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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
//#pragma GCC optimize("Ofast")
//#pragma GCC target("avx,avx2,fma")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,sse4a,avx,avx2,popcnt,tune=native")
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <bitset>
#include <array>
#include <climits>
using namespace std;

#ifdef LOCAL
	#define eprintf(...) {fprintf(stderr, __VA_ARGS__);fflush(stderr);}
#else
	#define eprintf(...) 42
#endif

using ll = long long;
using ld = long double;
using uint = unsigned int;
using ull = unsigned long long;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
ll myRand(ll B) {
	return (ull)rng() % B;
}

#define mp make_pair
#define all(x) (x).begin(),(x).end()

clock_t startTime;
double getCurrentTime() {
	return (double)(clock() - startTime) / CLOCKS_PER_SEC;
}

ll floor_div(ll x, ll y) {
	assert(y != 0);
	if (y < 0) {
		y = -y;
		x = -x;
	}
	if (x >= 0) return x / y;
	return (x + 1) / y - 1;
}
ll ceil_div(ll x, ll y) {
	assert(y != 0);
	if (y < 0) {
		y = -y;
		x = -x;
	}
	if (x <= 0) return x / y;
	return (x - 1) / y + 1;
}
template<typename T>
T sqr(T x) {
	return x * x;
}

struct Item {
	int x;
	int p;

	Item() : x(0), p(-1) {}
	Item(int _x, int _p) : x(_x), p(_p) {}

	Item& operator &= (const Item &I) {
		if (x < I.x) *this = I;
		return *this;
	}
	Item operator & (const Item &I) const {
		return Item(*this) &= I;
	}
};
template<typename T> struct SparseTable {
	vector<vector<T>> b;

	SparseTable() : b() {}
	SparseTable(vector<T> a) {
		int n = (int)a.size();
		int m = 1;
		while((1 << m) < n) m++;
		b = vector<vector<T>>(m);
		for (int k = 0; k < m; k++) {
			b[k] = a;
			int len = 1 << k;
			for (int mid = len; mid < n; mid += (len << 1)) {
				for (int i = mid - 2; i >= mid - len; i--)
					b[k][i] &= b[k][i + 1];
				for (int i = mid + 1; i < min(n, mid + len); i++)
					b[k][i] = b[k][i - 1] & b[k][i];
			}
		}
	}

	T get(int l, int r) const {
		if (l >= r) return T();
		r--;
		if (l == r) return b[0][l];
		int k = 31 - __builtin_clz(l ^ r);
		return b[k][l] & b[k][r];
	}
};

const int N = 400400;
int g[N][2];
int h[N];
int par[N];
int parDir[N];
int ancestorInDiffDir[N];
int deep[N];
int dp[N];
int n;
char s[N];
SparseTable<Item> ST;

int build(int l, int r, int H, int P, int PD) {
	if (l >= r) return -1;
	int v = ST.get(l, r).p;
	h[v] = H;
	par[v] = P;
	parDir[v] = PD;
	if (P == -1 || par[P] == -1) {
		ancestorInDiffDir[v] = -1;
	} else {
		if (parDir[v] != parDir[P]) {
			ancestorInDiffDir[v] = P;
		} else {
			ancestorInDiffDir[v] = ancestorInDiffDir[P];
		}
	}
	g[v][0] = build(l, v, H + 1, v, 0);
	g[v][1] = build(v + 1, r, H + 1, v, 1);
	dp[v] = 0;
	deep[v] = 0;
	for (int i = 0; i < 2; i++) {
		int u = g[v][i];
		if (u == -1) continue;
		deep[v] = max(deep[v], deep[u] + 1);
		int w = 0;
		if (g[v][i ^ 1] != -1) w = deep[g[v][i ^ 1]] + 1;
		w += deep[u];
		dp[u] = max(dp[u], w);
		if (ancestorInDiffDir[u] != -1) {
			int z = ancestorInDiffDir[u];
			dp[z] = max(dp[z], dp[u] + h[u] - h[z]);
		}
	}
	return v;
}

void solve() {
	scanf("%d", &n);
	vector<Item> a(n);
	for (int i = 0; i < n; i++) {
		int x;
		scanf("%d", &x);
		a[i] = Item(x, i);
	}
	scanf("%s", s);
	ST = SparseTable<Item>(a);
	int root = build(0, n, 0, -1, -1);
	int ans = 0;
	for (int v = 0; v < n; v++) if (v != root)
		ans = max(ans, dp[v] + 1);
	printf("%d\n", ans);
}

int main() {
	startTime = clock();
//	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);

	int t;
	scanf("%d", &t);
	for (int i = 1; i <= t; i++) {
		eprintf("--- Case #%d start ---\n", i);
		//printf("Case #%d: ", i);

		solve();

		//printf("%lld\n", (ll)solve());

		/*
		if (solve()) {
			printf("Yes\n");
		} else {
			printf("No\n");
		}
		*/

		eprintf("--- Case #%d end ---\n", i);
		eprintf("time = %.5lf\n", getCurrentTime());
		eprintf("++++++++++++++++++++\n");
	}

	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input for the given problem, we need to focus on edge cases that can expose potential bugs in the code. Here are the steps to think through this:

1. **Understand the Problem Constraints:**
   - The input consists of multiple test cases.
   - Each test case has a permutation `p` and a string `s` of the same length.
   - The string `s` only contains the character `?`.

2. **Identify Potential Bugs:**
   - The code might fail in handling very small or very large permutations.
   - The code might have issues with permutations that are already sorted or have a specific pattern.
   - The code might have issues with the sparse table construction or the tree building process.

3. **Generate Test Cases:**
   - **Small Permutations:** Test cases with very small `n` (e.g., 2, 3, 4).
   - **Sorted Permutations:** Test cases with a permutation that is already sorted.
   - **Reverse Sorted Permutations:** Test cases with a permutation that is sorted in reverse.
   - **Random Permutations:** Test cases with random permutations.
   - **Edge Cases:** Test cases with `n` at the upper limit (e.g., 400,000).

4. **Choose a Small-Scale Test Case:**
   - For simplicity, let's choose a small permutation that can expose potential bugs in the code.

Here is a small-scale test input that can be used to expose potential bugs:

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

This test input includes:
- A test case with a small permutation of length 3.
- A test case with a reverse sorted permutation of length 4.
- A test case with a permutation that has a specific pattern of length 5.

