**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Desert Rabbit
time_limit_ms: 3000
memory_limit_mb: 256
Description: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!

While trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.

The rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:

- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.
- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.

For each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5 \cdot 10^5$$$) — the number of trees.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$) — the height of the trees.

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

Output: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Examples:
input:
5
4
2 3 1 4
5
5 4 3 2 1
4
2 1 1 3
4
1 1 3 1
8
2 4 1 6 3 8 5 7
output:
3 3 3 4
5 5 5 5 5
2 2 2 3
1 1 3 3
8 8 8 8 8 8 8 8

Note: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.

- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.
- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.

In the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.

In the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.

Buggy Code:
#include<bits/stdc++.h>
#include <cmath>
#include <complex>
#include<string>
#include <ext/pb_ds/assoc_container.hpp>
//#include "testlib.h"
//freopen("orxor.in", "r", stdin);
using namespace std;
//using namespace __gnu_pbds;
//template <class T>
//using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define ll long long
#define pb push_back
#define F first
#define S second
#define fast ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define dcm(a) setprecision(a)<<fixed4
#define double long double
ll const md= 1e9+7;
double const EPS=1e-9;
int const LOG=22;
using uint = unsigned int;
using ull = unsigned long long;
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define clr(memo, a) memset(memo,a,sizeof(memo))
#define PI acos(-1)
#define endl '\n'
struct segtree
{
    vector<ll> value;
    vector<ll> op;
    ll NEUTRAL=0;
    ll NO_OP=0;
    int sz=1;
    void init(int n)
    {
        sz=1;
        while(sz<n)
            sz*=2;
        value.assign(2*sz,0);
        op.assign(2*sz,0);
    }
    ll modify_op(ll a,ll b,ll len)
    {
        if(b==NO_OP)
            return a;
        return (b*len);
    }
    ll calcval(ll a,ll b)
    {
        return max(a,b);
    }
    void apply_mod(ll &a,ll b,ll len)
    {
        a=modify_op(a,b,len);
    }
    void propagte(int x,int lx,int rx)
    {
        if(rx-lx==1)
            return;
        int m=(rx+lx)/2;
        apply_mod(value[2*x+1],op[x],m-lx);
        apply_mod(value[2*x+2],op[x],rx-m);
        apply_mod(op[2*x+1],op[x],1);
        apply_mod(op[2*x+2],op[x],1);
        op[x]=NO_OP;
    }
    void modify(int l,int r,ll v,int x,int lx,int rx)
    {
        propagte(x,lx,rx);
        if(lx>=r||l>=rx)return;
        if(lx>=l&&rx<=r)
        {
            apply_mod(op[x],v,1);
            apply_mod(value[x],v,rx-lx);
            return;
        }
        int m=(rx+lx)/2;
        modify(l,r,v,2*x+1,lx,m);
        modify(l,r,v,2*x+2,m,rx);
        value[x]=calcval(value[2*x+1],value[2*x+2]);

    }
    void modify(int l,int r,ll  v)
    {
        modify(l,r,v,0,0,sz);
    }
    ll calc(int l,int r,int x,int lx,int rx)
    {
        propagte(x,lx,rx);
        if(lx>=r||l>=rx)
            return NEUTRAL;
        if(lx>=l&&rx<=r)
        {
            return value[x];
        }
        int m=(rx+lx)/2;
        ll c1=calc(l,r,2*x+1,lx,m);
        ll c2=calc(l,r,2*x+2,m,rx);
        return calcval(c1,c2);
    }

    ll calc(int l,int r)
    {
        return calc(l,r,0,0,sz);
    }
    void build(vector<ll> &a,int x,int lx,int rx)
    {
        if(rx-lx==1)
        {

            if(lx<a.size())
                value[x]=a[lx];
            return;
        }
        int md=(lx+rx)/2;
        build(a,2*x+1,lx,md);
        build(a,2*x+2,md,rx);
        value[x]=calcval(value[2*x+1],value[2*x+2]);
    }
    void build(vector<ll> &a)
    {
        build(a,0,0,sz);
    }

};
void solve(int testCase)
{
    int n;
    cin>>n;
    segtree seg;
    seg.init(n);
    vector<int> ans(n+1);
    vector<vector<int> > adj(n+5);
    int mx=0;
    for(int i=0; i<n; i++)
    {
        int x;
        cin>>x;
        mx=max(mx,x);
        ans[i]=mx;
        adj[x].push_back(i);
    }
    for(int i=1; i<=n; i++)
    {
        for(int j:adj[i])
        {
            int s=seg.calc(j+1,n);
            ans[j]=max(ans[j],s);
        }
        for(int j:adj[i])
        {
            seg.modify(j,j+1,ans[j]);
            if(ans[j]>i){
                adj[ans[j]].push_back(j);
            }
        }
    }
    for(int i=0; i<n; i++)
        cout<<ans[i]<<" ";
    cout<<endl;
}

void precalc()
{

}

int main()
{

    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cerr.tie(0);
    int testCase=1;
    precalc();
    //freopen("permutations.in", "r", stdin);
    cin>>testCase;
    //freopen("test.txt", "w", stdout);
    for(int i=1; i<=testCase; i++)
        solve(i);

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
4
4 3 2 1
5
1 2 3 4 5
8
1 3 5 7 2 4 6 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Desert Rabbit
time_limit_ms: 3000
memory_limit_mb: 256
Description: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!

While trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.

The rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:

- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.
- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.

For each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5 \cdot 10^5$$$) — the number of trees.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$) — the height of the trees.

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

Output: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Examples:
input:
5
4
2 3 1 4
5
5 4 3 2 1
4
2 1 1 3
4
1 1 3 1
8
2 4 1 6 3 8 5 7
output:
3 3 3 4
5 5 5 5 5
2 2 2 3
1 1 3 3
8 8 8 8 8 8 8 8

Note: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.

- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.
- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.

In the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.

In the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.

Buggy Code:
#include <bits/stdc++.h>
#define ff cout.flush()
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define FAST ios_base::sync_with_stdio(0) , cin.tie(0) , cout.tie(0);
#define ll long long
#define str string
using ld = long double;
#define pb push_back
#define pf push_front
#define pob pop_back()
#define pof pop_front()
#define in insert
#define fi first
#define se second
#define begin begin()
#define end end()
#define size size()
#define back back()
#define rend rend()
#define rbegin rbegin()
#define ret return
#define ull unsigned long long
#define all(a) a.begin , a.end
#define gcd(a , b) __gcd(a , b)
#define lcm(a , b) (a * b) / gcd(a , b)
#define friopen freopen ("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
using namespace std;
using namespace __gnu_pbds;
typedef tree<long long, null_type, less_equal<long long>, rb_tree_tag, tree_order_statistics_node_update> pbds;

///                         ERASE
/*

str --  a.erase(index nachala , kolichestvo simvolov);

deque , vector -- a.erase(a.begin + index nachala , a.begin + index konca + 1);

*/

///------------

const ll mod = 1e9 + 7 , mod2 = 998244353;
const ll inf = 1e18;
const ld esp = 0.0000001 , Pi = 3.1415926535897932384626433832795;

ll binpow(ll x , ll y)
{
    if (y == 0) ret 1;
    else if (y == 1) ret x;
    else if (y % 2 == 1) ret binpow(x , y - 1) * x;
    else ret binpow(x , y / 2) * binpow(x , y / 2);
}

ll qpow(ll a , ll b , ll md){
	a = (a + md)%md;
	ll ans = 1;
	while(b > 0){
		if(b&1){
			ans = (ans*a)%md;
		}
		a = (a * a)%md;
		b>>=1;
	}
	return ans;
}

ll div(ll a, ll b, ll mod) {
    ret (a % mod) * (qpow(b, mod-2 , mod)) % mod;
}

ll sum_ (ll x)
{
    ret x * (x + 1) / 2;
}

ll ask(ll x , ll y , ll a , ll b)
{
	ret (x == a) + (y == b);
}

ld dist(ld x , ld y , ld a , ld b)
{
	ret sqrt(abs(x - a) * abs(x - a) + abs(y - b) * abs(y - b));
}

void kol_a()
{
	ll n , i , j , mn = 1e9 , mx = 0;
	cin >> n;
	ll a[n + 12] = {} , s[n + 12] = {} , d[n + 12] = {};
	for (i = 1 ; i <= n ; i ++)
	{
		cin >> a[i];
		d[i] = d[i - 1];
		if (a[i] > mx)
		{
			mx = a[i];
			d[i] = i;
		}
	}
	for (i = n ; i >= 1 ; i --)
	{
		s[i] = s[i + 1];
		if (a[i] < mn)
		{
			mn = a[i];
			s[i] = i;
		}
	}
	ll ans[n + 12] = {};
	for (i = 1 ; i <= n ; i ++)
	{
		if (ans[i])
		{
			cout << ans[i] << " ";
			continue;
		}
		map<ll , ll> us;
		ll ind = i;
		set<ll> v;
		while(us[ind] == 0 && a[ind] != mx)
		{
			v.in(ind);
			us[ind] ++;
			ll l = ind , r = n + 1 , m;
			while(l + 1 < r)
			{
				m = (l + r) / 2;
				if (a[ind] > a[s[m]]) l = m;
				else r = m;
			}
			ind = d[l];
		}
		cout << a[ind] << " ";
		for (auto x : v) ans[x] = a[ind];
	}
	cout << '\n';
}

main()
{
	//cout.flush()
	FAST
	// friopen
	ll _ = 1;
	cin >> _;
    while(_ --) kol_a();
}

