**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<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
const int N=500000;
int p[N],q[N],ls[N],rs[N],ans,dp1[N],dp2[N],dp3[N],dp4[N],dp5[N];
char s[N];
void MAX(int&x,int y)
{
	if (x<y) x=y;
}
void dfs(int u)
{
	int v,w;
	bool l,r;
	if (!u) return;
	v=ls[u];
	dfs(v);
	w=rs[u];
	dfs(w);
	l=s[u-1]!='R';
	r=s[u-1]!='L';
	dp1[u]=dp2[u]=dp3[u]=dp4[u]=dp5[u]=0;
	MAX(dp1[u],dp1[v]);
	if (l) MAX(dp1[u],dp2[v]+1);
	if (r) MAX(dp2[u],dp2[v]+1);
	if (l) MAX(dp1[u],dp1[w]+1);
	if (r) MAX(dp2[u],dp1[w]+1);
	MAX(dp2[u],dp2[w]);
	MAX(dp3[u],dp1[v]+dp1[w]);
	if (l) MAX(dp3[u],dp1[v]+dp1[w]+1);
	if (r) MAX(dp4[u],dp1[v]+dp1[w]+1);
	MAX(dp4[u],dp1[v]+dp2[w]);
	MAX(ans,dp2[v]+dp1[w]);
	MAX(dp5[u],dp2[v]+dp2[w]);
	if (l) MAX(dp4[u],dp2[v]+dp2[w]+1);
	if (r) MAX(dp5[u],dp2[v]+dp2[w]+1);
	MAX(dp3[u],dp3[v]);
	MAX(dp4[u],dp4[v]);
	if (l) MAX(dp3[u],dp4[v]+1);
	if (r) MAX(dp4[u],dp4[v]+1);
	MAX(dp5[u],dp5[v]);
	if (l) MAX(dp4[u],dp5[v]+1);
	if (r) MAX(dp5[u],dp5[v]+1);
	MAX(ans,dp5[v]);
	MAX(dp3[u],dp3[w]);
	if (l) MAX(dp3[u],dp3[w]+1);
	if (r) MAX(dp4[u],dp3[w]+1);
	MAX(ans,dp3[w]);
	MAX(dp4[u],dp4[w]);
	if (l) MAX(dp4[u],dp4[w]+1);
	if (r) MAX(dp5[u],dp4[w]+1);
	MAX(dp5[u],dp5[w]);
}
int main()
{
	int T,n,i,b,rt;
	for (scanf("%d",&T);T;T--)
	{
		scanf("%d",&n);
		for (i=1;i<=n;i++) scanf("%d",&p[i]);
		scanf("%s",s);
		b=0;
		for (i=1;i<=n;i++)
		{
			if (b&&p[q[b-1]]<p[i])
			{
				for (;b;b--) if (p[q[b-1]]>p[i]) break;
				ls[i]=q[b];
			}
			else ls[i]=0;
			rs[i]=0;
			if (b) rs[q[b-1]]=i;
			else if (p[i]<n&&s[i-1]=='L')
			{
				printf("-1\n");
				break;
			}
			q[b++]=i;
		}
		if (i<=n) continue;
		for (i=1;i<b;i++) if (s[q[i]-1]=='R')
		{
			printf("-1\n");
			break;
		}
		if (i<b) continue;
		rt=q[0];
		ans=0;
		dfs(rt);
		printf("%d\n",ans);
	}
	return 0;
}

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

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

Think step by step.

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


**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<iostream>
#include<fstream>
#include<algorithm>
#include<vector>
#include<cstring>
#define int long long
#define eb emplace_back
#define mp make_pair
//#define DEBUG
using namespace std;
const int P = 998244353, inf = 0x3f3f3f3f3f3f3f3f;
using pii = pair<int, int>;
namespace azus{
	int n;
	vector<int> a;
	vector<int> lc;
	vector<int> rc;
	int f[400005][3];
	string s;
	int ans, root;
	int dfs(int u){
		if(!lc[u] && !rc[u]){
			if(s[u] == '?' || s[u] == 'L') f[u][0] = 1;
			if(s[u] == '?' || s[u] == 'R') f[u][1] = 1;
			return 0;
		}
		dfs(lc[u]), dfs(rc[u]);
		f[u][0] = f[lc[u]][0], f[u][1] = f[rc[u]][1];
		f[u][2] = max(f[u][2], f[lc[u]][0] + f[rc[u]][1]);
		if(s[u] != 'R'){
			f[u][0] = max(f[u][0], max(f[lc[u]][1], f[rc[u]][0]) + 1);
			f[u][2] = max(f[u][2], max(f[lc[u]][1] + f[rc[u]][1], f[rc[u]][2]) + 1);
		}
		if(s[u] != 'L'){
			f[u][1] = max(f[u][1], max(f[lc[u]][1], f[rc[u]][0]) + 1);
			f[u][2] = max(f[u][2], max(f[lc[u]][0] + f[rc[u]][0], f[rc[u]][2]) + 1);
		}

		ans = max(ans, f[lc[u]][1] + f[rc[u]][0]);
		if(u != root){
			if(s[u] != 'R')
				ans = max(ans, max(f[lc[u]][0] + f[rc[u]][0], f[lc[u]][2]) + 1);
			if(s[u] != 'L')
				ans = max(ans, max(f[lc[u]][1] + f[rc[u]][1], f[rc[u]][2]) + 1);
		}

		return 0;
	}
	bool Flg;
	int main(){
		cin >> n;
		a.assign(n + 5, 0), lc.assign(n + 5, 0), rc.assign(n + 5, 0);
		for(int i = 1; i <= n; i ++)
			cin >> a[i];
		vector<int> st(n + 5, 0);
		int top = 0;
		for(int i = 1; i <= n; i ++){
			int nw = top;
			while(nw > 0 && a[st[nw]] < a[i])
				nw --;
			if(nw) rc[st[nw]] = i;
			if(nw < top) lc[i] = st[nw + 1];
			st[++ nw] = i, top = nw;
		}
		root = max_element(a.begin() + 1, a.begin() + n + 1) - a.begin();
		vector<int> cr(n + 5, -1);
		int u = lc[root];
		while(u) cr[u] = 0, u = lc[u];
		u = rc[root];
		while(u) cr[u] = 1, u = rc[u];
		cin >> s, s = " " + s;
		for(int i = 1; i <= n; i ++){
			if(s[i] == 'L' && cr[i] == 0) return cout << "-1\n", 0;
			if(s[i] == 'R' && cr[i] == 1) return cout << "-1\n", 0;
			if(cr[i] == 0) s[i] = 'R';
			if(cr[i] == 1) s[i] = 'L';
		}
		if(Flg){
			cout << n << s << "\n";
			return 0;
		}
		ans = 0;
		dfs(root);
		cout << ans << "\n";
		for(int i = 1; i <= n; i ++)
			f[i][0] = f[i][1] = f[i][2] = -inf;
		return 0;
	}
}
signed main(){
#ifndef DEBUG
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
#endif
	int T = 1;
	cin >> T;
	int x = 0;
	while(T --){
		x ++;
		if(x == 185)
			azus::Flg = 1;
		azus::main();
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
5
5 4 3 2 1
R?R?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<iostream>
#include<fstream>
#include<algorithm>
#include<vector>
#include<cstring>
#define int long long
#define eb emplace_back
#define mp make_pair
//#define DEBUG
using namespace std;
const int P = 998244353, inf = 0x3f3f3f3f3f3f3f3f;
using pii = pair<int, int>;
namespace azus{
	int n;
	vector<int> a;
	vector<int> lc;
	vector<int> rc;
	int f[400005][3];
	string s;
	int ans, root;
	int dfs(int u){
		if(!lc[u] && !rc[u]){
			if(s[u] == '?' || s[u] == 'L') f[u][0] = 1;
			if(s[u] == '?' || s[u] == 'R') f[u][1] = 1;
			return 0;
		}
		dfs(lc[u]), dfs(rc[u]);
		f[u][0] = f[lc[u]][0], f[u][1] = f[rc[u]][1];
		f[u][2] = max(f[u][2], f[lc[u]][0] + f[rc[u]][1]);
		if(s[u] != 'R'){
			f[u][0] = max(f[u][0], max(f[lc[u]][1], f[rc[u]][0]) + 1);
			f[u][2] = max(f[u][2], max(f[lc[u]][1] + f[rc[u]][1], f[rc[u]][2]) + 1);
		}
		if(s[u] != 'L'){
			f[u][1] = max(f[u][1], max(f[lc[u]][1], f[rc[u]][0]) + 1);
			f[u][2] = max(f[u][2], max(f[lc[u]][0] + f[rc[u]][0], f[rc[u]][2]) + 1);
		}

		ans = max(ans, f[lc[u]][1] + f[rc[u]][0]);
		if(u != root){
			if(s[u] != 'R')
				ans = max(ans, max(f[lc[u]][0] + f[rc[u]][0], f[lc[u]][2]) + 1);
			if(s[u] != 'L')
				ans = max(ans, max(f[lc[u]][1] + f[rc[u]][1], f[rc[u]][2]) + 1);
		}

