**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Satay Sticks
time_limit_ms: 1500
memory_limit_mb: 256
Description: Penchick and his friend Kohane are touring Indonesia, and their next stop is in Surabaya!

In the bustling food stalls of Surabaya, Kohane bought $$$n$$$ satay sticks and arranged them in a line, with the $$$i$$$-th satay stick having length $$$p_i$$$. It is given that $$$p$$$ is a permutation$$$^{\text{∗}}$$$ of length $$$n$$$.

Penchick wants to sort the satay sticks in increasing order of length, so that $$$p_i=i$$$ for each $$$1\le i\le n$$$. For fun, they created a rule: they can only swap neighboring satay sticks whose lengths differ by exactly $$$1$$$. Formally, they can perform the following operation any number of times (including zero):

- Select an index $$$i$$$ ($$$1\le i\le n-1$$$) such that $$$|p_{i+1}-p_i|=1$$$;
- Swap $$$p_i$$$ and $$$p_{i+1}$$$.

Determine whether it is possible to sort the permutation $$$p$$$, thus the satay sticks, by performing the above operation.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 2\cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of satay sticks.

The second line of each test case contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$) — the permutation $$$p$$$ representing the length of the satay sticks.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot 10^5$$$.

Output: For each test case, output "YES" if it is possible to sort permutation $$$p$$$ by performing the operation. Otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
2
4
2 1 3 4
4
4 2 3 1
output:
YES
NO

Note: In the first test case, we can sort permutation $$$p = [2, 1, 3, 4]$$$ by performing an operation on index $$$1$$$ ($$$|p_2 - p_1| = |1 - 2| = 1$$$), resulting in $$$p = [1, 2, 3, 4]$$$.

In the second test case, it can be proven that it is impossible to sort permutation $$$p = [4, 2, 3, 1]$$$ by performing the operation. Here is an example of a sequence of operations that can be performed on the permutation:

- Select $$$i = 2$$$ ($$$|p_3 - p_2| = |3 - 2| = 1$$$). This results in $$$p = [4, 3, 2, 1]$$$.
- Select $$$i = 1$$$ ($$$|p_2 - p_1| = |3 - 4| = 1$$$). This results in $$$p = [3, 4, 2, 1]$$$.
- Select $$$i = 3$$$ ($$$|p_4 - p_3| = |1 - 2| = 1$$$). This results in $$$p = [3, 4, 1, 2]$$$.

Unfortunately, permutation $$$p$$$ remains unsorted after performing the operations.

Buggy Code:
#include "bits/stdc++.h"
#define fastio std::ios_base::sync_with_stdio(0); std::cin.tie(0); std::cout.tie(0);
using namespace std;

const int ms = 500000;

int father[ms];
set<int> sets[ms];

int find(int x) {
    if(x == father[x]) return x;
    return father[x] = find(father[x]);
}

void join(int x, int y) {
    x = find(x);
    y = find(y);
    if(x == y) return;

    if(sets[x].size() > sets[y].size()) {
        swap(x, y);
    }

    father[x] = y;
    for(int w : sets[x])
        sets[y].insert(x);
}

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

    for(int i = 0; i < n; ++i) {
        father[i] = i;
        sets[i].clear();
        sets[i].insert(i);
    }

    vector<int> arr(n);
    for(int &w : arr) cin >> w, w--;

    for(int i = 1; i < n; ++i) {
        if(abs(arr[i] - arr[i - 1]) == 1) {
            join(arr[i], arr[i - 1]);
        }
    }

    set<int> toSkip;

    vector<int> v;

    for(int i = 0; i < n; ++i) {
        int l = find(arr[i]);
        if(toSkip.count(l)) continue;

        // cout << "entrou\n";
        for(auto it = sets[l].begin(); it != sets[l].end(); ++it){
            v.push_back(*it);
            // cout << *it << ' ';
        }
        // cout << endl;
        toSkip.insert(l);
    }

    // cout << endl;

    // for(int w : v)
    //     cout << w << ' ';
    // cout << endl;

    if(is_sorted(v.begin(), v.end()))
        cout << "YES\n";
    else
        cout << "NO\n";
}

int32_t main(){
    fastio;

    int t;
    cin >> t;
    for(int i = 1; i <= t; ++i) {
        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 or scenarios where the code might fail. The key points to consider are:

1. **Permutations with large gaps**: The code might fail if there are large gaps between elements that cannot be bridged by swaps of adjacent elements differing by 1.
2. **Edge cases with small permutations**: Small permutations can sometimes reveal issues that are not apparent in larger inputs.
3. **Special cases**: Permutations where the elements are almost sorted but have a few elements out of place.

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

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

### Explanation:
1. **First Test Case: `1 3 2 4 5`**
   - This permutation is almost sorted but has a small gap between `1` and `3`.
   - The elements `3` and `2` can be swapped to sort the permutation.
   - Expected output: `YES`

2. **Second Test Case: `5 4 3 2 1`**
   - This permutation is the reverse of the sorted order.
   - Each pair of adjacent elements differs by 1, but the overall permutation cannot be sorted by swapping adjacent elements.
   - Expected output: `NO`

This test input is small and valid, and it should expose the bug in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Satay Sticks
time_limit_ms: 1500
memory_limit_mb: 256
Description: Penchick and his friend Kohane are touring Indonesia, and their next stop is in Surabaya!

In the bustling food stalls of Surabaya, Kohane bought $$$n$$$ satay sticks and arranged them in a line, with the $$$i$$$-th satay stick having length $$$p_i$$$. It is given that $$$p$$$ is a permutation$$$^{\text{∗}}$$$ of length $$$n$$$.

Penchick wants to sort the satay sticks in increasing order of length, so that $$$p_i=i$$$ for each $$$1\le i\le n$$$. For fun, they created a rule: they can only swap neighboring satay sticks whose lengths differ by exactly $$$1$$$. Formally, they can perform the following operation any number of times (including zero):

- Select an index $$$i$$$ ($$$1\le i\le n-1$$$) such that $$$|p_{i+1}-p_i|=1$$$;
- Swap $$$p_i$$$ and $$$p_{i+1}$$$.

Determine whether it is possible to sort the permutation $$$p$$$, thus the satay sticks, by performing the above operation.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 2\cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of satay sticks.

The second line of each test case contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$) — the permutation $$$p$$$ representing the length of the satay sticks.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot 10^5$$$.

Output: For each test case, output "YES" if it is possible to sort permutation $$$p$$$ by performing the operation. Otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
2
4
2 1 3 4
4
4 2 3 1
output:
YES
NO

Note: In the first test case, we can sort permutation $$$p = [2, 1, 3, 4]$$$ by performing an operation on index $$$1$$$ ($$$|p_2 - p_1| = |1 - 2| = 1$$$), resulting in $$$p = [1, 2, 3, 4]$$$.

In the second test case, it can be proven that it is impossible to sort permutation $$$p = [4, 2, 3, 1]$$$ by performing the operation. Here is an example of a sequence of operations that can be performed on the permutation:

- Select $$$i = 2$$$ ($$$|p_3 - p_2| = |3 - 2| = 1$$$). This results in $$$p = [4, 3, 2, 1]$$$.
- Select $$$i = 1$$$ ($$$|p_2 - p_1| = |3 - 4| = 1$$$). This results in $$$p = [3, 4, 2, 1]$$$.
- Select $$$i = 3$$$ ($$$|p_4 - p_3| = |1 - 2| = 1$$$). This results in $$$p = [3, 4, 1, 2]$$$.

Unfortunately, permutation $$$p$$$ remains unsorted after performing the operations.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
void solve();
#define py cout<<"Yes\n"
#define pn cout<<"No\n"
#define ll long long
// int solve(int a,vector<int> b,int n);
#define debugv(a)for(auto it:a)cout<<it<<" ";cout<<endl;
#define debugvp(a)for(auto it:a)cout<<it.first<<" "<<it.second<<endl;
#define get(v)for(auto &it:v)cin>>it;
#define fori(i,a,b) for(ll i=a;i<b;i++)
#define vll vector<ll>
#define mll map<ll,ll>
#define pll pair<ll,ll>
#define vpll vector<pair<ll,ll>>
#define sll set<ll>
#define all(a) a.begin(),a.end()
#define f first
#define sc second
#define pb push_back
#define endl "\n"
const ll mod = 1e9 + 7;
// #define
vll p(1e5+1,0);
ll solve(ll n,ll m){
	return n/m+((n%m) != 0);
}

long long power(long long A, long long B)
{
    if (B == 0)
        return 1;

    long long res = power(A, B / 2);

    if (B % 2)
        return res * res * A;
    else
        return res * res;
}
ll cost=1e9+37;
bool pm(ll n){
	if(n<=1)
		return false;
	for(ll i=2;i*i<=n;i++){
		if(n%i==0)
			return false;
	}
	return true;
}

ll upper(ll a,ll b){
	return ((a/b)+((a%b)!=0));
}
ll bin(ll a,ll b){
	 if (b == 0)
        return 1;
    long long res = bin(a, b / 2);
    if (b % 2)
        return res * res * a;
    else
        return res * res;
}
#define MAXN 10000001
vector<ll> spf(MAXN + 1, 1);

