**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Make Majority
time_limit_ms: 1500
memory_limit_mb: 512
Description: You are given a sequence $[a_1,\ldots,a_n]$ where each element $a_i$ is either $0$ or $1$. You can apply several (possibly zero) operations to the sequence. In each operation, you select two integers $1\le l\le r\le |a|$ (where $|a|$ is the current length of $a$) and replace $[a_l,\ldots,a_r]$ with a single element $x$, where $x$ is the majority of $[a_l,\ldots,a_r]$.

Here, the majority of a sequence consisting of $0$ and $1$ is defined as follows: suppose there are $c_0$ zeros and $c_1$ ones in the sequence, respectively.

  * If $c_0\ge c_1$, the majority is $0$.    * If $c_0<c_1$, the majority is $1$.

For example, suppose $a=[1,0,0,0,1,1]$. If we select $l=1,r=2$, the resulting sequence will be $[0,0,0,1,1]$. If we select $l=4,r=6$, the resulting sequence will be $[1,0,0,1]$.

Determine if you can make $a=[1]$ with a finite number of operations.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 4\cdot 10^4$). Description of the test cases follows.

The first line of each testcase contains one integer $n$ ($1\le n\le 2\cdot 10^5$).

The second line of each testcase contains a string consisting of $0$ and $1$, describing the sequence $a$.

It's guaranteed that the sum of $n$ over all testcases does not exceed $2\cdot 10^5$.

Output: For each testcase, if it's possible to make $a=[1]$, print YES. Otherwise, print NO. You can output the answer in any case (upper or lower). For example, the strings yEs, yes, Yes, and YES will be recognized as positive responses.

Examples:
input:
5
1
0
1
1
2
01
9
100000001
9
000011000
output:
No
Yes
No
Yes
No

Note: In the fourth testcase of the example, initially $a=[1,0,0,0,0,0,0,0,1]$. A valid sequence of operations is:

  1. Select $l=2,r=8$ and apply the operation. $a$ becomes $[1,0,1]$.    2. Select $l=1,r=3$ and apply the operation. $a$ becomes $[1]$.

Buggy Code:
from math import ceil
from math import log2

def make(S):
    if (S[0]=="1" and S[-1]=="1"):
        print("YES")
        return

    if (("11" in S) and (S[-1]=="1")):
        print("YES")
        return

    if (("11" in S) and (S[0]=="1")):
        print("YES")
        return

    if ("111" in S):
        print("YES")
    print("NO")

case =int(input())
for i in range(case):
    input()
    make(input())

**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
1
0
1
1
2
01
6
100101
```

**Explanation:**
1. **Test Case 1:**
   - Input: `1\n0`
   - Expected Output: `NO`
   - Reason: The sequence is `[0]`, which cannot be reduced to `[1]`.

2. **Test Case 2:**
   - Input: `1\n1`
   - Expected Output: `YES`
   - Reason: The sequence is already `[1]`.

3. **Test Case 3:**
   - Input: `2\n01`
   - Expected Output: `NO`
   - Reason: The sequence is `[0, 1]`, and neither `0` nor `1` can be reduced to `[1]`.

4. **Test Case 4:**
   - Input: `6\n100101`
   - Expected Output: `NO`
   - Reason: The sequence is `[1, 0, 0, 1, 0, 1]`. Despite having `1` at the start and end, the majority operation cannot reduce this sequence to `[1]` because the majority of any subsequence does not allow it.

These test cases are designed to cover different scenarios:
- Single element sequences (`0` and `1`).
- Two-element sequences (`01`).
- A longer sequence with a mix of `0`s and `1`s (`100101`).

The buggy code has specific conditions that might not handle these edge cases correctly, especially when the sequence has a mix of `0`s and `1`s and does not start or end with `111`. The test case `100101` is particularly challenging as it does not fit any of the simple conditions in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Make Majority
time_limit_ms: 1500
memory_limit_mb: 512
Description: You are given a sequence $[a_1,\ldots,a_n]$ where each element $a_i$ is either $0$ or $1$. You can apply several (possibly zero) operations to the sequence. In each operation, you select two integers $1\le l\le r\le |a|$ (where $|a|$ is the current length of $a$) and replace $[a_l,\ldots,a_r]$ with a single element $x$, where $x$ is the majority of $[a_l,\ldots,a_r]$.

Here, the majority of a sequence consisting of $0$ and $1$ is defined as follows: suppose there are $c_0$ zeros and $c_1$ ones in the sequence, respectively.

  * If $c_0\ge c_1$, the majority is $0$.    * If $c_0<c_1$, the majority is $1$.

For example, suppose $a=[1,0,0,0,1,1]$. If we select $l=1,r=2$, the resulting sequence will be $[0,0,0,1,1]$. If we select $l=4,r=6$, the resulting sequence will be $[1,0,0,1]$.

Determine if you can make $a=[1]$ with a finite number of operations.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 4\cdot 10^4$). Description of the test cases follows.

The first line of each testcase contains one integer $n$ ($1\le n\le 2\cdot 10^5$).

The second line of each testcase contains a string consisting of $0$ and $1$, describing the sequence $a$.

It's guaranteed that the sum of $n$ over all testcases does not exceed $2\cdot 10^5$.

Output: For each testcase, if it's possible to make $a=[1]$, print YES. Otherwise, print NO. You can output the answer in any case (upper or lower). For example, the strings yEs, yes, Yes, and YES will be recognized as positive responses.

Examples:
input:
5
1
0
1
1
2
01
9
100000001
9
000011000
output:
No
Yes
No
Yes
No

Note: In the fourth testcase of the example, initially $a=[1,0,0,0,0,0,0,0,1]$. A valid sequence of operations is:

  1. Select $l=2,r=8$ and apply the operation. $a$ becomes $[1,0,1]$.    2. Select $l=1,r=3$ and apply the operation. $a$ becomes $[1]$.

Buggy Code:
import java.util.Scanner;

public class Q2 {
    public static boolean canReduceToOne(int n, String sequence) {
        int countOnes = 0;
        int countZeros = 0;

        for (char ch : sequence.toCharArray()) {
            if (ch == '1') {
                countOnes++;
            } else {
                countZeros++;
            }
        }

        // To reduce the sequence to [1], count of 1s must be strictly greater than count of 0s
        return countOnes > countZeros;
    }
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        for (int x = 0; x < t; x++) {
            int n=sc.nextInt();
            String s=sc.next();
            if(s.charAt(0)=='1' && s.charAt(s.length()-1)=='1') System.out.println("Yes");
            else System.out.println(canReduceToOne(n,s)?"Yes":"No");
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on the logic that checks if the sequence can be reduced to `[1]`. The buggy code has a logical flaw in the `canReduceToOne` method, which only checks if the count of `1`s is strictly greater than the count of `0`s. This is not sufficient to determine if the sequence can be reduced to `[1]`.

The bug is that the code incorrectly assumes that if the first and last characters are `1`, the sequence can always be reduced to `[1]`. However, this is not true in all cases. We need to create a test case where the sequence starts and ends with `1`, but it cannot be reduced to `[1]` due to the intermediate values.

Let's construct a small-scale test input that exposes this bug:

```plaintext
3
1
1
3
101
5
10001
```

### Explanation:
1. **Test Case 1:**
   - Input: `1 1`
   - Output: `Yes`
   - Explanation: The sequence is already `[1]`.

2. **Test Case 2:**
   - Input: `3 101`
   - Output: `Yes`
   - Explanation: The sequence can be reduced to `[1]` by selecting `l=1, r=3`.

3. **Test Case 3:**
   - Input: `5 10001`
   - Output: `No`
   - Explanation: The sequence starts and ends with `1`, but it cannot be reduced to `[1]` because the middle part has more `0`s than `1`s. The buggy code will incorrectly output `Yes` because it only checks the first and last characters.

This test input should expose the bug in the given code, as the third test case will produce the wrong output.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Make Majority
time_limit_ms: 1500
memory_limit_mb: 512
Description: You are given a sequence $[a_1,\ldots,a_n]$ where each element $a_i$ is either $0$ or $1$. You can apply several (possibly zero) operations to the sequence. In each operation, you select two integers $1\le l\le r\le |a|$ (where $|a|$ is the current length of $a$) and replace $[a_l,\ldots,a_r]$ with a single element $x$, where $x$ is the majority of $[a_l,\ldots,a_r]$.

Here, the majority of a sequence consisting of $0$ and $1$ is defined as follows: suppose there are $c_0$ zeros and $c_1$ ones in the sequence, respectively.

  * If $c_0\ge c_1$, the majority is $0$.    * If $c_0<c_1$, the majority is $1$.

For example, suppose $a=[1,0,0,0,1,1]$. If we select $l=1,r=2$, the resulting sequence will be $[0,0,0,1,1]$. If we select $l=4,r=6$, the resulting sequence will be $[1,0,0,1]$.

Determine if you can make $a=[1]$ with a finite number of operations.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 4\cdot 10^4$). Description of the test cases follows.

The first line of each testcase contains one integer $n$ ($1\le n\le 2\cdot 10^5$).

The second line of each testcase contains a string consisting of $0$ and $1$, describing the sequence $a$.

It's guaranteed that the sum of $n$ over all testcases does not exceed $2\cdot 10^5$.

Output: For each testcase, if it's possible to make $a=[1]$, print YES. Otherwise, print NO. You can output the answer in any case (upper or lower). For example, the strings yEs, yes, Yes, and YES will be recognized as positive responses.

Examples:
input:
5
1
0
1
1
2
01
9
100000001
9
000011000
output:
No
Yes
No
Yes
No

Note: In the fourth testcase of the example, initially $a=[1,0,0,0,0,0,0,0,1]$. A valid sequence of operations is:

  1. Select $l=2,r=8$ and apply the operation. $a$ becomes $[1,0,1]$.    2. Select $l=1,r=3$ and apply the operation. $a$ becomes $[1]$.

Buggy Code:
import java.util.*;
public class Main{
   static void solve(int n,String str){

        if(str.charAt(0)=='1' && str.charAt(n-1)=='1'){
            System.out.println("YES");
            return;

           }

           boolean flag=false;
           boolean flag2=false;
           int curr=0;
           for(int i=0;i<n;i++){
            if(str.charAt(i)=='1'){
                curr++;
            }
            else{
                curr=0;
            }
            if(curr==2){
                if(!flag) flag=true;
                else  flag2=true;
            }
           }
           if(str.charAt(0)=='1' || str.charAt(n-1)=='1'){
            if(flag){
                System.out.println("YES");
                return;

            }
           }
           if(flag2) {
               System.out.println("YES");
              return;

           }

               System.out.println("NO");

    }
    public static void main(String args[]){
        Scanner s=new Scanner(System.in);
       int t=s.nextInt();
       while(t-->0){
            int n=s.nextInt();
        String str=s.next();
            solve(n,str);

    }}
}

**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 buggy code, we need to focus on edge cases or scenarios that the code might handle incorrectly. The buggy code has a few key points to consider:

1. It checks if the first and last characters are both '1'.
2. It looks for sequences of two consecutive '1's.
3. It checks if there are two separate sequences of two consecutive '1's.

A good test case to expose the bug should:
- Have a sequence where the first and last characters are not both '1'.
- Have sequences of '1's that do not meet the criteria for two consecutive '1's but can still be reduced to `[1]` through valid operations.

Let's consider a sequence where the first and last characters are not both '1', but the sequence can still be reduced to `[1]` through valid operations. For example, a sequence like `0110110` can be reduced to `[1]` by the following operations:
1. Select `l=2, r=3` to get `0110110` -> `0110110` (no change since `1` is already the majority).
2. Select `l=4, r=5` to get `011110` (since `1` is the majority).
3. Select `l=1, r=6` to get `[1]`.

However, the buggy code will not correctly handle this sequence because it does not check for the possibility of reducing the sequence to `[1]` through valid operations when the first and last characters are not both '1'.

Here is the test input:

```plaintext
1
7
0110110
```

This test case should output `YES` because it is possible to reduce the sequence to `[1]` through valid operations, but the buggy code will likely output `NO` due to its flawed logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Make Majority
time_limit_ms: 1500
memory_limit_mb: 512
Description: You are given a sequence $[a_1,\ldots,a_n]$ where each element $a_i$ is either $0$ or $1$. You can apply several (possibly zero) operations to the sequence. In each operation, you select two integers $1\le l\le r\le |a|$ (where $|a|$ is the current length of $a$) and replace $[a_l,\ldots,a_r]$ with a single element $x$, where $x$ is the majority of $[a_l,\ldots,a_r]$.

Here, the majority of a sequence consisting of $0$ and $1$ is defined as follows: suppose there are $c_0$ zeros and $c_1$ ones in the sequence, respectively.

  * If $c_0\ge c_1$, the majority is $0$.    * If $c_0<c_1$, the majority is $1$.

For example, suppose $a=[1,0,0,0,1,1]$. If we select $l=1,r=2$, the resulting sequence will be $[0,0,0,1,1]$. If we select $l=4,r=6$, the resulting sequence will be $[1,0,0,1]$.

Determine if you can make $a=[1]$ with a finite number of operations.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 4\cdot 10^4$). Description of the test cases follows.

The first line of each testcase contains one integer $n$ ($1\le n\le 2\cdot 10^5$).

The second line of each testcase contains a string consisting of $0$ and $1$, describing the sequence $a$.

It's guaranteed that the sum of $n$ over all testcases does not exceed $2\cdot 10^5$.

Output: For each testcase, if it's possible to make $a=[1]$, print YES. Otherwise, print NO. You can output the answer in any case (upper or lower). For example, the strings yEs, yes, Yes, and YES will be recognized as positive responses.

Examples:
input:
5
1
0
1
1
2
01
9
100000001
9
000011000
output:
No
Yes
No
Yes
No

Note: In the fourth testcase of the example, initially $a=[1,0,0,0,0,0,0,0,1]$. A valid sequence of operations is:

  1. Select $l=2,r=8$ and apply the operation. $a$ becomes $[1,0,1]$.    2. Select $l=1,r=3$ and apply the operation. $a$ becomes $[1]$.

Buggy Code:
//
//    |  | |                                | |
//  __|  | |     __,   _  _    __,   ,   _  | |   __,   ,_
// /  |  |/ \   /  |  / |/ |  /  |  / \_|/  |/_) /  |  /  |
// \_/|_/|   |_/\_/|_/  |  |_/\_/|_/ \/ |__/| \_/\_/|_/   |_/(FIGLET)
#include<bits/stdc++.h>
using namespace std;

#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
typedef long long node;
typedef tree<node, null_type, less<node>,
            rb_tree_tag, tree_order_statistics_node_update> ordered_set;

typedef long long ll;
typedef vector<ll> vll;
typedef vector <vll> vvll;
typedef pair <ll,ll> pll;
typedef vector <pll> vpll;
#define forl(i,k,n) for(ll i=k;i<n;i++)
#define MOD 1000000007 //10^9 + 7
#define all(x) x.begin(),x.end()

#ifdef klatchu27
    #include "e:\cp\debug.h"
#endif

template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v)
{int cnt=0;for(auto it : v){cnt != v.size()-1? os << it << ' ':os << it;cnt++;}return os;}
template<typename T>
void print(T var1)
{    cout<<var1<<'\n';}
template<typename T,typename... Types>
void print(T var1,Types...var2)
{    cout<<var1<<' ';print(var2...); }
template<typename T>
void input(vector<T> &v) {for(auto& i: v)cin>>i;}

ll INF = 1000000000000000005LL;

void solve(int tt);
int main() {
    ios_base::sync_with_stdio(false);cin.tie(NULL);
    #ifdef klatchu27
        freopen("input.txt", "r", stdin); freopen("error.txt", "w", stderr);
        freopen("output.txt", "w", stdout);
    #else
        #define debug(...) 42
    #endif
    int t=1;
    cin>>t;
    forl(ii,1,t+1)
        solve(ii);
    //cerr<<"time taken : "<<(float)clock()/CLOCKS_PER_SEC<<" secs\n";return 0;
}

bool check(ll start,ll end,string& s){

    ll n = end-start+1;

    if(n==1)
        return (s[start]=='1');

    ll one_count = 0;
    forl(i,start,end+1)
        if(s[i]=='1')
            one_count++;

    if(one_count==1)
        return false;

    bool left_zero = false;
    bool right_zero = false;
    bool centre = false;

    ll first_one = 0;
    ll last_one = 0;

    ll count = 0;
    forl(i,start,end+1){
        if(s[i]=='1'){
            first_one = i;
            break;
        }
        else{
            count++;
        }
    }

    left_zero = (count>0);
    count = 0;

    for(int i=end;i>=start;i--){
        if(s[i]=='1'){
            last_one = i;
            break;
        }
        else{
            count++;
        }
    }

    right_zero = (count>0);

    if(first_one+1<=last_one-1)
        centre = check(first_one+1,last_one-1,s);
    else
        centre = false;

    if(centre)
        return true;

    if(left_zero && right_zero)
        return false;

    if(left_zero==false && right_zero==false)
        return true;

    if(first_one==last_one-1)
        return true;
    else{
        if(first_one==last_one-1)
            return false;
        return check(first_one,last_one-1,s)|check(first_one+1,last_one,s);
    }
}

void solve(int tt)
{
    ll n;
    cin>>n;

    string s;
    cin>>s;

    debug(n,s);

    bool ans = check(0,n-1,s);
    if(ans)
        print("YES");
    else
        print("NO");

}
//cd "e:\cp\" && g++ a.cpp -o 505 && "e:\cp\"505
//https://wumbogames.com/CFNotifications
//custom music url: https://youtu.be/FX9eEhoRZhY?t=7

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

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

Think step by step.

Response:
```plaintext
3
1
1
2
11
3
111
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Make Majority
time_limit_ms: 1500
memory_limit_mb: 512
Description: You are given a sequence $[a_1,\ldots,a_n]$ where each element $a_i$ is either $0$ or $1$. You can apply several (possibly zero) operations to the sequence. In each operation, you select two integers $1\le l\le r\le |a|$ (where $|a|$ is the current length of $a$) and replace $[a_l,\ldots,a_r]$ with a single element $x$, where $x$ is the majority of $[a_l,\ldots,a_r]$.

Here, the majority of a sequence consisting of $0$ and $1$ is defined as follows: suppose there are $c_0$ zeros and $c_1$ ones in the sequence, respectively.

  * If $c_0\ge c_1$, the majority is $0$.    * If $c_0<c_1$, the majority is $1$.

For example, suppose $a=[1,0,0,0,1,1]$. If we select $l=1,r=2$, the resulting sequence will be $[0,0,0,1,1]$. If we select $l=4,r=6$, the resulting sequence will be $[1,0,0,1]$.

Determine if you can make $a=[1]$ with a finite number of operations.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 4\cdot 10^4$). Description of the test cases follows.

The first line of each testcase contains one integer $n$ ($1\le n\le 2\cdot 10^5$).

The second line of each testcase contains a string consisting of $0$ and $1$, describing the sequence $a$.

It's guaranteed that the sum of $n$ over all testcases does not exceed $2\cdot 10^5$.

Output: For each testcase, if it's possible to make $a=[1]$, print YES. Otherwise, print NO. You can output the answer in any case (upper or lower). For example, the strings yEs, yes, Yes, and YES will be recognized as positive responses.

Examples:
input:
5
1
0
1
1
2
01
9
100000001
9
000011000
output:
No
Yes
No
Yes
No

Note: In the fourth testcase of the example, initially $a=[1,0,0,0,0,0,0,0,1]$. A valid sequence of operations is:

  1. Select $l=2,r=8$ and apply the operation. $a$ becomes $[1,0,1]$.    2. Select $l=1,r=3$ and apply the operation. $a$ becomes $[1]$.

Buggy Code:
// Created by r-o-s-t on 11/07/2024.

#include<bits/stdc++.h>
#ifdef LOCAL
#include <C:\algo\debug.h>
#else
#define debug(...)
#define endl '\n'
#endif

using ll = long long;
using ull = unsigned long long;
using namespace std;

void solve() {
    int n; cin >> n;
    string a; cin >> a;

    char last = '1', c0 = 0, c1 = 0;
    for (int i = 0; i < n; ++i) {
        if (a[i] == '1') c1++;
        else if (last == '1') c0++;
        last = a[i];
    }
    if (c1 > c0) cout << "Yes" << endl;
    else cout << "No" << endl;
}

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

    int t; cin >> t;
    while (t--) solve();
}

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

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

Think step by step.

