**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.io.*;
import java.util.*;

public class Main {
    static int N = 300010;
    static int[] a, cnt;

    public static void main(String[] args) throws IOException {
        FastScanner f = new FastScanner();
        PrintWriter w = new PrintWriter(System.out);
        int T = f.nextInt();
        int t = 0;
        while (T-- > 0) {
            ++t;
            int n = f.nextInt();
            if (t == 180) System.out.println(n);
            a = new int[n + 1];
            HashMap<Integer, Integer> map = new HashMap<>();
            //记录每个数的次数
            for (int i = 1; i <= n; i++) {
                a[i] = f.nextInt();
                if (t == 180) System.out.print(a[i] + " ");
                map.put(a[i], map.getOrDefault(a[i], 0) + 1);
            }
            //奇数应该越大越好 偶数越小越好
            //考虑要两个单调栈
            Deque<Integer> b = new ArrayDeque<>();
            Deque<Integer> c = new ArrayDeque<>();
            List<Integer> ans = new ArrayList<>();
            HashSet<Integer> set = new HashSet<>();
            int idx = 0;//当前要填奇数
            for (int i = 1; i <= n; i++) {
                //加入答案就可以不用选了
                if (set.contains(a[i])) continue;
                //出现次数大于1可以考虑装到栈里去备选
                if (map.get(a[i]) > 1) {
                    //维护单调栈
                    while (!b.isEmpty() && a[b.peekLast()] < a[i]) b.pollLast();
                    b.addLast(i);
                    while (!c.isEmpty() && a[c.peekLast()] > a[i]) c.pollLast();
                    c.addLast(i);
                    map.put(a[i], map.getOrDefault(a[i], 0) - 1);
                } else {
                    //如果当前数一定要选的话
                    //如果当前填入的是奇数个，要找一个最大的数
                    int pre = -1;
                    while (true) {
                        //奇数
                        while (!b.isEmpty() && (b.peekFirst() <= pre || set.contains(a[b.peekFirst()]))) b.pollFirst();
                        while (!c.isEmpty() && (c.peekFirst() <= pre || set.contains(a[c.peekFirst()]))) c.pollFirst();
                        if (idx == 0) {
                            if (!b.isEmpty() && a[b.peekFirst()] >= a[i]) {
                                pre = b.peekFirst();
                                ans.add(a[b.peekFirst()]);
                                set.add(a[b.pollFirst()]);
                                idx ^= 1;
                            } else {
                                b.clear();
                                break;
                            }
                            //偶数
                        } else {
                            if (!c.isEmpty() && a[c.peekFirst()] <= a[i]) {
                                pre = c.peekFirst();
                                ans.add(a[c.peekFirst()]);
                                set.add(a[c.pollFirst()]);
                                idx ^= 1;
                            } else {
                                c.clear();
                                break;
                            }
                        }
                        //下标小于已经选的数了就弹出
                    }
                    if (!set.contains(a[i])) {
                        ans.add(a[i]);
                        set.add(a[i]);
                        idx ^= 1;
                    }
                }
            }
            w.println(ans.size());
            for (int x : ans) w.print(x + " ");
            w.println();
        }
        w.close();
    }

    private static class FastScanner {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        private FastScanner() throws IOException {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        private short nextShort() throws IOException {
            short ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do ret = (short) (ret * 10 + c - '0');
            while ((c = read()) >= '0' && c <= '9');
            if (neg) return (short) -ret;
            return ret;
        }

        private int nextInt() throws IOException {
            int ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do ret = ret * 10 + c - '0';
            while ((c = read()) >= '0' && c <= '9');
            if (neg) return -ret;
            return ret;
        }

        public long nextLong() throws IOException {
            long ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do ret = ret * 10 + c - '0';
            while ((c = read()) >= '0' && c <= '9');
            if (neg) return -ret;
            return ret;
        }

        private char nextChar() throws IOException {
            byte c = read();
            while (c <= ' ') c = read();
            return (char) c;
        }

        private String nextString() throws IOException {
            StringBuilder ret = new StringBuilder();
            byte c = read();
            while (c <= ' ') c = read();
            do {
                ret.append((char) c);
            } while ((c = read()) > ' ');
            return ret.toString();
        }

        private void fillBuffer() throws IOException {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1) buffer[0] = -1;
        }

        private byte read() throws IOException {
            if (bufferPointer == bytesRead) fillBuffer();
            return buffer[bufferPointer++];
        }
    }
}

**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 provided code, we need to focus on scenarios where the code might fail to handle specific edge cases or where the logic might break down. Here are some key points to consider:

1. **Duplicate Elements**: The code should handle sequences with multiple duplicate elements correctly.
2. **Odd and Even Positions**: The code should correctly handle the requirement that odd positions should be maximized and even positions should be minimized.
3. **Lexicographical Order**: The code should ensure that the lexicographical order is correctly handled after multiplying terms at odd positions by -1.

Given these considerations, let's create a test input that includes a sequence with multiple duplicate elements and a mix of odd and even positions to test the logic thoroughly.

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

### Explanation:
- **Sequence**: `[3, 2, 3, 1, 2, 1]`
- **Duplicate Elements**: The sequence contains multiple `3`s, `2`s, and `1`s.
- **Odd and Even Positions**: The sequence has a mix of odd and even positions, which will test the code's ability to correctly place elements in the desired positions.

### Expected Output:
- The longest subsequence without duplicate elements is `[3, 2, 1]`.
- After multiplying terms at odd positions by -1, the sequence becomes `[-3, 2, -1]`.
- This sequence is lexicographically smaller than any other possible sequence of the same length.

This test input should be challenging enough to expose any issues in the provided buggy code.