// Calculating SPF (Smallest Prime Factor) for every
// number till MAXN.
// Time Complexity : O(nloglogn)
void sieve()
{
    // stores smallest prime factor for every number

    spf[0] = 0;
    for (ll i = 2; i <= MAXN; i++) {
        if (spf[i] == 1) { // if the number is prime ,mark
                           // all its multiples who havent
                           // gotten their spf yet
            for (int j = i; j <= MAXN; j += i) {
                if (spf[j]== 1) // if its smallest prime factor is
                          // 1 means its spf hasnt been
                          // found yet so change it to i
                    spf[j] = i;
            }
        }
    }
}
ll loga_to_baseB(ll a,ll b){
	return (log2(a)/log2(b));
}
ll lar(ll n,ll k){
	ll res=1;
	while(res*k<=n)
		res*=k;
	return res;
}
bool ok(ll mid,ll k,ll sm){
	ll tm=upper(sm,mid);
	return tm<=k;
}
double dis(ll x1,ll y1, ll x2,ll y2){
	return sqrt(((x2-x1)*(x2-x1)) + ((y2-y1)*(y2-y1)));
}
bool cmp(pair<int,int> a,pair<int,int> b){
	if(max(a.f,a.sc) != max(b.f,b.sc))
	return max(a.f,a.sc)<max(b.f,b.sc);
	return min(a.f,a.sc)<min(b.f,b.sc);
}
bool isp(string &s,string &t){
	reverse(all(t));
	return (s==t);
}
void s(){
	ll n;
	cin>>n;
	vll a(n);
	get(a);
	// if(a[0]!=1 && a[1] !=1)
	// {
	// 	pn;
	// 	return;
	// }
	for(ll i=0;i<n-1;i++){
		if(a[i] !=(i+1))
		{	if(i==0){
				if(a[i+1]!=(i+1))
				{
					pn;
					return;
				}
				else
					{swap(a[i],a[i+1]);}
				}
			else{
					if(a[i+1]!=(i+1) && (a[i-1] != (i+1))){
						pn;
						return;
					}
					else
						{
							if(a[i+1]== (i+1))
								swap(a[i],a[i+1]);
							else
								swap(a[i],a[i-1]);
						}
			}
		}
	}
	// debugv(a);
	py;
	}
int main()
{
	// sieve();
	ll t;
	cin>>t;
	// t=1;
	while(t--)
		s();

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Satay Sticks
time_limit_ms: 1500
memory_limit_mb: 256
Description: Penchick and his friend Kohane are touring Indonesia, and their next stop is in Surabaya!

In the bustling food stalls of Surabaya, Kohane bought $$$n$$$ satay sticks and arranged them in a line, with the $$$i$$$-th satay stick having length $$$p_i$$$. It is given that $$$p$$$ is a permutation$$$^{\text{∗}}$$$ of length $$$n$$$.

Penchick wants to sort the satay sticks in increasing order of length, so that $$$p_i=i$$$ for each $$$1\le i\le n$$$. For fun, they created a rule: they can only swap neighboring satay sticks whose lengths differ by exactly $$$1$$$. Formally, they can perform the following operation any number of times (including zero):

- Select an index $$$i$$$ ($$$1\le i\le n-1$$$) such that $$$|p_{i+1}-p_i|=1$$$;
- Swap $$$p_i$$$ and $$$p_{i+1}$$$.

Determine whether it is possible to sort the permutation $$$p$$$, thus the satay sticks, by performing the above operation.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 2\cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of satay sticks.

The second line of each test case contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$) — the permutation $$$p$$$ representing the length of the satay sticks.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot 10^5$$$.

Output: For each test case, output "YES" if it is possible to sort permutation $$$p$$$ by performing the operation. Otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
2
4
2 1 3 4
4
4 2 3 1
output:
YES
NO

Note: In the first test case, we can sort permutation $$$p = [2, 1, 3, 4]$$$ by performing an operation on index $$$1$$$ ($$$|p_2 - p_1| = |1 - 2| = 1$$$), resulting in $$$p = [1, 2, 3, 4]$$$.

In the second test case, it can be proven that it is impossible to sort permutation $$$p = [4, 2, 3, 1]$$$ by performing the operation. Here is an example of a sequence of operations that can be performed on the permutation:

- Select $$$i = 2$$$ ($$$|p_3 - p_2| = |3 - 2| = 1$$$). This results in $$$p = [4, 3, 2, 1]$$$.
- Select $$$i = 1$$$ ($$$|p_2 - p_1| = |3 - 4| = 1$$$). This results in $$$p = [3, 4, 2, 1]$$$.
- Select $$$i = 3$$$ ($$$|p_4 - p_3| = |1 - 2| = 1$$$). This results in $$$p = [3, 4, 1, 2]$$$.

Unfortunately, permutation $$$p$$$ remains unsorted after performing the operations.

Buggy Code:
for _ in range(int(input())):

    n=int(input())
    L=list(map(int,input().split()))
    S=L.copy()

    P=[]

    ans=True

    for i in range(n-1):
        if L[i]==i+1 and L[i+1]==i+2:
            continue
        elif L[i]==i+2 and L[i+1]==i+1:
            L[i],L[i+1]=L[i+1],L[i]
        else:
            ans=False
            print("NO")
            break

    if ans :
        print("YES")

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Satay Sticks
time_limit_ms: 1500
memory_limit_mb: 256
Description: Penchick and his friend Kohane are touring Indonesia, and their next stop is in Surabaya!

In the bustling food stalls of Surabaya, Kohane bought $$$n$$$ satay sticks and arranged them in a line, with the $$$i$$$-th satay stick having length $$$p_i$$$. It is given that $$$p$$$ is a permutation$$$^{\text{∗}}$$$ of length $$$n$$$.

Penchick wants to sort the satay sticks in increasing order of length, so that $$$p_i=i$$$ for each $$$1\le i\le n$$$. For fun, they created a rule: they can only swap neighboring satay sticks whose lengths differ by exactly $$$1$$$. Formally, they can perform the following operation any number of times (including zero):

- Select an index $$$i$$$ ($$$1\le i\le n-1$$$) such that $$$|p_{i+1}-p_i|=1$$$;
- Swap $$$p_i$$$ and $$$p_{i+1}$$$.

Determine whether it is possible to sort the permutation $$$p$$$, thus the satay sticks, by performing the above operation.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 2\cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of satay sticks.

The second line of each test case contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$) — the permutation $$$p$$$ representing the length of the satay sticks.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot 10^5$$$.

Output: For each test case, output "YES" if it is possible to sort permutation $$$p$$$ by performing the operation. Otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
2
4
2 1 3 4
4
4 2 3 1
output:
YES
NO

Note: In the first test case, we can sort permutation $$$p = [2, 1, 3, 4]$$$ by performing an operation on index $$$1$$$ ($$$|p_2 - p_1| = |1 - 2| = 1$$$), resulting in $$$p = [1, 2, 3, 4]$$$.

In the second test case, it can be proven that it is impossible to sort permutation $$$p = [4, 2, 3, 1]$$$ by performing the operation. Here is an example of a sequence of operations that can be performed on the permutation:

- Select $$$i = 2$$$ ($$$|p_3 - p_2| = |3 - 2| = 1$$$). This results in $$$p = [4, 3, 2, 1]$$$.
- Select $$$i = 1$$$ ($$$|p_2 - p_1| = |3 - 4| = 1$$$). This results in $$$p = [3, 4, 2, 1]$$$.
- Select $$$i = 3$$$ ($$$|p_4 - p_3| = |1 - 2| = 1$$$). This results in $$$p = [3, 4, 1, 2]$$$.

Unfortunately, permutation $$$p$$$ remains unsorted after performing the operations.

Buggy Code:
import sys, math
from collections import *
from bisect import *
from types import GeneratorType
from random import randrange

MOD = 1000000007
# MOD = 998244353

RANDOM = randrange(2**62)

def wrap(x):
    return x ^ RANDOM

def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
    return wrappedfunc

def linp():
	return list(map(int, sys.stdin.readline().split()))

def minp():
	return map(int, sys.stdin.readline().split())

def sinp():
	return sys.stdin.readline().rstrip('\n');

def inp():
	return int(sys.stdin.readline())

for t in range(inp()):
	n = inp()
	p = linp()
	possible = True
	for i in range(n):
		if p[i] - i > 2:
			possible = False
			break
	print("YES" if possible else "NO")

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on the logic that checks if the permutation can be sorted by swapping neighboring elements that differ by exactly 1. The current logic in the code is:

```python
for i in range(n):
    if p[i] - i > 2:
        possible = False
        break
```

This logic is incorrect because it checks if the difference between the value at position `i` and `i` is greater than 2. This is not the correct condition to determine if the permutation can be sorted by the allowed swaps. Instead, we need to check if the permutation can be sorted by ensuring that no two elements are more than 1 position apart when they should be.

To create a test input that exposes this bug, we can use a permutation where the difference between the value and its position is greater than 2, but the permutation can still be sorted by the allowed swaps. A good example is a permutation where elements are already in a nearly sorted order, but with a few elements out of place.

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

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

### Explanation:
1. **Test Case 1:**
   - Permutation: `[1, 3, 2, 4, 5]`
   - The elements are almost in order, but `3` and `2` are swapped. This can be fixed by swapping `3` and `2`, which are neighbors and differ by 1. The final sorted permutation would be `[1, 2, 3, 4, 5]`.
   - The current logic in the code will incorrectly determine that this permutation cannot be sorted because `p[1] - 1 = 3 - 1 = 2`, which is not greater than 2.

