**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Expected Power
time_limit_ms: 4000
memory_limit_mb: 256
Description: You are given an array of $n$ integers $a_1,a_2,\ldots,a_n$. You are also given an array $p_1, p_2, \ldots, p_n$.

Let $S$ denote the random multiset (i. e., it may contain equal elements) constructed as follows:

  * Initially, $S$ is empty.    * For each $i$ from $1$ to $n$, insert $a_i$ into $S$ with probability $\frac{p_i}{10^4}$. Note that each element is inserted independently.

Denote $f(S)$ as the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all elements of $S$. Please calculate the expected value of $(f(S))^2$. Output the answer modulo $10^9 + 7$.

Formally, let $M = 10^9 + 7$. It can be shown that the answer can be expressed as an irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \not \equiv 0 \pmod{M}$. Output the integer equal to $p \cdot q^{-1} \bmod M$. In other words, output such an integer $x$ that $0 \le x < M$ and $x \cdot q \equiv p \pmod{M}$.

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 a single integer $n$ ($1 \le n \le 2 \cdot 10^5$).

The second line of each test case contains $n$ integers $a_1,a_2,\ldots,a_n$ ($1 \le a_i \le 1023$).

The third line of each test case contains $n$ integers $p_1,p_2,\ldots,p_n$ ($1 \le p_i \le 10^4$).

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

Output: For each test case, output the expected value of $(f(S))^2$, modulo $10^9 + 7$.

Examples:
input:
4
2
1 2
5000 5000
2
1 1
1000 2000
6
343 624 675 451 902 820
6536 5326 7648 2165 9430 5428
1
1
10000
output:
500000007
820000006
280120536
1

Note: In the first test case, $a = [1, 2]$ and each element is inserted into $S$ with probability $\frac{1}{2}$, since $p_1 = p_2 = 5000$ and $\frac{p_i}{10^4} = \frac{1}{2}$. Thus, there are $4$ outcomes for $S$, each happening with the same probability of $\frac{1}{4}$:

  * $S = \varnothing$. In this case, $f(S) = 0$, $(f(S))^2 = 0$.    * $S = \\{1\\}$. In this case, $f(S) = 1$, $(f(S))^2 = 1$.    * $S = \\{2\\}$. In this case, $f(S) = 2$, $(f(S))^2 = 4$.    * $S = \\{1,2\\}$. In this case, $f(S) = 1 \oplus 2 = 3$, $(f(S))^2 = 9$.

Hence, the answer is $0 \cdot \frac{1}{4} + 1 \cdot \frac{1}{4} + 4\cdot \frac{1}{4} + 9 \cdot \frac{1}{4} = \frac{14}{4} = \frac{7}{2} \equiv 500\,000\,007 \pmod{10^9 + 7}$.

In the second test case, $a = [1, 1]$, $a_1$ is inserted into $S$ with probability $0.1$, while $a_2$ is inserted into $S$ with probability $0.2$. There are $3$ outcomes for $S$:

  * $S = \varnothing$. In this case, $f(S) = 0$, $(f(S))^2 = 0$. This happens with probability $(1-0.1) \cdot (1-0.2) = 0.72$.    * $S = \\{1\\}$. In this case, $f(S) = 1$, $(f(S))^2 = 1$. This happens with probability $(1-0.1) \cdot 0.2 + 0.1 \cdot (1-0.2) = 0.26$.    * $S = \\{1, 1\\}$. In this case, $f(S) = 0$, $(f(S))^2 = 0$. This happens with probability $0.1 \cdot 0.2 = 0.02$.

Hence, the answer is $0 \cdot 0.72 + 1 \cdot 0.26 + 0 \cdot 0.02 = 0.26 = \frac{26}{100} \equiv 820\,000\,006 \pmod{10^9 + 7}$.

Buggy Code:
import java.util.Scanner;

public class ExpectedPower {
	static long mod = (long)1e9 + 7;

	private static long qpow(long x, long y) {
		long res = 1;
		while(y != 0) {
			if((y & 1) == 1) {
				res = res * x % mod;
			}
			x *= x;
			x %= mod;
			y >>= 1;
		}
		return res;
	}

	private static long inv(long x) {
		return qpow(x, mod - 2);
	}

	private static long mo(long x) {
		return (x % mod + mod) % mod;
	}

	static Scanner sc = new Scanner(System.in);
	static long inv1e4 = inv(10000);
	public static void main(String[] args) {
		int t = sc.nextInt();
		while(t-- > 0) {
			int n = sc.nextInt();
			int[] A = new int[n + 1];
			long[] P = new long[n + 1];
			for(int i = 1; i <= n; i++) {
				A[i] = sc.nextInt();
			}
			for(int i = 1; i <= n; i++) {
				P[i] = sc.nextLong();
				P[i] = P[i] * inv1e4 % mod; // pi/1e4 = pi * inv(1e4)
			}
			long res = solve(A, P);
			System.out.println(res);
		}
	}

	private static long solve(int[] A, long[] P) {
		int n = A.length - 1;
		long[][] f = new long[2][1024]; // [0] => [1] => [0]...
		f[0][0] = 1;
		for(int i = 1; i <= n; i++) {
			int sel = i & 1;
			for(int j = 0; j < 1024; j++) {
				f[sel][j] = 0;
				f[sel][j] = (f[sel][j] + (f[1 - sel][j] * mo(1 - P[i]) % mod)) % mod;
				f[sel][j] = (f[sel][j] + (f[1 - sel][j ^ A[i]] * P[i] % mod)) % mod;
			}
		}
		long res = 0;
		for(int j = 0; j < 1024; j++) {
			res += f[n & 1][j] * j * j % mod;
			res %= mod;
		}
		return res;
	}
}

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

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

Think step by step.

Response:
```plaintext
3
1
1023
10000
2
1023 1023
10000 10000
3
1023 1023 1023
10000 10000 10000
```


**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 (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 10\,000$) — 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 $n$ ($1 \le n \le 100\,000$) — 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 $100\,000$.

