**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Move Back at a Cost
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array of integers $$$a$$$ of length $$$n$$$. You can perform the following operation zero or more times:

- In one operation choose an index $$$i$$$ ($$$1 \le i \le n$$$), assign $$$a_i := a_i + 1$$$, and then move $$$a_i$$$ to the back of the array (to the rightmost position). For example, if $$$a = [3, 5, 1, 9]$$$, and you choose $$$i = 2$$$, the array becomes $$$[3, 1, 9, 6]$$$.

Find the lexicographically smallest$$$^{\text{∗}}$$$ array you can get by performing these operations.

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

The first line contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$), the length of the array.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), the elements of the array.

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

Output: For each test case, print the lexicographically smallest array you can get.

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

Note:

Buggy Code:
from heapq import heappush, heappop

t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    lo = 0
    op = []
    while lo < len(a):
        mini = min(a[lo:])
        min_ind = a[lo:].index(mini) + lo
        op.append(a[min_ind])
        if min_ind == lo:
            lo += 1
            continue
        heap = []
        for i in range(lo, min_ind):
            heappush(heap, a[i])

        while heap:
            ele = heappop(heap)
            a.append(ele + 1)

        lo = min_ind + 1
    print(*op)

**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
2 2 1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Move Back at a Cost
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array of integers $$$a$$$ of length $$$n$$$. You can perform the following operation zero or more times:

- In one operation choose an index $$$i$$$ ($$$1 \le i \le n$$$), assign $$$a_i := a_i + 1$$$, and then move $$$a_i$$$ to the back of the array (to the rightmost position). For example, if $$$a = [3, 5, 1, 9]$$$, and you choose $$$i = 2$$$, the array becomes $$$[3, 1, 9, 6]$$$.

Find the lexicographically smallest$$$^{\text{∗}}$$$ array you can get by performing these operations.

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

The first line contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$), the length of the array.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), the elements of the array.

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

Output: For each test case, print the lexicographically smallest array you can get.

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

Note:

Buggy Code:
t = int(input())
for _ in range (t):
    n = int(input())
    l = list(map(int,input().split()))
    mark = [False] * n
    b = n-1
    mini = l[-1]
    while b >= 0:
        if l[b] <= mini:
            mini = l[b]
        else:
            mark[b] = True
        b -= 1
    l_m = None
    m_dep = max(l)
    while b < n:
        if mark[b]:
            l_m = b
            m_dep = min(m_dep, l[b])
        b += 1

    if l_m == None:
        print(*l)
    else:
        b = l_m + 1
        while b < n:
            if l[b] <= m_dep + 1:
                b += 1
            else:
                mark[b] = True
                b += 1
        n_l = []
        for i in range (n):
            if mark[i]:
                n_l.append(l[i]+1)
            else:
                n_l.append(l[i])
        n_l.sort()

        print(*n_l)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, let's consider the following steps:

1. **Identify the Bug**: The code seems to have issues with correctly identifying and marking elements that need to be moved to the back and incremented. It also has problems with sorting and maintaining the lexicographically smallest order.

2. **Edge Cases**: We need to focus on edge cases where:
   - The array is already sorted.
   - The array has repeated elements.
   - The array has a large difference between the smallest and largest elements.
   - The array has elements that are very close to each other.

3. **Small-Scale Test Input**: We will create a test input with a small number of test cases and small array sizes to ensure it is manageable and fits within the output length limit.

Given these considerations, here is a challenging test input:

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

### Explanation:
1. **Test Case 1**: `3 1 2`
   - The array is not sorted, and the elements are close to each other. The correct output should be `1 2 3`.

2. **Test Case 2**: `4 3 2 1`
   - The array is in reverse order. The correct output should be `1 2 3 4`.

3. **Test Case 3**: `1 3 2 4 5`
   - The array has a mix of elements that are close and far apart. The correct output should be `1 2 3 4 5`.

This test input should expose the issues with the marking and sorting logic 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: Move Back at a Cost
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array of integers $$$a$$$ of length $$$n$$$. You can perform the following operation zero or more times:

- In one operation choose an index $$$i$$$ ($$$1 \le i \le n$$$), assign $$$a_i := a_i + 1$$$, and then move $$$a_i$$$ to the back of the array (to the rightmost position). For example, if $$$a = [3, 5, 1, 9]$$$, and you choose $$$i = 2$$$, the array becomes $$$[3, 1, 9, 6]$$$.

Find the lexicographically smallest$$$^{\text{∗}}$$$ array you can get by performing these operations.

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

The first line contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$), the length of the array.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), the elements of the array.

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

Output: For each test case, print the lexicographically smallest array you can get.

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

Note:

Buggy Code:
import java.util.*;

public class Solution {

    public static class IndexValuePair implements Comparable<IndexValuePair> {
        int value, index;

        IndexValuePair(int val, int ind) {
            value = val;
            index = ind;
        }

        IndexValuePair() {}

        @Override
        public int compareTo(IndexValuePair o) {
            return Integer.compare(value, o.value);
        }
    }

    public static void main(String[] args) {
        int t,n;
        Scanner sc = new Scanner(System.in);
        LinkedList<LinkedList<Integer>> linkedList = new LinkedList<>();
        LinkedList<Integer>[] sortList = new LinkedList[100000];
        LinkedList<Integer> prevList = new LinkedList<>();
        int tmp, prev, smallest;
        t = sc.nextInt();
        while (t-- != 0) {
            linkedList.clear();
            n = sc.nextInt();
            prev = -1;
            for (int i = 0; i < n; i++) {
                tmp = sc.nextInt();
                if (prev == -1 || prev > tmp) {
                    prevList = new LinkedList<>();
                    linkedList.add((LinkedList<Integer>) prevList);
                    prevList.add(tmp);
                }
                else {
                    prevList.add(tmp);
                }
                prev = tmp;
            }

            for (int i = 0; i < n; i++) {
                smallest = linkedList.get(0).get(0);
                for (List<Integer> ls : linkedList) {
                    if (smallest > ls.get(0)) {
                        smallest = ls.get(0);
                    }
                }
                System.out.print(smallest+" ");
                int j = 0;
                while (linkedList.getFirst().get(0) != smallest) {
                    sortList[j++] = linkedList.getFirst();
                    linkedList.removeFirst();
                }
                LinkedList<Integer> smallestList = linkedList.getFirst();
                smallestList.removeFirst();
                if (smallestList.size() == 0) {
                    linkedList.removeFirst();
                }
                if (j > 0)
                    linkedList.add(sortByHeap(sortList, j));
            }
            System.out.println();
        }
    }

    private static LinkedList<Integer> sortByHeap(LinkedList<Integer>[] sortList, int n) {
        PriorityQueue<IndexValuePair> minHeap = new PriorityQueue<>();
        LinkedList<Integer> sortedList = new LinkedList<>();
        for (int i = 0; i < n; i++) {
            minHeap.add(new IndexValuePair(sortList[i].get(0), 0));
            sortList[i].removeFirst();
        }
        IndexValuePair top;
        int index;
        while (!minHeap.isEmpty()) {
            sortedList.add(minHeap.peek().value);
            top = minHeap.poll();
            index = top.index;
            if (sortList[index].size() != 0) {
                minHeap.add(new IndexValuePair(sortList[index].get(0), index));
                sortList[index].removeFirst();
            }
        }
        ListIterator<Integer> iterator = sortedList.listIterator();
        while (iterator.hasNext()) {
            int current = iterator.next();
            iterator.set(current + 1);
        }
        return sortedList;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on scenarios where the code's logic might fail. The key points to consider are:

1. **Lexicographical Order**: The code should produce the lexicographically smallest array.
2. **Operation Constraints**: The code should correctly handle the operations of incrementing and moving elements to the back.

### Step-by-Step Analysis

1. **Edge Cases**:
   - **Single Element Array**: This is trivial and should be handled correctly.
   - **Already Sorted Array**: The code should not perform any operations and return the array as is.
   - **Reverse Sorted Array**: This should be the most challenging case as it requires the maximum number of operations to sort the array.
   - **Array with Duplicates**: Handling duplicates correctly is crucial for lexicographical order.

2. **Challenging Scenarios**:
   - **Array with a Small Number of Elements**: To ensure the test input is small-scale.
   - **Array with Elements Close to Each Other**: This can expose issues in the logic that handles incremental operations.

### Test Input Generation

Based on the above analysis, a small-scale test input that can expose the bug is:

- **Test Case 1**: A reverse sorted array with a small number of elements.
- **Test Case 2**: An array with elements close to each other and a few duplicates.

### Chosen Test Input

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

### Explanation

1. **Test Case 1**:
   - **Input**: `4 4 3 2 1`
   - **Expected Output**: `1 2 3 5`
   - **Explanation**: The array is reverse sorted. The operations will increment and move elements to the back to achieve the lexicographically smallest order.

2. **Test Case 2**:
   - **Input**: `5 2 3 2 4 3`
   - **Expected Output**: `2 2 3 3 5`
   - **Explanation**: The array has elements close to each other and duplicates. The operations should handle the duplicates correctly to achieve the lexicographically smallest order.

This test input should effectively expose any bugs in the provided code by testing its ability to handle reverse sorted arrays and arrays with elements close to each other and duplicates.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Move Back at a Cost
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array of integers $$$a$$$ of length $$$n$$$. You can perform the following operation zero or more times:

- In one operation choose an index $$$i$$$ ($$$1 \le i \le n$$$), assign $$$a_i := a_i + 1$$$, and then move $$$a_i$$$ to the back of the array (to the rightmost position). For example, if $$$a = [3, 5, 1, 9]$$$, and you choose $$$i = 2$$$, the array becomes $$$[3, 1, 9, 6]$$$.

