each edge.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print "YES" (without quotes) if there is a palindrome route between any pair of nodes, and "NO" (without quotes) 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:
7
5
RRRRR
5
RRRRB
5
RBBRB
6
RBRBRB
6
RRBBRB
5
RBRBR
12
RRBRRBRRBRRB
output:
YES
YES
YES
NO
NO
YES
NO

Note: In the first test case, it is easy to show that there is a palindrome route between any two vertices.

In the second test case, for any two vertices, there exists a palindrome route with only red edges.

In the third test case, the cycle is as follows: $0\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}3\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}0$. Take $(i,j)=(0,3)$ as an example, then $0\color{red}{\overset{\texttt{R}}{\longrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longrightarrow}}3\color{red}{\overset{\texttt{R}}{\longrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longrightarrow}}0\color{blue}{\overset{\texttt{B}}{\longrightarrow}}4\color{red}{\overset{\texttt{R}}{\longrightarrow}}3$ is a palindrome route. Thus, the condition holds for $(i,j)=(0,3)$.

In the fourth test case, when $(i,j)=(0,2)$, there does not exist a palindrome route.

Buggy Code:
import sys
import os
from io import BytesIO, IOBase

# region fastio
BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

if sys.version_info[0] < 3:
    sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

file = sys.stdin
if os.environ.get('USER') == "loic":
    file = open("data.in")

line = lambda: file.readline().split()
ui = lambda: int(line()[0])
mi = lambda: map(int,line())
li = lambda: list(mi())
lsi = lambda: list(map(int,line()[0]))
#######################################################################

def solve():

    r = A.count('R')
    b = A.count('B')

    if r <= 1 or b <= 1:
        return "YES"

    if N % 2 == 1:
        alt = True
        for i in range(1,N):
            if A[i] == A[i-1]:
                alt = False
                break
        if alt:
            return "YES"

        if r == 2 or b == 2:
            val = 'R' if r == 2 else 'B'
            f = A.index(val)
            s = A.index(val, f + 1, N)
            if abs(s - f - (f + N - s)) == 1:
                return "YES"

    return "NO"

for test in range(1,ui()+1):
    N = ui()
    A = list(line()[0])

    if test == 126:
        print("_".join(v for v in A))

    print(solve())

file.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
RRRRR
5
RBBRB
6
RRBBRB
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Palindrome Everywhere
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a cycle with $n$ vertices numbered from $0$ to $n-1$. For each $0\le i\le n-1$, there is an undirected edge between vertex $i$ and vertex $((i+1)\bmod n)$ with the color $c_i$ ($c_i=\texttt{R}$ or $\texttt{B}$).

Determine whether the following condition holds for every pair of vertices $(i,j)$ ($0\le i<j\le n-1$):

  * There exists a palindrome route between vertex $i$ and vertex $j$. Note that the route may not be simple. Formally, there must exist a sequence $p=[p_0,p_1,p_2,\ldots,p_m]$ such that:     * $p_0=i$, $p_m=j$;      * For each $0\leq x\le m-1$, either $p_{x+1}=(p_x+1)\bmod n$ or $p_{x+1}=(p_{x}-1)\bmod n$;      * For each $0\le x\le y\le m-1$ satisfying $x+y=m-1$, the edge between $p_x$ and $p_{x+1}$ has the same color as the edge between $p_y$ and $p_{y+1}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer $n$ ($3\leq n\leq10^6$) — the number of vertices in the cycle.

The second line contains a string $c$ of length $n$ ($c_i=\texttt{R}$ or $\texttt{B}$) — the color of each edge.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print "YES" (without quotes) if there is a palindrome route between any pair of nodes, and "NO" (without quotes) 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:
7
5
RRRRR
5
RRRRB
5
RBBRB
6
RBRBRB
6
RRBBRB
5
RBRBR
12
RRBRRBRRBRRB
output:
YES
YES
YES
NO
NO
YES
NO

Note: In the first test case, it is easy to show that there is a palindrome route between any two vertices.

In the second test case, for any two vertices, there exists a palindrome route with only red edges.

In the third test case, the cycle is as follows: $0\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}3\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}0$. Take $(i,j)=(0,3)$ as an example, then $0\color{red}{\overset{\texttt{R}}{\longrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longrightarrow}}3\color{red}{\overset{\texttt{R}}{\longrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longrightarrow}}0\color{blue}{\overset{\texttt{B}}{\longrightarrow}}4\color{red}{\overset{\texttt{R}}{\longrightarrow}}3$ is a palindrome route. Thus, the condition holds for $(i,j)=(0,3)$.

In the fourth test case, when $(i,j)=(0,2)$, there does not exist a palindrome route.

Buggy Code:
//make sure to make new file!
import java.io.*;
import java.util.*;

public class F985{

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

      int t = Integer.parseInt(f.readLine());

      for(int q = 1; q <= t; q++){

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

         char[] array = f.readLine().toCharArray();
         //check all equal
         boolean allequal = true;
         int rfreq = 0;
         int bfreq = 0;
         for(int k = 1; k < n; k++){
            if(array[k] != array[k-1]){
               allequal = false;
            }
            if(array[k] == 'R') rfreq++;
            else bfreq++;
         }

         if(allequal || rfreq == 1 || bfreq == 1){
            out.println("Yes");
            continue;
         }

         //only way is if there are equal 4 groupings
         //one color only has groups of 1, other color has 1 even group and 1 odd group
         int[] psize = new int[n];
         psize[0] = 1;
         ArrayList<Integer> alist = new ArrayList<Integer>();
         for(int k = 1; k < n; k++){
            if(array[k] == array[k-1]) psize[k] = psize[k-1] + 1;
            else{
               psize[k] = 1;
               alist.add(psize[k-1]);
            }
         }

         if(array[0] == array[n-1]) alist.set(0,alist.get(0) + psize[n-1]);
         else alist.add(psize[n-1]);

         if(check(alist)){
            out.println("Yes");
         } else {
            out.println("No");
         }

      }

      out.close();
   }

   public static boolean check(ArrayList<Integer> alist){
      if(alist.size() != 4) return false;
      boolean b1 = alist.get(0) == 1 && alist.get(2) == 1 && ((alist.get(1)%2) != (alist.get(3)%2));
      boolean b2 = alist.get(1) == 1 && alist.get(3) == 1 && ((alist.get(0)%2) != (alist.get(2)%2));
      return b1 || b2;
   }

}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the given buggy code, we need to focus on edge cases where the code might fail. The buggy code has a specific check for the number of groups and their sizes, which might not cover all valid scenarios. Here are the steps to think through the problem:

1. **Check for All Equal Edges:**
   - The code correctly handles cases where all edges are the same color. We don't need to focus on this.