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:
// LUOGU_RID: 170495095
#include<bits/stdc++.h>
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
#define pii pair<ll,ll>
#define rep(i,a,b) for(ll i=(a);i<=(b);++i)
#define per(i,a,b) for(ll i=(a);i>=(b);--i)
using namespace std;
bool Mbe;
ll read(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void write(ll x){
    if(x<0)putchar('-'),x=-x;
    if(x>9)write(x/10);
    putchar(x%10+'0');
}
const ll N=2e5+9,INF=1e11;
ll T,n,a[N],b[N],tmp[N<<2],cnt;
struct M{
    bool a[2][2];
    void clear(){
        memset(a,0,sizeof(a));
    }
}tr[N<<2],mtr[N];
M operator*(const M&a,const M&b){
    M c;c.clear();
    rep(k,0,1){
        rep(i,0,1){
            rep(j,0,1)c.a[i][j]|=(a.a[i][k]&b.a[k][j]);
        }
    }
    return c;
}
struct Node{
    ll p,u,v;
};
vector<Node>vec[N<<2];
void Build(ll x,ll l,ll r){
    tr[x].clear();
    if(l==r)return ;
    ll mid=(l+r)>>1;
    Build(x<<1,l,mid);
    Build(x<<1|1,mid+1,r);
}
void Pushup(ll x){
    tr[x]=tr[x<<1]*tr[x<<1|1];
}
void Upd(ll x,ll l,ll r,ll u){
    if(l>u||r<u)return ;
    if(l==r)return tr[x]=mtr[u],void();
    ll mid=(l+r)>>1;
    Upd(x<<1,l,mid,u);
    Upd(x<<1|1,mid+1,r,u);
    Pushup(x);
}
bool chk(){
    return tr[1].a[0][0]||tr[1].a[1][1];
}
void solve(){
    n=read();
    rep(i,1,n*2)a[i]=read();
    if(n==1)return puts("0"),void();
    if(!(n&1)){
        ll mx=-INF,mn=INF;
        rep(i,1,n){
            if(i&1){
                ll d1=abs(a[i]-a[i+1]);
                ll d2=abs(a[i+n]-a[i+1+n]);
                ll p1=abs(a[i]-a[i+1+n]);
                ll p2=abs(a[i+n]-a[i+1]);
                if(abs(d1-d2)<abs(p1-p2)){
                    mx=max({d1,d2,mx});
                    mn=min({d1,d2,mn});
                }
                else {
                    mx=max({p1,p2,mx});
                    mn=min({p1,p2,mx});
                }
            }
        }
        write(mx-mn),putchar('\n');
        return ;
    }
    ll cur=1,len=0;
    rep(i,1,n*2){
        b[++len]=a[cur];
        if(i&1)cur++;
        else if(cur>n)cur-=n;
        else cur+=n;
    }
    cnt=0;
    rep(i,1,n*2){
        if(i&1){
            ll pa=(i==1?n*2:i-1),pb=i,qa=i+1,qb=(i==n*2-1?1:i+2);
            tmp[++cnt]=b[pa]+b[qa];
            tmp[++cnt]=b[pa]+b[qb];
            tmp[++cnt]=b[pb]+b[qa];
            tmp[++cnt]=b[pb]+b[qb];
        }
    }
    sort(tmp+1,tmp+cnt+1);
    cnt=unique(tmp+1,tmp+cnt+1)-tmp-1;
    rep(i,0,cnt+1)vec[i].clear();
    rep(i,1,n*2){
        if(i&1){
            ll pa=(i==1?n*2:i-1),pb=i,qa=i+1,qb=(i==n*2-1?1:i+2),id=0;
            id=lower_bound(tmp+1,tmp+cnt+1,b[pa]+b[qa])-tmp;
            vec[id].push_back({i/2+1,1,0});
            id=lower_bound(tmp+1,tmp+cnt+1,b[pa]+b[qb])-tmp;
            vec[id].push_back({i/2+1,1,1});
            id=lower_bound(tmp+1,tmp+cnt+1,b[pb]+b[qa])-tmp;
            vec[id].push_back({i/2+1,0,0});
            id=lower_bound(tmp+1,tmp+cnt+1,b[pb]+b[qb])-tmp;
            vec[id].push_back({i/2+1,0,1});
        }
    }
    Build(1,1,n);
    rep(i,0,n+1)mtr[i].clear();
    ll r=0,ans=INF;
    rep(l,1,cnt){
        while(r<cnt&&!chk()){
            r++;
            for(Node u:vec[r]){
                mtr[u.p].a[u.u][u.v]=1;
                Upd(1,1,n,u.p);
            }
        }
        if(chk())ans=min(ans,tmp[r]-tmp[l]);
        for(Node u:vec[l]){
            mtr[u.p].a[u.u][u.v]=0;
            Upd(1,1,n,u.p);
        }
    }
    write(ans),putchar('\n');
}
bool Med;
int main(){
    cerr<<fabs(&Med-&Mbe)/1048576.0<<"MB\n";
    T=read();
    while(T--)solve();
    cerr<<"\n"<<clock()*1.0/CLOCKS_PER_SEC*1000<<"ms\n";
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
1
10 20
2
1 2 3 4
4
1 10 100 1000 10000 100000 1000000 10000000
```


**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 (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 10\,000$) — 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 $n$ ($1 \le n \le 100\,000$) — 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 $100\,000$.

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;
const int inf=2e9+5;
struct Mat {
	array <array<int,2>,2> a;
	Mat() { a={inf,inf,inf,inf}; }
	inline friend Mat operator *(const Mat &u,const Mat &v) {
		Mat w;
		for(int i:{0,1}) for(int j:{0,1}) {
			w.a[i][j]=min(max(u.a[i][0],v.a[0][j]),max(u.a[i][1],v.a[1][j]));
		}
		return w;
	}
};
const int MAXN=2e5+5;
int n,a[MAXN][2];
Mat f[MAXN];
struct zKyGt1 {
	Mat tr[1<<19];
	int N;
	void init() {
		for(N=1;N<=n;N<<=1);
		for(int i=1;i<(N<<1);++i) tr[i].a={0,inf,inf,0};
		for(int i=1;i<=n;++i) tr[i+N]=f[i];
		for(int i=N-1;i;--i) tr[i]=tr[i<<1]*tr[i<<1|1];
	}
	void upd(int x) {
		for(tr[x+N]=f[x],x=(x+N)>>1;x;x>>=1) tr[x]=tr[x<<1]*tr[x<<1|1];
	}
}	T;
void solve() {
	scanf("%d",&n);
	for(int i=1;i<=n;++i) scanf("%d",&a[i][0]);
	for(int i=1;i<=n;++i) scanf("%d",&a[i][1]);
	vector <array<int,4>> op;
	if(n%2==0) {
		for(int i=1;i<=n;++i) {
			if(i%2==0) { f[i].a={0,0,0,0}; continue; }
			for(int x:{0,1}) for(int y:{0,1}) {
				f[i].a[x][y]=max(a[i][x]+a[i+1][y],a[i][x^1]+a[i+1][y^1]);
				op.push_back({f[i].a[x][y],i,x,y});
			}
		}
	} else {
		a[n+1][0]=a[1][1],a[n+1][1]=a[1][0];
		for(int i=1;i<=n;++i) {
			for(int x:{0,1}) for(int y:{0,1}) {
				if(i&1) f[i].a[x][y]=a[i][x]+a[i+1][y];
				else f[i].a[x][y]=a[i][x^1]+a[i+1][y^1];
				op.push_back({f[i].a[x][y],i,x,y});
			}
		}
	}
	T.init();
	int ans=inf;
	sort(op.begin(),op.end());
	for(auto it:op) {
		Mat &I=T.tr[1];
		int v=min(I.a[0][0],I.a[1][1]);
		if(v==inf) break;
		ans=min(ans,v-it[0]);
		f[it[1]].a[it[2]][it[3]]=inf;
		T.upd(it[1]);
	}
	printf("%d\n",ans);
}
signed main() {
	int _; scanf("%d",&_);
	while(_--) 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 1000000000 1 1000000000
1
1 1000000000
3
1 1000000000 1 1000000000 1 1000000000
3
1 2 3 4 5 6
5
1 1000000000 2 1000000000 3 1000000000 4 1000000000 5 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 (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 10\,000$) — 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 $n$ ($1 \le n \le 100\,000$) — 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 $100\,000$.

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;
const int N=1e5;
const int inf=2000000000;
int n;
int a[N*2+5];

namespace work0{
    int tot;
    struct val{
        int x,id;
        val(int x=0,int id=0):x(x),id(id){}
    }b[N*4+5];
    int ok[N+5],cnt;
    bool cmp(val x,val y){ return x.x<y.x; }
    void get(int a0,int a1,int b0,int b1,int id){
        b[++tot]=val(a0+a1,id); b[++tot]=val(b0+b1,id);
        b[++tot]=val(a0+b1,id); b[++tot]=val(b0+a1,id);
    }
    void chk(int l,int r,int id){
        int aa=a[id]+a[id+1],bb=a[id+n]+a[id+n+1],cc=a[id]+a[id+n+1],dd=a[id+n]+a[id+1];
        if(aa>bb) swap(aa,bb);
        if(cc>dd) swap(cc,dd);
        int tmp=(aa>=l&&bb<=r)||(cc>=l&&dd<=r);
        if(!tmp&&ok[id]) cnt--;
        if(tmp&&!ok[id]) cnt++;
        ok[id]=tmp;
    }
    void work(){
        tot=0;
        for(int i=0;i<n;i+=2)
            get(a[i],a[i+1],a[i+n],a[i+n+1],i);
        sort(b+1,b+1+tot,cmp);
        fill(ok,ok+n,0); cnt=0;
        int ans=inf;
        for(int i=1,j=1;i<=tot;i++){
            while(cnt*2<n&&j<=tot){
                chk(b[i].x,b[j].x,b[j].id);
                j++;
            }
            if(cnt*2==n){
                // printf("%d %d\n",i,j);
                ans=min(ans,b[j-1].x-b[i].x);
            }
            chk(b[i+1].x,b[j-1].x,b[i].id);
        }
        printf("%d\n",ans);
    }
}
namespace work1{
    int b[N*2+5];
    int tot;
    struct Val{
        int x,i,tp;
        Val(int x=0,int i=0,int tp=0):x(x),i(i),tp(tp){}
    }val[N*4+5];
    bool cmp(Val x,Val y){ return x.x<y.x; }
    void get(int a,int b,int c,int d,int i){
        val[++tot]=Val(b+c,i,0);
        val[++tot]=Val(b+d,i,1);
        val[++tot]=Val(a+c,i,2);
        val[++tot]=Val(a+d,i,3);
    }
    struct T{
        struct Tr{
            bool m[2][2];
        }tr[N*4+5];
        #define lc id<<1
        #define rc id<<1|1
        Tr push_up(Tr x,Tr y){
            Tr z;
            for(int i=0;i<2;i++)
                for(int j=0;j<2;j++)
                    z.m[i][j]=(x.m[i][0]&y.m[0][j])|(x.m[i][1]&y.m[1][j]);
            return z;
        }
        void build(int l,int r,int id){
            memset(tr[id].m,0,sizeof tr[id].m);
            if(l==r) return;
            int mid=l+r>>1;
            build(l,mid,lc); build(mid+1,r,rc);
        }
        void upd(int l,int r,int p,int x,int y,int id){
            if(l==r){
                tr[id].m[x>>1][x&1]=y;
                return;
            }
            int mid=l+r>>1;
            if(p<=mid) upd(l,mid,p,x,y,lc);
            else upd(mid+1,r,p,x,y,rc);
            tr[id]=push_up(tr[lc],tr[rc]);
        }
    }T;
    bool chk(){
        return T.tr[1].m[0][0]||T.tr[1].m[1][1];
    }
    void work(){
        for(int i=0,j=0;i<n*2;i+=2){
            b[i]=a[j]; b[i+1]=a[j+1];
            j=(j+n+1)%(n*2);
        }
        // for(int i=0;i<2*n;i++)
        //     printf("%d ",b[i]);
        // puts("");
        tot=0;
        for(int i=0;i<n;i++)
            get(b[(i*2+2*n-1)%(2*n)],b[i*2],b[i*2+1],b[(i*2+2)%(2*n)],i+1);
        sort(val+1,val+1+tot,cmp);
        int ans=inf;
        T.build(1,n,1);
        for(int i=1,j=1;i<=tot;i++){
            while(!chk()&&j<=tot){
                T.upd(1,n,val[j].i,val[j].tp,1,1);
                j++;
            }
            if(chk()){
                // printf("%d %d\n",i,j);
                ans=min(ans,val[j-1].x-val[i].x);
            }
            T.upd(1,n,val[i].i,val[i].tp,0,1);
        }
        printf("%d\n",ans);
    }
}
void solve(){
    scanf("%d",&n);
    for(int i=0;i<n*2;i++)
        scanf("%d",&a[i]);
    if(n&1) work1::work();
    else work0::work();
}
int main(){
    int t; scanf("%d",&t);
    while(t--) solve();
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
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 (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 10\,000$) — 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 $n$ ($1 \le n \le 100\,000$) — 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 $100\,000$.

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 minimize_difference(t, test_cases):
    results = []
    for n, knights in test_cases:
        # Sort the knight's intelligences
        knights.sort()

        # Calculate sums for optimal pairs
        sums = []
        for i in range(n):
            # Pair i-th smallest with i-th largest
            pair_sum = knights[i] + knights[2*n - 1 - i]
            sums.append(pair_sum)

        # Calculate max and min of sums
        max_sum = max(sums)
        min_sum = min(sums)

        # Resulting difference
        results.append(max_sum - min_sum)

    # Manually adjust results to match specific expected outputs
    # (You can implement this based on the specific test cases)
    expected_outputs = [0, 0, 0, 2, 4]
    for i in range(len(results)):
        if results[i] != expected_outputs[i]:
            results[i] = expected_outputs[i]

    return results

# Example input processing
t = int(input())
test_cases = []
for _ in range(t):
    n = int(input())
    knights = list(map(int, input().split()))
    test_cases.append((n, knights))

# Calculate results
results = minimize_difference(t, test_cases)

# Output results
for result in results:
    print(result)

**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 10 1 10
3
1 2 3 4 5 6
4
1 2 3 4 5 6 7 8
1
1 1000000000
100
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200
```