/*

*/

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases where the code might fail. The code is designed to determine the maximum height a rabbit can reach starting from each tree, given the conditions for jumping. Here are some scenarios to consider:

1. **All Trees of the Same Height**: If all trees have the same height, the rabbit cannot jump anywhere.
2. **Increasing Heights**: If the heights of the trees are strictly increasing, the rabbit can only jump backward to taller trees.
3. **Decreasing Heights**: If the heights of the trees are strictly decreasing, the rabbit can only jump forward to shorter trees.
4. **Mixed Heights**: A mix of increasing and decreasing heights can test the logic for both forward and backward jumps.

Let's construct a small-scale test input that combines these scenarios to maximize the failure probability:

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

Explanation:
- The first test case has all trees of the same height (1).
- The second test case has strictly increasing heights.
- The third test case has strictly decreasing heights.
- The fourth test case has a mix of increasing and decreasing heights.
- The fifth test case has a more complex mix of heights.

This input will help identify any issues with the logic for handling different height patterns and ensure the code correctly processes each scenario.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Desert Rabbit
time_limit_ms: 3000
memory_limit_mb: 256
Description: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!

While trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.

The rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:

- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.
- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.

For each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5 \cdot 10^5$$$) — the number of trees.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$) — the height of the trees.

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

Output: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Examples:
input:
5
4
2 3 1 4
5
5 4 3 2 1
4
2 1 1 3
4
1 1 3 1
8
2 4 1 6 3 8 5 7
output:
3 3 3 4
5 5 5 5 5
2 2 2 3
1 1 3 3
8 8 8 8 8 8 8 8

Note: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.

- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.
- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.

In the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.

In the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.

Buggy Code:
visitados = [0]
saida = [0]

def bfs(grafo, lista, i):
    if visitados[i] == 0:
        visitados[i] = 1
    else:
        return 0

    fila = [i]
    aux = [i]
    maior = lista[i]

    while fila:
        v_atual = fila[0]
        del fila[0]
        for vz in grafo[v_atual]:
            if visitados[vz] == 0:
                if lista[vz] > maior:
                    maior = lista[vz]
                fila.append(vz)
                aux.append(vz)
                visitados[vz] = 1

    for valor in aux:
        saida[valor] = maior

t = int(input())
for _ in range(t):
    n = int(input())
    visitados = [0] * n
    saida = [0] * n
    lista = list(map(int, input().split()))

    grafo = [[] for _ in range(n)]
    for i in range(n):
        for j in range(i, n):
            if lista[i] > lista[j]:
                grafo[i].append(j)
                grafo[j].append(i)

    for i in range(n):
        bfs(grafo, lista, i)
        print(saida[i], end=' ')
    print()

**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
1 2 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Desert Rabbit
time_limit_ms: 3000
memory_limit_mb: 256
Description: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!

While trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.

The rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:

- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.
- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.

For each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5 \cdot 10^5$$$) — the number of trees.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$) — the height of the trees.

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

Output: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Examples:
input:
5
4
2 3 1 4
5
5 4 3 2 1
4
2 1 1 3
4
1 1 3 1
8
2 4 1 6 3 8 5 7
output:
3 3 3 4
5 5 5 5 5
2 2 2 3
1 1 3 3
8 8 8 8 8 8 8 8

Note: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.

- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.
- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.

In the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.

In the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.

Buggy Code:
I = lambda : list(map(int, input().split(' ')))
R = lambda : (int(input()))

import sys
import math
from collections import deque, defaultdict
from bisect import bisect_left, bisect_right
from heapq import heappop, heapify, heappush
# from typing import List

def join_parents(tp1 : int, tp2 : int, grp_map, parents_map ):

    # print(tp1, tp2)
    tp1 = parents_map[tp1]
    tp2 = parents_map[tp2]

    if len(grp_map[tp1]) < len(grp_map[tp2]):
        tp1, tp2 = tp2, tp1

    if parents_map[tp1] == parents_map[tp2] : return

    for e in grp_map[tp2]:
        grp_map[tp1].add(e)
        parents_map[e] = parents_map[tp1]

    grp_map[tp2].clear()
    grp_map[tp2] = []

for kp in range(R()):

    n = R()
    a = I()

    max_height = []
    parents = []
    new_a = []

    for i in range(n):
        parents.append(i) ;
        max_height.append(set([i])) ;
        new_a.append([(a[i],i)])

    # if n == 500000 : print("kk") ; continue

    while len(new_a) > 1:

        temp_a = []

        for i in range(1,len(new_a),2):
            arr1 = new_a[i-1]
            arr2 = new_a[i]

            arr3 = []

            ind1 = -1
            ind2 = 0

            curr_val = arr2[ind2]

            for j in range(len(arr1)):
                if arr2[ind2][0] < arr1[j][0]: break
                else :
                    arr3.append(arr1[j])
                    if j == len(arr1) - 1: j += 1

            # updating the max_height of arr1
            for k in range(j,len(arr1)):
                # ind = arr1[k][1]
                # max_height[ind] = max(max_height[ind], arr1[-1][0])
                join_parents(arr1[k][1], arr1[-1][1], max_height, parents)

            ind1 = j

            while ind1 < len(arr1) and ind2 < len(arr2):

                if arr1[ind1][0] > arr2[ind2][0]:
                    arr3.append(arr2[ind2])

                    # updating the max_height
                    # curr_max_height = max( max_height[arr2[ind2][1]], max_height[arr1[ind1][1]] )
                    # max_height[arr2[ind2][1]] = curr_max_height
                    # max_height[arr1[ind1][1]] = curr_max_height
                    join_parents(arr2[ind2][1], arr1[ind1][1], max_height, parents)
                    # print(ind1, ind2, arr2[ind2],  arr1[ind1])
                    ind2 += 1

                else:
                    arr3.append(arr1[ind1])
                    ind1 += 1

            arr3 = arr3 + arr1[ind1:] + arr2[ind2:]
            temp_a.append(arr3)

            # print(arr3, max_height, parents)

        # print("new_a : ", new_a,'\n', "temp_a : ", temp_a, '\n')
        new_a = temp_a + new_a[i+1:]
        # del temp_a
        # del arr3
        # del arr1

    max_val = []
    for i in range(n):
        val = -1
        for j in max_height[i]: val = max(val, a[j])
        max_val.append(val)

    res = []
    for i in range(n):
        res.append(max_val[parents[i]])

    print(*res)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Desert Rabbit
time_limit_ms: 3000
memory_limit_mb: 256
Description: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!

While trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.

The rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:

- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.
- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.

For each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5 \cdot 10^5$$$) — the number of trees.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$) — the height of the trees.

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

Output: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Examples:
input:
5
4
2 3 1 4
5
5 4 3 2 1
4
2 1 1 3
4
1 1 3 1
8
2 4 1 6 3 8 5 7
output:
3 3 3 4
5 5 5 5 5
2 2 2 3
1 1 3 3
8 8 8 8 8 8 8 8

Note: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.

- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.
- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.

In the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.

In the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.

Buggy Code:
import java.util.*;

public class RabbitJump {
    public static void solve() {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();

        while (t-- > 0) {
            int n = sc.nextInt();
            int[] heights = new int[n];

            for (int i = 0; i < n; i++) {
                heights[i] = sc.nextInt();
            }

            // For each starting position
            for (int start = 0; start < n; start++) {
                boolean[] visited = new boolean[n];
                int maxHeight = dfs(start, heights, visited);
                System.out.print(maxHeight + " ");
            }
            System.out.println();
        }
    }

    private static int dfs(int current, int[] heights, boolean[] visited) {
        visited[current] = true;
        int maxHeight = heights[current];

        // Try all possible jumps
        for (int next = 0; next < heights.length; next++) {
            if (!visited[next]) {
                // Check jumping conditions
                if ((next < current && heights[next] > heights[current]) ||
                    (next > current && heights[next] < heights[current])) {
                    maxHeight = Math.max(maxHeight, dfs(next, heights, visited));
                }
            }
        }

        return maxHeight;
    }

    public static void main(String[] args) {
        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
5
4
1 2 3 4
4
4 3 2 1
5
1 2 3 4 5
5
5 4 3 2 1
6
3 2 4 1 5 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Desert Rabbit
time_limit_ms: 3000
memory_limit_mb: 256
Description: Dedicated to pushing himself to his limits, Penchick challenged himself to survive the midday sun in the Arabian Desert!

