**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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <queue>
//#define int long long
#define endl '\n'
#define lson i<<1
#define rson i<<1|1
#define debug(x) cerr << #x << " = " <<  x << "\n";
#define debug2(l,r) cerr << "[" #l "," #r "] = [" << l << "," << r << "]\n";
#define debug3(x,y,z) cerr << "{" #x "," #y "," #z "} = " << "{" << x << "," << y << "," << z << "}\n";
#define debugv(v) cerr << #v": size= " << v.size() << "\nelement: " ; for(auto p:v)  cerr << p << " ";
using namespace std;
const int N = 1e5 + 55, MOD = 1e9 + 7, INF = 2.1e9;
int T, n, m, ar[N * 2], a[N][2];
struct node {
    int v, x;
    bool i, j;
} b[N * 4];
bool cmp(node a, node b) {
    return a.v > b.v;
}
struct mat {
    int x, y, z, t;
} o = {INF, INF, INF, INF}, e = {0, INF, INF, 0};
mat operator *(mat a, mat b) {
    mat res = o;
    res.x = min(max(a.x, b.x), max(a.y, b.z));
    res.y = min(max(a.x, b.y), max(a.y, b.t));
    res.z = min(max(a.z, b.x), max(a.t, b.z));
    res.t = min(max(a.z, b.y), max(a.t, b.t));
    return res;
}
namespace Sgt {
    mat a[N * 4];
    void build(int i, int L, int R) {
        a[i] = o;
        if (L == R) return;
        int mid = (L + R) >> 1;
        build(lson, L, mid);
        build(rson, mid + 1, R);
    }
    void push_up(int i) {
        a[i] = a[lson] * a[rson];
    }
    void update(int i, int L, int R, int pos, bool i1, bool j1, int x) {
        if (L == R) {
            if (!i1 && !j1) a[i].x = x;
            else if (!i1 && j1) a[i].y = x;
            else if (i1 && !j1) a[i].z = x;
            else a[i].t = x;
            return;
        }
        int mid = (L + R) >> 1;
        if (mid >= pos) update(lson, L, mid, pos, i1, j1, x);
        else update(rson, mid + 1, R, pos, i1, j1, x);
        push_up(i);
    }
    mat query(int i, int L, int R, int l, int r) {
        if (l > r) return e;
        if (l <= L && R <= r) return a[i];
        int mid = (L + R) >> 1;
        mat t = e;
        if (mid >= l) t = t * query(lson, L, mid, l, r);
        if (mid < r) t = t * query(rson, mid + 1, R, l, r);
        return t;
    }
}
signed main() {
	cin.tie(0), ios_base::sync_with_stdio(false);
	cin >> T;
	while (T--) {
        cin >> n;
        m = n * 2;
        for (int i = 1; i <= m; i++) cin >> ar[i];
        if (n == 1) cout << 0 << endl;
        else if (n & 1) {
            Sgt::build(1, 1, n);
            for (int i = 1, j = 0; j < n; ) {
                a[++j][0] = ar[i], a[j][1] = ar[i + 1];
                if (i + 1 > n) i += 1 - n;
                else i += 1 + n;
            }
            for (int i = 1, r = 0; i <= n; i++) {
                int x = a[(i - 2 + n) % n + 1][1];
                int y = a[i][0], z = a[i][1];
                int t = a[i % n + 1][0];
                for (int j = 0; j < 2; j++) {
                    for (int k = 0; k < 2; k++) {
                        int u = j ? y : x, v = k ? z : t;
                        b[++r] = {u + v, i, j, k};
                    }
                }
            }
            sort(b + 1, b + n * 4 + 1, cmp);
            int ans = INF;
            for (int i = 1, j = 0; i <= n * 4; i++) {
                while (j + 1 <= n * 4 && b[j + 1].v == b[i].v)
                    j++, Sgt::update(1, 1, n, b[j].x, b[j].i, b[j].j, b[j].v);
                int x = b[i].x, y = 0;
                mat t = Sgt::query(1, 1, n, x + 1, n) * Sgt::query(1, 1, n, 1, x - 1);
                if (!b[i].j && !b[i].i) y = t.x;
                else if (!b[i].j && b[i].i) y = t.y;
                else if (b[i].j && !b[i].i) y = t.z;
                else y = t.t;
                ans = min(ans, y - b[i].v);
            }
            cout << ans << endl;
        } else {
            int mi = INF, ma = -INF;
            for (int i = 1; i <= n; i += 2) {
                int x = ar[i] + ar[i + 1], y = ar[i + n] + ar[i + n + 1];
                int z = ar[i + n] + ar[i + 1], t = ar[i] + ar[i + n + 1];
                if (x > y) swap(x, y);
                if (z > t) swap(z, t);
                if (x < z) swap(x, z), swap(y, t);
                mi = min(mi, x), ma = max(ma, y);
            }
            cout << ma - mi << endl;
        }
	}
	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Let Me Teach You a Lesson (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>

#define ll long long
#define pb push_back
#define f first
#define s second
#define sz(x) (int)(x).size()
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define ios ios_base::sync_with_stdio(false);cin.tie(NULL)
#define ld long double
#define li __int128

using namespace std;

mt19937 rng(time(NULL));

void test(){
	int n;
	cin >> n;
	vector<int> a(2*n);
	for(int i=0;i<2*n;i++){
		cin >> a[i];
	}
	if(n==1){
		printf("0\n");
		return;
	}
	vector<int> part(n);
	for(int i=0;i<n;i++){
		int mn=min(a[i],a[i+n]);
		part[i]=a[i]+a[i+n]-2*mn;
		a[i]=mn;
		a[i+n]=mn;
	}
	vector<int> mns;
	for(int i=0;i<2*n;i+=2){
		int x=a[i]+a[i+1];
		int A=part[min(i,(i+n)%(2*n))];
		int B=part[min(i+1,(i+1+n)%(2*n))];
		mns.pb(x);
		mns.pb(x+A);
		mns.pb(x+B);
		mns.pb(x+A+B);
	}
	sort(all(mns));
	mns.erase(unique(all(mns)),mns.end());
	vector<int> sm(n),pr(n);
	if(n%2==1){
		int i=0;
		int ind=0;
		while(1){
			sm[ind]=a[i]+a[i+1];
			pr[ind]=part[min(i,(i+n)%(2*n))];
			ind++;
			i=(i+n-1)%(2*n);
			if(i==0)break;
		}
	}
	auto getmx=[&](int mn){
		int mx;
		// O(n*4)
		if(n%2==0){
			mx=mn;
			for(int i=0;i<n;i+=2){
				int s1=a[i]+a[i+1];
				int s2=a[i+n]+a[i+n+1];
				int A=part[i],B=part[i+1];
				int mnmx=INT_MAX;
				for(int k1=0;k1<2;k1++){
					for(int k2=0;k2<2;k2++){
						int S1=s1,S2=s2;
						if(k1){
							S1+=A;
						}
						else{
							S2+=A;
						}
						if(k2){
							S1+=B;
						}
						else{
							S2+=B;
						}
						if(S1>S2)swap(S1,S2);
						if(S1>=mn)mnmx=min(mnmx,S2);
					}
				}
				mx=max(mx,mnmx);
			}
		}
		else{
			mx=INT_MAX;
			for(int k=0;k<2;k++){
				if(k==0)
					sm[0]+=pr[n-1];
				else
					sm[n-1]+=pr[n-1];

				// Dp check
				vector dp(n,pair<int,int>());
				// f last did take it, s last didnt take it
				dp[n-1].f=sm[n-1]>=mn?sm[n-1]:INT_MAX;
				dp[n-1].s=sm[n-1]+pr[n-2]>=mn?sm[n-1]+pr[n-2]:INT_MAX;
				for(int i=n-2;i>=0;i--){
					dp[i].f=
						min(sm[i]>=mn?max(sm[i],dp[i+1].s):INT_MAX,
						sm[i]+pr[i]>=mn?max(sm[i]+pr[i],dp[i+1].f):INT_MAX);
					if(i)
					dp[i].s=
						min(sm[i]+pr[i-1]>=mn?max(sm[i]+pr[i-1],dp[i+1].s):INT_MAX,
						sm[i]+pr[i-1]+pr[i]>=mn?max(sm[i]+pr[i-1]+pr[i],dp[i+1].f):INT_MAX);
				}
				mx=min(mx,dp[0].f);

				if(k==0)
					sm[0]-=pr[n-1];
				else
					sm[n-1]-=pr[n-1];
			}
		}
		return mx;
	};
	int ans=INT_MAX;
	bool end=0;
	int dir=-1;
	int lst=INT_MAX;
	int lstmx=INT_MAX;
	int lstmn=0;
	for(auto mn:mns){ // Ternary searchable??
		int mx=getmx(mn);
		if(mx!=INT_MAX){
			assert(!end);
			if(lstmx!=mx){
				int my=lstmx-lstmn;
				//printf("%i %i: %i\n",lstmn,lstmx,my);
				if(lstmx!=INT_MAX){
					if(my<lst){
						assert(dir==-1);
					}
					if(my>lst){
						dir=1;
					}
				}
				lst=my;
				lstmx=mx;
			}
			ans=min(ans,mx-mn);
		}
		else{
			if(lstmx!=mx){
				int my=lstmx-lstmn;
				if(lstmx!=INT_MAX){
					//printf("%i %i: %i\n",lstmn,lstmx,my);
					if(my<lst){
						assert(dir==-1);
					}
					if(my>lst){
						dir=1;
					}
				}
				lst=my;
				lstmx=mx;
			}
			end=1;
		}
		lstmn=mn;
	}
	printf("%i\n",ans);
}
int main()
{
	ios;
	int t;
	cin >> t;
	while(t--){
		test();
	}
	return 0;
}

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

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

Think step by step.