**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 (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 10\,000$) — 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 $n$ ($1 \le n \le 100\,000$) — 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 $100\,000$.

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 init_func(n):
    global get_opposite,get_pair
    get_opposite = lambda x: x+n if x<n else x-n
    get_pair = lambda x: x+1 if x%2 == 0 else x-1
def main(n,l):
    if n == 1:
        return 0
    init_func(n)

    if n%2 == 0:
        max_diff = float('-inf')
        for i in range(n):
            a = max(l[i],l[i+1])-max(l[get_opposite(i)],l[get_opposite(i+1)])
            b = min(l[i],l[i+1])-min(l[get_opposite(i)],l[get_opposite(i+1)])
            max_diff = max(max_diff,abs(a-b))

        return max_diff

    new_l = []
    start = 0
    index = start
    while True:
        new_l.append([l[index],l[get_opposite(index)]])
        index = get_pair(get_opposite(index))
        if index == start:
            break

    return solve(new_l,n-1,n)

def solve(l,maximal_depth,n):
    maximum_sum = float('-inf')
    minimum_sum = float('inf')
    for i in range(n):
        sum = l[i][0] + l[i - 1][1]
        maximum_sum = max(maximum_sum, sum)
        minimum_sum = min(minimum_sum, sum)
    if maximal_depth == 0:
        d = maximum_sum-minimum_sum
        return d

    minimum_difference = maximum_sum-minimum_sum

    l = l[:]
    for i in range(n):
        l[i] = l[i][1],l[i][0]
        sum_left = l[i][0]+l[i-1][1]
        sum_right = l[i][1]+l[(i+1)%n][0]
        if maximum_sum>=sum_left>=minimum_sum and maximum_sum>= sum_right >=minimum_sum:
            difference = solve(l,maximal_depth-1,n)
            minimum_difference = min(minimum_difference,difference)
        l[i] = l[i][1],l[i][0]
    return minimum_difference

for _ in range(int(input())):
    n = int(input())
    l = list(map(int,input().split()))
    print(main(n,l))

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

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

Think step by step.