While trekking along a linear oasis, Penchick spots a desert rabbit preparing to jump along a line of palm trees. There are $$$n$$$ trees, each with a height denoted by $$$a_i$$$.

The rabbit can jump from the $$$i$$$-th tree to the $$$j$$$-th tree if exactly one of the following conditions is true:

- $$$j < i$$$ and $$$a_j > a_i$$$: the rabbit can jump backward to a taller tree.
- $$$j > i$$$ and $$$a_j < a_i$$$: the rabbit can jump forward to a shorter tree.

For each $$$i$$$ from $$$1$$$ to $$$n$$$, determine the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

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

The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 5 \cdot 10^5$$$) — the number of trees.

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$) — the height of the trees.

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

Output: For each test case, output $$$n$$$ integers. The $$$i$$$-th integer should contain the maximum height among all trees that the rabbit can reach if it starts from the $$$i$$$-th tree.

Examples:
input:
5
4
2 3 1 4
5
5 4 3 2 1
4
2 1 1 3
4
1 1 3 1
8
2 4 1 6 3 8 5 7
output:
3 3 3 4
5 5 5 5 5
2 2 2 3
1 1 3 3
8 8 8 8 8 8 8 8

Note: In the first test case, the initial heights of trees are $$$a = [2, 3, 1, 4]$$$.

- If the rabbit starts from the first tree, it can jump to the third tree as $$$3 > 1$$$ and $$$1 < 2$$$. Then, the rabbit can jump to the second tree as $$$2 < 3$$$ and $$$3 > 1$$$. It can be proved that the rabbit cannot reach the fourth tree; hence, the maximum height of the tree that the rabbit can reach is $$$a_2 = 3$$$.
- If the rabbit starts from the fourth tree, it does not need to jump anywhere as it is already at the highest tree.

In the second test case, the rabbit can jump to the first tree regardless of which tree it starts from.

In the fifth test case, if the rabbit starts from the fifth tree, it can jump to the fourth tree. Then the rabbit can jump to the seventh tree and finally reach the sixth tree. Therefore, the maximum height of the tree that the rabbit can reach is $$$8$$$.

Buggy Code:
import java.util.*;
public class program{
	public static void main(String [] args) {
		Scanner sc=new Scanner(System.in);
		int t= sc.nextInt();
		while(t-->0) {
			int n=sc.nextInt();
			ArrayList<Integer> x=new ArrayList<>();
			int [] prevMax=new int[n];
			int max=0;
			for(int i=0;i<n;i++) {
				int a=sc.nextInt();
				if(a>max) {
					max=a;
				}
				x.add(a);
				prevMax[i]=max;
			}
			Map<Integer,Integer> map=new HashMap<>();
			int [] ans=new int[n];
			int k=n-1;
			for(int i=n-1;i>=0;i--) {
				int mxel=0;
				int a=prevMax[i];
				for(Map.Entry<Integer,Integer> y: map.entrySet()) {
					if(y.getKey()<a) {
						mxel=Math.max(mxel,y.getValue());
					}
				}
				int z=Math.max(mxel,a);
				map.put(x.get(i),z);
				ans[k--]=z;
			}
			for(int i=0;i<n;i++) {
				System.out.print(ans[i]+" ");
			}
			System.out.println();
		}
	}
}

**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
5
5
1 2 3 4 5
5
5 4 3 2 1
4
1 1 1 1
4
1 2 2 3
8
1 3 2 4 5 6 7 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Auto Complete
time_limit_ms: 4000
memory_limit_mb: 1024
Description: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types "App", your editor will magically suggest the word "Application"! Even better, users can personalize the words that auto-complete in your editor.

Your editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):

1. Add an auto complete pattern $$$p_i$$$.
2. Delete an auto complete pattern $$$p_i$$$.
3. Append a string $$$s$$$ to the end of $$$t$$$.
4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.

After each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:

1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.
2. If there are multiple $$$p_i$$$, pick the longest one.
3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.
4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.

For example, let us say we have three candidates: "alice", "bob", and "charlie", with ID 1, 2, and 3. At first, there is nothing on the screen, so "charlie" (3) should be suggested because it is the longest. Then, let us say the user types "b". You should suggest "bob" (2) because it is the only one that starts with "b". Finally, let us say the user type "body". You should print -1 because there is no matched pattern.

Input: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.

There are four types of actions:

1. $$$i$$$ $$$p_i$$$
2. $$$i$$$
3. $$$s$$$
4. $$$c$$$

- $$$1 \leq n \leq 10^6$$$
- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\times 10^6$$$.
- $$$1 \leq c \leq 2\times 10^6$$$
- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).
- The ID $$$i$$$ for each add operation is unique.
- The ID $$$i$$$ for each delete is guaranteed to be valid.
- Each ID $$$i$$$ satisfies $$$0\le i \le n$$$.

Output: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.

Examples:
input:
6
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
append 2_bobabc
backspace 3
output:
1
1
3
3
-1
2

input:
6
append pattern
add 1 pattern1_alice____
add 2 pattern2_bob______
add 3 pattern3_charlie__
delete 1
delete 2
output:
-1
1
1
1
2
3

Note:

Buggy Code:
/*
  Code by: linvg
  Created: 25.11.2024 14:07:16
*/

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

#define int long long
#define all(x) (x).begin(), (x).end()
#define sz(x) ((int)(x).size())
string to_upper(string a) { for (int i = 0;i < (int)a.size();++i) if (a[i] >= 'a' && a[i] <= 'z') a[i] -= 'a' - 'A'; return a; }
string to_lower(string a) { for (int i = 0;i < (int)a.size();++i) if (a[i] >= 'A' && a[i] <= 'Z') a[i] += 'a' - 'A'; return a; }

template<class T> T gcd(T a, T b) { T r; while (b != 0) { r = a % b; a = b; b = r; } return a; }
template<class T> T gcd(initializer_list<T> __l) { int a = 0; for (auto x : __l) { a = gcd(a, x); } return a; }
template<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; }
template<class T> T sqr(T x) { return x * x; }
template<class T> T cube(T x) { return x * x * x; }

// #define DBG

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

const int INF = 0x3f3f3f3f3f;
const int MOD = 1e9 + 7;

unordered_map<char, int> mapped;

#define MAXLEN 1000010
using namespace std;
constexpr uint64_t mod = (1ULL << 61) - 1;
const uint64_t seed = chrono::system_clock::now().time_since_epoch().count();
const uint64_t base = mt19937_64(seed)() % (mod / 3) + (mod / 3);
uint64_t base_pow[MAXLEN];
int64_t modmul(uint64_t a, uint64_t b) {
   uint64_t l1 = (uint32_t)a, h1 = a >> 32, l2 = (uint32_t)b, h2 = b >> 32;
   uint64_t l = l1 * l2, m = l1 * h2 + l2 * h1, h = h1 * h2;
   uint64_t ret = (l & mod) + (l >> 61) + (h << 3) + (m >> 29) + (m << 35 >> 3) + 1;
   ret = (ret & mod) + (ret >> 61);
   ret = (ret & mod) + (ret >> 61);
   return ret - 1;
}
void init() {
   base_pow[0] = 1;
   for (int i = 1; i < MAXLEN; i++) {
      base_pow[i] = modmul(base_pow[i - 1], base);
   }
}
struct PolyHash {
   /// Remove suff vector and usage if reverse hash is not required for more speed
   vector<int64_t> pref, suff;
   PolyHash() {}
   template <typename T>
   PolyHash(const vector<T>& ar) {
      if (!base_pow[0]) init();
      int n = ar.size();
      assert(n < MAXLEN);
      pref.resize(n + 3, 0), suff.resize(n + 3, 0);
      for (int i = 1; i <= n; i++) {
         pref[i] = modmul(pref[i - 1], base) + ar[i - 1] + 997;
         if (pref[i] >= mod) pref[i] -= mod;
      }
      for (int i = n; i >= 1; i--) {
         suff[i] = modmul(suff[i + 1], base) + ar[i - 1] + 997;
         if (suff[i] >= mod) suff[i] -= mod;
      }
   }
   PolyHash(const char* str)
      : PolyHash(vector<char>(str, str + strlen(str))) {
   }
   uint64_t get_hash(int l, int r) {
      int64_t h = pref[r + 1] - modmul(base_pow[r - l + 1], pref[l]);
      return h < 0 ? h + mod : h;
   }
   uint64_t rev_hash(int l, int r) {
      int64_t h = suff[l + 1] - modmul(base_pow[r - l + 1], suff[r + 2]);
      return h < 0 ? h + mod : h;
   }
   bool is_palindrome(int l, int r) {
      return get_hash(l, r) == rev_hash(l, r);
   }
};

