**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Remove Bridges
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices, numbered from $1$ to $n$. Vertex $1$ is the root. Additionally, the root only has one child.

You are asked to add exactly $k$ edges to the tree (possibly, multiple edges and/or edges already existing in the tree).

Recall that a bridge is such an edge that, after you remove it, the number of connected components in the graph increases. So, initially, all edges of the tree are bridges.

After $k$ edges are added, some original edges of the tree are still bridges and some are not anymore. You want to satisfy two conditions:

  * for every bridge, all tree edges in the subtree of the lower vertex of that bridge should also be bridges;    * the number of bridges is as small as possible.

Solve the task for all values of $k$ from $1$ to $n - 1$ and output the smallest number of bridges.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of testcases.

The first line of each testcase contains a single integer $n$ ($2 \le n \le 3 \cdot 10^5$) — the number of vertices of the tree.

Each of the next $n - 1$ lines contain two integers $v$ and $u$ ($1 \le v, u \le n$) — the description of the edges of the tree. It's guaranteed that the given edges form a valid tree.

Additional constraint on the input: the root (vertex $1$) has exactly one child.

The sum of $n$ over all testcases doesn't exceed $3 \cdot 10^5$.

Output: For each testcase, print $n - 1$ integers. For each $k$ from $1$ to $n - 1$ print the smallest number of bridges that can be left after you add $k$ edges to the tree.

Examples:
input:
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
output:
0
7 3 1 0 0 0 0 0 0 0 0
4 1 0 0 0 0 0
0 0 0 0

Note:

Buggy Code:
#include <bits/stdc++.h>
#define pii pair<int,int>
using namespace std;
const int N=1e6;
int T;
int n;
int h[N],ne[N],to[N],idx;
void add(int u,int v) {
	idx++;
	to[idx]=v;
	ne[idx]=h[u];
	h[u]=idx;
}
int dep[N],f[N],dfn[N],mdfn[N],rk[N],cnt;
int vst[N];
void dfs(int u,int fa) {
	cnt++;
	dfn[u]=cnt;
	rk[cnt]=u;
	mdfn[u]=dfn[u];
	for(int i=h[u];i;i=ne[i]) {
		int v=to[i];
		if(v==fa) continue;
		dep[v]=dep[u]+1;
		f[v]=u;
		dfs(v,u);
		mdfn[u]=max(mdfn[u],mdfn[v]);
	}
}
int val[4*N],mpos[4*N],tag[4*N];
void build(int p,int l,int r) {
	if(l==r) {
		val[p]=dep[rk[l]];
		mpos[p]=l;
		return;
	}
	int mid=(l+r)>>1;
	build(p*2,l,mid);
	build(p*2+1,mid+1,r);
	if(val[p*2]>=val[p*2+1]) {
		val[p]=val[p*2];
		mpos[p]=mpos[p*2];
	}
	else {
		val[p]=val[p*2+1];
		mpos[p]=mpos[p*2+1];
	}
}
void spread(int p,int l,int r) {
	if(tag[p]!=0) {
		val[p*2]+=tag[p];
		tag[p*2]+=tag[p];
		val[p*2+1]+=tag[p];
		tag[p*2+1]+=tag[p];
		tag[p]=0;
	}
}
void modify(int p,int l,int r,int x,int y,int d) {
	if(x<=l&&r<=y) {
		val[p]+=d;
		tag[p]+=d;
		return;
	}
	int mid=(l+r)>>1;
	spread(p,l,r);
	if(x<=mid) modify(p*2,l,mid,x,y,d);
	if(mid+1<=y) modify(p*2+1,mid+1,r,x,y,d);
	if(val[p*2]>=val[p*2+1]) {
		val[p]=val[p*2];
		mpos[p]=mpos[p*2];
	}
	else {
		val[p]=val[p*2+1];
		mpos[p]=mpos[p*2+1];
	}
}
pii query(int p,int l,int r,int x,int y) {
	if(x<=l&&r<=y) {
		return make_pair(val[p],mpos[p]);
	}
	int mid=(l+r)>>1;
	spread(p,l,r);
	pii ans=make_pair(0,0);
	if(x<=mid) {
		pii ans1=query(p*2,l,mid,x,y);
		if(ans1.first>ans.first) {
			ans=ans1;
		}
	}
	if(mid+1<=y) {
		pii ans1=query(p*2+1,mid+1,r,x,y);
		if(ans1.first>ans.first) {
			ans=ans1;
		}
	}
 	return ans;
}
int main() {
	scanf("%d",&T);
	while(T--) {
		memset(h,0,sizeof(h));
		memset(vst,0,sizeof(vst));
		memset(val,0,sizeof(val));
		memset(mpos,0,sizeof(mpos));
		memset(tag,0,sizeof(tag));
		idx=0;
		dep[1]=0;
		cnt=0;
		scanf("%d",&n);
		for(int i=1;i<=n-1;i++) {
			int u,v;
			scanf("%d%d",&u,&v);
			add(u,v);
			add(v,u);
		}
		dfs(1,0);
		build(1,1,n);
		int ans=n-1;
		vst[0]=1;
		for(int k=1;k<=2*n-3;k++) {
			if(ans==0) {
				if(k%2==1) printf("0 ");
				continue;
			}
			pii lans=query(1,1,n,1,n);
			ans-=lans.first;
			int u=rk[lans.second];
			stack<int> s;
			while(vst[u]==0) {
				s.push(u);
				vst[u]=1;
				u=f[u];
			}
			while(s.size()!=0) {
				int v=s.top();
				s.pop();
				if(v!=1) modify(1,1,n,dfn[v],mdfn[v],-1);
			}
			if(k%2==1) printf("%d ",ans);
		}
		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
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Remove Bridges
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices, numbered from $1$ to $n$. Vertex $1$ is the root. Additionally, the root only has one child.

You are asked to add exactly $k$ edges to the tree (possibly, multiple edges and/or edges already existing in the tree).

Recall that a bridge is such an edge that, after you remove it, the number of connected components in the graph increases. So, initially, all edges of the tree are bridges.

After $k$ edges are added, some original edges of the tree are still bridges and some are not anymore. You want to satisfy two conditions:

  * for every bridge, all tree edges in the subtree of the lower vertex of that bridge should also be bridges;    * the number of bridges is as small as possible.

Solve the task for all values of $k$ from $1$ to $n - 1$ and output the smallest number of bridges.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of testcases.

The first line of each testcase contains a single integer $n$ ($2 \le n \le 3 \cdot 10^5$) — the number of vertices of the tree.

Each of the next $n - 1$ lines contain two integers $v$ and $u$ ($1 \le v, u \le n$) — the description of the edges of the tree. It's guaranteed that the given edges form a valid tree.

Additional constraint on the input: the root (vertex $1$) has exactly one child.

The sum of $n$ over all testcases doesn't exceed $3 \cdot 10^5$.

Output: For each testcase, print $n - 1$ integers. For each $k$ from $1$ to $n - 1$ print the smallest number of bridges that can be left after you add $k$ edges to the tree.

Examples:
input:
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
output:
0
7 3 1 0 0 0 0 0 0 0 0
4 1 0 0 0 0 0
0 0 0 0

Note:

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

#define endl '\n'
#define fi first
#define se second
#define MOD(n,k) ( ( ((n) % abs(k)) + abs(k) ) % abs(k))
#define forn(i,n) for (int i = 0; i < int(n); i++)
#define forr(i,a,b) for (int i = int(a); i <= int(b); i++)
#define all(v) v.begin(), v.end()
#define pb push_back

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ii> vii;

const int MX = 3e5 + 5;
int n, ex[MX], tot, p[MX];
ii dp[MX];
vi adj[MX];

void dfs (int u) {
	dp[u] = {0, u};

	for (int v : adj[u])
		if (v != p[u]) {
			p[v] = u;
			dfs(v);
			dp[u] = max(dp[u], dp[v]);
		}

	dp[u].fi++;
}

priority_queue<ii> pq;

void erase (int u) {
	while (!ex[u]) {
		ex[u] = 1;
		tot--;

		for (int v : adj[u])
			if (v != p[u] && !ex[v])
				pq.push(dp[v]);

		u = p[u];
	}
}

void main_() {
	cin >> n;

	forr (i, 1, n) {
		adj[i].clear();
		ex[i] = 0;
		p[i] = -1;
	}

	tot = n - 1;
	ex[1] = 1;

	forn (i, n - 1) {
		int u, v;
		cin >> u >> v;
		adj[u].pb(v);
		adj[v].pb(u);
	}

	dfs(1);

	erase(dp[1].se);
	cout << tot << " ";

	forr (i, 2, n - 1) {
		vi v;
		while ((int)v.size() < 2 && pq.size()) {
			if (!ex[pq.top().se])
				v.pb(pq.top().se);
			pq.pop();
		}

		for (int u : v)
			erase(u);

		cout << tot << " ";
	}
	cout << endl;
}

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

	int t = 1;
	cin >> t;
	while (t--)
		main_();

	return 0;
}

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

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

Think step by step.

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

1. **Tree Structure**: The tree should have a structure that maximizes the number of bridges, which can be challenging to handle with the given algorithm.
2. **Root with One Child**: The root (vertex 1) must have exactly one child.
3. **Edge Cases**: Consider edge cases where the tree is a long chain or a star-like structure with a single long path from the root.
4. **Complex Subtree Structure**: Include subtrees that can test the algorithm's ability to handle multiple levels of children and bridges.

Let's create a test case where the tree is a long chain. This will ensure that the number of bridges is maximized, and the algorithm will have to handle a linear structure.

### Test Case
```plaintext
1
5
1 2
2 3
3 4
4 5
```

### Explanation
- **Number of Testcases**: 1
- **Number of Vertices**: 5
- **Edges**:
  - 1 -> 2
  - 2 -> 3
  - 3 -> 4
  - 4 -> 5

This test case forms a linear tree where each vertex has exactly one child, except for the root (vertex 1), which has one child (vertex 2). This structure will maximize the number of bridges, and the algorithm will need to handle the removal of bridges efficiently.

### Expected Output
For this test case, the expected output should be:
```plaintext
4 2 1 0
```

- **k = 1**: The smallest number of bridges is 4.
- **k = 2**: The smallest number of bridges is 2.
- **k = 3**: The smallest number of bridges is 1.
- **k = 4**: The smallest number of bridges is 0.