These cases can help in identifying issues with the code's handling of different types of permutations and the construction of the graph.


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
#include <bits/stdc++.h>
#define ff first
#define ss second
#define MAX 400001
#define INF LLONG_MAX
#define pb push_back
#define pii pair<int,int>
//#define int long long
using namespace std;
string s;
int n,a[MAX],pre[MAX],suf[MAX],pre1[MAX],suf1[MAX];
void solve(){
    for(int i=0;i<n;i++){
        pre[i]=suf[i]=pre1[i]=suf1[i]=0;
    }
    int res=0;
    vector<pii> b[4];
    int h=0;
    for(int i=0;i<n;i++){
        pii nxt={i,0};
        int h1=0;
        int zor=0;
        while(!b[zor].empty() && (a[b[zor].back().ff]<a[i])){
            pii p=b[zor].back();
            b[zor].pop_back();
            if(!b[zor].empty()){
                b[zor].back().ss=max(b[zor].back().ss,p.ss+1);
            }
            nxt.ss=max(nxt.ss,p.ss+1);
        }
        b[zor].pb(nxt);
        h=max(h,(int)b[zor].size()+nxt.ss);
        pre[i]=h;
    }
    h=0;
    for(int i=n-1;i>=0;i--){
        pii nxt={i,0};
        int h1=0;
        int zor=2;
        while(!b[zor].empty() && (a[b[zor].back().ff]<a[i])){
            pii p=b[zor].back();
            b[zor].pop_back();
            if(!b[zor].empty()){
                b[zor].back().ss=max(b[zor].back().ss,p.ss+1);
            }
            nxt.ss=max(nxt.ss,p.ss+1);
        }
        b[zor].pb(nxt);
        h=max(h,(int)b[zor].size()+nxt.ss);
        suf[i]=h;
    }
    h=0;
    for(int j=0;j<(int)b[0].size()-1;j++){
        for(int i=b[0][j].ff;i<b[0][j+1].ff;i++){
            pii nxt={i,0};
            int h1=0;
            int zor=1;
            while(!b[zor].empty() && (a[b[zor].back().ff]<a[i])){
                pii p=b[zor].back();
                b[zor].pop_back();
                if(!b[zor].empty()){
                    b[zor].back().ss=max(b[zor].back().ss,p.ss+1);
                }
                nxt.ss=max(nxt.ss,p.ss+1);
            }
            b[zor].pb(nxt);
            h=max(h,(int)b[zor].size()+nxt.ss);
            pre1[i]=h;
            res=max(res,pre1[i]+suf[i+1]);
        }
    }
    for(int j=0;j<(int)b[2].size()-1;j++){
        h=0;
        for(int i=b[2][j].ff;i>b[2][j+1].ff;i--){
            pii nxt={i,0};
            int h1=0;
            int zor=3;
            while(!b[zor].empty() && (a[b[zor].back().ff]<a[i])){
                pii p=b[zor].back();
                b[zor].pop_back();
                if(!b[zor].empty()){
                    b[zor].back().ss=max(b[zor].back().ss,p.ss+1);
                }
                nxt.ss=max(nxt.ss,p.ss+1);
            }
            b[zor].pb(nxt);
            h=max(h,(int)b[zor].size()+nxt.ss);
            suf1[i]=h;
            res=max(res,suf1[i]+pre[i-1]);
        }
    }
    res=max(res,pre[b[0][0].ff]+suf[b[0][0].ff]-1);
    cout << res-1 << "\n";
}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t=1;
    cin >> t;
    while(t--){
        cin >> n;
        for(int i=0;i<n;i++){
            cin >> a[i];
        }
        cin >> s;
        solve();
    }
}

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

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

Think step by step.

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

public class Main {
    private static BufferedReader br;
    private static StringTokenizer tokenizer;
    private static String next() throws IOException{
        if (br == null) br = new BufferedReader(new InputStreamReader(System.in));
        if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());
        while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());
        return tokenizer.nextToken();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }
    private static long nextLong() throws IOException{
        return Long.parseLong(next());
    }

    private static int[] arr = new int[400006];
    private static int[][] best = new int[2][400006];
    private static int[][] sers = new int[2][400006];
    private static int[] cnts = new int[2];
    private static Pair[] stack = new Pair[400006];
    private static class Pair {
        int ele;
        int v;
        Pair(int ele, int v) {
            this.ele = ele;
            this.v = v;
        }
        public static Pair of(int ele, int v) {
            return new Pair(ele, v);
        }
    }
    private static void swap(int[] arr, int start, int end) {
        while (start < end) {
            int tmp = arr[start];
            arr[start] = arr[end];
            arr[end] = tmp;
            ++start;
            --end;
        }
    }

    public static void main(String[] args)  throws IOException {
        int tt = nextInt();
        StringBuilder sb = new StringBuilder();
        for (int ig = 0; ig < tt; ++ig) {
            int n = nextInt();
            for (int i = 0; i < n; ++i) arr[i + 1] = nextInt();
            String s = next();
            for (int z = 0; z < 2; ++z) {
                int next = 0;
                int max = 0;
                for (int i = 1; i <= n; ++i) {
                    int v = 1;
                    while (next > 0 && arr[stack[next - 1].ele] < arr[i]) {
                        Pair pre = stack[next - 1];
                        --next;
                        v = Math.max(v, pre.v + 1);
                        if (next > 0) stack[next - 1].v = Math.max(stack[next - 1].v, pre.v + 1);
                    }
                    stack[next++] = Pair.of(i, v);
                    max = Math.max(max, v + next - 1);
                    best[z][i] = max;
                }
                cnts[z] = next;
                for (int i = 0; i < next; ++i) {
                    sers[z][i] = stack[i].ele;
                }

                swap(arr, 1, n);
            }

            int ans = best[0][sers[0][0]] + best[1][n + 1 - sers[0][0]] - 1;
            for (int z = 0; z < 2; ++z) {
                int size = cnts[z];
                for (int i = 0; i < size - 1; ++i) {
                    int next = 0;
                    int max = 0;
                    for (int j = sers[z][i]; j < sers[z][i + 1]; ++j) {
                        int v = 1;
                        while (next > 0 && arr[stack[next - 1].ele] < arr[j]) {
                            Pair pre = stack[next - 1];
                            --next;
                            v = Math.max(v, pre.v + 1);
                            if (next > 0) stack[next - 1].v = Math.max(stack[next - 1].v, pre.v + 1);
                        }
                        stack[next++] = Pair.of(j, v);
                        max = Math.max(max, v + next - 1);
                        ans = Math.max(ans, max + best[1 - z][n - j]);
                    }
                }
            }

            sb.append(ans - 1).append("\n");
        }
        System.out.println(sb);
    }
}

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

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

Think step by step.