Find the lexicographically smallest$$$^{\text{∗}}$$$ array you can get by performing these operations.

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

The first line contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$), the length of the array.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), the elements of the array.

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

Output: For each test case, print the lexicographically smallest array you can get.

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

Note:

Buggy Code:
import java.util.*;
public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-- > 0) {
            int n = sc.nextInt();
            int[] array = new int[n];
            int min = Integer.MAX_VALUE;
            for (int i = 0; i < n; i++) {
                array[i] = sc.nextInt();
                min = Math.min(min, array[i]);
            }
            int last = 0;
            for (int i = 0; i < n; i++) {
                if(array[i] == min){
                    last = i;
                    System.out.print(min + " ");
                }
            }
            int compare = min;
            min = Integer.MAX_VALUE;
            for(int i = 0; i < last; i++){
                if(array[i] != compare){
                    array[i]++;
                    min = Math.min(min,array[i]);
                }
            }
            boolean[] added = new boolean[n];
            for(int j = last + 1; j < n; j++){
                if(array[j] > min){
                    added[j] = true;
                    array[j]++;
                }
            }
            min = array[n - 1];
            for(int j = n - 2; j > last; j--){
                if(array[j] > min && !added[j]){
                    array[j]++;
                }
                else{
                    min = array[j];
                }
            }
            List<Integer> list = new ArrayList<>();
            for(int num : array){
                if(num != compare){
                    list.add(num);
                }
            }
            Collections.sort(list);
            for(int num : list){
                System.out.print(num + " ");
            }
            System.out.println();
        }
    }
}

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

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

Think step by step.

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

### Explanation:
1. **Test Case 1: All elements are the same**
   - Input: `3 1 1 1`
   - Expected Output: `1 1 1`
   - This tests the scenario where all elements are the same. The algorithm should not perform any operations and return the array as is.

2. **Test Case 2: Multiple elements with the same minimum value**
   - Input: `5 1 1 2 2 2`
   - Expected Output: `1 1 2 3 3`
   - This tests the scenario where there are multiple minimum values. The algorithm should move all minimum values to the front and then increment the other elements to achieve the lexicographically smallest array.

3. **Test Case 3: All elements are the same with a larger array**
   - Input: `6 1 1 1 1 1 1`
   - Expected Output: `1 1 1 1 1 1`
   - This tests the scenario where all elements are the same in a larger array. The algorithm should not perform any operations and return the array as is.

These test cases are designed to expose issues in the buggy code by focusing on edge cases where all elements are the same or where there are multiple minimum values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Move Back at a Cost
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array of integers $$$a$$$ of length $$$n$$$. You can perform the following operation zero or more times:

- In one operation choose an index $$$i$$$ ($$$1 \le i \le n$$$), assign $$$a_i := a_i + 1$$$, and then move $$$a_i$$$ to the back of the array (to the rightmost position). For example, if $$$a = [3, 5, 1, 9]$$$, and you choose $$$i = 2$$$, the array becomes $$$[3, 1, 9, 6]$$$.

Find the lexicographically smallest$$$^{\text{∗}}$$$ array you can get by performing these operations.

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

The first line contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$), the length of the array.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), the elements of the array.

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

Output: For each test case, print the lexicographically smallest array you can get.

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

Note:

Buggy Code:
#include <bits/stdc++.h>
#include <chrono>
#include <random>
#define lli long long int
using namespace std;
// #define mod 1000000007
#define mod 998244353
#define fastio                        \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL);
#define INF_INT 2000000000
#define INF_LLI 1000000000000000000
#define common cout << "Case #" << w + 1 << ": "

#ifdef LOCAL
#include "//debug.h"
#else
#define debug(x...)
#endif

template <class T>
void VecPrint(vector<T> &v, int len = -1)
{
    if (len == -1)
    {
        len = (int)v.size();
    }
    for (int i = 0; i < len; i++)
        cout << v[i] << ' ';
    cout << '\n';
}

void addHandleMinus1(int &a, int b) {
    if (a == -1) {
        a = b;
        return;
    }
    a += b;
    if (a >= mod) {
        a -= mod;
    }
}

void add(int &a, int b) {
    a += b;
    if (a >= mod) {
        a -= mod;
    }
}

void sub(int &a, int b)
{
    a -= b;
    if (a < 0) {
        a += mod;
    }
}

/*-------------------------------------------------------------------Code-----------------------------------------------------------------*/

void solve() {
    int n;
    cin >> n;
    vector<int> v(n);
    vector<pair<int, int> > imp(n);
    for (int i = 0; i < n; i++) {
        cin >> v[i];
        imp[i] = {v[i], i};
    }
    sort(imp.begin(), imp.end());
    vector<int> left;
    vector<int> ans{imp[0].first};
    int prev = imp[0].second;
    for (int i = 1; i < n; i++) {
        if (imp[i].second < prev) {
            left.push_back(imp[i].first);
            continue;
        }
        if (left.size()) {
            if (left[0] + 1 < imp[i].first) {
                for (int &ent: left) {
                    ans.push_back(ent + 1);
                }
                left.clear();
                for (int j = i; j < n; j++) {
                    ans.push_back(imp[j].first + 1);
                    break;
                }
            } else {
                ans.push_back(imp[i].first);
                prev = imp[i].second;
            }
        } else {
            ans.push_back(imp[i].first);
            prev = imp[i].second;
        }
    }
    for (int &ent: left) {
        ans.push_back(ent + 1);
    }
    VecPrint<int>(ans);
}

int main() {
    fastio;
    int te = 1;
    cin >> te;
    while (te--) {
        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 provided buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. Here are the steps to think through this:

1. **Identify the Bug**: The code seems to have issues with handling elements that need to be moved to the back after being incremented. Specifically, it might not correctly handle cases where the order of elements is significant.

2. **Edge Cases**:
   - **All elements are the same**: This can test how the code handles uniform inputs.
   - **Elements in descending order**: This can test how the code handles cases where the initial array is already in reverse order.
   - **Elements with a few out-of-order elements**: This can test how the code handles small disruptions in the order.

3. **Small-Scale Test Input**: We need a small input that can be processed within the constraints but is still challenging.

Let's create a test input that includes a mix of these cases:

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

### Explanation:

1. **Test Case 1**: `3 2 1`
   - This is a descending order array. The expected output should be `1 3 3` because the smallest element should be moved to the front, and the rest should be adjusted accordingly.

2. **Test Case 2**: `2 1 4 3`
   - This array has elements out of order. The expected output should be `1 2 4 4` because the smallest element should be moved to the front, and the next smallest elements should follow.

3. **Test Case 3**: `1 2 2 1 4`
   - This array has repeated elements and a small disruption in order. The expected output should be `1 1 3 3 5` because the smallest elements should be moved to the front, and the rest should be adjusted accordingly.

This test input should help expose any issues in the code's handling of element movement and order adjustments.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Move Back at a Cost
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array of integers $$$a$$$ of length $$$n$$$. You can perform the following operation zero or more times:

- In one operation choose an index $$$i$$$ ($$$1 \le i \le n$$$), assign $$$a_i := a_i + 1$$$, and then move $$$a_i$$$ to the back of the array (to the rightmost position). For example, if $$$a = [3, 5, 1, 9]$$$, and you choose $$$i = 2$$$, the array becomes $$$[3, 1, 9, 6]$$$.

Find the lexicographically smallest$$$^{\text{∗}}$$$ array you can get by performing these operations.

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

The first line contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$), the length of the array.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), the elements of the array.

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

Output: For each test case, print the lexicographically smallest array you can get.

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

Note:

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

void solve(){
    int n;
    cin>>n;
    vector<int> arr(n);
    for(int i=0;i<n;i++) cin>> arr[i];
    vector<int> arr1 = arr;
    sort(arr1.begin(),arr1.end());
    int ind=0;
    for(int i=0;i<n;i++){
        if(arr[i]==arr1[ind]) ind++;
    }
    int t=n;
    if(ind < n){
        int j=n-1;
        while(arr[j]== arr1[ind]+1 && j>0) j--;
        t= j+1;
    }
    for(int i=ind;i<t;i++) arr1[i]++;
    sort(arr1.begin(),arr1.end());
    for(int i=0;i<n;i++) cout<<arr1[i]<<" ";
    cout<<endl;
    return ;
}

int main() {
	// your code goes here
    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 provided buggy code, we need to consider edge cases that can cause the code to fail. The primary issues in the buggy code are:

1. **Incorrect Handling of Duplicates**: The code does not correctly handle cases where there are duplicate elements in the array.
2. **Incorrect Handling of Increment and Rearrange Operations**: The code does not correctly simulate the operation of incrementing an element and moving it to the back of the array.

Given these issues, let's construct a small-scale test input that can expose these bugs:

1. **Test Case with Duplicates**: An array with duplicate elements can cause the code to fail if it incorrectly handles the duplicates.
2. **Test Case with Specific Patterns**: An array that requires specific operations to achieve the lexicographically smallest array can expose the bug in the logic of incrementing and rearranging.

Here is a small-scale test input that includes these edge cases:

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

### Explanation:
1. **Test Case 1**:
   - **Input**: `4 1 1 2 2`
   - **Expected Output**: `1 1 2 2`
   - **Explanation**: The array is already in the lexicographically smallest form, so no operations are needed. The buggy code should handle this correctly, but it might fail if it incorrectly increments elements or handles duplicates.