This test case will help expose any issues in the algorithm's handling of linear structures and the efficient removal of bridges.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Remove Bridges
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices, numbered from $1$ to $n$. Vertex $1$ is the root. Additionally, the root only has one child.

You are asked to add exactly $k$ edges to the tree (possibly, multiple edges and/or edges already existing in the tree).

Recall that a bridge is such an edge that, after you remove it, the number of connected components in the graph increases. So, initially, all edges of the tree are bridges.

After $k$ edges are added, some original edges of the tree are still bridges and some are not anymore. You want to satisfy two conditions:

  * for every bridge, all tree edges in the subtree of the lower vertex of that bridge should also be bridges;    * the number of bridges is as small as possible.

Solve the task for all values of $k$ from $1$ to $n - 1$ and output the smallest number of bridges.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of testcases.

The first line of each testcase contains a single integer $n$ ($2 \le n \le 3 \cdot 10^5$) — the number of vertices of the tree.

Each of the next $n - 1$ lines contain two integers $v$ and $u$ ($1 \le v, u \le n$) — the description of the edges of the tree. It's guaranteed that the given edges form a valid tree.

Additional constraint on the input: the root (vertex $1$) has exactly one child.

The sum of $n$ over all testcases doesn't exceed $3 \cdot 10^5$.

Output: For each testcase, print $n - 1$ integers. For each $k$ from $1$ to $n - 1$ print the smallest number of bridges that can be left after you add $k$ edges to the tree.

Examples:
input:
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
output:
0
7 3 1 0 0 0 0 0 0 0 0
4 1 0 0 0 0 0
0 0 0 0

Note:

Buggy Code:
#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mkp make_pair
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 300100;

int n, a[maxn], f[maxn], fa[maxn], st[maxn], ed[maxn], tim, val[maxn], p[maxn];
vector<int> G[maxn];

void dfs(int u, int fa) {
	st[u] = ++tim;
	if (u > 1) {
		G[u].erase(find(G[u].begin(), G[u].end(), fa));
	}
	a[u] = 0;
	for (int v : G[u]) {
		::fa[v] = u;
		dfs(v, u);
		++a[u];
	}
	ed[u] = tim;
}

struct node {
	int u, d;
	node(int a = 0, int b = 0) : u(a), d(b) {}
};

inline bool operator < (const node &a, const node &b) {
	return a.d < b.d || (a.d == b.d && a.u < b.u);
}

namespace SGT {
	int a[maxn << 2];

	inline void pushup(int x) {
		a[x] = max(a[x << 1], a[x << 1 | 1]);
	}

	void build(int rt, int l, int r) {
		a[rt] = 0;
		if (l == r) {
			return;
		}
		int mid = (l + r) >> 1;
		build(rt << 1, l, mid);
		build(rt << 1 | 1, mid + 1, r);
	}

	void update(int rt, int l, int r, int x, int y) {
		if (l == r) {
			a[rt] = y;
			return;
		}
		int mid = (l + r) >> 1;
		(x <= mid) ? update(rt << 1, l, mid, x, y) : update(rt << 1 | 1, mid + 1, r, x, y);
		pushup(rt);
	}

	int query(int rt, int l, int r, int ql, int qr) {
		if (ql <= l && r <= qr) {
			return a[rt];
		}
		int mid = (l + r) >> 1, res = 0;
		if (ql <= mid) {
			res = max(res, query(rt << 1, l, mid, ql, qr));
		}
		if (qr > mid) {
			res = max(res, query(rt << 1 | 1, mid + 1, r, ql, qr));
		}
		return res;
	}
}

void solve() {
	tim = 0;
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		vector<int>().swap(G[i]);
	}
	for (int i = 1, u, v; i < n; ++i) {
		scanf("%d%d", &u, &v);
		G[u].pb(v);
		G[v].pb(u);
	}
	dfs(1, -1);
	set<node> S;
	int cnt = 0;
	for (int i = 1; i <= n; ++i) {
		if (!a[i]) {
			int x = i, y = 1;
			while (x > 1 && a[fa[x]] == 1) {
				x = fa[x];
				++y;
			}
			p[i] = x;
			val[i] = y;
			S.emplace(i, y);
			++cnt;
		}
	}
	SGT::build(1, 1, n);
	for (int i = 1; i <= n; ++i) {
		if (!a[i]) {
			SGT::update(1, 1, n, st[i], i);
		}
	}
	for (int i = 1; i <= cnt; ++i) {
		node u = *S.begin();
		S.erase(S.begin());
		f[i] = f[i - 1] + u.d;
		SGT::update(1, 1, n, st[u.u], 0);
		// printf("f[%d] = %d %d %d\n", i, f[i], u.u, u.d);
		if (i == cnt) {
			break;
		}
		if (a[fa[u.u]] == 1) {
			int x = u.u;
			while (x > 1 && a[fa[x]] == 1) {
				x = fa[x];
			}
			--a[fa[x]];
		} else {
			--a[fa[u.u]];
			if (a[fa[u.u]] == 1) {
				// printf("fa: %d\n", fa[u.u]);
				int x = SGT::query(1, 1, n, st[fa[u.u]], ed[fa[u.u]]);
				S.erase(node(x, val[x]));
				int t = p[x];
				while (t > 1 && a[fa[t]] == 1) {
					t = fa[t];
					++val[x];
				}
				S.emplace(x, val[x]);
			}
		}
	}
	for (int i = 1; i < n; ++i) {
		if (cnt - i * 2 + 1 <= 0) {
			printf("0 ");
		} else {
			printf("%d ", f[cnt - i * 2 + 1]);
		}
	}
	putchar('\n');
}

int main() {
	int T = 1;
	scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Remove Bridges
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices, numbered from $1$ to $n$. Vertex $1$ is the root. Additionally, the root only has one child.

You are asked to add exactly $k$ edges to the tree (possibly, multiple edges and/or edges already existing in the tree).

Recall that a bridge is such an edge that, after you remove it, the number of connected components in the graph increases. So, initially, all edges of the tree are bridges.

After $k$ edges are added, some original edges of the tree are still bridges and some are not anymore. You want to satisfy two conditions:

  * for every bridge, all tree edges in the subtree of the lower vertex of that bridge should also be bridges;    * the number of bridges is as small as possible.

Solve the task for all values of $k$ from $1$ to $n - 1$ and output the smallest number of bridges.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of testcases.

The first line of each testcase contains a single integer $n$ ($2 \le n \le 3 \cdot 10^5$) — the number of vertices of the tree.

Each of the next $n - 1$ lines contain two integers $v$ and $u$ ($1 \le v, u \le n$) — the description of the edges of the tree. It's guaranteed that the given edges form a valid tree.

Additional constraint on the input: the root (vertex $1$) has exactly one child.

The sum of $n$ over all testcases doesn't exceed $3 \cdot 10^5$.

Output: For each testcase, print $n - 1$ integers. For each $k$ from $1$ to $n - 1$ print the smallest number of bridges that can be left after you add $k$ edges to the tree.

Examples:
input:
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
output:
0
7 3 1 0 0 0 0 0 0 0 0
4 1 0 0 0 0 0
0 0 0 0

Note:

Buggy Code:
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;
// typedef tree<int, null_type, std::less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;

#include <bits/stdc++.h>

#define f first
#define s second
#define pb push_back

typedef long long int ll;
// typedef __int128 lll;
typedef unsigned long long int ull;
using namespace std;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

template<typename T> int die(T x) { cout << x << endl; return 0; }

#define mod_fft 998244353
#define mod_nfft 1000000007
#define INF 100000000000000
#define LNF 1e15
#define LOL 12345678912345719ll

struct LL {

    static const ll m = mod_fft;
    long long int val;

    LL(ll v) {val=reduce(v);};
    LL() {val=0;};
    ~LL(){};
    LL(const LL& l) {val=l.val;};
    LL& operator=(int l) {val=l; return *this;}
    LL& operator=(ll l) {val=l; return *this;}
    LL& operator=(LL l) {val=l.val; return *this;}

    static long long int reduce(ll x, ll md = m) {
        x %= md;
        while (x >= md) x-=md;
        while (x < 0) x+=md;
        return x;
    }

    bool operator<(const LL& b) { return val<b.val; }
    bool operator<=(const LL& b) { return val<=b.val; }
    bool operator==(const LL& b) { return val==b.val; }
    bool operator>=(const LL& b) { return val>=b.val; }
    bool operator>(const LL& b) { return val>b.val; }

    LL operator+(const LL& b) { return LL(val+b.val); }
    LL operator+(const ll& b) { return (*this+LL(b)); }
    LL& operator+=(const LL& b) { return (*this=*this+b); }
    LL& operator+=(const ll& b) { return (*this=*this+b); }

    LL operator-(const LL& b) { return LL(val-b.val); }
    LL operator-(const ll& b) { return (*this-LL(b)); }
    LL& operator-=(const LL& b) { return (*this=*this-b); }
    LL& operator-=(const ll& b) { return (*this=*this-b); }

    LL operator*(const LL& b) { return LL(val*b.val); }
    LL operator*(const ll& b) { return (*this*LL(b)); }
    LL& operator*=(const LL& b) { return (*this=*this*b); }
    LL& operator*=(const ll& b) { return (*this=*this*b); }

    static LL exp(const LL& x, const ll& y){
        ll z = y;
        z = reduce(z,m-1);
        LL ret = 1;
        LL w = x;
        while (z) {
            if (z&1) ret *= w;
            z >>= 1; w *= w;
        }
        return ret;
    }
    LL& operator^=(ll y) { return (*this=LL(val^y)); }

    LL operator/(const LL& b) { return ((*this)*exp(b,-1)); }
    LL operator/(const ll& b) { return (*this/LL(b)); }
    LL operator/=(const LL& b) { return ((*this)*=exp(b,-1)); }
    LL& operator/=(const ll& b) { return (*this=*this/LL(b)); }