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

**Explanation:**

- The permutation `4 1 3 2` is chosen to create a situation where the algorithm needs to carefully decide the placement of edges to maximize the diameter.
- The string `???` allows for flexibility in choosing the edges, which can expose any issues in the logic that determines the best edge to add.
- The permutation and string combination ensures that the algorithm must handle the case where the largest and smallest values are not at the ends, and the middle values are in a non-trivial order.
- This test case is small enough to be manageable but complex enough to potentially reveal bugs in the code.


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

public class Main {
    private static BufferedReader br;
    private static StringTokenizer tokenizer;
    private static String next() throws IOException{
        if (br == null) br = new BufferedReader(new InputStreamReader(System.in));
        if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());
        while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());
        return tokenizer.nextToken();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }
    private static long nextLong() throws IOException{
        return Long.parseLong(next());
    }

    private static int[] arr = new int[400006];
    private static Pair[][] mps = new Pair[2][400006];
    private static Pair[] stack = new Pair[400006];
    private static class Pair {
        int ele;
        int v;
        private Pair(int ele, int v) {
            this.ele = ele;
            this.v = v;
        }
        public static Pair of(int ele, int v) {
            return new Pair(ele, v);
        }
    }
    private static void swap(int[] arr, int start, int end) {
        while (start < end) {
            int tmp = arr[start];
            arr[start] = arr[end];
            arr[end] = tmp;
            ++start;
            --end;
        }
    }

    public static void main(String[] args)  throws IOException {
        int tt = nextInt();
        StringBuilder sb = new StringBuilder();
        for (int ig = 0; ig < tt; ++ig) {
            int n = nextInt();
            for (int i = 0; i < n; ++i) arr[i + 1] = nextInt();
            String s = next();
            if (ig == 90) {
                int[] outArr = new int[n];
                for (int i = 0; i < n; ++i) outArr[i] = arr[i + 1];
                System.out.println(Arrays.toString(outArr));
            }
            for (int z = 0; z < 2; ++z) {
                int mv = 0;
                int next = 0;
                for (int i = 1; i <= n; ++i) {
                    int cur = 1;
                    while (next > 0 && arr[stack[next - 1].ele] < arr[i]) {
                        cur = Math.max(cur, Math.max(stack[next - 1].v, mv - (next - 1)) + 1);
                        --next;
                    }
                    stack[next++] = Pair.of(i, cur);
                    mv = Math.max(mv, cur + next - 1);
                    mps[z][i] = Pair.of(stack[0].ele, Math.max(stack[0].v, mv));
                }

                swap(arr, 1, n);
            }
            int ans = Math.max(mps[0][n].v, mps[1][n].v);
            for (int z = 0; z < 2; ++z) {
                int next = 0;
                int mv = 0;
                for (int i = 1; i < n; ++i) {
                    Pair op = mps[1 - z][n - i];
                    int cur = 1;
                    while (next > 0 && arr[stack[next - 1].ele] < arr[i]) {
                        cur = Math.max(cur, Math.max(stack[next - 1].v, mv - (next - 1)) + 1);
                        --next;
                    }
                    stack[next++] = Pair.of(i, cur);
                    mv = Math.max(mv, cur + next - 1);
                    int l = 0, r = next;
                    while (l < r) {
                        int mid = l + ((r - l) >> 1);
                        if (arr[stack[mid].ele] > arr[n + 1 - op.ele]) {
                            l = mid + 1;
                        } else {
                            r = mid;
                        }
                    }
                    if (l > 0) ans = Math.max(ans, op.v + Math.max(stack[l - 1].v, mv - (l - 1)));
                }
                swap(arr, 1, n);
            }
            sb.append(ans - 1).append("\n");
        }
        System.out.println(sb);
    }
}

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

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

Think step by step.

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define lowbit(x) (x&-x)
const int inf=0x3f3f3f3f;

int T,n;

char sss[400005];

int a[400005];

int tr[400005],b[400005];

int stk[400005],tail;

int gp[400005],gtail;

void add(int pos,int d) {
	for(int i=pos;i<=n;i+=lowbit(i)) tr[i]+=d;
}

int sum(int pos) {
	int res=0;
	for(int i=pos;i>=1;i-=lowbit(i)) res+=tr[i];
	return res;
}

pair<int,int> tag[400005];

int main() {
	scanf("%d",&T);
	while(T--) {
		scanf("%d",&n);
		for(int i=1;i<=n;i++) scanf("%d",&a[i]);
		scanf("%s",sss+1);
		memset(tr,0,sizeof(int)*(n+1)),memset(b,0,sizeof(int)*(n+1));
		tail=0,gtail=0;
		int s=0;
		for(int i=1;i<=n;i++) {
			int w=-s;
			while(tail>=1 && a[stk[tail]]<a[i]) {
				if(b[tail]!=-1) gtail--;
				s-=b[tail],w=b[tail],tail--;
			}
			b[++tail]=w,stk[tail]=i,s+=b[tail];
			if(b[tail]!=-1) gp[++gtail]=tail;
			b[gp[gtail]]++,s++;
			if(b[gp[gtail]]==-1) gtail--;
			tag[i+1]={b[1],stk[1]};
		}
		int ans=b[1];
		memset(tr,0,sizeof(int)*(n+1)),memset(b,0,sizeof(int)*(n+1));
		tail=0,gtail=0,s=0;
		for(int i=n;i>=1;i--) {
			int w=-s;
			while(tail>=1 && a[stk[tail]]<a[i]) {
				if(b[tail]!=-1) gtail--;
				s-=b[tail],add(tail,-b[tail]),w=b[tail],tail--;
			}
			b[++tail]=w,add(tail,b[tail]),stk[tail]=i,s+=b[tail];
			if(b[tail]!=-1) gp[++gtail]=tail;
			b[gp[gtail]]++,add(gp[gtail],1),s++;
			if(b[gp[gtail]]==-1) gtail--;

			int L=1,R=tail;
			while(L<R) {
				int mid=(L+R+1)>>1;
				if(a[stk[mid]]>a[tag[i].second]) L=mid;
				else R=mid-1;
			}
			ans=max(ans,tag[i].first+sum(L));
		}
		ans=max(ans,b[1]);
		printf("%d\n",ans-1);
	}
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases that can maximize the failure probability. The code is complex and involves several data structures and algorithms, so we should consider scenarios where the logic might break down.