**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 <sys/resource.h>
//common file for PBDS
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
// including tree_order_statistics_node_update
using namespace std;
using namespace __gnu_pbds;
#define file freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);
#define Pi acos(-1.0)
#define ull unsigned long long
#define int long long
#define krish
#define float long double
#define setbits(x) __builtin_popcount(x)
#define cntbits(x) __builtin_clz(x)
#define bit(x, y) (((x) >> (y)) & 1)
#define mkp(a,b) make_pair(a,b)
#define endl '\n'
#define fix fixed<<setprecision(12);
#define l_b(x, y) lower_bound(x.begin(), x.end(), y)-x.begin();
#define u_b(x, y) upper_bound(x.begin(), x.end(), y)-x.begin();
#define pb push_back
#define ppb pop_back
#define pf push_front
#define ppf pop_front
#define vi vector<int>
#define vc vector<char>
#define vs vector<string>
#define vvi vector<vector<int>>
#define vvc vector<vector<char>>
#define vvii vector<vector<pair<int, int>>>
#define vii vector<pair<int, int>>
#define pqi priority_queue<int>
#define pqii priority_queue<pair<int, int>>
#define ini(x, i) memset(x, i, sizeof(x))
#define ff first
#define ss second
#define all(x) x.begin(), x.end()
#define mxa(x) *max_element(x.begin(), x.end())
typedef pair<int, int> pii;
typedef tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
// find_by_order order_of_key
const int MOD=1e9+7;
// const int MOD=998244353;
const int maxn=2e6+10;
int fact[maxn];
const int N=(1e6+5);
const int INF=LONG_LONG_MAX;
int POW(int a,int b){
    if(b==0) return 1;
    if(a==0) return 0;
    int temp=POW(a,b/2);
    if(b%2==0) return (temp*temp)%MOD;
    else return (((temp*temp)%MOD)*a)%MOD;
}
void factorial(){
    fact[0]=1;
    for(int i=1; i<maxn; i++){fact[i]=((i*fact[i-1])%MOD+MOD)%MOD;}
}
int factorial(int n){
    if(n==0)return 1;
    int ans=1;
    for(int i=1; i<=n; i++){ ans=((ans)*(i));}
    return ans;
}
bool isprime(int n){
    if(n==2) return 1;
    if(n==0 || n==1 || n%2==0) return 0;
    for(int i=3; i<=sqrt(n); i+=2){
        if(n%i==0){
            return 0;
        }
    }
    return 1;
}
int lcm(int a, int b){ return (a/__gcd(a, b))*b;}
int add(int a, int b, int m=MOD){ a=a%m, b=b%m;  return (((a+b)%m)+m)%m;}
int sub(int a, int b, int m=MOD){ a=a%m, b=b%m;  return (((a-b)%m)+m)%m;}
int mul(int a, int b, int m=MOD){ a=a%m, b=b%m;  return (((a*b)%m)+m)%m;}
int power(int a, int b) { int r = 1; while (b) { if (b & 1) r = mul(r, a); a = mul(a, a); b >>= 1; } return r; }
int ckdiv(int a, int b) { return mul(a, power(b, MOD - 2)); }
int inv(int n){ return power(n,MOD-2);}
int nCr(int n,int r){
    if(r>n || n<0 || r<0)return 0;
    if(n==r or r==0)return 1;
    int num=fact[n];
    int denm=(fact[r]*fact[n-r])%MOD;
    num = (num*inv(denm))%MOD;
    return num;
}
bool cmp(pair<string, int> &p1, pair<string, int> &p2){ if(p1.first.size()==p2.first.size()) return p1<p2; return p1.first.size()>p2.first.size();}
vector<bool> isPrime;
vi smf,pri;
void sieve(int n){
	isPrime.assign(n+1,true);
	smf.assign(n+1,0);
	isPrime[0]=isPrime[1]=false;
	pri={};
	for(int i=2;i<=n;i++){
		if(isPrime[i]){
			smf[i]=i;
			pri.pb(i);
			for(int j=i*i;j<=n;j+=i){
                if(isPrime[j]){
                    isPrime[j]=false;
                    smf[j]=i;
                }
            }
		}
	}
}
void solve(){
    int n; cin>>n;
    vi a(n);
    map<int, int> mp;
    map<int, deque<int>> idx;
    for(int i=0; i<n; i++){cin>>a[i], mp[a[i]]++; idx[a[i]].pb(i);}
    multiset<int> st;
    queue<int> pq;
    int j=0, cnt=1;
    vi ans;
    for(int i=0; i<n; i++){
        if(mp[a[i]]>0)st.insert(a[i]);
        if(mp[a[i]]==1)pq.push(i);
        if(pq.size()){
            auto it=0;
            if(cnt%2){
                it=*--st.end();
                if(pq.size()) it=min(idx[it][0], pq.front());
            }
            else{
                it=*st.begin();
                if(pq.size()) it=min(idx[it][0], pq.front());
            }
            ans.pb(a[it]);
            while(j<it){
                if(st.find(a[j])!=st.end()){
                    // mp[a[j]]--;
                    st.erase(st.find(a[j]));
                }
                idx[a[j]].ppf();
                j++;
            }
            if(st.size()){
                st.erase((a[j]));
                mp[a[j]]=0;
            }
            j++;
            if(pq.size() && a[it]==a[pq.front()]) pq.pop();
            idx[a[it]].ppf();
            cnt++;
        }
        mp[a[i]]--;
    }
    cout<<ans.size()<<endl;
    for(auto it: ans)cout<<it<<" ";
    cout<<endl;
}
signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    factorial();
    sieve(2e5);
    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
5
4
1 1 2 2
5
1 2 3 2 1
6
1 2 2 3 3 1
7
1 2 2 3 3 2 1
8
1 2 3 2 3 4 2 1
```

### Explanation:
1. **Test Case 1:**
   - **Input:** `1 1 2 2`
   - **Expected Output:** `2 1 2`
   - **Reason:** The longest subsequence without duplicates is `[1, 2]`. The lexicographical order after multiplying terms at odd positions by `-1` is `[-1, 2]`.

2. **Test Case 2:**
   - **Input:** `1 2 3 2 1`
   - **Expected Output:** `3 1 2 3`
   - **Reason:** The longest subsequence without duplicates is `[1, 2, 3]`. The lexicographical order after multiplying terms at odd positions by `-1` is `[-1, 2, -3]`.

3. **Test Case 3:**
   - **Input:** `1 2 2 3 3 1`
   - **Expected Output:** `3 1 2 3`
   - **Reason:** The longest subsequence without duplicates is `[1, 2, 3]`. The lexicographical order after multiplying terms at odd positions by `-1` is `[-1, 2, -3]`.

4. **Test Case 4:**
   - **Input:** `1 2 2 3 3 2 1`
   - **Expected Output:** `3 1 2 3`
   - **Reason:** The longest subsequence without duplicates is `[1, 2, 3]`. The lexicographical order after multiplying terms at odd positions by `-1` is `[-1, 2, -3]`.

5. **Test Case 5:**
   - **Input:** `1 2 3 2 3 4 2 1`
   - **Expected Output:** `4 1 2 3 4`
   - **Reason:** The longest subsequence without duplicates is `[1, 2, 3, 4]`. The lexicographical order after multiplying terms at odd positions by `-1` is `[-1, 2, -3, 4]`.

These test cases are designed to cover various edge cases, including repeated elements and sequences with multiple elements that can form the longest subsequence. They are small-scale to ensure they fit within the output length limit while still being challenging enough to expose potential bugs in the given code.


**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 fori(n) for(int i = 0; i<n; i++)
#define forj(n) for(int j = 0; j<n; j++)
#define ll long long

using namespace std;

void solve() {
    int n;
    cin>>n;
    int A[n];
    fori(n) cin>>A[i];

    int last[n+1];
    fori(n) last[A[i]] = i;
    bool added[n+1];
    fill(added, added+n+1, false);

    stack<int> s;

    fori(n) {

                while (!s.empty()) {
                    int v = s.top();
                    if (s.size()&1 && last[abs(v)] > i && -A[i] < v) {added[-v] = false; s.pop();}
                    else if (s.size()%2 == 0 && last[v] > i && A[i] < v) {added[v] = false; s.pop();}
                    else if (s.size() > 1 && i < last[abs(s.top())]) {
                        s.pop();
                        int prv = v;
                        added[prv] = false;
                        v = s.top();
                        if (s.size()&1 && last[abs(v)] > i && -A[i] < v) {added[-v] = false; s.pop();}
                        else if (s.size()%2 == 0 && last[v] > i && A[i] < v) {added[v] = false; s.pop();}
                        else {s.push(prv); added[prv] = true; break;}
                    }
                    else break;
                }
                if (added[A[i]]) continue;
                if (s.size()&1) s.push(A[i]);
                else s.push(-A[i]);
                added[A[i]] = true;

    }

    vector<int> ans;
    while (!s.empty()) {ans.push_back(s.top()); s.pop();}
    reverse(ans.begin(), ans.end());
    cout<<ans.size()<<"\n";
    for (auto &i : ans) cout<<abs(i)<<" ";
    cout<<"\n";

}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
	int t = 1;
	cin>>t;
	while (t) {
		t--;
		solve();
	}
}

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

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

Think step by step.

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

1. **Duplicate Elements**: The code must handle duplicate elements correctly. If the same element appears multiple times, the algorithm should still be able to find the correct subsequence.

2. **Lexicographical Order**: The code should correctly handle the lexicographical order after multiplying terms at odd positions by -1. This means that the sequence should be chosen not only based on its length but also on the lexicographical order of the transformed sequence.

3. **Edge Cases**: Consider edge cases such as sequences with all elements the same, sequences with a small number of distinct elements, and sequences with a large number of distinct elements.

Given these points, let's create a test input that includes a sequence with duplicate elements and a sequence where the lexicographical order after transformation is crucial.

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

### Explanation
1. **Test Case 1: [1, 2, 3, 2, 1]**
   - The set of subsequences without duplicates is: `{[1], [2], [3], [1, 2], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2], [1, 2, 3], [2, 1, 3], [3, 2, 1]}`.
   - The longest subsequences are `[1, 2, 3]`, `[2, 1, 3]`, and `[3, 2, 1]`.
   - After transforming by multiplying terms at odd positions by -1, the sequences become `[-1, 2, -3]`, `[-2, 1, -3]`, and `[-3, 2, -1]`.
   - The lexicographically smallest sequence is `[-3, 2, -1]`, which corresponds to `[3, 2, 1]`.

2. **Test Case 2: [1, 1, 1, 1]**
   - The only subsequence is `[1]`.
   - The output should be `1` and the sequence should be `[1]`.

3. **Test Case 3: [3, 2, 2, 1, 3, 2]**
   - The set of subsequences without duplicates is: `{[1], [2], [3], [2, 1], [2, 3], [3, 1], [3, 2], [2, 2, 1], [2, 2, 3], [3, 2, 1], [3, 2, 2], [2, 3, 1], [2, 3, 2], [3, 2, 1, 2]}`.
   - The longest subsequences are `[2, 3, 1]` and `[3, 2, 1]`.
   - After transforming by multiplying terms at odd positions by -1, the sequences become `[-2, 3, -1]` and `[-3, 2, -1]`.
   - The lexicographically smallest sequence is `[-3, 2, -1]`, which corresponds to `[3, 2, 1]`.

This test input should help in exposing any bugs in the given code, particularly in handling duplicate elements and the lexicographical order of the transformed sequences.


**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 solve(arr):
    counts = {}
    for i in arr:
        if i not in counts.keys():
            counts[i] = 0

        counts[i] += 1

    m = len(counts)
    stack = []
    taken = set()
    for i in arr:
        counts[i] -= 1
        if i in taken:
            continue

        if not stack:
            stack.append(i)
            taken.add(i)
        else:
            while stack:
                stack_len = len(stack)
                last = stack[-1]
                if len(stack) >= 2:
                    second_last = stack[-2]
                else:
                    if stack_len%2 == 0:
                        second_last = -float('inf')
                    else:
                        second_last = float('inf')

                if stack_len%2 == 0:
                    if i > second_last and counts[last] > 0 and counts.get(second_last, 1) > 0:
                        taken.remove(stack.pop())
                        if len(stack) >= 1:
                            taken.remove(stack.pop())
                    elif i < last and counts[last] > 0:
                        taken.remove(stack.pop())
                        break
                    else:
                        break
                else:
                    if len(stack) == 1 and i > stack[-1] and counts[last] > 0:
                        taken.remove(stack.pop())
                    elif len(stack) > 1 and i < second_last and counts[last] > 0 and counts.get(second_last, 1) > 0:
                        taken.remove(stack.pop())
                        if len(stack) >= 1:
                            taken.remove(stack.pop())
                    elif len(stack) > 1 and i > last and counts[last] > 0:
                        taken.remove(stack.pop())
                        break
                    else:
                        break

                #print(stack)

            stack.append(i)
            taken.add(i)

            #print(stack, i)

    return str(len(stack)) + "\n" + " ".join(map(str, stack))

def main():
    t = int(input())
    ans = []
    for i in range(t):
        n = int(input())
        arr = list(map(int, input().split()))
        ans.append(solve(arr))

    print("\n".join(map(str, ans)))

main()

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

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

Think step by step.

Response:
```plaintext
1
4
3 2 3 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 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(0,index)[0]
            temp=seg22.query(0,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
4
3
1 2 3
3
3 2 1
4
1 2 3 4
4
4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 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>
using namespace std;