}; ostream& operator<<(ostream& os, const LL& obj) { return os << obj.val; }

vector<vector<int>> adj;
priority_queue<pii, vector<pii>, less<pii>> PQ;
vector<bool> vis;
vector<vector<pii>> bucket;
int leafs;
int dfs(int v) {
    vis[v] = 1;
    if (adj[v].size() == 1 && v != 0) {
        ++leafs;
        return 0;
    }
    int depth = 0;
    for (int u : adj[v]) {
        if (!vis[u]) {
            int d = dfs(u) + 1;
            bucket[v].pb({d,u});
            depth = max(depth, d);
        }
    }
    sort(bucket[v].begin(), bucket[v].end());
    reverse(bucket[v].begin(), bucket[v].end());
    return depth;
}
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int T; cin >> T;
    while (T--) {
        int N; cin >> N;
        adj = vector<vector<int>>(N);
        PQ = priority_queue<pii, vector<pii>, less<pii>>();
        vis = vector<bool>(N, 0);
        bucket = vector<vector<pii>>(N);
        leafs = 1;
        for (int i = 1; i < N; i++) {
            int a, b; cin >> a >> b; --a; --b;
            adj[a].pb(b);
            adj[b].pb(a);
        }
        /*for (int v = 0; v < N; v++) {
            cout << v << ": ";
            for (pii x : bucket[v]) cout << "(" << x.f << ", " << x.s << ") ";
            cout << endl;
        }*/
        // do 1 by hand
        PQ.push({dfs(0), 0});
        int ans = N-1;
        for (int t = 1; t < N; t++) {
            // cout << "part " << t << endl;
            if (PQ.size() == 0 || 2 * t >= leafs) {
                ans = 0;
            } else if (PQ.size() == 1) {
                pii x = PQ.top();
                PQ.pop();
                // cout << "TOP (" << x.f << " " << x.s << ") " << endl;
                ans -= x.first;
                int cur = x.second;
                for (int q = 1; q < bucket[cur].size(); q++) {
                    PQ.push(bucket[cur][q]);
                    // cout << "pushed (" << bucket[cur][q].f << " " << bucket[cur][q].s << ")" << endl;
                }
                while (adj[cur].size() != 1 || cur == 0) {
                    cur = bucket[cur][0].second;
                    for (int q = 1; q < bucket[cur].size(); q++) {
                        PQ.push(bucket[cur][q]);
                        // cout << "pushed (" << bucket[cur][q].f << " " << bucket[cur][q].s << ")" << endl;
                    }

                }
            } else {
                pii x = PQ.top();
                PQ.pop();
                pii y = PQ.top();
                PQ.pop();
                // cout << "TOP (" << x.f << " " << x.s << ") " << endl;
                // cout << "TOP (" << y.f << " " << y.s << ") " << endl;
                ans -= x.first;
                int curx = x.second;
                ans -= y.first;
                int cury = y.second;
                for (int q = 1; q < bucket[curx].size(); q++) {
                    PQ.push(bucket[curx][q]);
                    // cout << "pushed (" << bucket[curx][q].f << " " << bucket[curx][q].s << ")" << endl;
                }
                for (int q = 1; q < bucket[cury].size(); q++) {
                    PQ.push(bucket[cury][q]);
                    // cout << "pushed (" << bucket[cury][q].f << " " << bucket[cury][q].s << ")" << endl;
                }
                while (adj[curx].size() != 1 || curx == 0) {
                    curx = bucket[curx][0].second;
                    for (int q = 1; q < bucket[curx].size(); q++) {
                        PQ.push(bucket[curx][q]);
                        // cout << "pushed (" << bucket[curx][q].f << " " << bucket[curx][q].s << ")" << endl;
                    }
                }
                while (adj[cury].size() != 1 || cury == 0) {
                    cury = bucket[cury][0].second;
                    for (int q = 1; q < bucket[cury].size(); q++) {
                        PQ.push(bucket[cury][q]);
                        // cout << "pushed (" << bucket[cury][q].f << " " << bucket[cury][q].s << ")" << endl;
                    }
                }
            }
            cout << ans << " ";
            // cout << endl;
        }
        cout << endl;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that are likely to cause issues. The code involves depth-first search (DFS), priority queues, and handling of tree structures. Here are some considerations:

1. **Tree Structure**: Ensure the tree has a structure that can challenge the algorithm's ability to correctly manage bridges and leaf nodes.
2. **Leaf Nodes**: The code specifically handles leaf nodes, so a tree with many leaf nodes might expose issues.
3. **Multiple Levels**: A tree with multiple levels and complex branching can test the code's ability to handle depth and breadth correctly.
4. **Edge Cases**: Consider edge cases like a tree with only two nodes, a linear tree, and a balanced tree.

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

- A tree with 6 nodes.
- The root has exactly one child.
- The tree has multiple levels and some leaf nodes.

Here is the test input:

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

### Explanation:
- **Test Case 1**:
  - **Number of Nodes (N)**: 6
  - **Edges**:
    - 1 (root) -> 2
    - 2 -> 3
    - 2 -> 4
    - 3 -> 5
    - 3 -> 6

This tree has:
- The root (1) with one child (2).
- Node 2 has two children (3 and 4).
- Node 3 has two children (5 and 6).
- Node 4 is a leaf.
- Node 5 and Node 6 are leaves.

This structure ensures that the algorithm needs to handle multiple levels and leaf nodes, which can help expose any issues in the code's logic for managing bridges and leaf nodes.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Remove Bridges
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices, numbered from $1$ to $n$. Vertex $1$ is the root. Additionally, the root only has one child.

You are asked to add exactly $k$ edges to the tree (possibly, multiple edges and/or edges already existing in the tree).

Recall that a bridge is such an edge that, after you remove it, the number of connected components in the graph increases. So, initially, all edges of the tree are bridges.

After $k$ edges are added, some original edges of the tree are still bridges and some are not anymore. You want to satisfy two conditions:

  * for every bridge, all tree edges in the subtree of the lower vertex of that bridge should also be bridges;    * the number of bridges is as small as possible.

Solve the task for all values of $k$ from $1$ to $n - 1$ and output the smallest number of bridges.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of testcases.

The first line of each testcase contains a single integer $n$ ($2 \le n \le 3 \cdot 10^5$) — the number of vertices of the tree.

Each of the next $n - 1$ lines contain two integers $v$ and $u$ ($1 \le v, u \le n$) — the description of the edges of the tree. It's guaranteed that the given edges form a valid tree.

Additional constraint on the input: the root (vertex $1$) has exactly one child.

The sum of $n$ over all testcases doesn't exceed $3 \cdot 10^5$.

Output: For each testcase, print $n - 1$ integers. For each $k$ from $1$ to $n - 1$ print the smallest number of bridges that can be left after you add $k$ edges to the tree.

Examples:
input:
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
output:
0
7 3 1 0 0 0 0 0 0 0 0
4 1 0 0 0 0 0
0 0 0 0

Note:

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

#define fastio ios_base::sync_with_stdio(0); cin.tie(0)
#define int long long

using namespace std;

void dbg_out() { cerr << endl; }
template <typename H, typename... T>
void dbg_out(H h, T... t) { cerr << ' ' << h; dbg_out(t...); }
#define dbg(...) { cerr << #__VA_ARGS__ << ':'; dbg_out(__VA_ARGS__); }

void solve() {
	int n; cin>>n;
	vector<vector<int>> adj(n);
	for (int i = 1; i < n; i++) {
		int u, v; cin>>u>>v; u -= 1, v -= 1;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}

	vector<int> h(n), best(n), p(n, -1);
	function<void(int)> dfs = [&](int u) {
		best[u] = u;
		for (int v: adj[u]) if (v != p[u]) {
			p[v] = u;

			dfs(v);

			if (h[v]+1 > h[u]) {
				best[u] = best[v];
				h[u] = h[v]+1;
			}
		}
	};
	dfs(0);

	int res = n;
	vector<bool> marked(n);
	priority_queue<pair<int,int>> pq;
	function<void(int)> remove_path = [&](int u) {
		if (marked[u] or u == -1) return;
		marked[u] = true;
		for (int v: adj[u]) if (v != p[u] and not marked[v]) {
			pq.emplace(h[v], best[v]);
		}
		res -= 1;
		remove_path(p[u]);
	};
	remove_path(best[0]);

	for (int k = 1; k <= n-1; k++) {
		cout<<res<<' ';
		if (pq.size() == 1) {
			res = 0;
			pq.pop();
		} else if (pq.size() >= 2) {
			int u = pq.top().second; pq.pop();
			int v = pq.top().second; pq.pop();
			remove_path(u);
			remove_path(v);
		}
	}
	cout<<'\n';
}

signed main(){
	fastio;
	int tt; cin>>tt;
	while (tt--) solve();
	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Remove Bridges
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices, numbered from $1$ to $n$. Vertex $1$ is the root. Additionally, the root only has one child.

You are asked to add exactly $k$ edges to the tree (possibly, multiple edges and/or edges already existing in the tree).

Recall that a bridge is such an edge that, after you remove it, the number of connected components in the graph increases. So, initially, all edges of the tree are bridges.

After $k$ edges are added, some original edges of the tree are still bridges and some are not anymore. You want to satisfy two conditions:

  * for every bridge, all tree edges in the subtree of the lower vertex of that bridge should also be bridges;    * the number of bridges is as small as possible.

Solve the task for all values of $k$ from $1$ to $n - 1$ and output the smallest number of bridges.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of testcases.

The first line of each testcase contains a single integer $n$ ($2 \le n \le 3 \cdot 10^5$) — the number of vertices of the tree.

Each of the next $n - 1$ lines contain two integers $v$ and $u$ ($1 \le v, u \le n$) — the description of the edges of the tree. It's guaranteed that the given edges form a valid tree.

Additional constraint on the input: the root (vertex $1$) has exactly one child.

The sum of $n$ over all testcases doesn't exceed $3 \cdot 10^5$.

Output: For each testcase, print $n - 1$ integers. For each $k$ from $1$ to $n - 1$ print the smallest number of bridges that can be left after you add $k$ edges to the tree.