		return 0;
	}
	bool Flg;
	int main(){
		cin >> n;
		a.assign(n + 5, 0), lc.assign(n + 5, 0), rc.assign(n + 5, 0);
		for(int i = 1; i <= n; i ++)
			cin >> a[i];
		if(Flg == 1){
			for(int i = 4; i <= n; i ++)
				cout << a[i] << " ";
			cout << "\n";
			return 0;
		}
		vector<int> st(n + 5, 0);
		int top = 0;
		for(int i = 1; i <= n; i ++){
			int nw = top;
			while(nw > 0 && a[st[nw]] < a[i])
				nw --;
			if(nw) rc[st[nw]] = i;
			if(nw < top) lc[i] = st[nw + 1];
			st[++ nw] = i, top = nw;
		}
		root = max_element(a.begin() + 1, a.begin() + n + 1) - a.begin();
		vector<int> cr(n + 5, -1);
		int u = lc[root];
		while(u) cr[u] = 0, u = lc[u];
		u = rc[root];
		while(u) cr[u] = 1, u = rc[u];
		cin >> s, s = " " + s;
		for(int i = 1; i <= n; i ++){
			if(s[i] == 'L' && cr[i] == 0) return cout << "-1\n", 0;
			if(s[i] == 'R' && cr[i] == 1) return cout << "-1\n", 0;
			if(cr[i] == 0) s[i] = 'R';
			if(cr[i] == 1) s[i] = 'L';
		}
		ans = 0;
		dfs(root);
		cout << ans << "\n";
		for(int i = 1; i <= n; i ++)
			f[i][0] = f[i][1] = f[i][2] = 0;
		return 0;
	}
}
signed main(){
#ifndef DEBUG
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
#endif
	int T = 1;
	cin >> T;
	int x = 0;
	while(T --){
		x ++;
		if(x == 185)
			azus::Flg = 1;
		azus::main();
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
3
3 1 2
L?R
4
4 2 3 1
?????
5
5 3 4 1 2
?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 <iostream>
#include <cstdio>
#include <set>
#include <map>
using namespace std;
const int maxn = 4e5+10;
int p[maxn],tmp[maxn],l[maxn],r[maxn],n; char s[maxn];
typedef pair<int,int> pii; pii dp[maxn][3];
map<pii,int> mp; int tot = 0;
int main() {
	cin.tie(0)->sync_with_stdio(false);
	int T; scanf("%d",&T);
	while(T--) {
		scanf("%d",&n); for(int i = 1; i <= n; i++) {
			scanf("%d",&p[i]); tmp[p[i]] = i;
		} scanf("%s",s+1); set<int> ss;
		ss.clear(); ss.insert(0),ss.insert(n+1);
		for(int i = n; i >= 1; i--) {
			auto x = ss.lower_bound(tmp[i]);
			r[tmp[i]] = *x; x--; l[tmp[i]] = *x;
			ss.insert(tmp[i]);
		} mp.clear(); tot = 0; int ans = 0;
		bool ok = true;
		for(int i = 1; i <= n; i++) {
			if(!l[i] && s[i] == 'L') {cout << -1 << endl; ok = false; break;}
			if(r[i] == n+1 && s[i] == 'R') {
				cout << -1 << endl; ok = false; break;
			}
			if(!mp[{l[i],i}]) mp[{l[i],i}] = ++tot;
			if(!mp[{i,r[i]}]) mp[{i,r[i]}] = ++tot;
			if(!mp[{l[i],r[i]}]) mp[{l[i],r[i]}] = ++tot;
		}
		for(int i = 1; i <= tot; i++) {
			dp[i][0] = {0,0}; dp[i][1] = {0,0}; dp[i][2] = {0,0};
		}
		if(!ok) continue;
		for(int i = 1; i <= n; i++) {
			int x = tmp[i],ll = l[x],rr = r[x];
			int now = mp[{ll,rr}],lll = mp[{ll,x}],rrr = mp[{x,rr}];
			dp[now][0] = dp[now][1] = dp[now][2] = {-1,-1};
			for(auto u:dp[lll]) {
				for(auto v:dp[rrr]) {
					ans = max(ans,u.second+v.first);
					if(s[x] != 'R' && ll >= 1) { //apply left
						ans = max(ans,u.first+max(u.second,v.first)+1);
						pii hom = {max(max(u.second,v.first)+1,u.first),v.second};
						if(hom >= dp[now][0]) dp[now][0] = hom;
						//if(hom.first > dp[now][1].first || (hom.second > dp[now][1].second && hom.first == dp[now][1].first)) dp[now][1] = hom;
						if(hom.second > dp[now][1].second || (hom.first > dp[now][1].first && hom.second == dp[now][1].second)) dp[now][1] = hom;
						if(hom.first+hom.second > dp[now][2].first+dp[now][2].second) dp[now][2] = hom;
						//cout << hom.first << ' ' << hom.second << endl;
					}
					if(s[x] != 'L' && rr <= n) { //apply right
						ans = max(ans,v.second+max(u.second,v.first)+1);
						pii hom = {u.first,max(max(u.second,v.first)+1,v.second)};
						if(hom >= dp[now][0]) dp[now][0] = hom;
						//if(hom.first > dp[now][1].first || (hom.second > dp[now][1].second && hom.first == dp[now][1].first)) dp[now][1] = hom;
						if(hom.second > dp[now][1].second || (hom.first > dp[now][1].first && hom.second == dp[now][1].second)) dp[now][1] = hom;
						if(hom.first+hom.second > dp[now][2].first+dp[now][2].second) dp[now][2] = hom;
						//cout << u.first << ' ' << u.second << ' ' << v.first << ' ' << v.second << endl;
						//cout << hom.first << ' ' << hom.second << endl;
					}
				}
			}
			//cout << l[i] << ' ' << r[i] << endl;
			//cout << dp[now][0].first << ' ' << dp[now][0].second << ' ' << dp[now][1].first << ' ' << dp[now][1].second << ' ' << dp[now][2].first << ' ' << dp[now][2].second << endl;
		}
		cout << ans << endl;
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
5
2 1 4 3 5
R?RL?
6
6 2 3 4 5 1
?LLRLL
8
1 7 5 6 2 8 4 3
?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;

const int N = 1e5 + 10;

int T, n, Rt, tmp, ans;
int p[N], log_2[N];
int ls[N], rs[N], Lf[N], Rf[N];
pair<int, int> st[N][20];
int h[N], e[N << 1], ne[N << 1], idx;
int f[N][3];
char s[N];

void add(int a, int b) {
	e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
}

pair<int, int> ST(int l, int r) {
	int k = log_2[r - l + 1];
	return max(st[l][k], st[r - (1 << k) + 1][k]);
}

int dfs(int l, int r) {
	if (l == r) {
		ls[l] = rs[l] = 0;
		return l;
	}
	if (l > r) return 0;
	int Rt = ST(l, r).second;
	ls[Rt] = dfs(l, Rt - 1);
	rs[Rt] = dfs(Rt + 1, r);
	return Rt;
}

void Dfs(int x, int L, int R) {
	if (!x) return;
	Lf[x] = L, Rf[x] = R;
	if (s[x] == 'L') {
		if (Lf[x]) tmp ++;
	}
	if (s[x] == 'R') {
		if (Rf[x]) tmp ++;
	}
	if (s[x] == '-') {
		if (!Lf[x] && Rf[x]) {
			s[x] = 'R';
			tmp ++;
		}
		if (Lf[x] && !Rf[x]) {
			s[x] = 'L';
			tmp ++;
		}
		if (Lf[x] && Rf[x]) {
			tmp ++;
		}
	}
	Dfs(ls[x], L, x);
	Dfs(rs[x], x, R);
}

void Dp(int Rt) {
	if (!Rt) return;
	if (ls[Rt]) Dp(ls[Rt]);
	if (rs[Rt]) Dp(rs[Rt]);

	f[Rt][0] = max(f[Rt][0], f[ls[Rt]][0]);
	f[Rt][1] = max(f[Rt][1], f[rs[Rt]][1]);
	f[Rt][2] = max(f[Rt][2], f[ls[Rt]][0] + f[rs[Rt]][1]);

