**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Let Me Teach You a Lesson (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of a problem. The only difference between an easy and a hard version is the constraints on $t$ and $n$. You can make hacks only if both versions of the problem are solved.

Arthur is giving a lesson to his famous $2 n$ knights. Like any other students, they're sitting at the desks in pairs, but out of habit in a circle. The knight $2 i - 1$ is sitting at the desk with the knight $2 i$.

Each knight has intelligence, which can be measured by an integer. Let's denote the intelligence of the $i$-th knight as $a_i$. Arthur wants the maximal difference in total intelligence over all pairs of desks to be as small as possible. More formally, he wants to minimize $\max\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i}) - \min\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i})$.

However, the Code of Chivalry only allows swapping the opposite knights in the circle, i.e., Arthur can simultaneously perform $a_i := a_{i + n}$, $a_{i + n} := a_i$ for any $1 \le i \le n$. Arthur can make any number of such swaps. What is the best result he can achieve?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$) — the number of desks.

The second line consists of $2n$ integers $a_1, a_2, \ldots, a_{2 n}$ ($1 \le a_i \le 10^9$) — the intelligence values of the knights.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, output a single line containing one integer — the minimal difference Arthur can achieve.

Examples:
input:
5
2
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
output:
0
0
0
2
4

Note: In the first test case, Arthur can swap the second and the fourth knights. Then the total intelligence at both desks will be $10$.

In the third test case, Arthur can make $0$ operations, which will result in the total intelligence of $11$ at each of the desks.

In the fourth test case, Arthur can swap knights with indices $2$ and $5$ and achieve the difference of $2$. It can be proven that he cannot improve his result any further.

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

#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define FORR2(x,y,arr) for(auto& [x,y]:arr)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
template<class T> bool chmax(T &a, const T &b) { if(a<b){a=b;return 1;}return 0;}
template<class T> bool chmin(T &a, const T &b) { if(a>b){a=b;return 1;}return 0;}
//-------------------------------------------------------

int T,N;
ll A[202020];
ll B[202020];

template<class V,int NV> class SegTree_1 {
public:
	vector<V> val;
	V comp(V l,V r){
		int a,b,c,v=0;
		FOR(a,2) FOR(b,2) FOR(c,2) {
			if((l&(1<<(a+b*2)))&&(r&(1<<(b+c*2)))) {
				v|=1<<(a+c*2);
			}
		}
		return v;
	};

	SegTree_1(){val=vector<V>(NV*2,9);};
	void update(int entry, V v) {
		entry += NV;
		val[entry]=v;
		while(entry>1) entry>>=1, val[entry]=comp(val[entry*2],val[entry*2+1]);
	}
};
SegTree_1<int,1<<20> st;

vector<int> ev[808080];

void update(int i,ll mi,ll ma) {
	ll a=B[i*2];
	ll b=B[i*2+1];
	ll c=B[(i*2-1+2*N)%(2*N)];
	ll d=B[(i*2+2+2*N)%(2*N)];
	int mask=0;
	if(a+b>=mi&&a+b<=ma) mask|=1;
	if(c+b>=mi&&c+b<=ma) mask|=2;
	if(a+d>=mi&&a+d<=ma) mask|=4;
	if(c+d>=mi&&c+d<=ma) mask|=8;
	//cout<<"#"<<i<<" "<<mi<<" "<<ma<<" "<<mask<<endl;
	st.update(i,mask);

}

void solve() {
	int i,j,k,l,r,x,y; string s;

	cin>>T;
	while(T--) {
		cin>>N;
		FOR(i,2*N) cin>>A[i];
		if(N==1) {
			cout<<0<<endl;
			continue;
		}
		if(N%2==0) {
			ll ma=1LL<<60;
			ll mi=0;
			FOR(i,N/2) {
				ll a=A[2*i];
				ll b=A[2*i+1];
				ll c=A[2*i+N];
				ll d=A[2*i+N+1];
				if(a>c) swap(a,c);
				if(b<d) swap(b,d);
				ma=min(ma,max(a+b,c+d));
				mi=max(mi,min(a+b,c+d));

			}
			cout<<ma-mi<<endl;
			continue;
		}

		FOR(i,N) st.update(i,0);
		x=0;
		FOR(i,N) {
			B[i*2]=A[x];
			x^=1;
			B[i*2+1]=A[x];
			x=(x+N)%(2*N);

			//cout<<B[i*2]<<" "<<B[i*2+1]<<" ";
		}
		//cout<<endl;
		vector<ll> cand;
		FOR(i,N) {
			ll a=B[i*2];
			ll b=B[i*2+1];
			ll c=B[(i*2-1+2*N)%(2*N)];
			ll d=B[(i*2+2+2*N)%(2*N)];
			cand.push_back(a+b);
			cand.push_back(c+b);
			cand.push_back(a+d);
			cand.push_back(c+d);
		}
		sort(ALL(cand));
		cand.erase(unique(ALL(cand)),cand.end());
		//FORR(a,cand) cout<<a<<":";
		//cout<<endl;
		FOR(i,cand.size()+4) ev[i].clear();
		FOR(i,N) {
			ll a=B[i*2];
			ll b=B[i*2+1];
			ll c=B[(i*2-1+2*N)%(2*N)];
			ll d=B[(i*2+2+2*N)%(2*N)];
			vector<ll> C={a+b,c+b,a+d,c+d};
			FORR(v,C) {
				x=lower_bound(ALL(cand),v)-cand.begin();
				if(x) ev[x-1].push_back(i);
				ev[x].push_back(i);
				ev[x+1].push_back(i);
			}
		}
		ll ret=1LL<<60;
		for(int L=0,R=0;L<cand.size();L++) {
			FORR(e,ev[L]) {
				if(L==R) {
					update(e,-1,-1);
				}
				else {
					update(e,cand[L],cand[R-1]);
				}
			}
			while(((st.val[1]&9)==0)&&R<cand.size()) {
				FORR(e,ev[R]) update(e,cand[L],cand[R]);
				R++;
			}
			//FOR(j,N) update(j,cand[L],cand[R-1]);
			//cout<<"!"<<cand[5]<<" "<<(st.val[1]&9)<<endl;
			if(st.val[1]&9) {
				//cout<<L<<" "<<R-1<<" "<<cand[L]<<" "<<cand[R-1]<<" "<<st.val[1]<<endl;
				ret=min(ret,cand[R-1]-cand[L]);
			}
		}
		cout<<ret<<endl;

		FOR(i,N) st.update(i,9);
	}

}

int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n'; FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	cout.tie(0); 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
5
2
1 2 2 1
1
10 20
3
1 2 3 4 5 6
3
1 1 10 10 10 1
5
1 10 2 20 3 30 4 40 5 50
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Let Me Teach You a Lesson (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of a problem. The only difference between an easy and a hard version is the constraints on $t$ and $n$. You can make hacks only if both versions of the problem are solved.

Arthur is giving a lesson to his famous $2 n$ knights. Like any other students, they're sitting at the desks in pairs, but out of habit in a circle. The knight $2 i - 1$ is sitting at the desk with the knight $2 i$.

Each knight has intelligence, which can be measured by an integer. Let's denote the intelligence of the $i$-th knight as $a_i$. Arthur wants the maximal difference in total intelligence over all pairs of desks to be as small as possible. More formally, he wants to minimize $\max\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i}) - \min\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i})$.

However, the Code of Chivalry only allows swapping the opposite knights in the circle, i.e., Arthur can simultaneously perform $a_i := a_{i + n}$, $a_{i + n} := a_i$ for any $1 \le i \le n$. Arthur can make any number of such swaps. What is the best result he can achieve?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$) — the number of desks.

The second line consists of $2n$ integers $a_1, a_2, \ldots, a_{2 n}$ ($1 \le a_i \le 10^9$) — the intelligence values of the knights.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, output a single line containing one integer — the minimal difference Arthur can achieve.

Examples:
input:
5
2
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
output:
0
0
0
2
4

Note: In the first test case, Arthur can swap the second and the fourth knights. Then the total intelligence at both desks will be $10$.

In the third test case, Arthur can make $0$ operations, which will result in the total intelligence of $11$ at each of the desks.

In the fourth test case, Arthur can swap knights with indices $2$ and $5$ and achieve the difference of $2$. It can be proven that he cannot improve his result any further.

Buggy Code:
def II():
    return(int(input()))
def LMI():
    return(list(map(int,input().split())))
def I():
    return(input())
def MII():
    return(map(int,input().split()))
import sys
input=sys.stdin.readline
# import io,os
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# from collections import Counter
# int(math.log(len(L)))
# import math
# from collections import defaultdict
# mod=10**9+7
# from collections import deque
import  math

# from functools import cache
# @cache
for _ in range(II()):
    n=II()
    L=LMI()
    alp=min(L)
    for i in range(2*n):
        L[i]-=alp
    if n==1:
        print(0)
        continue
    X=[]
    if n%2==0:
        for i in range(0,n,2):
            X.append(max(L[i],L[i+n])+min(L[i+1],L[i+n+1]))
            X.append(min(L[i],L[i+n])+max(L[i+1],L[i+n+1]))
        print(max(X)-min(X))
        continue
    X.append(1)
    X.append(2)
    for i in range(n-1):
        if X[-1]<n:
            X.append(n+X[-1])
            X.append(X[-1]+1)
        else:
            X.append(X[-1]-n)
            X.append(X[-1]+1)
    for i in range(2*n):
        X[i]=L[X[i]-1]
    arr=[]
    def sub(x):
        if x>=n:
            return x-n
        return x+n
    for i in range(0,2*n,2):
        arr.append(L[i]+L[i+1])
        arr.append(L[i]+L[sub(i+1)])
        arr.append(L[sub(i)]+L[sub(i+1)])
        arr.append(L[sub(i)]+L[i+1])
    arr.sort()
    ans=math.inf
    for i in arr:
        dp=[[math.inf]*2 for _ in range(2*n)]
        if X[0]+X[1]>=i:
            dp[1][0]=X[0]+X[1]
        if X[0]+X[2]>=i:
            dp[1][1]=X[0]+X[2]
        for j in range(2,2*n):
            if j==2*n-1:
                if X[-1]+X[-2]>=i:
                    dp[-1][0]=min(dp[-1][0],max(dp[j-1][0],X[-1]+X[-2]))
                if X[-1]+X[-3]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-2],dp[j-1][1]))
                continue
            if j%2:
                if X[j]+X[j-2]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-2],dp[j-1][1]))
                if X[j]+X[j-1]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-1],dp[j-1][0]))
                if X[j+1]+X[j-1]>=i:
                    dp[j][1]=min(dp[j][1],max(X[j+1]+X[j-1],dp[j-1][0]))
                if X[j+1]+X[j-2]>=i:
                    dp[j][1]=min(dp[j][1],max(X[j+1]+X[j-2],dp[j-1][1]))
            else:
                dp[j][0]=dp[j-1][0]
                dp[j][1]=dp[j-1][1]

        ans=min(ans,dp[-1][0]-i)
        # if ans==1:
        #     print(dp)
        #     break
        X[0],X[-1]=X[-1],X[0]
        dp=[[math.inf]*2 for _ in range(2*n)]
        if X[0]+X[1]>=i:
            dp[1][0]=X[0]+X[1]
        if X[0]+X[2]>=i:
            dp[1][1]=X[0]+X[2]
        for j in range(2,2*n):
            if j==2*n-1:
                if X[-1]+X[-2]>=i:
                    dp[-1][0]=min(dp[-1][0],max(dp[j-1][0],X[-1]+X[-2]))
                if X[-1]+X[-3]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-2],dp[j-1][1]))
                continue
            if j%2:
                if X[j]+X[j-2]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-2],dp[j-1][1]))
                if X[j]+X[j-1]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-1],dp[j-1][0]))
                if X[j+1]+X[j-1]>=i:
                    dp[j][1]=min(dp[j][1],max(X[j+1]+X[j-1],dp[j-1][0]))
                if X[j+1]+X[j-2]>=i:
                    dp[j][1]=min(dp[j][1],max(X[j+1]+X[j-2],dp[j-1][1]))
            else:
                dp[j][0]=dp[j-1][0]
                dp[j][1]=dp[j-1][1]
        ans=min(ans,dp[-1][0]-i)
        # if ans==1:
        #     print(dp)
        #     break
    print(ans)

#     return

# if __name__=="__main__":

#     for _ in range(II()):
#         t()
    # 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
3
1 2 3 4 5 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Let Me Teach You a Lesson (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of a problem. The only difference between an easy and a hard version is the constraints on $t$ and $n$. You can make hacks only if both versions of the problem are solved.