Response:
```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:
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
// #define int long long
using namespace std;
template<class T>inline bool cmax(T &a, T b) {return a < b ? a = b, 1 : 0;}
template<class T>inline bool cmin(T &a, T b) {return b < a ? a = b, 1 : 0;}
const int N = 1e5 + 10, Inf = 0x7fffffff;
int T, n, a[N * 2], tmp_a[N * 2], b[N][2][2];
struct
Matrix {
  int d[2][2];
  Matrix() {memset(d, 0, sizeof(d));}
  Matrix(int a[2][2]) {d[0][0] = a[0][0], d[0][1] = a[0][1], d[1][0] = a[1][0], d[1][1] = a[1][1];}
  inline int* operator [](int x) {return d[x];}
  inline Matrix operator *(const Matrix &B) const {
    Matrix ans;
    ans.d[0][0] = min(max(d[0][0], B.d[0][0]), max(d[0][1], B.d[1][0]));
    ans.d[0][1] = min(max(d[0][0], B.d[0][1]), max(d[0][1], B.d[1][1]));
    ans.d[1][0] = min(max(d[1][0], B.d[0][0]), max(d[1][1], B.d[1][0]));
    ans.d[1][1] = min(max(d[1][0], B.d[0][1]), max(d[1][1], B.d[1][1]));
    return ans;
  }
} tree[N << 2];
#define lc (p << 1)
#define rc (p << 1 | 1)
struct Map {int pos, id[2];};
vector<Map> c[N << 2];
inline void push_up(int p) {tree[p] = tree[lc] * tree[rc];}
inline void
build(int p, int l, int r) {
  if(l == r) {
    tree[p] = Matrix(b[l]);
    return ;
  }
  int mid = (l + r) / 2;
  build(lc, l, mid);
  build(rc, mid + 1, r);
  push_up(p);
}
inline void
update(int p, int l, int r, int x) {
  if(l == r) {
    tree[p] = Matrix(b[l]);
    return ;
  }
  int mid = (l + r) / 2;
  if(x <= mid) update(lc, l, mid, x);
  else update(rc, mid + 1, r, x);
  push_up(p);
}
inline void
solve() {
  scanf("%d", &n);
  for(int i = 1; i <= n * 2; i ++) scanf("%d", a + i);
  if(n == 1) {puts("0"); return ;}
  if(n % 2 == 0) {
    int mn = Inf, mx = 0;
    for(int i = 1; i <= n / 2; i ++) {
      int tmp1 = a[2 * i]  + a[2 * i - 1];
      int tmp2 = a[2 * i + n] + a[2 * i - 1 + n];
      int tmp3 = a[2 * i + n] + a[2 * i - 1];
      int tmp4 = a[2 * i] + a[2 * i - 1 + n];
      if(abs(tmp1 - tmp2) < abs(tmp3 - tmp4)) {
        cmax(mx, max(tmp1, tmp2));
        cmin(mn, min(tmp1, tmp2));
      } else {
        cmax(mx, max(tmp3, tmp4));
        cmin(mn, min(tmp3, tmp4));
      }
    }
    printf("%d\n", mx - mn);
    return ;
  }
  vector<int> num;
  memcpy(tmp_a, a, sizeof(int) * (n * 2 + 1));
  for(int i = 1; i <= n / 2; i ++) {
    a[4 * i - 3] = tmp_a[2 * i - 1];
    a[4 * i - 2] = tmp_a[2 * i];
    a[4 * i - 1] = tmp_a[2 * i + n];
    a[4 * i] = tmp_a[2 * i + n + 1];
  }
  a[n * 2 - 1] = tmp_a[n], a[n * 2] = tmp_a[n + 1];
  a[0] = a[n * 2], a[n * 2 + 1] = a[1];
  for(int i = 1; i <= n; i ++) {
    b[i][0][0] = a[2 * i - 1] + a[2 * i];
    b[i][1][0] = a[2 * i - 2] + a[2 * i];
    b[i][0][1] = a[2 * i - 1] + a[2 * i + 1];
    b[i][1][1] = a[2 * i - 2] + a[2 * i + 1];
    num.push_back(b[i][0][0]);
    num.push_back(b[i][1][0]);
    num.push_back(b[i][0][1]);
    num.push_back(b[i][1][1]);
  }
  sort(num.begin(), num.end());
  num.erase(unique(num.begin(), num.end()), num.end());
  for(int i = 0; i < num.size(); i ++) c[i].clear();
  for(int i = 1; i <= n; i ++) {
    int tmp1 = lower_bound(num.begin(), num.end(), b[i][0][0]) - num.begin();
    c[tmp1].push_back({i, 0, 0});
    int tmp2 = lower_bound(num.begin(), num.end(), b[i][1][0]) - num.begin();
    c[tmp2].push_back({i, 1, 0});
    int tmp3 = lower_bound(num.begin(), num.end(), b[i][0][1]) - num.begin();
    c[tmp3].push_back({i, 0, 1});
    int tmp4 = lower_bound(num.begin(), num.end(), b[i][1][1]) - num.begin();
    c[tmp4].push_back({i, 1, 1});
  }
  build(1, 1, n);
  int ans = Inf;
  for(int i = 0; i < num.size(); i ++) {
    /*int tmp[2][2];
    memcpy(tmp, b[1], sizeof(b[1]));
    b[1][0][0] = b[1][0][1] = b[1][1][0] = b[1][1][1] = Inf;
    if(tmp[0][0] >= num[i]) b[1][0][0] = tmp[0][0];
    if(tmp[0][1] >= num[i]) b[1][0][1] = tmp[0][1];
    update(1, 1, n, 1);
    cmin(ans, tree[1][0][0] - num[i]);
    b[1][0][0] = b[1][0][1] = Inf;
    if(tmp[1][0] >= num[i]) b[1][1][0] = tmp[1][0];
    if(tmp[1][1] >= num[i]) b[1][1][1] = tmp[1][1];
    update(1, 1, n, 1);
    cmin(ans, tree[1][1][1] - num[i]);
    memcpy(b[1], tmp, sizeof(b[1]));
    update(1, 1, n, 1);*/
    cmin(ans, min(tree[1][0][0], tree[1][1][1]) - num[i]);
    // cout << ans << "\n-----------------------" << endl;
    int watch = num[i];
    for(auto [j, id] : c[i]) {
      b[j][id[0]][id[1]] = Inf;
      update(1, 1, n, j);
    }
  }
  printf("%d\n", ans);
  return ;
}
int
main() {
  #ifndef ONLINE_JUDGE
  freopen("CF1995E2.in", "r", stdin);
  freopen("CF1995E2.out", "w", stdout);
  #endif
  scanf("%d", &T);
  // for(int i = 1; i <= T; i ++) cout << "Case " << i << ": \n", solve();
  while(T --) solve();
  return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. The code handles different cases based on whether `n` is even or odd, and it involves complex matrix operations and updates. Here are some steps to create a test input:

1. **Edge Case with Small `n`**: Test with the smallest possible value of `n` to ensure the code handles it correctly.
2. **Odd and Even `n`**: Test with both odd and even values of `n` to ensure the code handles both cases.
3. **Large Intelligence Values**: Use large values for `a_i` to ensure the code handles large integers correctly.
4. **Specific Patterns**: Use specific patterns in the intelligence values to test the logic of the code.

Let's create a test input with these considerations:

```plaintext
3
1
10 17
2
1 1000000000 1 1000000000
3
1 2 3 4 5 6
```

### Explanation:
1. **Test Case 1**:
   - `n = 1`
   - `a = [10, 17]`
   - This is a simple case where `n` is 1, and the code should output 0 because there is only one pair.

2. **Test Case 2**:
   - `n = 2`
   - `a = [1, 1000000000, 1, 1000000000]`
   - This tests the code with large integers and ensures it can handle the maximum possible value for `a_i`.

3. **Test Case 3**:
   - `n = 3`
   - `a = [1, 2, 3, 4, 5, 6]`
   - This tests the code with a sequence of integers to ensure it can handle a more complex case where the values are not equal and the code needs to perform swaps to minimize the difference.

This test input is designed to cover a range of scenarios and edge cases to expose any bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: 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:
/*
 *                                                     __----~~~~~~~~~~~------___
 *                                    .  .   ~~//====......          __--~ ~~
 *                    -.            \_|//     |||\\  ~~~~~~::::... /~
 *                 ___-==_       _-~o~  \/    |||  \\            _/~~-
 *         __---~~~.==~||\=_    -_--~/_-~|-   |\\   \\        _/~
 *     _-~~     .=~    |  \\-_    '-~7  /-   /  ||    \      /
 *   .~       .~       |   \\ -_    /  /-   /   ||      \   /
 *  /  ____  /         |     \\ ~-_/  /|- _/   .||       \ /
 *  |~~    ~~|--~~~~--_ \     ~==-/   | \~--===~~        .\
 *           '         ~-|      /|    |-~\~~       __--~~
 *                       |-~~-_/ |    |   ~\_   _-~            /\
 *                            /  \     \__   \/~                \__
 *                        _--~ _/ | .-~~____--~-/                  ~~==.
 *                       ((->/~   '.|||' -_|    ~~-/ ,              . _||
 *                                  -_     ~\      ~~---l__i__i__i--~~_/
 *                                  _-~-__   ~)  \--______________--~~
 *                                //.-~~~-~_--~- |-------~~~~~~~~
 *                                       //.-~~~--\
 *                       ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
 *
 *                               神兽保佑            永无BUG
 */

/*
 * @Description: I want to be the weakest vegetable in the world!
 * @Author: I.Z.
 */
#include<bits/stdc++.h>
using namespace std;
#define MOD         998244353
#define speMOD      2933256077ll
#define int         long long
#define pii         pair<int,int>
#define all(v)      v.begin(),v.end()
#define pb          push_back
#define REP(i,b,e)  for(int i=(b);i<(int)(e);++i)
#define over(x)     {cout<<(x)<<endl;return;}
#define lowbit(x)   ((x)&(-(x)))
#define cntbit(x)   __builtin_popcount(x)
#define rf(v)       shuffle(all(v),sd);
#define deal(v)     sort(all(v));v.erase(unique(v.begin(),v.end()),v.end())
#define lbound(v,x) lower_bound(all(v),x)-v.begin()
mt19937 sd(random_device{}());
int qpow(int a,int b,int m=MOD,int res=1){
	a%=m;
	while(b>0)res=(b&1)?(res*a%m):(res),a=a*a%m,b>>=1;
	return res;
}
string YN(bool x,string Y="Yes",string N="No") {
	if(x){
		return Y;
	} else {
		return N;
	}
}
int exgcd(int x,int y,int &a,int &b){
	if(y==0){
		a=1;b=0;
		return x;
	}
	int d=exgcd(y,x%y,a,b);
	int z=b;
	b=a-b*(x/y);
	a=z;
	return d;
}
int _x_,_y_;
inline int inverse(int x,int y=MOD){
	exgcd(x,y,_x_,_y_);
	return (_x_+y)%y;
}
int fac[2000005],inv[2000005];
void init(int n){
	fac[0]=inv[0]=1;
	REP(i,1,n+1)fac[i]=fac[i-1]*i%MOD;
	inv[n]=qpow(fac[n],MOD-2);
	for(int i=n-1;i>=1;--i)inv[i]=inv[i+1]*(i+1)%MOD;
}
int binom(int x,int y){
	return x<y||y<0? 0:fac[x]*inv[y]%MOD*inv[x-y]%MOD;
}
struct matrix{
	int a,b,c,d;
	matrix operator *(matrix e){
		return matrix{(a&e.a)|(b&e.c),(a&e.b)|(b&e.d),(c&e.a)|(d&e.c),(c&e.b)|(d&e.d)};
	}
}seg[800005];
int n;
int a[200005];
void build(int l,int r,int p){
	seg[p]={0,0,0,0};
	if(l==r)return;
	int m=(l+r)>>1;
	build(l,m,p*2+1);build(m+1,r,p*2+2);
}
void update(int pos,int l,int r,int p,int val,int op){
	if(l==r){
		if(op==0)seg[p].a=val;
		else if(op==1)seg[p].b=val;
		else if(op==2)seg[p].c=val;
		else seg[p].d=val;
		return;
	}
	int m=(l+r)>>1;
	if(m>=pos)update(pos,l,m,p*2+1,val,op);
	else update(pos,m+1,r,p*2+2,val,op);
	seg[p]=seg[p*2+1]*seg[p*2+2];
}
vector<int>t;
vector<pii>ps[400005];
void add(int x){
	for(auto [i,j]:ps[x])if(i<n-1)update(i+1,1,n-1,0,1,j);
}
void del(int x){
	for(auto [i,j]:ps[x])if(i<n-1)update(i+1,1,n-1,0,0,j);
}
bool check(int l,int r){
	return (seg[0].c&&a[n*2-1]+a[n]>=t[l]&&a[n*2-1]+a[n]<=t[r])||
	(seg[0].d&&a[n-1]+a[n]>=t[l]&&a[n-1]+a[n]<=t[r])||
	(seg[0].a&&a[n*2-1]+a[0]>=t[l]&&a[n*2-1]+a[0]<=t[r])||
	(seg[0].b&&a[n-1]+a[0]>=t[l]&&a[n-1]+a[0]<=t[r]);
}
void Main() {
	cin>>n;
	REP(i,0,n*2)cin>>a[i];
	t.clear();
	if(n%2==0){
		vector<int>v;
		REP(i,0,n/2){
			int b1=a[i*2]+a[i*2+1],b2=a[i*2+n]+a[i*2+1+n];
			int c1=a[i*2]+a[i*2+1+n],c2=a[i*2+n]+a[i*2+1];
			if(b1>b2)swap(b1,b2);
			if(c1>c2)swap(c1,c2);
			if(b1>=c1)v.pb(b1),v.pb(b2);
			else v.pb(c1),v.pb(c2);
		}
		sort(all(v));
		over(v[n-1]-v[0])
	}
	if(n==1)over(0)
	REP(i,0,n){
		REP(j,0,2){
			REP(k,0,2){
				t.pb(a[j*n+i]+a[k*n+(i+1)%n]);
			}
		}
	}
	deal(t);
	REP(i,0,t.size())ps[i].clear();
	REP(i,0,n){
		REP(j,0,2){
			REP(k,0,2){
				int x=lbound(t,a[j*n+i]+a[k*n+(i+1)%n]);
				ps[x].pb({i,((1-j)<<1)|k});
//				cout<<x<<' '<<i<<' '<<(((1-j)<<1)|k)<<endl;
			}
		}
	}
	build(1,n-1,0);
	int x=-1;
	int ans=t[t.size()-1]-t[0];
//	for(auto i:t)cout<<i<<' ';
//	cout<<endl;
	add(1);
//	cout<<seg[0].c<<' '<<seg[0].d<<' '<<check(1,1)<<endl;
	REP(i,0,t.size()){
		if(x<i)add(++x);
		while(x+1<(int)t.size()&&!check(i,x))add(++x);
		if(check(i,x))ans=min(ans,t[x]-t[i]);
//		cout<<i<<' '<<x<<' '<<check(i,x)<<endl;
		if(i==x)++x;
		else del(i);
	}
	over(ans)
}
void TC() {
    int tc=1;
    cin>>tc;
	while(tc--){
		Main();
		cout.flush();
	}
}
signed main() {
	return cin.tie(0),cout.tie(0),ios::sync_with_stdio(0),TC(),0;
}
/*
1. CLEAR the arrays (ESPECIALLY multitests)
2. DELETE useless output
 */

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to consider edge cases and scenarios that maximize the failure probability. The algorithm is designed to minimize the maximal difference in total intelligence over all pairs of desks. The bug might be in how it handles the swapping and the calculation of the minimal difference.

### Step-by-Step Analysis:

1. **Edge Cases:**
   - **Small `n` values:** Test with `n = 1` and `n = 2`.
   - **Equal Intelligence:** All knights have the same intelligence.
   - **Highly Skewed Intelligence:** One pair has very high intelligence, and the rest have very low intelligence.

2. **Scenarios:**
   - **Odd and Even `n`:** Test both odd and even values of `n`.
   - **Symmetrical Intelligence:** Test with symmetrical intelligence values around the circle.

### Test Input Generation:

Let's create a test input that includes these scenarios:

- **Test Case 1: `n = 1`**
  - All knights have different intelligence.
- **Test Case 2: `n = 2`**
  - One pair has very high intelligence, and the other has very low intelligence.
- **Test Case 3: `n = 3`**
  - Symmetrical intelligence values.
- **Test Case 4: `n = 4`**
  - All knights have the same intelligence.

### Test Input:

```plaintext
4
1
10 20
2
1 100 2 100
3
1 10 1 10 1 10
4
5 5 5 5 5 5 5 5
```

### Explanation:

1. **Test Case 1: `n = 1`**
   - Input: `10 20`
   - Expected Output: `0`
   - Explanation: The only pair has a total intelligence of 30, and there are no other pairs to compare.

2. **Test Case 2: `n = 2`**
   - Input: `1 100 2 100`
   - Expected Output: `0`
   - Explanation: The pairs are (1, 100) and (2, 100), both with a total intelligence of 101. Swapping the opposite pairs does not change the result.

3. **Test Case 3: `n = 3`**
   - Input: `1 10 1 10 1 10`
   - Expected Output: `0`
   - Explanation: The pairs are (1, 10), (1, 10), and (1, 10), all with a total intelligence of 11. Swapping the opposite pairs does not change the result.

4. **Test Case 4: `n = 4`**
   - Input: `5 5 5 5 5 5 5 5`
   - Expected Output: `0`
   - Explanation: All pairs have a total intelligence of 10. Swapping the opposite pairs does not change the result.

This test input should help expose any bugs in the algorithm, especially in handling edge cases and symmetrical intelligence values.


**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>
typedef long long ll;
using namespace std;
int T,n,tot;
ll val[100005],bel[200005],a[200005];
int rev(int x){
	return x<=n?x+n:x-n;
}
ll calc(int pos,int id){
	if(id)return a[rev(pos)];
	return a[pos];
}
struct node{
	ll val;
	int p,a,b;
}arr[400005];
bool cmp(node x,node y){
	return x.val<y.val;
}
struct matrix{
	ll a[2][2];
	void init(){
		a[0][0]=a[0][1]=a[1][0]=a[1][1]=2e18;
	}
}res,A;
matrix mul(matrix a,matrix b){
	res.a[0][0]=min(max(a.a[0][0],b.a[0][0]),max(a.a[0][1],b.a[1][0]));
	res.a[0][1]=min(max(a.a[0][0],b.a[0][1]),max(a.a[0][1],b.a[1][1]));
	res.a[1][0]=min(max(a.a[1][0],b.a[0][0]),max(a.a[1][1],b.a[1][0]));
	res.a[1][1]=min(max(a.a[1][0],b.a[0][1]),max(a.a[1][1],b.a[1][1]));
	return res;
}
struct Seg{
	int l,r;
	matrix val;
}Tree[200005];
ll vmn;
matrix init(int j){
	int p1=(j<<1)+1,p2=(j<<1)+2;
	res.init();
	for(int k=0;k<2;k++){
		if(val[bel[p2]]!=0){
			ll vi=calc(p1,k)+calc(p2,1);
			if(vi>=vmn){
				if(val[bel[p2+1]]!=0){
					ll vj=calc(rev(p2+1),1)+calc(rev(p2),1);
					if(vj>=vmn)res.a[k][1]=min(res.a[k][1],max(vi,vj));
				}
				if(val[bel[p2+1]]!=1){
					ll vj=calc(rev(p2+1),0)+calc(rev(p2),1);
					if(vj>=vmn)res.a[k][0]=min(res.a[k][0],max(vi,vj));
				}
			}
		}
		if(val[bel[p2]]!=1){
			ll vi=calc(p1,k)+calc(p2,0);
			if(vi>=vmn){
				if(val[bel[p2+1]]!=0){
					ll vj=calc(rev(p2+1),1)+calc(rev(p2),0);
					if(vj>=vmn)res.a[k][1]=min(res.a[k][1],max(vi,vj));
				}
				if(val[bel[p2+1]]!=1){
					ll vj=calc(rev(p2+1),0)+calc(rev(p2),0);
					if(vj>=vmn)res.a[k][0]=min(res.a[k][0],max(vi,vj));
				}
			}
		}
	}
	return res;
}
void ztree(int p,int l,int r){
	Tree[p].l=l,Tree[p].r=r;
	if(l==r){
		Tree[p].val=init(l-1);
		return;
	}
	int mid=l+r>>1;
	ztree(p*2,l,mid);
	ztree(p*2+1,mid+1,r);
	Tree[p].val=mul(Tree[p*2].val,Tree[p*2+1].val);
}
void update(int p,int l){
	if(Tree[p].l==Tree[p].r){
		Tree[p].val=init(l-1);
		return;
	}
	int mid=Tree[p].l+Tree[p].r>>1;
	if(l<=mid)update(p*2,l);
	else update(p*2+1,l);
	Tree[p].val=mul(Tree[p*2].val,Tree[p*2+1].val);
}
int vis[50005],tim;
void upd(int pos){
	if(1<=pos-1&&pos-1<=n/2&&vis[pos-1]!=tim)update(1,pos-1),vis[pos-1]=tim;
	if(1<=pos&&pos<=n/2&&vis[pos]!=tim)update(1,pos),vis[pos]=tim;
	if(1<=pos+1&&pos+1<=n/2&&vis[pos+1]!=tim)update(1,pos+1),vis[pos+1]=tim;
}
int main(){
	scanf("%d",&T);
//	T=1;
	while(T--){
		scanf("%d",&n);
//		n=99999;
		for(int i=1;i<=n*2;i++)scanf("%lld",&a[i]),bel[i]=i>n?(i-n):i;
		if(n==1){
			puts("0");
			continue;
		}
		if(n%2==0){
			for(int i=1;i<=n;i+=2){
				if(a[i]<a[i+n]&&a[i+1]<a[i+1+n])swap(a[i],a[i+n]);
				if(a[i]>a[i+n]&&a[i+1]>a[i+1+n])swap(a[i],a[i+n]);
			}
			ll mn=1e18,mx=0;
			for(int i=1;i<=n;i++){
				mn=min(mn,a[i*2-1]+a[i*2]);
				mx=max(mx,a[i*2-1]+a[i*2]);
			}
			cout<<mx-mn<<endl;
			continue;
		}
		ll ans=2e18;
		for(int i=1;i<=n;i++)val[i]=-1;
		tot=0;
		for(int i=1;i<=n;i++){
			for(int a=0;a<2;a++){
				for(int b=0;b<2;b++){
					ll va=calc(i*2-1,a)+calc(i*2,b);
					arr[++tot].a=a,arr[tot].b=b,arr[tot].p=i,arr[tot].val=va;
				}
			}
		}
		sort(arr+1,arr+1+tot,cmp);
		ztree(1,1,n/2);
		for(int i=1;i<=tot;i++){
			int cur=arr[i].p;
			vmn=arr[i].val;
			val[bel[cur*2-1]]=arr[i].a,val[bel[cur*2]]=arr[i].b;
			tim++;
			upd(bel[cur*2-1]+1>>1);
			upd(bel[cur*2]+1>>1);
			if(val[1]!=0){
				A.init();
				A.a[0][1]=0;
				A=mul(A,Tree[1].val);
				for(int j=0;j<2;j++){
					ll val=calc(n,j)+calc(n+1,1);
					if(val>=vmn)ans=min(ans,max(A.a[0][j],val)-vmn);
				}
			}
			if(val[1]!=1){
				A.init();
				A.a[0][0]=0;
				A=mul(A,Tree[1].val);
				for(int j=0;j<2;j++){
					ll val=calc(n,j)+calc(n+1,0);
					if(val>=vmn)ans=min(ans,max(A.a[0][j],val)-vmn);
				}
			}
			val[bel[cur*2-1]]=val[bel[cur*2]]=-1;
			vmn=arr[i].val+1;
			tim++;
			upd(bel[cur*2-1]+1>>1);
			upd(bel[cur*2]+1>>1);
		}
		cout<<ans<<endl;
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
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: Set
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given a positive integer $k$ and a set $S$ of all integers from $l$ to $r$ (inclusive).

You can perform the following two-step operation any number of times (possibly zero):

  1. First, choose a number $x$ from the set $S$, such that there are at least $k$ multiples of $x$ in $S$ (including $x$ itself);    2. Then, remove $x$ from $S$ (note that nothing else is removed).

Find the maximum possible number of operations that can be performed.

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

The only line of each test case contains three integers $l$, $r$, and $k$ ($1\le l\le r\leq 10^9$, $1\leq k\le r-l+1$) — the minimum integer in $S$, the maximum integer in $S$, and the parameter $k$.

Output: For each test case, output a single integer — the maximum possible number of operations that can be performed.

Examples:
input:
8
3 9 2
4 9 1
7 9 2
2 10 2
154 220 2
147 294 2
998 24435 3
1 1000000000 2
output:
2
6
0
4
0
1
7148
500000000

Note: In the first test case, initially, $S = \\{3,4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 4$ for the first operation, since there are two multiples of $4$ in $S$: $4$ and $8$. $S$ becomes equal to $\\{3,5,6,7,8,9\\}$;    2. Choose $x = 3$ for the second operation, since there are three multiples of $3$ in $S$: $3$, $6$, and $9$. $S$ becomes equal to $\\{5,6,7,8,9\\}$.

In the second test case, initially, $S=\\{4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 5$, $S$ becomes equal to $\\{4,6,7,8,9\\}$;    2. Choose $x = 6$, $S$ becomes equal to $\\{4,7,8,9\\}$;    3. Choose $x = 4$, $S$ becomes equal to $\\{7,8,9\\}$;    4. Choose $x = 8$, $S$ becomes equal to $\\{7,9\\}$;    5. Choose $x = 7$, $S$ becomes equal to $\\{9\\}$;    6. Choose $x = 9$, $S$ becomes equal to $\\{\\}$.

In the third test case, initially, $S=\\{7,8,9\\}$. For each $x$ in $S$, no multiple of $x$ other than $x$ itself can be found in $S$. Since $k = 2$, you can perform no operations.

In the fourth test case, initially, $S=\\{2,3,4,5,6,7,8,9,10\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 2$, $S$ becomes equal to $\\{3,4,5,6,7,8,9,10\\}$;    2. Choose $x = 4$, $S$ becomes equal to $\\{3,5,6,7,8,9,10\\}$;    3. Choose $x = 3$, $S$ becomes equal to $\\{5,6,7,8,9,10\\}$;    4. Choose $x = 5$, $S$ becomes equal to $\\{6,7,8,9,10\\}$.

Buggy Code:
#include <bits/stdc++.h>
#define fast_io ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
using namespace std;
#define ll long long
#define vl vector<ll>
const ll MOD = 1e9 + 7;

int main() {
    fast_io;
    ll t;
    cin >> t;
    while (t--) {
        ll k,l,r;cin>>l>>r>>k;
        ll sm=0;
        if(l*k>r){
            cout<<0<<endl;
            continue;
        }
        ll low=l,hi=1e18;
        ll tr=l;
        while(low<hi){
            ll mid=(low+hi-1)/2;
            if(mid*k<=r){
                tr=mid;
                low=mid+1;
            }
            else hi=mid;
        }
        cout<<tr-l+1<<endl;
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
1 100 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Set
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given a positive integer $k$ and a set $S$ of all integers from $l$ to $r$ (inclusive).

You can perform the following two-step operation any number of times (possibly zero):

  1. First, choose a number $x$ from the set $S$, such that there are at least $k$ multiples of $x$ in $S$ (including $x$ itself);    2. Then, remove $x$ from $S$ (note that nothing else is removed).

Find the maximum possible number of operations that can be performed.

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

The only line of each test case contains three integers $l$, $r$, and $k$ ($1\le l\le r\leq 10^9$, $1\leq k\le r-l+1$) — the minimum integer in $S$, the maximum integer in $S$, and the parameter $k$.

Output: For each test case, output a single integer — the maximum possible number of operations that can be performed.

Examples:
input:
8
3 9 2
4 9 1
7 9 2
2 10 2
154 220 2
147 294 2
998 24435 3
1 1000000000 2
output:
2
6
0
4
0
1
7148
500000000

Note: In the first test case, initially, $S = \\{3,4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 4$ for the first operation, since there are two multiples of $4$ in $S$: $4$ and $8$. $S$ becomes equal to $\\{3,5,6,7,8,9\\}$;    2. Choose $x = 3$ for the second operation, since there are three multiples of $3$ in $S$: $3$, $6$, and $9$. $S$ becomes equal to $\\{5,6,7,8,9\\}$.

In the second test case, initially, $S=\\{4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 5$, $S$ becomes equal to $\\{4,6,7,8,9\\}$;    2. Choose $x = 6$, $S$ becomes equal to $\\{4,7,8,9\\}$;    3. Choose $x = 4$, $S$ becomes equal to $\\{7,8,9\\}$;    4. Choose $x = 8$, $S$ becomes equal to $\\{7,9\\}$;    5. Choose $x = 7$, $S$ becomes equal to $\\{9\\}$;    6. Choose $x = 9$, $S$ becomes equal to $\\{\\}$.

In the third test case, initially, $S=\\{7,8,9\\}$. For each $x$ in $S$, no multiple of $x$ other than $x$ itself can be found in $S$. Since $k = 2$, you can perform no operations.

In the fourth test case, initially, $S=\\{2,3,4,5,6,7,8,9,10\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 2$, $S$ becomes equal to $\\{3,4,5,6,7,8,9,10\\}$;    2. Choose $x = 4$, $S$ becomes equal to $\\{3,5,6,7,8,9,10\\}$;    3. Choose $x = 3$, $S$ becomes equal to $\\{5,6,7,8,9,10\\}$;    4. Choose $x = 5$, $S$ becomes equal to $\\{6,7,8,9,10\\}$.

Buggy Code:
#include<iostream>
using namespace std;
int main()
{
	int t = 0;
	cin >> t;
	for (int i = 0; i < t; i++)
	{
		int l = 0, r = 0, k = 0;
		cin >> l >> r >> k;
		if ((r - k * l) / k + 1 < 0)
		{
			cout << 0 << endl;
		}
		else
		{
			cout << (r - k * l) / k + 1 << endl;
		}

	}

	return 0;
}

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

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

Think step by step.

Response:
```plaintext
5
1 10 1
1 10 2
10 10 1
10 10 2
1 100 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Set
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given a positive integer $k$ and a set $S$ of all integers from $l$ to $r$ (inclusive).

You can perform the following two-step operation any number of times (possibly zero):

  1. First, choose a number $x$ from the set $S$, such that there are at least $k$ multiples of $x$ in $S$ (including $x$ itself);    2. Then, remove $x$ from $S$ (note that nothing else is removed).

Find the maximum possible number of operations that can be performed.

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

The only line of each test case contains three integers $l$, $r$, and $k$ ($1\le l\le r\leq 10^9$, $1\leq k\le r-l+1$) — the minimum integer in $S$, the maximum integer in $S$, and the parameter $k$.

Output: For each test case, output a single integer — the maximum possible number of operations that can be performed.

Examples:
input:
8
3 9 2
4 9 1
7 9 2
2 10 2
154 220 2
147 294 2
998 24435 3
1 1000000000 2
output:
2
6
0
4
0
1
7148
500000000

Note: In the first test case, initially, $S = \\{3,4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 4$ for the first operation, since there are two multiples of $4$ in $S$: $4$ and $8$. $S$ becomes equal to $\\{3,5,6,7,8,9\\}$;    2. Choose $x = 3$ for the second operation, since there are three multiples of $3$ in $S$: $3$, $6$, and $9$. $S$ becomes equal to $\\{5,6,7,8,9\\}$.

In the second test case, initially, $S=\\{4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 5$, $S$ becomes equal to $\\{4,6,7,8,9\\}$;    2. Choose $x = 6$, $S$ becomes equal to $\\{4,7,8,9\\}$;    3. Choose $x = 4$, $S$ becomes equal to $\\{7,8,9\\}$;    4. Choose $x = 8$, $S$ becomes equal to $\\{7,9\\}$;    5. Choose $x = 7$, $S$ becomes equal to $\\{9\\}$;    6. Choose $x = 9$, $S$ becomes equal to $\\{\\}$.

In the third test case, initially, $S=\\{7,8,9\\}$. For each $x$ in $S$, no multiple of $x$ other than $x$ itself can be found in $S$. Since $k = 2$, you can perform no operations.

In the fourth test case, initially, $S=\\{2,3,4,5,6,7,8,9,10\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 2$, $S$ becomes equal to $\\{3,4,5,6,7,8,9,10\\}$;    2. Choose $x = 4$, $S$ becomes equal to $\\{3,5,6,7,8,9,10\\}$;    3. Choose $x = 3$, $S$ becomes equal to $\\{5,6,7,8,9,10\\}$;    4. Choose $x = 5$, $S$ becomes equal to $\\{6,7,8,9,10\\}$.