2. **Test Case 2**:
   - **Input**: `5 1 2 2 1 4`
   - **Expected Output**: `1 1 2 3 4`
   - **Explanation**: To achieve the lexicographically smallest array, we need to perform operations to move and increment elements. For example, we can increment the second `2` to `3` and move it to the back, resulting in `1 1 2 4 3`, and then sort it to get `1 1 2 3 4`. The buggy code might fail to correctly handle this sequence of operations.

This test input should help expose the bugs in the provided code by focusing on edge cases with duplicates and specific patterns that require careful handling of the operations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Move Back at a Cost
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array of integers $$$a$$$ of length $$$n$$$. You can perform the following operation zero or more times:

- In one operation choose an index $$$i$$$ ($$$1 \le i \le n$$$), assign $$$a_i := a_i + 1$$$, and then move $$$a_i$$$ to the back of the array (to the rightmost position). For example, if $$$a = [3, 5, 1, 9]$$$, and you choose $$$i = 2$$$, the array becomes $$$[3, 1, 9, 6]$$$.

Find the lexicographically smallest$$$^{\text{∗}}$$$ array you can get by performing these operations.

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

The first line contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$), the length of the array.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), the elements of the array.

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

Output: For each test case, print the lexicographically smallest array you can get.

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

Note:

Buggy Code:
# from collections import deque
# from dataclasses import dataclass
# @dataclass
# class Node:
#     next: "Node"=None
#     value: int
# ntc=int(input())
# for _ in range(ntc):
#     n=int(input())
#     a=map(int,input().split())
#     first=None
#     last=None
#     for i in a:
#         nd=Node(value=i)
#         if last is not None:
#             last.next=nd
#         else:
#             first = nd
#         last=nd

#     flag=True
#     while flag:
#         nd=first
#         while nd.next.next:
#             nd=nd.next
#             ndn=nd.next
#             if nd.value>ndn.value:

ntc = int(input())
for _ in range(ntc):
    n = int(input())
    a = list(map(int, input().split()))
    flag = True
    app_lst = [0]
    while app_lst:
        i = 0
        app_lst = []
        while i < len(a) - 1:
            if a[i] > a[i + 1]:
                value = a.pop(i)
                app_lst.append(value + 1)
            else:
                i += 1
        app_lst.sort()
        a.extend(app_lst)
    print(" ".join(map(str, a)))

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

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

Think step by step.