2. **Check for Single Different Edge:**
   - The code also handles cases where there is exactly one edge of a different color. This is correctly handled as well.

3. **Check for Groupings:**
   - The code has a specific check for exactly 4 groupings where one color has groups of 1, and the other color has one even and one odd group. This is a very specific condition and might not cover all valid scenarios.

4. **Edge Cases:**
   - We need to test cases where the number of groupings is more than 4 but still forms a valid palindrome route.
   - We also need to test cases where the groupings do not strictly follow the 4-group condition but still form valid palindrome routes.

Given these considerations, a good test input to expose the bug would be a cycle with more than 4 groupings that still forms valid palindrome routes. Here is a small-scale test input:

```plaintext
1
10
RBRBRBRBRB
```

### Explanation:
- **Number of Test Cases:** 1
- **Number of Vertices:** 10
- **Edge Colors:** RBRBRBRBRB

### Why This Input?
- This input has 10 vertices, and the edge colors form a repeating pattern of "RBRBRBRBRB".
- This pattern has 5 groups of 1 (R, B, R, B, R, B, R, B, R, B).
- According to the problem, this should be a valid palindrome route because any pair of vertices can be connected through a sequence that maintains the palindrome condition.
- However, the buggy code might fail to recognize this because it specifically checks for exactly 4 groupings, and this input has 10 groups.

This input should expose the bug in the code's logic for checking the number of groupings.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Palindrome Everywhere
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a cycle with $n$ vertices numbered from $0$ to $n-1$. For each $0\le i\le n-1$, there is an undirected edge between vertex $i$ and vertex $((i+1)\bmod n)$ with the color $c_i$ ($c_i=\texttt{R}$ or $\texttt{B}$).

Determine whether the following condition holds for every pair of vertices $(i,j)$ ($0\le i<j\le n-1$):

  * There exists a palindrome route between vertex $i$ and vertex $j$. Note that the route may not be simple. Formally, there must exist a sequence $p=[p_0,p_1,p_2,\ldots,p_m]$ such that:     * $p_0=i$, $p_m=j$;      * For each $0\leq x\le m-1$, either $p_{x+1}=(p_x+1)\bmod n$ or $p_{x+1}=(p_{x}-1)\bmod n$;      * For each $0\le x\le y\le m-1$ satisfying $x+y=m-1$, the edge between $p_x$ and $p_{x+1}$ has the same color as the edge between $p_y$ and $p_{y+1}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer $n$ ($3\leq n\leq10^6$) — the number of vertices in the cycle.

The second line contains a string $c$ of length $n$ ($c_i=\texttt{R}$ or $\texttt{B}$) — the color of each edge.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print "YES" (without quotes) if there is a palindrome route between any pair of nodes, and "NO" (without quotes) 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:
7
5
RRRRR
5
RRRRB
5
RBBRB
6
RBRBRB
6
RRBBRB
5
RBRBR
12
RRBRRBRRBRRB
output:
YES
YES
YES
NO
NO
YES
NO

Note: In the first test case, it is easy to show that there is a palindrome route between any two vertices.

In the second test case, for any two vertices, there exists a palindrome route with only red edges.

In the third test case, the cycle is as follows: $0\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}3\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}0$. Take $(i,j)=(0,3)$ as an example, then $0\color{red}{\overset{\texttt{R}}{\longrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longrightarrow}}3\color{red}{\overset{\texttt{R}}{\longrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longrightarrow}}0\color{blue}{\overset{\texttt{B}}{\longrightarrow}}4\color{red}{\overset{\texttt{R}}{\longrightarrow}}3$ is a palindrome route. Thus, the condition holds for $(i,j)=(0,3)$.

In the fourth test case, when $(i,j)=(0,2)$, there does not exist a palindrome route.

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

public class CF2029E extends PrintWriter {
	CF2029E() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF2029E o = new CF2029E(); o.main(); o.flush();
	}

	int[] ds;

	int find(int i) {
		return ds[i] < 0 ? i : (ds[i] = find(ds[i]));
	}

	void join(int i, int j) {
		i = find(i);
		j = find(j);
		if (i == j)
			return;
		if (ds[i] > ds[j])
			ds[i] = j;
		else {
			if (ds[i] == ds[j])
				ds[i]--;
			ds[j] = i;
		}
	}

	void main() {
		for (int t = sc.nextInt(); t-- > 0; ) {
			int n = sc.nextInt();
			byte[] bb = sc.next().getBytes();
			ds = new int[n * 2];
			Arrays.fill(ds, -1);
			for (int i = 0, j; i < n; i++) {
				j = (i + 1) % n;
				join(i << 1, j << 1 | 1);
				join(i << 1 | 1, j << 1);
				if (bb[i] == bb[(i + 1) % n]) {
					j = (i + 2) % n;
					join(i << 1, j << 1);
					join(i << 1 | 1, j << 1 | 1);
				}
			}
			int cnt = 0;
			for (int i = 0; i < n * 2; i++)
				if (ds[i] < 0)
					cnt++;
			println(cnt == 1 ? "YES" : "NO");
		}
	}
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on scenarios that might cause the algorithm to fail. The primary issue in the code seems to be in the union-find (disjoint set) operations and how it handles the palindrome condition. Specifically, the code might not correctly handle cases where the cycle has a specific pattern of colors that should theoretically allow for palindrome routes but are not recognized by the algorithm.

Let's consider a few edge cases:

1. **Uniform Color Cycle**: A cycle with all edges of the same color (e.g., all red or all blue) should always have a palindrome route between any two vertices.
2. **Alternating Color Cycle**: A cycle with alternating colors (e.g., RBRBRB) should also have a palindrome route between any two vertices.
3. **Mixed Color Cycle**: A cycle with a mix of colors that should theoretically allow for palindrome routes but might be misinterpreted by the algorithm.

Given these considerations, let's generate a small-scale test input that includes a mixed color cycle:

```plaintext
3
6
RBRBRB
6
RRBBRR
6
RBRRBR
```

Explanation:
- The first test case (`RBRBRB`) should be `YES` because it is a well-known alternating pattern that always allows for palindrome routes.
- The second test case (`RRBBRR`) should be `NO` because it does not allow for a palindrome route between certain vertices.
- The third test case (`RBRRBR`) should be `YES` because it is a mixed pattern that should theoretically allow for palindrome routes but might expose the bug if the algorithm is not correctly handling the union-find operations.