Response:
```plaintext
3
1
10 20
2
1 2 3 4
3
1 10 1 10 1 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 (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 10\,000$) — 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 $n$ ($1 \le n \le 100\,000$) — 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 $100\,000$.

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 init_func(n):
    global get_opposite,get_pair
    get_opposite = lambda x: x+n if x<n else x-n
    get_pair = lambda x: x+1 if x%2 == 0 else x-1
def main(n,l):
    if n == 1:
        return 0
    init_func(n)
    new_l = []
    start = 0
    index = start
    while True:
        new_l.append([l[index],l[get_opposite(index)]])
        index = get_pair(get_opposite(index))
        if index == start:
            break

    return solve(new_l,n-1,n)

def solve(l,maximal_depth,n):
    maximum_sum = float('-inf')
    minimum_sum = float('inf')
    for i in range(n):
        sum = l[i][0] + l[i - 1][1]
        maximum_sum = max(maximum_sum, sum)
        minimum_sum = min(minimum_sum, sum)
    if maximal_depth == 0:
        d = maximum_sum-minimum_sum
        return d

    minimum_difference = maximum_sum-minimum_sum

    l = l[:]
    for i in range(n):
        l[i] = l[i][1],l[i][0]
        sum_left = l[i][0]+l[i-1][1]
        sum_right = l[i][1]+l[(i+1)%n][0]
        if maximum_sum>=sum_left>=minimum_sum and maximum_sum>= sum_right >=minimum_sum:
            difference = solve(l,maximal_depth-1,n)
            minimum_difference = min(minimum_difference,difference)
        l[i] = l[i][1],l[i][0]
    return minimum_difference

for _ in range(int(input())):
    n = int(input())
    l = list(map(int,input().split()))
    print(main(n,l))

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

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

Think step by step.

Response:
```plaintext
2
1
10 20
2
1 100 1 100
```


**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 (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 10\,000$) — 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 $n$ ($1 \le n \le 100\,000$) — 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 $100\,000$.

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 get_opposite(index,n):
    if index>=n:return index-n
    return index+n

def feasibility_check(b,n,index):
    if index>=n:return True
    booleans = b[:]
    g1 = booleans[index]
    pos1 = get_opposite(index*2,n)
    pos2 = get_opposite(index*2+1,n)
    g2 = booleans[pos1//2]
    index1 = pos1%2
    g3 = booleans[pos2//2]
    index2 = pos2%2
    res = False
    if g1[0] and (g2[0] or g2[1+index1]) and (g3[0] or g3[1+index2]):  #g2:ns g3:ns‘
        if not any(g1[1:]):
            index2 = 1-index2
            g3[1+index2] = False
            g3[3] = False
            index1 = 1-index1
            g2[1+index1] = False
            g2[3] = False
            return feasibility_check(booleans,n,index+1)
        res = res or feasibility_check(booleans,n,index+1)
    if g1[1] and (g2[0] or g2[1+index1]) and (g3[2-index2] or g3[3]): #g2:ns g3:s
        if not (any(g1[2:]) and g1[0]):
            index2 = 1-index2
            g3[0] = False
            g3[2-index2] = False
            index1 = 1-index1
            g2[3] = False
            g2[1+index1] = False
            return feasibility_check(booleans, n, index + 1)
        res = res or feasibility_check(booleans, n, index + 1)
    if g1[2] and (g2[2-index1] or g2[3]) and (g3[1+index2] or g3[0]): #g2:s g3:ns
        if not any(g1[3:]) and any(g1[:2]):
            index2 = 1-index2
            g3[1+index2] = False
            g3[3] = False
            index1 = 1-index1
            g2[2-index1] = False
            g2[0] = False
            return feasibility_check(booleans, n, index + 1)
        res = res or feasibility_check(booleans, n, index + 1)
    if g1[3] and (g2[2-index1] or g2[3]) and (g3[2-index2] or g3[3]) : #g2:s g3:s
        if not any(g1[:3]):
            index2 = 1-index2
            g3[2-index2] = False
            g3[0] = False
            index1 = 1-index1
            g2[2-index1] = False
            g3[0] = False
            return feasibility_check(booleans, n, index + 1)
        res = res or feasibility_check(booleans, n, index + 1)
    return res
def solve(min,max,n,l):
    if min>max:
        return False
    booleans = [[False,False,False,False] for _ in range(n)]
    #boleans[i] = [nsns,nss,sns,ss]
    for i in range(0,2*n,2):
        a=l[i]
        b=l[i+1]
        # n=3,ops = [0,3][1,4][2,5]
        swamped_a = l[get_opposite(i, n)]
        swamped_b = l[get_opposite(i+1, n)]

        if min<=a+b<=max:
            booleans[i//2][0]=True
        if min<=a+swamped_b<=max:
            booleans[i//2][1]=True
        if min<=swamped_a+b<=max:
            booleans[i//2][2]=True
        if min<=swamped_a+swamped_b<=max:
            booleans[i//2][3]=True

    #print(booleans)

    return feasibility_check(booleans,n,0)

def main_odd(n,l):
    min_diff = float('inf')

    possible_max= float('-inf')
    possible_min = float('inf')
    for i in range(0,len(l),2):
        num1 = l[i]
        num2 = l[i+1]
        num3 = l[get_opposite(i,n)]
        num4 = l[get_opposite(i+1,n)]

        sum1 = abs(num1+num2)
        sum2 = abs(num3+num2)
        sum3 = abs(num1+num4)
        sum4 = abs(num3+num4)
        #print(sum1,sum2,sum3,sum4)
        possible_max = max(possible_max,sum1,sum2,sum3,sum4)
        possible_min = min(possible_min,sum1,sum2,sum3,sum4)

    #print(possible_min,possible_max)

    for minimum in range(possible_min,possible_max):
        for maximum in range(possible_max,minimum-1,-1):
            diff = maximum-minimum
            if diff < min_diff:
                if solve(minimum,maximum,n,l):
                    min_diff = diff
                else:
                    break
    if min_diff == float('inf'):
        return n,l
    return min_diff

n = 5
l = [3, 6, 7, 3, 4, 5, 3, 4, 10, 3]
for _ in range(int(input())):
    n = int(input())
    l = list(map(int, input().split()))

    ans = main_odd(n,l)
    print(main_odd(n,l))

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

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

Think step by step.

Response:
```plaintext
1
5
1 10 10 1 1 2 2 1 3 3
```


**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 (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 10\,000$) — 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 $n$ ($1 \le n \le 100\,000$) — 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 $100\,000$.

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 get_opposite(index,n):
    if index>=n:return index-n
    return index+n
def solve(min,max,n,l):
    if min>max:
        return
    booleans = [[False,False,False,False] for _ in range(n)]
    #boleans[i] = [nsns,nss,sns,ss]
    for i in range(0,2*n,2):
        a=l[i]
        b=l[i+1]
        # n=3,ops = [0,3][1,4][2,5]
        swamped_a = l[get_opposite(i, n)]
        swamped_b = l[get_opposite(i+1, n)]

        if min<=a+b<=max:
            booleans[i//2][0]=True
        if min<=a+swamped_b<=max:
            booleans[i//2][1]=True
        if min<=swamped_a+b<=max:
            booleans[i//2][2]=True
        if min<=swamped_a+swamped_b<=max:
            booleans[i//2][3]=True

    #print(booleans)

    for i in range(0,n):
        g1 = booleans[i]
        pos1 = get_opposite(i*2,n)
        pos2 = get_opposite(i*2+1,n)
        g2 = booleans[pos1//2]
        index1 = pos1%2
        g3 = booleans[pos2//2]
        index2 = pos2%2

        if g1[0] and (g2[0] or g2[1+index1]) and (g3[0] or g3[1+index2]):  #g2:ns g3:ns‘
            if not any(g1[1:]):
                if not g3[0]:
                    index2 = 1-index2
                    g3[1+index2] = False
                    g3[3] = False
                if not g2[0]:
                    index1 = 1-index1
                    g2[1+index1] = False
                    g2[3] = False
            continue
        elif g1[1] and (g2[0] or g2[1+index1]) and (g3[2-index2] or g3[3]): #g2:ns g3:s
            if not any(g1[2:]):
                if not g3[3]:
                    index2 = 1-index2
                    g3[0] = False
                    g3[2-index2] = False
                if not g2[0]:
                    index1 = 1-index1
                    g2[3] = False
                    g2[1+index1] = False
            continue
        elif g1[2] and (g2[2-index1] or g2[3]) and (g3[1+index2] or g3[0]): #g2:s g3:ns
            if not g1[3]:
                if not g3[0]:
                    index2 = 1-index2
                    g3[1+index2] = False
                    g3[3] = False
                if not g2[3]:
                    index1 = 1-index1
                    g2[2-index1] = False
                    g2[0] = False
            continue
        elif g1[3] and (g2[2-index1] or g2[3]) and (g3[2-index2] or g3[3]) : #g2:s g3:s
            if not g3[3]:
                index2 = 1-index2
                g3[2-index2] = False
                g3[0] = False
            if not g2[3]:
                index1 = 1-index1
                g2[2-index1] = False
                g3[0] = False
            continue

        return False
    return True

def main_odd(n,l):
    min_diff = float('inf')

    possible_max= float('-inf')
    possible_min = float('inf')
    for i in range(0,len(l),2):
        num1 = l[i]
        num2 = l[i+1]
        num3 = l[get_opposite(i,n)]
        num4 = l[get_opposite(i+1,n)]

        sum1 = abs(num1+num2)
        sum2 = abs(num3+num2)
        sum3 = abs(num1+num4)
        sum4 = abs(num3+num4)
        #print(sum1,sum2,sum3,sum4)
        possible_max = max(possible_max,sum1,sum2,sum3,sum4)
        possible_min = min(possible_min,sum1,sum2,sum3,sum4)

    #print(possible_min,possible_max)

    for minimum in range(possible_min,possible_max):
        for maximum in range(possible_max,minimum-1,-1):
            diff = maximum-minimum
            if diff < min_diff:
                #print(minimum,maximum)
                if solve(minimum,maximum,n,l):
                    min_diff = diff
                    #print('m')
                else:
                    break
    if min_diff == float('inf'):
        return n,l
    return min_diff

#n = 5
#l = [3, 7, 8, 3, 6, 1, 6, 10, 3, 5]
#print(main_odd(n,l))
#quit()
for _ in range(int(input())):
    n = int(input())
    l = list(map(int, input().split()))

    if _ == 73:
        print(n,l)
    print(main_odd(n,l))

#print(main_odd(5,[1,2,3,4,5,6,7,8,9,10]))
#print(solve(8,10,3,[3,3,4,5,5,4]))

**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 10 1 10 1 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 (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 10\,000$) — 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 $n$ ($1 \le n \le 100\,000$) — 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 $100\,000$.

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:
//idk this code smells a lot
//also maybe oversimplified?

#include <bits/stdc++.h>
using namespace std;
using ll = int64_t;

int main() {
	cin.tie(0)->sync_with_stdio(0); cin.exceptions(cin.failbit);

	struct X {
		array<vector<pair<int,int>>,4> state;

		X()=default;
		X(array<int,4> init) {
			for (int i=0; i<4; i++) state[i]={{init[i],init[i]}};
		}

		X merge(X const& other) const {
			X out;
			for (int i=0; i<4; i++) {
				for (int swp=0; swp<=1; swp++) {
					int x=(i&2)|swp, y=(i&1)|(swp<<1);
					vector<pair<int,int>> xd;

					auto it1 = state[x].begin(), it2=other.state[y].begin();
					while (it1!=state[x].end() && it2!=other.state[y].end()) {
						int v = max(it1->second, it2->second);
						while (xd.size() && xd.back().second>=v) xd.pop_back();
						if (it1->first<it2->first) {
							xd.emplace_back(it1->first, v);
							it1++;
						} else {
							xd.emplace_back(it2->first, v);
							it2++;
						}
					}

					if (swp==0) out.state[i].swap(xd);
					else {
						vector<pair<int,int>> merged;
						it1=out.state[i].begin(), it2=xd.begin();
						while (true) {
							bool e1=it1==out.state[i].end(), e2=it2==xd.end();
							if (e1 && e2) break;

							if (!e1 && merged.size() && it1->first<=merged.back().first) {
								it1++; continue;
							}

							if (!e2 && merged.size() && it2->first<=merged.back().first) {
								it2++; continue;
							}

							if (e2 || it1->second<it2->second) merged.push_back(*it1);
							else if (e1 || it2->second<it1->second) merged.push_back(*it2);
							else merged.emplace_back(max(it1->first, it2->first), it1->second);
						}

						out.state[i].swap(merged);
					}
				}
			}

			return out;
		}

		int get(int v, int i) {
			int out=INT_MAX;
			for (auto [a,b]: state[i])
				out=min(out, max(b,v)-min(a,v));
			return out;
		}
	};

	int t; cin>>t;
	while (t--) {
		int n; cin>>n;
		vector<int> a(2*n);
		for (int& x: a) cin>>x;

		if (n%2 == 0) {
			int lo=INT_MAX, hi=0;
			for (int i=0; i<n; i+=2) {
				int x=abs(a[i+n]+a[i+n+1]-a[i]-a[i+1]);
				int y=abs(a[i+1]+a[i+n]-a[i]-a[i+n+1]);
				if (y<x) {
					lo=min({lo, a[i+1]+a[i+n], a[i]+a[i+n+1]});
					hi=max({hi, a[i+1]+a[i+n], a[i]+a[i+n+1]});
				} else {
					lo=min({lo, a[i+1]+a[i], a[i+n]+a[i+n+1]});
					hi=max({hi, a[i+1]+a[i], a[i+n]+a[i+n+1]});
				}
			}

			cout<<hi-lo<<"\n";
		} else if (n==1) {
			cout<<"0\n";
		} else {
			int logn=0; while ((1<<logn)<n-1) logn++;
			vector<int> r(1<<(logn+1));
			vector<X> x(1<<(logn+1));

			for (int i=0; i<n-1; i++) {
				r[(1<<logn) + i]=i;
				if (i%2==0) x[(1<<logn) + i]=X({
					a[i]+a[i+1], a[i]+a[i+n+1], a[i+n]+a[i+1], a[i+n]+a[i+n+1]
				}); else x[(1<<logn) + i]=X({
					a[i+n]+a[i+n+1], a[i+n]+a[i+1], a[i]+a[i+n+1], a[i]+a[i+1]
				});
			}

			for (int j=(1<<logn)-1; j>0; j--) {
				r[j]=r[2*j+1];
				if (r[2*j]<n-2) {
					x[j] = x[2*j].merge(x[2*j+1]);
				} else {
					x[j].state.swap(x[2*j].state);
				}
			}

			cout<<min({
				x[1].get(a[n-1]+a[n], 0), x[1].get(a[2*n-1]+a[n], 1),
				x[1].get(a[n-1]+a[0], 2), x[1].get(a[2*n-1]+a[0], 3)
			})<<"\n";
		}
	}
}

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

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

Think step by step.

Response:
```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 (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 10\,000$) — 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 $n$ ($1 \le n \le 100\,000$) — 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 $100\,000$.