Response:
```plaintext
3
3
3 2 1
5
5 4 3 2 1
6
10 9 8 7 6 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Move Back at a Cost
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array of integers $$$a$$$ of length $$$n$$$. You can perform the following operation zero or more times:

- In one operation choose an index $$$i$$$ ($$$1 \le i \le n$$$), assign $$$a_i := a_i + 1$$$, and then move $$$a_i$$$ to the back of the array (to the rightmost position). For example, if $$$a = [3, 5, 1, 9]$$$, and you choose $$$i = 2$$$, the array becomes $$$[3, 1, 9, 6]$$$.

Find the lexicographically smallest$$$^{\text{∗}}$$$ array you can get by performing these operations.

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

The first line contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$), the length of the array.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), the elements of the array.

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

Output: For each test case, print the lexicographically smallest array you can get.

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

Note:

Buggy Code:
import queue

case = int(input())
for _ in range(case):
    n = int(input())
    a = list(map(int, input().split()))
    Q = queue.PriorityQueue()
    for i in range(n):
        Q.put((a[i], i))
    now = 0
    flag = 0
    while Q.qsize() > 0:
        x, y = Q.get()
        if y < now and flag == 0:
            Q.put((x + 1, n))
            continue
        if y == n:
            flag = 1
        print(x + (flag if y < n else 0), end=' ')
        if y < n:
            now = y + 1
    print()

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Move Back at a Cost
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array of integers $$$a$$$ of length $$$n$$$. You can perform the following operation zero or more times:

- In one operation choose an index $$$i$$$ ($$$1 \le i \le n$$$), assign $$$a_i := a_i + 1$$$, and then move $$$a_i$$$ to the back of the array (to the rightmost position). For example, if $$$a = [3, 5, 1, 9]$$$, and you choose $$$i = 2$$$, the array becomes $$$[3, 1, 9, 6]$$$.

Find the lexicographically smallest$$$^{\text{∗}}$$$ array you can get by performing these operations.

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

The first line contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$), the length of the array.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), the elements of the array.

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

Output: For each test case, print the lexicographically smallest array you can get.

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

Note:

Buggy Code:
import static java.lang.Math.*;

import java.util.*;

public class B {

	public Object solve () {
		int N = sc.nextInt();
		long [] A = sc.nextLongs();

		long [] B = A.clone();
		for (int i : sep(N-1))
			B[i] = min(B[i], B[i+1]);

		PriorityQueue<Long> Q = new PriorityQueue<>();
		boolean [] V = new boolean [N];

		for (int i : rep(N))
			if (B[i] < A[i]) {
				V[i] = true;
				Q.add(A[i] + 1);
			}

		for (int i : rep(N))
			if (!V[i] && A[i] > Q.peek()) {
				V[i] = true;
				Q.add(A[i] + 1);
			}

		List<Long> res = new ArrayList<>();
		for (int i : rep(N))
			if (!V[i])
				res.add(A[i]);

		while (!Q.isEmpty())
			res.add(Q.poll());

		return res;
	}

	private static final int CONTEST_TYPE = 2;

	private static void init () {
	}

	private static int [] rep (int N) { return rep(0, N); }
	private static int [] rep (int S, int T) { if (S >= T) return new int [0]; int [] res = new int [T-S]; for (int i = S; i < T; ++i) res[i-S] = i; return res; }
	private static int [] sep (int N) { return sep(0, N); }
	private static int [] sep (final int S, final int T) { if (S >= T) return new int [0]; int [] res = new int [T-S]; for (int i = S; i < T; ++i) res[T-1-i] = i; return res; }
	//////////////////////////////////////////////////////////////////////////////////// OFF
	private static IOUtils.MyScanner sc = new IOUtils.MyScanner();
	private static class IOUtils {
		public static class MyScanner {
			public String next () { newLine(); return line[index++]; }
			public int nextInt () { return Integer.parseInt(next()); }
			public String nextLine () { line = null; return readLine(); }
			public String [] nextStrings () { return split(nextLine()); }
			public long [] nextLongs () {
				String [] S = nextStrings(); int N = S.length;
				long [] res = new long [N];
				for (int i = 0; i < N; ++i)
					res[i] = Long.parseLong(S[i]);
				return res;
			}
			//////////////////////////////////////////////
			private boolean eol () { return index == line.length; }
			private String readLine () {
				try {
					int b;
					StringBuilder res = new StringBuilder();
					while ((b = System.in.read()) < ' ')
						continue;
					res.append((char)b);
					while ((b = System.in.read()) >= ' ')
						res.append((char)b);
					return res.toString();
				} catch (Exception e) {
					throw new Error (e);
				}
			}
			private MyScanner () {
				try {
					while (System.in.available() == 0)
						Thread.sleep(1);
					start();
				} catch (Exception e) {
					throw new Error(e);
				}
			}
			private String [] line;
			private int index;
			private void newLine () {
				if (line == null || eol()) {
					line = split(readLine());
					index = 0;
				}
			}
			private String [] split (String s) { return s.length() > 0 ? s.split(" ") : new String [0]; }
		}
		private static String build (Object o, Object ... A) { return buildDelim(" ", o, A); }
		private static String buildDelim (String delim, Object o, Object ... A) {
			java.util.List<String> str = new java.util.ArrayList<>();
			append(str, o);
			for (Object p : A)
				append(str, p);
			return String.join(delim, str.toArray(new String [0]));
		}
		//////////////////////////////////////////////////////////////////////////////////
		private static java.text.DecimalFormat formatter = new java.text.DecimalFormat("#.#########");
		private static void start () { if (t == 0) t = millis(); }
		private static void append (java.util.function.Consumer<Object> f, java.util.function.Consumer<Object> g, Object o) {
			if (o.getClass().isArray()) {
				int len = java.lang.reflect.Array.getLength(o);
				for (int i = 0; i < len; ++i)
					f.accept(java.lang.reflect.Array.get(o, i));
			}
			else if (o instanceof Iterable<?>)
				((Iterable<?>)o).forEach(f::accept);
			else
				g.accept(o instanceof Double ? formatter.format(o) : o);
		}
		private static void append (java.util.List<String> str, Object o) {
			append(x -> append(str, x), x -> str.add(x.toString()), o);
		}
		private static java.io.PrintWriter pw = new java.io.PrintWriter(System.out);
		private static void flush () { pw.flush(); System.out.flush(); }
		private static void print (Object o, Object ... A) {
			String res = build(o, A);
			if (DEBUG == 0) {
				String out = res.split(System.lineSeparator())[0];
				if (out.length() > 80)
					out = out.substring(0, 80);
				if (out.length() < res.length())
					out += " ...";
				err(out, '(', time(), ')');
			}
			if (DEBUG == 2)
				err(res, '(', time(), ')');
			if (res.length() > 0)
				pw.println(res);
			if (DEBUG == 1)
				flush();
		}
		private static void err (Object o, Object ... A) { System.err.println(build(o, A)); }
		private static int DEBUG = -1;
		private static void exit () {
			String end = "------------------" + System.lineSeparator() + time();
			switch(DEBUG) {
			case 1: print(end); break;
			case 0:
			case 2: err(end); break;
			}
			IOUtils.pw.close();
			System.out.flush();
			System.exit(0);
		}
		private static long t;
		private static long millis () { return System.currentTimeMillis(); }
		private static String time () { return "Time: " + (millis() - t) / 1000.0; }
		private static void run (int N) {
			try { DEBUG = Integer.parseInt(System.getProperty("DEBUG")); }
			catch (Throwable t) {}

			Runnable task = new Runnable() {
				@Override
				public void run() {
					for (int n = 1; n <= N; ++n) {
						Object res = new B().solve();
						if (res != null) {
							@SuppressWarnings("all")
							Object o = CONTEST_TYPE == 0 ? "Case #" + n + ": " + build(res) : res;
							print(o);
						}
					}
				}
			};
			Thread t = new Thread(null, task, "", 1L << 30);
			t.start();
			try { t.join(); } catch (Exception e) { throw new Error(e); }
			exit();
		}
	}
	////////////////////////////////////////////////////////////////////////////////////
	public static void main (String[] args) {
		@SuppressWarnings("all")
		int N = CONTEST_TYPE == 1 ? 1 : sc.nextInt();
		init();
		IOUtils.run(N);
	}
}

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

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

Think step by step.

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

1. **Understand the Algorithm**: The algorithm aims to transform the array into its lexicographically smallest form by performing operations that increase an element and move it to the back.

2. **Identify Potential Bugs**:
   - **Incorrect Handling of Minimum Values**: The algorithm might not correctly handle cases where the minimum value in the array needs to be moved.
   - **Queue Management**: The priority queue might not be managed correctly, leading to incorrect final results.
   - **Array Boundary Conditions**: The algorithm might have issues with the boundaries of the array, especially when the array is very small.

3. **Construct Edge Cases**:
   - **Single Element Array**: An array with a single element should not change.
   - **Array with All Elements Equal**: An array where all elements are the same should not change.
   - **Array with Elements in Descending Order**: This can test the algorithm's ability to handle a scenario where the smallest element needs to be moved to the back multiple times.
   - **Array with Elements in Ascending Order**: This is a straightforward case but can still expose issues if the algorithm incorrectly modifies the array.
   - **Array with Mixed Values**: This can test the algorithm's ability to handle a mix of values and ensure the lexicographically smallest form is achieved.

4. **Select a Small-Scale Test Input**:
   - We will choose a small array that can still expose the potential bugs.

Based on the above steps, a small-scale test input that can expose the bug is:

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

This input will test the algorithm's ability to handle an array in descending order, where the smallest element needs to be moved to the back multiple times. The expected output should be `[1, 3, 3]`, but the buggy code might not handle this correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Move Back at a Cost
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an array of integers $$$a$$$ of length $$$n$$$. You can perform the following operation zero or more times:

- In one operation choose an index $$$i$$$ ($$$1 \le i \le n$$$), assign $$$a_i := a_i + 1$$$, and then move $$$a_i$$$ to the back of the array (to the rightmost position). For example, if $$$a = [3, 5, 1, 9]$$$, and you choose $$$i = 2$$$, the array becomes $$$[3, 1, 9, 6]$$$.

Find the lexicographically smallest$$$^{\text{∗}}$$$ array you can get by performing these operations.

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

The first line contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$), the length of the array.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$), the elements of the array.

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

Output: For each test case, print the lexicographically smallest array you can get.

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

Note:

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

public class B_Move_Back_at_a_Cost {
    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {}
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }

        char[] nextCharArray() {
            try {
                String line = br.readLine();
                return line.toCharArray();
            } catch (IOException e) {
                e.printStackTrace();
                return new char[0];
            }
        }
    }

    public static void main(String[] args) {
        PrintWriter out = new PrintWriter(System.out);
        FastScanner s = new FastScanner();

        int t = s.nextInt();

        while(t-- != 0) {
            int n = s.nextInt();

            LinkedList<Integer> ll = new LinkedList<>();
            for(int i = 0; i < n; i++) {
                ll.add(s.nextInt());
            }

            PriorityQueue<Integer> pq = new PriorityQueue<>();

            int last = Integer.MIN_VALUE;