Response:
```plaintext
5
1
0
1
1
2
01
3
100
3
010
```

**Explanation:**
1. **Test Case 1:**
   - Input: `1 0`
   - Expected Output: `No`
   - Reason: The sequence is `[0]`, and it cannot be made `[1]` with any operations.

2. **Test Case 2:**
   - Input: `1 1`
   - Expected Output: `Yes`
   - Reason: The sequence is `[1]`, which is already `[1]`.

3. **Test Case 3:**
   - Input: `2 01`
   - Expected Output: `No`
   - Reason: The sequence is `[0, 1]`, and the only possible operation is to replace `[0, 1]` with `0`, resulting in `[0]`, which cannot be made `[1]`.

4. **Test Case 4:**
   - Input: `3 100`
   - Expected Output: `Yes`
   - Reason: The sequence is `[1, 0, 0]`. One possible operation is to replace `[1, 0, 0]` with `0`, resulting in `[0]`, which cannot be made `[1]`. However, another possible operation is to replace `[0, 0]` with `0`, resulting in `[1, 0]`. Then, replace `[1, 0]` with `1`, resulting in `[1]`.

5. **Test Case 5:**
   - Input: `3 010`
   - Expected Output: `No`
   - Reason: The sequence is `[0, 1, 0]`. Any operation will result in a sequence that cannot be made `[1]`.

These test cases are designed to cover edge cases and ensure that the buggy code is exposed. The buggy code incorrectly counts the number of `0`s and `1`s, leading to incorrect conclusions in some scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Longest Max Min Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer sequence $a_1, a_2, \ldots, a_n$. Let $S$ be the set of all possible non-empty subsequences of $a$ without duplicate elements. Your goal is to find the longest sequence in $S$. If there are multiple of them, find the one that minimizes lexicographical order after multiplying terms at odd positions by $-1$.

For example, given $a = [3, 2, 3, 1]$, $S = \\{[1], [2], [3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 3, 1], [3, 2, 1]\\}$. Then $[2, 3, 1]$ and $[3, 2, 1]$ would be the longest, and $[3, 2, 1]$ would be the answer since $[-3, 2, -1]$ is lexicographically smaller than $[-2, 3, -1]$.

A sequence $c$ is a subsequence of a sequence $d$ if $c$ can be obtained from $d$ by the deletion of several (possibly, zero or all) elements.

A sequence $c$ is lexicographically smaller than a sequence $d$ if and only if one of the following holds:

  * $c$ is a prefix of $d$, but $c \ne d$;   * in the first position where $c$ and $d$ differ, the sequence $c$ has a smaller element than the corresponding element in $d$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 5 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains an integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output the answer in the following format:

Output an integer $m$ in the first line — the length of $b$.

Then output $m$ integers $b_1, b_2, \ldots, b_m$ in the second line — the sequence $b$.

Examples:
input:
4
4
3 2 1 3
4
1 1 1 1
9
3 2 1 3 2 1 3 2 1
1
1
output:
3
3 2 1
1
1
3
3 1 2
1
1

input:
10
2
1 2
10
5 2 1 7 9 7 2 5 5 2
2
1 2
10
2 2 8 7 7 9 8 1 9 6
9
9 1 7 5 8 5 6 4 1
3
3 3 3
6
1 6 4 4 6 5
6
3 4 4 5 3 3
10
4 1 4 5 4 5 10 1 5 1
7
1 2 1 3 2 4 6
output:
2
1 2
5
5 1 9 7 2
2
1 2
6
2 7 9 8 1 6
7
9 1 7 5 8 6 4
1
3
4
1 4 6 5
3
4 5 3
4
5 4 10 1
5
2 1 3 4 6

Note: In the first example, $S = \\{[1], [2], [3], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 1, 3], [3, 2, 1]\\}$. Among them, $[2, 1, 3]$ and $[3, 2, 1]$ are the longest and $[-3, 2, -1]$ is lexicographical smaller than $[-2, 1, -3]$, so $[3, 2, 1]$ is the answer.

In the second example, $S = \\{[1]\\}$, so $[1]$ is the answer.

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

// t ={{max,maxi},{min,mini}}
template<class T>
struct SegmentTree
{
	int n;
	vector<T>it;
	SegmentTree(int nn)
	{
		n=nn;
		it.resize(4*n);
	}
	T combine(const T &l,const T &r)
	{
		T ans;
    if(l.first.first < r.first.first) ans.first = r.first;
    else ans.first = l.first;
    if(l.second.first >= r.second.first) ans.second = r.second;
    else ans.second = l.second;
		return ans;
	}
	void up(int x,int l,int r,const int &L,const int &R,const T &U)
	{
    // cout<<x<<" "<<l<<" "<<r<<" "<<L<<" "<<R<<" "<<U.first.first<<endl;
		if(L<=l && r<=R)
		{
			it[x]=U;
			return;
		}
		if(L<=(l+r)/2)up(x*2,l,(l+r)/2,L,R,U);
		if(R>(l+r)/2)up(x*2+1,(l+r)/2+1,r,L,R,U);
		it[x]=combine(it[x*2],it[x*2+1]);
	}
	void update(int l,int r,T u)
	{
		up(1,0,n-1,l,r,u);
	}
	T qu(int x,int l,int r,int L,int R)
	{
		if(L<=l && r<=R)return it[x];
		T ansl,ansr;
		int f=0;
		if(L<=(l+r)/2)
		{
			f|=1;
			ansl=qu(x*2,l,(l+r)/2,L,R);
		}
		if(R>(l+r)/2)
		{
			f|=2;
			ansr=qu(x*2+1,(l+r)/2+1,r,L,R);
		}
		if(f==1)return ansl;
		if(f==2)return ansr;
		return combine(ansl,ansr);
	}
	T query(int l,int r)
	{
		return qu(1,0,n-1,l,r);
	}
};
void solve()
{
  int n,i;
  cin>>n;
  SegmentTree<pair<pair<int,int>,pair<int,int>>> T(n);
  vector<int>a(n+1),br(n+1,0),used(n+1,0);
  vector<vector<int>>l(n+1);
  int o=0,c=0,r,le=0,p;
  for(i=0;i<n;i++)
  {
      cin>>a[i];
      l[a[i]].push_back(i);
      if(br[a[i]]++==0)++o;
      T.update(i,i,{{a[i],i},{a[i],i}});
  }
  br = vector<int>(n+1,0);
  for(i=n-1;i>=0;--i)
  {
      if(br[a[i]]++==0)++c;
      if(c==o)break;
  }
  r=i;
  vector<int>ans;
  for(i=0;i<o;i++)
  {
      // cout<<le<<" "<<r<<endl;
      auto x = T.query(le,r);
      if(i&1)
      {
        le = x.second.second+1;
        p = x.second.first;
      }
      else
      {
        le = x.first.second+1;
        p = x.first.first;
      }
      for(int x:l[p])
      {
        T.update(x,x,{{-n*2,x},{n*2,x}});
        used[x]=1;
      }
      while(used[r])
        r++;
      ans.push_back(p);
  }
  cout<<o<<'\n';
  for(int x:ans)
  cout<<x<<' ';
  cout<<'\n';
}
int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int t=1;
	cin>>t;
	while(t--)
		solve();
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Longest Max Min Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer sequence $a_1, a_2, \ldots, a_n$. Let $S$ be the set of all possible non-empty subsequences of $a$ without duplicate elements. Your goal is to find the longest sequence in $S$. If there are multiple of them, find the one that minimizes lexicographical order after multiplying terms at odd positions by $-1$.

For example, given $a = [3, 2, 3, 1]$, $S = \\{[1], [2], [3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 3, 1], [3, 2, 1]\\}$. Then $[2, 3, 1]$ and $[3, 2, 1]$ would be the longest, and $[3, 2, 1]$ would be the answer since $[-3, 2, -1]$ is lexicographically smaller than $[-2, 3, -1]$.

A sequence $c$ is a subsequence of a sequence $d$ if $c$ can be obtained from $d$ by the deletion of several (possibly, zero or all) elements.

A sequence $c$ is lexicographically smaller than a sequence $d$ if and only if one of the following holds:

  * $c$ is a prefix of $d$, but $c \ne d$;   * in the first position where $c$ and $d$ differ, the sequence $c$ has a smaller element than the corresponding element in $d$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 5 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains an integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output the answer in the following format:

Output an integer $m$ in the first line — the length of $b$.

Then output $m$ integers $b_1, b_2, \ldots, b_m$ in the second line — the sequence $b$.

Examples:
input:
4
4
3 2 1 3
4
1 1 1 1
9
3 2 1 3 2 1 3 2 1
1
1
output:
3
3 2 1
1
1
3
3 1 2
1
1

input:
10
2
1 2
10
5 2 1 7 9 7 2 5 5 2
2
1 2
10
2 2 8 7 7 9 8 1 9 6
9
9 1 7 5 8 5 6 4 1
3
3 3 3
6
1 6 4 4 6 5
6
3 4 4 5 3 3
10
4 1 4 5 4 5 10 1 5 1
7
1 2 1 3 2 4 6
output:
2
1 2
5
5 1 9 7 2
2
1 2
6
2 7 9 8 1 6
7
9 1 7 5 8 6 4
1
3
4
1 4 6 5
3
4 5 3
4
5 4 10 1
5
2 1 3 4 6

Note: In the first example, $S = \\{[1], [2], [3], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 1, 3], [3, 2, 1]\\}$. Among them, $[2, 1, 3]$ and $[3, 2, 1]$ are the longest and $[-3, 2, -1]$ is lexicographical smaller than $[-2, 1, -3]$, so $[3, 2, 1]$ is the answer.

In the second example, $S = \\{[1]\\}$, so $[1]$ is the answer.