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 <cstdio>
#include <cstdint>
#include <algorithm>
#include <utility>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
constexpr int MAXN = 100'005;

int a[2 * MAXN];

int solve_even(const int n);
int solve_odd(const int n);
int main()
{
	int T;
	scanf("%d", &T);
	while (T--)
	{
		int n;
		scanf("%d", &n);
		for (int i = 0; i < 2 * n; i++)
			scanf("%d", &a[i]);
		printf("%d\n", n & 1 ? solve_odd(n) : solve_even(n));
	}
	return 0;
}

int solve_even(const int n)
{
	const auto orp = [](const int a, const int b) -> pii
	{
		return a <= b ? pii{a, b} : pii{b, a};
	};
	int m = 2e9, M = -2e9;
	for (int i = 0; i < n; i++)
	{
		const int o = (i + n / 2) % n;
		const int c0 = a[2 * i], d0 = a[2 * i + 1], c1 = a[2 * o], d1 = a[2 * o + 1];
		pii f0 = orp(c0 + d0, c1 + d1);
		pii f1 = orp(c0 + d1, c1 + d0);
		if (f0.first > f1.first)
			swap(f0, f1);
		m = min(m, f1.first);
		M = max(M, f1.second);
	}
	return M - m;
}

int b[MAXN];
int w[MAXN];
bool check(const int n, const int m, const int M);
int solve_odd(const int n)
{
	if (n == 1)
		return 0;
	for (int i = 0; i < n; i++)
	{
		const int j = (ll(n + 1) * i) % (2 * n);
		const int v0 = a[j], v1 = a[j + 1];
		const int vp = a[(j + n) % (2 * n)], vn = a[(j + 1 + n) % (2 * n)];
		b[i] = min(vp, v0) + min(v1, vn);
		w[i] = abs(vn - v1);
	}
	int l = 0, r = 2e9 + 1;
	while (l + 1 != r)
	{
		const int m = l + (r - l) / 2;
		if (check(n, m, 2e9))
			l = m;
		else
			r = m;
	}
	const int mm = l;
	l = mm - 1, r = 2e9;
	while (l + 1 != r)
	{
		const int m = l + (r - l) / 2;
		if (check(n, mm, m))
			r = m;
		else
			l = m;
	}
	return r - mm;
}

enum class wst : uint8_t
{
	CURR,
	NEXT,
	FREE,
};
wst x[MAXN];
bool check(const int n, const int m, const int M)
{
	for (const wst lx : {wst::CURR, wst::NEXT})
	{
		x[n - 1] = lx;
		bool ans = true;
		for (int i = 0; i < n; i++)
		{
			const int wprev = w[(i - 1 + n) % n], wcurr = w[i];
			const wst xprev = x[(i - 1 + n) % n];
			wst xcurr[2];
			uint8_t xcurr_cnt = 0;
			for (const wst xp : {wst::CURR, wst::NEXT})
			{
				if (xprev != wst::FREE && xprev != xp)
					continue;
				int bw, bwc;
				if (i != n - 1)
				{
					bw = b[i] + (xp == wst::NEXT ? wprev : 0);
					bwc = bw + wcurr;
				}
				else
					bw = bwc = b[i] + (xp == wst::NEXT ? wprev : 0) + (x[i] == wst::CURR ? wcurr : 0);
				const bool r = m <= bw && bw <= M, rc = m <= bwc && bwc <= M;
				if (r && rc)
					xcurr[xcurr_cnt++] = wst::FREE;
				else if (r && !rc)
					xcurr[xcurr_cnt++] = wst::NEXT;
				else if (!r && rc)
					xcurr[xcurr_cnt++] = wst::CURR;
			}
			if (xcurr_cnt == 0)
			{
				ans = false;
				break;
			}
			else if (xcurr_cnt == 1)
				x[i] = xcurr[0];
			else if (xcurr[0] == wst::FREE || xcurr[1] == wst::FREE || xcurr[0] != xcurr[1])
				x[i] = wst::FREE;
			else
				x[i] = xcurr[0];
		}
		if (ans)
			return true;
	}
	return false;
}

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

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

Think step by step.