	if (s[Rt] != 'R') {
		f[Rt][0] = max(f[Rt][0], max(f[ls[Rt]][1], f[rs[Rt]][0]) + 1);
		f[Rt][2] = max(f[Rt][2], max(f[ls[Rt]][1] + f[rs[Rt]][1], f[rs[Rt]][2]) + 1);
		ans = max(ans, f[ls[Rt]][0] + f[rs[Rt]][0] + 1);
		ans = max(ans, f[ls[Rt]][2] + 1);
	}
	if (s[Rt] != 'L') {
		f[Rt][1] = max(f[Rt][1], max(f[ls[Rt]][1], f[rs[Rt]][0]) + 1);
		f[Rt][2] = max(f[Rt][2], max(f[ls[Rt]][0] + f[rs[Rt]][0], f[ls[Rt]][2]) + 1);
		ans = max(ans, f[ls[Rt]][1] + f[rs[Rt]][1] + 1);
		ans = max(ans, f[rs[Rt]][2] + 1);
	}
	if (!ls[Rt] && !rs[Rt]) f[Rt][0] = f[Rt][1] = f[Rt][2] = 0;
//	cout << Rt << '-' << s[Rt] << '-' << f[Rt][0] << '-' << f[Rt][1] << '-' << f[Rt][2] << endl;
	ans = max(ans, (f[ls[Rt]][1] ? f[ls[Rt]][1] + 1 : 0) + (f[rs[Rt]][0] ? f[rs[Rt]][0] + 1 : 0));
}

int main() {
	mt19937 Rnd(time(NULL));

	scanf("%d", &T);

	while (T --) {
		idx = 0;
		memset(h, -1, sizeof h);
		memset(f, 0, sizeof f);

		scanf("%d", &n);
		for (int i = 1; i <= n; i ++) scanf("%d", &p[i]);
		scanf("%s", s + 1);
		for (int i = 1; i <= n; i ++)
			if (s[i] == '?') s[i] = '-';

		for (int i = 2; i <= n; i ++) log_2[i] = log_2[i / 2] + 1;
		for (int i = 1; i <= n; i ++) st[i][0] = make_pair(p[i], i);
		for (int j = 1; j < 20; j ++)
			for (int i = 1; i + (1 << j) - 1 <= n; i ++)
				st[i][j] = max(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
		Rt = dfs(1, n);

		tmp = 0;
		Dfs(Rt, 0, 0);

		ans = 0;
		if (tmp < n - 1) puts("-1");
		else {
			f[0][0] = f[0][1] = f[0][2] = 0;
			Dp(Rt);
			printf("%d\n", ans);
		}
	}

	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
6
3 1 4 2 5 6
L?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>
using namespace std;
int p[400005];
char s[400005];
int st[400005];
int ls[400005],rs[400005],fa[400005],dep[400005];
int hb[400005];
int f[400005][3];
int ans;
void dfs(int x){
	if(ls[x])dfs(ls[x]);
	if(rs[x])dfs(rs[x]);
	f[x][0]=(s[x]!='R');
	f[x][1]=(s[x]!='L');
	int l=ls[x],r=rs[x];
	f[x][0]=max(f[x][0],f[l][0]);
	f[x][1]=max(f[x][1],f[r][1]);
	f[x][2]=max(f[x][2],f[l][0]+f[r][1]);
	if(s[x]!='R')f[x][0]=max(f[x][0],max(f[l][1],f[r][0])+1);
	if(s[x]!='L')f[x][1]=max(f[x][1],max(f[l][1],f[r][0])+1);
	if(s[x]!='L')f[x][2]=max(f[x][2],f[l][2]+1);
	if(s[x]!='R')f[x][2]=max(f[x][2],f[r][2]+1);
	if(s[x]!='L')f[x][2]=max(f[x][2],f[l][0]+1);
	if(s[x]!='R')f[x][2]=max(f[x][2],f[r][1]+1);
	if(s[x]!='R')f[x][2]=max(f[x][2],f[l][1]+f[r][1]+1);
	if(s[x]!='L')f[x][2]=max(f[x][2],f[l][0]+f[r][0]+1);
	ans=max(ans,max(0,f[l][1])+max(0,f[r][0]));
	if(fa[x]){
		if(x==ls[fa[x]]&&s[x]!='L'){
			ans=max(ans,f[r][2]+1);
			ans=max(ans,f[l][1]+f[r][1]+1);
		}
		if(x==rs[fa[x]]&&s[x]!='R'){
			ans=max(ans,f[l][2]+1);
			ans=max(ans,f[l][0]+f[r][0]+1);
		}
	}
}
int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		int n;
		scanf("%d",&n);
		for(int i=1;i<=n;++i)scanf("%d",&p[i]);
		scanf("%s",s+1);
		int tt=0;
		for(int i=1;i<=n;++i)ls[i]=rs[i]=fa[i]=dep[i]=0;
		for(int i=1;i<=n;++i){
			int ll=0;
			while(tt&&p[st[tt]]<p[i])ll=st[tt--];
			ls[i]=ll,fa[ll]=i;
			rs[st[tt]]=i,fa[i]=st[tt];
			st[++tt]=i;
		}
		int rt=st[1];
		stack<int>ss;
		while(ss.size())ss.pop();
		for(int i=1;i<=n;++i)hb[i]=-fa[i];
		for(int i=1;i<=n;++i){
			while(ss.size()&&p[ss.top()]<p[i])ss.pop();
			if(ss.size())hb[i]+=ss.top();
			ss.emplace(i);
		}
		while(ss.size())ss.pop();
		for(int i=n;i>=1;--i){
			while(ss.size()&&p[ss.top()]<p[i])ss.pop();
			if(ss.size())hb[i]+=ss.top();
			ss.emplace(i);
		}
		int fl=1;
		for(int i=1;i<=n;++i)if(i!=rt){
			if(!hb[i]){
				if(fa[i]<i){
					if(s[i]=='R')fl=0;
					s[i]='L';
				}else{
					if(s[i]=='L')fl=0;
					s[i]='R';
				}
			}
		}
		if(!fl){
			puts("-1");
			continue;
		}
		for(int i=0;i<=n;++i)for(int j=0;j<3;++j)f[i][j]=-1e9;
		ans=0;dfs(rt);
		printf("%d\n",ans);
	}
	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 3 1 4 2
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>
using namespace std;
const int N=1e5+5,inf=1e9;
int T,n,p[N],l[N],r[N],L[N],R[N],ans,w;
char s[N];
int f[N][3];
stack<int> S;
int c1[N],c2[N];
void add1(int x,int y) {
	while(x<=n) c1[x]=min(c1[x],y),x+=x&-x;
}
void add2(int x,int y) {
	while(x<=n) c2[x]=max(c2[x],y),x+=x&-x;
}
int Min(int x) {
	int res=inf;
	while(x) res=min(res,c1[x]),x-=x&-x;
	return res;
}
int Max(int x) {
	int res=0;
	while(x) res=max(res,c2[x]),x-=x&-x;
	return res;
}
void dfs(int x) {
	f[x][0]=f[x][1]=f[x][2]=0;
	if(!x) return;
	dfs(l[x]);dfs(r[x]);
	ans=max(ans,f[l[x]][1]+f[r[x]][0]);
	f[x][0]=max(f[x][0],f[l[x]][0]);
	f[x][1]=max(f[x][1],f[r[x]][1]);
	f[x][2]=max(f[x][2],f[l[x]][0]+f[r[x]][1]);
	if(x!=w) {
		if(s[x]!='R') {
			f[x][0]=max(f[x][0],max(f[l[x]][1],f[r[x]][0])+1);
			f[x][2]=max(f[x][2],max(f[r[x]][2],f[l[x]][1]+f[r[x]][1])+1);
			ans=max(ans,max(f[l[x]][2],f[l[x]][0]+f[r[x]][0])+1);
		}
		if(s[x]!='L') {
			f[x][1]=max(f[x][1],max(f[l[x]][1],f[r[x]][0])+1);
			f[x][2]=max(f[x][2],max(f[l[x]][2],f[l[x]][0]+f[r[x]][0])+1);
			ans=max(ans,max(f[r[x]][2],f[l[x]][1]+f[r[x]][1])+1);
		}
	}
}
int main() {
	scanf("%d",&T);
	while(T--) {
		bool fl=0;ans=0;
		scanf("%d",&n);
		for(int i=1;i<=n;i++) l[i]=0,r[i]=0,c1[i]=inf,c2[i]=0;
		for(int i=1;i<=n;i++) scanf("%d",&p[i]);
		scanf("%s",s+1);w=1;S.push(1);
		for(int i=2;i<=n;i++) {
			while(!S.empty()) {
				if(p[S.top()]>p[i]) break;
				S.pop();
			}
			if(S.empty()) {
				l[i]=w;
				w=i;
			} else {
				int x=S.top();
				l[i]=r[x];
				r[x]=i;
			}
			S.push(i);
		}
		while(!S.empty()) S.pop();
		for(int i=1;i<=n;i++) {
			L[i]=Max(n-p[i]+1);
			add2(n-p[i]+1,i);
		}
		for(int i=n;i>=1;i--) {
			R[i]=Min(n-p[i]+1);
			add1(n-p[i]+1,i);
		}
		for(int i=1;i<=n;i++) {
			if(p[i]==n) continue;
			if(s[i]=='L') {
				if(!L[i]) {
					fl=1;
					break;
				}
			} else if(s[i]=='R') {
				if(R[i]==inf) {
					fl=1;
					break;
				}
			}
		}
		if(fl) {
			puts("-1");
			continue;
		}
		dfs(w);
		printf("%d\n",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
1
5
3 1 4 2 5
R?RL?
```


**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 = 1e6 + 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' && i != rt) chk = 0; stk[++ top] = i;
  } top = 0;
  rrp (i, 1, n) {
    while (top && p[stk[top]] < p[i]) -- top;
    if (! top && s[i] == 'R' && i != rt) 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
2 1 4 3 5
R?RL?
```

### Explanation:
This test case is designed to expose issues in the buggy code by focusing on a permutation and a string that create a complex scenario for the algorithm to handle. Here's a breakdown of the input:

- **n = 5**: The length of the permutation and the string.
- **Permutation = [2, 1, 4, 3, 5]**: This permutation is chosen to create a non-trivial tree structure when the edges are added based on the string.
- **String = "R?RL?"**: The string contains a mix of 'R', 'L', and '?'. The '?' characters provide flexibility in edge addition, which can be used to test the algorithm's ability to maximize the diameter.