Arthur is giving a lesson to his famous $2 n$ knights. Like any other students, they're sitting at the desks in pairs, but out of habit in a circle. The knight $2 i - 1$ is sitting at the desk with the knight $2 i$.

Each knight has intelligence, which can be measured by an integer. Let's denote the intelligence of the $i$-th knight as $a_i$. Arthur wants the maximal difference in total intelligence over all pairs of desks to be as small as possible. More formally, he wants to minimize $\max\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i}) - \min\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i})$.

However, the Code of Chivalry only allows swapping the opposite knights in the circle, i.e., Arthur can simultaneously perform $a_i := a_{i + n}$, $a_{i + n} := a_i$ for any $1 \le i \le n$. Arthur can make any number of such swaps. What is the best result he can achieve?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$) — the number of desks.

The second line consists of $2n$ integers $a_1, a_2, \ldots, a_{2 n}$ ($1 \le a_i \le 10^9$) — the intelligence values of the knights.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, output a single line containing one integer — the minimal difference Arthur can achieve.

Examples:
input:
5
2
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
output:
0
0
0
2
4

Note: In the first test case, Arthur can swap the second and the fourth knights. Then the total intelligence at both desks will be $10$.

In the third test case, Arthur can make $0$ operations, which will result in the total intelligence of $11$ at each of the desks.

In the fourth test case, Arthur can swap knights with indices $2$ and $5$ and achieve the difference of $2$. It can be proven that he cannot improve his result any further.

Buggy Code:
def II():
    return(int(input()))
def LMI():
    return(list(map(int,input().split())))
def I():
    return(input())
def MII():
    return(map(int,input().split()))
import sys
input=sys.stdin.readline
# import io,os
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# from collections import Counter
# int(math.log(len(L)))
# import math
# from collections import defaultdict
# mod=10**9+7
# from collections import deque
import  math

# from functools import cache
# @cache
for _ in range(II()):
    n=II()
    L=LMI()
    alp=min(L)
    for i in range(2*n):
        L[i]-=alp
    if n==1:
        print(0)
        continue
    X=[]
    if n%2==0:
        for i in range(0,n,2):
            X.append(max(L[i],L[i+n])+min(L[i+1],L[i+n+1]))
            X.append(min(L[i],L[i+n])+max(L[i+1],L[i+n+1]))
        print(max(X)-min(X))
        continue
    X.append(1)
    X.append(2)
    for i in range(n-1):
        if X[-1]<n:
            X.append(n+X[-1])
            X.append(X[-1]+1)
        else:
            X.append(X[-1]-n)
            X.append(X[-1]+1)
    for i in range(2*n):
        X[i]=L[X[i]-1]
    arr=[]
    def sub(x):
        if x>=n:
            return x-n
        return x+n
    for i in range(0,2*n,2):
        arr.append(L[i]+L[i+1])
        arr.append(L[i]+L[sub(i+1)])
        arr.append(L[sub(i)]+L[sub(i+1)])
        arr.append(L[sub(i)]+L[i+1])

    # arr.sort()
    # new=[arr[0]]
    # for i in arr:
    #     if i!=new[-1]:
    #         new.append(i)
    # arr=new
    ans=math.inf
    for i in arr:
        dp=[[math.inf]*2 for _ in range(2*n)]
        if X[0]+X[1]>=i:
            dp[1][0]=X[0]+X[1]
        if X[0]+X[2]>=i:
            dp[1][1]=X[0]+X[2]
        for j in range(2,2*n):
            if j==2*n-1:
                if X[-1]+X[-2]>=i:
                    dp[-1][0]=min(dp[-1][0],max(dp[j-1][0],X[-1]+X[-2]))
                if X[-1]+X[-3]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-2],dp[j-1][1]))
                continue
            if j%2:
                if X[j]+X[j-2]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-2],dp[j-1][1]))
                if X[j]+X[j-1]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-1],dp[j-1][0]))
                if X[j+1]+X[j-1]>=i:
                    dp[j][1]=min(dp[j][1],max(X[j+1]+X[j-1],dp[j-1][0]))
                if X[j+1]+X[j-2]>=i:
                    dp[j][1]=min(dp[j][1],max(X[j+1]+X[j-2],dp[j-1][1]))
            else:
                dp[j][0]=dp[j-1][0]
                dp[j][1]=dp[j-1][1]

        ans=min(ans,dp[-1][0]-i)
        # if ans==1:
        #     print(dp)
        #     break
        X[0],X[-1]=X[-1],X[0]
        dp=[[math.inf]*2 for _ in range(2*n)]
        if X[0]+X[1]>=i:
            dp[1][0]=X[0]+X[1]
        if X[0]+X[2]>=i:
            dp[1][1]=X[0]+X[2]
        for j in range(2,2*n):
            if j==2*n-1:
                if X[-1]+X[-2]>=i:
                    dp[-1][0]=min(dp[-1][0],max(dp[j-1][0],X[-1]+X[-2]))
                if X[-1]+X[-3]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-2],dp[j-1][1]))
                continue
            if j%2:
                if X[j]+X[j-2]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-2],dp[j-1][1]))
                if X[j]+X[j-1]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-1],dp[j-1][0]))
                if X[j+1]+X[j-1]>=i:
                    dp[j][1]=min(dp[j][1],max(X[j+1]+X[j-1],dp[j-1][0]))
                if X[j+1]+X[j-2]>=i:
                    dp[j][1]=min(dp[j][1],max(X[j+1]+X[j-2],dp[j-1][1]))
            else:
                dp[j][0]=dp[j-1][0]
                dp[j][1]=dp[j-1][1]
        ans=min(ans,dp[-1][0]-i)
        # if ans==1:
        #     print(dp)
        #     break
    print(ans)

#     return

# if __name__=="__main__":

#     for _ in range(II()):
#         t()
    # 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
5
2
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Let Me Teach You a Lesson (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of a problem. The only difference between an easy and a hard version is the constraints on $t$ and $n$. You can make hacks only if both versions of the problem are solved.

Arthur is giving a lesson to his famous $2 n$ knights. Like any other students, they're sitting at the desks in pairs, but out of habit in a circle. The knight $2 i - 1$ is sitting at the desk with the knight $2 i$.

Each knight has intelligence, which can be measured by an integer. Let's denote the intelligence of the $i$-th knight as $a_i$. Arthur wants the maximal difference in total intelligence over all pairs of desks to be as small as possible. More formally, he wants to minimize $\max\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i}) - \min\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i})$.

However, the Code of Chivalry only allows swapping the opposite knights in the circle, i.e., Arthur can simultaneously perform $a_i := a_{i + n}$, $a_{i + n} := a_i$ for any $1 \le i \le n$. Arthur can make any number of such swaps. What is the best result he can achieve?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$) — the number of desks.

The second line consists of $2n$ integers $a_1, a_2, \ldots, a_{2 n}$ ($1 \le a_i \le 10^9$) — the intelligence values of the knights.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, output a single line containing one integer — the minimal difference Arthur can achieve.

Examples:
input:
5
2
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
output:
0
0
0
2
4

Note: In the first test case, Arthur can swap the second and the fourth knights. Then the total intelligence at both desks will be $10$.

In the third test case, Arthur can make $0$ operations, which will result in the total intelligence of $11$ at each of the desks.

In the fourth test case, Arthur can swap knights with indices $2$ and $5$ and achieve the difference of $2$. It can be proven that he cannot improve his result any further.

Buggy Code:
import java.util.ArrayList;
import java.util.Scanner;

public class Solution{

    public static void caseEven(int[] intellgence,int n){

        int maxPair = Integer.MIN_VALUE;
        int minPair = Integer.MAX_VALUE;
        for (int left = 1,right = 2; right <= n; left+=2,right+=2) {
            int pair1 = Math.max(intellgence[left],intellgence[left+n])
                + Math.min(intellgence[right],intellgence[right+n]);
            int pair2 = intellgence[left] + intellgence[left+n]
                + intellgence[right] + intellgence[right+n] - pair1;
            if (pair1 > maxPair) maxPair = pair1;
            if (pair1 < minPair) minPair = pair1;
            if (pair2 > maxPair) maxPair = pair2;
            if (pair2 < minPair) minPair = pair2;
        }

        System.out.println(maxPair - minPair);
    }

    //数组很适合
    // record PairInfo(int zeroZero,int zeroOne,int oneZero,int oneOne){};
    static final int zeroZero = 0;
    static final int zeroOne = 1;
    static final int oneZero = 2;
    static final int oneOne = 3;
    static int myMax(int x,int y){
        if (x == 0
            || y == 0)
        {
            return 0;
        }
        return Math.max(x, y);
    }
    static int myMin(int x,int y){
        if (x == 0
            || y == 0)
        {
            return Math.max(x, y);
        }
        return Math.min(x, y);
    }
    record SumInfo(int pairIndex,int sumIndex,int value){}

    static int[] merge(int[] p1,int[] p2){
        return new int[]{
            Math.min(Math.max(p1[zeroZero],p2[zeroZero]),Math.max(p1[zeroOne],p2[oneZero])),
            Math.min(Math.max(p1[zeroZero],p2[zeroOne]),Math.max(p1[zeroOne],p2[oneOne])),
            Math.min(Math.max(p1[oneZero],p2[zeroZero]),Math.max(p1[oneOne],p2[oneZero])),
            Math.min(Math.max(p1[oneZero],p2[zeroOne]), Math.max(p1[oneOne], p2[oneOne]))};
    }

    static void caseOdd(int[] knight,int n){

        int[][] pairArr = new int[n][0];
        ArrayList<Solution.SumInfo> sumList = new ArrayList<SumInfo>();
        for (int i = 1; i <= n; i++) {
            int left,right,anotherLeft,anotherRight;
            if (i % 2 == 1) {
                left = i;
                right = i+1;
                anotherLeft = i+n;
                anotherRight = (i+1+n)%(2*n);
            }
            else {
                anotherLeft = i;
                anotherRight = i+1;
                left = i+n;
                right = i+1+n;
            }

            int[] thisPair = new int[]{
                knight[left]+knight[right],
                knight[left]+knight[anotherRight],
                knight[anotherLeft]+knight[right],
                knight[anotherLeft]+knight[anotherRight]};
            for (int j = 0; j < 4;j++){
                sumList.add(new SumInfo(i-1, j, thisPair[j]));
            }
            pairArr[i-1] = thisPair;
        }
        sumList.sort((s1,s2)->(s1.value - s2.value));

        int minDiff = Integer.MAX_VALUE;
        int prevMinSum = -1;
        for (SumInfo info : sumList) {
            if (info.value == prevMinSum) {
                pairArr[info.pairIndex][info.sumIndex] = Integer.MAX_VALUE;
                continue;
            }