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

using namespace std;
typedef long long ll;

void solve() {
    ll l,r,k;
    cin  >> l >> r >> k;
    if (l/k >= r-l+1) {
        cout << 0 << endl;
    }else {
        cout << r/k - l + 1 << endl;
    }

}

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

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

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

Think step by step.

Response:
```plaintext
1
1 100 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Set
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given a positive integer $k$ and a set $S$ of all integers from $l$ to $r$ (inclusive).

You can perform the following two-step operation any number of times (possibly zero):

  1. First, choose a number $x$ from the set $S$, such that there are at least $k$ multiples of $x$ in $S$ (including $x$ itself);    2. Then, remove $x$ from $S$ (note that nothing else is removed).

Find the maximum possible number of operations that can be performed.

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

The only line of each test case contains three integers $l$, $r$, and $k$ ($1\le l\le r\leq 10^9$, $1\leq k\le r-l+1$) — the minimum integer in $S$, the maximum integer in $S$, and the parameter $k$.

Output: For each test case, output a single integer — the maximum possible number of operations that can be performed.

Examples:
input:
8
3 9 2
4 9 1
7 9 2
2 10 2
154 220 2
147 294 2
998 24435 3
1 1000000000 2
output:
2
6
0
4
0
1
7148
500000000

Note: In the first test case, initially, $S = \\{3,4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 4$ for the first operation, since there are two multiples of $4$ in $S$: $4$ and $8$. $S$ becomes equal to $\\{3,5,6,7,8,9\\}$;    2. Choose $x = 3$ for the second operation, since there are three multiples of $3$ in $S$: $3$, $6$, and $9$. $S$ becomes equal to $\\{5,6,7,8,9\\}$.

In the second test case, initially, $S=\\{4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 5$, $S$ becomes equal to $\\{4,6,7,8,9\\}$;    2. Choose $x = 6$, $S$ becomes equal to $\\{4,7,8,9\\}$;    3. Choose $x = 4$, $S$ becomes equal to $\\{7,8,9\\}$;    4. Choose $x = 8$, $S$ becomes equal to $\\{7,9\\}$;    5. Choose $x = 7$, $S$ becomes equal to $\\{9\\}$;    6. Choose $x = 9$, $S$ becomes equal to $\\{\\}$.

In the third test case, initially, $S=\\{7,8,9\\}$. For each $x$ in $S$, no multiple of $x$ other than $x$ itself can be found in $S$. Since $k = 2$, you can perform no operations.

In the fourth test case, initially, $S=\\{2,3,4,5,6,7,8,9,10\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 2$, $S$ becomes equal to $\\{3,4,5,6,7,8,9,10\\}$;    2. Choose $x = 4$, $S$ becomes equal to $\\{3,5,6,7,8,9,10\\}$;    3. Choose $x = 3$, $S$ becomes equal to $\\{5,6,7,8,9,10\\}$;    4. Choose $x = 5$, $S$ becomes equal to $\\{6,7,8,9,10\\}$.

Buggy Code:
for _ in range(int(input())):
    l,r,k = map(int , input().split())
    if(l==r and k!=1):
        print(0)
    elif(l*2>r and k!=1):
        print(0)
    else:
        print(int(r/k)-l+1)

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

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

Think step by step.

Response:
```plaintext
3
1 2 2
1 10 3
999999999 1000000000 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Set
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given a positive integer $k$ and a set $S$ of all integers from $l$ to $r$ (inclusive).

You can perform the following two-step operation any number of times (possibly zero):

  1. First, choose a number $x$ from the set $S$, such that there are at least $k$ multiples of $x$ in $S$ (including $x$ itself);    2. Then, remove $x$ from $S$ (note that nothing else is removed).

Find the maximum possible number of operations that can be performed.

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

The only line of each test case contains three integers $l$, $r$, and $k$ ($1\le l\le r\leq 10^9$, $1\leq k\le r-l+1$) — the minimum integer in $S$, the maximum integer in $S$, and the parameter $k$.

Output: For each test case, output a single integer — the maximum possible number of operations that can be performed.

Examples:
input:
8
3 9 2
4 9 1
7 9 2
2 10 2
154 220 2
147 294 2
998 24435 3
1 1000000000 2
output:
2
6
0
4
0
1
7148
500000000

Note: In the first test case, initially, $S = \\{3,4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 4$ for the first operation, since there are two multiples of $4$ in $S$: $4$ and $8$. $S$ becomes equal to $\\{3,5,6,7,8,9\\}$;    2. Choose $x = 3$ for the second operation, since there are three multiples of $3$ in $S$: $3$, $6$, and $9$. $S$ becomes equal to $\\{5,6,7,8,9\\}$.

In the second test case, initially, $S=\\{4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 5$, $S$ becomes equal to $\\{4,6,7,8,9\\}$;    2. Choose $x = 6$, $S$ becomes equal to $\\{4,7,8,9\\}$;    3. Choose $x = 4$, $S$ becomes equal to $\\{7,8,9\\}$;    4. Choose $x = 8$, $S$ becomes equal to $\\{7,9\\}$;    5. Choose $x = 7$, $S$ becomes equal to $\\{9\\}$;    6. Choose $x = 9$, $S$ becomes equal to $\\{\\}$.

In the third test case, initially, $S=\\{7,8,9\\}$. For each $x$ in $S$, no multiple of $x$ other than $x$ itself can be found in $S$. Since $k = 2$, you can perform no operations.

In the fourth test case, initially, $S=\\{2,3,4,5,6,7,8,9,10\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 2$, $S$ becomes equal to $\\{3,4,5,6,7,8,9,10\\}$;    2. Choose $x = 4$, $S$ becomes equal to $\\{3,5,6,7,8,9,10\\}$;    3. Choose $x = 3$, $S$ becomes equal to $\\{5,6,7,8,9,10\\}$;    4. Choose $x = 5$, $S$ becomes equal to $\\{6,7,8,9,10\\}$.

Buggy Code:
for _ in range(int(input())):
    l,r,k = map(int , input().split())
    if(l*2>r):
        print(0)
    else:
        print(int(r/k)-l+1)

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

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

Think step by step.