#define int long long int
#define endl '\n'

void solve(){
    int n;
    cin>>n;
    int arr[n];
    map<int,vector<int>>v;
    map<int,int>en,vis;
    int last=-1;
    multiset<int>st;
    int x=0;

    for(int i=0;i<n;i++){
        cin>>arr[i];
        v[arr[i]].push_back(i);
        en[arr[i]]=i;
    }
    vector<int>rtr;

    for(int i=0;i<n;i++){
        if(vis[arr[i]]==0) st.insert(arr[i]);

        if(en[arr[i]]==i && vis[arr[i]]==0){
            while(1){
                int ppp=0;
                if((x%2)==0){
                    auto itr=st.end();
                    itr--;
                    if((*itr)>arr[i]){
                        auto ub=upper_bound(v[(*itr)].begin(),v[(*itr)].end(),last);
                        for(int j=last+1;j<=(*ub);j++){
                            auto itrrrr=st.find(arr[j]);
                            if(itrrrr!=st.end()){
                                st.erase(itrrrr);
                            }
                        }
                        x++;
                        rtr.push_back((*itr));
                        vis[(*itr)]=1;
                        last=(*ub);
                        st.erase((*itr));
                    }else{
                        ppp++;
                    }
                }else{
                    auto itr=st.begin();
                    if((*itr)<arr[i]){
                        auto ub=upper_bound(v[(*itr)].begin(),v[(*itr)].end(),last);
                        for(int j=last+1;j<=(*ub);j++){
                            auto itrrrr=st.find(arr[j]);
                            if(itrrrr!=st.end()){
                                st.erase(itrrrr);
                            }
                        }
                        x++;
                        rtr.push_back((*itr));
                        vis[(*itr)]=1;
                        last=(*ub);
                        st.erase((*itr));
                    }else{
                        ppp++;
                    }
                }

                if(ppp>0) break;
            }

            rtr.push_back(arr[i]);
            vis[arr[i]]=1;
            auto ub=upper_bound(v[arr[i]].begin(),v[arr[i]].end(),last);
            for(int j=last+1;j<=(*ub);j++){
                auto itrrrr=st.find(arr[j]);
                if(itrrrr!=st.end()){
                    st.erase(itrrrr);
                }
            }
            last=(*ub);
            x++;
            st.erase(arr[i]);
        }
    }

    cout<<rtr.size()<<endl;

    for(auto i:rtr) cout<<i<<" ";
    cout<<endl;
}
int32_t main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);

    int _t; cin>>_t; while(_t--)
    solve();
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that might cause the code to fail. Here are some considerations:

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

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

- A sequence with multiple duplicate elements.
- A sequence where the longest subsequence is not immediately obvious.
- A sequence that tests the lexicographical order condition.

Here is a test input that meets these criteria:

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

### Explanation:
- **Sequence**: `[3, 2, 3, 1, 2]`
- **Expected Output**:
  - The longest subsequence without duplicates is `[3, 2, 1]`.
  - After multiplying terms at odd positions by -1, the sequence becomes `[-3, 2, -1]`.
  - The lexicographical order of `[-3, 2, -1]` is the smallest.

This input will help in verifying if the code correctly handles duplicate elements, finds the longest subsequence, and correctly applies the lexicographical order condition.