            //检测prevMinSum
            int minSum = info.value;
            int[] dp = pairArr[0];
            int[] dpWithMin = new int[]{Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE};
            for (int i = 0; i < 4; i++) {
                if (dp[i] == minSum) {
                    dpWithMin[i] = minSum;
                }
            }

            for (int i = 1; i < pairArr.length; i++) {
                int[] thisPair = pairArr[i];

                dpWithMin = merge(dpWithMin,thisPair);
                for (int j = 0; j < 4; j++) {
                    if (thisPair[j] != minSum) continue;
                    int[] auxPair = new int[]{
                        thisPair[0] == minSum ? minSum : Integer.MAX_VALUE,
                        thisPair[1] == minSum ? minSum : Integer.MAX_VALUE,
                        thisPair[2] == minSum ? minSum : Integer.MAX_VALUE,
                        thisPair[3] == minSum ? minSum : Integer.MAX_VALUE};
                    int[] auxDp = merge(dp,auxPair);
                    for (int k = 0; k < 4; k++) {
                        if (auxDp[k] < dpWithMin[k])
                            dpWithMin[k] = auxDp[k];
                    }
                }
                dp = merge(dp,thisPair);
            }
            minDiff = Math.min(minDiff,dpWithMin[0] - minSum);
            minDiff = Math.min(minDiff,dpWithMin[3] - minSum);

            prevMinSum = info.value;
            pairArr[info.pairIndex][info.sumIndex] = Integer.MAX_VALUE;
        }

        System.out.println(minDiff);
    }

    public static void slove(Scanner scanner){

        int t = scanner.nextInt();
        while (t > 0 ) {t--;
            int n = scanner.nextInt();
            int[] knight = new int[2*n+1];
            for (int i = 1; i < knight.length; i++) {
                knight[i] = scanner.nextInt();
            }
            if (n % 2 == 0) {
                caseEven(knight,n);
                continue;
            }
            if (n == 1){
                System.out.println(0);
                continue;
            }
            caseOdd(knight,n);
        }
    }

    public static void main(String[] args) {
        try (Scanner scanner = new Scanner(System.in);) {
            slove(scanner);
        }
    }
}

**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
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Let Me Teach You a Lesson (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of a problem. The only difference between an easy and a hard version is the constraints on $t$ and $n$. You can make hacks only if both versions of the problem are solved.

Arthur is giving a lesson to his famous $2 n$ knights. Like any other students, they're sitting at the desks in pairs, but out of habit in a circle. The knight $2 i - 1$ is sitting at the desk with the knight $2 i$.

Each knight has intelligence, which can be measured by an integer. Let's denote the intelligence of the $i$-th knight as $a_i$. Arthur wants the maximal difference in total intelligence over all pairs of desks to be as small as possible. More formally, he wants to minimize $\max\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i}) - \min\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i})$.

However, the Code of Chivalry only allows swapping the opposite knights in the circle, i.e., Arthur can simultaneously perform $a_i := a_{i + n}$, $a_{i + n} := a_i$ for any $1 \le i \le n$. Arthur can make any number of such swaps. What is the best result he can achieve?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$) — the number of desks.

The second line consists of $2n$ integers $a_1, a_2, \ldots, a_{2 n}$ ($1 \le a_i \le 10^9$) — the intelligence values of the knights.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, output a single line containing one integer — the minimal difference Arthur can achieve.

Examples:
input:
5
2
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
output:
0
0
0
2
4

Note: In the first test case, Arthur can swap the second and the fourth knights. Then the total intelligence at both desks will be $10$.

In the third test case, Arthur can make $0$ operations, which will result in the total intelligence of $11$ at each of the desks.

In the fourth test case, Arthur can swap knights with indices $2$ and $5$ and achieve the difference of $2$. It can be proven that he cannot improve his result any further.

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

using namespace std;

using ll = long long;
using ui = unsigned int;
using ull = unsigned long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<long long, long long>;

const int inf = 1e9;
const ll inf64 = 1e18;

struct output {
    int res;

    void print() {
        cout << res << "\n";
    }

    bool operator == (const output& o) const {
        throw;
    }
};

struct TwoSat {
    int n;
    vector<pii> a;

    TwoSat(int n): n(n) {}

    void add_or(int x, int y) {
        // x or y
        a.emplace_back(x, y);
    }

    void add_nor_and(int x, int y) {
        // not (x and y) = not(x) or not(y)
        add_or(nor(x), nor(y));
    }

    int nor(int x) const {
        return x < n ? x + n : x - n;
    }

    bool solve() {
        vector<vector<int>> g(2 * n), gr(2 * n);
        for (auto [x, y] : a) {
            g[nor(x)].push_back(y);
            g[nor(y)].push_back(x);
        }
        for (int v = 0; v < 2 * n; ++v) for (int to : g[v]) gr[to].push_back(v);
        vector<int> used(2 * n), tops;
        function<void(int)> dfs = [&](int v) {
            if (used[v]) return;
            used[v] = 1;
            for (int to : g[v]) dfs(to);
            tops.push_back(v);
        };
        for (int v = 0; v < 2 * n; ++v) dfs(v);
        reverse(tops.begin(), tops.end());
        vector<int> color(2 * n, -1);
        function<void(int, int)> go = [&](int v, int c) {
            if (color[v] != -1) return;
            color[v] = c;
            for (int to : gr[v]) go(to, c);
        };
        int C = 0;
        for (int v : tops) {
            if (color[v] == -1) {
                go(v, C);
                ++C;
            }
        }
        for (int v = 0; v < n; ++v) {
            if (color[v] == color[v + n]) return false;
        }
        return true;
    }

    bool solve_slow() {
        auto get_val = [&](int x, int mask) -> int {
            if (x < n) return (mask >> x) & 1;
            else return !((mask >> (x - n)) & 1);
        };
        for (int mask = 0; mask < (1 << n); ++mask) {
            int ok = 1;
            for (auto [x, y] : a) {
                if (get_val(x, mask) || get_val(y, mask)) {}
                else {
                    ok = 0;
                    break;
                }
            }
            if (ok) {
                return true;
            }
        }
        return false;
    }
};

struct input {
    int n;
    vector<int> a;

    input() = default;

    void read() {
        cin >> n;
        a.resize(2 * n);
        for (auto& x : a) cin >> x;
    }

    void print() {

    }

    void gen() {
        // use static
    }

    void gen_max_test() {

    }

    output fast() {
        auto check = [&](int lx, int rx) -> bool {
            if (lx > rx) return false;
            TwoSat two_sat(n);
            for (int i = 0; i + 1 < 2 * n; i += 2) {
                int x0 = i % n;
                int y0 = (i + 1) % n;
                for (int do_x : {0, 1}) {
                    for (int do_y : {0, 1}) {
                        int x = do_x ? x0 : two_sat.nor(x0);
                        int y = do_y ? y0 : two_sat.nor(y0);
                        int value = 0;
                        if (do_x) value += a[(i + n) % (2 * n)]; else value += a[i];
                        if (do_y) value += a[(i + 1 + n) % (2 * n)]; else value += a[i + 1];
                        if (value < lx || value > rx) {
                            two_sat.add_nor_and(x, y);
                        }
                    }
                }
            }
            return two_sat.solve();
        };
        vector<int> ar;
        for (int i = 0; i + 1 < 2 * n; i += 2) {
            for (int mask = 0; mask < (1 << 2); ++mask) {
                int value = 0;
                if ((mask >> 0) & 1) value += a[(i + n) % (2 * n)]; else value += a[i];
                if ((mask >> 1) & 1) value += a[(i + 1 + n) % (2 * n)]; else value += a[i + 1];
                ar.push_back(value);
            }
        }
        sort(ar.begin(), ar.end());
        ar.erase(unique(ar.begin(), ar.end()), ar.end());
        int res = 2 * inf + 5;
        for (int j = 0, i = 0; i < (int) ar.size(); ++i)  {
            if (!check(0, ar[i])) continue;
            while (j < (int) ar.size() && check(ar[j], ar[i])) ++j;
            res = min(res, ar[i] - ar[j - 1]);
        }
        // int mn, mx;
        // {
        //     int bl = 0, br = 2 * inf + 5, bm;
        //     while (br - bl > 1) {
        //         bm = bl + (br - bl) / 2;
        //         if (check(bm, 2 * inf + 5)) bl = bm;
        //         else br = bm;
        //     }
        //     mn = bl;
        // }
        // {
        //     int bl = 0, br = 2 * inf + 5, bm;
        //     while (br - bl > 1) {
        //         bm = bl + (br - bl) / 2;
        //         if (check(0, bm)) br = bm;
        //         else bl = bm;
        //     }
        //     mx = br;
        // }
        // int res = mx - mn;
        return output{res};
    }

#ifdef DEBUG
    output slow() {
        return output();
    }
#endif
};

void test_case() {
    input in;
    in.read();
    output res = in.fast();
    res.print();
}

void work() {
    int t = 1;
// #ifdef DEBUG
    cin >> t;
// #endif
    while (t--)
        test_case();
}

#ifdef DEBUG
void test() {
    for (int t = 1;;t++) {
        input in;
        in.gen();
        input in_fs = in;
        input in_sl = in;
        output fs = in_fs.fast();
        output sl = in_sl.slow();
        if (fs == sl) {
            cout << "OK" << endl;
            fs.print();
            cout << "\n=========" << endl;
        } else {
            cout << "WA " << t << "\n";
            cout << "exp\n";
            sl.print();
            cout << "\n=========\n";
            cout << "fnd\n";
            fs.print();
            cout << "\n=========\n";
            in.print();
            break;
        }
    }
}
#endif

#ifdef DEBUG
void max_test() {
    input in;
    in.gen_max_test();
    input in_fs = in;
    output fs = in_fs.fast();
    fs.print();
}
#endif

int main() {

#ifdef DEBUG
    freopen("input.txt", "r", stdin);
#endif

    ios_base::sync_with_stdio(0);
    cin.tie(0);

    work();
//    test();
//    max_test();

    return 0;
}

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

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

Think step by step.