### Key Points to Consider:
1. **Edge Cases for Permutations**: Permutations with specific patterns (e.g., increasing, decreasing, or specific intervals) can reveal bugs.
2. **Small Test Cases**: Since the output length is limited, we should focus on small-scale test cases that are complete and valid.
3. **String `s`**: The string `s` only contains the character `?`, which gives us flexibility in how the graph is formed.

### Challenging Test Input:
Let's consider a permutation where the elements are in a specific pattern that might cause the code to fail. For example, a permutation with a few large elements followed by smaller elements can be tricky.

### Test Case:
```plaintext
1
6
6 5 4 3 2 1
??????
```

### Explanation:
- **Permutation**: `6 5 4 3 2 1` is a strictly decreasing permutation.
- **String `s`**: `??????` allows for flexibility in choosing the edges.

### Why This Test Case is Challenging:
- **Decreasing Permutation**: In a strictly decreasing permutation, the values of `l_i` and `r_i` for each element will be very specific. For each `i`, `l_i` will be `i` itself, and `r_i` will be `i+1` (except for the last element).
- **Flexibility in Edge Choices**: The flexibility in choosing the edges (since `s` is all `?`) can lead to different connected graphs, and the code needs to handle this correctly to find the maximum possible diameter.

### Expected Output:
- The maximum possible diameter for this test case is `5`, as the graph can be formed such that the longest path has 5 edges.

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'

string S;
int N;
vector<int> arr;
vector<int> stk[2], longest_path[2];

vector<pair<int, int>> calc(int l, int r, vector<int>& L) {
    vector<pair<int, int>> vec;
    int res = 0;
    for (int i = l; i < r; i++) {
        pair<int, int> p = {i, 0};
        while (!vec.empty() && arr[vec.end()[-1].first] < arr[i]) {
            auto [_, v] = vec.end()[-1];
            vec.pop_back();
            if (!vec.empty()) vec.end()[-1].second = max(vec.end()[-1].second, v + 1);
            p.second = max(p.second, v + 1);
        }
        vec.push_back(p);
        res = max(res, vec[0].second + (int)vec.size());
        L[i] = res;
    }
    return vec;
}

void solve() {
    cin >> N;
    arr.resize(N);
    for (int i = 0; i < N; i++) {
        cin >> arr[i];
    }
    cin >> S;
    for (int i = 0; i < 2; i++) {
        stk[i].clear();
        longest_path[i].resize(N);
        vector<pair<int, int>> V = calc(0, N, longest_path[i]);
        for (auto [u, v] : V) stk[i].push_back(u);
        reverse(arr.begin(), arr.end());
    }
    // cout << longest_path[0][stk[0][0]] << " " << longest_path[1][N - stk[0][0] - 1] << " " << stk[0][0] << endl;
    int ans = longest_path[0][stk[0][0]] + longest_path[1][N - stk[0][0] - 1] - 2;
    // cout << "ans: " << ans << endl;
    for (int i = 0; i < 2; i++) {
        vector<int> cur(N);
        for (int j = 1; j < stk[i].size(); j++) {
            calc(stk[i][j - 1], stk[i][j], cur);
            for (int k = stk[i][j - 1]; k < stk[i][j]; k++) {
                // cout << k << " " << N - k - 2 << " " << cur[k] << " " << longest_path[i ^ 1][N - k - 2] << endl;
                ans = max(ans, cur[k] + longest_path[i ^ 1][N - k - 2] - 1);
            }
        }
        reverse(arr.begin(), arr.end());
    }
    cout << ans << endl;
}

signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int T;
    cin >> T;
    while (T--) {
        solve();
    }
    return 0;
}

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

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

Think step by step.

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
#include <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <iostream>
#include <queue>
#include <unordered_map>
typedef long long LL;
typedef unsigned long long ull;
using namespace std;
inline int read() {
    int num = 0 ,f = 1; char c = getchar();
    while (!isdigit(c)) f = c == '-' ? -1 : f ,c = getchar();
    while (isdigit(c)) num = (num << 1) + (num << 3) + (c ^ 48) ,c = getchar();
    return num * f;
}
const int N = 4e5 + 5 ,LG = 20 ,INF = 0x3f3f3f3f;
int arr[N] ,n ,final ,L[N] ,R[N];
struct ST {
    int f[LG][N] ,lg[N];
    inline void build(int n) {
        for (int i = 2; i <= n; i++) lg[i] = lg[i >> 1] + 1;
        for (int i = 1; i <= n; i++) f[0][i] = i;
        for (int i = 1; i <= lg[n]; i++)
            for (int j = 1; j + (1 << i) - 1 <= n; j++) {
                int l = f[i - 1][j] ,r = f[i - 1][j + (1 << (i - 1))];
                f[i][j] = arr[l] >= arr[r] ? l : r;
            }
    }
    inline int query(int l ,int r) {
    	if (l < 1 || r > n) return -INF;
        int p = lg[r - l + 1];
        int x = f[p][l] ,y = f[p][r - (1 << p) + 1];
        return arr[x] >= arr[y] ? x : y;
    }
}st;
unordered_map <ull ,int> mp;
inline int getdis(int x ,int y) {
	if (mp.find(1ull * x * y) != mp.end()) return mp[1ull * x * y];
	int res = 0 ,tx = x ,ty = y;
	if (arr[x] < arr[y] && L[x]) res = getdis(L[x] ,y) + 1;
	else if (arr[x] > arr[y] && R[y]) res = getdis(x ,R[y]) + 1;
	return mp[1ull * tx * ty] = res;
}
inline int solve(int l ,int r) {
    if (l > r) return 0;
    if (l == r) return 1;
    int p = st.query(l ,r);
    int a = solve(l ,p - 1) ,b = solve(p + 1 ,r);
    final = max(final ,a + b + 1);
    if (l != 1 || r != n) final = max(final ,a + b + 2);
    if (l != 1 && r != n)
    	final = max(final ,a + b + 3 + getdis(l - 1 ,r + 1));
    return max(a ,b) + 1;
}
int stk[N] ,top; char str[N];
inline void fuling_solve() {
    n = read();
    for (int i = 1; i <= n; i++) arr[i] = read();
    arr[0] = INF;
    int top = 0;
    for (int i = 1; i <= n; i++) {
        while (top && arr[stk[top]] < arr[i]) top--;
        L[i] = stk[top];
        stk[++top] = i;
    }
    top = 0;
    for (int i = n; i >= 1; i--) {
        while (top && arr[stk[top]] < arr[i]) top--;
        R[i] = stk[top];
        stk[++top] = i;
    }
    st.build(n);
    scanf("%s" ,str + 1);
    final = max(final ,solve(1 ,n));
    printf("%d\n" ,final - 1);
    final = 0;
    mp.clear();
}
signed main() {
    int t = read();
    while (t--) fuling_solve();
    return 0;
}

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

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

Think step by step.