**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>

using namespace std;

#define int long long
#define append push_back

void print(vector <int> a) {
    for (int i = 0; i < a.size(); i++) cout << a[i] << " ";
    cout << endl;
}

signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);

    int tt;
    cin >> tt;
    while (tt--) {
        int n;
        cin >> n;
        vector <int> a(n);
        set <int> ss, s;

        for (int i = 0; i < n;i++) cin >> a[i];
        for (int i = 0; i < n;i++) ss.insert(a[i]);
        int len = ss.size();
        vector <int> c(n+1, 0);
        vector <int> un(len+1, 0);
        for (int i = n-1; i >= 0; i--) {
            int previous_size = (int)s.size();
            s.insert(a[i]);
            if ((int)s.size() - previous_size == 1)
            un[(int)s.size()] = i;
            c[a[i]]++;
        }

        int curr_sum = 0;
        for (int i = 0; i < n + 1; i++){
            if (c[i] > 0) curr_sum += 1;
        }

        vector <int> ans;
        int last_index = 0;
        for (int i = 0; i < len; i++) {
            vector <int> temp = c;
            int temp_curr_sum = curr_sum;
            if (i != len - 1) {
                if (i % 2 == 0) {
                int mx = -1;
            int idx = -1;
            for (int j = last_index; j < un[len-i-1]; j++) {

                if (a[j] > mx && temp_curr_sum - 1 >= len - i - 1 && c[a[j]] > 0) {
                    mx = a[j];
                    idx = j;
                }

                if (temp[a[j]] == 1) {
                    temp[a[j]]--;
                    temp_curr_sum--;
                }
                else if (temp[a[j]] > 1) {
                    temp[a[j]]--;
                }
            }

            for (int j = last_index; j <= idx; j++) {
                if (j != idx){
                    if (c[a[j]] == 1) {
                    c[a[j]]--;
                    curr_sum--;
                }
                else if (c[a[j]] > 1) {
                    c[a[j]]--;
                }
                } else {
                    curr_sum--;
                    c[a[j]] = 0;
                }

            }

            ans.append(a[idx]);
            last_index = idx+1;
            }
            else {
                int mn = 1000000000000;
            int idx = -1;
            for (int j = last_index; j < un[len-i-1]; j++) {
                if (a[j] < mn && temp_curr_sum-1 >= len-i-1 && c[a[j]] > 0) {
                    mn = a[j];
                    idx = j;
                }
                 if (temp[a[j]] == 1) {
                    temp[a[j]]--;
                    temp_curr_sum--;
                }
                else if (temp[a[j]] > 1) {
                    temp[a[j]]--;
                }
            }

            for (int j = last_index; j <= idx; j++) {
                if (j != idx){
                    if (c[a[j]] == 1) {
                    c[a[j]]--;
                    curr_sum--;
                }
                else if (c[a[j]] > 1) {
                    c[a[j]]--;
                }
                } else {
                    curr_sum--;
                    c[a[j]] = 0;
                }

            }

            ans.append(a[idx]);
            last_index = idx+1;
            }
            } else {
                if (i % 2 == 0) {
                    int mx = -1;
                    int idx = -1;
                    for (int j = last_index; j<n; j++) {
                        if (c[a[j]] > 0 and a[j] > mx) {
                            mx = a[j];
                            idx = j;
                        }
                    }
                    ans.append(a[idx]);
                } else  {
                    int mn = 1000000000000;
                    int idx = -1;
                    for (int j = last_index; j<n; j++) {
                        if (c[a[j]] > 0 and a[j] < mn) {
                            mn = a[j];
                            idx = j;
                        }
                    }
                    ans.append(a[idx]);
                }
            }

        }
        cout << len << endl;
        for (int i = 0; i < len; i++) cout << ans[i] << " ";
        cout << endl;
    }

}

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

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

Think step by step.

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

This test input is designed to expose potential issues in the buggy code by:

1. **Including duplicate elements**: The sequence contains duplicate elements (2 and 3), which can challenge the logic for handling unique subsequences.
2. **Small scale**: The sequence is small enough to be manageable and to ensure that the output can be verified easily.
3. **Lexicographical order consideration**: The sequence is structured to test the lexicographical order requirement, especially after multiplying terms at odd positions by -1.

The expected output for this test input should be:
```plaintext
3
3 2 1
```

This is because the longest subsequence without duplicates is `[3, 2, 1]`, and after applying the lexicographical order rule, `[-3, 2, -1]` is the smallest.


**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.io.*;
import java.util.*;

public class longest_max_min_subsequence {

    public static void main(String[] args) throws IOException {

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);
        int test = Integer.parseInt(br.readLine());

        for (int t = 0; t<test; t++) {

            int length = Integer.parseInt(br.readLine());
            List<Integer> list = new ArrayList<>();

            String[] line = br.readLine().split(" ");
            for (String str : line) {
                list.add(Integer.parseInt(str));
            }

            Set<Integer> set = new HashSet<>(list);
            int n = set.size();
            pw.println(n);
            pw.flush();
            if (n==1) {
                pw.println(list.get(0));
            } else if (n==length) {
                for (int i : list) pw.print(i+" ");
                pw.println();
            } else {
                List<Integer> res = new ArrayList<>();
                res = findSub(n, list, res, false);
                for (int i : res) {
                    if (i!=0) pw.print(i+" ");
                }
                pw.println();
            }
            pw.flush();

        }
    }

    private static List<Integer> findSub(int remainder, List<Integer> list, List<Integer> res,
                                         boolean beMin) {

        while (remainder!=0) {
            int minMax;
            if (beMin) {
                minMax = Integer.MAX_VALUE;
            } else {
                minMax = 0;
            }

            int listSize = list.size();
            List<Integer> filtered = null;
            for (int i = 0; i<=listSize-remainder; i++) {
                int pointer = list.get(i);
                if (!beMin) {
                    if (pointer>minMax) {
                        List<Integer> temp = list.subList(i, listSize);
                        Set<Integer> remaining = new HashSet<>(temp);
                        remaining.remove(pointer);
                        if (remaining.size()>=remainder-1) {
                            minMax = list.get(i);
                            filtered = temp;
                        } else {
                            break;
                        }
                        // ! else sketchy
                    }
                } else {
                    if (pointer<minMax) {
                        List<Integer> temp = list.subList(i, listSize);
                        Set<Integer> remaining = new HashSet<>(temp);
                        remaining.remove(pointer);
                        if (remaining.size()>=remainder-1) {
                            minMax = list.get(i);
                            filtered = temp;
                        } else {
                            break;
                        }
                    }
                }
            }
            res.add(minMax);
            remainder--;
            int pointer = minMax;
            list = filtered;
            list.removeIf(i->i==pointer);
            beMin = !beMin;
        }