            int smallest_removed = Integer.MAX_VALUE;
            for(int i = 0; i < ll.size(); i++) {
                int cur = ll.get(i);
                if(cur < last) {
                    pq.add(last + 1);
                    smallest_removed = Math.min(smallest_removed, last + 1);
                    ll.remove(i-1);
                    last = i-2 >= 0 ? ll.get(i-2) : Integer.MIN_VALUE;
                    i = Math.max(-1, i-2);
                } else {
                    last = cur;
                }
            }

            while(smallest_removed < ll.getLast()) {
                pq.add(ll.getLast() + 1);
                ll.removeLast();
            }

            for(int i = 0; i < ll.size(); i++) {
                pq.add(ll.get(i));
            }

            while (!pq.isEmpty()) {
                out.print(pq.poll() + " ");
            }
            out.println();
        }
        out.close();
    }
}

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

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

Think step by step.

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

### Explanation:
1. **Test Case 1:**
   - Input: `3 2 1 3`
   - Expected Output: `1 3 3`
   - This test case checks if the algorithm can handle a small array and correctly move elements to the back to achieve the lexicographically smallest array.

2. **Test Case 2:**
   - Input: `5 4 3 2 1 5`
   - Expected Output: `1 2 3 4 5`
   - This test case is designed to check if the algorithm can handle a decreasing sequence and correctly rearrange it to the lexicographically smallest order.

3. **Test Case 3:**
   - Input: `6 1 2 3 6 5 4`
   - Expected Output: `1 2 3 4 6 7`
   - This test case is designed to check if the algorithm can handle a partially sorted array and correctly rearrange it to the lexicographically smallest order, including handling the increment and move operations.

These test cases are small-scale but cover a range of scenarios that can expose potential bugs in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Asesino (Easy Version)
time_limit_ms: 2500
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version, you can ask at most $$$n+69$$$ questions. You can make hacks only if both versions of the problem are solved.

This is an interactive problem.

It is a tradition in Mexico's national IOI trainings to play the game "Asesino", which is similar to "Among Us" or "Mafia".

Today, $$$n$$$ players, numbered from $$$1$$$ to $$$n$$$, will play "Asesino" with the following three roles:

- Knight: a Knight is someone who always tells the truth.
- Knave: a Knave is someone who always lies.
- Impostor: an Impostor is someone everybody thinks is a Knight, but is secretly a Knave.

Each player will be assigned a role in the game. There will be exactly one Impostor but there can be any (possible zero) number of Knights and Knaves.

As the game moderator, you have accidentally forgotten the roles of everyone, but you need to determine the player who is the Impostor.

To determine the Impostor, you will ask some questions. In each question, you will pick two players $$$i$$$ and $$$j$$$ ($$$1 \leq i, j \leq n$$$; $$$i \neq j$$$) and ask if player $$$i$$$ thinks that player $$$j$$$ is a Knight. The results of the question is shown in the table below.

KnightKnaveImpostorKnightYesNoYesKnaveNoYesNoImpostorNoYes— The response of the cell in row $$$a$$$ and column $$$b$$$ is the result of asking a question when $$$i$$$ has role $$$a$$$ and $$$j$$$ has row $$$b$$$. For example, the "Yes" in the top right cell belongs to row "Knight" and column "Impostor", so it is the response when $$$i$$$ is a Knight and $$$j$$$ is an Impostor.

Find the Impostor in at most $$$n + 69$$$ questions.

Note: the grader is adaptive: the roles of the players are not fixed in the beginning and may change depending on your questions. However, it is guaranteed that there exists an assignment of roles that is consistent with all previously asked questions under the constraints of this problem.

Input: The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$3 \le n \le 10^5$$$) — the number of people playing the game.

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

Output: None

Examples:
input:
2
7

1

0

0

1

1

0

0

1

4

0

1

1

1
output:
? 1 3

? 7 6

? 2 5

? 6 2

? 4 5

? 4 6

? 1 4

? 2 4

! 4

? 1 2

? 2 3

? 3 4

? 4 1

! 3

Note: Note that the example test cases do not represent an optimal strategy for asking questions and are only shown for the sake of demonstrating the interaction format. Specifically, we cannot determine which player is the Impostor from the questions asked in the examples.

In the first test case of the example, players at indices $$$2$$$ and $$$6$$$ are Knights, players at indices $$$1$$$, $$$3$$$, $$$5$$$, and $$$7$$$ are Knaves, and the Impostor is at index $$$4$$$. The following is an explanation of the questions asked:

- In the first query, player $$$i$$$ is a Knave and player $$$j$$$ is a Knave. The answer is "yes" since Knaves always lie.
- In the second query, player $$$i$$$ is a Knave and player $$$j$$$ is a Knight. The answer is "no" since Knaves always lie.
- In the third query, player $$$i$$$ is a Knight and player $$$j$$$ is a Knave. The answer is "no" since Knights always tell the truth.
- In the fourth query, player $$$i$$$ is a Knight and player $$$j$$$ is a Knight. The answer is "yes" since Knights always tell the truth.
- In the fifth query, player $$$i$$$ is a Impostor and player $$$j$$$ is a Knave. The answer is "yes" since the Impostor always lies.
- In the sixth query, player $$$i$$$ is a Impostor and player $$$j$$$ is a Knight. The answer is "no" since the Impostor always lies.
- In the seventh query, player $$$i$$$ is a Knave and player $$$j$$$ is a Impostor. The answer is "no" since Knaves always lie and Knaves thinks that the Impostor is a Knight.
- In the eighth query, player $$$i$$$ is a Knight and player $$$j$$$ is a Impostor. The answer is "yes" since Knights always tell the truth and Knights think that the Impostor is a Knight.

Buggy Code:
#include<iostream>
#include<vector>
using namespace std;
int ask(int i,int j){
    cout<<"?"<<" "<<i<<" "<<j<<endl;
    cout.flush();
    int res;
    cin>>res;
    return res;
}
int main(){
    int t;
    cin>>t;
    while(t>0){
        t--;
        int n;
        cin>>n;
        int answer;
        for(int i=1;i+1<=n;i+=2){
            if(ask(i,i+1)!=ask(i+1,i)){
                int j=(i-1>=1?i-1:i+2);
                if(ask(i,j)!=ask(j,i)){
                 answer=i;
                }
                else{
                    answer=i+1;
                }
            }
        }
        cout<<"!"<<" "<<answer<<endl;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases or scenarios where the code's logic might fail. The buggy code seems to rely on comparing the responses of pairs of players and then making a decision based on the results. The main issue in the code is that it does not handle all possible scenarios correctly, especially when the number of players is small or the players' roles are not straightforward.

Let's consider a small-scale test case with \( n = 3 \) players. This will allow us to test the code's logic thoroughly with a minimal number of players.

### Test Case Analysis
- **Players**: 3 (let's call them 1, 2, and 3)
- **Roles**:
  - Player 1: Knight
  - Player 2: Impostor
  - Player 3: Knave

### Expected Behavior
- **Question 1**: Ask Player 1 about Player 2
  - Player 1 (Knight) thinks Player 2 is a Knight (but Player 2 is actually the Impostor, so the response is "Yes").
- **Question 2**: Ask Player 2 about Player 1
  - Player 2 (Impostor) thinks Player 1 is a Knight (but Player 2 always lies, so the response is "No").
- **Question 3**: Ask Player 1 about Player 3
  - Player 1 (Knight) thinks Player 3 is a Knave (so the response is "No").
- **Question 4**: Ask Player 3 about Player 1
  - Player 3 (Knave) thinks Player 1 is a Knave (but Player 3 always lies, so the response is "Yes").

From these questions, we can see that:
- Player 1 and Player 2 give conflicting responses about each other.
- Player 1 and Player 3 give consistent responses (both "No").

Given this setup, the buggy code should ideally identify Player 2 as the Impostor. However, the code's logic might fail to handle this scenario correctly.

### Test Input
```plaintext
1
3
```

### Explanation
- **Test Case 1**: \( n = 3 \)
  - The roles are:
    - Player 1: Knight
    - Player 2: Impostor
    - Player 3: Knave

This test input will expose the bug in the code because:
- The code compares the responses of pairs of players.
- The responses of Player 1 and Player 2 will be different, which should lead to the correct identification of the Impostor.
- However, the code's logic might not handle the consistent responses from Player 1 and Player 3 correctly, leading to an incorrect identification of the Impostor.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Asesino (Easy Version)
time_limit_ms: 2500
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version, you can ask at most $$$n+69$$$ questions. You can make hacks only if both versions of the problem are solved.

This is an interactive problem.

It is a tradition in Mexico's national IOI trainings to play the game "Asesino", which is similar to "Among Us" or "Mafia".

Today, $$$n$$$ players, numbered from $$$1$$$ to $$$n$$$, will play "Asesino" with the following three roles:

- Knight: a Knight is someone who always tells the truth.
- Knave: a Knave is someone who always lies.
- Impostor: an Impostor is someone everybody thinks is a Knight, but is secretly a Knave.

Each player will be assigned a role in the game. There will be exactly one Impostor but there can be any (possible zero) number of Knights and Knaves.

As the game moderator, you have accidentally forgotten the roles of everyone, but you need to determine the player who is the Impostor.

To determine the Impostor, you will ask some questions. In each question, you will pick two players $$$i$$$ and $$$j$$$ ($$$1 \leq i, j \leq n$$$; $$$i \neq j$$$) and ask if player $$$i$$$ thinks that player $$$j$$$ is a Knight. The results of the question is shown in the table below.

KnightKnaveImpostorKnightYesNoYesKnaveNoYesNoImpostorNoYes— The response of the cell in row $$$a$$$ and column $$$b$$$ is the result of asking a question when $$$i$$$ has role $$$a$$$ and $$$j$$$ has row $$$b$$$. For example, the "Yes" in the top right cell belongs to row "Knight" and column "Impostor", so it is the response when $$$i$$$ is a Knight and $$$j$$$ is an Impostor.

Find the Impostor in at most $$$n + 69$$$ questions.

Note: the grader is adaptive: the roles of the players are not fixed in the beginning and may change depending on your questions. However, it is guaranteed that there exists an assignment of roles that is consistent with all previously asked questions under the constraints of this problem.

Input: The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$3 \le n \le 10^5$$$) — the number of people playing the game.

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

Output: None

Examples:
input:
2
7

1

0

0

1

1

0

0

1

4

0

1

1

1
output:
? 1 3

? 7 6

? 2 5

? 6 2

? 4 5

? 4 6

? 1 4

? 2 4

! 4

? 1 2

? 2 3

? 3 4

? 4 1

! 3

Note: Note that the example test cases do not represent an optimal strategy for asking questions and are only shown for the sake of demonstrating the interaction format. Specifically, we cannot determine which player is the Impostor from the questions asked in the examples.

In the first test case of the example, players at indices $$$2$$$ and $$$6$$$ are Knights, players at indices $$$1$$$, $$$3$$$, $$$5$$$, and $$$7$$$ are Knaves, and the Impostor is at index $$$4$$$. The following is an explanation of the questions asked:

- In the first query, player $$$i$$$ is a Knave and player $$$j$$$ is a Knave. The answer is "yes" since Knaves always lie.
- In the second query, player $$$i$$$ is a Knave and player $$$j$$$ is a Knight. The answer is "no" since Knaves always lie.
- In the third query, player $$$i$$$ is a Knight and player $$$j$$$ is a Knave. The answer is "no" since Knights always tell the truth.
- In the fourth query, player $$$i$$$ is a Knight and player $$$j$$$ is a Knight. The answer is "yes" since Knights always tell the truth.
- In the fifth query, player $$$i$$$ is a Impostor and player $$$j$$$ is a Knave. The answer is "yes" since the Impostor always lies.
- In the sixth query, player $$$i$$$ is a Impostor and player $$$j$$$ is a Knight. The answer is "no" since the Impostor always lies.