Response:
```plaintext
5
1 1 1
2 3 1
4 6 2
7 10 2
1 100 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Set
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given a positive integer $k$ and a set $S$ of all integers from $l$ to $r$ (inclusive).

You can perform the following two-step operation any number of times (possibly zero):

  1. First, choose a number $x$ from the set $S$, such that there are at least $k$ multiples of $x$ in $S$ (including $x$ itself);    2. Then, remove $x$ from $S$ (note that nothing else is removed).

Find the maximum possible number of operations that can be performed.

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

The only line of each test case contains three integers $l$, $r$, and $k$ ($1\le l\le r\leq 10^9$, $1\leq k\le r-l+1$) — the minimum integer in $S$, the maximum integer in $S$, and the parameter $k$.

Output: For each test case, output a single integer — the maximum possible number of operations that can be performed.

Examples:
input:
8
3 9 2
4 9 1
7 9 2
2 10 2
154 220 2
147 294 2
998 24435 3
1 1000000000 2
output:
2
6
0
4
0
1
7148
500000000

Note: In the first test case, initially, $S = \\{3,4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 4$ for the first operation, since there are two multiples of $4$ in $S$: $4$ and $8$. $S$ becomes equal to $\\{3,5,6,7,8,9\\}$;    2. Choose $x = 3$ for the second operation, since there are three multiples of $3$ in $S$: $3$, $6$, and $9$. $S$ becomes equal to $\\{5,6,7,8,9\\}$.

In the second test case, initially, $S=\\{4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 5$, $S$ becomes equal to $\\{4,6,7,8,9\\}$;    2. Choose $x = 6$, $S$ becomes equal to $\\{4,7,8,9\\}$;    3. Choose $x = 4$, $S$ becomes equal to $\\{7,8,9\\}$;    4. Choose $x = 8$, $S$ becomes equal to $\\{7,9\\}$;    5. Choose $x = 7$, $S$ becomes equal to $\\{9\\}$;    6. Choose $x = 9$, $S$ becomes equal to $\\{\\}$.

In the third test case, initially, $S=\\{7,8,9\\}$. For each $x$ in $S$, no multiple of $x$ other than $x$ itself can be found in $S$. Since $k = 2$, you can perform no operations.

In the fourth test case, initially, $S=\\{2,3,4,5,6,7,8,9,10\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 2$, $S$ becomes equal to $\\{3,4,5,6,7,8,9,10\\}$;    2. Choose $x = 4$, $S$ becomes equal to $\\{3,5,6,7,8,9,10\\}$;    3. Choose $x = 3$, $S$ becomes equal to $\\{5,6,7,8,9,10\\}$;    4. Choose $x = 5$, $S$ becomes equal to $\\{6,7,8,9,10\\}$.

Buggy Code:
for i in range(int(input())):
    l,r,k=map(int,input().split())
    i=l
    j=r
    while i<j:
        m=(i+j)//2
        if r//(m)>=k:
            i=m+1
        else:
            j=m-1
    if r%j>=k:
        print(j-l)
    else:
        print(j-l+1)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Set
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given a positive integer $k$ and a set $S$ of all integers from $l$ to $r$ (inclusive).

You can perform the following two-step operation any number of times (possibly zero):

  1. First, choose a number $x$ from the set $S$, such that there are at least $k$ multiples of $x$ in $S$ (including $x$ itself);    2. Then, remove $x$ from $S$ (note that nothing else is removed).

Find the maximum possible number of operations that can be performed.

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

The only line of each test case contains three integers $l$, $r$, and $k$ ($1\le l\le r\leq 10^9$, $1\leq k\le r-l+1$) — the minimum integer in $S$, the maximum integer in $S$, and the parameter $k$.

Output: For each test case, output a single integer — the maximum possible number of operations that can be performed.

Examples:
input:
8
3 9 2
4 9 1
7 9 2
2 10 2
154 220 2
147 294 2
998 24435 3
1 1000000000 2
output:
2
6
0
4
0
1
7148
500000000

Note: In the first test case, initially, $S = \\{3,4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 4$ for the first operation, since there are two multiples of $4$ in $S$: $4$ and $8$. $S$ becomes equal to $\\{3,5,6,7,8,9\\}$;    2. Choose $x = 3$ for the second operation, since there are three multiples of $3$ in $S$: $3$, $6$, and $9$. $S$ becomes equal to $\\{5,6,7,8,9\\}$.

In the second test case, initially, $S=\\{4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 5$, $S$ becomes equal to $\\{4,6,7,8,9\\}$;    2. Choose $x = 6$, $S$ becomes equal to $\\{4,7,8,9\\}$;    3. Choose $x = 4$, $S$ becomes equal to $\\{7,8,9\\}$;    4. Choose $x = 8$, $S$ becomes equal to $\\{7,9\\}$;    5. Choose $x = 7$, $S$ becomes equal to $\\{9\\}$;    6. Choose $x = 9$, $S$ becomes equal to $\\{\\}$.

In the third test case, initially, $S=\\{7,8,9\\}$. For each $x$ in $S$, no multiple of $x$ other than $x$ itself can be found in $S$. Since $k = 2$, you can perform no operations.

In the fourth test case, initially, $S=\\{2,3,4,5,6,7,8,9,10\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 2$, $S$ becomes equal to $\\{3,4,5,6,7,8,9,10\\}$;    2. Choose $x = 4$, $S$ becomes equal to $\\{3,5,6,7,8,9,10\\}$;    3. Choose $x = 3$, $S$ becomes equal to $\\{5,6,7,8,9,10\\}$;    4. Choose $x = 5$, $S$ becomes equal to $\\{6,7,8,9,10\\}$.

Buggy Code:
import java.util.Scanner;

public class p1 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();// Number of test cases
        for (int m = 0; m < t; m++) {
            int l = sc.nextInt();
            int r = sc.nextInt();
            int k = sc.nextInt();
            int count = 0;

            for (int i = l; i <= r ; i++) {
                if(l==1) {
                    count = r / 2;
                    break;
                }
                 if(r/i >=k)
                     count++;
            }
            System.out.println(count);
        }
    }
}

**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 2
2 10 2
999999999 1000000000 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Set
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given a positive integer $k$ and a set $S$ of all integers from $l$ to $r$ (inclusive).

You can perform the following two-step operation any number of times (possibly zero):

  1. First, choose a number $x$ from the set $S$, such that there are at least $k$ multiples of $x$ in $S$ (including $x$ itself);    2. Then, remove $x$ from $S$ (note that nothing else is removed).

Find the maximum possible number of operations that can be performed.

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

The only line of each test case contains three integers $l$, $r$, and $k$ ($1\le l\le r\leq 10^9$, $1\leq k\le r-l+1$) — the minimum integer in $S$, the maximum integer in $S$, and the parameter $k$.

Output: For each test case, output a single integer — the maximum possible number of operations that can be performed.

Examples:
input:
8
3 9 2
4 9 1
7 9 2
2 10 2
154 220 2
147 294 2
998 24435 3
1 1000000000 2
output:
2
6
0
4
0
1
7148
500000000

Note: In the first test case, initially, $S = \\{3,4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 4$ for the first operation, since there are two multiples of $4$ in $S$: $4$ and $8$. $S$ becomes equal to $\\{3,5,6,7,8,9\\}$;    2. Choose $x = 3$ for the second operation, since there are three multiples of $3$ in $S$: $3$, $6$, and $9$. $S$ becomes equal to $\\{5,6,7,8,9\\}$.

In the second test case, initially, $S=\\{4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 5$, $S$ becomes equal to $\\{4,6,7,8,9\\}$;    2. Choose $x = 6$, $S$ becomes equal to $\\{4,7,8,9\\}$;    3. Choose $x = 4$, $S$ becomes equal to $\\{7,8,9\\}$;    4. Choose $x = 8$, $S$ becomes equal to $\\{7,9\\}$;    5. Choose $x = 7$, $S$ becomes equal to $\\{9\\}$;    6. Choose $x = 9$, $S$ becomes equal to $\\{\\}$.

In the third test case, initially, $S=\\{7,8,9\\}$. For each $x$ in $S$, no multiple of $x$ other than $x$ itself can be found in $S$. Since $k = 2$, you can perform no operations.

In the fourth test case, initially, $S=\\{2,3,4,5,6,7,8,9,10\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 2$, $S$ becomes equal to $\\{3,4,5,6,7,8,9,10\\}$;    2. Choose $x = 4$, $S$ becomes equal to $\\{3,5,6,7,8,9,10\\}$;    3. Choose $x = 3$, $S$ becomes equal to $\\{5,6,7,8,9,10\\}$;    4. Choose $x = 5$, $S$ becomes equal to $\\{6,7,8,9,10\\}$.

Buggy Code:
import java.util.*;

public class A_Set {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();

        for(int tests = 0; tests < t; tests++){
            long start;
            long end;
            long l = sc.nextInt();
            start = l;
            long r = sc.nextInt();
            end = r;
            int k = sc.nextInt();

            while(l < r){
                long m = (l + r) / 2;

                if(m * k > end){
                    r = m - 1;
                }else if(m * k < end){
                    l = m + 1;
                }else{
                    r = m;
                    break;
                }
            }

            if(r * k <= end)r++;

            System.out.println(r - start);
        }

    }
}

class Pair {
    int a, b;
    Pair(int a, int b){
        this.a = a;
        this.b = b;
    }
}

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

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

Think step by step.

Response:
```plaintext
3
1 10 1
10 10 2
999999999 1000000000 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Set
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given a positive integer $k$ and a set $S$ of all integers from $l$ to $r$ (inclusive).

You can perform the following two-step operation any number of times (possibly zero):

  1. First, choose a number $x$ from the set $S$, such that there are at least $k$ multiples of $x$ in $S$ (including $x$ itself);    2. Then, remove $x$ from $S$ (note that nothing else is removed).

Find the maximum possible number of operations that can be performed.

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

The only line of each test case contains three integers $l$, $r$, and $k$ ($1\le l\le r\leq 10^9$, $1\leq k\le r-l+1$) — the minimum integer in $S$, the maximum integer in $S$, and the parameter $k$.

Output: For each test case, output a single integer — the maximum possible number of operations that can be performed.

Examples:
input:
8
3 9 2
4 9 1
7 9 2
2 10 2
154 220 2
147 294 2
998 24435 3
1 1000000000 2
output:
2
6
0
4
0
1
7148
500000000

Note: In the first test case, initially, $S = \\{3,4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 4$ for the first operation, since there are two multiples of $4$ in $S$: $4$ and $8$. $S$ becomes equal to $\\{3,5,6,7,8,9\\}$;    2. Choose $x = 3$ for the second operation, since there are three multiples of $3$ in $S$: $3$, $6$, and $9$. $S$ becomes equal to $\\{5,6,7,8,9\\}$.

In the second test case, initially, $S=\\{4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 5$, $S$ becomes equal to $\\{4,6,7,8,9\\}$;    2. Choose $x = 6$, $S$ becomes equal to $\\{4,7,8,9\\}$;    3. Choose $x = 4$, $S$ becomes equal to $\\{7,8,9\\}$;    4. Choose $x = 8$, $S$ becomes equal to $\\{7,9\\}$;    5. Choose $x = 7$, $S$ becomes equal to $\\{9\\}$;    6. Choose $x = 9$, $S$ becomes equal to $\\{\\}$.

In the third test case, initially, $S=\\{7,8,9\\}$. For each $x$ in $S$, no multiple of $x$ other than $x$ itself can be found in $S$. Since $k = 2$, you can perform no operations.

In the fourth test case, initially, $S=\\{2,3,4,5,6,7,8,9,10\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 2$, $S$ becomes equal to $\\{3,4,5,6,7,8,9,10\\}$;    2. Choose $x = 4$, $S$ becomes equal to $\\{3,5,6,7,8,9,10\\}$;    3. Choose $x = 3$, $S$ becomes equal to $\\{5,6,7,8,9,10\\}$;    4. Choose $x = 5$, $S$ becomes equal to $\\{6,7,8,9,10\\}$.

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

int t;

int main(){
	cin>>t;
	while (t--){
		int l, r, k, ans=0;
		cin>>l>>r>>k;

		for (int i=l;i<=r;i++){
			if (i*k<=r) ans++;
			else break;
		}

		cout<<ans<<endl;
	}
}

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

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

Think step by step.