// compare two strings (1: s < t, 0: s == t, -1: s > t)
int compare(PolyHash& a, PolyHash&b) {
   int len = min(a.pref.size(), b.pref.size()) - 1;
   int l = 0, r = len;
   while (l < r) {
      int mid = (l + r + 1) / 2;
      if (a.get_hash(0, mid - 1) == b.get_hash(0, mid - 1)) l = mid;
      else r = mid - 1;
   }
   if (l == len) {
      if (a.pref.size() == b.pref.size()) return 0;
      return a.pref.size() < b.pref.size() ? 1 : -1;
   }
   return a.pref[l] < b.pref[l] ? 1 : -1;
}

void concat(string& s, string& t, int& start) {
   if (start == 0) {
      s = t;
      start = sz(s);
   }
   else if (start == sz(s)) {
      s += t;
      start = sz(s);
   }
   else {
      int n = sz(s);
      for (int i = 0; i < t.size(); i++, start++) {
         if (start < sz(s)) s[start] = t[i];
         else s += t[i];
      }
   }
}

const int N = 1e6 + 7;

struct Hash {
   size_t operator()(const pair<int, int>& p) const {
      return p.first ^ p.second;
   }
};

unordered_map<int, PolyHash> mp;

struct Trie
{
   struct Node
   {
      unordered_map<int, Node*> next;
      unordered_set<pair<int, int>, Hash> comb;
   };

   Node* root;
   Trie() {
      root = new Node();
   }

   void add(string& s, int idx) {
      Node* cur = root;
      for (int i = 0; i < s.size(); i++) {
         int c = mapped[s[i]];
         if (cur->next.count(c) == 0) {
            cur->next[c] = new Node();
         }
         cur->comb.insert({ sz(s), N - idx });
         cur = cur->next[c];
      }
      cur->comb.insert({ sz(s), N - idx });
   }

   int find(string& t, int start) {
      Node* cur = root;
      int idx = -1, loop = 0, mxLen = 0;
      for (; loop < start; ++loop) {
         if (cur->next.count(mapped[t[loop]]) == 0 || cur->next[mapped[t[loop]]]->comb.size() == 0) return -1;
         cur = cur->next[mapped[t[loop]]];
      }
      if (cur->comb.size() > 0) {
         int c = -1;
         for (auto& p : cur->comb) {
            if (p.first > mxLen) {
               mxLen = p.first;
               idx = N - p.second;
               c = N - p.second;
            }
            else if (p.first == mxLen) {
               if (compare(mp[c], mp[N - p.second]) == -1) {
                  c = N - p.second;
                  idx = N - p.second;
               }
               else if (compare(mp[c], mp[N - p.second]) == 0) {
                  idx = min(idx, N - p.second);
               }
            }
         }
      }
      return idx;
   }

   void del(string& s, int idx) {
      Node* cur = root;
      // dbg("del", s, idx);
      for (int i = 0; i < s.size(); i++) {
         int c = mapped[s[i]];
         if (cur->next.count(c) == 0) return;
         cur->comb.erase({ sz(s), N - idx });
         cur = cur->next[c];
      }
      // dbg("3", cur->comb);
      cur->comb.erase({ sz(s), N - idx });
   }
};

void solve()
{
   for (int i = 33; i <= 126; i++) {
      mapped[(char)i] = i - 33;
   }

   int n;
   cin >> n;
   Trie trie;
   string t;
   int start = 0;
   unordered_map<int, string> cache;
   while (n--) {
      string op;
      cin >> op;
      // dbg("op", op);
      if (op == "add") {
         string pattern;
         int idx;
         cin >> idx >> pattern;
         vector<char> v(pattern.begin(), pattern.end());
         mp[idx] = PolyHash(v);
         cache[idx] = pattern;
         trie.add(pattern, idx);
      }
      else if (op == "append") {
         string pattern;
         cin >> pattern;
         concat(t, pattern, start);
      }
      else if (op == "delete") {
         int idx;
         cin >> idx;
         trie.del(cache[idx], idx);
      }
      else {
         int del;
         cin >> del;
         start -= min(start, del);
      }
      // dbg("t", t, start);
      cout << trie.find(t, start) << '\n';
   }
}

int32_t main() {
   ios_base::sync_with_stdio(0); cin.tie(NULL); cout.tie(NULL);
   // 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
6
add 1 pattern1
add 2 pattern12
add 3 pattern123
append pattern12
backspace 1
backspace 2
```

6
add 1 pattern1
add 2 pattern12
add 3 pattern123
append pattern12
backspace 1
backspace 2


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Auto Complete
time_limit_ms: 4000
memory_limit_mb: 1024
Description: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types "App", your editor will magically suggest the word "Application"! Even better, users can personalize the words that auto-complete in your editor.

Your editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):

1. Add an auto complete pattern $$$p_i$$$.
2. Delete an auto complete pattern $$$p_i$$$.
3. Append a string $$$s$$$ to the end of $$$t$$$.
4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.

After each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:

1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.
2. If there are multiple $$$p_i$$$, pick the longest one.
3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.
4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.

For example, let us say we have three candidates: "alice", "bob", and "charlie", with ID 1, 2, and 3. At first, there is nothing on the screen, so "charlie" (3) should be suggested because it is the longest. Then, let us say the user types "b". You should suggest "bob" (2) because it is the only one that starts with "b". Finally, let us say the user type "body". You should print -1 because there is no matched pattern.

Input: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.

There are four types of actions:

1. $$$i$$$ $$$p_i$$$
2. $$$i$$$
3. $$$s$$$
4. $$$c$$$

- $$$1 \leq n \leq 10^6$$$
- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\times 10^6$$$.
- $$$1 \leq c \leq 2\times 10^6$$$
- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).
- The ID $$$i$$$ for each add operation is unique.
- The ID $$$i$$$ for each delete is guaranteed to be valid.
- Each ID $$$i$$$ satisfies $$$0\le i \le n$$$.

Output: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.

Examples:
input:
6
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
append 2_bobabc
backspace 3
output:
1
1
3
3
-1
2

input:
6
append pattern
add 1 pattern1_alice____
add 2 pattern2_bob______
add 3 pattern3_charlie__
delete 1
delete 2
output:
-1
1
1
1
2
3

Note:

Buggy Code:
class TrieNode:
    def __init__(self):
        self.children = {}
        self.pattern_ids = []

class Trie:
    def __init__(self):
        self.root = TrieNode()

    def insert(self, pattern, pattern_id):
        node = self.root
        for char in pattern:
            if char not in node.children:
                node.children[char] = TrieNode()
            node = node.children[char]
        node.pattern_ids.append(pattern_id)
        node.pattern_ids.sort()

    def delete(self, pattern, pattern_id):
        node = self.root
        nodes = []
        chars = []
        for char in pattern:
            nodes.append(node)
            chars.append(char)
            node = node.children[char]
        node.pattern_ids.remove(pattern_id)
        if not node.children and not node.pattern_ids:
            for i in range(len(pattern)-1, -1, -1):
                char = chars[i]
                node = nodes[i]
                del node.children[char]
                if node.children or node.pattern_ids:
                    break

    def find_matching_patterns(self, prefix):
        node = self.root
        for char in prefix:
            if char not in node.children:
                return []
            node = node.children[char]
        return self._collect_all_patterns(node)

    def _collect_all_patterns(self, node):
        patterns = []
        stack = [(node, "")]
        while stack:
            current, prefix = stack.pop()
            if current.pattern_ids:
                for pattern_id in current.pattern_ids:
                    patterns.append((prefix, pattern_id))
            for char, child_node in current.children.items():
                stack.append((child_node, prefix + char))
        return patterns

def find_best_match(current_text, patterns):
    matching_patterns = trie.find_matching_patterns(current_text)
    if not matching_patterns:
        print(-1)
        return
    matching_patterns.sort(key=lambda x: (-len(x[0]), x[0], x[1]))
    best_match = matching_patterns[0][1]
    print(best_match)

import sys
input = sys.stdin.read
data = input().split()

n = int(data[0])
index = 1
trie = Trie()

current_text = ""

for _ in range(n):
    action = data[index]
    index += 1
    if action == "add":
        pattern_id = int(data[index])
        pattern = data[index + 1]
        index += 2
        trie.insert(pattern, pattern_id)
        find_best_match(current_text, trie)
    elif action == "delete":
        pattern_id = int(data[index])
        index += 1
        pattern = data[index]
        index += 1
        trie.delete(pattern, pattern_id)
        find_best_match(current_text, trie)
    elif action == "append":
        s = data[index]
        index += 1
        current_text += s
        find_best_match(current_text, trie)
    elif action == "backspace":
        c = int(data[index])
        index += 1
        current_text = current_text[:-c]
        find_best_match(current_text, trie)

# xufhzrDiUAqMIHvGLWC

**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
5
add 1 a
add 2 aa
add 3 aaa
append a
backspace 1
```

5
add 1 a
add 2 aa
add 3 aaa
append a
backspace 1


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Auto Complete
time_limit_ms: 4000
memory_limit_mb: 1024
Description: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types "App", your editor will magically suggest the word "Application"! Even better, users can personalize the words that auto-complete in your editor.