2. **Test Case 2:**
   - Permutation: `[2, 1, 4, 3, 5]`
   - The elements `2` and `1` can be swapped, and `4` and `3` can be swapped, resulting in the sorted permutation `[1, 2, 3, 4, 5]`.
   - The current logic in the code will incorrectly determine that this permutation cannot be sorted because `p[1] - 1 = 2 - 1 = 1`, which is not greater than 2.

This test input should help expose the bug in the code by showing that the current logic is not correctly determining whether the permutation can be sorted.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Satay Sticks
time_limit_ms: 1500
memory_limit_mb: 256
Description: Penchick and his friend Kohane are touring Indonesia, and their next stop is in Surabaya!

In the bustling food stalls of Surabaya, Kohane bought $$$n$$$ satay sticks and arranged them in a line, with the $$$i$$$-th satay stick having length $$$p_i$$$. It is given that $$$p$$$ is a permutation$$$^{\text{∗}}$$$ of length $$$n$$$.

Penchick wants to sort the satay sticks in increasing order of length, so that $$$p_i=i$$$ for each $$$1\le i\le n$$$. For fun, they created a rule: they can only swap neighboring satay sticks whose lengths differ by exactly $$$1$$$. Formally, they can perform the following operation any number of times (including zero):

- Select an index $$$i$$$ ($$$1\le i\le n-1$$$) such that $$$|p_{i+1}-p_i|=1$$$;
- Swap $$$p_i$$$ and $$$p_{i+1}$$$.

Determine whether it is possible to sort the permutation $$$p$$$, thus the satay sticks, by performing the above operation.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 2\cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of satay sticks.

The second line of each test case contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$) — the permutation $$$p$$$ representing the length of the satay sticks.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot 10^5$$$.

Output: For each test case, output "YES" if it is possible to sort permutation $$$p$$$ by performing the operation. Otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
2
4
2 1 3 4
4
4 2 3 1
output:
YES
NO

Note: In the first test case, we can sort permutation $$$p = [2, 1, 3, 4]$$$ by performing an operation on index $$$1$$$ ($$$|p_2 - p_1| = |1 - 2| = 1$$$), resulting in $$$p = [1, 2, 3, 4]$$$.

In the second test case, it can be proven that it is impossible to sort permutation $$$p = [4, 2, 3, 1]$$$ by performing the operation. Here is an example of a sequence of operations that can be performed on the permutation:

- Select $$$i = 2$$$ ($$$|p_3 - p_2| = |3 - 2| = 1$$$). This results in $$$p = [4, 3, 2, 1]$$$.
- Select $$$i = 1$$$ ($$$|p_2 - p_1| = |3 - 4| = 1$$$). This results in $$$p = [3, 4, 2, 1]$$$.
- Select $$$i = 3$$$ ($$$|p_4 - p_3| = |1 - 2| = 1$$$). This results in $$$p = [3, 4, 1, 2]$$$.

Unfortunately, permutation $$$p$$$ remains unsorted after performing the operations.

Buggy Code:
//TIP To <b>Run</b> code, press <shortcut actionId="Run"/> or
// click the <icon src="AllIcons.Actions.Execute"/> icon in the gutter.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class Main {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader()
        {
            br = new BufferedReader(
                    new InputStreamReader(System.in));
        }

        String next()
        {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

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

        String nextLine()
        {
            String str = "";
            try {
                if(st.hasMoreTokens()){
                    str = st.nextToken("\n");
                }
                else{
                    str = br.readLine();
                }
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    public static void main(String[] args) {
        FastReader sc = new FastReader();
        int t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            int n = sc.nextInt();
            int[]arr = new int[n];
            for (int j = 0; j < n; j++) {
                arr[j]= sc.nextInt();
            }
            boolean fl = false;
            for (int j = 0; j < n; j++) {
                if(arr[j]!=j+1 && arr[j]!=i+2 && arr[j]!=j){
                    fl = true;
                    break;
                }

            }
            if(fl){
                System.out.println("NO");
            }else{
                System.out.println("YES");
            }

        }
    }

}

**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 where the faulty logic in the code is likely to fail. The buggy code checks if each element in the array is either `j+1`, `i+2`, or `j`. This logic is incorrect and does not correctly determine whether the permutation can be sorted using the allowed operations.

Here are the key points to consider:
1. **Inversion Pairs**: The code should check if the permutation can be sorted by only swapping adjacent elements that differ by 1. This means we need to ensure that the number of inversion pairs (pairs `(i, j)` such that `i < j` and `p[i] > p[j]`) is even.
2. **Edge Cases**: We should test with permutations that have a small number of elements and are close to being sorted but have a few inversions.

Given these points, let's create a small-scale test input that exposes the bug:

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

### Explanation:
1. **Test Case 1**: `4 2 1 3 4`
   - This is the same as the example in the problem statement where the answer is "YES". The code should correctly output "YES".

2. **Test Case 2**: `4 4 2 3 1`
   - This is another example where the answer is "NO". The code should correctly output "NO".

3. **Test Case 3**: `5 3 1 2 4 5`
   - This is a permutation that is almost sorted but has a few inversions. The correct answer is "YES" because it can be sorted by performing the allowed operations. However, the buggy code might incorrectly output "NO" due to its flawed logic.

This test input should help expose the bug in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Penchick and Satay Sticks
time_limit_ms: 1500
memory_limit_mb: 256
Description: Penchick and his friend Kohane are touring Indonesia, and their next stop is in Surabaya!

In the bustling food stalls of Surabaya, Kohane bought $$$n$$$ satay sticks and arranged them in a line, with the $$$i$$$-th satay stick having length $$$p_i$$$. It is given that $$$p$$$ is a permutation$$$^{\text{∗}}$$$ of length $$$n$$$.

Penchick wants to sort the satay sticks in increasing order of length, so that $$$p_i=i$$$ for each $$$1\le i\le n$$$. For fun, they created a rule: they can only swap neighboring satay sticks whose lengths differ by exactly $$$1$$$. Formally, they can perform the following operation any number of times (including zero):

- Select an index $$$i$$$ ($$$1\le i\le n-1$$$) such that $$$|p_{i+1}-p_i|=1$$$;
- Swap $$$p_i$$$ and $$$p_{i+1}$$$.

Determine whether it is possible to sort the permutation $$$p$$$, thus the satay sticks, by performing the above operation.

Input: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 2\cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2\cdot 10^5$$$) — the number of satay sticks.

The second line of each test case contains $$$n$$$ integers $$$p_1, p_2, \ldots, p_n$$$ ($$$1 \le p_i \le n$$$) — the permutation $$$p$$$ representing the length of the satay sticks.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\cdot 10^5$$$.

Output: For each test case, output "YES" if it is possible to sort permutation $$$p$$$ by performing the operation. Otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
2
4
2 1 3 4
4
4 2 3 1
output:
YES
NO

Note: In the first test case, we can sort permutation $$$p = [2, 1, 3, 4]$$$ by performing an operation on index $$$1$$$ ($$$|p_2 - p_1| = |1 - 2| = 1$$$), resulting in $$$p = [1, 2, 3, 4]$$$.

In the second test case, it can be proven that it is impossible to sort permutation $$$p = [4, 2, 3, 1]$$$ by performing the operation. Here is an example of a sequence of operations that can be performed on the permutation:

- Select $$$i = 2$$$ ($$$|p_3 - p_2| = |3 - 2| = 1$$$). This results in $$$p = [4, 3, 2, 1]$$$.
- Select $$$i = 1$$$ ($$$|p_2 - p_1| = |3 - 4| = 1$$$). This results in $$$p = [3, 4, 2, 1]$$$.
- Select $$$i = 3$$$ ($$$|p_4 - p_3| = |1 - 2| = 1$$$). This results in $$$p = [3, 4, 1, 2]$$$.

Unfortunately, permutation $$$p$$$ remains unsorted after performing the operations.

Buggy Code:
import javax.swing.*;
import java.io.*;
import java.util.*;
import java.math.*;

public class NumTheory {

    static ArrayList<Integer> primes;
    static int[] isComposite;
    static HashMap<Integer, Integer> hm;

/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

    public static void main(String[] args) throws IOException {
//        Scanner sc = new Scanner(new FileReader("input.txt"));
//        PrintWriter pw = new PrintWriter("output.txt");
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);

        int te = 1;
        te = sc.nextInt();
        while (te-- > 0) {
            int n = sc.nextInt();
            int[] arr = sc.nextIntarr(n);
            boolean f = true;

            for (int i = 0; i < n; i++) {
                if (arr[i] - 1 != i) {
                    if ( i+1 ==n){
                        f=false;
                        break;
                    }
                    if  (  arr[i+1] - 1 != i ){
                        f=false;
                        break;
                    }else {
                        i++;
                    }
                }
            }
            pw.println(f? "yes":"no");
        }
        pw.flush();
    }

    // if you have ans n of Integer ( a0,a1,a2,a3......) if you want to know if there exist num factorial of all
    // this Integer gcd(a0,a1,a2,a3......) !=1

    /*
    x % m≡y % m
    x = mk1 + r , y = mk2 +r
    x - y = m( k1 -k2)
    x−y≡0  (mod m)
    meaning that the difference  x−y is divisible by m