### Expected Behavior:
- The algorithm should correctly determine the maximum possible diameter of the graph by considering all possible ways to add edges for the '?' characters.
- The correct output for this test case should be `3`, as the maximum diameter can be achieved by adding edges in a way that maximizes the distance between the vertices.

### Why This Input is Challenging:
- The permutation and string combination create a scenario where the algorithm must carefully consider the placement of edges to maximize the diameter.
- The '?' characters add complexity, as the algorithm must evaluate multiple possible edge configurations.
- The permutation and string are designed to test the algorithm's ability to handle cases where the root of the tree (the maximum element in the permutation) is not at the beginning or end of the sequence.


**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;

int T;

int n;
int p[1000001];
char ch[1000001];
struct ctre
{
	int ls,rs;
	int f[3];
}a[1000001];
stack<int> s;
inline int build()
{
	while(!s.empty())s.pop();
	for(int i=1;i<=n;i++)
	{
		a[i].ls=a[i].rs=0;
		if(i==1)
		{
			s.push(i);continue;
		}
		int x=0;
		while(!s.empty() and p[s.top()]<p[i])
		{
			x=s.top();s.pop();
		}
		if(!s.empty())a[s.top()].rs=i;
		s.push(i);a[i].ls=x;
	}
	while(s.size()>1)s.pop();
	return s.top();
}

int ans,root;

inline void dfs(int u)
{
	a[u].f[0]=a[u].f[1]=a[u].f[2]=0;
	if(a[u].ls)dfs(a[u].ls);
	if(a[u].rs)dfs(a[u].rs);

	a[u].f[0]=a[a[u].ls].f[0];
	a[u].f[1]=a[a[u].rs].f[1];
	a[u].f[2]=a[a[u].ls].f[0]+a[a[u].rs].f[1];

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

	if(u==root)return;

	if(ch[u]!='R')
	{
		a[u].f[0]=max(a[u].f[0],max(a[a[u].ls].f[1],a[a[u].rs].f[0])+1);
		a[u].f[2]=max(a[u].f[2],max(a[a[u].ls].f[1]+a[a[u].rs].f[1],a[a[u].rs].f[2])+1);

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

	if(ch[u]!='L')
	{
		a[u].f[1]=max(a[u].f[1],max(a[a[u].ls].f[1],a[a[u].rs].f[0])+1);

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

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

int L[1000001],R[1000001];

int main()
{
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n);
		for(int i=1;i<=n;i++)scanf("%d",&p[i]),L[i]=R[i]=0;
		scanf("%s",ch+1);
		while(!s.empty())s.pop();
		s.push(1);
		for(int i=2;i<=n;i++)
		{
			while(!s.empty())
			{
				if(p[s.top()]<=p[i])
				{
					s.pop();continue;
				}

				break;
			}

			if(!s.empty())
			{
				L[i]=s.top();
			}
			s.push(i);
		}
		while(!s.empty())s.pop();
		s.push(n);
		for(int i=n-1;i>=1;i--)
		{
			while(!s.empty())
			{
				if(p[s.top()]<=p[i])
				{
					s.pop();continue;
				}

				break;
			}

			if(!s.empty())
			{
				R[i]=s.top();
			}
			s.push(i);
		}

		int num1=0,num2=0;

		for(int i=1;i<=n;i++)
		{
			if(ch[i]=='L' and L[i]==0)num1++;
			if(ch[i]=='R' and R[i]==0)num1++;
			if(ch[i]=='?' and L[i]==0 and R[i]==0)num2++;
		}
		if(num1+num2>=2)
		{
			printf("-1\n");continue;
		}
		ans=0;
		root=build();
		dfs(root);

		printf("%d\n",ans);
	}
	return 0;
}

/*
1
6
6 2 3 4 5 1
-LLRLL

*/

**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 1 3 2 4
?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;

const int MAXN = 4e5 + 5;
vector<int> adj[MAXN];
bool visited[MAXN];
int dist[MAXN];

// Find furthest node and its distance using BFS
pair<int, int> bfs(int start, int n) {
    for(int i = 1; i <= n; i++) {
        visited[i] = false;
        dist[i] = 0;
    }

    queue<int> q;
    q.push(start);
    visited[start] = true;

    int maxDist = 0, furthestNode = start;
    while(!q.empty()) {
        int v = q.front();
        q.pop();

        for(int u : adj[v]) {
            if(!visited[u]) {
                visited[u] = true;
                dist[u] = dist[v] + 1;
                if(dist[u] > maxDist) {
                    maxDist = dist[u];
                    furthestNode = u;
                }
                q.push(u);
            }
        }
    }
    return {furthestNode, maxDist};
}

// Check if graph is connected and return diameter
int getDiameter(int n) {
    // Check connectivity
    for(int i = 1; i <= n; i++) visited[i] = false;
    queue<int> q;
    q.push(1);
    visited[1] = true;
    int count = 1;

    while(!q.empty()) {
        int v = q.front();
        q.pop();
        for(int u : adj[v]) {
            if(!visited[u]) {
                visited[u] = true;
                q.push(u);
                count++;
            }
        }
    }

    if(count != n) return -1;  // Graph is not connected

    // Find diameter using two BFS
    auto furthestFromOne = bfs(1, n);
    auto furthestFromFurthest = bfs(furthestFromOne.first, n);
    return furthestFromFurthest.second;
}

void solve() {
    int n;
    cin >> n;
    vector<int> p(n+1);
    for(int i = 1; i <= n; i++) {
        cin >> p[i];
    }
    string s;
    cin >> s;
    s = " " + s;  // 1-based indexing

    // Calculate l_i and r_i
    vector<int> l(n+1), r(n+1);
    for(int i = 1; i <= n; i++) {
        // Find l_i
        l[i] = i;
        for(int j = i-1; j >= 1; j--) {
            if(p[j] > p[i]) {
                l[i] = j;
                break;
            }
        }

        // Find r_i
        r[i] = i;
        for(int j = i+1; j <= n; j++) {
            if(p[j] > p[i]) {
                r[i] = j;
                break;
            }
        }
    }

    int qCount = count(s.begin()+1, s.end(), '?');
    int maxDiameter = -1;

    // Try all possible combinations for '?'
    for(int mask = 0; mask < (1 << qCount); mask++) {
        // Clear previous graph
        for(int i = 1; i <= n; i++) adj[i].clear();

        int qIdx = 0;
        for(int i = 1; i <= n; i++) {
            char direction = s[i];
            if(direction == '?') {
                direction = (mask & (1 << qIdx)) ? 'L' : 'R';
                qIdx++;
            }

            if(direction == 'L') {
                adj[i].push_back(l[i]);
                adj[l[i]].push_back(i);
            } else {
                adj[i].push_back(r[i]);
                adj[r[i]].push_back(i);
            }
        }

        int diameter = getDiameter(n);
        maxDiameter = max(maxDiameter, diameter);
    }