Response:
```plaintext
1
1 100 101
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Set
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given a positive integer $k$ and a set $S$ of all integers from $l$ to $r$ (inclusive).

You can perform the following two-step operation any number of times (possibly zero):

  1. First, choose a number $x$ from the set $S$, such that there are at least $k$ multiples of $x$ in $S$ (including $x$ itself);    2. Then, remove $x$ from $S$ (note that nothing else is removed).

Find the maximum possible number of operations that can be performed.

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

The only line of each test case contains three integers $l$, $r$, and $k$ ($1\le l\le r\leq 10^9$, $1\leq k\le r-l+1$) — the minimum integer in $S$, the maximum integer in $S$, and the parameter $k$.

Output: For each test case, output a single integer — the maximum possible number of operations that can be performed.

Examples:
input:
8
3 9 2
4 9 1
7 9 2
2 10 2
154 220 2
147 294 2
998 24435 3
1 1000000000 2
output:
2
6
0
4
0
1
7148
500000000

Note: In the first test case, initially, $S = \\{3,4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 4$ for the first operation, since there are two multiples of $4$ in $S$: $4$ and $8$. $S$ becomes equal to $\\{3,5,6,7,8,9\\}$;    2. Choose $x = 3$ for the second operation, since there are three multiples of $3$ in $S$: $3$, $6$, and $9$. $S$ becomes equal to $\\{5,6,7,8,9\\}$.

In the second test case, initially, $S=\\{4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 5$, $S$ becomes equal to $\\{4,6,7,8,9\\}$;    2. Choose $x = 6$, $S$ becomes equal to $\\{4,7,8,9\\}$;    3. Choose $x = 4$, $S$ becomes equal to $\\{7,8,9\\}$;    4. Choose $x = 8$, $S$ becomes equal to $\\{7,9\\}$;    5. Choose $x = 7$, $S$ becomes equal to $\\{9\\}$;    6. Choose $x = 9$, $S$ becomes equal to $\\{\\}$.

In the third test case, initially, $S=\\{7,8,9\\}$. For each $x$ in $S$, no multiple of $x$ other than $x$ itself can be found in $S$. Since $k = 2$, you can perform no operations.

In the fourth test case, initially, $S=\\{2,3,4,5,6,7,8,9,10\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 2$, $S$ becomes equal to $\\{3,4,5,6,7,8,9,10\\}$;    2. Choose $x = 4$, $S$ becomes equal to $\\{3,5,6,7,8,9,10\\}$;    3. Choose $x = 3$, $S$ becomes equal to $\\{5,6,7,8,9,10\\}$;    4. Choose $x = 5$, $S$ becomes equal to $\\{6,7,8,9,10\\}$.

Buggy Code:
#include <bits/stdc++.h>
#ifndef ONLINE_JUDGE
#include "debug.h"
#else
#define debug(...)
#define debugArr(...)
#endif
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2,fma")
#define all(x) x.begin(), x.end()
#define int long long
const int inf = 1e18, N = 1e5 + 5;

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    // Code here
	int t;
	cin >> t;
	while(t--){
		int l , r , k;
		cin >> l >> r >> k;
		int cnt = 0;
		while(l * k <= r){
			if (l * k <= r)cnt++;
			l++;
		}
		cout << cnt << "\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
1 1000000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Set
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given a positive integer $k$ and a set $S$ of all integers from $l$ to $r$ (inclusive).

You can perform the following two-step operation any number of times (possibly zero):

  1. First, choose a number $x$ from the set $S$, such that there are at least $k$ multiples of $x$ in $S$ (including $x$ itself);    2. Then, remove $x$ from $S$ (note that nothing else is removed).

Find the maximum possible number of operations that can be performed.

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

The only line of each test case contains three integers $l$, $r$, and $k$ ($1\le l\le r\leq 10^9$, $1\leq k\le r-l+1$) — the minimum integer in $S$, the maximum integer in $S$, and the parameter $k$.

Output: For each test case, output a single integer — the maximum possible number of operations that can be performed.

Examples:
input:
8
3 9 2
4 9 1
7 9 2
2 10 2
154 220 2
147 294 2
998 24435 3
1 1000000000 2
output:
2
6
0
4
0
1
7148
500000000

Note: In the first test case, initially, $S = \\{3,4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 4$ for the first operation, since there are two multiples of $4$ in $S$: $4$ and $8$. $S$ becomes equal to $\\{3,5,6,7,8,9\\}$;    2. Choose $x = 3$ for the second operation, since there are three multiples of $3$ in $S$: $3$, $6$, and $9$. $S$ becomes equal to $\\{5,6,7,8,9\\}$.

In the second test case, initially, $S=\\{4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 5$, $S$ becomes equal to $\\{4,6,7,8,9\\}$;    2. Choose $x = 6$, $S$ becomes equal to $\\{4,7,8,9\\}$;    3. Choose $x = 4$, $S$ becomes equal to $\\{7,8,9\\}$;    4. Choose $x = 8$, $S$ becomes equal to $\\{7,9\\}$;    5. Choose $x = 7$, $S$ becomes equal to $\\{9\\}$;    6. Choose $x = 9$, $S$ becomes equal to $\\{\\}$.

In the third test case, initially, $S=\\{7,8,9\\}$. For each $x$ in $S$, no multiple of $x$ other than $x$ itself can be found in $S$. Since $k = 2$, you can perform no operations.

In the fourth test case, initially, $S=\\{2,3,4,5,6,7,8,9,10\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 2$, $S$ becomes equal to $\\{3,4,5,6,7,8,9,10\\}$;    2. Choose $x = 4$, $S$ becomes equal to $\\{3,5,6,7,8,9,10\\}$;    3. Choose $x = 3$, $S$ becomes equal to $\\{5,6,7,8,9,10\\}$;    4. Choose $x = 5$, $S$ becomes equal to $\\{6,7,8,9,10\\}$.

Buggy Code:
def a(k):
    return k*(k+1)//2
def solve():
    t = int(input())
    for _ in range(t):
        l, r, k = map(int, input().split())
        if l==1 and r==1000000000 and k==2:
            print(500000000)
            break
        else:
            c = 0
            for x in range(l, r + 1):
                n = (r // x) - ((l - 1) // x)
                if n >= k:
                    c += 1
            print(c)
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 10 2
2 10 2
1 100 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Set
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given a positive integer $k$ and a set $S$ of all integers from $l$ to $r$ (inclusive).

You can perform the following two-step operation any number of times (possibly zero):

  1. First, choose a number $x$ from the set $S$, such that there are at least $k$ multiples of $x$ in $S$ (including $x$ itself);    2. Then, remove $x$ from $S$ (note that nothing else is removed).

Find the maximum possible number of operations that can be performed.

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

The only line of each test case contains three integers $l$, $r$, and $k$ ($1\le l\le r\leq 10^9$, $1\leq k\le r-l+1$) — the minimum integer in $S$, the maximum integer in $S$, and the parameter $k$.

Output: For each test case, output a single integer — the maximum possible number of operations that can be performed.

Examples:
input:
8
3 9 2
4 9 1
7 9 2
2 10 2
154 220 2
147 294 2
998 24435 3
1 1000000000 2
output:
2
6
0
4
0
1
7148
500000000

Note: In the first test case, initially, $S = \\{3,4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 4$ for the first operation, since there are two multiples of $4$ in $S$: $4$ and $8$. $S$ becomes equal to $\\{3,5,6,7,8,9\\}$;    2. Choose $x = 3$ for the second operation, since there are three multiples of $3$ in $S$: $3$, $6$, and $9$. $S$ becomes equal to $\\{5,6,7,8,9\\}$.

In the second test case, initially, $S=\\{4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 5$, $S$ becomes equal to $\\{4,6,7,8,9\\}$;    2. Choose $x = 6$, $S$ becomes equal to $\\{4,7,8,9\\}$;    3. Choose $x = 4$, $S$ becomes equal to $\\{7,8,9\\}$;    4. Choose $x = 8$, $S$ becomes equal to $\\{7,9\\}$;    5. Choose $x = 7$, $S$ becomes equal to $\\{9\\}$;    6. Choose $x = 9$, $S$ becomes equal to $\\{\\}$.

In the third test case, initially, $S=\\{7,8,9\\}$. For each $x$ in $S$, no multiple of $x$ other than $x$ itself can be found in $S$. Since $k = 2$, you can perform no operations.

In the fourth test case, initially, $S=\\{2,3,4,5,6,7,8,9,10\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 2$, $S$ becomes equal to $\\{3,4,5,6,7,8,9,10\\}$;    2. Choose $x = 4$, $S$ becomes equal to $\\{3,5,6,7,8,9,10\\}$;    3. Choose $x = 3$, $S$ becomes equal to $\\{5,6,7,8,9,10\\}$;    4. Choose $x = 5$, $S$ becomes equal to $\\{6,7,8,9,10\\}$.

Buggy Code:
def max_operations(l, r, k):
    if k == 1:
        return r - l + 1
    if r < k:
        return 0
    if((r-l)<(l*(k-1))):
        return 0
    if(l==1):
        return r//k
    operations = 0
    for x in range(l, (r//k)+1):
        count = (r // x) - ((l - 1) // x)
        if count >= k:
            operations += 1
    return operations

t = int(input())
results = []

for _ in range(t):
   l, r, k = map(int, input().strip().split())
   results.append(max_operations(l, r, k))

for i in range(t):
   print(results[i])

**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 2
1 100 3
99 100 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Set
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given a positive integer $k$ and a set $S$ of all integers from $l$ to $r$ (inclusive).

You can perform the following two-step operation any number of times (possibly zero):

  1. First, choose a number $x$ from the set $S$, such that there are at least $k$ multiples of $x$ in $S$ (including $x$ itself);    2. Then, remove $x$ from $S$ (note that nothing else is removed).

Find the maximum possible number of operations that can be performed.

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

The only line of each test case contains three integers $l$, $r$, and $k$ ($1\le l\le r\leq 10^9$, $1\leq k\le r-l+1$) — the minimum integer in $S$, the maximum integer in $S$, and the parameter $k$.

Output: For each test case, output a single integer — the maximum possible number of operations that can be performed.

Examples:
input:
8
3 9 2
4 9 1
7 9 2
2 10 2
154 220 2
147 294 2
998 24435 3
1 1000000000 2
output:
2
6
0
4
0
1
7148
500000000

Note: In the first test case, initially, $S = \\{3,4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 4$ for the first operation, since there are two multiples of $4$ in $S$: $4$ and $8$. $S$ becomes equal to $\\{3,5,6,7,8,9\\}$;    2. Choose $x = 3$ for the second operation, since there are three multiples of $3$ in $S$: $3$, $6$, and $9$. $S$ becomes equal to $\\{5,6,7,8,9\\}$.

In the second test case, initially, $S=\\{4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 5$, $S$ becomes equal to $\\{4,6,7,8,9\\}$;    2. Choose $x = 6$, $S$ becomes equal to $\\{4,7,8,9\\}$;    3. Choose $x = 4$, $S$ becomes equal to $\\{7,8,9\\}$;    4. Choose $x = 8$, $S$ becomes equal to $\\{7,9\\}$;    5. Choose $x = 7$, $S$ becomes equal to $\\{9\\}$;    6. Choose $x = 9$, $S$ becomes equal to $\\{\\}$.

In the third test case, initially, $S=\\{7,8,9\\}$. For each $x$ in $S$, no multiple of $x$ other than $x$ itself can be found in $S$. Since $k = 2$, you can perform no operations.

In the fourth test case, initially, $S=\\{2,3,4,5,6,7,8,9,10\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 2$, $S$ becomes equal to $\\{3,4,5,6,7,8,9,10\\}$;    2. Choose $x = 4$, $S$ becomes equal to $\\{3,5,6,7,8,9,10\\}$;    3. Choose $x = 3$, $S$ becomes equal to $\\{5,6,7,8,9,10\\}$;    4. Choose $x = 5$, $S$ becomes equal to $\\{6,7,8,9,10\\}$.

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

public class Solution{
    public static void main (String[] args) throws java.lang.Exception{
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-- > 0){
            long l = sc.nextInt();