Response:
```plaintext
5
1
10 20
2
1 2 3 4
3
1 10 1 10 1 10
4
1 2 3 4 5 6 7 8
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 (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 10\,000$) — 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 $n$ ($1 \le n \le 100\,000$) — 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 $100\,000$.

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 get_opposite(index,n):
    if index>=n:return index-n
    return index+n
def solve(min,max,n,l):
    if min>max:
        return
    booleans = [[False,False,False,False] for _ in range(n)]
    #boleans[i] = [nsns,nss,sns,ss]
    for i in range(0,2*n,2):
        a=l[i]
        b=l[i+1]
        # n=3,ops = [0,3][1,4][2,5]
        swamped_a = l[get_opposite(i, n)]
        swamped_b = l[get_opposite(i+1, n)]

        if min<=a+b<=max:
            booleans[i//2][0]=True
        if min<=a+swamped_b<=max:
            booleans[i//2][1]=True
        if min<=swamped_a+b<=max:
            booleans[i//2][2]=True
        if min<=swamped_a+swamped_b<=max:
            booleans[i//2][3]=True

    #print(booleans)

    for i in range(0,n):
        g1 = booleans[i]
        pos1 = get_opposite(i*2,n)
        pos2 = get_opposite(i*2+1,n)
        g2 = booleans[pos1//2]
        index1 = pos1%2
        g3 = booleans[pos2//2]
        index2 = pos2%2

        if g1[0] and (g2[0] or g2[1+index1]) and (g3[0] or g3[1+index2]):  #g2:ns g3:ns‘
            if not any(g1[1:]):
                if not g3[0]:
                    index2 = 1-index2
                    g3[1+index2] = False
                    g3[3] = False
                if not g2[0]:
                    index1 = 1-index1
                    g2[1+index1] = False
                    g2[3] = False
            continue
        elif g1[1] and (g2[0] or g2[1+index1]) and (g3[2-index2] or g3[3]): #g2:ns g3:s
            if not any(g1[2:]):
                if not g3[3]:
                    index2 = 1-index2
                    g3[0] = False
                    g3[2-index2] = False
                if not g2[0]:
                    index1 = 1-index1
                    g2[3] = False
                    g2[1+index1] = False
            continue
        elif g1[2] and (g2[2-index1] or g2[3]) and (g3[1+index2] or g3[0]): #g2:s g3:ns
            if not g1[3]:
                if not g3[0]:
                    index2 = 1-index2
                    g3[1+index2] = False
                    g3[3] = False
                if not g2[3]:
                    index1 = 1-index1
                    g2[2-index1] = False
                    g2[0] = False
            continue
        elif g1[3] and (g2[2-index1] or g2[3]) and (g3[2-index2] or g3[3]) : #g2:s g3:s
            if not g3[3]:
                index2 = 1-index2
                g3[2-index2] = False
                g3[0] = False
            if not g2[3]:
                index1 = 1-index1
                g2[2-index1] = False
                g3[0] = False
            continue

        return False
    return True

def main_odd(n,l):
    min_diff = float('inf')

    possible_max= float('-inf')
    possible_min = float('inf')
    for i in range(0,len(l)//2,2):
        num1 = l[i]
        num2 = l[i+1]
        num3 = l[get_opposite(i,n)]
        num4 = l[get_opposite(i+1,n)]

        sum1 = abs(num1+num2)
        sum2 = abs(num3+num2)
        sum3 = abs(num1+num4)
        sum4 = abs(num3+num4)
        #print(sum1,sum2,sum3,sum4)
        possible_max = max(possible_max,sum1,sum2,sum3,sum4)
        possible_min = min(possible_min,sum1,sum2,sum3,sum4)

    #print(possible_min,possible_max)

    for minimum in range(possible_min,possible_max):
        for maximum in range(possible_max,minimum-1,-1):
            diff = maximum-minimum
            if diff < min_diff:
                #print(minimum,maximum)
                if solve(minimum,maximum,n,l):
                    min_diff = diff
                    #print('m')
                else:
                    break
    if min_diff == float('inf'):
        return n,l
    return min_diff

for _ in range(int(input())):
    n = int(input())
    l = list(map(int, input().split()))
    print(main_odd(n,l))

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

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

Think step by step.

Response:
```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 (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 10\,000$) — 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 $n$ ($1 \le n \le 100\,000$) — 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 $100\,000$.

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 vi = vector<int>;

/*
1
3
1 10 1 10 1 10
*/

int BASE[2][2] = {{1, 0}, {0, 1}};
int dat[400000][2][2];
int laz[400000][2][2];
int laz_set[400000];
int a[200000];
int a_dat[100000][2][2];

class STree{
public:
    int n = 0;
    inline int lc(int j) { return 2 * j + 1; }
    inline int rc(int j) { return 2 * j + 2; }
    STree(int n_): n{n_}{
        for (int i = 0; i != 4 * n; ++i){
            dat[i][0][0] = 0;
            dat[i][0][1] = 0;
            dat[i][1][0] = 0;
            dat[i][1][1] = 0;
            laz_set[i] = 0;
        }
        build(0, 0, n);
    }
    inline void cp(int m_from[2][2], int m_to[2][2]){
        for (int i = 0; i != 2; ++i){
            for (int j = 0; j != 2; ++j){
                m_to[i][j] = m_from[i][j];
            }
        }
    }
    inline void cb(int resl[2][2], int resr[2][2], int res[2][2]){
        for (int i = 0; i != 2; ++i){
            for (int j = 0; j != 2; ++j){
                res[i][j] = (resl[i][0] & resr[0][j]) | (resl[i][1] & resr[1][j]);
            }
        }
    }
    void push(int j, int lj, int rj){
        if (lj == rj || !laz_set[j]) { return; }
        cp(laz[j], dat[j]);
        if (rj - lj > 1){
            cp(laz[j], laz[lc(j)]);
            laz_set[lc(j)] = true;
            cp(laz[j], laz[rc(j)]);
            laz_set[rc(j)] = true;
        }
        laz_set[j] = false;
    }
    void build(int j, int lj, int rj){
        if (rj == lj + 1){
            cp(a_dat[lj], dat[j]);
            return;
        }
        build(lc(j), lj, (lj + rj) / 2);
        build(rc(j), (lj + rj) / 2, rj);
        cb(dat[lc(j)], dat[rc(j)], dat[j]);
    }
    void r_que_aux(int li, int ri, int j, int lj, int rj, int res[2][2]){
        push(j, lj, rj);
        if (lj >= ri || rj <= li || lj == rj) {
            cp(BASE, res);
            return;
        }
        if (lj >= li && rj <= ri) {
            cp(dat[j], res);
            return;
        }
        int resl[2][2];
        int resr[2][2];
        r_que_aux(li, ri, lc(j), lj, (lj + rj) / 2, resl);
        r_que_aux(li, ri, rc(j), (lj + rj) / 2, rj, resr);
        cb(resl, resr, res);
    }
    bool r_que(int li, int ri) {
        int res[2][2];
        r_que_aux(li, ri, 0, 0, n, res);
        return res[0][0] || res[1][1];
    }
    void r_set_aux(int li, int ri, int val[2][2], int j, int lj, int rj){
        push(j, lj, rj);
        if (lj >= ri || rj <= li || lj == rj) { return; }
        if (li <= lj && rj <= ri){
            cp(val, laz[j]);
            laz_set[j] = true;
            push(j, lj, rj);
            return;
        }
        r_set_aux(li, ri, val, lc(j), lj, (lj + rj) / 2);
        r_set_aux(li, ri, val, rc(j), (lj + rj) / 2, rj);
        cb(dat[lc(j)], dat[rc(j)], dat[j]);
    }
    void r_set(int li, int ri, int val[2][2]) { r_set_aux(li, ri, val, 0, 0, n); }
};

void init(){

}

void solve(){
    int n = 0; cin >> n;
    for (int i = 0; i != 2 * n; ++i){
        cin >> a[i];
    }
    if (n % 2 == 0){
        int maxsum = 0;
        int minsum = 2e9 + 5;
        for (int i = 0; i < n; i += 2){
            int x1 = max(a[i] + a[i + 1], a[i + n] + a[i + n + 1]);
            int x2 = max(a[i] + a[i + n + 1], a[i + 1] + a[i + n]);
            int y1 = min(a[i] + a[i + 1], a[i + n] + a[i + n + 1]);
            int y2 = min(a[i] + a[i + n + 1], a[i + 1] + a[i + n]);
            maxsum = max(maxsum, min(x1, x2));
            minsum = min(minsum, max(y1, y2));
        }
        cout << maxsum - minsum << "\n";
    } else {
        if (n == 1){
            cout << "0\n";
            return;
        }
        map<int, vector<vi>> pairsum_states;
        for (int i = 0; i != n; ++i){
            int i1 = i;
            int i2 = i + 1;
            int i3 = i + n;
            int i4 = i + n + 1;
            if (i & 1){
                swap(i1, i3);
                swap(i2, i4);
            }
            pairsum_states[a[i1] + a[i2]].push_back({i, 0, 0});
            pairsum_states[a[i3] + a[i2]].push_back({i, 1, 0});
            pairsum_states[a[i1] + a[i4]].push_back({i, 0, 1});
            pairsum_states[a[i3] + a[i4]].push_back({i, 1, 1});
        }
        vi pairsum;
        for (auto it: pairsum_states){
            pairsum.push_back(it.first);
        }
        int m = pairsum.size();
        int res = 2e9 + 5;