This input should help identify if the algorithm correctly handles different color patterns and their implications for palindrome routes.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Palindrome Everywhere
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a cycle with $n$ vertices numbered from $0$ to $n-1$. For each $0\le i\le n-1$, there is an undirected edge between vertex $i$ and vertex $((i+1)\bmod n)$ with the color $c_i$ ($c_i=\texttt{R}$ or $\texttt{B}$).

Determine whether the following condition holds for every pair of vertices $(i,j)$ ($0\le i<j\le n-1$):

  * There exists a palindrome route between vertex $i$ and vertex $j$. Note that the route may not be simple. Formally, there must exist a sequence $p=[p_0,p_1,p_2,\ldots,p_m]$ such that:     * $p_0=i$, $p_m=j$;      * For each $0\leq x\le m-1$, either $p_{x+1}=(p_x+1)\bmod n$ or $p_{x+1}=(p_{x}-1)\bmod n$;      * For each $0\le x\le y\le m-1$ satisfying $x+y=m-1$, the edge between $p_x$ and $p_{x+1}$ has the same color as the edge between $p_y$ and $p_{y+1}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer $n$ ($3\leq n\leq10^6$) — the number of vertices in the cycle.

The second line contains a string $c$ of length $n$ ($c_i=\texttt{R}$ or $\texttt{B}$) — the color of each edge.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print "YES" (without quotes) if there is a palindrome route between any pair of nodes, and "NO" (without quotes) 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:
7
5
RRRRR
5
RRRRB
5
RBBRB
6
RBRBRB
6
RRBBRB
5
RBRBR
12
RRBRRBRRBRRB
output:
YES
YES
YES
NO
NO
YES
NO

Note: In the first test case, it is easy to show that there is a palindrome route between any two vertices.

In the second test case, for any two vertices, there exists a palindrome route with only red edges.

In the third test case, the cycle is as follows: $0\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}3\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}0$. Take $(i,j)=(0,3)$ as an example, then $0\color{red}{\overset{\texttt{R}}{\longrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longrightarrow}}3\color{red}{\overset{\texttt{R}}{\longrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longrightarrow}}0\color{blue}{\overset{\texttt{B}}{\longrightarrow}}4\color{red}{\overset{\texttt{R}}{\longrightarrow}}3$ is a palindrome route. Thus, the condition holds for $(i,j)=(0,3)$.

In the fourth test case, when $(i,j)=(0,2)$, there does not exist a palindrome route.

Buggy Code:
//~ while (clock()<=69*CLOCKS_PER_SEC)
//~ #pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("O3")
//~ #pragma GCC target ("avx2")
//~ #pragma GCC optimize("Ofast")
//~ #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//~ #pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;

template <typename T>
using ordered_set =
    tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

#define sim template < class c
#define ris return * this
#define dor > debug & operator <<
#define eni(x) sim > typename \
  enable_if<sizeof dud<c>(0) x 1, debug&>::type operator<<(c i) {
sim > struct rge { c b, e; };
sim > rge<c> range(c i, c j) { return rge<c>{i, j}; }
sim > auto dud(c* x) -> decltype(cerr << *x, 0);
sim > char dud(...);
struct debug {
#ifdef LOCAL
~debug() { cerr << endl; }
eni(!=) cerr << boolalpha << i; ris; }
eni(==) ris << range(begin(i), end(i)); }
sim, class b dor(pair < b, c > d) {
  ris << "(" << d.first << ", " << d.second << ")";
}
sim dor(rge<c> d) {
  *this << "[";
  for (auto it = d.b; it != d.e; ++it)
    *this << ", " + 2 * (it == d.b) << *it;
  ris << "]";
}
#else
sim dor(const c&) { ris; }
#endif
};
#define imie(...) " [" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "] "

#define shandom_ruffle random_shuffle

using ll=long long;
using pii=pair<int,int>;
using pll=pair<ll,ll>;
using vi=vector<int>;
using vll=vector<ll>;
const int nax=400*1007;

int n;
char wcz[nax];
int tab[nax];

void ans(int v)
{
	if (v)
		printf("YES\n");
	else
		printf("NO\n");
}

vi czysc(vi wek)
{
	vi pom;
	for (int i : wek)
	{
		pom.push_back(i);
		int r=pom.size();
		while(r>=3 && pom[r-1]==pom[r-2] && pom[r-1]==pom[r-3])
		{
			r-=2;
			pom.pop_back();
			pom.pop_back();
		}
	}
	int r=pom.size();
	while(r>=3 && pom[0]==pom[r-1] && (pom[1]==pom[0] || pom[r-2]==pom[r-1]))
	{
		pom.erase(pom.begin());
		pom.pop_back();
	}
	return pom;
}

void test()
{
	scanf("%d", &n);
	scanf("%s", wcz);
	for (int i=0; i<n; i++)
		tab[i]=(wcz[i]=='R');
	vi wek;
	for (int i=0; i<n; i++)
		wek.push_back(tab[i]);
	wek=czysc(wek);
	//~ debug() << wek;
	int ile=0;
	int r=wek.size();
	for (int i=0; i<r; i++)
		if (wek[i]==wek[(i+1)%r])
			ile++;
	if (ile>=2 || !ile)
	{
		ans(0);
		return;
	}
	ans(1);
}

int main()
{
	int t;
	scanf("%d", &t);
	while(t--)
		test();
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases or scenarios that are likely to cause the algorithm to fail. The key points to consider are:

1. **Edge Cases for Small `n`**: Test with the minimum value of `n` (i.e., 3) to ensure the algorithm handles small cycles correctly.
2. **Symmetry and Palindromic Properties**: Test with patterns that are close to being palindromic but have a small discrepancy.
3. **Repeating Patterns**: Test with repeating patterns that might cause the algorithm to incorrectly simplify the sequence.
4. **Single Color Cycles**: Test with cycles that are entirely one color to ensure the algorithm handles these cases correctly.
5. **Mixed Patterns**: Test with mixed patterns that are not palindromic but might be incorrectly identified as such.

Given the constraints and the need for a small-scale test input, let's construct a test input that includes a mix of these cases.

```plaintext
3
3
RRR
4
RBRB
6
RBRBRB
```

### Explanation:

1. **Test Case 1 (3 vertices, all red)**
   - `n = 3`
   - `c = "RRR"`
   - This is a simple case where all edges are the same color. The algorithm should correctly identify that there is a palindrome route between any pair of vertices.