Examples:
input:
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
output:
0
7 3 1 0 0 0 0 0 0 0 0
4 1 0 0 0 0 0
0 0 0 0

Note:

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

#ifdef LOCAL
#define dbg(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define dbg(...)
#endif

void __print(int32_t x) {cerr << x;}
void __print(int64_t x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(string x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
template<typename T>void __print(complex<T> x) {cerr << '{'; __print(x.real()); cerr << ','; __print(x.imag()); cerr << '}';}

template<typename T>
void __print(const T &x);
template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto it = x.begin() ; it != x.end() ; it++) cerr << (f++ ? "," : ""), __print(*it); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}

int32_t main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);

	auto __solve_testcase = [&](int test) {
		int N;	cin >> N;
		vector<vector<int>> adj(N);
		vector<set<pair<pair<int, int>, int>>> b(N);
		vector<int> rep(N), height(N), leaf(N);
		for(int i = 1 ; i < N ; ++i) {
			int u, v;	cin >> u >> v;
			adj[u - 1].push_back(v - 1);
			adj[v - 1].push_back(u - 1);
		}

		auto dfs = [&](auto &&dfs, int nd, int par) -> void {
			rep[nd] = nd;
			for(auto &u: adj[nd]) if(u != par) {
				dfs(dfs, u, nd);
				leaf[nd] += leaf[u];
				height[nd] = max(height[nd], height[u] + 1);
				b[nd].insert({{height[u], leaf[u]}, u});
			}
			if((int)b[nd].size() == 0) {
				leaf[nd] = 1;
			}
		};
		dfs(dfs, 0, -1);
		int ans = N - 1;

		vector<bool> vis(N);

		auto remove = [&]() {
			auto dfs = [&](auto &&dfs, int nd, int par, int x) -> void {
				if(b[rep[nd]].empty())
					return;
				vector<int> rem;
				auto to_rem = prev(b[rep[nd]].end());
				rem.push_back(to_rem -> second);
				b[rep[nd]].erase(to_rem);
				if(x) {
					to_rem = prev(b[rep[nd]].end());
					rem.push_back(to_rem -> second);
					b[rep[nd]].erase(to_rem);
				}

				for(auto &u: rem) {
					assert(!vis[u]);
					vis[u] = 1;
					dfs(dfs, u, nd, 0);
					--ans;

					int lr = rep[nd], sm = rep[u];
					if((int)b[lr].size() < (int)b[sm].size()) {
						swap(lr, sm);
					}
					for(auto &x: b[sm])	b[lr].insert(x);
					rep[nd] = lr;
				}
			};

			dfs(dfs, 0, -1, (int)b[rep[0]].size() > 1);
		};
		for(int i = 1 ; i < N ; ++i) {
			remove();
			cout << ans << " \n"[i == N - 1];
		}
	};

	int NumTest = 1;
	cin >> NumTest;
	for(int testno = 1; testno <= NumTest ; ++testno) {
		__solve_testcase(testno);
	}

	return 0;
}

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

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

Think step by step.

Response:
```plaintext
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Remove Bridges
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices, numbered from $1$ to $n$. Vertex $1$ is the root. Additionally, the root only has one child.

You are asked to add exactly $k$ edges to the tree (possibly, multiple edges and/or edges already existing in the tree).

Recall that a bridge is such an edge that, after you remove it, the number of connected components in the graph increases. So, initially, all edges of the tree are bridges.

After $k$ edges are added, some original edges of the tree are still bridges and some are not anymore. You want to satisfy two conditions:

  * for every bridge, all tree edges in the subtree of the lower vertex of that bridge should also be bridges;    * the number of bridges is as small as possible.

Solve the task for all values of $k$ from $1$ to $n - 1$ and output the smallest number of bridges.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of testcases.

The first line of each testcase contains a single integer $n$ ($2 \le n \le 3 \cdot 10^5$) — the number of vertices of the tree.

Each of the next $n - 1$ lines contain two integers $v$ and $u$ ($1 \le v, u \le n$) — the description of the edges of the tree. It's guaranteed that the given edges form a valid tree.

Additional constraint on the input: the root (vertex $1$) has exactly one child.

The sum of $n$ over all testcases doesn't exceed $3 \cdot 10^5$.

Output: For each testcase, print $n - 1$ integers. For each $k$ from $1$ to $n - 1$ print the smallest number of bridges that can be left after you add $k$ edges to the tree.

Examples:
input:
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
output:
0
7 3 1 0 0 0 0 0 0 0 0
4 1 0 0 0 0 0
0 0 0 0

Note:

Buggy Code:
import sys
input = sys.stdin.readline

sys.setrecursionlimit(300020)

def dfs(u,p):
    mx=0
    for to in E[u]:
        if to!=p:
            t=dfs(to,u)+1

            if mx<t:
                if mx>0:
                    P.append(mx)
                mx=t
            else:
                P.append(t)
    return mx

t=int(input())
for tests in range(t):
    n=int(input())

    E=[[] for i in range(n)]
    for i in range(n-1):
        x,y=map(int,input().split())
        x-=1
        y-=1
        E[x].append(y)
        E[y].append(x)

    P=[]
    sc=dfs(0,-1)

    #print(ANS,P)

    P.sort(reverse=True)
    if len(P)%2==1:
        P.append(0)

    ANS=[n-sc-1]
    for i in range(0,len(P),2):
        ANS.append(ANS[-1]-P[i]-P[i+1])

    while len(ANS)<n-1:
        ANS.append(0)

    print(*ANS)

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

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

Think step by step.

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

1. **Tree Structure**: The tree should be structured in a way that the bug in the DFS logic can be easily exposed. Specifically, the tree should have multiple levels and branches to ensure that the edge cases are tested.

2. **Root with One Child**: The root (vertex 1) should have exactly one child, as specified in the problem constraints.

3. **Edge Cases**: The test input should include scenarios where the addition of edges can significantly change the number of bridges, especially in subtrees.

Given these considerations, let's construct a small-scale test input:

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

### Explanation:
- **Tree Structure**:
  - The root (vertex 1) has one child (vertex 2).
  - Vertex 2 has one child (vertex 3).
  - Vertex 3 has one child (vertex 4).
  - Vertex 4 has three children (vertices 5, 6, and 7).

- **Edge Cases**:
  - The tree is deep and has a single path from the root to vertex 4, followed by a branch with three children.
  - This structure ensures that the addition of edges can significantly change the number of bridges, especially when adding edges between the children of vertex 4.

This test input should help expose the bug in the DFS logic, particularly in how it handles the maximum depth and the addition of edges.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Remove Bridges
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given a rooted tree, consisting of $n$ vertices, numbered from $1$ to $n$. Vertex $1$ is the root. Additionally, the root only has one child.

You are asked to add exactly $k$ edges to the tree (possibly, multiple edges and/or edges already existing in the tree).

Recall that a bridge is such an edge that, after you remove it, the number of connected components in the graph increases. So, initially, all edges of the tree are bridges.

After $k$ edges are added, some original edges of the tree are still bridges and some are not anymore. You want to satisfy two conditions:

  * for every bridge, all tree edges in the subtree of the lower vertex of that bridge should also be bridges;    * the number of bridges is as small as possible.

Solve the task for all values of $k$ from $1$ to $n - 1$ and output the smallest number of bridges.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of testcases.

The first line of each testcase contains a single integer $n$ ($2 \le n \le 3 \cdot 10^5$) — the number of vertices of the tree.

Each of the next $n - 1$ lines contain two integers $v$ and $u$ ($1 \le v, u \le n$) — the description of the edges of the tree. It's guaranteed that the given edges form a valid tree.

Additional constraint on the input: the root (vertex $1$) has exactly one child.

The sum of $n$ over all testcases doesn't exceed $3 \cdot 10^5$.

Output: For each testcase, print $n - 1$ integers. For each $k$ from $1$ to $n - 1$ print the smallest number of bridges that can be left after you add $k$ edges to the tree.

Examples:
input:
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
output:
0
7 3 1 0 0 0 0 0 0 0 0
4 1 0 0 0 0 0
0 0 0 0

Note:

Buggy Code:
import sys
input = sys.stdin.readline

sys.setrecursionlimit(300020)

def dfs(u,p):
    mx=0
    for to in E[u]:
        if to!=p:
            t=dfs(to,u)+1

            if mx<t:
                if mx>0:
                    P.append(mx)
                mx=t
            else:
                P.append(t)
    return mx

t=int(input())
for tests in range(t):
    n=int(input())

    E=[[] for i in range(n)]
    for i in range(n-1):
        x,y=map(int,input().split())
        x-=1
        y-=1
        E[x].append(y)
        E[y].append(x)

    P=[]
    sc=dfs(0,-1)

    #print(ANS,P)

    P.sort(reverse=True)
    if len(P)%2==1:
        P.append(0)

    ANS=[n-sc-1]
    for i in range(0,len(P),2):
        ANS.append(ANS[-1]-P[i]-P[i+1])

    while len(ANS)<n-1:
        ANS.append(0)

    print(*ANS)

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

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

Think step by step.

Response:
```plaintext
4
2
1 2
12
4 10
5 12
12 11
3 6
9 6
1 6
12 7
11 6
2 11
10 9
10 8
8
1 2
2 3
2 4
3 5
3 6
4 7
4 8
5
1 2
2 3
3 4
4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Spinning Round (Hard Version)
time_limit_ms: 7000
memory_limit_mb: 1024
Description: This is the hard version of the problem. The only difference between the two versions are the allowed characters in $s$. 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 $n$, where each character is either L, R, or ?.

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 over all connected$^{\text{∗}}$ 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 characters L, R, and ?.

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
R?RL?
2
1 2
LR
3
3 1 2
L?R
7
5 3 1 6 4 2 7
?R?R?R?
5
5 2 1 3 4
?????
6
6 2 3 4 5 1
?LLRLL
8
1 7 5 6 2 8 4 3
?R??????
12
6 10 7 1 8 5 12 2 11 3 4 9
????????????
output:
3
-1
-1
4
4
3
5
8

Note: In the first test case, there are two connected graphs (the labels are indices):

![](CDN_BASE_URL/4561c9025a4357785c7aff0bb1bbff9a)| ![](CDN_BASE_URL/a680f9e3bffa38fd90dd7bbc4c3e5f3b)   ---|---      The graph on the left has a diameter of $2$, while the graph on the right has a diameter of $3$, so the answer is $3$.