     */

    static boolean isSqrt(Long n) {
        long root = (long) Math.sqrt(n);
        if (root * root == n)
            return true;
        return false;
    }

    public static boolean isPowerOf(int n, int x) {
        if (n < 1 || x < 1) {
            return false;
        }
        while (n > 1) {
            if (n % x != 0) {
                return false;
            }
            n /= x;
        }
        return true;
    }

    /*
     * 2. Fast Exponentiation
     */
    static int modPow(int a, int e, int mod)    // O(log e)
    {
        a %= mod;
        long res = 1;
        while (e > 0) {
            if ((e & 1) == 1)
                res = (res * a) % mod;
            a = (int) ((a * 1l * a) % mod);
            e >>= 1;
        }
        return (int) res;
    }

    // mark primes as true in composite boolean ans till N
    static void sieve(int N)    // O(N log log N) // ِAbout 1e7
    {
        isComposite = new int[N + 1];
        isComposite[0] = isComposite[1] = 1;            // 0 indicates a prime number
        primes = new ArrayList<>();

        for (int i = 2; i <= N; ++i)                    //can loop till i*i <= N if primes arr is not needed O(N log log sqrt(N))
            if (isComposite[i] == 0)                    //can loop in 2 and odd integers for slightly better performance
            {
                primes.add(i);
                if (1l * i * i <= N)
                    for (int j = i * i; j <= N; j += i)    // j = i * 2 will not affect performance
                        // too much, may alter in modified sieve
                        isComposite[j] = 1;
            }
    }

    public static ArrayList<Integer> primeFactors(int N)        // O(sqrt(N) / ln sqrt(N))
    {
        ArrayList<Integer> factors = new ArrayList<Integer>();        //take abs(N) in case of -ve integers
        int idx = 0, p = primes.get(idx);
        while (p * p <= N) {
            while (N % p == 0) {
                factors.add(p);
                N /= p;
            }
            p = primes.get(++idx);
        }
        if (N != 1)                        // last prime factor may be > sqrt(N)
            factors.add(N);                // for integers whose largest prime factor has a power of 1
        return factors;
    }

    public static ArrayList<Long> primeFactors(long N)        // O(sqrt(N) / ln sqrt(N))
    {
        ArrayList<Long> factors = new ArrayList<>();        //take abs(N) in case of -ve integers
        int idx = 0, p = primes.get(idx);
        while (1l * p * p <= N) {
            while (N % p == 0) {
                factors.add(1l * p);
                N /= p;
            }
            if (idx + 1 >= primes.size()) break;
            p = primes.get(++idx);
        }
        if (N != 1)                        // last prime factor may be > sqrt(N)
            factors.add(N);                // for integers whose largest prime factor has a power of 1
        return factors;
    }

    static ArrayList<Long> factors(long n) {  // O(sqrt(N)
        ArrayList<Long> fac = new ArrayList<Long>();
        for (long i = 2; i * i <= n; i++) {
            while (n % i == 0) {
                n /= i;
                fac.add(i);
            }
        }
        if (n > 1) {
            fac.add(n);
        }
        return fac;
    }

    static boolean isPrime(int N) {
        if (N < isComposite.length) return isComposite[N] == 0;
        for (int p : primes)                    //may stop if p * p > N
            if (N % p == 0) return false;
        return true;
    }

    public static boolean isPrime2(int num) {
        if (num == 2) return true;
        if (num < 2 || num % 2 == 0) return false;
        // Check from 2 to the square root of the number
        for (int i = 3; i * i <= num; i += 2) {
            if (num % i == 0) return false;
        }
        return true; // Number is prime
    }

    public static int firstOnePosition(int num) {
        int position = 0;
        while (num > 0) {
            if ((num & 1) == 1) {
                return position;
            }
            num >>= 1;
            position++;
        }
        return -1;  // This case will only occur if the number is 0
    }

    static int lcm(int a, int b) {
        return (a / gcd(a, b)) * b;
    }

    static int gcd(int a, int b) {
        if (b == 0) return a;
        return gcd(b, a % b);
    }

    static class Scanner {

        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(String file) throws IOException {
            br = new BufferedReader(new FileReader(file));
        }

        public Scanner(FileReader r) {
            br = new BufferedReader(r);
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public String readAllLines(BufferedReader reader) throws IOException {
            StringBuilder content = new StringBuilder();
            String line;
            while ((line = reader.readLine()) != null) {
                content.append(line);
                content.append(System.lineSeparator());
            }
            return content.toString();
        }

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

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

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++)
                if (x.charAt(i) == '.') {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec) f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

        public long[] nextlongarr(int n) throws IOException {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        public Long[] nextLongarr(int n) throws IOException {
            Long[] a = new Long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        public int[] nextIntarr(int n) throws IOException {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public Integer[] nextIntegerarr(int n) throws IOException {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public boolean ready() throws IOException {
            return br.ready();
        }

    }

}

**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 where the code's logic might fail. The code provided has a logic that checks if each element in the array is in its correct position or if the next element can be swapped to correct it. However, this logic is flawed and does not correctly handle all permutations.

### Key Points to Consider:
1. **Permutations with Large Gaps**: The code might fail if there are large gaps between elements that cannot be bridged by adjacent swaps of 1.
2. **Edge Cases**: Small permutations where the elements are in a specific order that cannot be sorted by the given operation.
3. **Special Cases**: Permutations where the elements are already in a sorted or nearly sorted order but have a few elements out of place.

### Challenging Test Input:
Let's create a test case where the permutation has a large gap that cannot be bridged by the allowed operations. For example, a permutation where the smallest and largest elements are adjacent to each other, making it impossible to sort the array.

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

### Explanation:
- **Input**: The test case consists of a single permutation of length 5.
- **Permutation**: `[5, 1, 2, 3, 4]`
- **Why it's challenging**:
  - The element `5` is at the beginning, and the element `1` is next to it.
  - To sort the array, `5` needs to be moved to the end, but it can only be swapped with `1` (a difference of 4), which is not allowed by the operation.
  - Therefore, it is impossible to sort this permutation using the given operation.

This test case should expose the bug in the code by demonstrating a scenario where the permutation cannot be sorted as expected.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum Multiple Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an integer $n$, find an integer $x$ such that:

  * $2 \leq x \leq n$.    * The sum of multiples of $x$ that are less than or equal to $n$ is maximized. Formally, $x + 2x + 3x + \dots + kx$ where $kx \leq n$ is maximized over all possible values of $x$.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

Each test case contains a single integer $n$ ($2 \leq n \leq 100$).

Output: For each test case, output an integer, the optimal value of $x$. It can be shown there is only one unique answer.

Examples:
input:
2
3
15
output:
3
2

Note: For $n = 3$, the possible values of $x$ are $2$ and $3$. The sum of all multiples of $2$ less than or equal to $n$ is just $2$, and the sum of all multiples of $3$ less than or equal to $n$ is $3$. Therefore, $3$ is the optimal value of $x$.

For $n = 15$, the optimal value of $x$ is $2$. The sum of all multiples of $2$ less than or equal to $n$ is $2 + 4 + 6 + 8 + 10 + 12 + 14 = 56$, which can be proven to be the maximal over all other possible values of $x$.

Buggy Code:
#include <iostream>
#include<math.h>
using namespace std;
int main ()
{
int t;
cin>>t;
while (t--)
{
int n=2;
cin >>n;

if (n>3)
{
cout<<"2"<<endl;
}
else
{
cout <<"3"<<endl;
}
}
return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
4
5
100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum Multiple Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an integer $n$, find an integer $x$ such that:

  * $2 \leq x \leq n$.    * The sum of multiples of $x$ that are less than or equal to $n$ is maximized. Formally, $x + 2x + 3x + \dots + kx$ where $kx \leq n$ is maximized over all possible values of $x$.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

Each test case contains a single integer $n$ ($2 \leq n \leq 100$).

Output: For each test case, output an integer, the optimal value of $x$. It can be shown there is only one unique answer.

Examples:
input:
2
3
15
output:
3
2

Note: For $n = 3$, the possible values of $x$ are $2$ and $3$. The sum of all multiples of $2$ less than or equal to $n$ is just $2$, and the sum of all multiples of $3$ less than or equal to $n$ is $3$. Therefore, $3$ is the optimal value of $x$.

For $n = 15$, the optimal value of $x$ is $2$. The sum of all multiples of $2$ less than or equal to $n$ is $2 + 4 + 6 + 8 + 10 + 12 + 14 = 56$, which can be proven to be the maximal over all other possible values of $x$.

Buggy Code:
#include <bits/stdc++.h>
// god is great man i love my life
using namespace std;
typedef long long ll;

// jai siya ram
#define modsum(a, b) ((a % m) + (b % m)) % m
#define modminus(a, b) ((a % m) - (b % m)) % m
#define modmul(a, b) ((a % m) * (b % m)) % m
#define gaurav ios::sync_with_stdio(false);
#define kumar cin.tie(0);

ll MOD = (7 + (1e9));
ll modulo(ll n)
{
    if (n >= 0)
    {
        return n % MOD;
    }
    // negative num ke lie remainder
    return MOD - ((1ll * (-1) * n) % MOD);
}

int main()
{
    gaurav kumar
        ll t;
    cin >> t;
    while (t--)
    {
        ll n;
        cin >> n;
        if (n < 5 && n != 2)
            cout << '3' << '\n';
        else
            cout << 2 << '\n';
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
4
10
12
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum Multiple Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an integer $n$, find an integer $x$ such that:

  * $2 \leq x \leq n$.    * The sum of multiples of $x$ that are less than or equal to $n$ is maximized. Formally, $x + 2x + 3x + \dots + kx$ where $kx \leq n$ is maximized over all possible values of $x$.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

Each test case contains a single integer $n$ ($2 \leq n \leq 100$).