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

#define ii pair<int,int>
#define fi first
#define se second

#define pub push_back
#define pob pop_back

#define rep(x,start,end) for(int x=(start)-((start)>(end));x!=(end)-((start)>(end));((start)<(end)?x++:x--))
#define sz(x) (int)(x).size()

int n;
int arr[400005];
string s;

int ans[2][400005];
int t[400005];
vector<int> stk[2];

vector<ii> process(int l,int r,int val[]){
	vector<ii> v;
	int curr=0;

	rep(x,l,r+1){
		ii res={x,0};
		while (!v.empty() && arr[v.back().fi]<arr[x]){
			int t=v.back().se; v.pob();
			if (!v.empty()) v.back().se=max(v.back().se,t+1);
			res.se=max(res.se,t+1);
		}
		v.pub(res);

		curr=max(curr,res.se+sz(v));
        // cout << "cur: " << curr << " " << res.fi << " " << res.se << endl;
        // for (auto [a, b] : v) {
        //     cout << a << " " << b << endl;
        // }
		val[x]=curr;
	}

	return v;
}

signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin.exceptions(ios::badbit | ios::failbit);

	int TC;
	cin>>TC;
	while (TC--){
		cin>>n;
		rep(x,1,n+1) cin>>arr[x];
		cin>>s;

		rep(z,0,2){
			auto v=process(1,n,ans[z]);
            // for (auto [a, b] : v) {
            //     cout << a << " " << b << endl;
            // }
			stk[z].clear();
			for (auto [a,b]:v) stk[z].pub(a);
            // for (int i : stk[z]) {
            //     cout << i << " ";
            // }
            // cout << endl;
			reverse(arr+1,arr+n+1);
		}
        // int mx = 0;
		int mx=ans[0][stk[0][0]]+ans[1][n-stk[0][0]+1]-1;
        // cout << stk[0][0] << " " << ans[0][stk[0][0]] << " " << ans[1][n - stk[0][0] + 1] << " " << mx - 1 << endl;
		// for (int i = 0; i < 2; i++) {
        //     for (int j = 1; j <= n; j++) {
        //         cout << ans[i][j] << " ";
        //     }
        //     cout << endl;
        // }
        // rep(z, 0, 2) {
        //     if (stk[z].size() > 1) {
        //         process(stk[z][0], stk[z][1], t);
        //         rep(y, stk[z][0], stk[z][1]) {
        //             // cout << y << " " << t[y] << " " << ans[z ^ 1][n - y] << endl;
        //             mx = max(mx, t[y] + ans[z ^ 1][n - y]);
        //         }
        //     }
        //     reverse(arr + 1, arr + n + 1);
        // }

		rep(z,0,2){
			rep(x,0,sz(stk[z])-1){
				process(stk[z][x],stk[z][x+1],t);
                // for (int i = 0; i <= n; i++) {
                //     cout << t[i] << " ";
                // }
                // cout << endl;
                int y = stk[z][x + 1] - 1;
                mx = max(mx, t[y] + ans[z ^ 1][n - y]);
                // mx = max(mx, t[stk[z][x + 1] - 1] + ans[z ^ 1][n - stk[z][x + 1] + 1] - 1);
				// rep(y,stk[z][x],stk[z][x+1]) {
                //     // cout << "y: " << y << " " << t[y] << " " << ans[z^1][n-y] << endl;
                //     mx=max(mx,t[y]+ans[z^1][n-y]);
                // }
			}

			reverse(arr+1,arr+n+1);
		}

		cout<<mx-1<<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 given buggy code, we need to focus on edge cases and scenarios that can cause the algorithm to fail. Here are the steps to create such a test input:

1. **Permutations with Specific Patterns**: The code processes permutations and strings to form graphs. We need to create a permutation that can cause the algorithm to misinterpret the structure of the graph.

2. **String of '?'**: Since the string only contains '?', we need to ensure that the algorithm's choice of adding edges (either to \(l_i\) or \(r_i\)) can lead to incorrect graph formations.

3. **Edge Cases**: Consider edge cases such as small permutations, permutations with a single maximum or minimum value, and permutations with alternating maximum and minimum values.

4. **Graph Formation**: The goal is to create a permutation and string that can lead to a disconnected graph or a graph with a smaller diameter than expected.

Given these considerations, let's create a test input:

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

### Explanation:
- **Permutation**: `1 5 2 4 3`
  - The permutation has a clear structure with a peak in the middle (`5`), followed by a descending sequence (`2 4 3`).
  - This structure can cause the algorithm to form a graph with a small diameter if it incorrectly processes the edges.

- **String**: `?????`
  - Since the string is all '?', the algorithm has the flexibility to choose either \(l_i\) or \(r_i\) for each index. This flexibility can be exploited to form a graph that does not maximize the diameter.

### Expected Behavior:
- The algorithm should correctly identify the largest and smallest indices for each \(p_i\) and form a graph that maximizes the diameter.
- For the given permutation and string, the maximum possible diameter should be 4.

This test input is small-scale and complete, and it is designed to challenge the buggy code by forcing it to make choices that can lead to an incorrect graph formation.