In the second test case, there are no connected graphs, so the answer is $-1$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
const int N = 1e6+5;
int n,rt,ans,top,p[N],stk[N],ls[N],rs[N],fa[N],L[N],R[N],f[N][3];
char s[N];
void clear(){
    top = ans = rt = 0;
    for(int i=0;i<=n;i++) ls[i] = rs[i] = fa[i] = stk[i] = 0,L[i] = R[i] = -1,f[i][0] = f[i][1] = f[i][2] = -1e9;
}
void get_LR(int u,int l,int r){
    if(!u) return ;
    L[u] = l,R[u] = r;
    get_LR(ls[u],l,u),get_LR(rs[u],u,r);
}
void get_f(int u){
    if(!u) return ;
    if(s[u] != 'R') f[u][0] = 1;
    if(s[u] != 'L') f[u][1] = 1;
    if(!ls[u] && !rs[u]) return ;
    get_f(ls[u]),get_f(rs[u]);
    if(!ls[u]){
        if(s[u] != 'R') f[u][0] = f[rs[u]][0]+1,f[u][2] = f[rs[u]][2]+1;
        f[u][1] = f[rs[u]][1];
        if(s[u] != 'L') f[u][1] = max(f[u][1],f[rs[u]][0]+1);
        return ;
    }else if(!rs[u]){
        f[u][0] = f[ls[u]][0];
        if(s[u] != 'R') f[u][0] = max(f[u][0],f[ls[u]][1]+1);
        if(s[u] != 'L') f[u][1] = f[ls[u]][1]+1,f[u][2] = f[ls[u]][2]+1;
        return ;
    }
    f[u][0] = f[ls[u]][0],f[u][1] = f[rs[u]][1],f[u][2] = f[u][0]+f[u][1];
    if(s[u] != 'R') f[u][0] = max(f[u][0],max(f[ls[u]][1],f[rs[u]][0])+1),f[u][2] = max(f[u][2],max(f[rs[u]][2],f[ls[u]][1]+f[rs[u]][1])+1);
    if(s[u] != 'L') f[u][1] = max(f[u][1],max(f[ls[u]][1],f[rs[u]][0])+1),f[u][2] = max(f[u][2],max(f[ls[u]][2],f[ls[u]][0]+f[rs[u]][0])+1);
}
void get_ans(int u){
    if(!u) return ;
    ans = max(ans,f[ls[u]][1]+f[rs[u]][0]);
    if(u != rt) {
        ans = max(ans,max(f[u][0],f[u][1]));
        if(s[u] != 'R'){
            if(rs[u]) ans = max(ans,max(f[ls[u]][2],f[ls[u]][0]+f[rs[u]][0])+1);
            else ans = max(ans,f[ls[u]][0]+1);
        }
        if(s[u] != 'L'){
            if(ls[u]) ans = max(ans,max(f[rs[u]][2],f[rs[u]][1]+f[ls[u]][1])+1);
            else ans = max(ans,f[rs[u]][1]+1);
        }
    }
    get_ans(ls[u]),get_ans(rs[u]);
}
void solve(){
    cin >> n;
    clear();
    for(int i=1;i<=n;i++) cin >> p[i];
    for(int i=1;i<=n;i++) cin >> s[i];
    rt = 1,stk[++top] = 1;
    for(int i=2;i<=n;i++){
        int k = top;
        while(k && p[stk[k]]<p[i]) k--;
        if(k) rs[stk[k]] = i,fa[i] = stk[k];
        else rt = i;
        if(k+1<=top) ls[i] = stk[k+1],fa[stk[k+1]] = i;
        stk[++k] = i,top = k;
    }
    get_LR(rt,-1,-1);
    int chk = 1;
    for(int i=1;i<=n;i++){
        if(i == rt) continue;
        if(s[i] == '?'){
            if(L[i] == -1) s[i] = 'R';
            else if(R[i] == -1) s[i] = 'L';
        }else{
            if(L[i] == -1 && s[i] == 'L') {
                cout << -1 << '\n';
                return ;
            }
            if(R[i] == -1 && s[i] == 'R'){
                cout << -1 << '\n';
                return ;
            }
        }
    }
    get_f(rt),get_ans(rt);
    cout << ans << '\n';
}
int main(){
    cin.tie(0),cout.tie(0);
    ios::sync_with_stdio(0);
    int T;cin >> T;
    while(T--) solve();
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Spinning Round (Hard Version)
time_limit_ms: 7000
memory_limit_mb: 1024
Description: This is the hard version of the problem. The only difference between the two versions are the allowed characters in $s$. 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 $n$, where each character is either L, R, or ?.

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 over all connected$^{\text{∗}}$ 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 characters L, R, and ?.

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
R?RL?
2
1 2
LR
3
3 1 2
L?R
7
5 3 1 6 4 2 7
?R?R?R?
5
5 2 1 3 4
?????
6
6 2 3 4 5 1
?LLRLL
8
1 7 5 6 2 8 4 3
?R??????
12
6 10 7 1 8 5 12 2 11 3 4 9
????????????
output:
3
-1
-1
4
4
3
5
8

Note: In the first test case, there are two connected graphs (the labels are indices):

![](CDN_BASE_URL/4561c9025a4357785c7aff0bb1bbff9a)| ![](CDN_BASE_URL/a680f9e3bffa38fd90dd7bbc4c3e5f3b)   ---|---      The graph on the left has a diameter of $2$, while the graph on the right has a diameter of $3$, so the answer is $3$.

In the second test case, there are no connected graphs, so the answer is $-1$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using i128 = __int128;
using u128 = unsigned __int128;
template <typename T>
void chkmax(T &x, const T &y) {
  x = max(x, y);
}
template <typename T>
void chkmin(T &x, const T &y) {
  x = min(x, y);
}
constexpr int MAXN = 4e5 + 10, INF = 0x1f3f3f3f;
int n, a[MAXN], typ[MAXN], ch[MAXN][2], dp[MAXN], fa[MAXN], go[MAXN], ans;
int dmx[MAXN][2], up1[MAXN], up2[MAXN], fgo[MAXN], dep[MAXN];
bool can1[MAXN], can2[MAXN];
int get_dir(int u) { return ch[fa[u]][1] == u; }
void dfs(int u) {
  dp[u] = 0;
  can1[u] = (typ[u] != get_dir(u) && fa[u]);
  can2[u] = (typ[u] != !get_dir(u) && fa[go[u]]);
  fgo[u] = (can1[u] ? fgo[fa[u]] : u);
  if (dep[fgo[u]] < dep[go[u]]) fgo[u] = go[u];
  if (!can1[u] && !can2[u] && fa[u]) {
    ans = -1;
    return;
  }
  if (fa[u]) {
    up1[u] = max({can1[u] ? 1 : -INF, can2[u] ? (go[u] == fa[u] ? up1[fa[u]] : up2[fa[u]]) + 1 : -INF});
  } else {
    up1[u] = -INF;
  }
  if (fa[go[u]]) {
    if (fgo[u] == go[u]) {
      up2[u] = up1[go[u]] + dep[u] - dep[go[u]];
    } else {
      up2[u] = dep[u] - dep[fgo[u]] + 1;
    }
  } else {
    up2[u] = -INF;
  }
  for (int t = 0, v; t < 2; ++t) {
    v = ch[u][t];
    if (!v) continue;
    fa[v] = u;
    dep[v] = dep[u] + 1;
    go[v] = (fa[u] && t == get_dir(u) ? go[u] : u);
    dfs(v);
    if (ans == -1) return;
  }
  if (can1[u]) chkmax(dp[fa[u]], dp[u] + 1);
  if (can2[u]) chkmax(dp[fa[go[u]]], dp[u] + 1);
  for (int t = 0, v; t < 2; ++t) {
    v = ch[u][t];
    dmx[u][t] = -INF;
    if (v) {
      chkmax(dmx[u][t], dmx[v][t]);
      if (can2[v]) chkmax(dmx[u][t], dp[v]);
    }
  }
  if (ch[u][0] && ch[u][1]) {
    chkmax(ans, max(can1[ch[u][0]] ? dp[ch[u][0]] : -INF, dmx[ch[u][0]][1]) +
                max(can1[ch[u][1]] ? dp[ch[u][1]] : -INF, dmx[ch[u][1]][0]) + 2);
    if (fa[u]) {
      chkmax(ans, dmx[u][0] + dmx[u][1] + 2 + (go[u] == fa[u] ? up1[fa[u]] : up2[fa[u]]));
      for (int t = 0; t < 2; ++t) {
        chkmax(ans, max(can1[ch[u][t]] ? dp[ch[u][t]] : -INF, dmx[ch[u][t]][!t]) + dmx[u][!t] +
                    (get_dir(u) == t ? up1[u] : up2[u]) + 2);
        chkmax(ans, dmx[u][t] + (get_dir(u) == t ? up2[u] : up1[u]) + 1);
      }
    }
  }
  chkmax(ans, dp[u]);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int cas;
  cin >> cas;
  while (cas--) {
    memset(dmx, 0, sizeof(dmx));
    memset(up1, 0, sizeof(up1));
    static int stk[MAXN], tp;
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int i = 1; i <= n; ++i) {
      char c;
      cin >> c;
      typ[i] = (c == '?' ? -1 : (c == 'R'));
    }
    tp = 0;
    for (int i = 1; i <= n; ++i) {
      ch[i][0] = ch[i][1] = 0;
      while (tp && a[i] > a[stk[tp]]) ch[i][0] = stk[tp--];
      if (tp) ch[stk[tp]][1] = i;
      stk[++tp] = i;
    }
    int rt = stk[1];
    fa[rt] = 0;
    go[rt] = rt;
    dep[rt] = 0;
    ans = 0;
    dfs(rt);
    cout << ans << "\n";
  }
  return 0;
}
/*
g++ G.cpp -o G -std=c++14 -O2 -Wall -Wextra -Wshadow -g -fsanitize=address,undefined
*/
/*
8
1 7 5 6 2 8 4 3
?R??????
*/

**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 that can cause the algorithm to fail. Here are the steps to think through this:

1. **Understanding the Bug**: The code seems to have issues with handling the connections and the maximum diameter calculation, especially when there are multiple choices for the edges (i.e., when the character is `?`).

2. **Edge Cases**:
   - **Small Permutations**: Test with small permutations to ensure the logic is correct.
   - **Permutations with Multiple `?`**: Test with permutations where there are multiple `?` characters to see if the code correctly handles the choices.
   - **Permutations with No Valid Connected Graphs**: Test with permutations that cannot form a connected graph to ensure the code outputs `-1`.

3. **Test Input**:
   - We will use a small permutation with multiple `?` characters to see if the code can handle the choices correctly.
   - We will also test with a permutation that cannot form a connected graph to ensure the code outputs `-1`.

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

```plaintext
3
5
2 1 4 3 5
R?RL?
3
3 1 2
L?R
4
4 3 2 1
?????
```

Explanation:
- **Test Case 1**: `5 2 1 4 3 5 R?RL?` - This should output `3` as the maximum diameter.
- **Test Case 2**: `3 3 1 2 L?R` - This should output `-1` because it is not possible to form a connected graph.
- **Test Case 3**: `4 4 3 2 1 ????` - This should output `3` as the maximum diameter.

This test input covers a range of scenarios to ensure the code is robust and can handle different permutations and choices correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Spinning Round (Hard Version)
time_limit_ms: 7000
memory_limit_mb: 1024
Description: This is the hard version of the problem. The only difference between the two versions are the allowed characters in $s$. 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 $n$, where each character is either L, R, or ?.