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

1. **Understand the Problem Constraints:**
   - \(1 \le t \le 1000\)
   - \(1 \le n \le 2000\)
   - \(1 \le a_i \le 10^9\)
   - The sum of \(n\) over all test cases does not exceed 2000.

2. **Identify Potential Bugs:**
   - The code uses a 2SAT solver to determine the feasibility of certain conditions.
   - The 2SAT solver might fail in cases where the constraints are tight or where the intelligence values are very large or very small.
   - The code might also fail in cases where the number of swaps needed to achieve the minimum difference is not correctly calculated.

3. **Generate a Small-Scale Test Input:**
   - We need a test case that is small enough to fit within the constraints but challenging enough to expose the bug.
   - Consider a case where the intelligence values are such that the 2SAT solver might not correctly determine the feasibility of the conditions.

4. **Construct the Test Case:**
   - Let's create a test case with \(t = 1\) and \(n = 3\).
   - Choose intelligence values that are close to each other but not identical, to make the 2SAT solver work harder to find a solution.

Here is the test input:

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

This test case is small and complete. It has \(t = 1\) test case with \(n = 3\), and the intelligence values are \(1, 2, 3, 4, 5, 6\). This setup is designed to challenge the 2SAT solver by requiring it to carefully consider the constraints and find the optimal swaps to minimize the difference in total intelligence over all pairs of desks.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Let Me Teach You a Lesson (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of a problem. The only difference between an easy and a hard version is the constraints on $t$ and $n$. You can make hacks only if both versions of the problem are solved.

Arthur is giving a lesson to his famous $2 n$ knights. Like any other students, they're sitting at the desks in pairs, but out of habit in a circle. The knight $2 i - 1$ is sitting at the desk with the knight $2 i$.

Each knight has intelligence, which can be measured by an integer. Let's denote the intelligence of the $i$-th knight as $a_i$. Arthur wants the maximal difference in total intelligence over all pairs of desks to be as small as possible. More formally, he wants to minimize $\max\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i}) - \min\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i})$.

However, the Code of Chivalry only allows swapping the opposite knights in the circle, i.e., Arthur can simultaneously perform $a_i := a_{i + n}$, $a_{i + n} := a_i$ for any $1 \le i \le n$. Arthur can make any number of such swaps. What is the best result he can achieve?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$) — the number of desks.

The second line consists of $2n$ integers $a_1, a_2, \ldots, a_{2 n}$ ($1 \le a_i \le 10^9$) — the intelligence values of the knights.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, output a single line containing one integer — the minimal difference Arthur can achieve.

Examples:
input:
5
2
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
output:
0
0
0
2
4

Note: In the first test case, Arthur can swap the second and the fourth knights. Then the total intelligence at both desks will be $10$.

In the third test case, Arthur can make $0$ operations, which will result in the total intelligence of $11$ at each of the desks.

In the fourth test case, Arthur can swap knights with indices $2$ and $5$ and achieve the difference of $2$. It can be proven that he cannot improve his result any further.

Buggy Code:
#include <bits/stdc++.h>
#define endl '\n'
using namespace std;
const int MOD = 998244353;
#define int long long
const int maxN = 1e6+5;
const int INF = 2e12+1;

 mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
 int rnd(int B){
 	return (unsigned long long)rng()%B;
 }
 int power(int a, int b){
 	if(b==0)return 1;
 	if(b%2)return a*power(a,b-1)%MOD;
 	return power(a*a%MOD,b/2);
 }
signed main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);

	int t;
	cin>>t;
	while(t--){
		int n;
		cin>>n;
		vector<int>a(2*n);
		for(int i = 0;i<2*n;i++){
			cin>>a[i];
		}
		if(n%2==0){
			int ans1 = 0;
			int ans2 = INF;
			for(int i = 0;i<2*n;i+=2){
				int a1 = a[i];
				int a2 = a[i+1];
				int a3 = a[(i+n)%(2*n)];
				int a4 = a[(i+1+n)%(2*n)];
				int d1 = abs(a1+a2 -a3-a4);
				int d2 = abs(a1+a4-a2-a3);
				if(d1>d2){
					ans1 = max(ans1, max(a1+a4,a2+a3));
					ans2 = min(ans2,min(a1+a4,a2+a3));
				}
				else{
					ans1 = max(ans1, max(a1+a2,a3+a4));
					ans2 = min(ans2,min(a1+a2,a3+a4));
				}
			}
			cout<<ans1-ans2<<endl;
		}
		else{
			unordered_set<int>minimums;
			for(int i = 0;i<2*n;i+=2){
				int a1 = a[i];
				int a2 = a[i+1];
				int a3 = a[(i+n)%(2*n)];
				int a4 = a[(i+1+n)%(2*n)];
				minimums.insert(a1+a2);
				minimums.insert(a1+a4);
				minimums.insert(a2+a3);
				minimums.insert(a3+a4);
			}
			int ans = INF;
			for(auto m : minimums){
				//cout<<m<<" ";
				vector<vector<int>>dp0(n,vector<int>(2,INF));
				vector<vector<int>>dp1(n,vector<int>(2,INF));
				int v = 0;
				int c1 = INF;
				int c2 = INF;
				int c3 = INF;
				int c4 = INF;
				if(a[(v+n)%(2*n)]+a[v^1]>=m){
					c1 = a[(v+n)%(2*n)]+a[v^1];
				}
				if( a[v]+a[v^1]>=m){
					c2 =  a[v]+a[v^1];
				}
				if(a[(v+n)%(2*n)]+a[((v^1)+n)%(2*n)]>=m){
					c3 = a[(v+n)%(2*n)]+a[((v^1)+n)%(2*n)];
				}
				if( a[v]+a[((v^1)+n)%(2*n)]>=m){
					c4 =  a[v]+a[((v^1)+n)%(2*n)];
				}
				dp0[0][0] = c2;
				dp1[0][0] = c1;

				dp0[0][1] = c4;
				dp1[0][1] =c3;

				for(int i = 1;i<n;i++){
					v^=1;
					v = (n+v)%(2*n);
					int c1 = INF;
					int c2 = INF;
					int c3 = INF;
					int c4 = INF;
					if(a[(v+n)%(2*n)]+a[v^1]>=m){
						c1 = a[(v+n)%(2*n)]+a[v^1];
					}
					if( a[v]+a[v^1]>=m){
						c2 =  a[v]+a[v^1];
					}
					if(a[(v+n)%(2*n)]+a[((v^1)+n)%(2*n)]>=m){
						c3 = a[(v+n)%(2*n)]+a[((v^1)+n)%(2*n)];
					}
					if( a[v]+a[((v^1)+n)%(2*n)]>=m){
						c4 =  a[v]+a[((v^1)+n)%(2*n)];
					}
					dp0[i][0] = min(max(dp0[i-1][1],c1), max(dp0[i-1][0],c2));
					dp0[i][1] = min(max(dp0[i-1][1],c3), max(dp0[i-1][0], c4));

					dp1[i][0] = min(max(dp1[i-1][1],c1), max(dp1[i-1][0], c2));
					dp1[i][1] = min(max(dp1[i-1][1],c3), max(dp1[i-1][0], c4));
				}
				//cout<< min(dp1[n-1][1],dp0[n-1][0])<<" "<<m<<endl;
				//cout<<dp0[n-1][0]<<" "<<dp1
				ans = min(ans, min(dp1[n-1][1],dp0[n-1][0])-m);
			}
			//cout<<endl;
			cout<<ans<<endl;

		}
	}

}

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

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

Think step by step.