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ls u<<1
#define rs u<<1|1
#define ll long long
#define ii pair<int,int>
#define fi first
#define se second
#define endl '\n'
#define debug(x) cout << #x << ": " << x << endl
#define pub push_back
#define pob pop_back
#define puf push_front
#define pof pop_front
#define lb lower_bound
#define ub upper_bound
#define rep(x,start,end) for(int x=(start)-((start)>(end));x!=(end)-((start)>(end));((start)<(end)?x++:x--))
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()
ll rd()
{
    ll a=0;int f=0;char p=getchar();
    while(!isdigit(p)){f|=p=='-';p=getchar();}
    while(isdigit(p)){a=(a<<3)+(a<<1)+(p^48);p=getchar();}
    return f?-a:a;
}
const int INF=998244353;
const int P=1e9+7;
const int N=1e6+5;
int n;
int a[N];
char s[N];
int f[N],g[N];
int t[N],top;
int len[N];
void solve()
{
	n=rd();
	for(int i=1;i<=n;++i)	a[i]=rd();
	cin>>s;
	//for(int p=0;p<2;++p)
	{
		reverse(a+1,a+n+1);
		top=0;
		for(int i=1;i<=n;++i)
		{
			int mx=0;
			while(top&&t[top]<a[i])
			{
				mx=max(mx,len[top]);
				top--;
			}
			++top;	t[top]=a[i];	len[top]=mx+1;
			for(int j=1;j<top;++j)	len[j]=max(len[j],len[top]+top-j);
			f[i]=len[1];
		}
		reverse(a+1,a+n+1);
		top=0;
		for(int i=1;i<=n;++i)
		{
			int mx=0;
			while(top&&t[top]<a[i])
			{
				mx=max(mx,len[top]);
				top--;
			}
			++top;	t[top]=a[i];	len[top]=mx+1;
			for(int j=1;j<top;++j)	len[j]=max(len[j],len[top]+top-j);
			g[i]=len[1];
		}

	}
	int ans=0;
	for(int i=1;i<n;++i)	ans=max(ans,f[i]+g[n-i]);
	cout<<ans-1<<endl;
}
int main()
{
	int T=rd();
	while(T--)	solve();
    return 0;
}

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

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

Think step by step.

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

### Explanation:
This test case is designed to expose potential issues in the buggy code by providing a permutation that is in strictly descending order. This scenario can be particularly challenging for algorithms that rely on finding elements that are greater than the current element to the left or right, as all elements to the left and right are smaller.

- **Permutation:** `5 4 3 2 1`
- **String:** `?????`

In this case, for each element `i`:
- `l_i` will always be `i` because there is no element to the left that is greater.
- `r_i` will always be `i` because there is no element to the right that is greater.

The algorithm must handle this edge case correctly, and the expected output should be the maximum possible diameter of the graph that can be formed. For this input, the maximum diameter is 4, as the graph can be formed in such a way that the longest path between any two nodes is 4 edges.


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

MAX_N = 400009
MAX_ST = 4000009

def dfs(n, dp, ma, op):
    global ans, ls, rs

    if n == 0:
        return 0

    a = dfs(ls[n], dp + 1, ma + 1 if op == 1 else ma, 1 if op == 1 else -1)
    b = dfs(rs[n], dp + 1, ma + 1 if op == 0 else ma, 0 if op == 0 else -1)
    ans = max(ans, a + b + dp - ma)
    return max(a + 1, b + 1)

def main():
    global ans, ls, rs

    input_data = input().split()
    index = 0

    T = int(input_data[index])
    index += 1

    results = []

    for _ in range(T):
        N = int(input_data[index])
        index += 1

        a = [0] * (N + 1)
        ls = [0] * (N + 1)
        rs = [0] * (N + 1)

        for i in range(1, N + 1):
            a[i] = int(input_data[index])
            index += 1

        s = input_data[index]
        index += 1

        tp = 0
        st = [0] * (MAX_ST + 1)

        for i in range(1, N + 1):
            k = tp
            while k > 0 and a[st[k]] < a[i]:
                k -= 1
            if k > 0:
                rs[st[k]] = i
            if k < tp:
                ls[i] = st[k + 1]
            st[k + 1] = i
            tp = k + 1

        ans = 0
        a = dfs(ls[st[1]], 2, 0, 1)
        b = dfs(rs[st[1]], 2, 0, 0)
        ans = max(ans, a + b + 1)

        results.append(ans - 1)

    for result in results:
        print(result)