    cout << maxDiameter << "\n";
}

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

    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
6
6 2 3 4 5 1
?LLRLL
```

**Explanation:**

This test case is designed to challenge the given buggy code in several ways:

1. **Permutation and String Combination**: The permutation is chosen such that it has a clear pattern, making it easy to understand the positions of `l_i` and `r_i`. The string `?LLRLL` includes a mix of `L`, `R`, and `?`, which requires the code to handle the `?` characters by considering all possible combinations.

2. **Edge Case for `?` Handling**: The presence of a single `?` in the string means that the code must correctly handle the two possible scenarios: choosing `L` or `R` for the `?` character. This tests whether the code correctly iterates through all combinations and updates the graph accordingly.

3. **Graph Connectivity**: The permutation and string combination are chosen to ensure that the graph can be connected in both scenarios (choosing `L` or `R` for the `?`). This tests whether the code correctly checks for graph connectivity and calculates the diameter.

4. **Diameter Calculation**: The permutation and string combination are designed to result in a graph with a specific diameter, which is 3 in this case. This tests whether the code correctly calculates the diameter using the BFS approach.

By using this test case, we can effectively expose any issues in the code's handling of `?` characters, graph connectivity, and diameter calculation.


**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 = 4e5 + 10;

int T, n, Rt, tmp, ans;
int p[N], log_2[N];
int ls[N], rs[N], Lf[N], Rf[N];
pair<int, int> st[N][20];
int h[N], e[N << 1], ne[N << 1], idx;
int f[N][3];
char s[N];

inline void add(int a, int b) {
	e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
}

inline pair<int, int> ST(int l, int r) {
	int k = log_2[r - l + 1];
	return max(st[l][k], st[r - (1 << k) + 1][k]);
}

int dfs(int l, int r) {
	if (l == r) {
		ls[l] = rs[l] = 0;
		return l;
	}
	if (l > r) return 0;
	int Rt = ST(l, r).second;
	ls[Rt] = dfs(l, Rt - 1);
	rs[Rt] = dfs(Rt + 1, r);
	return Rt;
}

void Dfs(int x, int L, int R) {
	if (!x) return;
	Lf[x] = L, Rf[x] = R;
	if (s[x] == 'L') {
		if (Lf[x]) tmp ++;
	}
	if (s[x] == 'R') {
		if (Rf[x]) tmp ++;
	}
	if (s[x] == '-') {
		if (!Lf[x] && Rf[x]) {
			s[x] = 'R';
			tmp ++;
		}
		if (Lf[x] && !Rf[x]) {
			s[x] = 'L';
			tmp ++;
		}
		if (Lf[x] && Rf[x]) {
			tmp ++;
		}
	}
	Dfs(ls[x], L, x);
	Dfs(rs[x], x, R);
}

void Dp(int Rt, int RRR) {
	if (!Rt) return;
	if (ls[Rt]) Dp(ls[Rt], RRR);
	if (rs[Rt]) Dp(rs[Rt], RRR);

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

	if (s[Rt] != 'R' && RRR != Rt) {
		f[Rt][0] = max(f[Rt][0], max(f[ls[Rt]][1], f[rs[Rt]][0]) + 1);
		f[Rt][2] = max(f[Rt][2], max(f[ls[Rt]][1] + f[rs[Rt]][1], f[rs[Rt]][2]) + 1);
		ans = max(ans, f[ls[Rt]][0] + f[rs[Rt]][0] + 1);
		ans = max(ans, f[ls[Rt]][2] + 1);
	}
	if (s[Rt] != 'L' && RRR != Rt) {
		f[Rt][1] = max(f[Rt][1], max(f[ls[Rt]][1], f[rs[Rt]][0]) + 1);
		f[Rt][2] = max(f[Rt][2], max(f[ls[Rt]][0] + f[rs[Rt]][0], f[ls[Rt]][2]) + 1);
		ans = max(ans, f[ls[Rt]][1] + f[rs[Rt]][1] + 1);
		ans = max(ans, f[rs[Rt]][2] + 1);
	}
//	cout << Rt << '-' << s[Rt] << '-' << f[Rt][0] << '-' << f[Rt][1] << '-' << f[Rt][2] << endl;
}

int main() {
	scanf("%d", &T);

	while (T --) {
		idx = 0;
		memset(h, -1, sizeof h);
		memset(f, 0, sizeof f);

		scanf("%d", &n);
		for (int i = 1; i <= n; i ++) scanf("%d", &p[i]);
		scanf("%s", s + 1);
		for (int i = 1; i <= n; i ++)
			if (s[i] == '?') s[i] = '-';

		for (int i = 2; i <= n; i ++) log_2[i] = log_2[i / 2] + 1;
		for (int i = 1; i <= n; i ++) st[i][0] = make_pair(p[i], i);
		for (int j = 1; j < 20; j ++)
			for (int i = 1; i + (1 << j) - 1 <= n; i ++)
				st[i][j] = max(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
		Rt = dfs(1, n);

		tmp = 0;
		Dfs(Rt, 0, 0);

		ans = 0;
		if (tmp < n - 1) puts("-1");
		else {
			f[0][0] = f[0][1] = f[0][2] = 0;
			Dp(Rt, Rt);
			printf("%d\n", ans);
		}
	}//

	return 0;
}

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

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

Think step by step.

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


**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;
typedef long long ll;
const int N=1e5+10;
int n,a[N],q[N],hh,ls[N],rs[N],L[N],R[N],f[N][3],ans;
char s[N];
bool fg;
inline void dfs(int u){
	if(fg||!u) return ;
	dfs(ls[u]),dfs(rs[u]);
	L[u]=ls[u]?L[ls[u]]:u;
	R[u]=rs[u]?R[rs[u]]:u;
	f[u][0]=f[ls[u]][0];
	f[u][1]=f[rs[u]][1];
	f[u][2]=f[ls[u]][0]+f[rs[u]][1];
	ans=max(ans,f[ls[u]][1]+f[rs[u]][0]);
	if(u==q[1]) return ;
	if((s[u]=='L'&&L[u]==1)||(s[u]=='R'&&R[u]==n)) return fg=1,void();
	if(L[u]==1) s[u]='R';
	if(R[u]==n) s[u]='L';
	if(s[u]!='R'){
		f[u][0]=max(f[u][0],1+max(f[ls[u]][1],f[rs[u]][0]));
		f[u][2]=max(f[u][2],1+max(f[ls[u]][1]+f[rs[u]][1],f[rs[u]][2]));
		ans=max(ans,1+max(f[ls[u]][0]+f[rs[u]][0],f[ls[u]][2]));
	}
	if(s[u]!='L'){
		f[u][1]=max(f[u][1],1+max(f[ls[u]][1],f[rs[u]][0]));
		f[u][2]=max(f[u][2],1+max(f[ls[u]][0]+f[rs[u]][0],f[ls[u]][2]));
		ans=max(ans,1+max(f[ls[u]][1]+f[rs[u]][1],f[rs[u]][2]));
	}
}
inline int solve(){
	scanf("%d",&n);
	hh=0;
	ans=0;
	fg=0;
	for(int i=1;i<=n;++i){
		scanf("%d",&a[i]);
		while(hh&&a[q[hh]]<a[i]) ls[i]=q[hh--];
		if(hh) rs[q[hh]]=i;
		q[++hh]=i;
	}
	scanf("%s",s+1);
	dfs(q[1]);
	return fg?-1:ans;
}
int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		printf("%d\n",solve());
		for(int i=1;i<=n;++i) ls[i]=rs[i]=L[i]=R[i]=0;
	}
	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Range Minimum Sum
time_limit_ms: 4000
memory_limit_mb: 512
Description: For an array $[a_1,a_2,\ldots,a_n]$ of length $n$, define $f(a)$ as the sum of the minimum element over all subsegments. That is, $$f(a)=\sum_{l=1}^n\sum_{r=l}^n \min_{l\le i\le r}a_i.$$

A permutation is a sequence of integers from $1$ to $n$ of length $n$ containing each number exactly once. You are given a permutation $[a_1,a_2,\ldots,a_n]$. For each $i$, solve the following problem independently:

  * Erase $a_i$ from $a$, concatenating the remaining parts, resulting in $b = [a_1,a_2,\ldots,a_{i-1},\;a_{i+1},\ldots,a_{n}]$.    * Calculate $f(b)$.

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

The first line of each test case contains an integer $n$ ($1\le n\le 5\cdot 10^5$).

The second line of each test case contains $n$ distinct integers $a_1,\ldots,a_n$ ($1\le a_i\le n$).

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

Output: For each test case, print one line containing $n$ integers. The $i$-th integer should be the answer when erasing $a_i$.

Examples:
input:
4
1
1
3
3 1 2
5
4 2 1 5 3
8
8 1 4 6 7 3 5 2
output:
0
4 7 5
19 21 27 17 19
79 100 72 68 67 80 73 80

Note: In the second test case, $a=[3,1,2]$.

  * When removing $a_1$, $b=[1,2]$. $f(b)=1+2+\min\\{1,2\\}=4$.    * When removing $a_2$, $b=[3,2]$. $f(b)=3+2+\min\\{3,2\\}=7$.    * When removing $a_3$, $b=[3,1]$. $f(b)=3+1+\min\\{3,1\\}=5$.

Buggy Code:
#include <bits/stdc++.h>
#pragma GCC target("avx")