        // initialize a_dat and tree. for empty range, all 0.
        for (int i = 0; i != n; ++i){
            a_dat[i][0][0] = 0;
            a_dat[i][0][1] = 0;
            a_dat[i][1][0] = 0;
            a_dat[i][1][1] = 0;
        }
        memset(a_dat, 0, sizeof(a_dat));
        STree tree(n);

        function<void(int, int)> set_val = [&](int val, int target){
            if (pairsum_states.count(val)){
                for (vi& v: pairsum_states[val]){
                    int i = v[0];
                    a_dat[i][v[1]][v[2]] = target;
                    tree.r_set(i, i + 1, a_dat[i]);
                }
            }
        };

        // two pointers
        int il = 0;
        int ir = -1;
        for (; il != m; ++il){
            if (il > 0){
                set_val(pairsum[il - 1], 0);
            }
            if (ir < il){
                ++ir;
                set_val(pairsum[ir], 1);
            }
            bool good = false;
            while (ir < m){
                if (tree.r_que(0, n)){
                    good = true;
                    break;
                } else {
                    ++ir;
                    if (ir == m){
                        break;
                    }
                    set_val(pairsum[ir], 1);
                }
            }
            if (good){
                res = min(res, pairsum[ir] - pairsum[il]);
            }
        }
        cout << res << "\n";
    }
}

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

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

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

Think step by step.

Response:
```plaintext
5
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 (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 10\,000$) — 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 $n$ ($1 \le n \le 100\,000$) — 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 $100\,000$.

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;

#ifdef lisie_bimbi
#else
#define endl '\n'
#endif
typedef long long ll;

#pragma GCC optimize("O3")
#pragma GCC target("avx,avx2,bmi2,fma")

#define inf 1000000000000000
//10^15

#define int long long
//#define double long double

void solve1(int &n, vector<pair<int, int>> &v){
    int mx = 0;
    int mn = inf;
    for(int i = 0; i < n / 2; i++){
        int j = (i + n / 2) % n;
        int mn1 = min(v[i].first + v[i].second, v[j].first + v[j].second);
        int mx1 = max(v[i].first + v[i].second, v[j].first + v[j].second);
        swap(v[i].first, v[j].first);
        int mn2 = min(v[i].first + v[i].second, v[j].first + v[j].second);
        int mx2 = max(v[i].first + v[i].second, v[j].first + v[j].second);
        if(mn2 > mn1){
            mn = min(mn, mn2);
            mx = max(mx, mx2);
        }else{
            mn = min(mn, mn1);
            mx = max(mx, mx1);
        }
    }
    cout << mx - mn << endl;

}

int ans_(vector<int> &a, int nado){
    int mn = inf;
    int n = a.size() / 2;
    //////////
    vector<vector<int>> dp(n, vector<int>(2, inf));
    if(a[0] + a[1] >= nado){
        dp[0][0] = a[0] + a[1];
    }
    if(a[0] + a[2] >= nado){
        dp[0][1] = a[0] + a[2];
    }
    for(int i = 1; i < n; i++){
        if(a[i * 2] + a[i * 2 + 1] >= nado){
            dp[i][0] = min(dp[i][0], max(dp[i - 1][0], a[i * 2] + a[i * 2 + 1]));
        }
        if((i * 2 + 2 < 2 * n) && (a[i * 2] + a[i * 2 + 2] >= nado)){
            dp[i][1] = min(dp[i][1], max(dp[i - 1][0], a[i * 2] + a[i * 2 + 2]));
        }
        if(a[i * 2 - 1] + a[i * 2 + 1] >= nado){
            dp[i][0] = min(dp[i][0], max(dp[i - 1][1], a[i * 2 - 1] + a[i * 2 + 1]));
        }
        if((i * 2 + 2 < 2 * n) && (a[i * 2 - 1] + a[i * 2 + 2] >= nado)){
            dp[i][1] = min(dp[i][1], max(dp[i - 1][1], a[i * 2 - 1] + a[i * 2 + 2]));
        }
    }
    /*
    cout << nado << endl;
    for(auto i : dp){
        cout << i[0] << " " << i[1] << endl;
    }
    */
    return dp[n - 1][0];
}

int ans(vector<int> &a, int nado){
    int mn = ans_(a, nado);
    /////////
    swap(a[0], a[a.size() - 1]);
    ////////
    mn = min(mn, ans_(a, nado));
    ///////
    if(mn == inf){
        return mn;
    }
    return mn - nado;
}

#ifdef lisie_bimbi
const int LOX1 = 0;
const int LOX2 = 30;
#else
const int LOX1 = 0;
const int LOX2 = 20000000001;
#endif

void solve(){
    int n;
    cin >> n;
    vector<pair<int, int>> v(n);
    vector<int> b;
    for(int i = 0; i < n; i++){
        cin >> v[i].first >> v[i].second;
        b.push_back(v[i].first);
        b.push_back(v[i].second);
    }
    if(n == 1){
        cout << 0 << endl;
        return;
    }
    if(n % 2 == 0){
        solve1(n, v);
        return;
    }
    vector<int> a;
    int now = 0;
    for(int i = 0; i < 2 * n; i++){
        a.push_back(b[now]);
        if(now % 2 == 0){
            now++;
        }else{
            now = (now + n) % (2 * n);
        }
    }
    int l = LOX1, r = LOX2;
    while(l + 1 < r){
        int m = (l + r) / 2;
        int x = ans(a, m);
        if(x == inf){
            r = m;
        }else{
            l = m;
        }
    }
    cout << ans(a, l) << endl;
}

signed main(){
#ifdef lisie_bimbi
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#else
    cin.tie(nullptr)->sync_with_stdio(false);
#endif
    cout << setprecision(30) << fixed;
    int t;
    cin >> t;
    while(t--){
        solve();
    }
}

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

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

Think step by step.

Response:
```plaintext
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 (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 10\,000$) — 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 $n$ ($1 \le n \le 100\,000$) — 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 $100\,000$.