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
1
5
5 4 3 2 1
?????
```


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

public class Main {
    private static BufferedReader br;
    private static StringTokenizer tokenizer;
    private static String next() throws IOException{
        if (br == null) br = new BufferedReader(new InputStreamReader(System.in));
        if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());
        while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());
        return tokenizer.nextToken();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }
    private static long nextLong() throws IOException{
        return Long.parseLong(next());
    }

    private static int[] arr = new int[400006];
    private static Pair[][] mps = new Pair[2][400006];
    private static Pair[] stack = new Pair[400006];
    private static class Pair {
        int ele;
        int v;
        private Pair(int ele, int v) {
            this.ele = ele;
            this.v = v;
        }
        public static Pair of(int ele, int v) {
            return new Pair(ele, v);
        }
    }
    private static void swap(int[] arr, int start, int end) {
        while (start < end) {
            int tmp = arr[start];
            arr[start] = arr[end];
            arr[end] = tmp;
            ++start;
            --end;
        }
    }

    public static void main(String[] args)  throws IOException {
        int tt = nextInt();
        StringBuilder sb = new StringBuilder();
        for (int ig = 0; ig < tt; ++ig) {
            int n = nextInt();
            for (int i = 0; i < n; ++i) arr[i + 1] = nextInt();
            String s = next();
            if (ig == 35) {
                int[] outArr = new int[n];
                for (int i = 0; i < n; ++i) outArr[i] = i;
                System.out.println(Arrays.toString(outArr));
            }
            for (int z = 0; z < 2; ++z) {
                int mv = 0;
                int next = 0;
                for (int i = 1; i <= n; ++i) {
                    int cur = 1;
                    while (next > 0 && arr[stack[next - 1].ele] < arr[i]) {
                        cur = Math.max(cur, Math.max(stack[next - 1].v, mv - (next - 1)) + 1);
                        --next;
                    }
                    stack[next++] = Pair.of(i, cur);
                    mv = Math.max(mv, cur + next - 1);
                    mps[z][i] = Pair.of(stack[0].ele, Math.max(stack[0].v, mv));
                }

                swap(arr, 1, n);
            }
            int ans = Math.max(mps[0][n].v, mps[1][n].v);
            for (int z = 0; z < 2; ++z) {
                int next = 0;
                int mv = 0;
                for (int i = 1; i < n; ++i) {
                    Pair op = mps[1 - z][n - i];
                    int cur = 1;
                    while (next > 0 && arr[stack[next - 1].ele] < arr[i]) {
                        cur = Math.max(cur, Math.max(stack[next - 1].v, mv - (next - 1)) + 1);
                        --next;
                    }
                    stack[next++] = Pair.of(i, cur);
                    mv = Math.max(mv, cur + next - 1);
                    int l = 0, r = next;
                    while (l < r) {
                        int mid = l + ((r - l) >> 1);
                        if (arr[stack[mid].ele] > arr[op.ele]) {
                            l = mid + 1;
                        } else {
                            r = mid;
                        }
                    }
                    if (l > 0) ans = Math.max(ans, op.v + Math.max(stack[l - 1].v, mv - (l - 1)));
                }
                swap(arr, 1, n);
            }
            sb.append(ans - 1).append("\n");
        }
        System.out.println(sb);
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that are likely to cause issues. Here are some steps to think through the problem:

1. **Understanding the Bug**: The buggy code seems to have issues with handling the permutation and the string `s` in a way that might not correctly compute the maximum diameter of the graph. Specifically, the code uses a stack to manage the values and their positions, and it might have issues with the order of operations or the way it calculates the maximum values.

2. **Edge Cases**:
   - **Small Permutations**: Test with small permutations to ensure the basic logic works.
   - **Large Permutations**: Test with large permutations to stress the code.
   - **Special Permutations**: Test with permutations that have specific patterns, such as sorted, reverse sorted, or with a few large elements.
   - **Single Element Permutations**: Test with permutations of length 2 or 3 to see if the code handles small cases correctly.

3. **String `s` Handling**:
   - Since `s` only contains `?`, the code should be able to choose the best edge for each position. This means we need to test with permutations that can create different graphs depending on the choice of edges.

4. **Test Case Generation**:
   - Generate a permutation that can create a large diameter when edges are added in a specific way.
   - Ensure the permutation and string `s` are valid and within the constraints.

Based on these considerations, here is a challenging test input:

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

Explanation:
- **Permutation**: `5 4 3 2 1` is a reverse sorted permutation.
- **String `s`**: `?????` allows the code to choose the best edge for each position.
- **Expected Behavior**: The code should handle this permutation and string correctly, but if there are issues with the stack management or the way it calculates the maximum values, this test case will likely expose them.

This test input is small and complete, making it easy to validate the behavior of the buggy code.


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

public class Main {
    private static BufferedReader br;
    private static StringTokenizer tokenizer;
    private static String next() throws IOException{
        if (br == null) br = new BufferedReader(new InputStreamReader(System.in));
        if (tokenizer == null) tokenizer = new StringTokenizer(br.readLine());
        while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(br.readLine());
        return tokenizer.nextToken();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }
    private static long nextLong() throws IOException{
        return Long.parseLong(next());
    }

    private static int[] arr = new int[400006];
    private static Pair[][] mps = new Pair[2][400006];
    private static Pair[] stack = new Pair[400006];
    private static class Pair {
        int ele;
        int v;
        private Pair(int ele, int v) {
            this.ele = ele;
            this.v = v;
        }
        public static Pair of(int ele, int v) {
            return new Pair(ele, v);
        }
    }
    private static void swap(int[] arr, int start, int end) {
        while (start < end) {
            int tmp = arr[start];
            arr[start] = arr[end];
            arr[end] = tmp;
            ++start;
            --end;
        }
    }

    public static void main(String[] args)  throws IOException {
        int tt = nextInt();
        StringBuilder sb = new StringBuilder();
        for (int ig = 0; ig < tt; ++ig) {
            int n = nextInt();
            for (int i = 0; i < n; ++i) arr[i + 1] = nextInt();
            String s = next();
            for (int z = 0; z < 2; ++z) {
                int mv = 0;
                int next = 0;
                for (int i = 1; i <= n; ++i) {
                    int cur = 1;
                    while (next > 0 && arr[stack[next - 1].ele] < arr[i]) {
                        cur = Math.max(cur, Math.max(stack[next - 1].v, mv - (next - 1)) + 1);
                        --next;
                    }
                    stack[next++] = Pair.of(i, cur);
                    mv = Math.max(mv, cur + next - 1);
                    mps[z][i] = Pair.of(stack[0].ele, Math.max(stack[0].v, mv));