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
using namespace std;
#define ll long long
int n;
ll sum=0;
int a[500005];
int sps[20][500005];
ll t[2000005];
int t1[2000005];
ll cont[500005];
pair<int,int> p[500005];
vector<vector<int>>leftp;
vector<vector<int>>rightp;
void build(int v,int tl,int tr){
	if(tl==tr){
		t[v]=0;return;
	}
	build(2*v,tl,((tl+tr)>>1));
	build(2*v+1,((tl+tr)>>1)+1,tr);
	t[v]=t[2*v]+t[2*v+1];
}
void update(int v,int tl,int tr,int l,int r,ll d){
	if(l>r){
		return;
	}
	else if(tl==l&&tr==r){
		t[v]+=d;return;
	}
	int tm=((tl+tr)>>1);
	update(2*v,tl,tm,l,min(r,tm),d);
	update(2*v+1,tm+1,tr,max(tm+1,l),r,d);
}
ll get(int v,int tl,int tr,int idx){
	if(tl==tr){
		return t[v];
	}
	int tm=((tl+tr)>>1);
	if(idx<=tm){
		return t[v]+get(2*v,tl,tm,idx);
	}
	else{
		return t[v]+get(2*v+1,tm+1,tr,idx);
	}
}
void build1(int v,int tl,int tr){
	if(tl==tr){
		t1[v]=tl;return;
	}
	build1(2*v,tl,((tl+tr)>>1));
	build1(2*v+1,((tl+tr)>>1)+1,tr);
	t1[v]=a[t1[2*v]]>a[t1[2*v+1]]?t1[2*v+1]:t1[2*v];
}
void update1(int v,int tl,int tr,int idx){
	if(tl>tr){
		return;
	}
	if(tl==tr){
		t1[v]=idx;return;
	}
	int tm=((tl+tr)>>1);
	if(idx<=tm){
		update1(2*v,tl,tm,idx);
	}
	else{
		update1(2*v+1,tm+1,tr,idx);
	}
	t1[v]=a[t1[2*v]]>a[t1[2*v+1]]?t1[2*v+1]:t1[2*v];
}
int get1(int v,int tl,int tr,int l,int r){
	if(l>r){
		return -1;
	}
	if(tl==l&&tr==r){
		return t1[v];
	}
	int tm=((tl+tr)>>1);
	int left=get1(2*v,tl,tm,l,min(tm,r));
	int right=get1(2*v+1,tm+1,tr,max(tm+1,l),r);
	if(left==-1){
		return right;
	}
	if(right==-1){
		return left;
	}
	return a[left]>a[right]?right:left;
}
int left(int idx){
	int lo=0;int hi=idx;int valid;
	while(lo<=hi){
		int mid=(lo+hi)/2;
		if(a[get1(1,0,n+1,mid,idx)]<a[idx]){
			lo=mid+1;valid=mid;
		}
		else{
			hi=mid-1;
		}
	}
	return valid;
}
int right(int idx){
	int lo=idx;int hi=n+1;int valid;
	while(lo<=hi){
		int mid=(lo+hi)/2;
		if(a[get1(1,0,n+1,idx,mid)]<a[idx]){
			hi=mid-1;valid=mid;
		}
		else{
			lo=mid+1;
		}
	}
	return valid;
}
void calcmin(){
	for(int i=1;i<=n;++i){
		p[i]={left(i),right(i)};
		if(p[i].first>=0){
			leftp[p[i].first].push_back(i);
		}
		if(p[i].second<=n){
			rightp[p[i].second].push_back(i);
		}
		ll l=i-p[i].first-1;
		ll r=p[i].second-1-i;
		cont[i]=(1+l*r+l+r)*a[i];
		sum+=cont[i];
	}
}
void calcupdate(){
	for(int i=1;i<=n;++i){
		for(auto q:leftp[i]){
			ll tmp=a[i];
			a[i]=1e9;
			update1(1,0,n+1,i);
			ll l=left(q);
			ll r=p[q].second;
			a[i]=tmp;
			l=q-l-2;
			r=r-1-q;
			ll delta=(1+l*r+l+r)*a[q]-cont[q];
			update(1,1,n,i,i,delta);
		}
		for(auto q:rightp[i]){
			ll tmp=a[i];
			a[i]=1e9;
			update1(1,0,n+1,i);
			ll l=p[q].first;
			ll r=right(q);
			a[i]=tmp;
			l=q-l-1;
			r=r-2-q;
			ll delta=(1+l*r+l+r)*a[q]-cont[q];
			update(1,1,n,i,i,delta);
		}
	}
	for(int i=1;i<=n;++i){
		update(1,1,n,i,i,-cont[i]);
		ll l=i-p[i].first-1;
		ll r=p[i].second-1-i;
		update(1,1,n,p[i].first+1,i-1,(1+(l-1)*r+l-1+r)*a[i]-cont[i]);
		update(1,1,n,i+1,p[i].second-1,(1+l*(r-1)+l+r-1)*a[i]-cont[i]);
	}
}
int main() {
	int t;cin>>t;
	while(t--){
		sum=0;
		cin>>n;
		leftp.clear();
		rightp.clear();
		leftp.resize(n+1);
		rightp.resize(n+1);
		a[0]=-1e9;
		for(int i=1;i<=n;++i){
			cin>>a[i];
		}
		a[n+1]=-1e9;
		build(1,1,n);
		build1(1,0,n+1);
		calcmin();
		calcupdate();
		for(int i=1;i<=n;++i){
			cout<<sum+get(1,1,n,i)<<' ';
		}
		cout<<'\n';
	}
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Range Minimum Sum
time_limit_ms: 4000
memory_limit_mb: 512
Description: For an array $[a_1,a_2,\ldots,a_n]$ of length $n$, define $f(a)$ as the sum of the minimum element over all subsegments. That is, $$f(a)=\sum_{l=1}^n\sum_{r=l}^n \min_{l\le i\le r}a_i.$$

A permutation is a sequence of integers from $1$ to $n$ of length $n$ containing each number exactly once. You are given a permutation $[a_1,a_2,\ldots,a_n]$. For each $i$, solve the following problem independently:

  * Erase $a_i$ from $a$, concatenating the remaining parts, resulting in $b = [a_1,a_2,\ldots,a_{i-1},\;a_{i+1},\ldots,a_{n}]$.    * Calculate $f(b)$.

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

The first line of each test case contains an integer $n$ ($1\le n\le 5\cdot 10^5$).

The second line of each test case contains $n$ distinct integers $a_1,\ldots,a_n$ ($1\le a_i\le n$).

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

Output: For each test case, print one line containing $n$ integers. The $i$-th integer should be the answer when erasing $a_i$.

Examples:
input:
4
1
1
3
3 1 2
5
4 2 1 5 3
8
8 1 4 6 7 3 5 2
output:
0
4 7 5
19 21 27 17 19
79 100 72 68 67 80 73 80

Note: In the second test case, $a=[3,1,2]$.

  * When removing $a_1$, $b=[1,2]$. $f(b)=1+2+\min\\{1,2\\}=4$.    * When removing $a_2$, $b=[3,2]$. $f(b)=3+2+\min\\{3,2\\}=7$.    * When removing $a_3$, $b=[3,1]$. $f(b)=3+1+\min\\{3,1\\}=5$.

Buggy Code:
#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
using namespace std;
long long t,n,a[555555],st[555555][33],st2[555555][33],l,r,mid,fa,ls,rs,jian[555555],cf[555555],j2[555555];
void stcsh(long long n)
{
	for(long long i=1;i<=n;i++)
	{
		st[i][0]=a[i];
		st2[i][0]=0x3f3f3f3f3f3f3f3f;
		st[i][1]=min(a[i],a[i+1]);
		st2[i][1]=max(a[i],a[i+1]);
	}
	long long _=log(n)/log(2)+1;
	for(long long j=2;j<_;j++)
	{
		for(long long i=1;i<=n-(1<<j)+1;i++)
		{
			st[i][j]=min(st[i][j-1],st[i+(1<<(j-1))][j-1]);
			st2[i][j]=0x3f3f3f3f3f3f3f3f;
			if(st[i][j-1]>st[i][j])
			{
				st2[i][j]=min(st2[i][j],st[i][j-1]);
			}
			if(st[i+(1<<(j-1))][j-1]>st[i][j])
			{
				st2[i][j]=min(st2[i][j],st[i+(1<<(j-1))][j-1]);
			}
			if(st2[i][j-1]>st[i][j])
			{
				st2[i][j]=min(st2[i][j],st2[i][j-1]);
			}
			if(st2[i+(1<<(j-1))][j-1]>st[i][j])
			{
				st2[i][j]=min(st2[i][j],st2[i+(1<<(j-1))][j-1]);
			}
		}
	}
}
long long query(long long l,long long r)
{
	long long k=__lg(r-l+1)/__lg(2);
	return min(st[l][k],st[r-(1<<k)+1][k]);
}
long long query2(long long l,long long r)
{
	long long k=__lg(r-l+1)/__lg(2);
	long long res=0x3f3f3f3f3f3f3f3f,mx=query(l,r);
	if(st[l][k]>mx)
	{
		res=min(res,st[l][k]);
	}
	if(st[r-(1<<k)+1][k]>mx)
	{
		res=min(res,st[r-(1<<k)+1][k]);
	}
	if(st2[l][k]>mx)
	{
		res=min(res,st2[l][k]);
	}
	if(st2[r-(1<<k)+1][k]>mx)
	{
		res=min(res,st2[r-(1<<k)+1][k]);
	}

	return res;
}
int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);

	cin>>t;
	for(long long qwer=1;qwer<=t;qwer++)
	{
		cin>>n;
		for(long long i=1;i<=n;i++)
		{
			cin>>a[i];
			jian[i]=0;
			cf[i]=0;
		}
		stcsh(n);
		fa=0;
		for(long long i=1;i<=n;i++)
		{
			l=1;
			r=i;
			while(l<r)
			{
				mid=(l+r)/2;
				if(query(mid,i)<a[i])
				{
					l=mid+1;
				}
				else
				{
					r=mid;
				}
			}
			ls=l;
			l=i;
			r=n;
			while(l<r)
			{
				mid=ceil((l+r)/2.0);
				if(query(i,mid)<a[i])
				{
					r=mid-1;
				}
				else
				{
					l=mid;
				}
			}
			rs=l;
			fa+=(i-ls+1)*(rs-i+1)*a[i];
			cf[ls]+=a[i]*(rs-i+1);
			cf[i+1]-=a[i]*(rs-i+1);
			cf[i]+=a[i]*(i-ls+1);
			cf[rs+1]-=a[i]*(i-ls+1);
			if(ls>1)
			{
				l=1;
				r=ls-1;
				while(l<r)
				{
					mid=(l+r)/2;
					if(query2(mid,i)<a[i])
					{
						l=mid+1;
					}
					else
					{
						r=mid;
					}
				}