Response:
```plaintext
3
1
1 1000000000
2
1 1000000000 1 1000000000
3
1 1000000000 1 1000000000 1 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Let Me Teach You a Lesson (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of a problem. The only difference between an easy and a hard version is the constraints on $t$ and $n$. You can make hacks only if both versions of the problem are solved.

Arthur is giving a lesson to his famous $2 n$ knights. Like any other students, they're sitting at the desks in pairs, but out of habit in a circle. The knight $2 i - 1$ is sitting at the desk with the knight $2 i$.

Each knight has intelligence, which can be measured by an integer. Let's denote the intelligence of the $i$-th knight as $a_i$. Arthur wants the maximal difference in total intelligence over all pairs of desks to be as small as possible. More formally, he wants to minimize $\max\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i}) - \min\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i})$.

However, the Code of Chivalry only allows swapping the opposite knights in the circle, i.e., Arthur can simultaneously perform $a_i := a_{i + n}$, $a_{i + n} := a_i$ for any $1 \le i \le n$. Arthur can make any number of such swaps. What is the best result he can achieve?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$) — the number of desks.

The second line consists of $2n$ integers $a_1, a_2, \ldots, a_{2 n}$ ($1 \le a_i \le 10^9$) — the intelligence values of the knights.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, output a single line containing one integer — the minimal difference Arthur can achieve.

Examples:
input:
5
2
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
output:
0
0
0
2
4

Note: In the first test case, Arthur can swap the second and the fourth knights. Then the total intelligence at both desks will be $10$.

In the third test case, Arthur can make $0$ operations, which will result in the total intelligence of $11$ at each of the desks.

In the fourth test case, Arthur can swap knights with indices $2$ and $5$ and achieve the difference of $2$. It can be proven that he cannot improve his result any further.

Buggy Code:
def II():
    return(int(input()))
def LMI():
    return(list(map(int,input().split())))
def I():
    return(input())
def MII():
    return(map(int,input().split()))
import sys
input=sys.stdin.readline
# import io,os
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# from collections import Counter
# int(math.log(len(L)))
# import math
# from collections import defaultdict
# mod=10**9+7
# from collections import deque
import  math

# from functools import cache
# @cache
for _ in range(II()):
    n=II()
    L=LMI()
    V=sorted(L)
    for i in range(2*n):
        L[i]-=V[n//2]
    if n==1:
        print(0)
        continue
    X=[]
    if n%2==0:
        for i in range(0,n,2):
            X.append(max(L[i],L[i+n])+min(L[i+1],L[i+n+1]))
            X.append(min(L[i],L[i+n])+max(L[i+1],L[i+n+1]))
        print(max(X)-min(X))
        continue
    X.append(1)
    X.append(2)
    for i in range(n-1):
        if X[-1]<n:
            X.append(n+X[-1])
            X.append(X[-1]+1)
        else:
            X.append(X[-1]-n)
            X.append(X[-1]+1)
    for i in range(2*n):
        X[i]=L[X[i]-1]
    arr=[]
    def sub(x):
        if x>=n:
            return x-n
        return x+n
    for i in range(0,2*n,2):
        arr.append(L[i]+L[i+1])
        arr.append(L[i]+L[sub(i+1)])
        arr.append(L[sub(i)]+L[sub(i+1)])
        arr.append(L[sub(i)]+L[i+1])

    arr=list(set(arr))
    ans=math.inf
    for i in arr:
        dp=[[math.inf]*2 for _ in range(2*n)]
        if X[0]+X[1]>=i:
            dp[1][0]=X[0]+X[1]
        if X[0]+X[2]>=i:
            dp[1][1]=X[0]+X[2]
        for j in range(2,2*n):
            if j==2*n-1:
                if X[-1]+X[-2]>=i:
                    dp[-1][0]=min(dp[-1][0],max(dp[j-1][0],X[-1]+X[-2]))
                if X[-1]+X[-3]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-2],dp[j-1][1]))
                continue
            if j%2:
                if X[j]+X[j-2]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-2],dp[j-1][1]))
                if X[j]+X[j-1]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-1],dp[j-1][0]))
                if X[j+1]+X[j-1]>=i:
                    dp[j][1]=min(dp[j][1],max(X[j+1]+X[j-1],dp[j-1][0]))
                if X[j+1]+X[j-2]>=i:
                    dp[j][1]=min(dp[j][1],max(X[j+1]+X[j-2],dp[j-1][1]))
            else:
                dp[j][0]=dp[j-1][0]
                dp[j][1]=dp[j-1][1]

        ans=min(ans,dp[-1][0]-i)
        # if ans==1:
        #     print(dp)
        #     break
        X[0],X[-1]=X[-1],X[0]
        dp=[[math.inf]*2 for _ in range(2*n)]
        if X[0]+X[1]>=i:
            dp[1][0]=X[0]+X[1]
        if X[0]+X[2]>=i:
            dp[1][1]=X[0]+X[2]
        for j in range(2,2*n):
            if j==2*n-1:
                if X[-1]+X[-2]>=i:
                    dp[-1][0]=min(dp[-1][0],max(dp[j-1][0],X[-1]+X[-2]))
                if X[-1]+X[-3]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-2],dp[j-1][1]))
                continue
            if j%2:
                if X[j]+X[j-2]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-2],dp[j-1][1]))
                if X[j]+X[j-1]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-1],dp[j-1][0]))
                if X[j+1]+X[j-1]>=i:
                    dp[j][1]=min(dp[j][1],max(X[j+1]+X[j-1],dp[j-1][0]))
                if X[j+1]+X[j-2]>=i:
                    dp[j][1]=min(dp[j][1],max(X[j+1]+X[j-2],dp[j-1][1]))
            else:
                dp[j][0]=dp[j-1][0]
                dp[j][1]=dp[j-1][1]
        ans=min(ans,dp[-1][0]-i)
        # if ans==1:
        #     print(dp)
        #     break
    print(ans)

#     return

# if __name__=="__main__":

#     for _ in range(II()):
#         t()
    # 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
5
2
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Let Me Teach You a Lesson (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of a problem. The only difference between an easy and a hard version is the constraints on $t$ and $n$. You can make hacks only if both versions of the problem are solved.

Arthur is giving a lesson to his famous $2 n$ knights. Like any other students, they're sitting at the desks in pairs, but out of habit in a circle. The knight $2 i - 1$ is sitting at the desk with the knight $2 i$.

Each knight has intelligence, which can be measured by an integer. Let's denote the intelligence of the $i$-th knight as $a_i$. Arthur wants the maximal difference in total intelligence over all pairs of desks to be as small as possible. More formally, he wants to minimize $\max\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i}) - \min\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i})$.

However, the Code of Chivalry only allows swapping the opposite knights in the circle, i.e., Arthur can simultaneously perform $a_i := a_{i + n}$, $a_{i + n} := a_i$ for any $1 \le i \le n$. Arthur can make any number of such swaps. What is the best result he can achieve?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$) — the number of desks.

The second line consists of $2n$ integers $a_1, a_2, \ldots, a_{2 n}$ ($1 \le a_i \le 10^9$) — the intelligence values of the knights.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, output a single line containing one integer — the minimal difference Arthur can achieve.

Examples:
input:
5
2
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
output:
0
0
0
2
4

Note: In the first test case, Arthur can swap the second and the fourth knights. Then the total intelligence at both desks will be $10$.

In the third test case, Arthur can make $0$ operations, which will result in the total intelligence of $11$ at each of the desks.

In the fourth test case, Arthur can swap knights with indices $2$ and $5$ and achieve the difference of $2$. It can be proven that he cannot improve his result any further.

Buggy Code:
def II():
    return(int(input()))
def LMI():
    return(list(map(int,input().split())))
def I():
    return(input())
def MII():
    return(map(int,input().split()))
import sys
input=sys.stdin.readline
# import io,os
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# from collections import Counter
# int(math.log(len(L)))
# import math
# from collections import defaultdict
# mod=10**9+7
# from collections import deque
import  math

# from functools import cache
# @cache
for _ in range(II()):
    n=II()
    L=LMI()
    alp=min(L)
    for i in range(2*n):
        L[i]-=alp
    if n==1:
        print(0)
        continue
    X=[]
    if n%2==0:
        for i in range(0,n,2):
            X.append(max(L[i],L[i+n])+min(L[i+1],L[i+n+1]))
            X.append(min(L[i],L[i+n])+max(L[i+1],L[i+n+1]))
        print(max(X)-min(X))
        continue
    X.append(1)
    X.append(2)
    for i in range(n-1):
        if X[-1]<n:
            X.append(n+X[-1])
            X.append(X[-1]+1)
        else:
            X.append(X[-1]-n)
            X.append(X[-1]+1)
    for i in range(2*n):
        X[i]=L[X[i]-1]
    arr=[]
    def sub(x):
        if x>=n:
            return x-n
        return x+n
    for i in range(0,2*n,2):
        arr.append(L[i]+L[i+1])
        arr.append(L[i]+L[sub(i+1)])
        arr.append(L[sub(i)]+L[sub(i+1)])
        arr.append(L[sub(i)]+L[i+1])

    arr.sort()
    new=[arr[0]]
    for i in arr:
        if i!=new[-1]:
            new.append(i)
    arr=new
    ans=math.inf
    for i in arr:
        dp=[[math.inf]*2 for _ in range(2*n)]
        if X[0]+X[1]>=i:
            dp[1][0]=X[0]+X[1]
        if X[0]+X[2]>=i:
            dp[1][1]=X[0]+X[2]
        for j in range(2,2*n):
            if j==2*n-1:
                if X[-1]+X[-2]>=i:
                    dp[-1][0]=min(dp[-1][0],max(dp[j-1][0],X[-1]+X[-2]))
                if X[-1]+X[-3]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-2],dp[j-1][1]))
                continue
            if j%2:
                if X[j]+X[j-2]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-2],dp[j-1][1]))
                if X[j]+X[j-1]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-1],dp[j-1][0]))
                if X[j+1]+X[j-1]>=i:
                    dp[j][1]=min(dp[j][1],max(X[j+1]+X[j-1],dp[j-1][0]))
                if X[j+1]+X[j-2]>=i:
                    dp[j][1]=min(dp[j][1],max(X[j+1]+X[j-2],dp[j-1][1]))
            else:
                dp[j][0]=dp[j-1][0]
                dp[j][1]=dp[j-1][1]

        ans=min(ans,dp[-1][0]-i)
        # if ans==1:
        #     print(dp)
        #     break
        X[0],X[-1]=X[-1],X[0]
        dp=[[math.inf]*2 for _ in range(2*n)]
        if X[0]+X[1]>=i:
            dp[1][0]=X[0]+X[1]
        if X[0]+X[2]>=i:
            dp[1][1]=X[0]+X[2]
        for j in range(2,2*n):
            if j==2*n-1:
                if X[-1]+X[-2]>=i:
                    dp[-1][0]=min(dp[-1][0],max(dp[j-1][0],X[-1]+X[-2]))
                if X[-1]+X[-3]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-2],dp[j-1][1]))
                continue
            if j%2:
                if X[j]+X[j-2]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-2],dp[j-1][1]))
                if X[j]+X[j-1]>=i:
                    dp[j][0]=min(dp[j][0],max(X[j]+X[j-1],dp[j-1][0]))
                if X[j+1]+X[j-1]>=i:
                    dp[j][1]=min(dp[j][1],max(X[j+1]+X[j-1],dp[j-1][0]))
                if X[j+1]+X[j-2]>=i:
                    dp[j][1]=min(dp[j][1],max(X[j+1]+X[j-2],dp[j-1][1]))
            else:
                dp[j][0]=dp[j-1][0]
                dp[j][1]=dp[j-1][1]
        ans=min(ans,dp[-1][0]-i)
        # if ans==1:
        #     print(dp)
        #     break
    print(ans)

#     return

# if __name__=="__main__":

#     for _ in range(II()):
#         t()
    # 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:
To generate a challenging test input that exposes the bug in the provided buggy code, let's consider the following steps:

1. **Understand the Bug**: The code seems to have issues with handling certain edge cases, especially when the number of desks is odd. The logic for handling odd `n` seems overly complex and might not be correctly minimizing the difference in total intelligence.

2. **Edge Cases**:
   - Small `n`: Test with `n = 1` and `n = 2`.
   - Odd `n`: Test with `n = 3` and `n = 5`.
   - Large `n`: Test with `n = 2000` (though this is not necessary for a small-scale test).

3. **Input Characteristics**:
   - Use a mix of high and low intelligence values to create a scenario where the difference in total intelligence is maximized.
   - Ensure that the values are not uniformly distributed to challenge the code's ability to handle varied inputs.

4. **Generate the Test Input**:
   - For `n = 3`, use a combination of values that might expose the bug in the odd `n` handling.
   - For `n = 5`, use a similar strategy to ensure the code handles larger odd `n` correctly.

Here is the small-scale test input:

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

This test input includes:
- A test case with `n = 3` and a pattern of intelligence values that might expose issues in the odd `n` handling.
- A test case with `n = 5` and a sequence of increasing intelligence values to challenge the code's ability to minimize the difference in total intelligence.

This input should help identify 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: Let Me Teach You a Lesson (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of a problem. The only difference between an easy and a hard version is the constraints on $t$ and $n$. You can make hacks only if both versions of the problem are solved.

Arthur is giving a lesson to his famous $2 n$ knights. Like any other students, they're sitting at the desks in pairs, but out of habit in a circle. The knight $2 i - 1$ is sitting at the desk with the knight $2 i$.

Each knight has intelligence, which can be measured by an integer. Let's denote the intelligence of the $i$-th knight as $a_i$. Arthur wants the maximal difference in total intelligence over all pairs of desks to be as small as possible. More formally, he wants to minimize $\max\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i}) - \min\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i})$.

However, the Code of Chivalry only allows swapping the opposite knights in the circle, i.e., Arthur can simultaneously perform $a_i := a_{i + n}$, $a_{i + n} := a_i$ for any $1 \le i \le n$. Arthur can make any number of such swaps. What is the best result he can achieve?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$) — the number of desks.

The second line consists of $2n$ integers $a_1, a_2, \ldots, a_{2 n}$ ($1 \le a_i \le 10^9$) — the intelligence values of the knights.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, output a single line containing one integer — the minimal difference Arthur can achieve.

Examples:
input:
5
2
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
output:
0
0
0
2
4

Note: In the first test case, Arthur can swap the second and the fourth knights. Then the total intelligence at both desks will be $10$.

In the third test case, Arthur can make $0$ operations, which will result in the total intelligence of $11$ at each of the desks.

In the fourth test case, Arthur can swap knights with indices $2$ and $5$ and achieve the difference of $2$. It can be proven that he cannot improve his result any further.

Buggy Code:
import java.util.ArrayList;
import java.util.Scanner;

public class Solution{

    public static void caseEven(int[] intellgence,int n){

        int maxPair = Integer.MIN_VALUE;
        int minPair = Integer.MAX_VALUE;
        for (int left = 1,right = 2; right <= n; left+=2,right+=2) {
            int pair1 = Math.max(intellgence[left],intellgence[left+n])
                + Math.min(intellgence[right],intellgence[right+n]);
            int pair2 = intellgence[left] + intellgence[left+n]
                + intellgence[right] + intellgence[right+n] - pair1;
            if (pair1 > maxPair) maxPair = pair1;
            if (pair1 < minPair) minPair = pair1;
            if (pair2 > maxPair) maxPair = pair2;
            if (pair2 < minPair) minPair = pair2;
        }

        System.out.println(maxPair - minPair);
    }

    //数组很适合
    // record PairInfo(int zeroZero,int zeroOne,int oneZero,int oneOne){};
    static final int zeroZero = 0;
    static final int zeroOne = 1;
    static final int oneZero = 2;
    static final int oneOne = 3;
    static int myMax(int x,int y){
        if (x == 0
            || y == 0)
        {
            return 0;
        }
        return Math.max(x, y);
    }
    static int myMin(int x,int y){
        if (x == 0
            || y == 0)
        {
            return Math.max(x, y);
        }
        return Math.min(x, y);
    }
    record SumInfo(int pairIndex,int sumIndex,int value){}

    static int[] merge(int[] p1,int[] p2){
        return new int[]{
            Math.min(Math.max(p1[zeroZero],p2[zeroZero]),Math.max(p1[zeroOne],p2[oneZero])),
            Math.min(Math.max(p1[zeroZero],p2[zeroOne]),Math.max(p1[zeroOne],p2[oneOne])),
            Math.min(Math.max(p1[oneZero],p2[zeroZero]),Math.max(p1[oneOne],p2[oneZero])),
            Math.min(Math.max(p1[oneZero],p2[zeroOne]), Math.max(p1[oneOne], p2[oneOne]))};
    }

    static boolean checkPass(int[] pair){
        if (pair[0] == Integer.MAX_VALUE
            && pair[1] == Integer.MAX_VALUE
            && pair[2] == Integer.MAX_VALUE
            && pair[3] == Integer.MAX_VALUE)
        {
            return false;
        }

        return true;
    }

    static void caseOdd(int[] knight,int n){

        int[][] pairArr = new int[n][0];
        ArrayList<Solution.SumInfo> sumList = new ArrayList<SumInfo>();
        for (int i = 1; i <= n; i++) {
            int left,right,anotherLeft,anotherRight;
            if (i % 2 == 1) {
                left = i;
                right = i+1;
                anotherLeft = i+n;
                anotherRight = (i+1+n)%(2*n);
            }
            else {
                anotherLeft = i;
                anotherRight = i+1;
                left = i+n;
                right = i+1+n;
            }

            int[] thisPair = new int[]{
                knight[left]+knight[right],
                knight[left]+knight[anotherRight],
                knight[anotherLeft]+knight[right],
                knight[anotherLeft]+knight[anotherRight]};
            for (int j = 0; j < 4;j++){
                sumList.add(new SumInfo(i-1, j, thisPair[j]));
            }
            pairArr[i-1] = thisPair;
        }
        sumList.sort((s1,s2)->(s1.value - s2.value));

        int minDiff = Integer.MAX_VALUE;
        int prevMinSum = -1;
        for (SumInfo info : sumList) {
            if (info.value == prevMinSum) {
                pairArr[info.pairIndex][info.sumIndex] = Integer.MAX_VALUE;
                if (!checkPass(pairArr[info.pairIndex])) {
                    break;
                }
                continue;
            }

            //检测prevMinSum
            int minSum = info.value;
            int[] dp = pairArr[0];
            int[] dpWithMin = new int[]{Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE};
            for (int i = 0; i < 4; i++) {
                if (dp[i] == minSum) {
                    dpWithMin[i] = minSum;
                }
            }

            for (int i = 1; i < pairArr.length; i++) {
                int[] thisPair = pairArr[i];

                dpWithMin = merge(dpWithMin,thisPair);
                for (int j = 0; j < 4; j++) {
                    if (thisPair[j] != minSum) continue;
                    int[] auxPair = new int[]{
                        thisPair[0] == minSum ? minSum : Integer.MAX_VALUE,
                        thisPair[1] == minSum ? minSum : Integer.MAX_VALUE,
                        thisPair[2] == minSum ? minSum : Integer.MAX_VALUE,
                        thisPair[3] == minSum ? minSum : Integer.MAX_VALUE};
                    int[] auxDp = merge(dp,auxPair);
                    for (int k = 0; k < 4; k++) {
                        if (auxDp[k] < dpWithMin[k])
                            dpWithMin[k] = auxDp[k];
                    }
                }
                dp = merge(dp,thisPair);
            }
            if (!checkPass(dpWithMin)) {
                continue;
            }
            minDiff = Math.min(minDiff,dpWithMin[0] - minSum);
            minDiff = Math.min(minDiff,dpWithMin[3] - minSum);

            prevMinSum = info.value;
            pairArr[info.pairIndex][info.sumIndex] = Integer.MAX_VALUE;
            if (!checkPass(pairArr[info.pairIndex])) {
                break;
            }
        }

        System.out.println(minDiff);
    }

    public static void slove(Scanner scanner){

        int t = scanner.nextInt();
        while (t > 0 ) {t--;
            int n = scanner.nextInt();
            int[] knight = new int[2*n+1];
            for (int i = 1; i < knight.length; i++) {
                knight[i] = scanner.nextInt();
            }
            if (n % 2 == 0) {
                caseEven(knight,n);
                continue;
            }
            if (n == 1){
                System.out.println(0);
                continue;
            }
            caseOdd(knight,n);
        }
    }

    public static void main(String[] args) {
        try (Scanner scanner = new Scanner(System.in);) {
            slove(scanner);
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases or scenarios that might cause the code to fail. The code has two main parts: one for even `n` and one for odd `n`. Let's consider the following points:

1. **Even `n`**:
   - The code assumes that pairs are formed by adjacent elements and their opposite elements. However, it might not handle cases where the intelligence values are not uniformly distributed correctly.

2. **Odd `n`**:
   - The code uses dynamic programming to find the minimum difference. It might fail in cases where the intelligence values are highly skewed or where the optimal solution requires a specific arrangement that the code does not handle correctly.

Given these points, let's create a small-scale test input that could expose potential issues:

### Test Input
```plaintext
3
2
1 10 10 1
3
1 10 1 10 1 10
5
1 2 3 4 5 6 7 8 9 10
```

### Explanation
1. **Test Case 1: `n = 2`**
   - Intelligence values: `1 10 10 1`
   - This case tests the even `n` part of the code. The optimal solution should be to swap the second and fourth knights, resulting in pairs `(1, 10)` and `(10, 1)`, both with a sum of 11, leading to a minimal difference of 0.

2. **Test Case 2: `n = 3`**
   - Intelligence values: `1 10 1 10 1 10`
   - This case tests the odd `n` part of the code. The optimal solution should be to keep the pairs as they are, resulting in pairs `(1, 10)`, `(1, 10)`, and `(1, 10)`, all with a sum of 11, leading to a minimal difference of 0.

3. **Test Case 3: `n = 5`**
   - Intelligence values: `1 2 3 4 5 6 7 8 9 10`
   - This case tests the odd `n` part of the code with a more complex set of intelligence values. The optimal solution should be to minimize the difference between the sums of the pairs, which should result in a minimal difference of 4.

This test input is designed to cover a range of scenarios that could expose issues in the code, particularly in how it handles different distributions of intelligence values and the logic for both even and odd `n`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Let Me Teach You a Lesson (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the easy version of a problem. The only difference between an easy and a hard version is the constraints on $t$ and $n$. You can make hacks only if both versions of the problem are solved.