Buggy Code:
#include<bits/stdc++.h>
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pi;
const int N=2e5+10;
bool st[N];
ll a[N];
void solved()
{
	ll n;
	cin>>n;
	map<ll,ll> mp;

	for(ll i=0;i<=n;i++)
	st[i]=false;
	ll last=0;
	priority_queue<pi> qmax;
	priority_queue<pi,vector<pi>,greater<pi>> qmin;
	for(ll i=1;i<=n;i++)
	{
		cin>>a[i];
		mp[a[i]]++;
	}
	bool flag=true;
	vector<ll> ans;

	for(ll i=1;i<=n;i++)
	{

		mp[a[i]]--;
		if(mp[a[i]]==0&&st[a[i]]==false)
		{
			ll t,pos;
			while(1)
			{
				if(flag)
				{
					if(qmax.empty())
					{
						st[a[i]]=true;
						last=a[i];
						ans.push_back(a[i]);
						flag=!flag;
						break;
					}
					t=qmax.top().x;
					pos=qmax.top().y;
					if(st[t]||pos<=last)
					{
					qmax.pop();
					continue;
					}
					if(a[i]>=t)
					{
						st[a[i]]=true;
						if(a[i]==t)
						last=pos;
						else
						last=i;
						ans.push_back(a[i]);
						flag=!flag;
						break;
					}
					else
					{
						st[t]=true;
						last=pos;
						ans.push_back(t);
						flag=!flag;
						continue;
					}
				}
				else
				{
					if(qmin.empty())
					{
						st[a[i]]=true;
						last=i;
						ans.push_back(a[i]);
						flag=!flag;
						break;
					}
					t=qmin.top().x;
					pos=qmin.top().y;
					if(st[t]||pos<=last)
					{
					qmin.pop();
					continue;
					}
					if(a[i]<=t)
					{
						st[a[i]]=true;
						if(a[i]==t)
						last=pos;
						else
						last=i;
						ans.push_back(a[i]);
						flag=!flag;
						break;
					}
					else
					{
						st[t]=true;
						last=pos;
						ans.push_back(t);
						flag=!flag;
						continue;
					}
				}
			}
		}
		qmax.push({a[i],i});
		qmin.push({a[i],i});
	}
	cout<<ans.size()<<endl;
	for(ll i=0;i<ans.size();i++)
	cout<<ans[i]<<" ";
	cout<<endl;
}
int main()
{
	ll t;
	cin>>t;
	while(t--)
	{
		solved();
	}
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Longest Max Min Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer sequence $a_1, a_2, \ldots, a_n$. Let $S$ be the set of all possible non-empty subsequences of $a$ without duplicate elements. Your goal is to find the longest sequence in $S$. If there are multiple of them, find the one that minimizes lexicographical order after multiplying terms at odd positions by $-1$.

For example, given $a = [3, 2, 3, 1]$, $S = \\{[1], [2], [3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 3, 1], [3, 2, 1]\\}$. Then $[2, 3, 1]$ and $[3, 2, 1]$ would be the longest, and $[3, 2, 1]$ would be the answer since $[-3, 2, -1]$ is lexicographically smaller than $[-2, 3, -1]$.

A sequence $c$ is a subsequence of a sequence $d$ if $c$ can be obtained from $d$ by the deletion of several (possibly, zero or all) elements.

A sequence $c$ is lexicographically smaller than a sequence $d$ if and only if one of the following holds:

  * $c$ is a prefix of $d$, but $c \ne d$;   * in the first position where $c$ and $d$ differ, the sequence $c$ has a smaller element than the corresponding element in $d$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 5 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains an integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output the answer in the following format:

Output an integer $m$ in the first line — the length of $b$.

Then output $m$ integers $b_1, b_2, \ldots, b_m$ in the second line — the sequence $b$.

Examples:
input:
4
4
3 2 1 3
4
1 1 1 1
9
3 2 1 3 2 1 3 2 1
1
1
output:
3
3 2 1
1
1
3
3 1 2
1
1

input:
10
2
1 2
10
5 2 1 7 9 7 2 5 5 2
2
1 2
10
2 2 8 7 7 9 8 1 9 6
9
9 1 7 5 8 5 6 4 1
3
3 3 3
6
1 6 4 4 6 5
6
3 4 4 5 3 3
10
4 1 4 5 4 5 10 1 5 1
7
1 2 1 3 2 4 6
output:
2
1 2
5
5 1 9 7 2
2
1 2
6
2 7 9 8 1 6
7
9 1 7 5 8 6 4
1
3
4
1 4 6 5
3
4 5 3
4
5 4 10 1
5
2 1 3 4 6

Note: In the first example, $S = \\{[1], [2], [3], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 1, 3], [3, 2, 1]\\}$. Among them, $[2, 1, 3]$ and $[3, 2, 1]$ are the longest and $[-3, 2, -1]$ is lexicographical smaller than $[-2, 1, -3]$, so $[3, 2, 1]$ is the answer.

In the second example, $S = \\{[1]\\}$, so $[1]$ is the answer.

Buggy Code:
#define _USE_MATH_DEFINES

#include <iostream>
#include <fstream>
#include <cmath>
#include <iomanip>
#include <vector>
#include <string>
#include <algorithm>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <cstring>
#include <deque>
#include <queue>
#include <stack>
#include <cctype>
#include <set>
#include <numeric>
#include <functional>
#include <regex>

using namespace std;

typedef long long ll;

int main() {
#ifdef LOCALIC
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#else
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
#endif

	int tt = 1;
	cin >> tt;
	while (tt--) {
		int n;
		cin >> n;
		vector<ll> a(n);
		for (int i = 0; i < n; i++) cin >> a[i];

		set<int> s;
		for (int i = 0; i < n; i++) s.insert(a[i]);
		int len = s.size();

		map<int, vector<int>> indexes;
		set<int> bounds;
		for (int r = n - 1; r >= 0; r--) {
			if (s.count(a[r])) {
				bounds.insert(r);
			}
			s.erase(a[r]);
			indexes[a[r]].push_back(r);
		}

		multiset<int> m;

		int l = 0;
		int r = 0;
		vector<int> ans;
		for (int i = 0; i < len; i++) {
			for (; r <= *bounds.begin(); r++) m.insert(a[r]);
			while (*m.begin() == -1) m.erase(m.begin());

			int val;
			if (i % 2 == 0) val = *m.rbegin();
			else val = *m.begin();
			int ind = lower_bound(indexes[val].begin(), indexes[val].end(), l) - indexes[val].begin();

			m.erase(val);
			for (int x : indexes[val]) {
				bounds.erase(x);
				a[x] = -1;
			}
			ans.push_back(val);
			for (; l <= ind; l++) if (m.find(a[l]) != m.end()) m.erase(m.find(a[l]));
		}

		cout << len << endl;
		for (int x : ans) cout << x << " ";
		cout << endl;
	}
}

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

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

Think step by step.

Response:
```plaintext
4
4
3 2 3 1
4
1 1 1 1
9
3 2 1 3 2 1 3 2 1
1
1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Longest Max Min Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer sequence $a_1, a_2, \ldots, a_n$. Let $S$ be the set of all possible non-empty subsequences of $a$ without duplicate elements. Your goal is to find the longest sequence in $S$. If there are multiple of them, find the one that minimizes lexicographical order after multiplying terms at odd positions by $-1$.

For example, given $a = [3, 2, 3, 1]$, $S = \\{[1], [2], [3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 3, 1], [3, 2, 1]\\}$. Then $[2, 3, 1]$ and $[3, 2, 1]$ would be the longest, and $[3, 2, 1]$ would be the answer since $[-3, 2, -1]$ is lexicographically smaller than $[-2, 3, -1]$.

A sequence $c$ is a subsequence of a sequence $d$ if $c$ can be obtained from $d$ by the deletion of several (possibly, zero or all) elements.

A sequence $c$ is lexicographically smaller than a sequence $d$ if and only if one of the following holds:

  * $c$ is a prefix of $d$, but $c \ne d$;   * in the first position where $c$ and $d$ differ, the sequence $c$ has a smaller element than the corresponding element in $d$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 5 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains an integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output the answer in the following format:

Output an integer $m$ in the first line — the length of $b$.

Then output $m$ integers $b_1, b_2, \ldots, b_m$ in the second line — the sequence $b$.

Examples:
input:
4
4
3 2 1 3
4
1 1 1 1
9
3 2 1 3 2 1 3 2 1
1
1
output:
3
3 2 1
1
1
3
3 1 2
1
1

input:
10
2
1 2
10
5 2 1 7 9 7 2 5 5 2
2
1 2
10
2 2 8 7 7 9 8 1 9 6
9
9 1 7 5 8 5 6 4 1
3
3 3 3
6
1 6 4 4 6 5
6
3 4 4 5 3 3
10
4 1 4 5 4 5 10 1 5 1
7
1 2 1 3 2 4 6
output:
2
1 2
5
5 1 9 7 2
2
1 2
6
2 7 9 8 1 6
7
9 1 7 5 8 6 4
1
3
4
1 4 6 5
3
4 5 3
4
5 4 10 1
5
2 1 3 4 6

Note: In the first example, $S = \\{[1], [2], [3], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 1, 3], [3, 2, 1]\\}$. Among them, $[2, 1, 3]$ and $[3, 2, 1]$ are the longest and $[-3, 2, -1]$ is lexicographical smaller than $[-2, 1, -3]$, so $[3, 2, 1]$ is the answer.

In the second example, $S = \\{[1]\\}$, so $[1]$ is the answer.

Buggy Code:
class D:
    def solve(self):
        t = int(input())
        for i in range(t):
            n = int(input())
            li = list(map(int, input().split()))
            val = set()
            unique = [0]
            for i in range(n-1, -1, -1):
                if li[i] not in val:
                    unique.insert(0, unique[0]+1)
                    val.add(li[i])
                else:
                    unique.insert(0, unique[0])
            ret = []
            ret1 = set()
            i = 0
            j = len(val)
            cnt = 0
            while(i < n):
                if(li[i] in ret1):
                    i+=1
                    continue
                minimax = li[i]
                minimin = li[i]
                ind = i
                while(unique[i] == j):
                    if(li[i] in ret1):
                        i+=1
                        continue
                    if(cnt%2 == 0):
                        if(minimax < li[i]):
                            minimax = li[i]
                            ind = i
                    else:
                        if(minimin > li[i]):
                            minimin = li[i]
                            ind = i
                    i+=1
                if(cnt%2 == 0):
                    ret.append(minimax)
                    ret1.add(minimax)
                else:
                    ret.append(minimin)
                    ret1.add(minimin)
                i = ind+1
                cnt+=1

            print(len(ret))
            for i in ret:
                print(i, end=" ")
            print()

if __name__ == "__main__":
    d = D()
    d.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:
To generate a challenging test input that exposes the bug in the provided code, we need to consider edge cases and scenarios where the code might fail. Here are some key points to consider:

1. **Duplicate Elements**: The code should handle sequences with duplicate elements correctly.
2. **Lexicographical Order**: The code should correctly handle the lexicographical order after multiplying terms at odd positions by -1.
3. **Longest Subsequence**: The code should find the longest subsequence without duplicates and then choose the one with the smallest lexicographical order.

Given these points, a good test input would be a sequence with multiple duplicate elements and a structure that challenges the lexicographical order handling. Here is a small-scale test input:

```plaintext
1
6
3 3 1 2 2 1
```

Explanation:
- **Input**: The sequence is `[3, 3, 1, 2, 2, 1]`.
- **Expected Output**: The longest subsequence without duplicates is `[3, 1, 2]`. After multiplying terms at odd positions by -1, the sequence becomes `[-3, 1, -2]`, which is the lexicographically smallest among all possible longest subsequences.

This input should expose any issues with handling duplicates and lexicographical order correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Longest Max Min Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer sequence $a_1, a_2, \ldots, a_n$. Let $S$ be the set of all possible non-empty subsequences of $a$ without duplicate elements. Your goal is to find the longest sequence in $S$. If there are multiple of them, find the one that minimizes lexicographical order after multiplying terms at odd positions by $-1$.

For example, given $a = [3, 2, 3, 1]$, $S = \\{[1], [2], [3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 3, 1], [3, 2, 1]\\}$. Then $[2, 3, 1]$ and $[3, 2, 1]$ would be the longest, and $[3, 2, 1]$ would be the answer since $[-3, 2, -1]$ is lexicographically smaller than $[-2, 3, -1]$.

A sequence $c$ is a subsequence of a sequence $d$ if $c$ can be obtained from $d$ by the deletion of several (possibly, zero or all) elements.

A sequence $c$ is lexicographically smaller than a sequence $d$ if and only if one of the following holds:

  * $c$ is a prefix of $d$, but $c \ne d$;   * in the first position where $c$ and $d$ differ, the sequence $c$ has a smaller element than the corresponding element in $d$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 5 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains an integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output the answer in the following format:

Output an integer $m$ in the first line — the length of $b$.

Then output $m$ integers $b_1, b_2, \ldots, b_m$ in the second line — the sequence $b$.

Examples:
input:
4
4
3 2 1 3
4
1 1 1 1
9
3 2 1 3 2 1 3 2 1
1
1
output:
3
3 2 1
1
1
3
3 1 2
1
1

input:
10
2
1 2
10
5 2 1 7 9 7 2 5 5 2
2
1 2
10
2 2 8 7 7 9 8 1 9 6
9
9 1 7 5 8 5 6 4 1
3
3 3 3
6
1 6 4 4 6 5
6
3 4 4 5 3 3
10
4 1 4 5 4 5 10 1 5 1
7
1 2 1 3 2 4 6
output:
2
1 2
5
5 1 9 7 2
2
1 2
6
2 7 9 8 1 6
7
9 1 7 5 8 6 4
1
3
4
1 4 6 5
3
4 5 3
4
5 4 10 1
5
2 1 3 4 6

Note: In the first example, $S = \\{[1], [2], [3], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 1, 3], [3, 2, 1]\\}$. Among them, $[2, 1, 3]$ and $[3, 2, 1]$ are the longest and $[-3, 2, -1]$ is lexicographical smaller than $[-2, 1, -3]$, so $[3, 2, 1]$ is the answer.

In the second example, $S = \\{[1]\\}$, so $[1]$ is the answer.

Buggy Code:
def get_last_occurrences(sequence):
    """
    Returns a dictionary mapping each unique element to its last occurrence index.
    """
    last_occurrence = {}
    for index, num in enumerate(sequence):
        last_occurrence[num] = index
    return last_occurrence

def get_modified_value(num, position):
    """
    Returns the modified value by multiplying by -1 if the position is odd.
    """
    return -num if position % 2 == 1 else num

def build_subsequence(sequence, last_occurrence):
    """
    Builds the desired subsequence based on the greedy approach.
    """
    stack = []
    seen = set()
    for index, num in enumerate(sequence):
        if num in seen:
            continue
        while stack:
            current_position = len(stack)  # Next position is current length
            last_num = stack[-1]
            modified_last = get_modified_value(last_num, current_position)
            modified_current = get_modified_value(num, current_position)
            if (modified_current < modified_last) and (last_occurrence[last_num] > index):
                seen.remove(stack.pop())
            else:
                break
        stack.append(num)
        seen.add(num)
    return stack

def process_test_case(n, sequence):
    """
    Processes a single test case and returns the result subsequence.
    """
    last_occurrence = get_last_occurrences(sequence)
    subsequence = build_subsequence(sequence, last_occurrence)
    return subsequence

def read_input():
    """
    Reads all input at once and returns it as a list of tokens.
    """
    import sys
    return sys.stdin.read().split()

def main():
    """
    Main function to handle input/output and process all test cases.
    """
    tokens = read_input()
    ptr = 0
    t = int(tokens[ptr])
    ptr += 1
    results = []
    for _ in range(t):
        n = int(tokens[ptr])
        ptr += 1
        sequence = list(map(int, tokens[ptr:ptr + n]))
        ptr += n
        subsequence = process_test_case(n, sequence)
        results.append((len(subsequence), subsequence))
    for length, subseq in results:
        print(length)
        print(' '.join(map(str, subseq)))

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:
To create a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases that can cause the algorithm to fail. Specifically, we need to consider the following scenarios:

1. **Sequences with repeated elements**: The code might struggle with correctly handling sequences that have multiple occurrences of the same element.
2. **Sequences with elements in decreasing order**: This can test the code's ability to handle sequences where the lexicographical order is critical.
3. **Sequences with elements in increasing order**: This can test the code's ability to handle sequences where the lexicographical order is critical.
4. **Sequences with a mix of increasing and decreasing elements**: This can test the code's ability to handle more complex sequences.

Given these points, let's create a small-scale test input that includes these edge cases.

### Test Input
```plaintext
4
5
1 2 3 4 5
5
5 4 3 2 1
6
1 2 3 3 2 1
6
3 1 2 2 1 3
```

### Explanation
1. **Sequence 1: [1, 2, 3, 4, 5]**
   - This sequence is in increasing order. The expected output is the sequence itself, as it is already the longest and lexicographically smallest.

2. **Sequence 2: [5, 4, 3, 2, 1]**
   - This sequence is in decreasing order. The expected output is the sequence itself, as it is already the longest and lexicographically smallest after applying the odd position multiplication.

3. **Sequence 3: [1, 2, 3, 3, 2, 1]**
   - This sequence has repeated elements. The expected output is [1, 2, 3, 2, 1], which is the longest and lexicographically smallest after applying the odd position multiplication.

4. **Sequence 4: [3, 1, 2, 2, 1, 3]**
   - This sequence has repeated elements and a mix of increasing and decreasing elements. The expected output is [3, 1, 2, 1, 3], which is the longest and lexicographically smallest after applying the odd position multiplication.

This test input should help expose any issues in the buggy code, particularly in handling repeated elements and maintaining the correct lexicographical order.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Longest Max Min Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer sequence $a_1, a_2, \ldots, a_n$. Let $S$ be the set of all possible non-empty subsequences of $a$ without duplicate elements. Your goal is to find the longest sequence in $S$. If there are multiple of them, find the one that minimizes lexicographical order after multiplying terms at odd positions by $-1$.

For example, given $a = [3, 2, 3, 1]$, $S = \\{[1], [2], [3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 3, 1], [3, 2, 1]\\}$. Then $[2, 3, 1]$ and $[3, 2, 1]$ would be the longest, and $[3, 2, 1]$ would be the answer since $[-3, 2, -1]$ is lexicographically smaller than $[-2, 3, -1]$.

A sequence $c$ is a subsequence of a sequence $d$ if $c$ can be obtained from $d$ by the deletion of several (possibly, zero or all) elements.

A sequence $c$ is lexicographically smaller than a sequence $d$ if and only if one of the following holds:

  * $c$ is a prefix of $d$, but $c \ne d$;   * in the first position where $c$ and $d$ differ, the sequence $c$ has a smaller element than the corresponding element in $d$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 5 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains an integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output the answer in the following format:

Output an integer $m$ in the first line — the length of $b$.

Then output $m$ integers $b_1, b_2, \ldots, b_m$ in the second line — the sequence $b$.

Examples:
input:
4
4
3 2 1 3
4
1 1 1 1
9
3 2 1 3 2 1 3 2 1
1
1
output:
3
3 2 1
1
1
3
3 1 2
1
1

input:
10
2
1 2
10
5 2 1 7 9 7 2 5 5 2
2
1 2
10
2 2 8 7 7 9 8 1 9 6
9
9 1 7 5 8 5 6 4 1
3
3 3 3
6
1 6 4 4 6 5
6
3 4 4 5 3 3
10
4 1 4 5 4 5 10 1 5 1
7
1 2 1 3 2 4 6
output:
2
1 2
5
5 1 9 7 2
2
1 2
6
2 7 9 8 1 6
7
9 1 7 5 8 6 4
1
3
4
1 4 6 5
3
4 5 3
4
5 4 10 1
5
2 1 3 4 6

Note: In the first example, $S = \\{[1], [2], [3], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 1, 3], [3, 2, 1]\\}$. Among them, $[2, 1, 3]$ and $[3, 2, 1]$ are the longest and $[-3, 2, -1]$ is lexicographical smaller than $[-2, 1, -3]$, so $[3, 2, 1]$ is the answer.

In the second example, $S = \\{[1]\\}$, so $[1]$ is the answer.