				jian[ls-1]+=((ls-2)-l+1)*(rs-i+1)*(a[ls-1]-a[i]);
			}

			if(rs<n)
			{
				l=rs+1;
				r=n;
				while(l<r)
				{
					mid=ceil((l+r)/2.0);
					if(query2(i,mid)<a[i])
					{
						r=mid-1;
					}
					else
					{
						l=mid;
					}
				}
				jian[rs+1]+=(i-ls+1)*(l-(rs+2)+1)*(a[rs+1]-a[i]);
			}

		}
		for(long long i=1;i<=n;i++)
		{
			j2[i]=j2[i-1]+cf[i];
		}

//		cout<<fa<<endl;
		for(long long i=1;i<=n;i++)
		{
			cout<<fa-jian[i]-j2[i]+a[i]<<' ';
		}
		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 buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. Here are some considerations:

1. **Edge Cases:**
   - Single element arrays.
   - Arrays where the minimum element is at the beginning or end.
   - Arrays with a large number of elements, but we need to keep it small-scale.

2. **Scenarios that Maximize Failure Probability:**
   - Arrays with elements in a specific order that might cause the binary search or query functions to fail.
   - Arrays with repeated patterns or sequences that might confuse the algorithm.

Given these considerations, let's create a small-scale test input that is complete and valid:

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

### Explanation:
1. **Single Element Array:**
   - Test case: `1 1`
   - This is a simple edge case where the array has only one element. The expected output is `0`.

2. **Array with Minimum Element at the Beginning:**
   - Test case: `3 3 1 2`
   - The minimum element `1` is at the beginning. This can test how the algorithm handles the minimum element being the first element.

3. **Array with Minimum Element at the End:**
   - Test case: `5 4 2 1 5 3`
   - The minimum element `1` is at the end. This can test how the algorithm handles the minimum element being the last element.

4. **Array with Increasing Elements:**
   - Test case: `6 1 2 3 4 5 6`
   - This is a simple increasing sequence. This can test how the algorithm handles a sorted array.

These test cases cover a range of scenarios that can help expose potential bugs in the algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Range Minimum Sum
time_limit_ms: 4000
memory_limit_mb: 512
Description: For an array $[a_1,a_2,\ldots,a_n]$ of length $n$, define $f(a)$ as the sum of the minimum element over all subsegments. That is, $$f(a)=\sum_{l=1}^n\sum_{r=l}^n \min_{l\le i\le r}a_i.$$

A permutation is a sequence of integers from $1$ to $n$ of length $n$ containing each number exactly once. You are given a permutation $[a_1,a_2,\ldots,a_n]$. For each $i$, solve the following problem independently:

  * Erase $a_i$ from $a$, concatenating the remaining parts, resulting in $b = [a_1,a_2,\ldots,a_{i-1},\;a_{i+1},\ldots,a_{n}]$.    * Calculate $f(b)$.

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

The first line of each test case contains an integer $n$ ($1\le n\le 5\cdot 10^5$).

The second line of each test case contains $n$ distinct integers $a_1,\ldots,a_n$ ($1\le a_i\le n$).

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

Output: For each test case, print one line containing $n$ integers. The $i$-th integer should be the answer when erasing $a_i$.

Examples:
input:
4
1
1
3
3 1 2
5
4 2 1 5 3
8
8 1 4 6 7 3 5 2
output:
0
4 7 5
19 21 27 17 19
79 100 72 68 67 80 73 80

Note: In the second test case, $a=[3,1,2]$.

  * When removing $a_1$, $b=[1,2]$. $f(b)=1+2+\min\\{1,2\\}=4$.    * When removing $a_2$, $b=[3,2]$. $f(b)=3+2+\min\\{3,2\\}=7$.    * When removing $a_3$, $b=[3,1]$. $f(b)=3+1+\min\\{3,1\\}=5$.

Buggy Code:
#include<bits/stdc++.h>
#define int long long
#define x first
#define y second
#define il inline
#define debug() puts("-----")
using namespace std;
typedef pair<int,int> pii;
il int read(){
    int x;
    scanf("%lld",&x);
    return x;
}
const int N=5e5+10;
int n;
int a[N];
int l[N],r[N];
int L[N],R[N];
int s[N],top=0;
vector<int> vec[N];
int fl[N],fr[N],g[N];
il int get(int i,int x){
    if(vec[i].empty()) return -1;
    int l=0,r=vec[i].size()-1,res=-1;
    while(l<=r){
        int mid=l+r>>1;
        if(a[vec[i][mid]]<x) r=mid-1,res=vec[i][mid];
        else l=mid+1;
    } return res;
}
struct Node{
    int l,r,w;
}tr[N<<2];
il void pushup(int u){
    tr[u].w=tr[u<<1].w+tr[u<<1|1].w;
}
il void build(int u,int l,int r){
    tr[u]={l,r}; tr[u].w=0;
    if(l==r) return ;
    int mid=l+r>>1;
    build(u<<1,l,mid),build(u<<1|1,mid+1,r);
}
il void modify(int u,int x,int w){
    if(tr[u].l==tr[u].r){
        tr[u].w+=w;
        return ;
    } int mid=tr[u].l+tr[u].r>>1;
    if(x<=mid) modify(u<<1,x,w);
    else modify(u<<1|1,x,w);
    pushup(u);
}
il int query(int u,int l,int r){
    if(l<=tr[u].l&&tr[u].r<=r) return tr[u].w;
    int w=0,mid=tr[u].l+tr[u].r>>1;
    if(l<=mid) w=query(u<<1,l,r);
    if(r>mid) w+=query(u<<1|1,l,r);
    return w;
}
il void solve(){
    n=read(); s[top=0]=0;
    for(int i=1;i<=n;i++) a[i]=read(),fl[i]=fr[i]=g[i]=0,s[i]=0;
    for(int i=1;i<=n;i++){
        vec[i].clear();
        while(top>0&&a[s[top]]>a[i]) vec[i].push_back(s[top]),top--;
        if(top){
            l[i]=s[top]+1;
            int p=get(s[top],a[i]);
            if(p==-1){
                if(top>1) L[i]=s[top-1]+1;
                else L[i]=1;
            } else L[i]=p+1;
        } else l[i]=1,L[i]=0;
        s[++top]=i;
    } top=0;
    for(int i=n;i>=1;i--){
        vec[i].clear();
        while(top>0&&a[s[top]]>a[i]) vec[i].push_back(s[top]),top--;
        if(top){
            r[i]=s[top]-1;
            int p=get(s[top],a[i]);
            if(p==-1){
                if(top>1) R[i]=s[top-1]-1;
                else R[i]=n;
            } else R[i]=p-1;
        } else r[i]=n,R[i]=n+1;
        s[++top]=i;
    } int sum=0;
    for(int i=1;i<=n;i++){
        sum+=(i-l[i]+1)*(r[i]-i+1)*a[i];
        if(l[i]!=L[i]) fl[l[i]]+=(l[i]-L[i]-1)*a[i]*(r[i]-i+1);
        if(r[i]!=R[i]) fr[r[i]]+=(R[i]-r[i]-1)*a[i]*(i-l[i]+1);
    } build(1,1,n);
    for(int i=1;i<=n;i++){
        g[i]=query(1,i,n);
        modify(1,r[i],(i-l[i]+1)*a[i]);
    } build(1,1,n);
    for(int i=n;i>=1;i--){
        g[i]+=query(1,1,i);
        modify(1,l[i],(r[i]-i+1)*a[i]);
    } for(int i=1;i<=n;i++){
        int ans=sum-a[i]*(i-l[i]+1)*(r[i]-i+1);
        ans+=(fr[i-1]+fl[i+1]-g[i]); printf("%lld ",ans);
    } puts("");
}
signed main(){
    int T=read();
    while(T--) solve();
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
1
1
3
2 3 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: Range Minimum Sum
time_limit_ms: 4000
memory_limit_mb: 512
Description: For an array $[a_1,a_2,\ldots,a_n]$ of length $n$, define $f(a)$ as the sum of the minimum element over all subsegments. That is, $$f(a)=\sum_{l=1}^n\sum_{r=l}^n \min_{l\le i\le r}a_i.$$

A permutation is a sequence of integers from $1$ to $n$ of length $n$ containing each number exactly once. You are given a permutation $[a_1,a_2,\ldots,a_n]$. For each $i$, solve the following problem independently:

  * Erase $a_i$ from $a$, concatenating the remaining parts, resulting in $b = [a_1,a_2,\ldots,a_{i-1},\;a_{i+1},\ldots,a_{n}]$.    * Calculate $f(b)$.

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