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 over all connected$^{\text{∗}}$ 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 characters L, R, and ?.

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
R?RL?
2
1 2
LR
3
3 1 2
L?R
7
5 3 1 6 4 2 7
?R?R?R?
5
5 2 1 3 4
?????
6
6 2 3 4 5 1
?LLRLL
8
1 7 5 6 2 8 4 3
?R??????
12
6 10 7 1 8 5 12 2 11 3 4 9
????????????
output:
3
-1
-1
4
4
3
5
8

Note: In the first test case, there are two connected graphs (the labels are indices):

![](CDN_BASE_URL/4561c9025a4357785c7aff0bb1bbff9a)| ![](CDN_BASE_URL/a680f9e3bffa38fd90dd7bbc4c3e5f3b)   ---|---      The graph on the left has a diameter of $2$, while the graph on the right has a diameter of $3$, so the answer is $3$.

In the second test case, there are no connected graphs, so the answer is $-1$.

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

using namespace std;

const int N = 100000 + 5, INF = 1000000000;

inline int Read() {int res; return scanf("%d", &res), res; }

int T, n, a[N], ls[N], rs[N];
int q[N], hh = 0;
int f[N][3], ans = 0, gg = 1;
char s[N];

void DFS(int u, int l, int r) {
	f[u][2] = -INF, f[u][0] = f[u][1] = -INF;

	if(ls[u]) {
		DFS(ls[u], l, u - 1);

		f[u][0] = max(f[u][0], f[ls[u]][0]);
		if(s[u] != 'L') f[u][1] = max(f[u][1], f[ls[u]][1] + 1);
		if(s[u] != 'R') f[u][0] = max(f[u][0], f[ls[u]][1] + 1);
		if(s[u] != 'L') f[u][2] = max(f[u][2], f[ls[u]][2] + 1);
		if(s[u] != 'R' && a[u] != n) ans = max(ans, f[ls[u]][2] + 3);
		ans = max(ans, f[ls[u]][1] + 1);
		if(s[u] != 'R' && a[u] != n) ans = max(ans, f[ls[u]][0] + 2);
	}

	if(rs[u]) {
		DFS(rs[u], u + 1, r);

		f[u][1] = max(f[u][1], f[rs[u]][1]);
		if(s[u] != 'L') f[u][1] = max(f[u][1], f[rs[u]][0] + 1);
		if(s[u] != 'R') f[u][0] = max(f[u][0], f[rs[u]][0] + 1);
		if(s[u] != 'R') f[u][2] = max(f[u][2], f[rs[u]][2] + 1);
		if(s[u] != 'L' && a[u] != n) ans = max(ans, f[rs[u]][2] + 3);
		ans = max(ans, f[rs[u]][0] + 1);
		if(s[u] != 'L' && a[u] != n) ans = max(ans, f[rs[u]][1] + 2);
	}

	if(s[u] == 'L' && l == 1 && a[u] != n) gg = 0;
	if(s[u] == 'R' && r == n && a[u] != n) gg = 0;

	if(ls[u] && rs[u]) {
		f[u][2] = max(f[u][2], f[ls[u]][0] + f[rs[u]][1]);

		if(s[u] != 'R') {
			f[u][2] = max(f[u][2], f[ls[u]][1] + f[rs[u]][1] + 1);
			if(a[u] != n) ans = max(ans, f[ls[u]][0] + f[rs[u]][0] + 3);
		}
		if(s[u] != 'L') {
			f[u][2] = max(f[u][2], f[ls[u]][0] + f[rs[u]][0] + 1);
			if(a[u] != n) ans = max(ans, f[ls[u]][1] + f[rs[u]][1] + 3);
		}

		ans = max(ans, f[ls[u]][1] + f[rs[u]][0] + 2);
	}

	if(s[u] != 'R' && !ls[u] && !rs[u]) f[u][0] = 0;
	if(s[u] != 'L' && !ls[u] && !rs[u]) f[u][1] = 0;

//	cout << u<< " " << f[u][0] << " " << f[u][1] << " " << f[u][2] << endl;
}

int main() {
//	freopen("17.in", "r", stdin);
//	freopen("walk.out", "w", stdout);

	T = Read();

	while(T --) {
		n = Read();
		for(int i = 1; i <= n; ++ i) a[i] = Read(), ls[i] = 0, rs[i]
		= 0;
		scanf("%s", s + 1);

		int tot = 0;
		for(int i = 2; i <= n; ++ i) tot += (a[i - 1] < a[i]);

		if(tot == n - 1) {
			int isok = 1;
			for(int i = 1; i < n; ++ i) {
				if(s[i] == 'L') isok = 0;
			}
			if(isok) printf("%d\n", n - 1);
			else puts("-1");

			continue;
		}

		q[hh = 1] = 1;
		for(int i = 2; i <= n; ++ i) {
			int las = 0;
			while(hh && a[i] > a[q[hh]]) las = q[hh], hh --;
			ls[i] = las;
			if(hh) rs[q[hh]] = i;
			q[++ hh] = i;
		}

		ans = 0, gg = 1;
		DFS(q[1], 1, n);

		if(gg) printf("%d\n", ans);
		else puts("-1");
	}

	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Spinning Round (Hard Version)
time_limit_ms: 7000
memory_limit_mb: 1024
Description: This is the hard version of the problem. The only difference between the two versions are the allowed characters in $s$. 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 $n$, where each character is either L, R, or ?.

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 over all connected$^{\text{∗}}$ 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 characters L, R, and ?.

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
R?RL?
2
1 2
LR
3
3 1 2
L?R
7
5 3 1 6 4 2 7
?R?R?R?
5
5 2 1 3 4
?????
6
6 2 3 4 5 1
?LLRLL
8
1 7 5 6 2 8 4 3
?R??????
12
6 10 7 1 8 5 12 2 11 3 4 9
????????????
output:
3
-1
-1
4
4
3
5
8

Note: In the first test case, there are two connected graphs (the labels are indices):

![](CDN_BASE_URL/4561c9025a4357785c7aff0bb1bbff9a)| ![](CDN_BASE_URL/a680f9e3bffa38fd90dd7bbc4c3e5f3b)   ---|---      The graph on the left has a diameter of $2$, while the graph on the right has a diameter of $3$, so the answer is $3$.

In the second test case, there are no connected graphs, so the answer is $-1$.

Buggy Code:
// created:  2024-08-25 16:01:23
#include<cstdio>
#include<cctype>
#include<algorithm>
#define F(i,l,r) for(int i=(l),i##_end=(r);i<i##_end;++i)
#define I128 //||is_same<T,__int128_t>::value||is_same<T,__uint128_t>::value
using namespace std;
template<typename T>enable_if_t<is_integral<T>::value I128,void> readmain(T &x)
{
	bool neg=false;int c=getchar();
	for(;!isdigit(c);c=getchar())if(c=='-')neg=true;
	for(x=0;isdigit(c);c=getchar())x=(T)(10*x+(c-'0'));
	if(neg)x=-x;
}
template<typename T>T& read(T &x){readmain(x);return x;}
template<typename T,typename ...Tr>void read(T &x,Tr&... r){readmain(x);read(r...);}
constexpr int N=4e5+5,INF=0xfffffff;
int n,p[N],fa[N],ch[N][2],s[N],t,f[N][4],ans,h[N];
char o[N];
void dfs(int u)
{
	int ls=ch[u][0],rs=ch[u][1];
	f[u][0]=0;
	if(o[u]!='R')f[u][1]=1;
	if(o[u]!='L')f[u][2]=1;
	if(~ls)
	{
		dfs(ls);
		f[u][1]=max(f[u][1],f[ls][1]);
		ans=max(ans,f[ls][2]);
		if(o[u]!='R')
		{
			f[u][1]=max(f[u][1],f[ls][2]+1);
			if(h[u]&1)ans=max(ans,f[ls][3]+1);
		}
		if(o[u]!='L')
		{
			f[u][2]=max(f[u][2],f[ls][2]+1);
			f[u][3]=max(f[u][3],f[ls][3]+1);
		}
	}
	if(~rs)
	{
		dfs(rs);
		f[u][2]=max(f[u][2],f[rs][2]);
		ans=max(ans,f[rs][1]);
		if(o[u]!='L')
		{
			f[u][2]=max(f[u][2],f[rs][1]+1);
			if(h[u]&2)ans=max(ans,f[rs][3]+1);
		}
		if(o[u]!='R')
		{
			f[u][1]=max(f[u][1],f[rs][1]+1);
			f[u][3]=max(f[u][3],f[rs][3]+1);
		}
	}
	if(~ls&&~rs)
	{
		ans=max(ans,f[ls][2]+f[rs][1]);
		if(o[u]!='R')
		{
			if(h[u]&1)ans=max(ans,f[ls][1]+f[rs][1]+1);
			f[u][3]=max(f[u][3],f[ls][2]+f[rs][2]+1);
		}
		if(o[u]!='L')
		{
			if(h[u]&2)ans=max(ans,f[ls][2]+f[rs][2]+1);
			f[u][3]=max(f[u][3],f[ls][1]+f[rs][1]+1);
		}
		f[u][3]=max(f[u][3],f[ls][1]+f[rs][2]);
	}
}
void solve()
{
	read(n);
	p[n]=n;
	F(i,0,n)--read(p[i]),fa[i]=n,h[i]=0;
	scanf("%s",o);
	t=0;
	F(i,0,n)
	{
		while(t&&p[s[t-1]]<p[i])--t;
		if(t&&p[s[t-1]]<p[fa[i]])fa[i]=s[t-1];
		if(t)h[i]|=1;
		if(p[i]!=n-1&&o[i]=='L'&&!t)
		{
			puts("-1");
			return;
		}
		s[t++]=i;
	}
	t=0;
	for(int i=n;~--i;)
	{
		while(t&&p[s[t-1]]<p[i])--t;
		if(t&&p[s[t-1]]<p[fa[i]])fa[i]=s[t-1];
		if(t)h[i]|=2;
		if(p[i]!=n-1&&o[i]=='R'&&!t)
		{
			puts("-1");
			return;
		}
		s[t++]=i;
	}
	F(i,0,n)ch[i][0]=ch[i][1]=-1;
	F(i,0,n)if(fa[i]!=n)ch[fa[i]][fa[i]<i]=i;
	F(i,0,n)F(j,0,4)f[i][j]=-INF;
	F(i,0,n)if(fa[i]==n)
	{
		ans=0;
		dfs(i);
		printf("%d\n",ans);
	}
}
int main()
{
	int tt;read(tt);
	while(tt--)solve();
	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Spinning Round (Hard Version)
time_limit_ms: 7000
memory_limit_mb: 1024
Description: This is the hard version of the problem. The only difference between the two versions are the allowed characters in $s$. 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 $n$, where each character is either L, R, or ?.