                }

                swap(arr, 1, n);
            }
            int ans = Math.max(mps[0][n].v, mps[1][n].v);
            for (int z = 0; z < 2; ++z) {
                int next = 0;
                int mv = 0;
                for (int i = 1; i < n; ++i) {
                    Pair op = mps[1 - z][n - i];
                    int cur = 1;
                    while (next > 0 && arr[stack[next - 1].ele] < arr[i]) {
                        cur = Math.max(cur, Math.max(stack[next - 1].v, mv - (next - 1)) + 1);
                        --next;
                    }
                    stack[next++] = Pair.of(i, cur);
                    mv = Math.max(mv, cur + next - 1);
                    int l = 0, r = next;
                    while (l < r) {
                        int mid = l + ((r - l) >> 1);
                        if (arr[stack[mid].ele] > arr[op.ele]) {
                            l = mid + 1;
                        } else {
                            r = mid;
                        }
                    }
                    if (l > 0) ans = Math.max(ans, op.v + Math.max(stack[l - 1].v, mv - (l - 1)));
                }
                swap(arr, 1, n);
            }
            sb.append(ans - 1).append("\n");
        }
        System.out.println(sb);
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases or scenarios that might cause the algorithm to fail. Here are some potential scenarios to consider:

1. **Permutations with a specific order**: Permutations where the elements are in a specific order, such as sorted or reverse sorted, can sometimes cause issues in algorithms that rely on relative positions.
2. **Edge cases in the permutation**: Permutations with a small number of elements, especially edge cases like `n = 2` or `n = 3`.
3. **Special characters in the string `s`**: Since the string `s` only contains `?`, we need to ensure that the algorithm handles this correctly.

Given these considerations, let's create a small-scale test input that might expose the bug:

### Test Input
```plaintext
5
2
1 2
??
3
3 1 2
???
5
5 2 1 3 4
?????
6
6 2 3 4 5 1
??????
8
1 7 5 6 2 8 4 3
????????
```

### Explanation
1. **Test Case 1**: `n = 2`, `p = [1, 2]`, `s = ??`
   - This is the smallest possible case. The only connected graph possible is a single edge between the two vertices, resulting in a diameter of 1.
2. **Test Case 2**: `n = 3`, `p = [3, 1, 2]`, `s = ???`
   - This case involves a small permutation with a specific order. The maximum possible diameter is 2.
3. **Test Case 3**: `n = 5`, `p = [5, 2, 1, 3, 4]`, `s = ????`
   - This case has a permutation with a more complex order. The maximum possible diameter is 4.
4. **Test Case 4**: `n = 6`, `p = [6, 2, 3, 4, 5, 1]`, `s = ??????`
   - This case involves a larger permutation with a specific order. The maximum possible diameter is 5.
5. **Test Case 5**: `n = 8`, `p = [1, 7, 5, 6, 2, 8, 4, 3]`, `s = ????????`
   - This case involves a larger permutation with a more complex order. The maximum possible diameter is 8.

These test cases cover a range of permutation lengths and orders, which should help in identifying any issues in the given buggy code.


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Buggy Code:
#include <set>
#include <map>
#include <list>
#include <queue>
#include <cmath>
#include <time.h>
#include <random>
#include <bitset>
#include <vector>
#include <cstdio>
#include <stdio.h>
#include <iomanip>
#include <assert.h>
#include <stdlib.h>
#include <memory.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <unordered_set>
#include <unordered_map>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;

#define mk make_pair
#define fi first
#define se second

char buf[1 << 20],*p1,*p2;
inline char gc(){
	if (p1 == p2){
		p1 = buf;
		p2 = buf + fread(buf,1,1 << 20,stdin);
	}
	return p1 == p2 ? EOF : *p1++;
	// return getchar();
}

inline int read(){
	int t = 0,f = 1;
	register char c = gc();
	while (c < 48 || c > 57) f = (c == '-') ? (-1) : (f),c = gc();
	while (c >= 48 && c <= 57) t = (t << 1) + (t << 3) + (c ^ 48),c = gc();
	return f * t;
}

inline void reads(){
	register char c = gc();
	while (c != '?') c = gc();
	while (c == '?') c = gc();
}

const int N = 4e5 + 3;
int T,n,stop,rt,ans,p[N],ls[N],rs[N],fa[N],stk[N],dep[N],mxdep[N],top[N],d[N];

void dfs(int u){
	if (!u) return ;
	mxdep[u] = dep[u] = dep[fa[u]] + 1;
	d[u] = rs[fa[u]] == u;if (u == rt) d[rt] = -1;
	if (d[u] != d[fa[u]]) top[u] = u;
	else top[u] = top[fa[u]];
	dfs(ls[u]),dfs(rs[u]);
	mxdep[u] = max(mxdep[u],max(mxdep[ls[u]],mxdep[rs[u]]));
}

void calc(int u){
	if (!u) return ;
	calc(ls[u]);
	calc(rs[u]);
	ans = max(ans,max(mxdep[ls[u]],mxdep[rs[u]]) - dep[u]);
	if (!ls[u] || !rs[u]) return ;
	ans = max(ans,mxdep[ls[u]] + mxdep[rs[u]] - dep[u] * 2);
	if (u == rt) return ;
	ans = max(ans,mxdep[ls[u]] + mxdep[rs[u]] - dep[u] * 2 + 1);
	if (top[u] != rt){
		if (fa[top[u]] == rt) ans = max(ans,mxdep[ls[u]] + mxdep[rs[u]] - dep[u] * 2 + dep[u] - dep[top[u]] + 1);
		else ans = max(ans,mxdep[ls[u]] + mxdep[rs[u]] - dep[u] * 2 + dep[u] - dep[top[u]] + 2);
	}
}

void solve(){
	n = read();ans = 0;
	for (int i = 1;i <= n;i++) p[i] = read(),ls[i] = rs[i] = fa[i] = 0;
	reads();
	stop = rt = 0;
	for (int i = 1;i <= n;i++){
		while (stop && p[stk[stop]] < p[i]){
			int A = ls[i],s = stk[stop];
			rs[s] = A,fa[A] = s,ls[i] = s,fa[s] = i;
			--stop;
		}
		if (!stop) rt = i;
		else rs[stk[stop]] = i,fa[i] = stk[stop];
		stk[++stop] = i;
	}
	// printf("ls[] = ");for (int i = 1;i <= n;i++) printf("%d ",ls[i]);puts("");
	// printf("rs[] = ");for (int i = 1;i <= n;i++) printf("%d ",rs[i]);puts("");
	d[0] = -1;
	dfs(rt);
	calc(rt);
	printf("%d\n",ans);
}

int main(){
#ifndef ONLINE_JUDGE
	freopen("in.in","r",stdin);
	// freopen("out.out","w",stdout);
#endif
	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:
To expose the bug in the provided code, we need to consider scenarios that challenge the logic of the algorithm, particularly in how it handles the construction of the graph and the calculation of the diameter. Here are some key points to consider:

1. **Permutation with a Long Chain**: A permutation where the elements form a long chain can challenge the algorithm's ability to correctly determine the left and right indices and build the graph.

2. **Permutation with a Single Maximum Element**: A permutation where there is a single maximum element and the rest are in a specific order can expose issues in how the algorithm handles the `l_i` and `r_i` definitions.

3. **Permutation with Alternating Elements**: A permutation where elements alternate in a specific pattern can challenge the algorithm's ability to handle complex relationships between elements.

4. **Small Permutation with Specific Order**: A small permutation with a specific order can be used to ensure the algorithm's logic is correct in simpler cases.

Given these considerations, let's generate a test input that combines these elements to maximize the probability of exposing the bug.

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

### Explanation:
1. **First Test Case**:
   - Permutation: `5 4 3 2 1`
   - String: `?????`
   - This permutation forms a long chain where each element is greater than the next. The algorithm should correctly determine the left and right indices and build the graph. The expected diameter is 4.

2. **Second Test Case**:
   - Permutation: `6 5 4 3 2 1`
   - String: `??????`
   - Similar to the first test case, but with a longer chain. The expected diameter is 5.

3. **Third Test Case**:
   - Permutation: `4 2 3 1`
   - String: `???`
   - This permutation has a more complex structure with a single maximum element and a mix of other elements. The expected diameter is 3.

These test cases are designed to challenge the algorithm's ability to correctly determine the left and right indices and build the graph, thereby maximizing the probability of exposing any bugs in the code.