Your editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):

1. Add an auto complete pattern $$$p_i$$$.
2. Delete an auto complete pattern $$$p_i$$$.
3. Append a string $$$s$$$ to the end of $$$t$$$.
4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.

After each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:

1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.
2. If there are multiple $$$p_i$$$, pick the longest one.
3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.
4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.

For example, let us say we have three candidates: "alice", "bob", and "charlie", with ID 1, 2, and 3. At first, there is nothing on the screen, so "charlie" (3) should be suggested because it is the longest. Then, let us say the user types "b". You should suggest "bob" (2) because it is the only one that starts with "b". Finally, let us say the user type "body". You should print -1 because there is no matched pattern.

Input: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.

There are four types of actions:

1. $$$i$$$ $$$p_i$$$
2. $$$i$$$
3. $$$s$$$
4. $$$c$$$

- $$$1 \leq n \leq 10^6$$$
- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\times 10^6$$$.
- $$$1 \leq c \leq 2\times 10^6$$$
- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).
- The ID $$$i$$$ for each add operation is unique.
- The ID $$$i$$$ for each delete is guaranteed to be valid.
- Each ID $$$i$$$ satisfies $$$0\le i \le n$$$.

Output: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.

Examples:
input:
6
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
append 2_bobabc
backspace 3
output:
1
1
3
3
-1
2

input:
6
append pattern
add 1 pattern1_alice____
add 2 pattern2_bob______
add 3 pattern3_charlie__
delete 1
delete 2
output:
-1
1
1
1
2
3

Note:

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

public class Main{

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());

        Map<String, TreeMap<Integer, String>> patterns = new HashMap<>();
        StringBuilder t = new StringBuilder();
        StringBuilder result = new StringBuilder();

        for (int i = 0; i < n; i++) {
            String[] parts = br.readLine().split(" ");
            if (parts[0].equals("add")) {
                int id = Integer.parseInt(parts[1]);
                String pattern = parts[2];
                patterns
                    .computeIfAbsent(pattern, k -> new TreeMap<>())
                    .put(id, pattern);
            } else if (parts[0].equals("delete")) {
                int id = Integer.parseInt(parts[1]);
                patterns.forEach((key, value) -> value.remove(id));
            } else if (parts[0].equals("append")) {
                t.append(parts[1]);
            } else if (parts[0].equals("backspace")) {
                int c = Integer.parseInt(parts[1]);
                int len = t.length();
                int toDelete = Math.min(c, len);
                t.delete(len - toDelete, len);
            }

            // Find the best match
            int bestId = findBestMatch(patterns, t.toString());
            result.append(bestId).append('\n');
        }

        System.out.print(result.toString());
    }

    private static int findBestMatch(Map<String, TreeMap<Integer, String>> patterns, String currentText) {
        int bestId = -1;
        String bestPattern = null;

        for (Map.Entry<String, TreeMap<Integer, String>> entry : patterns.entrySet()) {
            String pattern = entry.getKey();
            if (pattern.startsWith(currentText) &&
                (bestPattern == null ||
                 pattern.length() > bestPattern.length() ||
                 (pattern.length() == bestPattern.length() && pattern.compareTo(bestPattern) < 0))) {
                bestPattern = pattern;
            }
        }

        if (bestPattern != null) {
            bestId = patterns.get(bestPattern).firstKey();
        }

        return bestId;
    }
}

// 7oldVP1puzOiWmcQeXF

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

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

Think step by step.

Response:
```plaintext
7
add 1 alice
add 2 bob
add 3 charlie
add 4 charlie
delete 4
append al
append bob
```

7
add 1 alice
add 2 bob
add 3 charlie
add 4 charlie
delete 4
append al
append bob


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Auto Complete
time_limit_ms: 4000
memory_limit_mb: 1024
Description: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types "App", your editor will magically suggest the word "Application"! Even better, users can personalize the words that auto-complete in your editor.

Your editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):

1. Add an auto complete pattern $$$p_i$$$.
2. Delete an auto complete pattern $$$p_i$$$.
3. Append a string $$$s$$$ to the end of $$$t$$$.
4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.

After each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:

1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.
2. If there are multiple $$$p_i$$$, pick the longest one.
3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.
4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.

For example, let us say we have three candidates: "alice", "bob", and "charlie", with ID 1, 2, and 3. At first, there is nothing on the screen, so "charlie" (3) should be suggested because it is the longest. Then, let us say the user types "b". You should suggest "bob" (2) because it is the only one that starts with "b". Finally, let us say the user type "body". You should print -1 because there is no matched pattern.

Input: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.

There are four types of actions:

1. $$$i$$$ $$$p_i$$$
2. $$$i$$$
3. $$$s$$$
4. $$$c$$$

- $$$1 \leq n \leq 10^6$$$
- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\times 10^6$$$.
- $$$1 \leq c \leq 2\times 10^6$$$
- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).
- The ID $$$i$$$ for each add operation is unique.
- The ID $$$i$$$ for each delete is guaranteed to be valid.
- Each ID $$$i$$$ satisfies $$$0\le i \le n$$$.

Output: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.

Examples:
input:
6
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
append 2_bobabc
backspace 3
output:
1
1
3
3
-1
2

input:
6
append pattern
add 1 pattern1_alice____
add 2 pattern2_bob______
add 3 pattern3_charlie__
delete 1
delete 2
output:
-1
1
1
1
2
3

Note:

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

public class Main{
    static class TrieNode {
        Map<Character, TrieNode> children;
        int id;
        boolean isEndOfPattern;

        public TrieNode() {
            children = new HashMap<>();
            id = -1;
            isEndOfPattern = false;
        }
    }

    static class Pattern implements Comparable<Pattern> {
        int id;
        String pattern;

        public Pattern(int id, String pattern) {
            this.id = id;
            this.pattern = pattern;
        }

        @Override
        public int compareTo(Pattern other) {
            if (pattern.length() != other.pattern.length()) {
                return Integer.compare(other.pattern.length(), pattern.length());
            }
            if (!pattern.equals(other.pattern)) {
                return pattern.compareTo(other.pattern);
            }
            return Integer.compare(id, other.id);
        }
    }

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        TrieNode root = new TrieNode();
        Set<Pattern> patterns = new TreeSet<>();
        StringBuilder currentText = new StringBuilder();

        for (int i = 0; i < n; i++) {
            String[] parts = br.readLine().split(" ");
            if (parts[0].equals("add")) {
                int id = Integer.parseInt(parts[1]);
                String pattern = parts[2];
                addPattern(root, id, pattern, patterns);
            } else if (parts[0].equals("delete")) {
                int id = Integer.parseInt(parts[1]);
                deletePattern(root, id, patterns);
            } else if (parts[0].equals("append")) {
                String s = parts[1];
                append(root, currentText, s, patterns);
            } else {
                int c = Integer.parseInt(parts[1]);
                backspace(root, currentText, c, patterns);
            }
            printSuggestedPattern(patterns);
        }
    }

    private static void addPattern(TrieNode root, int id, String pattern, Set<Pattern> patterns) {
        TrieNode node = root;
        for (char c : pattern.toCharArray()) {
            if (!node.children.containsKey(c)) {
                node.children.put(c, new TrieNode());
            }
            node = node.children.get(c);
        }
        node.id = id;
        node.isEndOfPattern = true;
        patterns.add(new Pattern(id, pattern));
    }

    private static void deletePattern(TrieNode root, int id, Set<Pattern> patterns) {
        TrieNode node = root;
        for (char c : getPattern(root, id).toCharArray()) {
            node = node.children.get(c);
        }
        node.id = -1;
        node.isEndOfPattern = false;
        patterns.remove(new Pattern(id, getPattern(root, id)));
    }