        return 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
5
4
1 2 2 3
4
2 2 2 2
6
3 2 1 3 2 1
3
1 1 2
5
1 2 3 4 5
```

### Explanation:
1. **Test Case 1: [1, 2, 2, 3]**
   - This test case includes duplicate elements and a sequence that can be used to form the longest subsequence with unique elements.
   - The expected output should be `[1, 2, 3]` because it is the longest subsequence with unique elements and is lexicographically smallest after multiplying odd-position elements by -1.

2. **Test Case 2: [2, 2, 2, 2]**
   - This test case includes only one unique element.
   - The expected output should be `[2]` because it is the only subsequence possible.

3. **Test Case 3: [3, 2, 1, 3, 2, 1]**
   - This test case includes multiple duplicates and a sequence that can be used to form the longest subsequence with unique elements.
   - The expected output should be `[3, 1, 2]` because it is the longest subsequence with unique elements and is lexicographically smallest after multiplying odd-position elements by -1.

4. **Test Case 4: [1, 1, 2]**
   - This test case includes duplicate elements and a sequence that can be used to form the longest subsequence with unique elements.
   - The expected output should be `[1, 2]` because it is the longest subsequence with unique elements and is lexicographically smallest after multiplying odd-position elements by -1.

5. **Test Case 5: [1, 2, 3, 4, 5]**
   - This test case includes a sequence with all unique elements.
   - The expected output should be `[1, 2, 3, 4, 5]` because it is the longest subsequence with unique elements and is lexicographically smallest after multiplying odd-position elements by -1.

These test cases cover a variety of scenarios, including sequences with duplicates, sequences with all unique elements, and sequences that require careful handling of the 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 java.util.*;
import java.io.*;

public class cp {
    static class Datamax implements Comparable<Datamax>{
        int idx;
        int val;
        public Datamax(int val,int idx){
            this.idx = idx;
            this.val = val;
        }
        @Override
        public int compareTo(Datamax p1){
            if(this.val<p1.val) return 1;
            else if(this.val>p1.val) return -1;
            if(this.idx>p1.idx) return 1;
            return -1;
        }
    }
    static class Datamin implements Comparable<Datamin>{
        int idx;
        int val;
        public Datamin(int val,int idx){
            this.idx = idx;
            this.val = val;
        }
        @Override
        public int compareTo(Datamin p1){
            if(this.val>p1.val) return 1;
            else if(this.val<p1.val) return -1;
            if(this.idx>p1.idx) return 1;
            return -1;
        }
    }
    // static class Edge {
    //     int dst;
    //     int dir;
    //     public Edge(int dst,int dir){
    //         this.dir = dir;
    //         this.dst = dst;
    //     }
    // }
    public static PP.Reader cs = new PP.Reader(System.in);
    public static BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));
    public static void main(String[] args) throws IOException{
        int tests = cs.nextInt();
        while(tests>0){
            tests--;
            solve();
        }
        // writer.flush();
        writer.close();
    }
    public static long mod = (long)Math.pow(10,9) + 7;
    public static long MOD = 998244353;
    // public static int ptr = 0;
    public static void solve() throws IOException{
        // long n = cs.nextLong();
        // int it =(int)n;
        int n = cs.nextInt();
        //int m = cs.nextInt();
        // long k = cs.nextLong();
        // int k = cs.nextInt();
        //String str = cs.next();
        int[] arr = new int[n];
        for(int i = 0; i < n; i++){
            arr[i] = cs.nextInt();
        }
        int[][] lstocc = new int[n+1][2];
        for(int i = 1; i <= n; i++) lstocc[i][0] = i;
        for(int i=0;i<n;i++){
            lstocc[arr[i]][1] = i+1;
            //System.out.println(arr[i]+" "+lstocc[arr[i]][1]);
        }
        //System.out.println(lstocc[3][1]);
        Arrays.sort(lstocc,Comparator.comparingDouble(o -> o[1]));
        PriorityQueue<Datamax> pqmax = new PriorityQueue<Datamax>();
        PriorityQueue<Datamin> pqmin = new PriorityQueue<Datamin>();
        HashSet<Integer> taken = new HashSet<Integer>();
        HashSet<Integer> hs = new HashSet<Integer>();
        int ptr = 0;
        int l = 0;
        boolean max = true;
        for(int i=1;i<=n;i++){
            if(taken.contains(lstocc[i][0])|| lstocc[i][1] == 0) continue;
            int temp = lstocc[i][1];
        //    System.out.println(lstocc[i][1]+" temp");
            while(ptr<temp){
                if(!taken.contains(arr[ptr]) && !hs.contains(arr[ptr])){
                    pqmax.add(new Datamax(arr[ptr], ptr+1));
                    pqmin.add(new Datamin(arr[ptr], ptr+1));
                    //hs.add(arr[ptr]);
                    // System.out.println(arr[ptr]);
                }
                ptr++;
            }
            while(!taken.contains(lstocc[i][0])){
                if(max){
                    while(taken.contains(pqmax.peek().val) || pqmax.peek().idx<=l) pqmax.remove();
                    Datamax curr = pqmax.remove();
                    writer.write(curr.val+" ");
                    //System.out.print(curr.val+" max ");
                    l = curr.idx;
                    taken.add(curr.val);
                    max = false;
                }
                else{
                    while(taken.contains(pqmin.peek().val) || pqmin.peek().idx<=l) pqmin.remove();
                    Datamin curr = pqmin.remove();
                    writer.write(curr.val+" ");
                   // System.out.print(curr.val+" min ");
                    l = curr.idx;
                    taken.add(curr.val);
                    max = true;
                }
            }
        }
        writer.write("\n");
        System.out.println(taken.size());
        writer.flush();
    }
    //Write fxn here

    public static  int binary(int[] arr,int start,int end,int key){
        int mid = (int)start;
        int st = start;
        int en = end;
        while(start<=end){
            mid = (int)(start+end)/2;
            if(arr[mid]==key) return mid;
            else if(arr[mid]>key) end = mid-1;
            else if(arr[mid]<key) start = mid+1;
        }
        //if(end<start){
            if(end<st) end++;
            // if(start>en) start--;
            // if(Math.abs(arr[end]-key)<=Math.abs(arr[start]-key-1)){
                // return start;
            // }
            return end;
        //}
        //return mid+1;
    }
    public static long invmod(long A, long M)
    {

        for (int X = 1; X < (int)M; X++)
            if (((A % M) * (X % M)) % M == 1)
                return X;
        return 1;
    }
    public static long gcd(long a,long b){
        long c = Math.min(a,b);
        long d = Math.max(a,b);
        if(d%c==0) return c;
        return gcd(c,d%c);
    }
    public static long divmod(long a,long b,long m){   //claculate (a/b)%m
        // a = a%m;b = b%m  // Include if not working
        long inv =  powermod(b, m-2, m);
        return (inv*a)%m;
    }
    static long powermod(long x, long y, long p) //calculate (x^y)%p
    {
        long res = 1;
        x = x % p;
        while (y > 0) {
            if ((y & 1) > 0)
                res = (res * x) % p;
            y = y >> 1;
            x = (x * x) % p;
        }
        return res;
    }
}

class PP {
    static String IN = "%s";
    static String OUT = "%s";

    static class Reader {
        BufferedReader br;
        StringTokenizer st;

        Reader(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

        String next() {
            try {
                while (st == null || !st.hasMoreTokens())
                    st = new StringTokenizer(br.readLine());
                return st.nextToken();
            } catch (Exception ignored) {
            }
            return null;
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that might cause the algorithm to fail. Here are the steps to think through the problem:

1. **Edge Cases for Input Size**:
   - Test with the minimum input size (1 element).
   - Test with the maximum input size (300,000 elements).

2. **Edge Cases for Sequence Content**:
   - Test with all elements being the same.
   - Test with a sequence that has all unique elements.
   - Test with a sequence that has repeated elements but in a pattern that might cause issues with the algorithm.

3. **Edge Cases for Subsequence Length**:
   - Test with sequences where the longest subsequence is very short.
   - Test with sequences where the longest subsequence is very long.

4. **Edge Cases for Lexicographical Order**:
   - Test with sequences that have multiple subsequences of the same length but different lexicographical orders.

5. **Edge Cases for Odd and Even Position Multiplication**:
   - Test with sequences that have elements that, when multiplied by -1 at odd positions, might cause unexpected behavior.

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

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

### Explanation:
1. **Test Case 1: `[1, 2, 3, 4]`**