Arthur is giving a lesson to his famous $2 n$ knights. Like any other students, they're sitting at the desks in pairs, but out of habit in a circle. The knight $2 i - 1$ is sitting at the desk with the knight $2 i$.

Each knight has intelligence, which can be measured by an integer. Let's denote the intelligence of the $i$-th knight as $a_i$. Arthur wants the maximal difference in total intelligence over all pairs of desks to be as small as possible. More formally, he wants to minimize $\max\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i}) - \min\limits_{1 \le i \le n} (a_{2 i - 1} + a_{2 i})$.

However, the Code of Chivalry only allows swapping the opposite knights in the circle, i.e., Arthur can simultaneously perform $a_i := a_{i + n}$, $a_{i + n} := a_i$ for any $1 \le i \le n$. Arthur can make any number of such swaps. What is the best result he can achieve?

Input: Each test consists of several test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. It is followed by descriptions of the test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 2000$) — the number of desks.

The second line consists of $2n$ integers $a_1, a_2, \ldots, a_{2 n}$ ($1 \le a_i \le 10^9$) — the intelligence values of the knights.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2000$.

Output: For each test case, output a single line containing one integer — the minimal difference Arthur can achieve.

Examples:
input:
5
2
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
output:
0
0
0
2
4

Note: In the first test case, Arthur can swap the second and the fourth knights. Then the total intelligence at both desks will be $10$.

In the third test case, Arthur can make $0$ operations, which will result in the total intelligence of $11$ at each of the desks.

In the fourth test case, Arthur can swap knights with indices $2$ and $5$ and achieve the difference of $2$. It can be proven that he cannot improve his result any further.

Buggy Code:
import java.util.ArrayList;
import java.util.Scanner;

public class Solution{

    public static void caseEven(int[] intellgence,int n){

        int maxPair = Integer.MIN_VALUE;
        int minPair = Integer.MAX_VALUE;
        for (int left = 1,right = 2; right <= n; left+=2,right+=2) {
            int pair1 = Math.max(intellgence[left],intellgence[left+n])
                + Math.min(intellgence[right],intellgence[right+n]);
            int pair2 = intellgence[left] + intellgence[left+n]
                + intellgence[right] + intellgence[right+n] - pair1;
            if (pair1 > maxPair) maxPair = pair1;
            if (pair1 < minPair) minPair = pair1;
            if (pair2 > maxPair) maxPair = pair2;
            if (pair2 < minPair) minPair = pair2;
        }

        System.out.println(maxPair - minPair);
    }

    //数组很适合
    // record PairInfo(int zeroZero,int zeroOne,int oneZero,int oneOne){};
    static final int zeroZero = 0;
    static final int zeroOne = 1;
    static final int oneZero = 2;
    static final int oneOne = 3;
    static int myMax(int x,int y){
        if (x == 0
            || y == 0)
        {
            return 0;
        }
        return Math.max(x, y);
    }
    static int myMin(int x,int y){
        if (x == 0
            || y == 0)
        {
            return Math.max(x, y);
        }
        return Math.min(x, y);
    }
    record SumInfo(int pairIndex,int sumIndex,int value){}

    static int[] merge(int[] p1,int[] p2){
        return new int[]{
            Math.min(Math.max(p1[zeroZero],p2[zeroZero]),Math.max(p1[zeroOne],p2[oneZero])),
            Math.min(Math.max(p1[zeroZero],p2[zeroOne]),Math.max(p1[zeroOne],p2[oneOne])),
            Math.min(Math.max(p1[oneZero],p2[zeroZero]),Math.max(p1[oneOne],p2[oneZero])),
            Math.min(Math.max(p1[oneZero],p2[zeroOne]), Math.max(p1[oneOne], p2[oneOne]))};
    }

    static void caseOdd(int[] knight,int n){

        int[][] pairArr = new int[n][0];
        ArrayList<Solution.SumInfo> sumList = new ArrayList<SumInfo>();
        for (int i = 1; i <= n; i++) {
            int left,right,anotherLeft,anotherRight;
            if (i % 2 == 1) {
                left = i;
                right = i+1;
                anotherLeft = i+n;
                anotherRight = (i+1+n)%(2*n);
            }
            else {
                anotherLeft = i;
                anotherRight = i+1;
                left = i+n;
                right = i+1+n;
            }

            int[] thisPair = new int[]{
                knight[left]+knight[right],
                knight[left]+knight[anotherRight],
                knight[anotherLeft]+knight[right],
                knight[anotherLeft]+knight[anotherRight]};
            for (int j = 0; j < 4;j++){
                sumList.add(new SumInfo(i-1, j, thisPair[j]));
            }
            pairArr[i-1] = thisPair;
        }
        sumList.sort((s1,s2)->(s1.value - s2.value));

        int minDiff = Integer.MAX_VALUE;
        int prevMinSum = -1;
        for (SumInfo info : sumList) {
            if (info.value == prevMinSum) {
                pairArr[info.pairIndex][info.sumIndex] = Integer.MAX_VALUE;
                continue;
            }

            //检测prevMinSum
            int minSum = info.value;
            int[] dp = pairArr[0];
            int[] dpWithMin = new int[]{Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE,Integer.MAX_VALUE};
            for (int i = 0; i < 4; i++) {
                if (dp[i] == minSum) {
                    dpWithMin[i] = minSum;
                }
            }

            for (int i = 1; i < pairArr.length; i++) {
                int[] thisPair = pairArr[i];

                dpWithMin = merge(dpWithMin,thisPair);
                for (int j = 0; j < 4; j++) {
                    if (thisPair[j] != minSum) continue;
                    int[] auxPair = new int[]{
                        thisPair[0] == minSum ? minSum : Integer.MAX_VALUE,
                        thisPair[1] == minSum ? minSum : Integer.MAX_VALUE,
                        thisPair[2] == minSum ? minSum : Integer.MAX_VALUE,
                        thisPair[3] == minSum ? minSum : Integer.MAX_VALUE};
                    int[] auxDp = merge(dp,auxPair);
                    for (int k = 0; k < 4; k++) {
                        if (auxDp[k] < dpWithMin[k])
                            dpWithMin[k] = auxDp[k];
                    }
                }
                dp = merge(dp,thisPair);
            }
            if (dpWithMin[0] == Integer.MAX_VALUE
                && dpWithMin[1] == Integer.MAX_VALUE
                && dpWithMin[2] == Integer.MAX_VALUE
                && dpWithMin[3] == Integer.MAX_VALUE)
            {
                continue;
            }
            minDiff = Math.min(minDiff,dpWithMin[0] - minSum);
            minDiff = Math.min(minDiff,dpWithMin[3] - minSum);

            prevMinSum = info.value;
            pairArr[info.pairIndex][info.sumIndex] = Integer.MAX_VALUE;
        }

        System.out.println(minDiff);
    }

    public static void slove(Scanner scanner){

        int t = scanner.nextInt();
        while (t > 0 ) {t--;
            int n = scanner.nextInt();
            int[] knight = new int[2*n+1];
            for (int i = 1; i < knight.length; i++) {
                knight[i] = scanner.nextInt();
            }
            if (n % 2 == 0) {
                caseEven(knight,n);
                continue;
            }
            if (n == 1){
                System.out.println(0);
                continue;
            }
            caseOdd(knight,n);
        }
    }

    public static void main(String[] args) {
        try (Scanner scanner = new Scanner(System.in);) {
            slove(scanner);
        }
    }
}

**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
6 6 4 4
1
10 17
3
1 10 1 10 1 10
3
3 3 4 5 5 4
5
1 2 3 4 5 6 7 8 9 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Adjust The Presentation (Hard Version)
time_limit_ms: 5000
memory_limit_mb: 256
Description: This is the hard version of the problem. In the two versions, the constraints on $$$q$$$ and the time limit are different. In this version, $$$0 \leq q \leq 2 \cdot 10^5$$$. You can make hacks only if all the versions of the problem are solved.

A team consisting of $$$n$$$ members, numbered from $$$1$$$ to $$$n$$$, is set to present a slide show at a large meeting. The slide show contains $$$m$$$ slides.

There is an array $$$a$$$ of length $$$n$$$. Initially, the members are standing in a line in the order of $$$a_1, a_2, \ldots, a_n$$$ from front to back. The slide show will be presented in order from slide $$$1$$$ to slide $$$m$$$. Each section will be presented by the member at the front of the line. After each slide is presented, you can move the member at the front of the line to any position in the lineup (without changing the order of the rest of the members). For example, suppose the line of members is $$$[\color{red}{3},1,2,4]$$$. After member $$$3$$$ presents the current slide, you can change the line of members into either $$$[\color{red}{3},1,2,4]$$$, $$$[1,\color{red}{3},2,4]$$$, $$$[1,2,\color{red}{3},4]$$$ or $$$[1,2,4,\color{red}{3}]$$$.

There is also an array $$$b$$$ of length $$$m$$$. The slide show is considered good if it is possible to make member $$$b_i$$$ present slide $$$i$$$ for all $$$i$$$ from $$$1$$$ to $$$m$$$ under these constraints.

However, your annoying boss wants to make $$$q$$$ updates to the array $$$b$$$. In the $$$i$$$-th update, he will choose a slide $$$s_i$$$ and a member $$$t_i$$$ and set $$$b_{s_i} := t_i$$$. Note that these updates are persistent, that is changes made to the array $$$b$$$ will apply when processing future updates.

For each of the $$$q+1$$$ states of array $$$b$$$, the initial state and after each of the $$$q$$$ updates, determine if the slideshow is good.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$ and $$$q$$$ ($$$1 \le n, m \le 2 \cdot 10^5$$$; $$$0 \leq q \leq 2 \cdot 10^5$$$) — the number of members and the number of sections.

The second line of each test case contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le n$$$) — the initial order of the members from front to back. It is guaranteed that each integer from $$$1$$$ to $$$n$$$ appears exactly once in $$$a$$$.

The third line of each test case contains $$$m$$$ integers $$$b_1, b_2, \ldots, b_m$$$ ($$$1 \le b_i \le n$$$) — the members who should present each section.

Each of the next $$$q$$$ lines contains two integers $$$s_i$$$ and $$$t_i$$$ ($$$1 \le s_i \le m$$$, $$$1 \le t_i \le n$$$) — parameters of an update.

It is guaranteed that the sum of $$$n$$$, the sum of $$$m$$$ and the sum of $$$q$$$ over all test cases do not exceed $$$2 \cdot 10^5$$$ respectively.

Output: For each test case, output $$$q+1$$$ lines corresponding to the $$$q+1$$$ states of the array $$$b$$$. Output "YA" if the slide show is good, and "TIDAK" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yA", "Ya", "ya", and "YA" will be recognized as positive responses.

Examples:
input:
3
4 2 2
1 2 3 4
1 1
1 2
1 1
3 6 2
1 2 3
1 1 2 3 3 2
3 3
2 2
4 6 2
3 1 4 2
3 1 1 2 3 4
3 4
4 2
output:
YA
TIDAK
YA
YA
TIDAK
YA
TIDAK
YA
YA

Note: For the first test case, you do not need to move the members as both slides are presented by member $$$1$$$, who is already at the front of the line. After that, set $$$b_1 := 2$$$, now slide $$$1$$$ must be presented by member $$$2$$$ which is impossible as member $$$1$$$ will present slide $$$1$$$ first. Then, set $$$b_1 = 1$$$, the $$$b$$$ is the same as the initial $$$b$$$, making a good presentation possible.

Buggy Code:
#include <algorithm>
#include <iostream>
#include <queue>
#include <set>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <vector>

using namespace std;

int n, m, qq;
vector<int> A;
vector<int> B;

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

  // vector<int> ans(t);
  for (int q = 0; q < t; q++) {
    cin >> n >> m >> qq;
    A.resize(n);
    B.resize(m);
    vector<set<int>> ss(n);
    for (int i = 0; i < n; i++) {
      cin >> A[i];
    }
    for (int i = 0; i < m; i++) {