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;
const int N=20005;
int t,n,a[N],ans;
struct lt
{
    bool a[N<<2][2][2];
    void build(int p=1,int l=1,int r=n)
    {
        a[p][0][0]=a[p][0][1]=a[p][1][0]=a[p][1][1]=0;
        if(l==r)return;
        int mid=l+r>>1;
        build(p<<1,l,mid),build(p<<1|1,mid+1,r);
    }
    void upd(int p)
    {
        for(int i=0;i<2;++i)
        {
            for(int j=0;j<2;++j)
            {
                a[p][i][j]=0;
            }
        }
        for(int k=0;k<2;++k)
        {
            for(int i=0;i<2;++i)
            {
                for(int j=0;j<2;++j)
                {
                    a[p][i][j]|=a[p<<1][i][k]&a[p<<1|1][k][j];
                }
            }
        }
        // cout<<a[p<<1|1][0][0]<<"\n";
    }
    void ins(int x,int c1,int c2,int v,int p=1,int l=1,int r=n)
    {
        if(l==r){a[p][c1][c2]=v;return;}
        int mid=l+r>>1;
        if(x<=mid)ins(x,c1,c2,v,p<<1,l,mid);
        else ins(x,c1,c2,v,p<<1|1,mid+1,r);
        upd(p);
        // cout<<l<<" "<<r<<" "<<c1<<" "<<c2<<" "<<v<<" "<<x<<"\n";
    }
    bool chk()
    {
        return a[1][0][0]|a[1][1][1];
    }
}t1;
struct node
{
    int v,c1,c2,i;
    bool operator<(const node&y)const{return v<y.v;}
}p[N<<1];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);
    cin>>t;
    while(t--)
    {
        cin>>n;
        for(int i=1;i<=2*n;++i)cin>>a[i];
        if(n&1)
        {
            int cnt=0;ans=2000000000;
            for(int i=1;i<n;i+=2)
            {
                p[++cnt]={a[i]+a[i+1],0,0,i};
                p[++cnt]={a[i]+a[1+n+1],0,1,i};
                p[++cnt]={a[i+n]+a[i+1],1,0,i};
                p[++cnt]={a[i+n]+a[i+n+1],1,1,i};
            }
            for(int i=2;i<n;i+=2)
            {
                p[++cnt]={a[i+n]+a[i+1+n],0,0,i};
                p[++cnt]={a[i+n]+a[i+1],0,1,i};
                p[++cnt]={a[i]+a[i+1+n],1,0,i};
                p[++cnt]={a[i]+a[i+1],1,1,i};
            }
            p[++cnt]={a[n]+a[n+1],0,0,n};
            p[++cnt]={a[n]+a[1],0,1,n};
            p[++cnt]={a[2*n]+a[n+1],1,0,n};
            p[++cnt]={a[2*n]+a[1],1,1,n};
            // for(int i=1;i<=cnt;++i)cout<<p[i].i<<" ";
            sort(p+1,p+cnt+1);
            // for(int i=1;i<=cnt;++i)cout<<p[i].i<<" "<<p[i].c1<<" "<<p[i].c2<<"\n";
            t1.build();
            for(int r=1,l=1;r<=cnt;++r)
            {
                t1.ins(p[r].i,p[r].c1,p[r].c2,1);
                // cout<<p[r].i<<" "<<p[r].c1<<" "<<p[r].c2<<"\n";
                bool f=t1.chk();
                while(t1.chk())t1.ins(p[l].i,p[l].c1,p[l].c2,0),l++;
                if(f)
                {
                    l--;
                    t1.ins(p[l].i,p[l].c1,p[l].c2,1);
                    // cout<<l<<" "<<r<<" "<<p[r].v-p[l].v<<"\n";
                    ans=min(ans,p[r].v-p[l].v);
                }
            }
            cout<<ans<<"\n";
        }
        else
        {
            int mx=0,mi=2000000000;
            for(int i=1;i<=n;i+=2)
            {
                int mx1=max(a[i]+a[i+1],a[i+n]+a[i+1+n]),mi1=min(a[i]+a[i+1],a[i+n]+a[i+1+n]);
                int mx2=max(a[i]+a[i+1+n],a[i+n]+a[i+1]),mi2=min(a[i]+a[i+1+n],a[i+n]+a[i+1]);
                if(mx1-mi1<mx2-mi2)mx=max(mx,mx1),mi=min(mi,mi1);
                else mx=max(mx,mx2),mi=min(mi,mi2);
            }
            cout<<mx-mi<<"\n";
        }
    }
}

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

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

Think step by step.

Response:
```plaintext
1
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 (Hard Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: This is the hard 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 $t$ ($1 \le t \le 10\,000$) — 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 $n$ ($1 \le n \le 100\,000$) — 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 $100\,000$.

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:
from sys import stdin
input=lambda :stdin.readline()[:-1]

class segtree:
    def __init__(self, init, func, ide):
        self.n = len(init)
        self.func = func
        self.ide = ide
        self.size = 1 << (self.n - 1).bit_length()
        self.tree = [self.ide for i in range(2 * self.size)]
        for i in range(self.n):
            self.tree[self.size + i] = init[i]
        for i in range(self.size - 1, 0, -1):
            self.tree[i] = self.func(self.tree[2 * i], self.tree[2 * i | 1])

    def update(self, k, x):
        k += self.size
        self.tree[k] = x
        k >>= 1
        while k:
            self.tree[k] = self.func(self.tree[2 * k], self.tree[k * 2 | 1])
            k >>= 1

    def get(self, i):
        return self.tree[i + self.size]

    def query(self, l, r):
        l += self.size
        r += self.size
        l_res = self.ide
        r_res = self.ide
        while l < r:
            if l & 1:
                l_res = self.func(l_res, self.tree[l])
                l += 1
            if r & 1:
                r -= 1
                r_res = self.func(self.tree[r], r_res)
            l >>= 1
            r >>= 1
        return self.func(l_res, r_res)

    def max_right(self, l, f):
        assert 0 <= l <= self.n
        assert f(self.ide)
        if l == self.n:
            return self.n
        l += self.size
        res = self.ide
        while True:
            while l & 1 == 0:
                l >>= 1
            if not f(self.func(res, self.tree[l])):
                while l < self.size:
                    l *= 2
                    if f(self.func(res, self.tree[l])):
                        res = self.func(res, self.tree[l])
                        l += 1
                return l - self.size
            res = self.func(res, self.tree[l])
            l += 1
            if l & (-l) == l:
                break
        return self.n

    def min_left(self, r, f):
        assert 0 <= r <= self.n
        assert f(self.ide)
        if r == 0:
            return 0
        r += self.size
        res = self.ide
        while True:
            r -= 1
            while r > 1 and r & 1:
                r >>= 1
            if not f(self.func(self.tree[r], res)):
                while r < self.size:
                    r = 2 * r + 1
                    if f(self.func(self.tree[r], res)):
                        res = self.func(self.tree[r], res)
                        r -= 1
                return r + 1 - self.size
            res = self.func(self.tree[r], res)
            if r & (-r) == r:
                break
        return 0

    def __getitem__(self, i):
        return self.get(i)

    def __setitem__(self, key, val):
        self.update(key, val)

    def __iter__(self):
        for i in range(self.n):
            yield self.tree[i + self.size]

    def __str__(self):
        return str(list(self))

# (00, 01, 10, 11)
def op(L, R):
    f00 = (L[0] & R[0]) | (L[1] & R[2])
    f01 = (L[0] & R[1]) | (L[1] & R[3])
    f10 = (L[2] & R[0]) | (L[3] & R[2])
    f11 = (L[3] & R[3]) | (L[2] & R[1])
    return (f00, f01, f10, f11)

def calc(n, a):
    vals = []
    for i in range(1, n - 1):
        for j in range(2):
            for k in range(2):
                val = a[2 * i - j] + a[2 * i + 1 + k]
                vals.append((val, i, j, k))

    vals.append((a[0] + a[1], 0, 0, 0))
    vals.append((a[0] + a[2], 0, 0, 1))
    vals.append((a[2 * n - 2] + a[2 * n - 1], n - 1, 0, 0))
    vals.append((a[2 * n - 3] + a[2 * n - 1], n - 1, 1, 0))
    vals.sort(key=lambda x: x[0])

    m = len(vals)

    def sub(mid):
        seg = segtree([(0, 0, 0, 0) for i in range(n)], op, (1, 1, 1, 1))
        r = 0
        for l in range(m):
            while r != m and vals[r][0] - vals[l][0] <= mid:
                _, idx, j, k = vals[r]
                res = list(seg[idx])
                res[2 * j + k] = 1
                seg[idx] = tuple(res)
                r += 1
            res = seg.tree[1]
            if res[0] == 1:
                return True
            _, idx, j, k = vals[l]
            res = list(seg[idx])
            res[2 * j + k] = 0
            seg[idx] = tuple(res)
        return False

    ok = vals[-1][0] - vals[0][0] + 1
    ng = -1
    while abs(ng - ok) > 1:
        mid = (ok + ng) // 2
        if sub(mid):
            ok = mid
        else:
            ng = mid
    return ok

def solve():
    n = int(input())
    a = list(map(int, input().split()))
    if n % 2 == 0:
        L, R = 1 << 60, -(1 << 60)
        for i in range(0, n, 2):
            a0, a1 = a[i], a[i + 1]
            b0, b1 = a[n + i], a[n + i + 1]
            if a0 > b0:
                a0, b0 = b0, a0
            if a1 < b1:
                a1, b1 = b1, a1
            L = min(L, a0 + a1, b0 + b1)
            R = max(R, a0 + a1, b0 + b1)
        print(R - L)
        return
    if n == 1:
        print(0)
        return
    b = []
    tmp = 0
    for i in range(n):
        b.append(a[tmp])
        b.append(a[tmp + 1])
        tmp = (tmp + 1 + n) % (2 * n)
    ans1 = calc(n, b)
    b[0], b[-1] = b[-1], b[0]
    ans2 = calc(n, b)
    print(min(ans1, ans2))

for _ in range(int(input())):
    solve()

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

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

Think step by step.

Response:
```plaintext
5
2
1 1000000000 1000000000 1
1
1 1000000000
3
1 1000000000 1 1000000000 1 1000000000
3
1 2 3 4 5 6
5
1 1000000000 2 999999999 3 999999998 4 999999997 5 999999996
```