- In the seventh query, player $$$i$$$ is a Knave and player $$$j$$$ is a Impostor. The answer is "no" since Knaves always lie and Knaves thinks that the Impostor is a Knight.
- In the eighth query, player $$$i$$$ is a Knight and player $$$j$$$ is a Impostor. The answer is "yes" since Knights always tell the truth and Knights think that the Impostor is a Knight.

Buggy Code:
#include <iostream>
#include <vector>
#include <array>
#include <string>
#include <algorithm>
#include <numeric>
#include <unordered_map>
#include <unordered_set>
#include <queue>
#include <climits>
#include <cmath>
#include <functional>
#include <type_traits>
#include <bitset>

#define int long long
using namespace std;

bool solve() {
    int n;
    cin >> n;

    int i = 1;
    bool found = false;
    while (!found) {
        int a, b;
        cout << "? " << i << " " << i+1 << endl;
        cin >> a;
        if (a == -1)
            return true;
        cout << "? " << i+1 << " " << i << endl;
        cin >> b;
        if (b == -1)
            return true;

        if (a != b)
        {
            found = true;
        }
        i += 2;
    }
    i -= 2;

    int independent = i == 1 ? i+2 : i-1;
    int a, b;
    cout << "? " << i << " " << independent << endl;
    cin >> a;
    if (a == -1)
        return true;
    cout << "? " << independent << " " << i << endl;
    cin >> b;
    if (b == -1)
        return true;
    if (a != b)
    {
        cout << "! " << i << endl;
    }
    else
    {
        cout << "! " << i+1 << endl;
    }

    return false;
}

signed main()
{
    //ios::sync_with_stdio(0);
    //cin.tie(0);

    int t = 1;

    cin >> t;

    for (int i = 0; i < t; i++)
        if (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 buggy code, we need to consider edge cases and scenarios that maximize the failure probability in faulty solutions. The buggy code has a specific logic to identify the Impostor, but it may fail in certain configurations of players and their responses.

### Analysis of the Buggy Code:
1. **Pairwise Comparison**: The code asks questions about pairs of players (i, i+1) and (i+1, i) to find a pair where the responses are different.
2. **Independent Player**: After finding such a pair, it asks questions about an independent player to determine which of the two players is the Impostor.

### Potential Edge Cases:
1. **All Players Are Knights or Knaves**: If all players are either Knights or Knaves, the code might not find a pair where the responses are different.
2. **Impostor at the End**: If the Impostor is at the end of the list, the code might not correctly identify it.
3. **Small Number of Players**: The code might fail with a small number of players due to fewer possible pairs.

### Chosen Test Input:
Let's choose a small-scale test input with 4 players (n = 4) and place the Impostor at the end (player 4). This will test the code's ability to handle a small number of players and the Impostor at the end.

### Test Input:
```plaintext
1
4
```

### Explanation:
- **Test Case 1**: There are 4 players.
- **Player Roles**:
  - Player 1: Knight
  - Player 2: Knight
  - Player 3: Knave
  - Player 4: Impostor

### Expected Interaction:
1. **First Query**: `? 1 2`
   - Response: `1` (Player 1 is a Knight, and Player 2 is a Knight, so the response is "yes").
2. **Second Query**: `? 2 1`
   - Response: `1` (Player 2 is a Knight, and Player 1 is a Knight, so the response is "yes").
3. **Third Query**: `? 3 4`
   - Response: `1` (Player 3 is a Knave, and Player 4 is an Impostor, so the response is "yes" because the Knave lies).
4. **Fourth Query**: `? 4 3`
   - Response: `0` (Player 4 is an Impostor, and Player 3 is a Knave, so the response is "no" because the Impostor lies).

The code should correctly identify Player 4 as the Impostor.

### Output:
```plaintext
1
4
```

This test input should expose the bug in the buggy code by placing the Impostor at the end and ensuring that the code's logic is thoroughly tested.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Asesino (Easy Version)
time_limit_ms: 2500
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version, you can ask at most $$$n+69$$$ questions. You can make hacks only if both versions of the problem are solved.

This is an interactive problem.

It is a tradition in Mexico's national IOI trainings to play the game "Asesino", which is similar to "Among Us" or "Mafia".

Today, $$$n$$$ players, numbered from $$$1$$$ to $$$n$$$, will play "Asesino" with the following three roles:

- Knight: a Knight is someone who always tells the truth.
- Knave: a Knave is someone who always lies.
- Impostor: an Impostor is someone everybody thinks is a Knight, but is secretly a Knave.

Each player will be assigned a role in the game. There will be exactly one Impostor but there can be any (possible zero) number of Knights and Knaves.

As the game moderator, you have accidentally forgotten the roles of everyone, but you need to determine the player who is the Impostor.

To determine the Impostor, you will ask some questions. In each question, you will pick two players $$$i$$$ and $$$j$$$ ($$$1 \leq i, j \leq n$$$; $$$i \neq j$$$) and ask if player $$$i$$$ thinks that player $$$j$$$ is a Knight. The results of the question is shown in the table below.

KnightKnaveImpostorKnightYesNoYesKnaveNoYesNoImpostorNoYes— The response of the cell in row $$$a$$$ and column $$$b$$$ is the result of asking a question when $$$i$$$ has role $$$a$$$ and $$$j$$$ has row $$$b$$$. For example, the "Yes" in the top right cell belongs to row "Knight" and column "Impostor", so it is the response when $$$i$$$ is a Knight and $$$j$$$ is an Impostor.

Find the Impostor in at most $$$n + 69$$$ questions.

Note: the grader is adaptive: the roles of the players are not fixed in the beginning and may change depending on your questions. However, it is guaranteed that there exists an assignment of roles that is consistent with all previously asked questions under the constraints of this problem.

Input: The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$3 \le n \le 10^5$$$) — the number of people playing the game.

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

Output: None

Examples:
input:
2
7

1

0

0

1

1

0

0

1

4

0

1

1

1
output:
? 1 3

? 7 6

? 2 5

? 6 2

? 4 5

? 4 6

? 1 4

? 2 4

! 4

? 1 2

? 2 3

? 3 4

? 4 1

! 3

Note: Note that the example test cases do not represent an optimal strategy for asking questions and are only shown for the sake of demonstrating the interaction format. Specifically, we cannot determine which player is the Impostor from the questions asked in the examples.

In the first test case of the example, players at indices $$$2$$$ and $$$6$$$ are Knights, players at indices $$$1$$$, $$$3$$$, $$$5$$$, and $$$7$$$ are Knaves, and the Impostor is at index $$$4$$$. The following is an explanation of the questions asked:

- In the first query, player $$$i$$$ is a Knave and player $$$j$$$ is a Knave. The answer is "yes" since Knaves always lie.
- In the second query, player $$$i$$$ is a Knave and player $$$j$$$ is a Knight. The answer is "no" since Knaves always lie.
- In the third query, player $$$i$$$ is a Knight and player $$$j$$$ is a Knave. The answer is "no" since Knights always tell the truth.
- In the fourth query, player $$$i$$$ is a Knight and player $$$j$$$ is a Knight. The answer is "yes" since Knights always tell the truth.
- In the fifth query, player $$$i$$$ is a Impostor and player $$$j$$$ is a Knave. The answer is "yes" since the Impostor always lies.
- In the sixth query, player $$$i$$$ is a Impostor and player $$$j$$$ is a Knight. The answer is "no" since the Impostor always lies.
- In the seventh query, player $$$i$$$ is a Knave and player $$$j$$$ is a Impostor. The answer is "no" since Knaves always lie and Knaves thinks that the Impostor is a Knight.
- In the eighth query, player $$$i$$$ is a Knight and player $$$j$$$ is a Impostor. The answer is "yes" since Knights always tell the truth and Knights think that the Impostor is a Knight.

Buggy Code:
#include<iostream>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cmath>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<random>
#include<set>
#include<queue>
#include<unordered_set>
#include<unordered_map>
#include<tuple>
#include <utility>
#include<vector>
#include<random>
#include<string>
#include<complex>
#include<cstdio>
#include<tuple>
//#include<immintrin.h> rehbnm

//#pragma Glst optimize("Ofast")
//#pragma Glst optimize("unroll-loops")
//#pragma Glst target("avx,avx2")

#define girlfriend friend
#define int int_fast64_t
#define ftype complex<double>
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()

typedef long double ld;

template<typename T>
bool assign_min(T &a, T b) {
    if (b < a) {
        a = b;
        return true;
    }
    return false;
}

template<typename T>
bool assign_max(T &a, T b) {
    if (b > a) {
        a = b;
        return true;
    }
    return false;
}

template<typename T>
void pref_calc(std::vector<T> &a, std::vector<T> &b) {
    for (int i = 0; i < (int) a.size(); ++i) {
        b[i] = (i == 0 ? a[i] : b[i - 1] + a[i]);
    }
}

template<typename T1, typename T2>
std::istream &operator>>(std::istream &in, std::pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}

template<typename T1, typename T2>
std::ostream &operator<<(std::ostream &out, std::pair<T1, T2> &p) {
    out << p.first << " " << p.second;
    return out;
}

template<typename T>
void input(std::vector<T> &a, int l = 0, int r = -1) {
    for (int i = l; i <= (r == -1 ? (int) a.size() - 1 : r); ++i) std::cin >> a[i];
}

template<typename T>
void print(std::vector<T> &a, int l = 0, int r = -1) {
    for (int i = l; i <= (r == -1 ? (int) a.size() - 1 : r); ++i) { std::cout << a[i] << " "; }
    std::cout << '\n';
}

template<typename T>
std::istream &operator>>(std::istream &in, std::vector<T> &a) {
    for (int i = 0; i < (int) a.size(); ++i) {
        in >> a[i];
    }
    return in;
}

template<typename T>
std::ostream &operator<<(std::ostream &out, std::vector<T> &a) {
    for (int i = 0; i < (int) a.size(); ++i) {
        out << a[i] << " ";
    }
    return out;
}

using namespace std;
mt19937 rnd(69);
const int inf = (int) 1e18, MAXN = (int) 1e5, MAXM = (int) 1e6 + 10, MAXK = 101, MAXC = (int) 2e5 + 100, MAXQ = (int) 1e6 + 10, MAXSZ =
        (int) 36e6 + 10, ln = 20, mod = 998244353, mod1 = 1e9 + 7, mod2 =
        1e9 + 7, b1 = 31, b2 = 29, g_fft = 31, K = 500;
const long double eps = 1e-15, add_len = 1e6, Cnst = 139.228, PI = acos(-1);

void solve() {
    int n; cin >> n;
    for (int i = 1; i<=n; i+=2) {
        cout << "? " << i << " " << i+1 << endl;
        int val1; cin >> val1;
        cout << "? " << i+1 << " " << i << endl;
        int val2; cin >> val2;
        if (val1 != val2) {
            int j = (i == 1 ? 3 : 1);
            cout << "? " << i << " " << j << endl;
            int val3;
            cin >> val3;
            cout << "? " << i+1 << " " << j << endl;
            int val4; cin >> val4;
            if (val3 == val4) {
                if (val1) {
                    cout << "! " << i << endl;
                    return;
                }
                cout << "! " << i+1 << endl;
                return;
            }
            if (val1) {
                cout << "! " << i+1 << endl;
                return;
            }
            cout << "! " << i << endl;
            return;
        }
    }
    cout << "! " << n << endl;
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int t = 1; 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 provided buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. Here are the steps to consider:

1. **Understand the Buggy Code**:
   - The code asks pairs of questions (i, i+1) and (i+1, i) for each odd index i.
   - It then asks additional questions to determine the Impostor based on the responses.

2. **Identify Potential Bugs**:
   - The code assumes that the responses from (i, i+1) and (i+1, i) will be different if one of them is the Impostor.
   - It also assumes that the third and fourth questions will help determine the Impostor.

3. **Create a Test Case**:
   - We need a small-scale test case with a specific configuration of roles (Knight, Knave, Impostor) that can expose the bug.
   - A good test case would be one where the Impostor is not easily identifiable by the initial pairs of questions.

4. **Test Case Configuration**:
   - Let's consider a small number of players, say 5.
   - Assign roles such that the initial pairs of questions do not immediately reveal the Impostor.

Here is a small-scale test input that can expose the bug:

```plaintext
1
5
0
0
0
0
0
0
0
0
0
0
```

**Explanation**:
- The test case has 5 players.
- The responses are all 0, which means "No" for every question.
- This configuration can confuse the code because it will not immediately see a difference in the responses from the initial pairs of questions (i, i+1) and (i+1, i).

The code might fail to identify the Impostor correctly in this scenario because it relies on the assumption that the responses will differ if one of the players is the Impostor. In this test case, the responses are all the same, making it difficult for the code to determine the Impostor.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Asesino (Easy Version)
time_limit_ms: 2500
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version, you can ask at most $$$n+69$$$ questions. You can make hacks only if both versions of the problem are solved.

This is an interactive problem.

It is a tradition in Mexico's national IOI trainings to play the game "Asesino", which is similar to "Among Us" or "Mafia".

Today, $$$n$$$ players, numbered from $$$1$$$ to $$$n$$$, will play "Asesino" with the following three roles:

- Knight: a Knight is someone who always tells the truth.
- Knave: a Knave is someone who always lies.
- Impostor: an Impostor is someone everybody thinks is a Knight, but is secretly a Knave.

Each player will be assigned a role in the game. There will be exactly one Impostor but there can be any (possible zero) number of Knights and Knaves.

As the game moderator, you have accidentally forgotten the roles of everyone, but you need to determine the player who is the Impostor.

To determine the Impostor, you will ask some questions. In each question, you will pick two players $$$i$$$ and $$$j$$$ ($$$1 \leq i, j \leq n$$$; $$$i \neq j$$$) and ask if player $$$i$$$ thinks that player $$$j$$$ is a Knight. The results of the question is shown in the table below.

KnightKnaveImpostorKnightYesNoYesKnaveNoYesNoImpostorNoYes— The response of the cell in row $$$a$$$ and column $$$b$$$ is the result of asking a question when $$$i$$$ has role $$$a$$$ and $$$j$$$ has row $$$b$$$. For example, the "Yes" in the top right cell belongs to row "Knight" and column "Impostor", so it is the response when $$$i$$$ is a Knight and $$$j$$$ is an Impostor.

Find the Impostor in at most $$$n + 69$$$ questions.

Note: the grader is adaptive: the roles of the players are not fixed in the beginning and may change depending on your questions. However, it is guaranteed that there exists an assignment of roles that is consistent with all previously asked questions under the constraints of this problem.

Input: The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$3 \le n \le 10^5$$$) — the number of people playing the game.

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

Output: None

Examples:
input:
2
7

1

0

0

1

1

0

0

1

4

0

1

1

1
output:
? 1 3

? 7 6

? 2 5

? 6 2

? 4 5

? 4 6

? 1 4

? 2 4

! 4

? 1 2

? 2 3

? 3 4

? 4 1

! 3

Note: Note that the example test cases do not represent an optimal strategy for asking questions and are only shown for the sake of demonstrating the interaction format. Specifically, we cannot determine which player is the Impostor from the questions asked in the examples.

In the first test case of the example, players at indices $$$2$$$ and $$$6$$$ are Knights, players at indices $$$1$$$, $$$3$$$, $$$5$$$, and $$$7$$$ are Knaves, and the Impostor is at index $$$4$$$. The following is an explanation of the questions asked:

- In the first query, player $$$i$$$ is a Knave and player $$$j$$$ is a Knave. The answer is "yes" since Knaves always lie.
- In the second query, player $$$i$$$ is a Knave and player $$$j$$$ is a Knight. The answer is "no" since Knaves always lie.
- In the third query, player $$$i$$$ is a Knight and player $$$j$$$ is a Knave. The answer is "no" since Knights always tell the truth.
- In the fourth query, player $$$i$$$ is a Knight and player $$$j$$$ is a Knight. The answer is "yes" since Knights always tell the truth.
- In the fifth query, player $$$i$$$ is a Impostor and player $$$j$$$ is a Knave. The answer is "yes" since the Impostor always lies.
- In the sixth query, player $$$i$$$ is a Impostor and player $$$j$$$ is a Knight. The answer is "no" since the Impostor always lies.
- In the seventh query, player $$$i$$$ is a Knave and player $$$j$$$ is a Impostor. The answer is "no" since Knaves always lie and Knaves thinks that the Impostor is a Knight.
- In the eighth query, player $$$i$$$ is a Knight and player $$$j$$$ is a Impostor. The answer is "yes" since Knights always tell the truth and Knights think that the Impostor is a Knight.

Buggy Code:
import sys,math,random
from heapq import heappush,heappop
from bisect import bisect_right,bisect_left
from collections import Counter,deque,defaultdict
from itertools import permutations

# functions #
MOD = 998244353
MOD = 10**9 + 7
RANDOM = random.randrange(2**62)
def gcd(a,b):
    if a%b==0:
        return b
    else:
        return gcd(b,a%b)
def lcm(a,b):
    return a//gcd(a,b)*b
def w(x):
    return x ^ RANDOM
##

#String hashing : sh, fenwick sortedlist : fsortl, Number : numtheory, SparseTable : SparseTable
#bucket sorted list : bsortl, segment tree(lazy propogation) : SegmentTree, bootstrap : bootstrap
#binary indexed tree : BIT, segment tree(point updates) : SegmentPoint, Convex Hull : hull
#Combinatorics : pnc, Diophantine Equations : dpheq
#Template : https://github.com/OmAmar106/Template-for-Competetive-Programming

def solve():
    n = int(sys.stdin.readline().strip())
    on = n
    for i in range(0,n,2):
        print("? ",i+1,i+2,flush = True)
        n = int(input())
        print("? ",i+2,i+1,flush = True)
        n1 = int(input())
        if n!=n1:
            if i+3<=on:
                print("? ",i+1,i+3,flush = True)
                n = int(input())
                print("? ",i+3,i+1,flush = True)
                n1 = int(input())
                if n!=n1:
                    print("! ",i+1,flush=True)
                    return
                else:
                    print("! ",i+2,flush=True)
                    return
            else:
                print("? ",i,i+1,flush = True)
                n = int(input())
                print("? ",i+1,i,flush = True)
                n1 = int(input())
                if n!=n1:
                    print("! ",i+1,flush=True)
                    return
                else:
                    print("! ",i+2,flush=True)
                    return
    print("! ",on)
    return
    #L1 = list(map(int, sys.stdin.readline().split()))
    #st = sys.stdin.readline().strip()
for _ in range(int(sys.stdin.readline().strip())):
    solve()

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

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

Think step by step.

Response:
```plaintext
2
7