Output: For each test case, output an integer, the optimal value of $x$. It can be shown there is only one unique answer.

Examples:
input:
2
3
15
output:
3
2

Note: For $n = 3$, the possible values of $x$ are $2$ and $3$. The sum of all multiples of $2$ less than or equal to $n$ is just $2$, and the sum of all multiples of $3$ less than or equal to $n$ is $3$. Therefore, $3$ is the optimal value of $x$.

For $n = 15$, the optimal value of $x$ is $2$. The sum of all multiples of $2$ less than or equal to $n$ is $2 + 4 + 6 + 8 + 10 + 12 + 14 = 56$, which can be proven to be the maximal over all other possible values of $x$.

Buggy Code:
// In the name of Allah the most gracious, most merciful.
#include <bits/stdc++.h>
using namespace std;
#define endl "\n"
#define ll long long
void action()
{
    int n;
    cin >> n;
    int sum = 0, maxsum = 0, ans, m;
    for (int i = 2; i <= n; i++)
    {
        sum = 0, m = 1;
        while ((i * m) <= n)
        {
            sum += (i * m);
            m++;
        }
        if (sum > n)
            sum -= (i * m);
        if (sum > maxsum)
        {
            ans = i;
            maxsum = sum;
        }
    }
    cout << ans << endl;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--)
    {
        action();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
2
100
99
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum Multiple Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an integer $n$, find an integer $x$ such that:

  * $2 \leq x \leq n$.    * The sum of multiples of $x$ that are less than or equal to $n$ is maximized. Formally, $x + 2x + 3x + \dots + kx$ where $kx \leq n$ is maximized over all possible values of $x$.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

Each test case contains a single integer $n$ ($2 \leq n \leq 100$).

Output: For each test case, output an integer, the optimal value of $x$. It can be shown there is only one unique answer.

Examples:
input:
2
3
15
output:
3
2

Note: For $n = 3$, the possible values of $x$ are $2$ and $3$. The sum of all multiples of $2$ less than or equal to $n$ is just $2$, and the sum of all multiples of $3$ less than or equal to $n$ is $3$. Therefore, $3$ is the optimal value of $x$.

For $n = 15$, the optimal value of $x$ is $2$. The sum of all multiples of $2$ less than or equal to $n$ is $2 + 4 + 6 + 8 + 10 + 12 + 14 = 56$, which can be proven to be the maximal over all other possible values of $x$.

Buggy Code:
import sys

data = sys.stdin.read()
lines = data.splitlines()

n = int(lines[0])

for i in range(1, n+1):
    x = int(lines[i])
    if x<=3:
        print(3)
    else:
        print(2)

**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 where the code's logic fails. The buggy code incorrectly prints `3` for all values of `x` less than or equal to 3 and `2` for all other values of `x`. This logic is incorrect because it does not consider the actual sum of multiples of `x` that are less than or equal to `n`.

### Steps to Generate the Test Input:
1. **Test Case for Small `n`**: The code fails for `n = 2` because the correct answer is `2`, but the code prints `3`.
2. **Test Case for `n = 3`**: The code correctly prints `3`, but it's still an edge case.
3. **Test Case for `n = 4`**: The correct answer is `2` (sum of multiples: 2 + 4 = 6), but the code prints `3`.
4. **Test Case for `n = 5`**: The correct answer is `2` (sum of multiples: 2 + 4 = 6), but the code prints `3`.
5. **Test Case for `n = 6`**: The correct answer is `2` (sum of multiples: 2 + 4 + 6 = 12), but the code prints `3`.

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

### Explanation:
- For `n = 2`, the correct answer is `2` (sum of multiples: 2 = 2).
- For `n = 3`, the correct answer is `3` (sum of multiples: 3 = 3).
- For `n = 4`, the correct answer is `2` (sum of multiples: 2 + 4 = 6).
- For `n = 5`, the correct answer is `2` (sum of multiples: 2 + 4 = 6).
- For `n = 6`, the correct answer is `2` (sum of multiples: 2 + 4 + 6 = 12).

This test input will expose the bug in the buggy code where it incorrectly prints `3` for `n = 4`, `n = 5`, and `n = 6`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum Multiple Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an integer $n$, find an integer $x$ such that:

  * $2 \leq x \leq n$.    * The sum of multiples of $x$ that are less than or equal to $n$ is maximized. Formally, $x + 2x + 3x + \dots + kx$ where $kx \leq n$ is maximized over all possible values of $x$.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

Each test case contains a single integer $n$ ($2 \leq n \leq 100$).

Output: For each test case, output an integer, the optimal value of $x$. It can be shown there is only one unique answer.

Examples:
input:
2
3
15
output:
3
2

Note: For $n = 3$, the possible values of $x$ are $2$ and $3$. The sum of all multiples of $2$ less than or equal to $n$ is just $2$, and the sum of all multiples of $3$ less than or equal to $n$ is $3$. Therefore, $3$ is the optimal value of $x$.

For $n = 15$, the optimal value of $x$ is $2$. The sum of all multiples of $2$ less than or equal to $n$ is $2 + 4 + 6 + 8 + 10 + 12 + 14 = 56$, which can be proven to be the maximal over all other possible values of $x$.

Buggy Code:
def solve(input_data=None):
    import sys
    if input_data is None:
        data = sys.stdin.read().split()
    else:
        data = input_data.split()

    t = int(data[0])  # Number of test cases
    index = 1  # Index to track the current position in the input data
    found = 0
    for _ in range(t):
        n = int(data[index])
        index += 1  # Move the index to the next test case

        for i in range(n - 1, 3, -1):
            if found:
                break
            for j in range(2, i//2):
                if i % j == 0:
                    found = 1
                    n = j
                    break
        print(n)

if __name__ == "__main__":
    solve()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios where the code is likely to fail. The buggy code has a nested loop that attempts to find a divisor of `i` and then sets `n` to that divisor, which is incorrect. The correct approach should be to find the value of `x` that maximizes the sum of its multiples up to `n`.

### Analysis of the Buggy Code:
1. **Incorrect Logic in Inner Loop**: The inner loop is trying to find a divisor of `i` and setting `n` to that divisor, which is not the correct approach.
2. **No Proper Sum Calculation**: The code does not calculate the sum of multiples of `x` and compare it to find the maximum sum.
3. **Incorrect Break Conditions**: The `found` flag is used incorrectly, leading to premature termination of the loop.

### Challenging Test Input:
We need to provide a test case where the buggy code's incorrect logic will lead to an incorrect result. A good candidate is a value of `n` where the correct `x` is not a divisor of `n-1` or any other value in the range `n-1` to `3`.

### Test Input:
```plaintext
2
4
14
```

### Explanation:
1. **For `n = 4`**:
   - The correct answer is `2` because the sum of multiples of `2` (2 + 4 = 6) is greater than the sum of multiples of `3` (3 = 3).
   - The buggy code might incorrectly set `n` to a divisor of `3` or `2`, which is not the correct approach.

2. **For `n = 14`**:
   - The correct answer is `2` because the sum of multiples of `2` (2 + 4 + 6 + 8 + 10 + 12 + 14 = 56) is greater than the sum of multiples of any other value.
   - The buggy code might incorrectly set `n` to a divisor of `13` or any other value in the range `13` to `3`, leading to an incorrect result.

This test input should expose the bug in the provided code by showing that it does not correctly find the value of `x` that maximizes the sum of its multiples up to `n`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum Multiple Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an integer $n$, find an integer $x$ such that:

  * $2 \leq x \leq n$.    * The sum of multiples of $x$ that are less than or equal to $n$ is maximized. Formally, $x + 2x + 3x + \dots + kx$ where $kx \leq n$ is maximized over all possible values of $x$.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

Each test case contains a single integer $n$ ($2 \leq n \leq 100$).

Output: For each test case, output an integer, the optimal value of $x$. It can be shown there is only one unique answer.

Examples:
input:
2
3
15
output:
3
2

Note: For $n = 3$, the possible values of $x$ are $2$ and $3$. The sum of all multiples of $2$ less than or equal to $n$ is just $2$, and the sum of all multiples of $3$ less than or equal to $n$ is $3$. Therefore, $3$ is the optimal value of $x$.

For $n = 15$, the optimal value of $x$ is $2$. The sum of all multiples of $2$ less than or equal to $n$ is $2 + 4 + 6 + 8 + 10 + 12 + 14 = 56$, which can be proven to be the maximal over all other possible values of $x$.

Buggy Code:
def ia():
    return [int(i) for i in input().split()]
def ii():
    return int(input())
def is_():
    return [i for i in input().split()]
def isc():
    return [i for i in input()]
def solve():
    n = ii()
    res = 0
    maxn = 0
    for i in range(2, n+1):
        sum_ = i
        t = i
        while(t <= n):
            t += i
            sum_ += t
        if(sum_>=res):
            res = sum_
            maxn = i
    print(maxn)

for i in range(ii()):
    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
3
2
3
4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum Multiple Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an integer $n$, find an integer $x$ such that:

  * $2 \leq x \leq n$.    * The sum of multiples of $x$ that are less than or equal to $n$ is maximized. Formally, $x + 2x + 3x + \dots + kx$ where $kx \leq n$ is maximized over all possible values of $x$.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

Each test case contains a single integer $n$ ($2 \leq n \leq 100$).