    private static String getPattern(TrieNode root, int id) {
        StringBuilder sb = new StringBuilder();
        Queue<TrieNode> queue = new LinkedList<>();
        queue.add(root);
        while (!queue.isEmpty()) {
            TrieNode node = queue.poll();
            if (node.id == id) {
                return sb.toString();
            }
            for (Map.Entry<Character, TrieNode> entry : node.children.entrySet()) {
                sb.append(entry.getKey());
                queue.add(entry.getValue());
                sb.deleteCharAt(sb.length() - 1);
            }
        }
        return "";
    }

    private static void append(TrieNode root, StringBuilder currentText, String s, Set<Pattern> patterns) {
        currentText.append(s);
        TrieNode node = root;
        for (char c : currentText.toString().toCharArray()) {
            if (!node.children.containsKey(c)) {
                node.children.put(c, new TrieNode());
            }
            node = node.children.get(c);
        }
        patterns.clear();
        dfs(node, currentText.toString(), patterns);
    }

    private static void backspace(TrieNode root, StringBuilder currentText, int c, Set<Pattern> patterns) {
        currentText.delete(currentText.length() - c, currentText.length());
        TrieNode node = root;
        for (char c1 : currentText.toString().toCharArray()) {
            if (!node.children.containsKey(c1)) {
                node.children.put(c1, new TrieNode());
            }
            node = node.children.get(c1);
        }
        patterns.clear();
        dfs(node, currentText.toString(), patterns);
    }

    private static void dfs(TrieNode node, String prefix, Set<Pattern> patterns) {
        if (node.isEndOfPattern) {
            patterns.add(new Pattern(node.id, prefix));
        }
        for (Map.Entry<Character, TrieNode> entry : node.children.entrySet()) {
            dfs(entry.getValue(), prefix + entry.getKey(), patterns);
        }
    }

    private static void printSuggestedPattern(Set<Pattern> patterns) {
        if (patterns.isEmpty()) {
            System.out.println(-1);
        } else {
            System.out.println(patterns.iterator().next().id);
        }
    }
}

// oG6DPBNkctuX49HFQEI

**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
11
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append a
append l
append i
append c
append e
delete 1
backspace 5
backspace 3
```

11
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append a
append l
append i
append c
append e
delete 1
backspace 5
backspace 3


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Auto Complete
time_limit_ms: 4000
memory_limit_mb: 1024
Description: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types "App", your editor will magically suggest the word "Application"! Even better, users can personalize the words that auto-complete in your editor.

Your editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):

1. Add an auto complete pattern $$$p_i$$$.
2. Delete an auto complete pattern $$$p_i$$$.
3. Append a string $$$s$$$ to the end of $$$t$$$.
4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.

After each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:

1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.
2. If there are multiple $$$p_i$$$, pick the longest one.
3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.
4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.

For example, let us say we have three candidates: "alice", "bob", and "charlie", with ID 1, 2, and 3. At first, there is nothing on the screen, so "charlie" (3) should be suggested because it is the longest. Then, let us say the user types "b". You should suggest "bob" (2) because it is the only one that starts with "b". Finally, let us say the user type "body". You should print -1 because there is no matched pattern.

Input: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.

There are four types of actions:

1. $$$i$$$ $$$p_i$$$
2. $$$i$$$
3. $$$s$$$
4. $$$c$$$

- $$$1 \leq n \leq 10^6$$$
- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\times 10^6$$$.
- $$$1 \leq c \leq 2\times 10^6$$$
- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).
- The ID $$$i$$$ for each add operation is unique.
- The ID $$$i$$$ for each delete is guaranteed to be valid.
- Each ID $$$i$$$ satisfies $$$0\le i \le n$$$.

Output: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.

Examples:
input:
6
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
append 2_bobabc
backspace 3
output:
1
1
3
3
-1
2

input:
6
append pattern
add 1 pattern1_alice____
add 2 pattern2_bob______
add 3 pattern3_charlie__
delete 1
delete 2
output:
-1
1
1
1
2
3

Note:

Buggy Code:
#include <cstdio>
#include <algorithm>
#include <utility>
#include <cstring>
#include <set>

using namespace std;

int T;
char op[10],s[2000005];

struct Node
{
	Node* np[127];
	Node* pre;
	pair<int,int> lgst;//Length,-Id.
	set<pair<int,int> >* ids;

	Node()
	{
		for(int i=0;i<127;i++)
			np[i] = NULL;
		pre = NULL;
		ids = NULL;
		lgst = make_pair(0,0);
	}
	Node(Node* pre):pre(pre)
	{
		for(int i=0;i<127;i++)
			np[i] = NULL;
		ids = NULL;
		lgst = make_pair(0,0);
	}
};

typedef Node* LPNode;

LPNode pl[1000005],nwat,head,header;
int len[1000005];

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

	header = new Node();
	head = new Node();
	header->pre = header;
	head->pre = header;
	nwat = head;

	while(T--)
	{
		scanf("%s",op);
		if(op[1]=='d')//add
		{
			int id,n;
			scanf("%d%s",&id,s);

			LPNode t = head;
			n = strlen(s);

			for(int i=0;i<n;i++)
			{
				if(t->np[s[i]]==NULL)
				{
					t->np[s[i]] = new Node(t);
				}
				t = t->np[s[i]];
			}
			pl[id] = t;
			len[id] = n;

			if(t->ids==NULL)
			{
				t->ids = new set<pair<int,int> >();
			}
			t->ids->insert(make_pair(n,-id));
			while(t->pre!=t)
			{
				t->lgst = make_pair(0,0);
				if(t->ids!=NULL)
				{
					for(pair<int,int> v:*(t->ids))
					{
						t->lgst = max(t->lgst,v);
					}
				}
				for(int i=0;i<127;i++)
				{
					if(t->np[i]==NULL)
						continue;
					t->lgst = max(t->lgst,t->np[i]->lgst);
				}
				t = t->pre;
			}
		}
		else if(op[1]=='e')//delete
		{
			int id;
			scanf("%d",&id);

			LPNode t = pl[id];

			t->ids->erase(make_pair(len[id],-id));

			while(t->pre!=t)
			{
				t->lgst = make_pair(0,0);
				if(t->ids!=NULL)
				{
					for(pair<int,int> v:*(t->ids))
					{
						t->lgst = max(t->lgst,v);
					}
				}
				for(int i=0;i<127;i++)
				{
					if(t->np[i]==NULL)
						continue;
					t->lgst = max(t->lgst,t->np[i]->lgst);
				}
				t = t->pre;
			}
		}
		else if(op[1]=='p')//append
		{
			scanf("%s",s);

			int n = strlen(s);

			for(int i=0;i<n;i++)
			{
				if(nwat->np[s[i]]==NULL)
				{
					nwat->np[s[i]] = new Node(nwat);
				}
				nwat = nwat->np[s[i]];
			}

			LPNode t = nwat;

			while(t->pre!=t)
			{
				t->lgst = make_pair(0,0);
				if(t->ids!=NULL)
				{
					for(pair<int,int> v:*(t->ids))
					{
						t->lgst = max(t->lgst,v);
					}
				}
				for(int i=0;i<127;i++)
				{
					if(t->np[i]==NULL)
						continue;
					t->lgst = max(t->lgst,t->np[i]->lgst);
				}
				t = t->pre;
			}
		}
		else //op[1]=='a' backspace
		{
			int n;
			scanf("%d",&n);

			while(n--)
				nwat = nwat->pre;
			if(nwat==header)
				nwat = head;
		}

		if(nwat->lgst.first)//怎么有id为0的。。。注意。。。
		{
			printf("%d\n",-nwat->lgst.second);
		}
		else
		{
			puts("-1");
		}
	}

	return 0;
}

/*

Why????????

*/

**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
6
add 1 aaaaa
add 2 aaaa
add 3 aaa
add 4 aa
add 5 a
backspace 5
```

6
add 1 aaaaa
add 2 aaaa
add 3 aaa
add 4 aa
add 5 a
backspace 5


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Auto Complete
time_limit_ms: 4000
memory_limit_mb: 1024
Description: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types "App", your editor will magically suggest the word "Application"! Even better, users can personalize the words that auto-complete in your editor.

Your editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):

1. Add an auto complete pattern $$$p_i$$$.
2. Delete an auto complete pattern $$$p_i$$$.
3. Append a string $$$s$$$ to the end of $$$t$$$.
4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.

After each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:

1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.
2. If there are multiple $$$p_i$$$, pick the longest one.
3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.
4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.

For example, let us say we have three candidates: "alice", "bob", and "charlie", with ID 1, 2, and 3. At first, there is nothing on the screen, so "charlie" (3) should be suggested because it is the longest. Then, let us say the user types "b". You should suggest "bob" (2) because it is the only one that starts with "b". Finally, let us say the user type "body". You should print -1 because there is no matched pattern.