Buggy Code:
import bisect
import math
import sys
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
class SegmentTreeMin:
    def __init__(self, data):
        self.n = len(data)
        self.tree = [(float('inf'), -1)] * (2 * self.n)
        # Build the tree
        for i in range(self.n):
            self.tree[self.n + i] = (data[i], i)
        for i in range(self.n - 1, 0, -1):
            self.tree[i] = min(self.tree[2 * i], self.tree[2 * i + 1], key=lambda x: (x[0], x[1]))

    def update(self, index, value):
        index += self.n
        self.tree[index] = (value, index - self.n)
        while index > 1:
            index //= 2
            self.tree[index] = min(self.tree[2 * index], self.tree[2 * index + 1], key=lambda x: (x[0], x[1]))

    def query(self, left, right):
        left += self.n
        right += self.n + 1
        min_value = (float('inf'), -1)
        while left < right:
            if left % 2 == 1:
                min_value = min(min_value, self.tree[left], key=lambda x: (x[0], x[1]))
                left += 1
            if right % 2 == 1:
                right -= 1
                min_value = min(min_value, self.tree[right], key=lambda x: (x[0], x[1]))
            left //= 2
            right //= 2
        return min_value

class SegmentTreeMax:
    def __init__(self, data):
        self.n = len(data)
        self.tree = [(-float('inf'), -1)] * (2 * self.n)
        # Build the tree
        for i in range(self.n):
            self.tree[self.n + i] = (data[i], i)
        for i in range(self.n - 1, 0, -1):
            self.tree[i] = max(self.tree[2 * i], self.tree[2 * i + 1], key=lambda x: (x[0], -x[1]))

    def update(self, index, value):
        index += self.n
        self.tree[index] = (value, index - self.n)
        while index > 1:
            index //= 2
            self.tree[index] = max(self.tree[2 * index], self.tree[2 * index + 1], key=lambda x: (x[0], -x[1]))

    def query(self, left, right):
        left += self.n
        right += self.n + 1
        max_value = (-float('inf'), -1)
        while left < right:
            if left % 2 == 1:
                max_value = max(max_value, self.tree[left], key=lambda x: (x[0], -x[1]))
                left += 1
            if right % 2 == 1:
                right -= 1
                max_value = max(max_value, self.tree[right], key=lambda x: (x[0], -x[1]))
            left //= 2
            right //= 2
        return max_value
def sortremoveDuplicates(arr, n):
    if n == 0 or n == 1:
        return arr

    temp = []
    j = 0
    for i in range(0, n-1):
        if arr[i] != arr[i+1]:
            temp += [arr[i]]
            j += 1
    temp += [arr[n-1]]
    return temp
def bs(a,l1):
    low = 0
    high =len(l1)-1
    mid = 0

    while low <= high:

        mid = (high + low) // 2

        if l1[mid] <a :
            low = mid + 1
        elif l1[mid] > a:
            high = mid - 1
        else:
            return mid
    return -1
t=int(input())
for i in range(t):
    n=int(input())
    a=[int(num) for num in input().split()]
    l=a[:]
    l.sort()
    l=sortremoveDuplicates(l,len(l))
    ind=[]
    #print(l,a)
    for i in range(len(l)):
        ind.append([])
    maxind=[-1]*(n+1)
    for i in range(len(a)):
        maxind[a[i]]=max(maxind[a[i]],i)
        num=a[i]
        x=bs(num,l)
        #print(x,"hi")
        ind[x].append(i)
    #print(ind)
    maxi=[]
    mini=[]
    for i in range(len(a)):
        mini.append([i,a[i]])
    for i in range(len(ind)):
        maxi.append([ind[i][-1],l[i]])
        #mini.append([ind[i][0],l[i]])
    mini.sort()
    maxi.sort()
    mi=[]
    ma=[]
    indi=[]
    i1=[]
    for i in range(len(mini)):
        mi.append(mini[i][1])
        # ma.append(maxi[i][1])
        # indi.append(maxi[i][0])
        i1.append(mini[i][0])
    dct={}
    for i in range(len(maxi)):
        ma.append(maxi[i][1])
        indi.append(maxi[i][0])
        dct[maxi[i][1]]=i
    seg1=SegmentTreeMin(ma)
    seg2=SegmentTreeMin(mi)
    seg=SegmentTreeMin(indi)
    seg11=SegmentTreeMax(ma)
    seg22=SegmentTreeMax(mi)
    segg=SegmentTreeMax(indi)
    ans=[]
    prev=0
    for i in range(len(l)):
        num=seg.query(0,len(l)-1)[0]
        index=num
        #print(index,num)
        if(i%2==0):
            m=seg22.query(prev,index)[0]
            temp=seg22.query(prev,index)[1]
            index=temp
        else:
            m=seg2.query(0,index)[0]
            temp=seg2.query(0,index)[1]
            index=temp
        ans.append(m)
        #for j in range(index,prev-1,-1):
            #seg2.update(j,2*n)
            #seg22.update(j,-1)
        x=bs(m,l)
        for j in range(len(ind[x])):
            seg2.update(ind[x][j],2*n)
            seg22.update(ind[x][j],-1)
        i2=dct[m]
        seg.update(i2,2*n)
        segg.update(i2,-1)
        prev=index
    print(len(ans))
    print(*ans)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Longest Max Min Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer sequence $a_1, a_2, \ldots, a_n$. Let $S$ be the set of all possible non-empty subsequences of $a$ without duplicate elements. Your goal is to find the longest sequence in $S$. If there are multiple of them, find the one that minimizes lexicographical order after multiplying terms at odd positions by $-1$.

For example, given $a = [3, 2, 3, 1]$, $S = \\{[1], [2], [3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 3, 1], [3, 2, 1]\\}$. Then $[2, 3, 1]$ and $[3, 2, 1]$ would be the longest, and $[3, 2, 1]$ would be the answer since $[-3, 2, -1]$ is lexicographically smaller than $[-2, 3, -1]$.

A sequence $c$ is a subsequence of a sequence $d$ if $c$ can be obtained from $d$ by the deletion of several (possibly, zero or all) elements.

A sequence $c$ is lexicographically smaller than a sequence $d$ if and only if one of the following holds:

  * $c$ is a prefix of $d$, but $c \ne d$;   * in the first position where $c$ and $d$ differ, the sequence $c$ has a smaller element than the corresponding element in $d$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 5 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains an integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output the answer in the following format:

Output an integer $m$ in the first line — the length of $b$.

Then output $m$ integers $b_1, b_2, \ldots, b_m$ in the second line — the sequence $b$.

Examples:
input:
4
4
3 2 1 3
4
1 1 1 1
9
3 2 1 3 2 1 3 2 1
1
1
output:
3
3 2 1
1
1
3
3 1 2
1
1

input:
10
2
1 2
10
5 2 1 7 9 7 2 5 5 2
2
1 2
10
2 2 8 7 7 9 8 1 9 6
9
9 1 7 5 8 5 6 4 1
3
3 3 3
6
1 6 4 4 6 5
6
3 4 4 5 3 3
10
4 1 4 5 4 5 10 1 5 1
7
1 2 1 3 2 4 6
output:
2
1 2
5
5 1 9 7 2
2
1 2
6
2 7 9 8 1 6
7
9 1 7 5 8 6 4
1
3
4
1 4 6 5
3
4 5 3
4
5 4 10 1
5
2 1 3 4 6

Note: In the first example, $S = \\{[1], [2], [3], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 1, 3], [3, 2, 1]\\}$. Among them, $[2, 1, 3]$ and $[3, 2, 1]$ are the longest and $[-3, 2, -1]$ is lexicographical smaller than $[-2, 1, -3]$, so $[3, 2, 1]$ is the answer.

In the second example, $S = \\{[1]\\}$, so $[1]$ is the answer.

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

public class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0)
		{
		    int n=sc.nextInt();
		    int arr[]=new int[n],cnt[]=new int[n+1];
		    int max=0;
		    for(int i=0;i<n;i++){
		    arr[i]=sc.nextInt();cnt[arr[i]]++;}
		    int in[]=new int[n+1];
		    int ptr=0;int ans[]=new int[n];
		    for(int i=0;i<n;i++){
		        cnt[arr[i]]--;
		        if(ptr==0){ ans[ptr++]=arr[i];in[arr[i]]=1;}
		        else{
		            if(in[arr[i]]==1) continue;
		            while((ptr!=0 && cnt[arr[ptr-1]]!=0 && better(ptr-1,arr[i],ans[ptr-1])) ||
		            (ptr>1 && cnt[arr[ptr-1]]!=0 && cnt[arr[ptr-2]]!=0 && better(ptr-2,arr[i],ans[ptr-2])))
		            {
		                if((ptr!=0 && cnt[arr[ptr-1]]!=0 && better(ptr-1,arr[i],ans[ptr-1]))){
		                    ptr--;in[arr[ptr]]=0;
		                }
		                else{
		                    ptr--;in[arr[ptr]]=0;
		                    ptr--;in[arr[ptr]]=0;
		                }
		            }
		            ans[ptr++]=arr[i]; in[arr[i]]=1;
		        }
		    }
		    System.out.println(ptr);
		    for(int i=0;i<ptr;i++)
		    System.out.print(ans[i]+" ");
		    System.out.println();

		}

	}
	public static boolean better(int pos,int val,int val2){
	    if(pos%2==0) return val>val2;
	    else return val<val2;
	}
}

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

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

Think step by step.

Response:
```plaintext
4
4
3 2 1 3
4
1 1 1 1
9
3 2 1 3 2 1 3 2 1
1
1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Longest Max Min Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer sequence $a_1, a_2, \ldots, a_n$. Let $S$ be the set of all possible non-empty subsequences of $a$ without duplicate elements. Your goal is to find the longest sequence in $S$. If there are multiple of them, find the one that minimizes lexicographical order after multiplying terms at odd positions by $-1$.