      cin >> B[i];
    }

    unordered_map<int, int> sortt;
    for (int i = 0; i < n; i++) {
      sortt[A[i]] = i;
    }

    for (int i = 0; i < m; i++) {
      ss[sortt[B[i]]].insert(i);
    }

    vector<int> first(n);
    for (int i = 0; i < n; i++) {
      if (!ss[i].empty()) {
        first[i] = *ss[i].begin();
      } else {
        first[i] = m + 1;
      }
    }

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

    if (c == n - 1) {
      cout << "YA" << endl;
    } else {
      cout << "TIDAK" << endl;
    }

    for (int qqq = 0; qqq < qq; qqq++) {
      int s, t;
      cin >> s >> t;
      ss[sortt[B[s - 1]]].erase(s - 1);
      first[sortt[B[s - 1]]] = *ss[sortt[B[s - 1]]].begin();
      ss[sortt[t]].insert(s - 1);
      first[sortt[t]] = *ss[sortt[t]].begin();
      B[s - 1] = t;

      // for (int i = 0; i < n; i++) {
      //   if (!ss[i].empty()) {
      //     first[i] = *ss[i].begin();
      //   } else {
      //     first[i] = m + 10;
      //   }
      // }

      bool y = true;
      for (int i = 0; i < n - 1; i++) {
        if (first[i] > first[i + 1]) {
          y = false;
          break;
        }
      }
      if (y) {
        cout << "YA" << endl;
      } else {
        cout << "TIDAK" << endl;
      }
      // for (int i = 0; i < n; i++) {
      //   cout << i << ": ";
      //   for (auto it = ss[i].begin(); it != ss[i].end(); it++) {
      //     cout << *it << " ";
      //   }
      //   cout << 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
4 4 2
1 2 3 4
1 2 3 4
1 3
2 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Adjust The Presentation (Hard Version)
time_limit_ms: 5000
memory_limit_mb: 256
Description: This is the hard version of the problem. In the two versions, the constraints on $$$q$$$ and the time limit are different. In this version, $$$0 \leq q \leq 2 \cdot 10^5$$$. You can make hacks only if all the versions of the problem are solved.

A team consisting of $$$n$$$ members, numbered from $$$1$$$ to $$$n$$$, is set to present a slide show at a large meeting. The slide show contains $$$m$$$ slides.

There is an array $$$a$$$ of length $$$n$$$. Initially, the members are standing in a line in the order of $$$a_1, a_2, \ldots, a_n$$$ from front to back. The slide show will be presented in order from slide $$$1$$$ to slide $$$m$$$. Each section will be presented by the member at the front of the line. After each slide is presented, you can move the member at the front of the line to any position in the lineup (without changing the order of the rest of the members). For example, suppose the line of members is $$$[\color{red}{3},1,2,4]$$$. After member $$$3$$$ presents the current slide, you can change the line of members into either $$$[\color{red}{3},1,2,4]$$$, $$$[1,\color{red}{3},2,4]$$$, $$$[1,2,\color{red}{3},4]$$$ or $$$[1,2,4,\color{red}{3}]$$$.

There is also an array $$$b$$$ of length $$$m$$$. The slide show is considered good if it is possible to make member $$$b_i$$$ present slide $$$i$$$ for all $$$i$$$ from $$$1$$$ to $$$m$$$ under these constraints.

However, your annoying boss wants to make $$$q$$$ updates to the array $$$b$$$. In the $$$i$$$-th update, he will choose a slide $$$s_i$$$ and a member $$$t_i$$$ and set $$$b_{s_i} := t_i$$$. Note that these updates are persistent, that is changes made to the array $$$b$$$ will apply when processing future updates.

For each of the $$$q+1$$$ states of array $$$b$$$, the initial state and after each of the $$$q$$$ updates, determine if the slideshow is good.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$ and $$$q$$$ ($$$1 \le n, m \le 2 \cdot 10^5$$$; $$$0 \leq q \leq 2 \cdot 10^5$$$) — the number of members and the number of sections.

The second line of each test case contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le n$$$) — the initial order of the members from front to back. It is guaranteed that each integer from $$$1$$$ to $$$n$$$ appears exactly once in $$$a$$$.

The third line of each test case contains $$$m$$$ integers $$$b_1, b_2, \ldots, b_m$$$ ($$$1 \le b_i \le n$$$) — the members who should present each section.

Each of the next $$$q$$$ lines contains two integers $$$s_i$$$ and $$$t_i$$$ ($$$1 \le s_i \le m$$$, $$$1 \le t_i \le n$$$) — parameters of an update.

It is guaranteed that the sum of $$$n$$$, the sum of $$$m$$$ and the sum of $$$q$$$ over all test cases do not exceed $$$2 \cdot 10^5$$$ respectively.

Output: For each test case, output $$$q+1$$$ lines corresponding to the $$$q+1$$$ states of the array $$$b$$$. Output "YA" if the slide show is good, and "TIDAK" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yA", "Ya", "ya", and "YA" will be recognized as positive responses.

Examples:
input:
3
4 2 2
1 2 3 4
1 1
1 2
1 1
3 6 2
1 2 3
1 1 2 3 3 2
3 3
2 2
4 6 2
3 1 4 2
3 1 1 2 3 4
3 4
4 2
output:
YA
TIDAK
YA
YA
TIDAK
YA
TIDAK
YA
YA

Note: For the first test case, you do not need to move the members as both slides are presented by member $$$1$$$, who is already at the front of the line. After that, set $$$b_1 := 2$$$, now slide $$$1$$$ must be presented by member $$$2$$$ which is impossible as member $$$1$$$ will present slide $$$1$$$ first. Then, set $$$b_1 = 1$$$, the $$$b$$$ is the same as the initial $$$b$$$, making a good presentation possible.

Buggy Code:
#include<iostream>

#define int long long
using namespace std;
const int N=2e5+10;
int a[N],b[N];
int st[N];
int et[N];

signed main()
{
	cin.tie(0),cout.tie(0),ios::sync_with_stdio(false);
	int t;
	cin>>t;
	while(t--)
	{
		int n,m,q;
		cin>>n>>m>>q;
		for(int i=1;i<=n;i++)cin>>a[i],et[a[i]]=i;
		for(int i=1;i<=m;i++)cin>>b[i];
		int es=0;
		bool f=true;
		int ett=-1;
		for(int i=1;i<=m;i++)
		{
			if(et[b[i]]<=es+1)
			{
				es=max(es,et[b[i]]);
				st[i]=es;
				continue;
			}
			ett=i;
			f=false;
			break;
		}
		if(f) cout<<"YA"<<'\n',ett=n+1;
		else cout<<"TIDAK"<<'\n';
		while(q--)
		{
			int x,y;
			bool ef=false;
			f=false;
			cin>>x>>y;
			if(x<ett)
			{
				if(et[y]<=st[x]&&ett>n) f=true;
				if(st[x-1]==st[x]&&et[y]<=st[x]+1&&ett>n)f=true;
				if(ett<=n)
				{
					b[x]=y;
					ef=true;
					if(st[x-1]==st[x]&&st[x]+1>=et[y]) st[x]=max(et[y],st[x]),es=max(es,et[y]);
					int sst=x;
					while(sst<=n)
					{
						if(et[b[sst]]<=es+1)
						{
							es=max(et[b[sst]],es);
							st[sst]=max(st[sst],es);
							sst++;
							continue;
						}
						break;
					}
					if(sst==n+1) f=true;
					ett=max(sst,ett);
				}
			}
			else if(x==ett)
			{
				ef=true;
				f=true;;
				b[x]=y;
				while(ett<=n)
				{
					if(et[b[ett]]<=es+1)
					{
						es=max(es,et[b[ett]]);
						st[ett]=es;
						ett++;
						continue;
					}
					f=false;
					break;
				}
			}
			if(!ef)b[x]=y,st[x]=max(st[x],et[y]);
			if(f) cout<<"YA"<<'\n';
			else cout<<"TIDAK"<<'\n';
		}
	}
	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Adjust The Presentation (Hard Version)
time_limit_ms: 5000
memory_limit_mb: 256
Description: This is the hard version of the problem. In the two versions, the constraints on $$$q$$$ and the time limit are different. In this version, $$$0 \leq q \leq 2 \cdot 10^5$$$. You can make hacks only if all the versions of the problem are solved.

A team consisting of $$$n$$$ members, numbered from $$$1$$$ to $$$n$$$, is set to present a slide show at a large meeting. The slide show contains $$$m$$$ slides.

There is an array $$$a$$$ of length $$$n$$$. Initially, the members are standing in a line in the order of $$$a_1, a_2, \ldots, a_n$$$ from front to back. The slide show will be presented in order from slide $$$1$$$ to slide $$$m$$$. Each section will be presented by the member at the front of the line. After each slide is presented, you can move the member at the front of the line to any position in the lineup (without changing the order of the rest of the members). For example, suppose the line of members is $$$[\color{red}{3},1,2,4]$$$. After member $$$3$$$ presents the current slide, you can change the line of members into either $$$[\color{red}{3},1,2,4]$$$, $$$[1,\color{red}{3},2,4]$$$, $$$[1,2,\color{red}{3},4]$$$ or $$$[1,2,4,\color{red}{3}]$$$.

There is also an array $$$b$$$ of length $$$m$$$. The slide show is considered good if it is possible to make member $$$b_i$$$ present slide $$$i$$$ for all $$$i$$$ from $$$1$$$ to $$$m$$$ under these constraints.

However, your annoying boss wants to make $$$q$$$ updates to the array $$$b$$$. In the $$$i$$$-th update, he will choose a slide $$$s_i$$$ and a member $$$t_i$$$ and set $$$b_{s_i} := t_i$$$. Note that these updates are persistent, that is changes made to the array $$$b$$$ will apply when processing future updates.

For each of the $$$q+1$$$ states of array $$$b$$$, the initial state and after each of the $$$q$$$ updates, determine if the slideshow is good.

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

The first line of each test case contains three integers $$$n$$$, $$$m$$$ and $$$q$$$ ($$$1 \le n, m \le 2 \cdot 10^5$$$; $$$0 \leq q \leq 2 \cdot 10^5$$$) — the number of members and the number of sections.

The second line of each test case contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$1 \le a_i \le n$$$) — the initial order of the members from front to back. It is guaranteed that each integer from $$$1$$$ to $$$n$$$ appears exactly once in $$$a$$$.

The third line of each test case contains $$$m$$$ integers $$$b_1, b_2, \ldots, b_m$$$ ($$$1 \le b_i \le n$$$) — the members who should present each section.

Each of the next $$$q$$$ lines contains two integers $$$s_i$$$ and $$$t_i$$$ ($$$1 \le s_i \le m$$$, $$$1 \le t_i \le n$$$) — parameters of an update.

It is guaranteed that the sum of $$$n$$$, the sum of $$$m$$$ and the sum of $$$q$$$ over all test cases do not exceed $$$2 \cdot 10^5$$$ respectively.

Output: For each test case, output $$$q+1$$$ lines corresponding to the $$$q+1$$$ states of the array $$$b$$$. Output "YA" if the slide show is good, and "TIDAK" otherwise.

You can output the answer in any case (upper or lower). For example, the strings "yA", "Ya", "ya", and "YA" will be recognized as positive responses.

Examples:
input:
3
4 2 2
1 2 3 4
1 1
1 2
1 1
3 6 2
1 2 3
1 1 2 3 3 2
3 3
2 2
4 6 2
3 1 4 2
3 1 1 2 3 4
3 4
4 2
output:
YA
TIDAK
YA
YA
TIDAK
YA
TIDAK
YA
YA

Note: For the first test case, you do not need to move the members as both slides are presented by member $$$1$$$, who is already at the front of the line. After that, set $$$b_1 := 2$$$, now slide $$$1$$$ must be presented by member $$$2$$$ which is impossible as member $$$1$$$ will present slide $$$1$$$ first. Then, set $$$b_1 = 1$$$, the $$$b$$$ is the same as the initial $$$b$$$, making a good presentation possible.

Buggy Code:
#include<iostream>
#include<algorithm>
#include<queue>
#include<cstring>
#include<vector>
#include<unordered_map>
#include<deque>
#include<cmath>
using namespace std;

typedef long long ll;
typedef pair<int, int> PII;
#define x first
#define y second

const int N = 2e5, INF = 1e9;

ll a[N],b[N];
int n, m, q;
unordered_map<int, int> itoi,itoj;

void check()
{
    int j = 1, i = 0;
    while (i <= n && j <= m)
    {
        while (a[i] == b[j] || itoi[b[j]]) j++;
        if (a[++i] != b[j]) break;
        else {
            itoi[a[i]] = 1;
            itoj[a[i]] = j;//a[i]从j之后就自由，否则不自由
        }
    }
    if (i > n || j > m) cout << "ya" << endl;
    else cout << "TIDAK" << endl;

}

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        itoi.clear();
        cin >> n >> m >> q;
        for (int i = 1; i <= n; i++) cin >> a[i];
        for (int i = 1; i <= m; i++) cin >> b[i];
        check();
        for (int i = 1; i <= q; i++)
        {
            itoi.clear();
            int x, y;
            cin >> x >> y;
            b[x] = y;
            if (itoj[y] && itoj[y] < x)
            {
                cout << "ya" << endl;
                continue;
            }

            check();
        }
    }
    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
4 4 2
1 2 3 4
1 2 3 4
1 2
3 1
```