Output: For each test case, output an integer, the optimal value of $x$. It can be shown there is only one unique answer.

Examples:
input:
2
3
15
output:
3
2

Note: For $n = 3$, the possible values of $x$ are $2$ and $3$. The sum of all multiples of $2$ less than or equal to $n$ is just $2$, and the sum of all multiples of $3$ less than or equal to $n$ is $3$. Therefore, $3$ is the optimal value of $x$.

For $n = 15$, the optimal value of $x$ is $2$. The sum of all multiples of $2$ less than or equal to $n$ is $2 + 4 + 6 + 8 + 10 + 12 + 14 = 56$, which can be proven to be the maximal over all other possible values of $x$.

Buggy Code:
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();  // read the number of test cases

        for (int i = 0; i < t; i++) {
            int n = scanner.nextInt();  // read the value of n for each test case
            System.out.println(findOptimalX(n));
        }

        scanner.close();
    }

    private static int findOptimalX(int n) {
        if (n % 2 == 0) {
            return 2;
        }
        for (int i = 3; i * i <= n; i += 2) {
            if (n % i == 0) {
                return 2;
            }
        }
        return n;  // if n is prime
    }
}

**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
100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum Multiple Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an integer $n$, find an integer $x$ such that:

  * $2 \leq x \leq n$.    * The sum of multiples of $x$ that are less than or equal to $n$ is maximized. Formally, $x + 2x + 3x + \dots + kx$ where $kx \leq n$ is maximized over all possible values of $x$.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

Each test case contains a single integer $n$ ($2 \leq n \leq 100$).

Output: For each test case, output an integer, the optimal value of $x$. It can be shown there is only one unique answer.

Examples:
input:
2
3
15
output:
3
2

Note: For $n = 3$, the possible values of $x$ are $2$ and $3$. The sum of all multiples of $2$ less than or equal to $n$ is just $2$, and the sum of all multiples of $3$ less than or equal to $n$ is $3$. Therefore, $3$ is the optimal value of $x$.

For $n = 15$, the optimal value of $x$ is $2$. The sum of all multiples of $2$ less than or equal to $n$ is $2 + 4 + 6 + 8 + 10 + 12 + 14 = 56$, which can be proven to be the maximal over all other possible values of $x$.

Buggy Code:
import java.util.*;

public class Main {
    public static void main(String[] args) {
        new Problem().simulateProblem();
    }
}

class Problem {
    private int testCases;
    int n;

    public Problem() {
        this.testCases = 0;
    }

    private void askForTestCasesNumber(Scanner scanner) {
        this.testCases = Integer.parseInt(scanner.nextLine().trim());
    }

//    private void getArr(Scanner scanner) {
//        scanner.nextLine();
//        String[] strings = scanner.nextLine().trim().split(" ");
//        arr = new int[strings.length];
//        for (int i = 0; i < arr.length; i++) {
//            arr[i] = Integer.parseInt(strings[i]);
//        }
//    }

    public void solve() {
        if( n / 2 > 2) System.out.println(2);
        else {
            System.out.println(n);
        }
    }

    public void simulateProblem() {
        Scanner scanner = new Scanner(System.in);
        askForTestCasesNumber(scanner);
        while (this.testCases > 0) {
            n = scanner.nextInt();
            solve();
            this.testCases--;
        }
        scanner.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
2
4
10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum Multiple Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an integer $n$, find an integer $x$ such that:

  * $2 \leq x \leq n$.    * The sum of multiples of $x$ that are less than or equal to $n$ is maximized. Formally, $x + 2x + 3x + \dots + kx$ where $kx \leq n$ is maximized over all possible values of $x$.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

Each test case contains a single integer $n$ ($2 \leq n \leq 100$).

Output: For each test case, output an integer, the optimal value of $x$. It can be shown there is only one unique answer.

Examples:
input:
2
3
15
output:
3
2

Note: For $n = 3$, the possible values of $x$ are $2$ and $3$. The sum of all multiples of $2$ less than or equal to $n$ is just $2$, and the sum of all multiples of $3$ less than or equal to $n$ is $3$. Therefore, $3$ is the optimal value of $x$.

For $n = 15$, the optimal value of $x$ is $2$. The sum of all multiples of $2$ less than or equal to $n$ is $2 + 4 + 6 + 8 + 10 + 12 + 14 = 56$, which can be proven to be the maximal over all other possible values of $x$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
int main()
{
	int t;
	cin >> t;
	while(t--)
	{
		int n;
		cin >> n;
		int mmax = -1;
		int w = 0;
		for(int i = 2 ; i <= n ; i++)
		{
			int p = i;
			int sum = 0;
			for(int j = 1 ; p <= n ; j++)
			{
				p = p * j;
				sum = sum + p;
			}
			if(sum > mmax)
			{
				mmax = sum;
				w = i;
			}
		}
		cout << w << endl;
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum Multiple Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an integer $n$, find an integer $x$ such that:

  * $2 \leq x \leq n$.    * The sum of multiples of $x$ that are less than or equal to $n$ is maximized. Formally, $x + 2x + 3x + \dots + kx$ where $kx \leq n$ is maximized over all possible values of $x$.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

Each test case contains a single integer $n$ ($2 \leq n \leq 100$).

Output: For each test case, output an integer, the optimal value of $x$. It can be shown there is only one unique answer.

Examples:
input:
2
3
15
output:
3
2

Note: For $n = 3$, the possible values of $x$ are $2$ and $3$. The sum of all multiples of $2$ less than or equal to $n$ is just $2$, and the sum of all multiples of $3$ less than or equal to $n$ is $3$. Therefore, $3$ is the optimal value of $x$.

For $n = 15$, the optimal value of $x$ is $2$. The sum of all multiples of $2$ less than or equal to $n$ is $2 + 4 + 6 + 8 + 10 + 12 + 14 = 56$, which can be proven to be the maximal over all other possible values of $x$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ahmd ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define endl '\n'
//double pi = acos(-1);
void solve(){
 int n; cin>>n;
 if(n>3) cout<<2<<endl;
 else cout<<3<<endl;
}
int main() {
  ahmd;
  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:
```plaintext
3
4
10
11
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum Multiple Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an integer $n$, find an integer $x$ such that:

  * $2 \leq x \leq n$.    * The sum of multiples of $x$ that are less than or equal to $n$ is maximized. Formally, $x + 2x + 3x + \dots + kx$ where $kx \leq n$ is maximized over all possible values of $x$.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

Each test case contains a single integer $n$ ($2 \leq n \leq 100$).

Output: For each test case, output an integer, the optimal value of $x$. It can be shown there is only one unique answer.

Examples:
input:
2
3
15
output:
3
2

Note: For $n = 3$, the possible values of $x$ are $2$ and $3$. The sum of all multiples of $2$ less than or equal to $n$ is just $2$, and the sum of all multiples of $3$ less than or equal to $n$ is $3$. Therefore, $3$ is the optimal value of $x$.

For $n = 15$, the optimal value of $x$ is $2$. The sum of all multiples of $2$ less than or equal to $n$ is $2 + 4 + 6 + 8 + 10 + 12 + 14 = 56$, which can be proven to be the maximal over all other possible values of $x$.

Buggy Code:
for each in range(int(input())):
	n = int(input())
	if n!=2 and n<6:
		print(3)
	else:
		print(2)

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

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

Think step by step.

Response:
```plaintext
3
2
5
10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum Multiple Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an integer $n$, find an integer $x$ such that:

  * $2 \leq x \leq n$.    * The sum of multiples of $x$ that are less than or equal to $n$ is maximized. Formally, $x + 2x + 3x + \dots + kx$ where $kx \leq n$ is maximized over all possible values of $x$.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

Each test case contains a single integer $n$ ($2 \leq n \leq 100$).

Output: For each test case, output an integer, the optimal value of $x$. It can be shown there is only one unique answer.

Examples:
input:
2
3
15
output:
3
2

Note: For $n = 3$, the possible values of $x$ are $2$ and $3$. The sum of all multiples of $2$ less than or equal to $n$ is just $2$, and the sum of all multiples of $3$ less than or equal to $n$ is $3$. Therefore, $3$ is the optimal value of $x$.

For $n = 15$, the optimal value of $x$ is $2$. The sum of all multiples of $2$ less than or equal to $n$ is $2 + 4 + 6 + 8 + 10 + 12 + 14 = 56$, which can be proven to be the maximal over all other possible values of $x$.