2. **Test Case 2 (4 vertices, alternating colors)**
   - `n = 4`
   - `c = "RBRB"`
   - This is a case where the colors alternate. The algorithm should correctly identify that there is no palindrome route between certain pairs of vertices, such as (0, 2).

3. **Test Case 3 (6 vertices, alternating colors)**
   - `n = 6`
   - `c = "RBRBRB"`
   - This is a case where the colors alternate. The algorithm should correctly identify that there is no palindrome route between certain pairs of vertices, such as (0, 2).

These test cases are designed to cover a range of scenarios, including small cycles, repeating patterns, and alternating patterns, to ensure the algorithm is robust and handles these cases correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Palindrome Everywhere
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a cycle with $n$ vertices numbered from $0$ to $n-1$. For each $0\le i\le n-1$, there is an undirected edge between vertex $i$ and vertex $((i+1)\bmod n)$ with the color $c_i$ ($c_i=\texttt{R}$ or $\texttt{B}$).

Determine whether the following condition holds for every pair of vertices $(i,j)$ ($0\le i<j\le n-1$):

  * There exists a palindrome route between vertex $i$ and vertex $j$. Note that the route may not be simple. Formally, there must exist a sequence $p=[p_0,p_1,p_2,\ldots,p_m]$ such that:     * $p_0=i$, $p_m=j$;      * For each $0\leq x\le m-1$, either $p_{x+1}=(p_x+1)\bmod n$ or $p_{x+1}=(p_{x}-1)\bmod n$;      * For each $0\le x\le y\le m-1$ satisfying $x+y=m-1$, the edge between $p_x$ and $p_{x+1}$ has the same color as the edge between $p_y$ and $p_{y+1}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer $n$ ($3\leq n\leq10^6$) — the number of vertices in the cycle.

The second line contains a string $c$ of length $n$ ($c_i=\texttt{R}$ or $\texttt{B}$) — the color of each edge.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print "YES" (without quotes) if there is a palindrome route between any pair of nodes, and "NO" (without quotes) 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:
7
5
RRRRR
5
RRRRB
5
RBBRB
6
RBRBRB
6
RRBBRB
5
RBRBR
12
RRBRRBRRBRRB
output:
YES
YES
YES
NO
NO
YES
NO

Note: In the first test case, it is easy to show that there is a palindrome route between any two vertices.

In the second test case, for any two vertices, there exists a palindrome route with only red edges.

In the third test case, the cycle is as follows: $0\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}3\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}0$. Take $(i,j)=(0,3)$ as an example, then $0\color{red}{\overset{\texttt{R}}{\longrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longrightarrow}}3\color{red}{\overset{\texttt{R}}{\longrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longrightarrow}}0\color{blue}{\overset{\texttt{B}}{\longrightarrow}}4\color{red}{\overset{\texttt{R}}{\longrightarrow}}3$ is a palindrome route. Thus, the condition holds for $(i,j)=(0,3)$.

In the fourth test case, when $(i,j)=(0,2)$, there does not exist a palindrome route.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int kmaxn=1e5+10;
int T,n;
string a;
signed main(){
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin>>T;
	while(T--){
		cin>>n>>a;
		if(n==2){
			cout<<"Yes\n";
			continue;
		}
		int A[2]={0,0},B[2]={0,0};
		for(int i=0;i<n;i++){
			int r=(i+1)%n;
			if(a[i]=='R')B[0]=1;
			else B[1]=1;
			if(a[i]==a[r]){
				if(a[i]=='R')A[0]=1;
				else A[1]=1;
			}
		}
		cout<<(((A[0]&&A[1])||n%2==0)&&(B[0]+B[1]>1)?"No\n":"Yes\n");
	}
	return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to create a test input that will cause the code to produce an incorrect output. The code seems to have a logic issue in determining whether a palindrome route exists between any two vertices.

### Analysis of the Buggy Code:
1. **Edge Cases:**
   - The code incorrectly handles the case where `n == 2`. It outputs "Yes" for any input with `n == 2`, which is incorrect because a cycle with 2 vertices does not form a valid cycle as per the problem's constraints.
   - The code uses arrays `A` and `B` to track the presence of certain conditions, but the logic for determining the final output is flawed.

2. **Key Conditions:**
   - The code checks if both 'R' and 'B' colors are present in the cycle (`B[0]+B[1]>1`).
   - It also checks if there are edges of both colors that are consecutive (`A[0]&&A[1]`).
   - Finally, it checks if the number of vertices is even (`n%2==0`).

### Creating a Test Case:
To create a test case that exposes the bug, we need a cycle that:
- Has both 'R' and 'B' colors.
- Does not satisfy the condition where there are edges of both colors that are consecutive.
- Has an even number of vertices.

### Test Case:
```plaintext
1
4
RBRB
```

### Explanation:
- **Input:**
  - `t = 1` (one test case)
  - `n = 4` (four vertices)
  - `c = "RBRB"` (colors of the edges)

- **Expected Output:**
  - "NO" (because there is no palindrome route between certain pairs of vertices, e.g., (0, 2))

### Why This Test Case Exposes the Bug:
- The cycle "RBRB" has both 'R' and 'B' colors.
- There are no consecutive edges of the same color, so `A[0]` and `A[1]` are both 0.
- The number of vertices is even (`n = 4`).

The buggy code will incorrectly output "Yes" because it does not correctly handle the condition where there are no consecutive edges of the same color. The correct output should be "NO" because there is no palindrome route between certain pairs of vertices.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Palindrome Everywhere
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a cycle with $n$ vertices numbered from $0$ to $n-1$. For each $0\le i\le n-1$, there is an undirected edge between vertex $i$ and vertex $((i+1)\bmod n)$ with the color $c_i$ ($c_i=\texttt{R}$ or $\texttt{B}$).

Determine whether the following condition holds for every pair of vertices $(i,j)$ ($0\le i<j\le n-1$):

  * There exists a palindrome route between vertex $i$ and vertex $j$. Note that the route may not be simple. Formally, there must exist a sequence $p=[p_0,p_1,p_2,\ldots,p_m]$ such that:     * $p_0=i$, $p_m=j$;      * For each $0\leq x\le m-1$, either $p_{x+1}=(p_x+1)\bmod n$ or $p_{x+1}=(p_{x}-1)\bmod n$;      * For each $0\le x\le y\le m-1$ satisfying $x+y=m-1$, the edge between $p_x$ and $p_{x+1}$ has the same color as the edge between $p_y$ and $p_{y+1}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer $n$ ($3\leq n\leq10^6$) — the number of vertices in the cycle.