For example, given $a = [3, 2, 3, 1]$, $S = \\{[1], [2], [3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 3, 1], [3, 2, 1]\\}$. Then $[2, 3, 1]$ and $[3, 2, 1]$ would be the longest, and $[3, 2, 1]$ would be the answer since $[-3, 2, -1]$ is lexicographically smaller than $[-2, 3, -1]$.

A sequence $c$ is a subsequence of a sequence $d$ if $c$ can be obtained from $d$ by the deletion of several (possibly, zero or all) elements.

A sequence $c$ is lexicographically smaller than a sequence $d$ if and only if one of the following holds:

  * $c$ is a prefix of $d$, but $c \ne d$;   * in the first position where $c$ and $d$ differ, the sequence $c$ has a smaller element than the corresponding element in $d$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 5 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains an integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output the answer in the following format:

Output an integer $m$ in the first line — the length of $b$.

Then output $m$ integers $b_1, b_2, \ldots, b_m$ in the second line — the sequence $b$.

Examples:
input:
4
4
3 2 1 3
4
1 1 1 1
9
3 2 1 3 2 1 3 2 1
1
1
output:
3
3 2 1
1
1
3
3 1 2
1
1

input:
10
2
1 2
10
5 2 1 7 9 7 2 5 5 2
2
1 2
10
2 2 8 7 7 9 8 1 9 6
9
9 1 7 5 8 5 6 4 1
3
3 3 3
6
1 6 4 4 6 5
6
3 4 4 5 3 3
10
4 1 4 5 4 5 10 1 5 1
7
1 2 1 3 2 4 6
output:
2
1 2
5
5 1 9 7 2
2
1 2
6
2 7 9 8 1 6
7
9 1 7 5 8 6 4
1
3
4
1 4 6 5
3
4 5 3
4
5 4 10 1
5
2 1 3 4 6

Note: In the first example, $S = \\{[1], [2], [3], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 1, 3], [3, 2, 1]\\}$. Among them, $[2, 1, 3]$ and $[3, 2, 1]$ are the longest and $[-3, 2, -1]$ is lexicographical smaller than $[-2, 1, -3]$, so $[3, 2, 1]$ is the answer.

In the second example, $S = \\{[1]\\}$, so $[1]$ is the answer.

Buggy Code:
import java.util.Scanner;
import java.util.HashSet;
import java.util.HashMap;

public class Solution {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for (int j = 0; j < t; j++) {
            int n = sc.nextInt();
            int[] a = new int[n];

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

            HashSet<Integer> temp = new HashSet<>();
            HashMap<Integer, Integer> availableDup = new HashMap<>();
            HashSet<Integer> removedDup = new HashSet<>();

            for (int i = 0; i < n; i++) {
                if (temp.contains(a[i])) {
                    availableDup.put(a[i], i);
                } else {
                    temp.add(a[i]);
                }
            }

            int[] b = new int[n];
            int k = 0;

            int i = 0;
            while (i < n) {
                if (removedDup.contains(a[i])) {
                    i++;
                    continue;
                }
                if (!availableDup.containsKey(a[i]) || i == availableDup.get(a[i])) {
                    b[k] = a[i];
                    k++;
                    i++;
                    continue;
                }
                int limit = availableDup.get(a[i]);
                int max = a[i];
                int maxi = i;
                int min = a[i];
                int mini = i;
                while (i < n && i < limit && availableDup.containsKey(a[i])) {
                    limit = Math.min(limit, availableDup.get(a[i]));

                    if (a[i] > max) {
                        max = a[i];
                        maxi = i;
                    }
                    if (a[i] < min) {
                        min = a[i];
                        mini = i;
                    }
                    i++;
                }
                b[k] = k % 2 == 0 ? max : min;
                if (k % 2 == 0) {
                    i = maxi + 1;
                    availableDup.remove(max);
                    removedDup.add(max);
                } else {
                    i = mini + 1;
                    availableDup.remove(min);
                    removedDup.add(min);
                }
                k++;
            }

            System.out.println(k);
            for (int m = 0; m < k; m++) {
                System.out.print(b[m] + " ");
            }
            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
4
4
3 2 3 1
5
2 2 3 3 1
6
1 2 2 3 3 1
7
1 1 2 2 3 3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Longest Max Min Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer sequence $a_1, a_2, \ldots, a_n$. Let $S$ be the set of all possible non-empty subsequences of $a$ without duplicate elements. Your goal is to find the longest sequence in $S$. If there are multiple of them, find the one that minimizes lexicographical order after multiplying terms at odd positions by $-1$.

For example, given $a = [3, 2, 3, 1]$, $S = \\{[1], [2], [3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 3, 1], [3, 2, 1]\\}$. Then $[2, 3, 1]$ and $[3, 2, 1]$ would be the longest, and $[3, 2, 1]$ would be the answer since $[-3, 2, -1]$ is lexicographically smaller than $[-2, 3, -1]$.

A sequence $c$ is a subsequence of a sequence $d$ if $c$ can be obtained from $d$ by the deletion of several (possibly, zero or all) elements.

A sequence $c$ is lexicographically smaller than a sequence $d$ if and only if one of the following holds:

  * $c$ is a prefix of $d$, but $c \ne d$;   * in the first position where $c$ and $d$ differ, the sequence $c$ has a smaller element than the corresponding element in $d$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 5 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains an integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output the answer in the following format:

Output an integer $m$ in the first line — the length of $b$.

Then output $m$ integers $b_1, b_2, \ldots, b_m$ in the second line — the sequence $b$.

Examples:
input:
4
4
3 2 1 3
4
1 1 1 1
9
3 2 1 3 2 1 3 2 1
1
1
output:
3
3 2 1
1
1
3
3 1 2
1
1

input:
10
2
1 2
10
5 2 1 7 9 7 2 5 5 2
2
1 2
10
2 2 8 7 7 9 8 1 9 6
9
9 1 7 5 8 5 6 4 1
3
3 3 3
6
1 6 4 4 6 5
6
3 4 4 5 3 3
10
4 1 4 5 4 5 10 1 5 1
7
1 2 1 3 2 4 6
output:
2
1 2
5
5 1 9 7 2
2
1 2
6
2 7 9 8 1 6
7
9 1 7 5 8 6 4
1
3
4
1 4 6 5
3
4 5 3
4
5 4 10 1
5
2 1 3 4 6

Note: In the first example, $S = \\{[1], [2], [3], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 1, 3], [3, 2, 1]\\}$. Among them, $[2, 1, 3]$ and $[3, 2, 1]$ are the longest and $[-3, 2, -1]$ is lexicographical smaller than $[-2, 1, -3]$, so $[3, 2, 1]$ is the answer.

In the second example, $S = \\{[1]\\}$, so $[1]$ is the answer.

Buggy Code:
import sys
input = sys.stdin.readline
from bisect import insort

def ii():
    return int(input())

def lii():
    return [int(x) for x in input().split()]

MOD = 10**9+7

global total
total = 0

def solve():
    global total
    n = ii()
    a = lii()
    idxs = {}
    for i in range(n):
        if a[i] not in idxs:
            idxs[a[i]] = 1
        else:
            idxs[a[i]] += 1

    negative = True
    i = 0
    startidx = 0
    output = []
    used = set()
    while i < n:
        if a[i] in used:
            i += 1
            continue
        idxs[a[i]] -= 1
        if idxs[a[i]] == 0:
            bigger = set()
            smaller = set()
            equal = 0
            for j in range(startidx, i):
                if a[j] in used:
                    continue
                if a[j] > a[i]:
                    bigger.add(a[j])
                elif a[j] < a[i]:
                    smaller.add(a[j])
                else:
                    equal += 1
            bigger = sorted(list(bigger))
            smaller = sorted(list(smaller))
            breakcountdown = False
            breakcount = 0
            for j in range(startidx, i):
                if negative:
                    if not bigger:
                        if not breakcountdown:
                            if equal > 0:
                                output.append(a[i])
                                used.add(a[i])
                                breakcountdown = True
                                breakcount += 1
                                negative = not negative
                            else:
                                break

                    elif a[j] == bigger[-1]:
                        output.append(bigger.pop())
                        used.add(a[j])
                        negative = not negative
                        if breakcountdown:
                            breakcount += 1
                            if breakcount == 2:
                                break
                else:
                    if not smaller:
                        if not breakcountdown:
                            if equal > 0:
                                output.append(a[i])
                                used.add(a[i])
                                breakcountdown = True
                                breakcount += 1
                                negative = not negative
                            else:
                                break

                    elif a[j] == smaller[0]:
                        output.append(smaller.pop(0))
                        used.add(a[j])
                        negative = not negative

                        if breakcountdown:
                            breakcount += 1
                            if breakcount == 2:
                                break
                if not bigger and not smaller:
                    break
            if a[i] not in used:
                output.append(a[i])
                used.add(a[i])
                negative = not negative

            if breakcount == 2:
                for k in range(j, i):
                    idxs[a[k]] += 1
                i = j

            currarr = []
            startidx = i+1

        i += 1

    if a[-1] not in used:
        output.append(a[-1])

    total += 1 + len(output)
    if total >= 7289:
        return a
    return f"{len(output)}\n {' '.join(map(str, output))}"

t = ii()
for _ in range(t):
    print(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
3 2 3 1
4
1 1 1 1
9
3 2 1 3 2 1 3 2 1
1
1
10
5 2 1 7 9 7 2 5 5 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Longest Max Min Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer sequence $a_1, a_2, \ldots, a_n$. Let $S$ be the set of all possible non-empty subsequences of $a$ without duplicate elements. Your goal is to find the longest sequence in $S$. If there are multiple of them, find the one that minimizes lexicographical order after multiplying terms at odd positions by $-1$.

For example, given $a = [3, 2, 3, 1]$, $S = \\{[1], [2], [3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 3, 1], [3, 2, 1]\\}$. Then $[2, 3, 1]$ and $[3, 2, 1]$ would be the longest, and $[3, 2, 1]$ would be the answer since $[-3, 2, -1]$ is lexicographically smaller than $[-2, 3, -1]$.

A sequence $c$ is a subsequence of a sequence $d$ if $c$ can be obtained from $d$ by the deletion of several (possibly, zero or all) elements.

A sequence $c$ is lexicographically smaller than a sequence $d$ if and only if one of the following holds:

  * $c$ is a prefix of $d$, but $c \ne d$;   * in the first position where $c$ and $d$ differ, the sequence $c$ has a smaller element than the corresponding element in $d$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 5 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains an integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output the answer in the following format:

Output an integer $m$ in the first line — the length of $b$.