   - This is a sequence with all unique elements in increasing order.
   - The longest subsequence is `[1, 2, 3, 4]`.

2. **Test Case 2: `[4, 3, 2, 1]`**
   - This is a sequence with all unique elements in decreasing order.
   - The longest subsequence is `[4, 3, 2, 1]`.

3. **Test Case 3: `[3, 3, 3, 3]`**
   - This is a sequence with all elements being the same.
   - The longest subsequence is `[3]`.

4. **Test Case 4: `[1, 1, 2, 2]`**
   - This is a sequence with repeated elements.
   - The longest subsequence is `[1, 2]`.

These test cases cover a range of scenarios, including sequences with unique elements, sequences with repeated elements, and sequences in both increasing and decreasing order. They should help expose any bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ksyusha and the Loaded Set
time_limit_ms: 3000
memory_limit_mb: 512
Description: Ksyusha decided to start a game development company. To stand out among competitors and achieve success, she decided to write her own game engine. The engine must support a set initially consisting of $n$ distinct integers $a_1, a_2, \ldots, a_n$.

The set will undergo $m$ operations sequentially. The operations can be of the following types:

  * Insert element $x$ into the set;    * Remove element $x$ from the set;    * Report the $k$-load of the set.

The $k$-load of the set is defined as the minimum positive integer $d$ such that the integers $d, d + 1, \ldots, d + (k - 1)$ do not appear in this set. For example, the $3$-load of the set $\\{3, 4, 6, 11\\}$ is $7$, since the integers $7, 8, 9$ are absent from the set, and no smaller value fits.

Ksyusha is busy with management tasks, so you will have to write the engine. Implement efficient support for the described operations.

Input: The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

The following lines describe the test cases.

The first line contains an integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the initial size of the set.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_1 < a_2 < \ldots < a_n \le 2 \cdot 10^6$) — the initial state of the set.

The third line contains an integer $m$ ($1 \le m \le 2 \cdot 10^5$) — the number of operations.

The next $m$ lines contain the operations. The operations are given in the following format:

  * + $x$ ($1 \le x \le 2 \cdot 10^6$) — insert element $x$ into the set (it is guaranteed that $x$ is not in the set);    * - $x$ ($1 \le x \le 2 \cdot 10^6$) — remove element $x$ from the set (it is guaranteed that $x$ is in the set);    * ? $k$ ($1 \le k \le 2 \cdot 10^6$) — output the value of the $k$-load of the set.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2 \cdot 10^5$, and the same holds for $m$.

Output: For each test case, output the answers to the operations of type "?".

Examples:
input:
3
5
1 2 5 905 2000000
15
- 2
? 2
? 1
- 1
? 1
+ 4
+ 2
? 2
+ 6
- 4
+ 7
? 2
? 3
? 4
? 2000000
5
3 4 5 6 8
9
? 5
- 5
? 5
+ 1
? 2
- 6
- 8
+ 6
? 5
5
6 7 8 9 10
10
? 5
- 6
? 4
- 10
+ 5
- 8
+ 3
+ 2
- 3
+ 10
output:
2 2 1 6 3 8 8 2000001
9 9 9 7
1 1

Note:

Buggy Code:
//#pragma GCC optimize(3)

#include <bits/stdc++.h>
// #include <bits/extc++.h>
// namespace ext {
//     using namespace __gnu_cxx;
//     using namespace __gnu_pbds;
// }
using namespace std;

#define endl '\n'
using ll = long long;
using ull = unsigned long long;
using lld = long double;
#define bigint __int128
#define STACK() int size(512 << 20); __asm__ ( "movq %0, %%rsp\n"::"r"((char *) malloc(size) + size));
const ll inf = 1e8, infl = 0x7fffffffffffffff, mod = 998244353;
//const ll mod = 1E9 + 7;
#define green    "\033[97;32m"
#define white    "\033[90;32m"
#define yellow   "\033[90;33m"
#define red      "\033[97;31m"
#define blue     "\033[97;34m"
#define magenta  "\033[97;35m"
#define cyan     "\033[97;36m"
#define reset    "\033[0m"
#define debug    cout << red

ll power(ll a, ll b, ll m = mod) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans *= a, ans %= m;
        a *= a, a %= m, b >>= 1;
    }
    return ans;
}

const int X = 2;
ll pw[X], fac[X], facinv[X], iv[X];

ll C(ll low, ll high, ll m = mod) {
    return (fac[high] * facinv[high] % mod) * facinv[high - low] % mod;
}

ll exgcd(ll a, ll b, ll &x, ll &y) {
    if (b == 0) {
        x = 1, y = 0;
        return a;
    }
    ll d = exgcd(b, a % b, y, x);
    y -= a / b * x;
    return d;
}

ll ex_inv(ll a, ll m = mod) { //求a在mod下的逆元，不存在逆元返回-1
    ll x, y, d = exgcd(a, m, x, y);
    return d == 1 ? (x % m + m) % m : -1;
}

ll inv(ll x, ll m = mod) { return power(x, m - 2); }

void init() {
    fac[0] = 1;
    for (int i = 1; i < X; i++) {
        fac[i] = fac[i - 1] * i % mod;
    }
    facinv[X - 1] = inv(fac[X - 1]);
    for (int i = X - 2; i >= 0; i--) {
        facinv[i] = facinv[i + 1] * i % mod;
    }
    for (int i = 1; i < X; i++) {
        iv[i] = facinv[i] * fac[i - 1] % mod;
    }
#ifdef endl
    ios::sync_with_stdio(0), cin.tie(0);
#endif
}

void solve() {
    int n;
    cin >> n;
    vector<int> v(n);
    map<int, int> s;
    for (int i = 0; i < n; i++) cin >> v[i], s[v[i]]++;
    int m;
    cin >> m;
    vector<int> ans;

    for (int i = 0; i < m; i++) {
        char c;
        int x;
        cin >> c >> x;
        //
        if (c == '+') {
            s[x]++;
        }
        if (c == '-') {
            s[x]--;
            if (s[x] == 0) {
                s.erase(x);
            }
        }
        if (c == '?') {
            auto it = s.begin();
            auto it2 = s.begin();
            if (it->first > x) {
                ans.push_back(1);
                continue;
            }
            it2++;
            int ok = 0;
            for (; it2 != s.end();) {
                if (it2->first - it->first > x) {
                    ans.push_back(it->first + 1);
                    ok = 1;
                    break;
                }
                it++;
                it2++;
            }
            if (!ok) {
                ans.push_back({it->first + 1});
            }
        }

    }
    for (int i = 0; i < ans.size(); i++) {
        cout << ans[i] << " \n"[i + 1 == ans.size()];
    }

}

void solve2() {
    int n;
    cin >> n;
    vector<int> v(n);
    map<ll, set<ll>> s;
    map<ll, int> mp;
    mp[0]++;
    mp[inf]++;
    s[inf].insert(0);
    for (int i = 0; i < n; i++) cin >> v[i];
    int m;
    cin >> m;
    vector<ll> ans;

    auto insert = [&](ll x) {
        if (mp[x]++ == 0) {
            auto it = mp.find(x);
            auto nx = next(it);
            auto pre = prev(it);
            auto gap = nx->first - (pre->first);
            auto gap1 = x - (pre->first);
            auto gap2 = nx->first - (x);
            s[gap].erase(pre->first);
            if (s[gap].size() == 0) {
                s.erase(gap);
            }
            s[gap1].insert(pre->first);
            s[gap2].insert(x);
        }
    };
    auto erase = [&](ll x) {
        if (--mp[x] == 0) {
            auto it = mp.find(x);
            auto nx = next(it);
            auto pre = prev(it);
            auto gap = nx->first - (pre->first);
            auto gap1 = x - (pre->first);
            auto gap2 = nx->first - (x);
            s[gap].insert(pre->first);
            s[gap1].erase(pre->first);
            s[gap2].erase(x);
            if (s[gap2].size() == 0) {
                s.erase(gap2);
            }

            if (s[gap1].size() == 0) {
                s.erase(gap1);
            }
            mp.erase(it);
        }
    };
    for (int i = 0; i < n; i++) {
        insert(v[i]);
    }

    for (int i = 0; i < m; i++) {
        char c;
        int x;
        cin >> c >> x;
        if (c == '+') {
            insert(x);
        }
        if (c == '-') {
            erase(x);
        }
        if (c == '?') {
            auto it1 = s.upper_bound(x);
            auto it2 = *it1->second.begin();
            ans.push_back(it2 + 1);
        }

    }
    for (int i = 0; i < ans.size(); i++) {
        cout << ans[i] << " \n"[i + 1 == ans.size()];
    }
}

int main() {
    init();
    int T = 1;
    cin >> T;
    auto begin = clock();
    while (T--) {
        solve2();
    }
#ifndef ONLINE_JUDGE
    auto end = clock();
    cerr << "run for " << end - begin << "ms" << endl;
#endif
    return 0;
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes potential bugs in the provided code, we need to focus on edge cases and scenarios that can stress the algorithm. Here are the key points to consider:

1. **Initial Set and Operations:**
   - Ensure the initial set is small but contains elements that can create gaps.
   - Include operations that insert and remove elements in a way that creates and modifies gaps in the set.

2. **Edge Cases:**
   - Test with an initial set that has a single element.
   - Test with an initial set that is a sequence of consecutive numbers.
   - Test with an initial set that has large gaps.

3. **Complex Operations:**
   - Include a mix of insert and remove operations that create and destroy gaps.
   - Include a mix of query operations to test the correctness of the $k$-load calculation.

4. **Boundary Conditions:**
   - Test with the maximum and minimum possible values for $k$-load queries.
   - Test with operations that insert or remove elements at the boundaries of the set.

Based on these considerations, here is a small-scale test input that should expose potential bugs:

```plaintext
3
5
1 2 3 4 5
10
+ 6
? 1
? 2
- 5
? 1
? 2
+ 7
? 3
- 4
? 3
5
1 2 1000000 1000001 2000000
5
? 2
- 1000000
? 2
+ 1000002
? 3
5
1 1000000 1000001 1000002 2000000
10
? 5
- 1000001
? 5
+ 1000003
? 5
- 2000000
? 5
+ 2000001
? 5
+ 2000002
? 5
```

This test input includes:
1. A small initial set with consecutive elements.
2. Operations that create and modify gaps.
3. Queries for $k$-load with different values.
4. Large and small values for elements and $k$-load queries to test boundary conditions.

This should help in identifying any issues in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ksyusha and the Loaded Set
time_limit_ms: 3000
memory_limit_mb: 512
Description: Ksyusha decided to start a game development company. To stand out among competitors and achieve success, she decided to write her own game engine. The engine must support a set initially consisting of $n$ distinct integers $a_1, a_2, \ldots, a_n$.

The set will undergo $m$ operations sequentially. The operations can be of the following types:

  * Insert element $x$ into the set;    * Remove element $x$ from the set;    * Report the $k$-load of the set.

The $k$-load of the set is defined as the minimum positive integer $d$ such that the integers $d, d + 1, \ldots, d + (k - 1)$ do not appear in this set. For example, the $3$-load of the set $\\{3, 4, 6, 11\\}$ is $7$, since the integers $7, 8, 9$ are absent from the set, and no smaller value fits.

Ksyusha is busy with management tasks, so you will have to write the engine. Implement efficient support for the described operations.

Input: The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

The following lines describe the test cases.

The first line contains an integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the initial size of the set.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_1 < a_2 < \ldots < a_n \le 2 \cdot 10^6$) — the initial state of the set.

The third line contains an integer $m$ ($1 \le m \le 2 \cdot 10^5$) — the number of operations.

The next $m$ lines contain the operations. The operations are given in the following format:

  * + $x$ ($1 \le x \le 2 \cdot 10^6$) — insert element $x$ into the set (it is guaranteed that $x$ is not in the set);    * - $x$ ($1 \le x \le 2 \cdot 10^6$) — remove element $x$ from the set (it is guaranteed that $x$ is in the set);    * ? $k$ ($1 \le k \le 2 \cdot 10^6$) — output the value of the $k$-load of the set.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2 \cdot 10^5$, and the same holds for $m$.

Output: For each test case, output the answers to the operations of type "?".

Examples:
input:
3
5
1 2 5 905 2000000
15
- 2
? 2
? 1
- 1
? 1
+ 4
+ 2
? 2
+ 6
- 4
+ 7
? 2
? 3
? 4
? 2000000
5
3 4 5 6 8
9
? 5
- 5
? 5
+ 1
? 2
- 6
- 8
+ 6
? 5
5
6 7 8 9 10
10
? 5
- 6
? 4
- 10
+ 5
- 8
+ 3
+ 2
- 3
+ 10
output:
2 2 1 6 3 8 8 2000001
9 9 9 7
1 1

Note:

Buggy Code:
#include <bits/stdc++.h>
#pragma GCC optimize(3,"Ofast","inline","unroll-loops")
using namespace std;
typedef unsigned uint;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<ll,ll> pll;
typedef pair<int,int> pii;
const int MAX=2e6+5,MOD=998244353;
int T,n,m;
bool b[MAX];
set<int> s;
struct node{
	int lv,rv,v,len;
	node(){lv=rv=v=len=0;}
	void merge(node a,node b){
		v=max(a.v,b.v);
		len=a.len+b.len;
		lv=a.lv; rv=b.rv;
		if(a.lv==a.len) lv=a.lv+b.lv;
		if(b.rv==b.len) rv=a.rv+b.rv;
		v=max(v,max(lv,rv));
	}
};
struct segtree{
    node a[MAX<<2];
    void init(int u,int L,int R){
    	if(L==R){
    		a[u].len=1;
    		if(b[L]) a[u].lv=a[u].rv=a[u].v=1;
    		else a[u].lv=a[u].rv=a[u].v=0;
    		return ;
		}
		int mid=L+R>>1;
		init(u*2,L,mid);
		init(u*2+1,mid+1,R);
		pushup(u,L,R);
	}
    void pushup(int u,int L,int R){
    	a[u].merge(a[u*2],a[u*2+1]);
	}
    void modify(int u,int L,int R,int v){
        if(L==R){
        	b[v]^=1;
        	if(b[v]) a[u].lv=a[u].rv=a[u].v=1;
        	else a[u].lv=a[u].rv=a[u].v=0;
			return ;
		}
        int mid=L+R>>1;
        if(v<=mid) modify(u*2,L,mid,v);
        else modify(u*2+1,mid+1,R,v);
        pushup(u,L,R);
    }
    int query(int u,int L,int R,int v){
    	if(a[u].len==v) return L;
    	//cerr<<u<<' '<<L<<' '<<R<<' '<<v<<' '<<a[u].v<<' '<<a[u].lv<<' '<<a[u].rv<<'\n';
    	int mid=L+R>>1;
    	if(a[u*2].v>=v) return query(u*2,L,mid,v);
    	if(a[u*2].rv+a[u*2+1].lv>=v){
    		//cerr<<a[u*2].rv<<' '<<a[u*2+1].lv<<'\n';
    		return mid-a[u*2].rv+1;
		}
		return query(u*2+1,mid+1,R,v);
    }
}seg;
void write(int x){
	if(x>=10) write(x/10);
	putchar(x%10+'0');
}
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin>>T;
	memset(b,1,sizeof(b));
	while(T--){
		for(int u:s) b[u]=1;
		s.clear();
		cin>>n;
		for(int i=1;i<=n;i++){
			int x; cin>>x;
			b[x]=0;
			s.insert(x);
		}
		seg.init(1,1,2e6);
		cin>>m;
		for(int i=1;i<=m;i++){
			char opt; int x;
			cin>>opt>>x;
			if(opt=='?'){
				if(seg.a[1].v<x){
					cout<<2000001-seg.a[1].rv<<' ';
					continue;
				}
				cout<<seg.query(1,1,2e6,x)<<' ';
			}
			else{
				if(b[x]) s.insert(x);
				else s.erase(x);
				seg.modify(1,1,2e6,x);
			}
		}
		cout<<'\n';
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
5
1 2 3 4 5
10
? 1
? 2
? 3
? 4
? 5
- 1
? 1
? 2
? 3
? 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ksyusha and the Loaded Set
time_limit_ms: 3000
memory_limit_mb: 512
Description: Ksyusha decided to start a game development company. To stand out among competitors and achieve success, she decided to write her own game engine. The engine must support a set initially consisting of $n$ distinct integers $a_1, a_2, \ldots, a_n$.

The set will undergo $m$ operations sequentially. The operations can be of the following types:

  * Insert element $x$ into the set;    * Remove element $x$ from the set;    * Report the $k$-load of the set.

The $k$-load of the set is defined as the minimum positive integer $d$ such that the integers $d, d + 1, \ldots, d + (k - 1)$ do not appear in this set. For example, the $3$-load of the set $\\{3, 4, 6, 11\\}$ is $7$, since the integers $7, 8, 9$ are absent from the set, and no smaller value fits.

Ksyusha is busy with management tasks, so you will have to write the engine. Implement efficient support for the described operations.

Input: The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases.

The following lines describe the test cases.

The first line contains an integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the initial size of the set.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_1 < a_2 < \ldots < a_n \le 2 \cdot 10^6$) — the initial state of the set.

The third line contains an integer $m$ ($1 \le m \le 2 \cdot 10^5$) — the number of operations.

The next $m$ lines contain the operations. The operations are given in the following format:

  * + $x$ ($1 \le x \le 2 \cdot 10^6$) — insert element $x$ into the set (it is guaranteed that $x$ is not in the set);    * - $x$ ($1 \le x \le 2 \cdot 10^6$) — remove element $x$ from the set (it is guaranteed that $x$ is in the set);    * ? $k$ ($1 \le k \le 2 \cdot 10^6$) — output the value of the $k$-load of the set.

It is guaranteed that the sum of $n$ across all test cases does not exceed $2 \cdot 10^5$, and the same holds for $m$.

Output: For each test case, output the answers to the operations of type "?".

Examples:
input:
3
5
1 2 5 905 2000000
15
- 2
? 2
? 1
- 1
? 1
+ 4
+ 2
? 2
+ 6
- 4
+ 7
? 2
? 3
? 4
? 2000000
5
3 4 5 6 8
9
? 5
- 5
? 5
+ 1
? 2
- 6
- 8
+ 6
? 5
5
6 7 8 9 10
10
? 5
- 6
? 4
- 10
+ 5
- 8
+ 3
+ 2
- 3
+ 10
output:
2 2 1 6 3 8 8 2000001
9 9 9 7
1 1

Note:

Buggy Code:
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define db double
#define ldb long double
#define max(A,B) (A>B? A:B)
#define min(A,B) (A<B? A:B)
#define yes cout<<"Yes\n"
#define no cout<<"No\n"
#define pir(X) pair<X,X>
#define mpr(A,B) make_pair(A,B)
#define fr first
#define sc second
using namespace std;

const int N=4e6+10;

int n,a[N];
set<int> st;

struct node{
    int len,mx,l_mx,r_mx;
    void init(int flg,int l){
        len=l;
        if(flg) mx=l_mx=r_mx=0;
        else mx=l_mx=r_mx=1;
    }
    node operator + (node o){
        node ret;
        ret.len=len+o.len;
        if(l_mx==len) ret.l_mx=len+o.l_mx;
        else ret.l_mx=l_mx;
        if(o.r_mx==o.len) ret.r_mx=o.r_mx+r_mx;
        else ret.r_mx=o.r_mx;
        ret.mx=max(max(mx,o.mx),r_mx+o.l_mx);
        return ret;
    }
};

struct seg_T{
    #define root 1,1,N-10
    #define lson (pos<<1)
    #define rson (lson|1)
    #define mid ((l+r)>>1)
    #define l_son lson,l,mid
    #define r_son rson,mid+1,r
    #define This pos,l,r
    #define len (r-l+1)
    node data[N<<2];
    void upd(int pos){
        data[pos]=data[lson]+data[rson];
    }
    void update(int pos,int l,int r,int ip,int v){
        if(l==r){
            data[pos].init(v,len);
            return;
        }
        if(ip<=mid) update(l_son,ip,v);
        if(ip>=mid+1) update(r_son,ip,v);
        upd(pos);
    }
    node qry(int pos,int l,int r,int L,int R){
        if(L<=l&&r<=R) return data[pos];
        if(R<=mid) return qry(l_son,L,R);
        if(L>=mid+1) return qry(r_son,L,R);
        return qry(l_son,L,R)+qry(r_son,L,R);
    }
    void build(int pos,int l,int r){
        if(l==r){
            data[pos].init(a[l],len);
            return;
        }
        build(l_son),build(r_son),upd(pos);
    }
}T;

void slv(){
    cin>>n;
    st.clear();
    for(int i=1;i<=n;i++){
        int x;
        cin>>x;
        a[x]=1;
        st.insert(x);
    }
    T.build(root);
    int m;
    cin>>m;
    auto func=[](int x,int v){
        // cerr<<" func: x="<<x<<" v="<<v<<'\n';
        a[x]+=v;
        if(v==1&&a[x]==1) st.insert(x),T.update(root,x,1);
        if(v==-1&&a[x]==0) st.erase(x),T.update(root,x,0);
    };
    while(m--){
        char opt;
        int x;
        cin>>opt>>x;
        if(opt=='-') func(x,-1);
        if(opt=='+') func(x,1);
        if(opt=='?'){
            int l=1,r=N-10,md,rec=N;
            while(l<=r){
                md=(l+r)>>1;
                // cerr<<" qry: md="<<md<<" result="<<T.qry(root,1,md).mx<<'\n';
                if(T.qry(root,1,md).mx>=x){
                    rec=md;
                    r=md-1;
                }
                else l=md+1;
            }
            cout<<rec-x+1<<' ';
        }
    }
    cout<<'\n';
    for(int x:st) a[x]=0;
    return;
}

void fst_IO(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
}

int main(){
    fst_IO();
    int T;
    cin>>T;
    while(T--){
        slv();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that can challenge the algorithm's correctness and efficiency. Here are some steps to think about:

1. **Initial Set with Gaps**: Create an initial set with gaps to test the $k$-load calculation.
2. **Insert and Remove Operations**: Include a mix of insert and remove operations to test the dynamic changes in the set.
3. **Large $k$-load Values**: Test with large $k$-load values to ensure the algorithm handles large queries correctly.
4. **Edge Cases**: Include edge cases such as the smallest and largest possible values for the set elements and $k$-load.

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