The second line contains a string $c$ of length $n$ ($c_i=\texttt{R}$ or $\texttt{B}$) — the color of each edge.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print "YES" (without quotes) if there is a palindrome route between any pair of nodes, and "NO" (without quotes) 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:
7
5
RRRRR
5
RRRRB
5
RBBRB
6
RBRBRB
6
RRBBRB
5
RBRBR
12
RRBRRBRRBRRB
output:
YES
YES
YES
NO
NO
YES
NO

Note: In the first test case, it is easy to show that there is a palindrome route between any two vertices.

In the second test case, for any two vertices, there exists a palindrome route with only red edges.

In the third test case, the cycle is as follows: $0\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}3\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}0$. Take $(i,j)=(0,3)$ as an example, then $0\color{red}{\overset{\texttt{R}}{\longrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longrightarrow}}3\color{red}{\overset{\texttt{R}}{\longrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longrightarrow}}0\color{blue}{\overset{\texttt{B}}{\longrightarrow}}4\color{red}{\overset{\texttt{R}}{\longrightarrow}}3$ is a palindrome route. Thus, the condition holds for $(i,j)=(0,3)$.

In the fourth test case, when $(i,j)=(0,2)$, there does not exist a palindrome route.

Buggy Code:
import re
import sys

input = lambda: sys.stdin.readline().strip()
printerr = lambda *args, **kwargs: print("\u001B[31m", *args, "\u001B[0m", file=sys.stderr, **kwargs)

def main():
    n = int(input())
    s = input()
    r = s.count('R')
    b = s.count('B')
    if r<=1 or b<=1:
        print('YES')
        return
    ri = s.find('R')
    bi = s.find('B')
    rc = sum(len(m) % 2 == 0 for m in re.findall('R+', s[bi:] + s[:bi]))
    bc = sum(len(m) % 2 == 0 for m in re.findall('B+', s[ri:] + s[:ri]))
    print(['NO','YES']['BB' not in s and rc==1 or 'RR' not in s and bc==1])

tcs = int(input())
for tc in range(tcs):
    main()

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

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

Think step by step.

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

1. **Edge Cases:**
   - When the cycle is very small (e.g., `n = 3`).
   - When the cycle is very large (e.g., `n = 10^6`).
   - When the cycle has only one type of color (e.g., all `R` or all `B`).
   - When the cycle has alternating colors (e.g., `RBRBRB`).

2. **Specific Patterns:**
   - Patterns where the cycle has a single pair of adjacent edges of the same color.
   - Patterns where the cycle has multiple pairs of adjacent edges of the same color.

3. **Test Case Structure:**
   - Each test case should be small enough to be manageable and valid.

Given these considerations, let's construct a test input that includes various edge cases and specific patterns:

```plaintext
5
3
RRR
3
RBR
4
RBBR
6
RBRBRB
10
RRBBRRBBRR
```

### Explanation of Test Cases:
1. **Test Case 1:**
   - `n = 3`
   - `c = "RRR"`
   - This case should return `YES` because all edges are the same color, and any route is a palindrome.

2. **Test Case 2:**
   - `n = 3`
   - `c = "RBR"`
   - This case should return `YES` because the cycle has alternating colors, and any route is a palindrome.

3. **Test Case 3:**
   - `n = 4`
   - `c = "RBBR"`
   - This case should return `NO` because there is no palindrome route between some pairs of vertices (e.g., from 0 to 2).

4. **Test Case 4:**
   - `n = 6`
   - `c = "RBRBRB"`
   - This case should return `NO` because there is no palindrome route between some pairs of vertices (e.g., from 0 to 3).

5. **Test Case 5:**
   - `n = 10`
   - `c = "RRBBRRBBRR"`
   - This case should return `NO` because there is no palindrome route between some pairs of vertices (e.g., from 0 to 5).

This test input covers a variety of edge cases and specific patterns, which should help in exposing the bug in the buggy code.

5
3
RRR
3
RBR
4
RBBR
6
RBRBRB
10
RRBBRRBBRR


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Palindrome Everywhere
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a cycle with $n$ vertices numbered from $0$ to $n-1$. For each $0\le i\le n-1$, there is an undirected edge between vertex $i$ and vertex $((i+1)\bmod n)$ with the color $c_i$ ($c_i=\texttt{R}$ or $\texttt{B}$).

Determine whether the following condition holds for every pair of vertices $(i,j)$ ($0\le i<j\le n-1$):

  * There exists a palindrome route between vertex $i$ and vertex $j$. Note that the route may not be simple. Formally, there must exist a sequence $p=[p_0,p_1,p_2,\ldots,p_m]$ such that:     * $p_0=i$, $p_m=j$;      * For each $0\leq x\le m-1$, either $p_{x+1}=(p_x+1)\bmod n$ or $p_{x+1}=(p_{x}-1)\bmod n$;      * For each $0\le x\le y\le m-1$ satisfying $x+y=m-1$, the edge between $p_x$ and $p_{x+1}$ has the same color as the edge between $p_y$ and $p_{y+1}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer $n$ ($3\leq n\leq10^6$) — the number of vertices in the cycle.

The second line contains a string $c$ of length $n$ ($c_i=\texttt{R}$ or $\texttt{B}$) — the color of each edge.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print "YES" (without quotes) if there is a palindrome route between any pair of nodes, and "NO" (without quotes) 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:
7
5
RRRRR
5
RRRRB
5
RBBRB
6
RBRBRB
6
RRBBRB
5
RBRBR
12
RRBRRBRRBRRB
output:
YES
YES
YES
NO
NO
YES
NO

Note: In the first test case, it is easy to show that there is a palindrome route between any two vertices.

In the second test case, for any two vertices, there exists a palindrome route with only red edges.

In the third test case, the cycle is as follows: $0\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}3\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}0$. Take $(i,j)=(0,3)$ as an example, then $0\color{red}{\overset{\texttt{R}}{\longrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longrightarrow}}3\color{red}{\overset{\texttt{R}}{\longrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longrightarrow}}0\color{blue}{\overset{\texttt{B}}{\longrightarrow}}4\color{red}{\overset{\texttt{R}}{\longrightarrow}}3$ is a palindrome route. Thus, the condition holds for $(i,j)=(0,3)$.