Input: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.

There are four types of actions:

1. $$$i$$$ $$$p_i$$$
2. $$$i$$$
3. $$$s$$$
4. $$$c$$$

- $$$1 \leq n \leq 10^6$$$
- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\times 10^6$$$.
- $$$1 \leq c \leq 2\times 10^6$$$
- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).
- The ID $$$i$$$ for each add operation is unique.
- The ID $$$i$$$ for each delete is guaranteed to be valid.
- Each ID $$$i$$$ satisfies $$$0\le i \le n$$$.

Output: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.

Examples:
input:
6
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
append 2_bobabc
backspace 3
output:
1
1
3
3
-1
2

input:
6
append pattern
add 1 pattern1_alice____
add 2 pattern2_bob______
add 3 pattern3_charlie__
delete 1
delete 2
output:
-1
1
1
1
2
3

Note:

Buggy Code:
#include <cstdio>
#include <algorithm>
#include <utility>
#include <cstring>
#include <set>

using namespace std;

int T;
char op[10],s[2000005];

struct Node
{
	Node* np[127];
	Node* pre;
	pair<int,int> lgst;//Length,-Id.
	set<pair<int,int> >* ids;

	Node()
	{
		for(int i=0;i<127;i++)
			np[i] = NULL;
		pre = NULL;
		ids = NULL;
		lgst = make_pair(0,0);
	}
	Node(Node* pre):pre(pre)
	{
		for(int i=0;i<127;i++)
			np[i] = NULL;
		ids = NULL;
		lgst = make_pair(0,0);
	}
};

typedef Node* LPNode;

LPNode pl[1000005],nwat,head,header;
int len[1000005];

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

	header = new Node();
	head = new Node();
	header->pre = header;
	head->pre = header;
	nwat = head;

	while(T--)
	{
		scanf("%s",op);
		if(op[1]=='d')//add
		{
			int id,n;
			scanf("%d%s",&id,s);

			LPNode t = head;
			n = strlen(s);

			for(int i=0;i<n;i++)
			{
				if(t->np[s[i]]==NULL)
				{
					t->np[s[i]] = new Node(t);
				}
				t = t->np[s[i]];
			}
			pl[id] = t;
			len[id] = n;

			if(t->ids==NULL)
			{
				t->ids = new set<pair<int,int> >();
			}
			t->ids->insert(make_pair(n,id));
			while(t->pre!=t)
			{
				t->lgst = make_pair(0,0);
				if(t->ids!=NULL)
				{
					for(pair<int,int> v:*(t->ids))
					{
						if(t->lgst.first<v.first)
						{
							t->lgst = v;
						}
					}
				}
				for(int i=0;i<127;i++)
				{
					if(t->np[i]==NULL)
						continue;
					if(t->lgst.first<t->np[i]->lgst.first)
					{
						t->lgst = t->np[i]->lgst;
					}
				}
				t = t->pre;
			}
		}
		else if(op[1]=='e')//delete
		{
			int id;
			scanf("%d",&id);

			LPNode t = pl[id];

			t->ids->erase(make_pair(len[id],id));
			if(t->ids->empty())
				delete t->ids;

			while(t->pre!=t)
			{
				t->lgst = make_pair(0,0);
				if(t->ids!=NULL)
				{
					for(pair<int,int> v:*(t->ids))
					{
						if(t->lgst.first<v.first)
						{
							t->lgst = v;
						}
					}
				}
				for(int i=0;i<127;i++)
				{
					if(t->np[i]==NULL)
						continue;
					if(t->lgst.first<t->np[i]->lgst.first)
					{
						t->lgst = t->np[i]->lgst;
					}
				}

				t = t->pre;

				LPNode tr = t;
				int f = 0;
				if(tr==head || tr==header)
					f = 1;
				if(tr->ids!=NULL)
					f = 1;
				if(f)
					continue;
				for(int i=0;i<127;i++)
				{
					if(tr->np[i]!=NULL)
					{
						f = 1;
						break;
					}
				}
				if(f)
					continue;
				delete tr;
			}
		}
		else if(op[1]=='p')//append
		{
			scanf("%s",s);

			int n = strlen(s);

			for(int i=0;i<n;i++)
			{
				if(nwat->np[s[i]]==NULL)
				{
					nwat->np[s[i]] = new Node(nwat);
				}
				nwat = nwat->np[s[i]];
			}
		}
		else //op[1]=='a' backspace
		{
			int n;
			scanf("%d",&n);

			int f = 0;
			while(n--)
			{
				LPNode tr = nwat;
				nwat = nwat->pre;

				if(tr==head || tr==header)
					f = 1;
				if(tr->ids!=NULL)
					f = 1;
				if(f)
					continue;
				for(int i=0;i<127;i++)
				{
					if(tr->np[i]!=NULL)
					{
						f = 1;
						break;
					}
				}
				if(f)
					continue;
				delete tr;
			}
			if(nwat==header)
				nwat = head;
		}

		if(nwat->lgst.first)//怎么有id为0的。。。注意。。。
		{
			printf("%d\n",nwat->lgst.second);
		}
		else
		{
			puts("-1");
		}
	}

	return 0;
}

/*

纯唐了。没有管字典序，只管了长度和id。
还是很菜。还是得多练。还是得多磨。

S703:?4Pd"N%ez/Vv7WJ`gNV5=]5"E...xSh<x:llEh"S:%hU\y{Zn>iamH_uEV)
S161:c'A;;f=ry]3F>y/?'(J+N.1jS...)z&LGdi,;*?O}x.b)_9<,kc)9%$jzSW

*/

/*

Why????????

*/

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

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

Think step by step.

Response:
```plaintext
7
add 1 pattern1_a
add 2 pattern2_b
add 3 pattern3_c
append a
backspace 1
append b
backspace 1
```

7
add 1 pattern1_a
add 2 pattern2_b
add 3 pattern3_c
append a
backspace 1
append b
backspace 1


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Auto Complete
time_limit_ms: 4000
memory_limit_mb: 1024
Description: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types "App", your editor will magically suggest the word "Application"! Even better, users can personalize the words that auto-complete in your editor.

Your editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):

1. Add an auto complete pattern $$$p_i$$$.
2. Delete an auto complete pattern $$$p_i$$$.
3. Append a string $$$s$$$ to the end of $$$t$$$.
4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.

After each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:

1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.
2. If there are multiple $$$p_i$$$, pick the longest one.
3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.
4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.

For example, let us say we have three candidates: "alice", "bob", and "charlie", with ID 1, 2, and 3. At first, there is nothing on the screen, so "charlie" (3) should be suggested because it is the longest. Then, let us say the user types "b". You should suggest "bob" (2) because it is the only one that starts with "b". Finally, let us say the user type "body". You should print -1 because there is no matched pattern.

Input: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.

There are four types of actions:

1. $$$i$$$ $$$p_i$$$
2. $$$i$$$
3. $$$s$$$
4. $$$c$$$

- $$$1 \leq n \leq 10^6$$$
- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\times 10^6$$$.
- $$$1 \leq c \leq 2\times 10^6$$$
- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).
- The ID $$$i$$$ for each add operation is unique.
- The ID $$$i$$$ for each delete is guaranteed to be valid.
- Each ID $$$i$$$ satisfies $$$0\le i \le n$$$.

Output: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.

Examples:
input:
6
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
append 2_bobabc
backspace 3
output:
1
1
3
3
-1
2

input:
6
append pattern
add 1 pattern1_alice____
add 2 pattern2_bob______
add 3 pattern3_charlie__
delete 1
delete 2
output:
-1
1
1
1
2
3

Note:

Buggy Code:
class TrieNode:
    def __init__(self, parent=None):
        self.children = {}
        self.best_id = -1
        self.max_length = 0
        self.is_end = False
        self.pattern_id = None
        self.parent = parent

class Trie:
    def __init__(self, n):
        self.root = TrieNode()
        self.patterns = [None] * (n + 1)
        self.lengths = [0] * (n + 1)

    def add(self, pattern, id):
        self.patterns[id] = pattern
        self.lengths[id] = len(pattern)
        node = self.root
        for char in pattern:
            if char not in node.children:
                node.children[char] = TrieNode(parent=node)
            node = node.children[char]
        node.is_end = True
        node.pattern_id = id
        self._update_best_id(node, id)

    def _update_best_id(self, node, id):
        while node:
            current_best_id = node.best_id
            current_max_length = node.max_length
            if current_best_id == -1:
                node.best_id = id
                node.max_length = self.lengths[id]
            elif self.lengths[id] > current_max_length:
                node.best_id = id
                node.max_length = self.lengths[id]
            elif self.lengths[id] == current_max_length:
                if self.patterns[id] < self.patterns[current_best_id]:
                    node.best_id = id
                elif self.patterns[id] == self.patterns[current_best_id]:
                    if id < current_best_id:
                        node.best_id = id
            node = node.parent

    def delete(self, id):
        pattern = self.patterns[id]
        node = self.root
        for char in pattern:
            if char in node.children:
                node = node.children[char]
            else:
                raise ValueError("Pattern not found")
        node.is_end = False
        node.pattern_id = None
        self.patterns[id] = None
        self.lengths[id] = 0
        self._update_best_id_after_delete(node)

    def _update_best_id_after_delete(self, node):
        while node:
            best_id = -1
            max_length = 0
            for child in node.children.values():
                if child.best_id != -1:
                    if best_id == -1 or \
                       (child.max_length > max_length) or \
                       (child.max_length == max_length and self.patterns[child.best_id] < self.patterns[best_id]) or \
                       (child.max_length == max_length and self.patterns[child.best_id] == self.patterns[best_id] and child.best_id < best_id):
                        best_id = child.best_id
                        max_length = child.max_length
            node.best_id = best_id
            node.max_length = max_length
            node = node.parent

    def search(self, text_list, index):
        node = self.root
        for i in range(index):
            char = text_list[i]
            if char in node.children:
                node = node.children[char]
            else:
                return -1
        return node.best_id

def main():
    import sys
    input = sys.stdin.read().splitlines()
    n = int(input[0])
    trie = Trie(n)
    current_text = []
    index = 0
    operation_index = 1
    for _ in range(n):
        line = input[operation_index].strip()
        operation_index += 1
        parts = line.split()
        if parts[0] == "add":
            i = int(parts[1])
            p_i = ' '.join(parts[2:])
            trie.add(p_i, i)
        elif parts[0] == "delete":
            i = int(parts[1])
            trie.delete(i)
        elif parts[0] == "append":
            s = ' '.join(parts[1:])
            current_text.extend(s)
            index += len(s)
        else:  # backspace
            c = int(parts[1])
            index = max(0, index - c)
        # After each operation, search for the best pattern ID
        best_id = trie.search(current_text, index)
        print(best_id)

if __name__ == "__main__":
    main()

# 3XiOGYCRkFpbQE8UM2l

**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
5
add 1 a
add 2 ab
add 3 abc
append a
backspace 1
```

5
add 1 a
add 2 ab
add 3 abc
append a
backspace 1


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Auto Complete
time_limit_ms: 4000
memory_limit_mb: 1024
Description: You are designing a snazzy new text editor, and you want to add a nifty auto-complete feature to help users save time. Here is how it will work: if a user types "App", your editor will magically suggest the word "Application"! Even better, users can personalize the words that auto-complete in your editor.

Your editor will support 4 kinds of operations (Let's say the current text in your editor is $$$t$$$):

1. Add an auto complete pattern $$$p_i$$$.
2. Delete an auto complete pattern $$$p_i$$$.
3. Append a string $$$s$$$ to the end of $$$t$$$.
4. Delete $$$c$$$ characters from the end of $$$t$$$. Note that if $$$c$$$ is larger then the length of $$$t$$$, delete all the characters from $$$t$$$.

After each action, your editor should suggest an auto-complete candidate $$$i$$$ that matches the following criteria:

1. The string $$$p_i$$$ has a prefix equal to $$$t$$$.
2. If there are multiple $$$p_i$$$, pick the longest one.
3. If there are still multiple $$$p_i$$$, pick the one with the smallest lexicographic order.
4. If there are still multiple $$$p_i$$$, pick the one with the smallest ID.

For example, let us say we have three candidates: "alice", "bob", and "charlie", with ID 1, 2, and 3. At first, there is nothing on the screen, so "charlie" (3) should be suggested because it is the longest. Then, let us say the user types "b". You should suggest "bob" (2) because it is the only one that starts with "b". Finally, let us say the user type "body". You should print -1 because there is no matched pattern.

Input: The first line contains an integer $$$n$$$, followed by $$$n$$$ lines, with each line containing an action.

There are four types of actions:

1. $$$i$$$ $$$p_i$$$
2. $$$i$$$
3. $$$s$$$
4. $$$c$$$

- $$$1 \leq n \leq 10^6$$$
- The total number of characters in all $$$p_i$$$ and $$$s$$$ does not exceed $$$2\times 10^6$$$.
- $$$1 \leq c \leq 2\times 10^6$$$
- The strings $$$p_i$$$ and $$$s$$$ may contain any printable characters, excluding all space characters (ASCII number in the range from $$$33$$$ to $$$126$$$).
- The ID $$$i$$$ for each add operation is unique.
- The ID $$$i$$$ for each delete is guaranteed to be valid.
- Each ID $$$i$$$ satisfies $$$0\le i \le n$$$.

Output: The program should output $$$n$$$ lines. For each action, output an integer $$$i$$$, which means that after the action, $$$p_i$$$ is the suggested auto complete candidate. If there is no $$$p_i$$$ that matches the requirement, output -1.

Examples:
input:
6
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append pattern
append 2_bobabc
backspace 3
output:
1
1
3
3
-1
2

input:
6
append pattern
add 1 pattern1_alice____
add 2 pattern2_bob______
add 3 pattern3_charlie__
delete 1
delete 2
output:
-1
1
1
1
2
3

Note:

Buggy Code:
from bisect import bisect_left

def find_best_match(current_text, patterns):
    best_match_id = -1
    best_length = -1

    for length, pattern, pattern_id in patterns:
        if length < len(current_text):
            continue
        if pattern.startswith(current_text):
            if length > best_length or (length == best_length and pattern_id < best_match_id):
                best_length = length
                best_match_id = pattern_id

    return best_match_id

def main():
    import sys
    input = sys.stdin.read
    data = input().split()

    n = int(data[0])
    idx = 1

    patterns = []
    pattern_dict = {}
    current_text = ""

    for _ in range(n):
        if data[idx] == 'add':
            pattern_id = int(data[idx+1])
            pattern = data[idx+2]
            pattern_dict[pattern_id] = pattern
            patterns.append((len(pattern), pattern, pattern_id))
            patterns.sort()
            idx += 3
        elif data[idx] == 'delete':
            pattern_id = int(data[idx+1])
            pattern = pattern_dict.pop(pattern_id)
            patterns.remove((len(pattern), pattern, pattern_id))
            idx += 2
        elif data[idx] == 'append':
            current_text += data[idx+1]
            idx += 2
        elif data[idx] == 'backspace':
            c = int(data[idx+1])
            current_text = current_text[:-c] if c < len(current_text) else ""
            idx += 2

        best_match_id = find_best_match(current_text, patterns)
        print(best_match_id)

if __name__ == "__main__":
    main()

# gybjniB6VlfuDqwCHmL

**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 create a challenging test input that exposes potential bugs in the given buggy code, we need to focus on edge cases and scenarios that can cause issues. Here are some key points to consider:

1. **Adding and Deleting Patterns**: Ensure that adding and deleting patterns in a way that affects the sorting and lookup mechanisms.
2. **Appending and Backspacing**: Test scenarios where appending and backspacing characters can lead to edge cases, such as backspacing all characters or appending very long strings.
3. **Multiple Patterns with the Same Prefix**: Test cases where multiple patterns have the same prefix and need to be compared based on length, lexicographic order, and ID.
4. **Empty Current Text**: Ensure that the program handles the case where the current text is empty correctly.

Given these considerations, here is a small-scale test input that should challenge the buggy code:

```plaintext
9
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append a
backspace 1
append b
delete 2
append c
append d
```

### Explanation:
1. **Add Patterns**:
   - Add "pattern1_alice" with ID 1.
   - Add "pattern2_bob" with ID 2.
   - Add "pattern3_charlie" with ID 3.
2. **Append 'a'**:
   - Current text: "a"
   - Expected output: 1 (since "pattern1_alice" starts with "a")
3. **Backspace 1**:
   - Current text: ""
   - Expected output: 3 (since "pattern3_charlie" is the longest)
4. **Append 'b'**:
   - Current text: "b"
   - Expected output: 2 (since "pattern2_bob" starts with "b")
5. **Delete Pattern with ID 2**:
   - Remove "pattern2_bob"
6. **Append 'c'**:
   - Current text: "bc"
   - Expected output: -1 (since no pattern starts with "bc")
7. **Append 'd'**:
   - Current text: "bcd"
   - Expected output: -1 (since no pattern starts with "bcd")

This test input should expose issues with the sorting, lookup, and handling of empty or non-matching text scenarios in the buggy code.

9
add 1 pattern1_alice
add 2 pattern2_bob
add 3 pattern3_charlie
append a
backspace 1
append b
delete 2
append c
append d