Buggy Code:
from collections import defaultdict, Counter, deque
from heapq import heapify, heappop, heappush
import math
DIRECT4 = [(1, 0), (0, 1), (-1, 0), (0, -1)]
DIRECT8 = [(1, 0), (0, 1), (-1, 0), (0, -1),
           (-1, -1), (-1, 1), (1, -1), (1, 1)]
INF = float("inf")
MOD = 1000000007

def mp(ty):
    if ty == 'int':
        array = list(map(int, input().split()))
        return tuple(array) if len(array) > 1 else array[0]
    elif ty == "str":
        return input()
    elif ty == 'lint':
        return list(map(int, input().split()))
    elif ty == 'lflo':
        return list(map(float, input().split()))
    elif ty == 'lstr':
        return list(map(str, input().split()))
    elif ty == 'lcha':
        return list(input())

def algo(condition) -> None:
    if condition == "m":
        main()
    else:
        tests = mp('int')
        while tests:
            main()
            tests -= 1

def main() -> None:
    n = mp("int")
    mx = -INF
    a = None
    for x in range(2, n+1):
        aa = x
        i = 1
        c = 0
        while x <= n:
            x = aa*i
            c += x
            i += 1
        if mx < c:
            mx = c
            a = aa
    print(a)

algo("t")

**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
100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum Multiple Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an integer $n$, find an integer $x$ such that:

  * $2 \leq x \leq n$.    * The sum of multiples of $x$ that are less than or equal to $n$ is maximized. Formally, $x + 2x + 3x + \dots + kx$ where $kx \leq n$ is maximized over all possible values of $x$.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

Each test case contains a single integer $n$ ($2 \leq n \leq 100$).

Output: For each test case, output an integer, the optimal value of $x$. It can be shown there is only one unique answer.

Examples:
input:
2
3
15
output:
3
2

Note: For $n = 3$, the possible values of $x$ are $2$ and $3$. The sum of all multiples of $2$ less than or equal to $n$ is just $2$, and the sum of all multiples of $3$ less than or equal to $n$ is $3$. Therefore, $3$ is the optimal value of $x$.

For $n = 15$, the optimal value of $x$ is $2$. The sum of all multiples of $2$ less than or equal to $n$ is $2 + 4 + 6 + 8 + 10 + 12 + 14 = 56$, which can be proven to be the maximal over all other possible values of $x$.

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

public class Main {

    public static void solve(Kattio sc) {
        int n = sc.nextInt();
        int s1 = 0;
        int s2 = 0;
        for (int i = 2; i <=n; i+=2) {
            s1+=i;
        }
        for (int i = 2; i <=n; i+=3) {
            s2+=i;
        }
        System.out.println(s1>s2?2:3);

    }

    public static void main(String[] args) {
        Kattio sc = new Kattio();
        int t = sc.nextInt();
        while (t-- > 0) solve(sc);
        sc.close();
    }

    static class Kattio extends PrintWriter {
        private BufferedReader r;
        private StringTokenizer st;

        public Kattio() {
            this(System.in, System.out);
        }

        public Kattio(InputStream i, OutputStream o) {
            super(o);
            r = new BufferedReader(new InputStreamReader(i));
        }

        public Kattio(String intput, String output) throws IOException {
            super(output);
            r = new BufferedReader(new FileReader(intput));
        }

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

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

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

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

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on the logic that the code uses to determine the optimal value of \( x \). The code currently only considers \( x = 2 \) and \( x = 3 \) and calculates the sum of their multiples up to \( n \). However, the problem requires finding the optimal \( x \) from \( 2 \) to \( n \), not just 2 and 3.

To create a challenging test input, we should select a value of \( n \) where the optimal \( x \) is neither 2 nor 3, but a different value. This will highlight the incorrect logic in the code.

Let's consider \( n = 14 \):

- For \( x = 2 \):
  - Multiples: 2, 4, 6, 8, 10, 12, 14
  - Sum: \( 2 + 4 + 6 + 8 + 10 + 12 + 14 = 56 \)

- For \( x = 3 \):
  - Multiples: 3, 6, 9, 12
  - Sum: \( 3 + 6 + 9 + 12 = 30 \)

- For \( x = 7 \):
  - Multiples: 7, 14
  - Sum: \( 7 + 14 = 21 \)

In this case, the optimal \( x \) is 2, but the code will only compare 2 and 3, and it will incorrectly choose 2 as the optimal value. However, we can see that 7 is a better candidate for \( n = 14 \).

To further challenge the code, we can use a value of \( n \) where the optimal \( x \) is not 2 or 3. For example, \( n = 12 \):

- For \( x = 2 \):
  - Multiples: 2, 4, 6, 8, 10, 12
  - Sum: \( 2 + 4 + 6 + 8 + 10 + 12 = 42 \)

- For \( x = 3 \):
  - Multiples: 3, 6, 9, 12
  - Sum: \( 3 + 6 + 9 + 12 = 30 \)

- For \( x = 4 \):
  - Multiples: 4, 8, 12
  - Sum: \( 4 + 8 + 12 = 24 \)

- For \( x = 6 \):
  - Multiples: 6, 12
  - Sum: \( 6 + 12 = 18 \)

Here, the optimal \( x \) is 2, but the code will only consider 2 and 3, and it will correctly choose 2. However, if we use \( n = 10 \):

- For \( x = 2 \):
  - Multiples: 2, 4, 6, 8, 10
  - Sum: \( 2 + 4 + 6 + 8 + 10 = 30 \)

- For \( x = 3 \):
  - Multiples: 3, 6, 9
  - Sum: \( 3 + 6 + 9 = 18 \)

- For \( x = 5 \):
  - Multiples: 5, 10
  - Sum: \( 5 + 10 = 15 \)

Here, the optimal \( x \) is 2, but the code will only consider 2 and 3, and it will correctly choose 2. However, if we use \( n = 11 \):

- For \( x = 2 \):
  - Multiples: 2, 4, 6, 8, 10
  - Sum: \( 2 + 4 + 6 + 8 + 10 = 30 \)

- For \( x = 3 \):
  - Multiples: 3, 6, 9
  - Sum: \( 3 + 6 + 9 = 18 \)

- For \( x = 5 \):
  - Multiples: 5, 10
  - Sum: \( 5 + 10 = 15 \)

- For \( x = 11 \):
  - Multiples: 11
  - Sum: \( 11 = 11 \)

Here, the optimal \( x \) is 2, but the code will only consider 2 and 3, and it will correctly choose 2. However, if we use \( n = 13 \):

- For \( x = 2 \):
  - Multiples: 2, 4, 6, 8, 10, 12
  - Sum: \( 2 + 4 + 6 + 8 + 10 + 12 = 42 \)

- For \( x = 3 \):
  - Multiples: 3, 6, 9, 12
  - Sum: \( 3 + 6 + 9 + 12 = 30 \)

- For \( x = 4 \):
  - Multiples: 4, 8, 12
  - Sum: \( 4 + 8 + 12 = 24 \)

- For \( x = 13 \):
  - Multiples: 13
  - Sum: \( 13 = 13 \)

Here, the optimal \( x \) is 2, but the code will only consider 2 and 3, and it will correctly choose 2. However, if we use \( n = 10 \):

- For \( x = 2 \):
  - Multiples: 2, 4, 6, 8, 10
  - Sum: \( 2 + 4 + 6 + 8 + 10 = 30 \)

- For \( x = 3 \):
  - Multiples: 3, 6, 9
  - Sum: \( 3 + 6 + 9 = 18 \)

- For \( x = 5 \):
  - Multiples: 5, 10
  - Sum: \( 5 + 10 = 15 \)

Here, the optimal \( x \) is 2, but the code will only consider 2 and 3, and it will correctly choose 2.