In the fourth test case, when $(i,j)=(0,2)$, there does not exist a palindrome route.

Buggy Code:
import sys
input=sys.stdin.readline
Q=int(input())
for _ in range(Q):
  N=int(input())
  S=input()[:-1]
  x=0
  y=0
  for i in range(1,2*N):
    a=S[i%N]
    b=S[(i-1)%N]
    if a==b:
      if a=='R':
        x=1
      else:
        y=1
  if (x==1 and y==1) or (x==0 and y==0):
    print('NO')
    continue
  L=[]
  if x==1:
    p='B'
  else:
    p='R'
  pos=-1
  for i in range(N):
    if S[i]==p:
      pos=i
      break
  if pos==-1:
    print('YES')
    continue
  L=[]
  x=0
  for i in range(pos+1,pos+N):
    if S[i%N]==S[(i-1)%N]:
      x+=1
    else:
      if S[i%N]!=p:
        x=1
      else:
        L.append(x)
        x=0
  count=0
  if x>0:
    L.append(x)
  for y in L:
    if y%2==0:
      count+=1
  if count<=1:
    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
6
RBRBRB
```

**Explanation:**
The provided test input is designed to expose the bug in the given buggy code. The input consists of a single test case with 6 vertices and the edge colors `RBRBRB`.

The cycle in this case is as follows:
- Vertex 0 to Vertex 1: R
- Vertex 1 to Vertex 2: B
- Vertex 2 to Vertex 3: R
- Vertex 3 to Vertex 4: B
- Vertex 4 to Vertex 5: R
- Vertex 5 to Vertex 0: B

This input is particularly challenging because it forms a cycle where the colors alternate between R and B, which should theoretically allow for palindrome routes between any two vertices. However, the buggy code may fail to correctly handle this alternating pattern, especially when it tries to find a specific color (R or B) and then checks for the number of consecutive edges of the same color.

The correct output for this test case should be `NO` because there is no palindrome route between any two vertices in this cycle. The buggy code, however, may incorrectly output `YES` due to its flawed logic in handling the alternating colors.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Palindrome Everywhere
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a cycle with $n$ vertices numbered from $0$ to $n-1$. For each $0\le i\le n-1$, there is an undirected edge between vertex $i$ and vertex $((i+1)\bmod n)$ with the color $c_i$ ($c_i=\texttt{R}$ or $\texttt{B}$).

Determine whether the following condition holds for every pair of vertices $(i,j)$ ($0\le i<j\le n-1$):

  * There exists a palindrome route between vertex $i$ and vertex $j$. Note that the route may not be simple. Formally, there must exist a sequence $p=[p_0,p_1,p_2,\ldots,p_m]$ such that:     * $p_0=i$, $p_m=j$;      * For each $0\leq x\le m-1$, either $p_{x+1}=(p_x+1)\bmod n$ or $p_{x+1}=(p_{x}-1)\bmod n$;      * For each $0\le x\le y\le m-1$ satisfying $x+y=m-1$, the edge between $p_x$ and $p_{x+1}$ has the same color as the edge between $p_y$ and $p_{y+1}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer $n$ ($3\leq n\leq10^6$) — the number of vertices in the cycle.

The second line contains a string $c$ of length $n$ ($c_i=\texttt{R}$ or $\texttt{B}$) — the color of each edge.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print "YES" (without quotes) if there is a palindrome route between any pair of nodes, and "NO" (without quotes) 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:
7
5
RRRRR
5
RRRRB
5
RBBRB
6
RBRBRB
6
RRBBRB
5
RBRBR
12
RRBRRBRRBRRB
output:
YES
YES
YES
NO
NO
YES
NO

Note: In the first test case, it is easy to show that there is a palindrome route between any two vertices.

In the second test case, for any two vertices, there exists a palindrome route with only red edges.

In the third test case, the cycle is as follows: $0\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}3\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}0$. Take $(i,j)=(0,3)$ as an example, then $0\color{red}{\overset{\texttt{R}}{\longrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longrightarrow}}3\color{red}{\overset{\texttt{R}}{\longrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longrightarrow}}0\color{blue}{\overset{\texttt{B}}{\longrightarrow}}4\color{red}{\overset{\texttt{R}}{\longrightarrow}}3$ is a palindrome route. Thus, the condition holds for $(i,j)=(0,3)$.

In the fourth test case, when $(i,j)=(0,2)$, there does not exist a palindrome route.

Buggy Code:
import os
import math
import sys
from io import BytesIO, IOBase

_str = str
str = lambda x=b"": x if type(x) is bytes else _str(x).encode()

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

prime = [1]*1000
for i in range(2, 1000):
    if prime[i] == 0:
        continue
    j = i*2
    while j < 1000:
        prime[j] = 0
        j += i
primes = [j for j in range(2, 1000) if prime[j]]

tests = int(input())
for _ in range(tests):
    n = int(input())
    c = list(input())
    ans = 1
    found_b = 0
    found_r = 0
    for i in range(n-1):
        if c[i] == "R" and c[i+1] == "R":
            found_r = 1
        if c[i] == "B" and c[i+1] == "B":
            found_b = 1
    if c[n-1] == "R" and c[0] == "R":
        found_r = 1
    if c[n-1] == "B" and c[0] == "B":
        found_b = 1
    if found_b and found_r:
        print("NO")
        continue
    if found_b == 0 and found_r == 0:
        print("NO")
        continue
    if found_b:
        for i in range(n):
            if c[i] == "R":
                c[i] = "B"
            else:
                c[i] = "R"
    pivot = 0
    for i in range(n):
        if c[i] == "B":
            pivot = i
    c = c[pivot:]+c[:pivot]
    streak = 0
    streaks = []
    for i in range(1,n):
        if c[i] == "B":
            streaks.append(streak)
            streak = 0
        else:
            streak += 1
    streaks.append(streak)
    evens = 0
    for i in streaks:
        if i%2==0:
            evens+=1
    if evens != 1:
        print("NO")
    else:
        print("YES")

**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
7
5
RRRRR
5
RRRRB
5
RBBRB
6
RBRBRB
6
RRBBRB
5
RBRBR
12
RRBRRBRRBRRB
```

7
5
RRRRR
5
RRRRB
5
RBBRB
6
RBRBRB
6
RRBBRB
5
RBRBR
12
RRBRRBRRBRRB


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Palindrome Everywhere
time_limit_ms: 2000
memory_limit_mb: 512
Description: You are given a cycle with $n$ vertices numbered from $0$ to $n-1$. For each $0\le i\le n-1$, there is an undirected edge between vertex $i$ and vertex $((i+1)\bmod n)$ with the color $c_i$ ($c_i=\texttt{R}$ or $\texttt{B}$).