Then output $m$ integers $b_1, b_2, \ldots, b_m$ in the second line — the sequence $b$.

Examples:
input:
4
4
3 2 1 3
4
1 1 1 1
9
3 2 1 3 2 1 3 2 1
1
1
output:
3
3 2 1
1
1
3
3 1 2
1
1

input:
10
2
1 2
10
5 2 1 7 9 7 2 5 5 2
2
1 2
10
2 2 8 7 7 9 8 1 9 6
9
9 1 7 5 8 5 6 4 1
3
3 3 3
6
1 6 4 4 6 5
6
3 4 4 5 3 3
10
4 1 4 5 4 5 10 1 5 1
7
1 2 1 3 2 4 6
output:
2
1 2
5
5 1 9 7 2
2
1 2
6
2 7 9 8 1 6
7
9 1 7 5 8 6 4
1
3
4
1 4 6 5
3
4 5 3
4
5 4 10 1
5
2 1 3 4 6

Note: In the first example, $S = \\{[1], [2], [3], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 1, 3], [3, 2, 1]\\}$. Among them, $[2, 1, 3]$ and $[3, 2, 1]$ are the longest and $[-3, 2, -1]$ is lexicographical smaller than $[-2, 1, -3]$, so $[3, 2, 1]$ is the answer.

In the second example, $S = \\{[1]\\}$, so $[1]$ is the answer.

Buggy Code:
from math import *
import copy

def solve():
    n=int(input())
    l=list(map(int,input().split()))
    s=set(l)
    x=[0]*(n+1)
    for i in range(n):
        x[l[i]]+=1

    for i in range(1,n+1):
        if x[i]==1:
            x[i]=-1
    if n==1:
        jj.append(1)
        jj.append(l)
        return

    amm=[0 for _ in range(n+1)]
    le=len(s)-1
    re= [set() for _ in range(le+2)]
    m3=0
    am=0
    xx=copy.copy(x)
    p=[0]*(le+1)
    for i in range(0,n):

        if xx[l[i]]==0 or x[l[i]]==0:
            continue
        if x[l[i]]==1:

            for j in range(m3,min(m3+am+1,le+1)):
                re[j].add(l[i])
            am-=1

            if m3%2==1:
                m=min(re[m3])
                while xx[m]==0 or x[m]==0:
                    re[m3].discard(m)
                    m=min(re[m3])
                p[m3]=m
                xx[m]==0
                x[m]=0
            else:
                m=max(re[m3])

                while xx[m]==0  or x[m]==0:
                    re[m3].discard(m)
                    m=max(re[m3])
                p[m3]=m
                xx[m]=1
                x[m]=0

            if m!=l[i]:
                m3+=1
                p[m3]=l[i]
                x[l[i]]=0
                xx[l[i]]=0
                am-=1
            x[l[i]]=1
            m3+=1
        elif x[l[i]]==-1:
            #print(i,l[i],m3,m3+am+1)
            for j in range(m3,min(m3+am+1,le+1)):
                re[j].add(l[i])

            if m3%2==1:
                m=min(re[m3])
                while xx[m]==0  or x[m]==0:
                    re[m3].discard(m)
                    m=min(re[m3])
                p[m3]=m
                xx[m]==0
                x[m]=0
            else:

                m=max(re[m3])
                while xx[m]==0  or x[m]==0:
                    re[m3].discard(m)
                    m=max(re[m3])
                p[m3]=m
                xx[m]=0
                x[m]=0
            if m!=l[i]:
                m3+=1
                p[m3]=l[i]
                x[l[i]]=0
                am-=1
                xx[l[i]]=0

            m3+=1
            x[l[i]]=1
        else:
            for j in range(m3,min(m3+am+1,le+1)):
                re[j].add(l[i])
            if amm[l[i]]==0:
                am+=1
                amm[l[i]]=1

        x[l[i]]-=1

    jj.append(le+1)
    jj.append(p)
    if len(jj)==1177*2:
        print(1)
        print(le+1)
        print(*p)
    #print(jj)

jj=[]
for w in range(int(input())):

    solve()

#print(jj)
for i in jj:
    if type(i)==int:
        print(i)
    else:
        print(*i)

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

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

Think step by step.

Response:
```plaintext
4
4
3 2 1 3
4
1 1 1 1
9
3 2 1 3 2 1 3 2 1
1
1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Longest Max Min Subsequence
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer sequence $a_1, a_2, \ldots, a_n$. Let $S$ be the set of all possible non-empty subsequences of $a$ without duplicate elements. Your goal is to find the longest sequence in $S$. If there are multiple of them, find the one that minimizes lexicographical order after multiplying terms at odd positions by $-1$.

For example, given $a = [3, 2, 3, 1]$, $S = \\{[1], [2], [3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 3, 1], [3, 2, 1]\\}$. Then $[2, 3, 1]$ and $[3, 2, 1]$ would be the longest, and $[3, 2, 1]$ would be the answer since $[-3, 2, -1]$ is lexicographically smaller than $[-2, 3, -1]$.

A sequence $c$ is a subsequence of a sequence $d$ if $c$ can be obtained from $d$ by the deletion of several (possibly, zero or all) elements.

A sequence $c$ is lexicographically smaller than a sequence $d$ if and only if one of the following holds:

  * $c$ is a prefix of $d$, but $c \ne d$;   * in the first position where $c$ and $d$ differ, the sequence $c$ has a smaller element than the corresponding element in $d$.

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 5 \cdot 10^4$). The description of the test cases follows.

The first line of each test case contains an integer $n$ ($1 \le n \le 3 \cdot 10^5$) — the length of $a$.

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

It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

Output: For each test case, output the answer in the following format:

Output an integer $m$ in the first line — the length of $b$.

Then output $m$ integers $b_1, b_2, \ldots, b_m$ in the second line — the sequence $b$.

Examples:
input:
4
4
3 2 1 3
4
1 1 1 1
9
3 2 1 3 2 1 3 2 1
1
1
output:
3
3 2 1
1
1
3
3 1 2
1
1

input:
10
2
1 2
10
5 2 1 7 9 7 2 5 5 2
2
1 2
10
2 2 8 7 7 9 8 1 9 6
9
9 1 7 5 8 5 6 4 1
3
3 3 3
6
1 6 4 4 6 5
6
3 4 4 5 3 3
10
4 1 4 5 4 5 10 1 5 1
7
1 2 1 3 2 4 6
output:
2
1 2
5
5 1 9 7 2
2
1 2
6
2 7 9 8 1 6
7
9 1 7 5 8 6 4
1
3
4
1 4 6 5
3
4 5 3
4
5 4 10 1
5
2 1 3 4 6

Note: In the first example, $S = \\{[1], [2], [3], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 1, 3], [3, 2, 1]\\}$. Among them, $[2, 1, 3]$ and $[3, 2, 1]$ are the longest and $[-3, 2, -1]$ is lexicographical smaller than $[-2, 1, -3]$, so $[3, 2, 1]$ is the answer.

In the second example, $S = \\{[1]\\}$, so $[1]$ is the answer.

Buggy Code:
import java.util.*;
public class LongestMaxMinSubsequence {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int tc = sc.nextInt();
        while(tc-->0){
            solve(sc);
        }
    }
    public static void solve(Scanner sc){
        int n = sc.nextInt();
        int[] a = new int[n];
        int lastElement = 0;
        for(int i=0;i<n;i++){
            a[i] = sc.nextInt();
            lastElement = Math.max(a[i],lastElement);
        }
        int[] lastIndex = new int[lastElement+1];
        Arrays.fill(lastIndex,-1);
        for(int i=0;i<n;i++){
            lastIndex[a[i]] = i;
        }
        PriorityQueue<Integer> pq1 = new PriorityQueue<>();
        for(int i:lastIndex){
            if(i == -1) continue;
            pq1.offer(i);
        }

        int[] ans = new int[pq1.size()];

        Set<Integer> vis = new HashSet<>();

        int i = 0, curr = 0;
        while (i<n && !pq1.isEmpty()) {
            int ind = pq1.poll();
            int min = Integer.MAX_VALUE, max = Integer.MIN_VALUE;
            int minIndex = -1, maxIndex = -1;
            for(int k=curr;k<=ind;k++){
                if(vis.contains(a[k])) continue;
                if(min>a[k]){
                    min = a[k];
                    minIndex = k;
                }
                if(max<a[k]){
                    max = a[k];
                    maxIndex = k;
                }
            }
            if((i+1)%2 == 0){
                // We want as minimum as possible
                if(min == Integer.MAX_VALUE){
                    curr = ind+1;
                    continue;
                }
                ans[i] = min;
                vis.add(min);
                curr = minIndex + 1;
            }else{
                // We want maximum as possible
                if(max == Integer.MIN_VALUE){
                    curr = ind+1;
                    continue;
                }
                ans[i] = max;
                vis.add(max);
                curr = maxIndex + 1;
            }
            if(curr<=ind){
                pq1.offer(ind);
            }
            i++;
        }
        System.out.println(ans.length);
        for(int il: ans){
            System.out.print(il+" ");
        }
        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:
To expose the bug in the provided code, we need to focus on edge cases or scenarios that can maximize the failure probability. The code seems to have issues with handling sequences that have repeated elements and ensuring the correct lexicographical order after the transformation.

Let's consider the following steps to generate a test input:

1. **Repeated Elements**: The code might struggle with sequences that have repeated elements, especially when the repeated elements are the maximum or minimum values.
2. **Lexicographical Order**: The code needs to handle the lexicographical order correctly after multiplying the odd-indexed elements by -1.

Given these considerations, a small-scale test input that can expose the bug is:

```plaintext
1
6
3 2 2 1 3 2
```

### Explanation:
- **Input**: The sequence `[3, 2, 2, 1, 3, 2]` contains repeated elements, which can test how the code handles duplicates.
- **Expected Output**: The longest subsequence without duplicates is `[3, 2, 1]`. After transforming the sequence by multiplying the odd-indexed elements by -1, the sequence becomes `[-3, 2, -1]`, which is lexicographically smaller than other possible sequences.

This input should help expose any issues in the code related to handling repeated elements and ensuring the correct lexicographical order.