To ensure the code fails, we can use a value of \( n \) where the optimal \( x \) is not 2 or 3, such as \( n = 14 \):

```plaintext
1
14
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum Multiple Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an integer $n$, find an integer $x$ such that:

  * $2 \leq x \leq n$.    * The sum of multiples of $x$ that are less than or equal to $n$ is maximized. Formally, $x + 2x + 3x + \dots + kx$ where $kx \leq n$ is maximized over all possible values of $x$.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

Each test case contains a single integer $n$ ($2 \leq n \leq 100$).

Output: For each test case, output an integer, the optimal value of $x$. It can be shown there is only one unique answer.

Examples:
input:
2
3
15
output:
3
2

Note: For $n = 3$, the possible values of $x$ are $2$ and $3$. The sum of all multiples of $2$ less than or equal to $n$ is just $2$, and the sum of all multiples of $3$ less than or equal to $n$ is $3$. Therefore, $3$ is the optimal value of $x$.

For $n = 15$, the optimal value of $x$ is $2$. The sum of all multiples of $2$ less than or equal to $n$ is $2 + 4 + 6 + 8 + 10 + 12 + 14 = 56$, which can be proven to be the maximal over all other possible values of $x$.

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

/**
 * @author Tushar Rastogi
 **/

public class Main {
    /**
     * Global Variables
     */
    FastReader sc = new FastReader();
    PrintWriter out = new PrintWriter(System.out);

    /**
     * Global Variables
     */

    public static void main(String[] args) throws IOException {
        new Main().run();
    }

    // Program kickstart
    void run() throws IOException {
        int t;
        t = 1;
        t = sc.nextInt();
        for (int i = 1; i <= t; i++) {
            // level sabke niklenge

            solve();
        }
        out.flush();
    }

    public int gcd(int a, int b) {

        if (b == 0) return a;

        return gcd(b, a % b);

    }

    public void solve() throws IOException {

         int n=sc.nextInt();

          int sum1=0,sum2=0;

            int a=2,b=3;

             while(sum1<=n){
                 sum1+=a;
                 a+=2;
             }

        while(sum2<=n){
            sum2+=b;
            b+=3;
        }

          if(sum1>sum2)
              out.println("2");

          else
              out.println("3");

    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(
                    new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

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

        String nextLine() {
            String str = "";
            try {
                if (st.hasMoreTokens()) {
                    str = st.nextToken("\n");
                } else {
                    str = br.readLine();
                }
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

}

/*

If the remainder of  number is -ve ,add the number from which the number is divided to its remainder to get the ans
e.g remainder of -36 % 5  is -1
 ans = -1 +5 = 4
 -a % b =  r + b
 */

**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
10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum Multiple Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an integer $n$, find an integer $x$ such that:

  * $2 \leq x \leq n$.    * The sum of multiples of $x$ that are less than or equal to $n$ is maximized. Formally, $x + 2x + 3x + \dots + kx$ where $kx \leq n$ is maximized over all possible values of $x$.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

Each test case contains a single integer $n$ ($2 \leq n \leq 100$).

Output: For each test case, output an integer, the optimal value of $x$. It can be shown there is only one unique answer.

Examples:
input:
2
3
15
output:
3
2

Note: For $n = 3$, the possible values of $x$ are $2$ and $3$. The sum of all multiples of $2$ less than or equal to $n$ is just $2$, and the sum of all multiples of $3$ less than or equal to $n$ is $3$. Therefore, $3$ is the optimal value of $x$.

For $n = 15$, the optimal value of $x$ is $2$. The sum of all multiples of $2$ less than or equal to $n$ is $2 + 4 + 6 + 8 + 10 + 12 + 14 = 56$, which can be proven to be the maximal over all other possible values of $x$.

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

int t;
void solve()
{
	int n;
	cin >> n;

	int answer = 0;
	int maxSum = 0;
	for (int x = n; x >= 2; --x)
	{
		for (int k = 0; k * x <= n; ++k)
		{
			int sum = (k * x + (1 + k)) / 2;
			if (sum <= n)
			{
				if (maxSum < sum)
				{
					maxSum = sum;
					answer = x;
				}
			}
		}
	}

	cout << answer << '\n';
}

int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);

	cin >> t;
	while (t-- > 0)
	{
		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 expose the bug in the provided code, we need to focus on edge cases and scenarios where the code's logic might fail. The code has a nested loop that iterates over possible values of \( x \) and \( k \), but the inner loop's calculation of the sum of multiples is incorrect. Specifically, the sum calculation should be the sum of an arithmetic series, but the code uses a formula that doesn't correctly represent this.

Let's consider the following edge cases:
1. **Small values of \( n \)**: The code might not handle small values of \( n \) correctly.
2. **Large values of \( n \)**: The code might fail to find the correct maximum sum for larger values of \( n \).
3. **Values of \( n \) that are close to a multiple of \( x \)**: The code might not correctly handle cases where \( n \) is just slightly larger than a multiple of \( x \).

Given the constraints \( 2 \leq n \leq 100 \), let's focus on a small-scale test input that can expose the bug. A good candidate is \( n = 4 \).

For \( n = 4 \):
- The possible values of \( x \) are 2 and 3.
- For \( x = 2 \), the sum of multiples is \( 2 + 4 = 6 \).
- For \( x = 3 \), the sum of multiples is \( 3 \).

The correct answer should be \( x = 2 \) because the sum of multiples of 2 (which is 6) is greater than the sum of multiples of 3 (which is 3).

Here is the test input:

```plaintext
1
4
```

This input should expose the bug in the code, as the incorrect sum calculation will likely lead to an incorrect answer.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum Multiple Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an integer $n$, find an integer $x$ such that:

  * $2 \leq x \leq n$.    * The sum of multiples of $x$ that are less than or equal to $n$ is maximized. Formally, $x + 2x + 3x + \dots + kx$ where $kx \leq n$ is maximized over all possible values of $x$.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

Each test case contains a single integer $n$ ($2 \leq n \leq 100$).

Output: For each test case, output an integer, the optimal value of $x$. It can be shown there is only one unique answer.

Examples:
input:
2
3
15
output:
3
2

Note: For $n = 3$, the possible values of $x$ are $2$ and $3$. The sum of all multiples of $2$ less than or equal to $n$ is just $2$, and the sum of all multiples of $3$ less than or equal to $n$ is $3$. Therefore, $3$ is the optimal value of $x$.

For $n = 15$, the optimal value of $x$ is $2$. The sum of all multiples of $2$ less than or equal to $n$ is $2 + 4 + 6 + 8 + 10 + 12 + 14 = 56$, which can be proven to be the maximal over all other possible values of $x$.

Buggy Code:
#include <bits/stdc++.h>

using namespace std;

int main()
{
  int t,ans;
  cin >> t;

  for (int j = 0; j < t; j++)
  {
    int n;
    cin>>n;
    if(n>3) {
     ans=2;
    }
    else ans=3;
    cout<<ans<<endl;
  }
  return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
4
5
10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum Multiple Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an integer $n$, find an integer $x$ such that:

  * $2 \leq x \leq n$.    * The sum of multiples of $x$ that are less than or equal to $n$ is maximized. Formally, $x + 2x + 3x + \dots + kx$ where $kx \leq n$ is maximized over all possible values of $x$.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

Each test case contains a single integer $n$ ($2 \leq n \leq 100$).

Output: For each test case, output an integer, the optimal value of $x$. It can be shown there is only one unique answer.

Examples:
input:
2
3
15
output:
3
2

Note: For $n = 3$, the possible values of $x$ are $2$ and $3$. The sum of all multiples of $2$ less than or equal to $n$ is just $2$, and the sum of all multiples of $3$ less than or equal to $n$ is $3$. Therefore, $3$ is the optimal value of $x$.

For $n = 15$, the optimal value of $x$ is $2$. The sum of all multiples of $2$ less than or equal to $n$ is $2 + 4 + 6 + 8 + 10 + 12 + 14 = 56$, which can be proven to be the maximal over all other possible values of $x$.

Buggy Code:
kolvo = int(input())
osnova = []
vihod = []
for i in range (kolvo):
    chislo = int(input())
    summassiv = []
    for g in range(2,chislo+1):
        #print('g',g)
        summa = 0
        f = 2
        r = 0
        osnova.append(g)
        while f <= chislo:
            f+=g
            r+=1
        summa = ((r)*f)/2
        #print('r',r)
        #print('f',f)
        #print('summa', summa)
        summassiv.append(summa)
    result = summassiv.index(max(summassiv))
    #print(result)
    ossnova = osnova[result]
    vihod.append(ossnova)
#print('summassiv', summassiv)
#print('osnova', osnova)
for h in range (len(vihod)):
    print(vihod[h])
# Sat Jul 06 2024 14:57:55 GMT+0300 (Moscow Standard Time)

**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 are likely to cause issues. Here are the steps to think through the process:

1. **Understand the Bug**:
   - The bug in the code is in the calculation of the sum of multiples. The formula used in the code is incorrect. The correct formula for the sum of multiples of \( x \) up to \( n \) is:
     \[
     \text{sum} = x \times \left(1 + 2 + 3 + \dots + k\right) = x \times \frac{k \times (k + 1)}{2}
     \]
     where \( k = \left\lfloor \frac{n}{x} \right\rfloor \).

2. **Identify Edge Cases**:
   - Small values of \( n \) where the optimal \( x \) is close to \( n \).
   - Values of \( n \) where the optimal \( x \) is a small number (e.g., 2 or 3).
   - Values of \( n \) that are perfect multiples of small numbers.

3. **Create Test Cases**:
   - Test case 1: \( n = 3 \) (optimal \( x = 3 \))
   - Test case 2: \( n = 15 \) (optimal \( x = 2 \))
   - Test case 3: \( n = 4 \) (optimal \( x = 2 \))
   - Test case 4: \( n = 6 \) (optimal \( x = 3 \))
   - Test case 5: \( n = 10 \) (optimal \( x = 2 \))
   - Test case 6: \( n = 20 \) (optimal \( x = 4 \))

4. **Combine Test Cases**:
   - Create a small-scale test input with these values to ensure the code is tested thoroughly.

Here is the test input:

```plaintext
6
3
15
4
6
10
20
```

This input includes a mix of small and larger values of \( n \) to ensure that the bug in the sum calculation is exposed.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximum Multiple Sum
time_limit_ms: 1000
memory_limit_mb: 256
Description: Given an integer $n$, find an integer $x$ such that:

  * $2 \leq x \leq n$.    * The sum of multiples of $x$ that are less than or equal to $n$ is maximized. Formally, $x + 2x + 3x + \dots + kx$ where $kx \leq n$ is maximized over all possible values of $x$.

Input: The first line contains $t$ ($1 \leq t \leq 100$) — the number of test cases.

Each test case contains a single integer $n$ ($2 \leq n \leq 100$).

Output: For each test case, output an integer, the optimal value of $x$. It can be shown there is only one unique answer.

Examples:
input:
2
3
15
output:
3
2

Note: For $n = 3$, the possible values of $x$ are $2$ and $3$. The sum of all multiples of $2$ less than or equal to $n$ is just $2$, and the sum of all multiples of $3$ less than or equal to $n$ is $3$. Therefore, $3$ is the optimal value of $x$.

For $n = 15$, the optimal value of $x$ is $2$. The sum of all multiples of $2$ less than or equal to $n$ is $2 + 4 + 6 + 8 + 10 + 12 + 14 = 56$, which can be proven to be the maximal over all other possible values of $x$.

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())
    if n == 3 or (n ** (1.0/3.0)) % 3 == 0:
        print(3)
    else:
        print(2)

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

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

Think step by step.

Response:
```plaintext
5
2
3
4
14
15
```