Determine whether the following condition holds for every pair of vertices $(i,j)$ ($0\le i<j\le n-1$):

  * There exists a palindrome route between vertex $i$ and vertex $j$. Note that the route may not be simple. Formally, there must exist a sequence $p=[p_0,p_1,p_2,\ldots,p_m]$ such that:     * $p_0=i$, $p_m=j$;      * For each $0\leq x\le m-1$, either $p_{x+1}=(p_x+1)\bmod n$ or $p_{x+1}=(p_{x}-1)\bmod n$;      * For each $0\le x\le y\le m-1$ satisfying $x+y=m-1$, the edge between $p_x$ and $p_{x+1}$ has the same color as the edge between $p_y$ and $p_{y+1}$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer $n$ ($3\leq n\leq10^6$) — the number of vertices in the cycle.

The second line contains a string $c$ of length $n$ ($c_i=\texttt{R}$ or $\texttt{B}$) — the color of each edge.

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$.

Output: For each test case, print "YES" (without quotes) if there is a palindrome route between any pair of nodes, and "NO" (without quotes) 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:
7
5
RRRRR
5
RRRRB
5
RBBRB
6
RBRBRB
6
RRBBRB
5
RBRBR
12
RRBRRBRRBRRB
output:
YES
YES
YES
NO
NO
YES
NO

Note: In the first test case, it is easy to show that there is a palindrome route between any two vertices.

In the second test case, for any two vertices, there exists a palindrome route with only red edges.

In the third test case, the cycle is as follows: $0\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}3\color{red}{\overset{\texttt{R}}{\longleftrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longleftrightarrow}}0$. Take $(i,j)=(0,3)$ as an example, then $0\color{red}{\overset{\texttt{R}}{\longrightarrow}}1\color{blue}{\overset{\texttt{B}}{\longrightarrow}}2\color{blue}{\overset{\texttt{B}}{\longrightarrow}}3\color{red}{\overset{\texttt{R}}{\longrightarrow}}4\color{blue}{\overset{\texttt{B}}{\longrightarrow}}0\color{blue}{\overset{\texttt{B}}{\longrightarrow}}4\color{red}{\overset{\texttt{R}}{\longrightarrow}}3$ is a palindrome route. Thus, the condition holds for $(i,j)=(0,3)$.

In the fourth test case, when $(i,j)=(0,2)$, there does not exist a palindrome route.

Buggy Code:
import os
import math
import sys
from io import BytesIO, IOBase

_str = str
str = lambda x=b"": x if type(x) is bytes else _str(x).encode()

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

prime = [1]*1000
for i in range(2, 1000):
    if prime[i] == 0:
        continue
    j = i*2
    while j < 1000:
        prime[j] = 0
        j += i
primes = [j for j in range(2, 1000) if prime[j]]

tests = int(input())
for _ in range(tests):
    n = int(input())
    c = list(input())
    ans = 1
    found_b = 0
    found_r = 0
    for i in range(n-1):
        if c[i] == "R" and c[i+1] == "R":
            found_r = 1
        if c[i] == "B" and c[i+1] == "B":
            found_b = 1
    if c[n-1] == "R" and c[0] == "R":
        found_r = 1
    if c[n-1] == "B" and c[0] == "B":
        found_b = 1
    if found_b and found_r:
        print("NO")
        continue
    if found_b == 0 and found_r == 0:
        if n > 2:
            print("NO")
        else:
            print("YES")
        continue
    if found_b:
        for i in range(n):
            if c[i] == "R":
                c[i] = "B"
            else:
                c[i] = "R"
    pivot = 0
    for i in range(n-1,-1,-1):
        if c[i] == "B":
            pivot = i
    c = c[pivot:]+c[:pivot]
    streak = 0
    streaks = []
    for i in range(1,n):
        if c[i] == "B":
            streaks.append(streak)
            streak = 0
        else:
            streak += 1
    streaks.append(streak)
    evens = 0
    for i in streaks:
        if i%2==0:
            evens+=1
    if evens > 1:
        print("NO")
    else:
        print("YES")

**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 create a test case that challenges the logic used to determine if there is a palindrome route between any pair of vertices. The buggy code has a few key points to consider:

1. **Detection of Consecutive Colors**: The code checks for consecutive "R" or "B" colors to determine if both colors are present in the cycle. If both colors are found, it immediately prints "NO".
2. **Rotation and Streak Calculation**: The code rotates the cycle to align with a specific pivot and then calculates the lengths of streaks of "R" colors. It checks if there are more than one even-length streaks and prints "NO" if so.

To create a challenging test case, we need to construct a cycle where the logic of the buggy code fails to correctly identify the presence of palindrome routes. Here’s a step-by-step approach to generating such a test case:

1. **Create a Cycle with Consecutive Colors**: Ensure that there are consecutive "R" and "B" colors to trigger the initial check.
2. **Ensure the Cycle Has a Valid Palindrome Route**: Despite the presence of consecutive colors, ensure that the cycle still has valid palindrome routes for all pairs of vertices.

Let’s construct a small-scale test case that meets these criteria:

- **Test Case**:
  - **Number of vertices**: 6
  - **Edge colors**: "RBRBRB"

This test case has consecutive "R" and "B" colors, but it should still have valid palindrome routes for all pairs of vertices. The cycle "RBRBRB" is a valid cycle where each pair of vertices can be connected by a palindrome route.

Here is the test input:

```plaintext
1
6
RBRBRB
```

This input will challenge the buggy code to correctly determine the presence of palindrome routes, despite the presence of consecutive "R" and "B" colors.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Spinning Round (Easy Version)
time_limit_ms: 7000
memory_limit_mb: 1024
Description: This is the easy version of the problem. The only difference between the two versions are the allowed characters in $s$. In the easy version, $s$ only contains the character ?. You can make hacks only if both versions of the problem are solved.

You are given a permutation $p$ of length $n$. You are also given a string $s$ of length $n$, consisting only of the character ?.

For each $i$ from $1$ to $n$:

  * Define $l_i$ as the largest index $j < i$ such that $p_j > p_i$. If there is no such index, $l_i := i$.    * Define $r_i$ as the smallest index $j > i$ such that $p_j > p_i$. If there is no such index, $r_i := i$.