1

0

0

1

1

0

0

1

4

0

1

1

1
```

This test input is designed to challenge the buggy code by providing a scenario where the roles of the players are such that the current implementation of the code may fail to correctly identify the Impostor. The input includes two test cases with 7 and 4 players respectively, and the responses are carefully chosen to simulate a mix of Knights, Knaves, and the Impostor, which should expose any logical flaws in the code's questioning strategy.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Asesino (Easy Version)
time_limit_ms: 2500
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version, you can ask at most $$$n+69$$$ questions. You can make hacks only if both versions of the problem are solved.

This is an interactive problem.

It is a tradition in Mexico's national IOI trainings to play the game "Asesino", which is similar to "Among Us" or "Mafia".

Today, $$$n$$$ players, numbered from $$$1$$$ to $$$n$$$, will play "Asesino" with the following three roles:

- Knight: a Knight is someone who always tells the truth.
- Knave: a Knave is someone who always lies.
- Impostor: an Impostor is someone everybody thinks is a Knight, but is secretly a Knave.

Each player will be assigned a role in the game. There will be exactly one Impostor but there can be any (possible zero) number of Knights and Knaves.

As the game moderator, you have accidentally forgotten the roles of everyone, but you need to determine the player who is the Impostor.

To determine the Impostor, you will ask some questions. In each question, you will pick two players $$$i$$$ and $$$j$$$ ($$$1 \leq i, j \leq n$$$; $$$i \neq j$$$) and ask if player $$$i$$$ thinks that player $$$j$$$ is a Knight. The results of the question is shown in the table below.

KnightKnaveImpostorKnightYesNoYesKnaveNoYesNoImpostorNoYes— The response of the cell in row $$$a$$$ and column $$$b$$$ is the result of asking a question when $$$i$$$ has role $$$a$$$ and $$$j$$$ has row $$$b$$$. For example, the "Yes" in the top right cell belongs to row "Knight" and column "Impostor", so it is the response when $$$i$$$ is a Knight and $$$j$$$ is an Impostor.

Find the Impostor in at most $$$n + 69$$$ questions.

Note: the grader is adaptive: the roles of the players are not fixed in the beginning and may change depending on your questions. However, it is guaranteed that there exists an assignment of roles that is consistent with all previously asked questions under the constraints of this problem.

Input: The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$3 \le n \le 10^5$$$) — the number of people playing the game.

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

Output: None

Examples:
input:
2
7

1

0

0

1

1

0

0

1

4

0

1

1

1
output:
? 1 3

? 7 6

? 2 5

? 6 2

? 4 5

? 4 6

? 1 4

? 2 4

! 4

? 1 2

? 2 3

? 3 4

? 4 1

! 3

Note: Note that the example test cases do not represent an optimal strategy for asking questions and are only shown for the sake of demonstrating the interaction format. Specifically, we cannot determine which player is the Impostor from the questions asked in the examples.

In the first test case of the example, players at indices $$$2$$$ and $$$6$$$ are Knights, players at indices $$$1$$$, $$$3$$$, $$$5$$$, and $$$7$$$ are Knaves, and the Impostor is at index $$$4$$$. The following is an explanation of the questions asked:

- In the first query, player $$$i$$$ is a Knave and player $$$j$$$ is a Knave. The answer is "yes" since Knaves always lie.
- In the second query, player $$$i$$$ is a Knave and player $$$j$$$ is a Knight. The answer is "no" since Knaves always lie.
- In the third query, player $$$i$$$ is a Knight and player $$$j$$$ is a Knave. The answer is "no" since Knights always tell the truth.
- In the fourth query, player $$$i$$$ is a Knight and player $$$j$$$ is a Knight. The answer is "yes" since Knights always tell the truth.
- In the fifth query, player $$$i$$$ is a Impostor and player $$$j$$$ is a Knave. The answer is "yes" since the Impostor always lies.
- In the sixth query, player $$$i$$$ is a Impostor and player $$$j$$$ is a Knight. The answer is "no" since the Impostor always lies.
- In the seventh query, player $$$i$$$ is a Knave and player $$$j$$$ is a Impostor. The answer is "no" since Knaves always lie and Knaves thinks that the Impostor is a Knight.
- In the eighth query, player $$$i$$$ is a Knight and player $$$j$$$ is a Impostor. The answer is "yes" since Knights always tell the truth and Knights think that the Impostor is a Knight.

Buggy Code:
import sys
input = sys.stdin.readline
for _ in range(int(input())):
    n = int(input())

    sus = [x+1 for x in range(n%3)]
    for i in range(n%3,n,3):
        a = 0
        print("?",i+3,i+1,flush=1)
        a += int(input())
        print("?",i+2,i+1,flush=1)
        a += int(input())
        print("?",i+2,i+3,flush=1)
        a += int(input())
        if a == 2 or a == 0:
            sus = [i+1,i+2,i+3]
            break

    while len(sus) < 3:
        for x in range(1,n):
            if x not in sus:
                sus.append(x)
                break

    print("?",sus[0],sus[1],flush=1)
    a12 = int(input())
    print("?",sus[1],sus[0],flush=1)
    a21 = int(input())
    print("?",sus[0],sus[2],flush=1)
    a13 = int(input())
    print("?",sus[2],sus[0],flush=1)
    a31 = int(input())
    print("?",sus[1],sus[2],flush=1)
    a23 = int(input())
    print("?",sus[2],sus[1],flush=1)
    a32 = int(input())

    if (a12 != a21 and a23 != a32):
        print("!", sus[1],flush=1)
        continue
    elif (a12 != a21 and a13 != a31):
        print("!", sus[0],flush=1)
        continue
    elif a23 != a32 and a13 != a31:
        print("!",sus[2],flush=1)
        continue
    print("!",1)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code has several issues, including incorrect handling of the groups and the logic for determining the Impostor.

Here are the key points to consider:
1. **Edge Case with Small `n`**: The code might fail when `n` is small, such as 3 or 4.
2. **Edge Case with Specific Grouping**: The code groups players in sets of 3 and processes them. We need to test with a number of players that is not a multiple of 3.
3. **Edge Case with Specific Roles**: The code might fail if the roles are arranged in a way that the initial assumptions do not hold.

Let's create a test input with `n = 4` and a specific arrangement of roles that should expose the bug in the code.

### Test Input
```plaintext
1
4
0
1
1
1
```

### Explanation
- **Number of Test Cases**: 1
- **Number of Players**: 4
- **Role Assignments**:
  - Player 1: Knave (0)
  - Player 2: Impostor (1)
  - Player 3: Knight (1)
  - Player 4: Knight (1)

### Expected Interaction
- **Query 1**: `? 1 2` (Knave asks about Impostor) -> Expected Response: `1` (Knave lies)
- **Query 2**: `? 2 3` (Impostor asks about Knight) -> Expected Response: `1` (Impostor lies)
- **Query 3**: `? 3 4` (Knight asks about Knight) -> Expected Response: `1` (Knight tells the truth)
- **Query 4**: `? 4 1` (Knight asks about Knave) -> Expected Response: `0` (Knight tells the truth)

### Expected Output
- The correct output should be `! 2` (Player 2 is the Impostor).

This test input should expose the bug in the code by creating a scenario where the code's assumptions about the groupings and role assignments are incorrect.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Asesino (Easy Version)
time_limit_ms: 2500
memory_limit_mb: 256
Description: This is the easy version of the problem. In this version, you can ask at most $$$n+69$$$ questions. You can make hacks only if both versions of the problem are solved.

This is an interactive problem.

It is a tradition in Mexico's national IOI trainings to play the game "Asesino", which is similar to "Among Us" or "Mafia".

Today, $$$n$$$ players, numbered from $$$1$$$ to $$$n$$$, will play "Asesino" with the following three roles:

- Knight: a Knight is someone who always tells the truth.
- Knave: a Knave is someone who always lies.
- Impostor: an Impostor is someone everybody thinks is a Knight, but is secretly a Knave.

Each player will be assigned a role in the game. There will be exactly one Impostor but there can be any (possible zero) number of Knights and Knaves.

As the game moderator, you have accidentally forgotten the roles of everyone, but you need to determine the player who is the Impostor.

To determine the Impostor, you will ask some questions. In each question, you will pick two players $$$i$$$ and $$$j$$$ ($$$1 \leq i, j \leq n$$$; $$$i \neq j$$$) and ask if player $$$i$$$ thinks that player $$$j$$$ is a Knight. The results of the question is shown in the table below.

KnightKnaveImpostorKnightYesNoYesKnaveNoYesNoImpostorNoYes— The response of the cell in row $$$a$$$ and column $$$b$$$ is the result of asking a question when $$$i$$$ has role $$$a$$$ and $$$j$$$ has row $$$b$$$. For example, the "Yes" in the top right cell belongs to row "Knight" and column "Impostor", so it is the response when $$$i$$$ is a Knight and $$$j$$$ is an Impostor.

Find the Impostor in at most $$$n + 69$$$ questions.

Note: the grader is adaptive: the roles of the players are not fixed in the beginning and may change depending on your questions. However, it is guaranteed that there exists an assignment of roles that is consistent with all previously asked questions under the constraints of this problem.

Input: The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$3 \le n \le 10^5$$$) — the number of people playing the game.

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

Output: None

Examples:
input:
2
7

1

0

0

1

1

0

0

1

4

0

1

1

1
output:
? 1 3

? 7 6

? 2 5

? 6 2

? 4 5

? 4 6

? 1 4

? 2 4

! 4

? 1 2

? 2 3

? 3 4

? 4 1

! 3

Note: Note that the example test cases do not represent an optimal strategy for asking questions and are only shown for the sake of demonstrating the interaction format. Specifically, we cannot determine which player is the Impostor from the questions asked in the examples.

In the first test case of the example, players at indices $$$2$$$ and $$$6$$$ are Knights, players at indices $$$1$$$, $$$3$$$, $$$5$$$, and $$$7$$$ are Knaves, and the Impostor is at index $$$4$$$. The following is an explanation of the questions asked:

- In the first query, player $$$i$$$ is a Knave and player $$$j$$$ is a Knave. The answer is "yes" since Knaves always lie.
- In the second query, player $$$i$$$ is a Knave and player $$$j$$$ is a Knight. The answer is "no" since Knaves always lie.
- In the third query, player $$$i$$$ is a Knight and player $$$j$$$ is a Knave. The answer is "no" since Knights always tell the truth.
- In the fourth query, player $$$i$$$ is a Knight and player $$$j$$$ is a Knight. The answer is "yes" since Knights always tell the truth.
- In the fifth query, player $$$i$$$ is a Impostor and player $$$j$$$ is a Knave. The answer is "yes" since the Impostor always lies.
- In the sixth query, player $$$i$$$ is a Impostor and player $$$j$$$ is a Knight. The answer is "no" since the Impostor always lies.
- In the seventh query, player $$$i$$$ is a Knave and player $$$j$$$ is a Impostor. The answer is "no" since Knaves always lie and Knaves thinks that the Impostor is a Knight.
- In the eighth query, player $$$i$$$ is a Knight and player $$$j$$$ is a Impostor. The answer is "yes" since Knights always tell the truth and Knights think that the Impostor is a Knight.

Buggy Code:
import sys
input = sys.stdin.readline
for _ in range(int(input())):
    n = int(input())

    sus = [x+1 for x in range(n%3)]
    for i in range(n%3,n,3):
        a = 0
        print("?",i+3,i+1,flush=1)
        a += int(input())
        print("?",i+2,i+1,flush=1)
        a += int(input())
        print("?",i+2,i+3,flush=1)
        a += int(input())
        if a == 2 or a == 0:
            sus = [i+1,i+2,i+3]
            break
    temp = set(sus)
    if len(sus) == 1:
        sus.append(2)
    if len(sus) == 2:
        sus.append(3)

    print("?",sus[0],sus[1],flush=1)
    a12 = int(input())
    print("?",sus[1],sus[0],flush=1)
    a21 = int(input())
    print("?",sus[0],sus[2],flush=1)
    a13 = int(input())
    print("?",sus[2],sus[0],flush=1)
    a31 = int(input())
    print("?",sus[1],sus[2],flush=1)
    a23 = int(input())
    print("?",sus[2],sus[1],flush=1)
    a32 = int(input())

    if (a12 != a21 and a23 != a32):
        print("!", sus[1],flush=1)
        continue
    elif (a12 != a21 and a13 != a31):
        print("!", sus[0],flush=1)
        continue
    elif a23 != a32 and a13 != a31:
        print("!",sus[2],flush=1)
        continue
    print("!",1,flush =1)

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

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

Think step by step.

Response:
```plaintext
2
3
4
```