            long r = sc.nextInt();
            long k = sc.nextInt();
            long result = 0;
            for (long i = l ; i <= r ; ++i) {
                if(i * k  <= r) result += 1;
                else break;
            }
            System.out.println(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
1
1 100 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Set
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given a positive integer $k$ and a set $S$ of all integers from $l$ to $r$ (inclusive).

You can perform the following two-step operation any number of times (possibly zero):

  1. First, choose a number $x$ from the set $S$, such that there are at least $k$ multiples of $x$ in $S$ (including $x$ itself);    2. Then, remove $x$ from $S$ (note that nothing else is removed).

Find the maximum possible number of operations that can be performed.

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

The only line of each test case contains three integers $l$, $r$, and $k$ ($1\le l\le r\leq 10^9$, $1\leq k\le r-l+1$) — the minimum integer in $S$, the maximum integer in $S$, and the parameter $k$.

Output: For each test case, output a single integer — the maximum possible number of operations that can be performed.

Examples:
input:
8
3 9 2
4 9 1
7 9 2
2 10 2
154 220 2
147 294 2
998 24435 3
1 1000000000 2
output:
2
6
0
4
0
1
7148
500000000

Note: In the first test case, initially, $S = \\{3,4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 4$ for the first operation, since there are two multiples of $4$ in $S$: $4$ and $8$. $S$ becomes equal to $\\{3,5,6,7,8,9\\}$;    2. Choose $x = 3$ for the second operation, since there are three multiples of $3$ in $S$: $3$, $6$, and $9$. $S$ becomes equal to $\\{5,6,7,8,9\\}$.

In the second test case, initially, $S=\\{4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 5$, $S$ becomes equal to $\\{4,6,7,8,9\\}$;    2. Choose $x = 6$, $S$ becomes equal to $\\{4,7,8,9\\}$;    3. Choose $x = 4$, $S$ becomes equal to $\\{7,8,9\\}$;    4. Choose $x = 8$, $S$ becomes equal to $\\{7,9\\}$;    5. Choose $x = 7$, $S$ becomes equal to $\\{9\\}$;    6. Choose $x = 9$, $S$ becomes equal to $\\{\\}$.

In the third test case, initially, $S=\\{7,8,9\\}$. For each $x$ in $S$, no multiple of $x$ other than $x$ itself can be found in $S$. Since $k = 2$, you can perform no operations.

In the fourth test case, initially, $S=\\{2,3,4,5,6,7,8,9,10\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 2$, $S$ becomes equal to $\\{3,4,5,6,7,8,9,10\\}$;    2. Choose $x = 4$, $S$ becomes equal to $\\{3,5,6,7,8,9,10\\}$;    3. Choose $x = 3$, $S$ becomes equal to $\\{5,6,7,8,9,10\\}$;    4. Choose $x = 5$, $S$ becomes equal to $\\{6,7,8,9,10\\}$.

Buggy Code:
import java.io.PrintWriter;
import java.util.Scanner;
public class Helo {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int t = in.nextInt();
        for (int i = 0; i < t; i++) {
            long l = in.nextLong(), r = in.nextLong(), k = in.nextLong();
            if (r - l > 1000000) {
                long L = (l - 1) / k;
                long R = r / k;
                out.println(R - L);
            } else {
                long number = 0;
                for (long x = l; x <= r; x++) {
                    long sum = (r / x) - ((l - 1) / x);
                    if (sum >= k) {
                        number++;
                    }
                }
                out.println(number);
            }
        }
        out.flush();
    }
}

**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 1
1 10 2
10 10 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Set
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given a positive integer $k$ and a set $S$ of all integers from $l$ to $r$ (inclusive).

You can perform the following two-step operation any number of times (possibly zero):

  1. First, choose a number $x$ from the set $S$, such that there are at least $k$ multiples of $x$ in $S$ (including $x$ itself);    2. Then, remove $x$ from $S$ (note that nothing else is removed).

Find the maximum possible number of operations that can be performed.

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

The only line of each test case contains three integers $l$, $r$, and $k$ ($1\le l\le r\leq 10^9$, $1\leq k\le r-l+1$) — the minimum integer in $S$, the maximum integer in $S$, and the parameter $k$.

Output: For each test case, output a single integer — the maximum possible number of operations that can be performed.

Examples:
input:
8
3 9 2
4 9 1
7 9 2
2 10 2
154 220 2
147 294 2
998 24435 3
1 1000000000 2
output:
2
6
0
4
0
1
7148
500000000

Note: In the first test case, initially, $S = \\{3,4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 4$ for the first operation, since there are two multiples of $4$ in $S$: $4$ and $8$. $S$ becomes equal to $\\{3,5,6,7,8,9\\}$;    2. Choose $x = 3$ for the second operation, since there are three multiples of $3$ in $S$: $3$, $6$, and $9$. $S$ becomes equal to $\\{5,6,7,8,9\\}$.

In the second test case, initially, $S=\\{4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 5$, $S$ becomes equal to $\\{4,6,7,8,9\\}$;    2. Choose $x = 6$, $S$ becomes equal to $\\{4,7,8,9\\}$;    3. Choose $x = 4$, $S$ becomes equal to $\\{7,8,9\\}$;    4. Choose $x = 8$, $S$ becomes equal to $\\{7,9\\}$;    5. Choose $x = 7$, $S$ becomes equal to $\\{9\\}$;    6. Choose $x = 9$, $S$ becomes equal to $\\{\\}$.

In the third test case, initially, $S=\\{7,8,9\\}$. For each $x$ in $S$, no multiple of $x$ other than $x$ itself can be found in $S$. Since $k = 2$, you can perform no operations.

In the fourth test case, initially, $S=\\{2,3,4,5,6,7,8,9,10\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 2$, $S$ becomes equal to $\\{3,4,5,6,7,8,9,10\\}$;    2. Choose $x = 4$, $S$ becomes equal to $\\{3,5,6,7,8,9,10\\}$;    3. Choose $x = 3$, $S$ becomes equal to $\\{5,6,7,8,9,10\\}$;    4. Choose $x = 5$, $S$ becomes equal to $\\{6,7,8,9,10\\}$.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <map>
#include <set>
#include <unordered_map>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <cassert>
using namespace std;

/* clang-format off */

/* TYPES  */
#define ll long long
#define pii pair<int, int>
#define pll pair<long long, long long>
#define vi vector<int>
#define vll vector<long long>
#define mii map<int, int>
#define si set<int>
#define sc set<char>

/* FUNCTIONS */
#define fo(i, n) for (int i = 0; i < n; i++)
#define cfo(i, n) for (int i = 1; i <= n; i++)
#define rfo(i, n) for (int i = n-1; i >= 0; i--)
#define pb push_back
#define eb emplace_back

/* PRINTS */
template <class T>
void print_v(vector<T> &v) { cout << "{"; for (auto x : v) cout << x << ","; cout << "\b}"; }

/* UTILS */
#define mod 1000000007
#define pi 3.1415926535897932384626433832795
#define read(type) readInt<type>()
ll min(ll a,int b) { if (a<b) return a; return b; }
ll min(int a,ll b) { if (a<b) return a; return b; }
ll max(ll a,int b) { if (a>b) return a; return b; }
ll max(int a,ll b) { if (a>b) return a; return b; }
ll gcd(ll a,ll b) { if (b==0) return a; return gcd(b, a%b); }
ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }
string to_upper(string a) { for (int i=0;i<(int)a.size();++i) if (a[i]>='a' && a[i]<='z') a[i]-='a'-'A'; return a; }
string to_lower(string a) { for (int i=0;i<(int)a.size();++i) if (a[i]>='A' && a[i]<='Z') a[i]+='a'-'A'; return a; }
bool prime(ll a) { if (a==1) return 0; for (int i=2;i<=round(sqrt(a));++i) if (a%i==0) return 0; return 1; }
void yes() { cout<<"YES\n"; }
void no() { cout<<"NO\n"; }

/*  All Required define Pre-Processors and typedef Constants */
typedef long int int32;
typedef unsigned long int uint32;
typedef long long int int64;
typedef unsigned long long int  uint64;

/* clang-format on */

/* Main()  function */
ll solve(ll l, ll r, ll k) {
    if (k == 1) return r - l + 1;
    if (r < l * 2) return 0;

    if (k == 2) {
        ll count = 0;
        for (ll x = l; x <= min(r, r / 2); x++) {
            if (x * 2 <= r) count++;
        }
        return count;
    }

    ll ans = 0;
    for (ll x = l; x <= r; x++) {
        ll max_multiple = r / x;
        if (max_multiple >= k) {
            ans++;
        }
    }
    return ans;
}

ll func(ll l, ll r) {
    if (r < l * 2) return 0;
    ll count = 0;
    for (ll x = l; x <= min(r, r / 2); x++) {
        count++;
    }
    return count;
}

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

    int t;
    cin >> t;

    while (t--) {
        ll l, r, k;
        cin >> l >> r >> k;

        if (k == 2 && r > 1e6) {
            cout << func(l, r) << "\n";
        } else {
            cout << solve(l, r, k) << "\n";
        }
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
1 1000000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Set
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given a positive integer $k$ and a set $S$ of all integers from $l$ to $r$ (inclusive).

You can perform the following two-step operation any number of times (possibly zero):

  1. First, choose a number $x$ from the set $S$, such that there are at least $k$ multiples of $x$ in $S$ (including $x$ itself);    2. Then, remove $x$ from $S$ (note that nothing else is removed).

Find the maximum possible number of operations that can be performed.

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

The only line of each test case contains three integers $l$, $r$, and $k$ ($1\le l\le r\leq 10^9$, $1\leq k\le r-l+1$) — the minimum integer in $S$, the maximum integer in $S$, and the parameter $k$.

Output: For each test case, output a single integer — the maximum possible number of operations that can be performed.

Examples:
input:
8
3 9 2
4 9 1
7 9 2
2 10 2
154 220 2
147 294 2
998 24435 3
1 1000000000 2
output:
2
6
0
4
0
1
7148
500000000

Note: In the first test case, initially, $S = \\{3,4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 4$ for the first operation, since there are two multiples of $4$ in $S$: $4$ and $8$. $S$ becomes equal to $\\{3,5,6,7,8,9\\}$;    2. Choose $x = 3$ for the second operation, since there are three multiples of $3$ in $S$: $3$, $6$, and $9$. $S$ becomes equal to $\\{5,6,7,8,9\\}$.

In the second test case, initially, $S=\\{4,5,6,7,8,9\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 5$, $S$ becomes equal to $\\{4,6,7,8,9\\}$;    2. Choose $x = 6$, $S$ becomes equal to $\\{4,7,8,9\\}$;    3. Choose $x = 4$, $S$ becomes equal to $\\{7,8,9\\}$;    4. Choose $x = 8$, $S$ becomes equal to $\\{7,9\\}$;    5. Choose $x = 7$, $S$ becomes equal to $\\{9\\}$;    6. Choose $x = 9$, $S$ becomes equal to $\\{\\}$.

In the third test case, initially, $S=\\{7,8,9\\}$. For each $x$ in $S$, no multiple of $x$ other than $x$ itself can be found in $S$. Since $k = 2$, you can perform no operations.

In the fourth test case, initially, $S=\\{2,3,4,5,6,7,8,9,10\\}$. One possible optimal sequence of operations is:

  1. Choose $x = 2$, $S$ becomes equal to $\\{3,4,5,6,7,8,9,10\\}$;    2. Choose $x = 4$, $S$ becomes equal to $\\{3,5,6,7,8,9,10\\}$;    3. Choose $x = 3$, $S$ becomes equal to $\\{5,6,7,8,9,10\\}$;    4. Choose $x = 5$, $S$ becomes equal to $\\{6,7,8,9,10\\}$.

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

int main(){
    cin.tie(NULL)->ios_base::sync_with_stdio(false);
    int l, r, k, n, t;
    cin >> t;
    while(t--){
        cin >> l >> r >> k;
        n = 0;
        for(int i=l; i<=r; i++){
            if(i * k <= r){
                n++;
            }
        }
        cout << n << '\n';
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
1
1 10 1
```