```plaintext
3
5
1 3 5 7 9
10
- 3
? 2
? 1
+ 4
- 5
+ 6
? 2
+ 8
? 3
? 4
5
2 4 6 8 10
7
? 5
- 4
? 5
+ 1
? 2
- 8
? 5
5
1 2 3 4 5
8
? 5
- 1
? 4
- 5
+ 6
- 3
+ 1
? 5
```

### Explanation of the Test Input:
1. **Test Case 1**:
   - Initial set: `{1, 3, 5, 7, 9}`
   - Operations:
     - Remove 3
     - Query 2-load (expected: 4)
     - Query 1-load (expected: 2)
     - Insert 4
     - Remove 5
     - Insert 6
     - Query 2-load (expected: 7)
     - Insert 8
     - Query 3-load (expected: 10)
     - Query 4-load (expected: 10)

2. **Test Case 2**:
   - Initial set: `{2, 4, 6, 8, 10}`
   - Operations:
     - Query 5-load (expected: 11)
     - Remove 4
     - Query 5-load (expected: 9)
     - Insert 1
     - Query 2-load (expected: 3)
     - Remove 8
     - Query 5-load (expected: 7)

3. **Test Case 3**:
   - Initial set: `{1, 2, 3, 4, 5}`
   - Operations:
     - Query 5-load (expected: 6)
     - Remove 1
     - Query 4-load (expected: 6)
     - Remove 5
     - Insert 6
     - Remove 3
     - Insert 1
     - Query 5-load (expected: 4)

This test input covers a range of scenarios including gaps in the initial set, dynamic insert and remove operations, and large $k$-load values. It should help expose any bugs in the provided code.