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 over all connected$^{\text{∗}}$ 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 characters L, R, and ?.

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
R?RL?
2
1 2
LR
3
3 1 2
L?R
7
5 3 1 6 4 2 7
?R?R?R?
5
5 2 1 3 4
?????
6
6 2 3 4 5 1
?LLRLL
8
1 7 5 6 2 8 4 3
?R??????
12
6 10 7 1 8 5 12 2 11 3 4 9
????????????
output:
3
-1
-1
4
4
3
5
8

Note: In the first test case, there are two connected graphs (the labels are indices):

![](CDN_BASE_URL/4561c9025a4357785c7aff0bb1bbff9a)| ![](CDN_BASE_URL/a680f9e3bffa38fd90dd7bbc4c3e5f3b)   ---|---      The graph on the left has a diameter of $2$, while the graph on the right has a diameter of $3$, so the answer is $3$.

In the second test case, there are no connected graphs, so the answer is $-1$.

Buggy Code:
#include <bits/stdc++.h>
#define rep(i, l, r) for (int i (l); i <= r; ++ i)
#define rrp(i, l, r) for (int i (r); i >= l; -- i)
#define pii pair <int, int>
#define eb emplace_back
#define ls p << 1
#define rs ls | 1
#define inf 1000000000
using namespace std;
constexpr int N = 4e5 + 5, B = 71, P = 998244353;
typedef unsigned long long ull;
inline int rd () {
  int x = 0, f = 1;
  char ch = getchar ();
  while (! isdigit (ch)) {
    if (ch == '-') f = -1;
    ch = getchar ();
  }
  while (isdigit (ch)) {
    x = (x << 1) + (x << 3) + ch - 48;
    ch = getchar ();
  }
  return x * f;
}
int qpow (int x, int y) {
  int ret = 1;
  for (; y; y >>= 1, x = x * x % P) if (y & 1) ret = ret * x % P;
  return ret;
}
int n, rt, p[N];
int f[N][3];
int l[N], r[N];
int ans;
char s[N];
void dfs (int u) {
  f[u][0] = f[u][1] = f[u][2] = 0;
  if (! u) return ;
  dfs (l[u]), dfs (r[u]);
  f[u][0] = f[l[u]][0];
  f[u][1] = f[r[u]][1];
  f[u][2] = f[l[u]][0] + f[r[u]][1];
  ans = max (ans, f[l[u]][1] + f[r[u]][0]);
  if (u ^ rt) {
    if (s[u] != 'R') {
      f[u][0] = max (f[u][0], max (f[l[u]][1], f[r[u]][0]) + 1);
      f[u][2] = max (f[u][2], max (f[r[u]][1] + f[l[u]][1] + 1, f[r[u]][2] + 1));
      ans = max (ans, max (f[l[u]][2] + 1, f[l[u]][0] + f[r[u]][0] + 1));
    }
    if (s[u] != 'L') {
      f[u][1] = max (f[u][1], max (f[l[u]][1], f[r[u]][0]) + 1);
      f[u][2] = max (f[u][2], max (f[r[u]][0] + f[l[u]][0] + 1, f[l[u]][2] + 1));
      ans = max (ans, max (f[r[u]][2] + 1, f[r[u]][1] + f[l[u]][1] + 1));
    }
  }
}
void solve () {
  ans = 0;
  n = rd ();
  rep (i, 1, n) p[i] = rd (), l[i] = r[i] = 0;
  scanf ("%s", s + 1);
  rt = 0;
  static int stk[N], top;
  rep (i, 1, n) {
    if (p[i] == n) rt = i;
    bool flag = 0;
    while (top && p[stk[top]] < p[i]) -- top, flag = 1;
    r[stk[top]] = i;
    if (flag) l[i] = stk[top + 1];
    stk[++ top] = i;
  } top = 0;
  bool chk = 1;
  rep (i, 1, n) {
    while (top && p[stk[top]] < p[i]) -- top;
    if (! top && s[i] == 'L') chk = 0; stk[++ top] = i;
  } top = 0;
  rrp (i, 1, n) {
    while (top && p[stk[top]] < p[i]) -- top;
    if (! top && s[i] == 'R') chk = 0; stk[++ top] = i;
  } top = 0;
  if (! chk) return void ( puts ("-1") );
  dfs (rt);
  printf ("%d\n", ans);
}
int main () {
  // freopen ("1.in", "r", stdin);
  // freopen ("1.out", "w", stdout);
  for (int T (rd ()); T; -- T) solve ();
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Spinning Round (Hard Version)
time_limit_ms: 7000
memory_limit_mb: 1024
Description: This is the hard version of the problem. The only difference between the two versions are the allowed characters in $s$. 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 $n$, where each character is either L, R, or ?.

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 over all connected$^{\text{∗}}$ 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 characters L, R, and ?.

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
R?RL?
2
1 2
LR
3
3 1 2
L?R
7
5 3 1 6 4 2 7
?R?R?R?
5
5 2 1 3 4
?????
6
6 2 3 4 5 1
?LLRLL
8
1 7 5 6 2 8 4 3
?R??????
12
6 10 7 1 8 5 12 2 11 3 4 9
????????????
output:
3
-1
-1
4
4
3
5
8

Note: In the first test case, there are two connected graphs (the labels are indices):

![](CDN_BASE_URL/4561c9025a4357785c7aff0bb1bbff9a)| ![](CDN_BASE_URL/a680f9e3bffa38fd90dd7bbc4c3e5f3b)   ---|---      The graph on the left has a diameter of $2$, while the graph on the right has a diameter of $3$, so the answer is $3$.

In the second test case, there are no connected graphs, so the answer is $-1$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define int long long
#define ff first
#define ss second
#define pb push_back
#define vi vector<int>
#define pii pair<int,int>
#define vii vector<pii>
#define yes cout<<"YES"<<'\n';
#define no cout<<"NO"<<'\n';
#define minit(v, x...) v = min({v, x})
#define maxit(v, x...) v = max({v, x})
const ll N = 2e5 + 5, M = 1e7 + 5, MOD = 1e9 + 7, oo = 1e18;

int32_t main() {
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin),
    freopen("output.txt", "w", stdout);
    #endif
    ios_base::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    cout << "3\n"
            "-1\n"
            "-1\n"
            "4\n"
            "4\n"
            "3\n"
            "5\n"
            "8" ;

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Spinning Round (Hard Version)
time_limit_ms: 7000
memory_limit_mb: 1024
Description: This is the hard version of the problem. The only difference between the two versions are the allowed characters in $s$. 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 $n$, where each character is either L, R, or ?.

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 over all connected$^{\text{∗}}$ 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 characters L, R, and ?.

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
R?RL?
2
1 2
LR
3
3 1 2
L?R
7
5 3 1 6 4 2 7
?R?R?R?
5
5 2 1 3 4
?????
6
6 2 3 4 5 1
?LLRLL
8
1 7 5 6 2 8 4 3
?R??????
12
6 10 7 1 8 5 12 2 11 3 4 9
????????????
output:
3
-1
-1
4
4
3
5
8

Note: In the first test case, there are two connected graphs (the labels are indices):

![](CDN_BASE_URL/4561c9025a4357785c7aff0bb1bbff9a)| ![](CDN_BASE_URL/a680f9e3bffa38fd90dd7bbc4c3e5f3b)   ---|---      The graph on the left has a diameter of $2$, while the graph on the right has a diameter of $3$, so the answer is $3$.

In the second test case, there are no connected graphs, so the answer is $-1$.

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

#define vi vector<int>
#define pb push_back
#define pii pair<int,int>
#define mkp make_pair

int rd() {
	int x = 0, f = 1;
	char ch = getchar();
	while (!('0' <= ch && ch <= '9')) {
		if (ch == '-') f = -1; ch = getchar();
	}
	while ('0' <= ch && ch <= '9') {
		x = (x << 1) + (x << 3) + (ch ^ 48); ch = getchar();
	}
	return x * f;
}

void wr(int x) {
	if (x < 0) putchar('-'), x = -x;
	if (x >= 10) wr(x / 10); putchar(x % 10 + '0');
}

const int N = 4e5 + 10;
const int inf = 1e9 + 1;

struct RMQ {
	int f[20][N];
	void init(int n) {
		for (int k = 1; k <= 20; ++k) {
			for (int i = 1; i <= n - (1 << k) + 1; ++i)
				f[k][i] = max(f[k - 1][i], f[k - 1][i + (1 << (k - 1))]);
		}
	}
	int get_mx(int l, int r) {
		int k = log2(r - l + 1);
		return max(f[k][l], f[k][r - (1 << k) + 1]);
	}
} T;

int ans;
string s;

int cm[N];
int faing[N], jump[N];
int p[N], q[N], f[N], g[N];

int ds[N][2];
int son[N][2], dis[N], disp[N], up[N], upp[N];

void Max (int &x, int y) {
	x = max(x, y);
}

int get_tree (int l, int r, int com) {
	if (l > r) return 0;
	int x = T.get_mx(l, r); x = q[x];
	cm[x] = com;
	son[x][0] = get_tree(l, x - 1, 0);
	son[x][1] = get_tree(x + 1, r, 1);
	return x;
}

void dfs1 (int u, int fa, int com, int tp) {
	int s = 0;
	dis[u] = disp[u] = ds[u][0] = ds[u][1] = 0;
	for (int p = 0; p < 2; ++p) {
		if (!son[u][p]) continue;
		int v = son[u][p];
		if (p == com || com == -1) dfs1 (v, u, p, tp);
		else dfs1(v, u, p, fa);
	}
	ans = max(ans, s);
	if (faing[u]) {
		dis[fa] = max(dis[fa], dis[u] + 1);
		ds[fa][com] = max(ds[fa][com], dis[u] + 1);
	}
	if (jump[u] && tp > 0) {
		dis[tp] = max(dis[tp], dis[u] + 1);
		disp[tp] = max(disp[tp], dis[u] + 1);
		ds[tp][com ^ 1] = max(ds[tp][com ^ 1], dis[u] + 1);
	}
	if (fa > 0 && tp < 0 && !faing[u]) ans = inf;
	ans = max(ans, ds[u][0] + ds[u][1]);
//	cout << u << " " << dis[u] << endl;
}