Initially, you have an undirected graph with $n$ vertices (numbered from $1$ to $n$) and no edges. Then, for each $i$ from $1$ to $n$, add one edge to the graph:

  * If $s_i =$ L, add the edge $(i, l_i)$ to the graph.    * If $s_i =$ R, add the edge $(i, r_i)$ to the graph.    * If $s_i =$ ?, either add the edge $(i, l_i)$ or the edge $(i, r_i)$ to the graph at your choice.

Find the maximum possible diameter$^{\text{∗}}$ over all connected graphs that you can form. Output $-1$ if it is not possible to form any connected graphs.

$^{\text{∗}}$ Let $d(s, t)$ denote the smallest number of edges on any path between $s$ and $t$.

The diameter of the graph is defined as the maximum value of $d(s, t)$ over all pairs of vertices $s$ and $t$.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 4 \cdot 10^5$) — the length of the permutation $p$.

The second line of each test case contains $n$ integers $p_1,p_2,\ldots, p_n$ ($1 \le p_i \le n$) — the elements of $p$, which are guaranteed to form a permutation.

The third line of each test case contains a string $s$ of length $n$. It is guaranteed that it consists only of the character ?.

It is guaranteed that the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, output the maximum possible diameter over all connected graphs that you form, or $-1$ if it is not possible to form any connected graphs.

Examples:
input:
8
5
2 1 4 3 5
?????
2
1 2
??
3
3 1 2
???
7
5 3 1 6 4 2 7
???????
5
5 2 1 3 4
?????
6
6 2 3 4 5 1
??????
8
1 7 5 6 2 8 4 3
????????
12
6 10 7 1 8 5 12 2 11 3 4 9
????????????
output:
4
1
2
6
4
5
5
8

Note: In the first test case, here are some possible connected graphs that you can form (the labels are indices):

![](CDN_BASE_URL/4561c9025a4357785c7aff0bb1bbff9a)| ![](CDN_BASE_URL/a680f9e3bffa38fd90dd7bbc4c3e5f3b)| ![](CDN_BASE_URL/bb06c0111d994ef8c2f7e8e1cb93e921)   ---|---|---      In the second test case, the only connected graph has a diameter of $1$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
typedef pair<int,int> pint;
const int N=4e5+114,K=20,Inf=0x3f3f3f3f3f3f3f3f;
int n,a[N],st[K][N],lg[N],fa[N],fa2[N],ans;
vector<int> G[N],G2[N];
int get(int l,int r){
	int k=lg[r-l+1];
	return a[st[k][l]]>a[st[k][r-(1<<k)+1]]?st[k][l]:st[k][r-(1<<k)+1];
}
void dfs1(int l,int r){
	if(l>r) return;
	int k=get(l,r);
//	cout << l << ' ' << r << ' ' << k << endl;
	if(l==1){
		if(r==n) fa[k]=fa2[k]=0;
		else fa[k]=r+1,fa2[k]=0;
	}
	else{
		if(r==n) fa[k]=l-1,fa2[k]=0;
		else fa[k]=(a[l-1]<a[r+1]?l-1:r+1),fa2[k]=(a[l-1]>a[r+1]?l-1:r+1);
	}
	dfs1(l,k-1);
	dfs1(k+1,r);
}

int f[N],pf[N],cf[N],pcf[N];
int fa_[K][N],g[K][N],dep[N];

int rt[N],dfn[N],tot;

void dfs2(int u){
	dep[u]=dep[fa[u]]+1;
	fa_[0][u]=fa[u];
	for(int i=1;i<K;++i) fa_[i][u]=fa_[i-1][fa_[i-1][u]];
	f[u]=0,pf[u]=0,cf[u]=0,pcf[u]=0;
	for(int v:G[u]){
		dfs2(v);
		if(f[v]+1>f[u]) cf[u]=f[u],pcf[u]=pf[u],f[u]=f[v]+1,pf[u]=v;
		else if(f[v]+1>cf[u]) cf[u]=f[v]+1,pcf[u]=v;
	}
}

void dfs3(int u,int rt_){
	g[0][u]=cf[u];
	for(int i=1;i<K;++i) g[i][u]=max(g[i-1][u]+(1<<(i-1)),g[i-1][fa_[i-1][u]]);
	rt[u]=rt_;
	dfn[u]=++tot;
	if(pf[u]) dfs3(pf[u],rt_);
	for(int v:G[u]){
		if(v!=pf[u]) dfs3(v,v);
	}
}

void dfs4(int u){
	ans=max(ans,f[u]+cf[u]+1);
	for(int v:G2[u]){
		if(dfn[u]==dfn[v]-dep[v]+dep[u]){
			int tu=fa[v],tmp=-Inf;
			for(int i=K-1;i>=0;--i){
				if(dep[tu]-(1<<i)>=dep[u]){
					tmp=max(tmp+(1<<i),g[i][tu]);
					tu=fa_[i][tu];
				}
			}
			tmp=max(tmp+1,cf[u]);
			ans=max(ans,tmp+f[v]+2);
		}
		else{
			ans=max(ans,f[u]+f[v]+2);
		}
	}
}

signed main(){
	for(int i=2;i<N;++i) lg[i]=lg[i>>1]+1;
	std::ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int t;
	cin >> t;
	while(t--){
		cin >> n;
		for(int i=1;i<=n;++i) cin >> a[i];
		string s;
		cin >> s;
		for(int i=1;i<=n;++i) st[0][i]=i;
		for(int i=1;i<K;++i){
			for(int j=1;j<=n-(1<<i)+1;++j){
				st[i][j]=(a[st[i-1][j]]>a[st[i-1][j+(1<<i-1)]]?st[i-1][j]:st[i-1][j+(1<<i-1)]);
			}
		}
//		cout << endl;
		dfs1(1,n);
//		for(int i=1;i<=n;++i) cout << fa[i] << ' ' << fa2[i] << endl;
		for(int i=1;i<=n;++i) G[i]=G2[i]=vector<int>();
		for(int i=1;i<=n;++i){
			if(fa[i]) G[fa[i]].push_back(i);
			if(fa2[i]) G2[fa2[i]].push_back(i);
		}
		ans=0;
		tot=0;
		int rt=get(1,n);
		dfs2(rt);
		dfs3(rt,rt);
		dfs4(rt);
		cout << ans-1 << '\n';
//		cout << endl;
	}
	return 0;
}
/*

 */

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

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

Think step by step.

Response:
```plaintext
1
5
2 1 4 3 5
?????
```