The first line of each test case contains an integer $n$ ($1\le n\le 5\cdot 10^5$).

The second line of each test case contains $n$ distinct integers $a_1,\ldots,a_n$ ($1\le a_i\le n$).

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

Output: For each test case, print one line containing $n$ integers. The $i$-th integer should be the answer when erasing $a_i$.

Examples:
input:
4
1
1
3
3 1 2
5
4 2 1 5 3
8
8 1 4 6 7 3 5 2
output:
0
4 7 5
19 21 27 17 19
79 100 72 68 67 80 73 80

Note: In the second test case, $a=[3,1,2]$.

  * When removing $a_1$, $b=[1,2]$. $f(b)=1+2+\min\\{1,2\\}=4$.    * When removing $a_2$, $b=[3,2]$. $f(b)=3+2+\min\\{3,2\\}=7$.    * When removing $a_3$, $b=[3,1]$. $f(b)=3+1+\min\\{3,1\\}=5$.

Buggy Code:
import sys
input=lambda:sys.stdin.readline().rstrip()
inp=lambda:map(int,input().split())
from collections import deque
sys.setrecursionlimit(200002)
#ind,left,right,indparent1,indparentid2
def cart(l,r):
    if l==r:
        return [l,None,None,-1,len(x)]
    mid=(l+r)//2
    return merge(cart(l,mid),cart(mid+1,r))
def merge(a,b):
    if a==None:return b
    if b==None:return a
    if x[a[0]]<x[b[0]]:
        return [a[0],a[1],merge(a[2],b),a[3],a[4]]
    else:
        return [b[0],merge(a,b[1]),b[2],b[3],b[4]]
def nval(val,o,lmove,po,llmove):
    #print("old",o[0],val)
    val-=contrib(o)
    v1,v2,v3=po[0],po[3],po[4]
    i=o[0]
    if v1>i:v1-=1
    if v2>i:v2-=1
    if v3!=None and v3>i:v3-=1
    res[i]=mergeval(val,o[1],o[2],[v1,po[1],po[2],v2,v3],lmove,llmove,i)
    if o[1]:
        v1,v2=o[3],o[4]
        i=o[1][0]
        if v1>i:v1-=1
        if v2>i:v2-=1
        nval(val+abs(v1-o[0]+1)*abs(v2-o[0]+1)*x[o[0]],o[1],"l",o,lmove)
    if o[2]:
        v1,v2=o[3],o[4]
        i=o[2][0]
        if v1>i:v1-=1
        if v2>i:v2-=1
        nval(val+abs(v1-o[0])*abs(v2-o[0])*x[o[0]],o[2],"r",o,lmove)
def mergeval(val,a,b,c,lmove,llmove,i):

    if a==None or b==None:
        #if i==1:print(a,b)
        return val
    #if i==1:print(a[0],b[0],val)
    v1,v2,v3=c[0],c[3],c[4]
    if x[a[0]]<x[b[0]]:
        val-=contrib(a)
        if lmove!=llmove:
            #if i==1:print("a new",a[0],v1,v2)
            val+=contrib((a[0],None,None,v1,v2))
            na=[a[0]-(a[0]>i),a[1],a[2],v1,v2]
            return mergeval(val,a[2],b,na,"r",lmove,i)
        else:
            #if i==1:print("a new",a[0],v1,v3)
            val+=contrib((a[0],None,None,v1,v3))
            na=[a[0]-(a[0]>i),a[1],a[2],v1,v3]
            return mergeval(val,a[2],b,na,"r",lmove,i)
    else:
        val-=contrib(b)
        if lmove!=llmove:
            #if i==1:print("b new",b[0],v1,v2)
            val+=abs(v1-b[0]+1)*abs(v2-b[0]+1)*x[b[0]]
            nb=[b[0]-(b[0]>i),b[1],b[2],v1,v2]
            #val+=contrib((b[0],None,None,v1,v2))
            return mergeval(val,a,b[1],nb,"l",lmove,i)
        else:
            #if i==1:print("b new",b[0],v1,v3)
            val+=abs(v1-b[0]+1)*abs(v3-b[0]+1)*x[b[0]]
            nb=[b[0]-(b[0]>i),b[1],b[2],v1,v3]
            #val+=contrib((b[0],None,None,v1,v3))
            return mergeval(val,a,b[1],nb,"l",lmove,i)

def contrib(u):
    return abs(u[3]-u[0])*abs(u[4]-u[0])*x[u[0]]

t=int(input())
for _ in range(t):
    n=int(input())
    *x,=inp()
    o=cart(0,n-1)
    q=deque([(o,"r")])
    val=0
    while q:
        u,lmove=q.popleft()
        val+=contrib(u)
        for v,move in[(u[1],"l"),(u[2],"r")]:
            if v==None:continue
            v[3]=u[0]
            v[4]=u[4-(lmove!=None and move!=lmove)]
            q.append((v,move))
    po=[-1,None,None,n,None]
    res=[None]*n
    #print(val)
    #print(nval(val,o,1,"r",po,"l"))
    nval(val,o,"r",po,"l")
    print(*res)

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

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

Think step by step.

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

This test case is designed to expose issues in the buggy code by using a small permutation that involves a specific sequence of values where the minimum element is not at the beginning or end of the array. This can help in identifying problems with how the code handles the removal of elements and the calculation of the sum of minimums over subsegments.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Range Minimum Sum
time_limit_ms: 4000
memory_limit_mb: 512
Description: For an array $[a_1,a_2,\ldots,a_n]$ of length $n$, define $f(a)$ as the sum of the minimum element over all subsegments. That is, $$f(a)=\sum_{l=1}^n\sum_{r=l}^n \min_{l\le i\le r}a_i.$$

A permutation is a sequence of integers from $1$ to $n$ of length $n$ containing each number exactly once. You are given a permutation $[a_1,a_2,\ldots,a_n]$. For each $i$, solve the following problem independently:

  * Erase $a_i$ from $a$, concatenating the remaining parts, resulting in $b = [a_1,a_2,\ldots,a_{i-1},\;a_{i+1},\ldots,a_{n}]$.    * Calculate $f(b)$.

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

The first line of each test case contains an integer $n$ ($1\le n\le 5\cdot 10^5$).

The second line of each test case contains $n$ distinct integers $a_1,\ldots,a_n$ ($1\le a_i\le n$).

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

Output: For each test case, print one line containing $n$ integers. The $i$-th integer should be the answer when erasing $a_i$.

Examples:
input:
4
1
1
3
3 1 2
5
4 2 1 5 3
8
8 1 4 6 7 3 5 2
output:
0
4 7 5
19 21 27 17 19
79 100 72 68 67 80 73 80

Note: In the second test case, $a=[3,1,2]$.

  * When removing $a_1$, $b=[1,2]$. $f(b)=1+2+\min\\{1,2\\}=4$.    * When removing $a_2$, $b=[3,2]$. $f(b)=3+2+\min\\{3,2\\}=7$.    * When removing $a_3$, $b=[3,1]$. $f(b)=3+1+\min\\{3,1\\}=5$.

Buggy Code:
def read_input():
    import sys
    return sys.stdin.read()

def parse_test_cases(input_data):
    lines = input_data.strip().split('\n')
    t = int(lines[0])
    test_cases = []
    idx = 1
    for _ in range(t):
        n = int(lines[idx])
        a = list(map(int, lines[idx + 1].split()))
        test_cases.append((n, a))
        idx += 2
    return test_cases

def remove_element(a, index):
    return a[:index] + a[index+1:]

def calculate_f(b):
    total = 0
    n = len(b)
    for l in range(n):
        current_min = b[l]
        for r in range(l, n):
            current_min = min(current_min, b[r])
            total += current_min
    return total

def process_test_case(n, a):
    results = []
    for i in range(n):
        b = remove_element(a, i)
        f_b = calculate_f(b)
        results.append(str(f_b))
    return ' '.join(results)

def solve(test_cases):
    outputs = []
    for n, a in test_cases:
        if n == 1:
            outputs.append('0')
            continue
        output = process_test_case(n, a)
        outputs.append(output)
    return outputs

def main():
    input_data = read_input()
    test_cases = parse_test_cases(input_data)
    results = solve(test_cases)
    for res in results:
        print(res)

if __name__ == "__main__":
    main()

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

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

Think step by step.

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