void dfs2 (int u, int fa, int com, int tp) {
	for (int p = 0; p < 2; ++p) {
		if (!son[u][p]) continue;
		int v = son[u][p];
		if (p == com || com == -1) {
			if (faing[v]) Max(f[v], 1 + f[u]), Max(g[v], 1);
			if (jump[v] && tp > 0) Max(f[v], 1), Max(g[v], 1 + f[u]);
			dfs2(v, u, p, tp);
		} else {
			if (faing[v]) Max(f[v], 1 + g[u]), Max(g[v], 1);
			if (jump[v]) Max(f[v], 1), Max(g[v], 1 + g[u]);
			dfs2(v, u, p, fa);
		}
	}
//	cout << p[u] << " " << f[u] << " " << g[u] << endl;
}

void dfs3 (int u, int fa, int com, int tp) {
	if (jump[u]) {
		ans = max(ans, dis[u] + 1 + up[u]);
		ans = max(ans, ds[u][com] + 1 + upp[u]);
	}
//	cout << p[u] << " " << dis[u] << " " << up[u] << " " << upp[u] << endl;
//	cout << ds[u][com] << endl;
	for (int p = 0; p < 2; ++p) {
		if (!son[u][p]) continue;
		int v = son[u][p];
		if (p == com || com == -1) {
			int w1 = disp[u];
			int w2 = ds[u][p ^ 1];
			up[v] = max(max(up[u], upp[u]), w2 + f[u]);
			upp[v] = max(up[v], w1 + f[u]);
			dfs3 (v, u, p, tp);
		} else {
			int w1 = disp[u];
			int w2 = ds[u][p ^ 1];
			up[v] = max(max(up[u], upp[u]), w2 + g[u]);
			upp[v] = max(up[v], w1 + g[u]);
			dfs3 (v, u, p, fa);
		}
	}
}

void solve() {
	int n;
	cin >> n; ans = 0;
	for (int i = 1; i <= n; ++i)
		cin >> p[i], q[p[i]] = i, up[i] = -inf, upp[i] = -inf;
	for (int i = 1; i <= n; ++i)
		f[i] = -inf, g[i] = -inf, jump[i] = faing[i] = 1;
	cin >> s; s = " " + s;
	for (int i = 1; i <= n; ++i)
		T.f[0][i] = p[i];
	T.init(n);
	int rt = get_tree(1, n, -1);
	for (int i = 1; i <= n; ++i) {
		if (s[i] == 'L') {
			if (cm[i] == 0) faing[i] = 0;
			else jump[i] = 0;
		} else if (s[i] == 'R') {
			if (cm[i] == 0) jump[i] = 0;
			else faing[i] = 0;
		}
	}
	g[rt] = 0;
	dfs1(rt, 0, -1, -1);
	dfs2(rt, 0, -1, -1);
	dfs3(rt, 0, -1, -1);
	if (ans == inf) ans = -1;
	cout << ans << endl;
}

int main() {
	int T;
	cin >> T;
	while (T--) solve();
	return 0;
}
//

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided code, we need to focus on edge cases or scenarios that can lead to incorrect behavior. The code is complex and involves multiple steps, including tree construction, depth-first searches, and dynamic programming. We should aim to create a test case that:

1. Forces the code to handle special characters (L, R, ?) in the string `s` in a way that can lead to incorrect graph construction.
2. Tests the edge conditions for connectedness and the calculation of the diameter.
3. Ensures that the permutation and string `s` are valid and within the constraints.

Let's consider the following test case:

- A permutation of length 5.
- A string `s` that includes L, R, and ? characters to test the handling of different edge cases.

Here is a small-scale test input that should expose potential issues:

```plaintext
1
5
2 1 4 3 5
R?RL?
```

### Explanation:

- **Permutation**: `2 1 4 3 5`
  - This permutation is chosen to have a mix of increasing and decreasing segments, which can make the calculation of `l_i` and `r_i` more complex.
- **String `s`**: `R?RL?`
  - The string includes a mix of L, R, and ? characters.
  - The '?' character allows for flexibility in edge addition, which can lead to different graph structures.
  - The 'R' and 'L' characters force specific edge additions, which can help test the correctness of the graph construction.

### Expected Output:
- The expected output for this test case is `3`.

This test case is designed to be small and manageable, yet it should be sufficient to expose any issues in the code's handling of the permutation and string `s`. If the code fails to produce the correct output, it indicates a potential bug in the algorithm or implementation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Spinning Round (Hard Version)
time_limit_ms: 7000
memory_limit_mb: 1024
Description: This is the hard version of the problem. The only difference between the two versions are the allowed characters in $s$. 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 $n$, where each character is either L, R, or ?.

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 over all connected$^{\text{∗}}$ 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 characters L, R, and ?.

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
R?RL?
2
1 2
LR
3
3 1 2
L?R
7
5 3 1 6 4 2 7
?R?R?R?
5
5 2 1 3 4
?????
6
6 2 3 4 5 1
?LLRLL
8
1 7 5 6 2 8 4 3
?R??????
12
6 10 7 1 8 5 12 2 11 3 4 9
????????????
output:
3
-1
-1
4
4
3
5
8

Note: In the first test case, there are two connected graphs (the labels are indices):

![](CDN_BASE_URL/4561c9025a4357785c7aff0bb1bbff9a)| ![](CDN_BASE_URL/a680f9e3bffa38fd90dd7bbc4c3e5f3b)   ---|---      The graph on the left has a diameter of $2$, while the graph on the right has a diameter of $3$, so the answer is $3$.

In the second test case, there are no connected graphs, so the answer is $-1$.

Buggy Code:
// LUOGU_RID: 167888596
#include<bits/stdc++.h>
using namespace std;
int read(){
	int x;char c;
	while((c=getchar())<'0'||c>'9');
	x=c^48;
	while((c=getchar())>='0'&&c<='9')x=(x<<3)+(x<<1)+(c^48);
	return x;
}
typedef long long ll;
const int maxn=400020;
int LOG[maxn];
int n,p[maxn],s[maxn],tag[maxn];
int mi[maxn][20];
int maxer(int x,int y){
	if(p[x]>p[y])return x;
	return y;
}
void st_init(){
	for(int i=1;i<=n;i++)
		mi[i][0]=i;
	int g=LOG[n],m;
	for(int i=1;i<=g;i++){
		m=n-(1<<i)+1;
		for(int j=1;j<=m;j++)
			mi[j][i]=maxer(mi[j][i-1],mi[j+(1<<(i-1))][i-1]);
	}
}
int query(int l,int r){
	int g=LOG[r-l];
	return maxer(mi[l][g],mi[r-(1<<g)+1][g]);
}
struct ANSWER{
	int al,ar,ac;
	ANSWER(){}
	ANSWER(int _al,int _ar,int _ac):al(_al),ar(_ar),ac(_ac){}
};
int maxs(int st,...){
	va_list x;
	va_start(x,st);
	int s=st,v;
	while(true){
		v=va_arg(x,int);
		if(v==-1)break;
		if(s<v)s=v;
	}
	va_end(x);
	return s;
}
ANSWER qr(int l,int r){
	if(l>r)return ANSWER(0,0,0);
	int d=query(l,r);
	ANSWER a=qr(l,d-1),b=qr(d+1,r);
	if(s[d]==0)return ANSWER(maxs(a.al,a.ar+1,b.al+1,-1),b.ar,maxs(a.ac,b.ac,a.al+b.al+1,a.al+b.ar+1,a.ar+b.al,a.ar+b.ar+2,-1));
	if(s[d]==1)return ANSWER(a.al,maxs(a.ar+1,b.al+1,b.ar,-1),maxs(a.ac,b.ac,a.al+b.al+2,a.al+b.ar+1,a.ar+b.al,a.ar+b.ar+1,-1));
	return ANSWER(maxs(a.al,a.ar+1,b.al+1,-1),maxs(a.ar+1,b.al+1,b.ar,-1),maxs(a.ac,b.ac,a.al+b.al+2,a.al+b.ar+1,a.ar+b.al,a.ar+b.ar+2,-1));
}
vector<ANSWER> qs;vector<int> qd,qh;
int solve(){
	n=read();
	for(int i=1;i<=n;i++)
		p[i]=read();
	for(int i=1;i<=n;i++){
		int c;
		while((c=getchar())!='L'&&c!='R'&&c!='?');
		if(c=='L')s[i]=0;
		else if(c=='R')s[i]=1;
		else s[i]=2;
		tag[i]=0;
	}
	st_init();
	int g=query(1,n),l=g;
	tag[g]=1;
	while(g!=n){
		g=query(g+1,n);
		if(s[g]==1)return -1;
		tag[g]=1;
	}
	while(l!=1){
		l=query(1,l-1);
		if(s[l]==0)return -1;
		tag[l]=1;
	}
	qs.clear();
	for(int i=2;i<=n;i++)
		if(tag[i]==1)
			qs.push_back(qr(l+1,i-1)),l=i;
	int ans=0,m=qs.size();
	qd.clear();qd.resize(m+1,0);
	for(int i=0;i<m;i++){
		if(ans<qs[i].ac)
			ans=qs[i].ac;
		if(qd[i]<qs[i].al)
			qd[i]=qs[i].al;
		if(qd[i+1]<qs[i].ar)
			qd[i+1]=qs[i].ar;
	}
	for(int i=1;i<m;i++)
		ans=maxs(ans,qs[i].al+qs[i-1].ar,qs[i].al+qs[i-1].al+1,qs[i].ar+qs[i-1].ar+1,-1);
	qh.clear();qh=qd;
	for(int i=1;i<=m;i++)
		if(qd[i]<qd[i-1]+1)
			qd[i]=qd[i-1]+1;
	for(int i=m-1;i>=0;i--)
		if(qh[i]<qh[i+1]+1)
			qh[i]=qh[i+1]+1;
	for(int i=1;i<m;i++)
		if(ans<qd[i-1]+qh[i+1]+2)
			ans=qd[i-1]+qh[i+1]+2;
	return ans;
}
int main(){
	LOG[0]=LOG[1]=0;
	for(int i=2;i<=400000;i++)
		LOG[i]=LOG[i>>1]+1;
	int t=read();
	while(t--)
		printf("%d\n",solve());
	return 0;
}

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

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

Think step by step.

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