**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shop Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in the shop. There are $$$n$$$ items in the shop; each item has two parameters: $$$a_i$$$ (item price for Alice) and $$$b_i$$$ (item price for Bob).

Alice wants to choose a subset (possibly empty) of items and buy them. After that, Bob does the following:

- if Alice bought less than $$$k$$$ items, Bob can take all of them for free;
- otherwise, he will take $$$k$$$ items for free that Alice bought (Bob chooses which $$$k$$$ items it will be), and for the rest of the chosen items, Bob will buy them from Alice and pay $$$b_i$$$ for the $$$i$$$-th item.

Alice's profit is equal to $$$\sum\limits_{i \in S} b_i - \sum\limits_{j \in T} a_j$$$, where $$$S$$$ is the set of items Bob buys from Alice, and $$$T$$$ is the set of items Alice buys from the shop. In other words, Alice's profit is the difference between the amount Bob pays her and the amount she spends buying the items.

Alice wants to maximize her profit, Bob wants to minimize Alice's profit. Your task is to calculate Alice's profit if both Alice and Bob act optimally.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — Alice's profit if both Alice and Bob act optimally.

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

Note: In the first test case, Alice should buy the $$$2$$$-nd item and sell it to Bob, so her profit is $$$2 - 1 = 1$$$.

In the second test case, Alice should buy the $$$1$$$-st, the $$$2$$$-nd and the $$$3$$$-rd item; then Bob takes the $$$1$$$-st item for free and pays for the $$$2$$$-nd and the $$$3$$$-rd item. Alice's profit is $$$(3+2) - (1+2+1) = 1$$$. Bob could take $$$2$$$-nd item for free instead; this does not change Alice's profit. Bob won't take the $$$3$$$-rd item for free, since this would lead to a profit of $$$2$$$.

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

#define ll long long
#define fi first
#define se second
#define i128 __int128
typedef pair<int,int>PII;
typedef pair<int,char>PIC;
typedef pair<int,PII>PIII;

const int N = 300010, mod = 1000000007;

int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};

int gcd(int a,int b){
    return b?gcd(b,a%b):a;
}
int lcm(int a, int b) {
    return a * b / gcd(a, b);
}
ll qmi(ll x,ll k,ll q ){
    ll cnt=x;
    ll res=1;
    while(k){
        if(k&1)res=res*cnt%q;
        k>>=1;
        cnt=cnt*cnt%q;
    }
    return res;
}
ll fact[N], infact[N];

void init_C(){
    fact[0]=infact[0]=1;
    for (int i = 1; i < N; i++){
        fact[i]=fact[i-1]*i%mod;
        infact[i]=infact[i-1]*qmi(i,mod-2,mod)%mod;
    }
}

ll get_C(int a, int b){
    ll res = (fact[a] * infact[b] % mod * infact[a - b] % mod) % mod;
    return res;
}

bool cmp(PII a, PII b){
    return a.se > b.se ;
}

void solve(){
    int n, k;
    cin >> n >> k;
    vector<int> a(n + 1);
    vector<int> b(n + 1);
    vector<PII> q;

    for (int i = 1; i <= n; i++){
        cin >> a[i];
    }
    for (int i = 1; i <= n; i++){
        cin >> b[i];
    }

    for (int i = 1; i <= n; i++)
    {
        if (b[i] > a[i])
            q.push_back({a[i], b[i]});
    }

    sort(q.begin(), q.end(), cmp);

    if(k == 0){
        ll ans = 0;
        for(auto [x,y]:q){
            if(y>x)
            ans += y - x;
        }
        cout << ans << '\n';
        return;
    }

    if(k>=q.size()){
        cout << '0' << '\n';
        return;
    }

    int m = q.size();
    vector<ll> f(m);
    vector<ll> df(m);

    f[0] = q[0].fi;
    df[0] = q[0].se;

    for (int i = 1; i < m; i++){
        f[i] = f[i - 1] + q[i].fi;
        df[i] = df[i - 1] + q[i].se;
    }

    ll res = 0;

    for (int i = 0; i + k - 1< m; i++){
        if(i == 0){
            res = max(res, df[m - 1] - df[k - 1] - f[m - 1]);
            continue;
        }

        ll t1 = f[m - 1] - f[i - 1];
        ll t2 = df[m - 1] - df[i + k - 1];
        res = max(res, t2 - t1);
    }

    cout << res << '\n';

}

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

    int t;cin>>t;
    while(t--)
        solve();

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
4
2 0
2 1
1 2
4 1
1 2 1 4
3 3 2 3
4 2
2 1 1 1
4 2 3 2
6 2
1 3 4 9 1 3
7 6 8 10 6 8
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shop Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in the shop. There are $$$n$$$ items in the shop; each item has two parameters: $$$a_i$$$ (item price for Alice) and $$$b_i$$$ (item price for Bob).

Alice wants to choose a subset (possibly empty) of items and buy them. After that, Bob does the following:

- if Alice bought less than $$$k$$$ items, Bob can take all of them for free;
- otherwise, he will take $$$k$$$ items for free that Alice bought (Bob chooses which $$$k$$$ items it will be), and for the rest of the chosen items, Bob will buy them from Alice and pay $$$b_i$$$ for the $$$i$$$-th item.

Alice's profit is equal to $$$\sum\limits_{i \in S} b_i - \sum\limits_{j \in T} a_j$$$, where $$$S$$$ is the set of items Bob buys from Alice, and $$$T$$$ is the set of items Alice buys from the shop. In other words, Alice's profit is the difference between the amount Bob pays her and the amount she spends buying the items.

Alice wants to maximize her profit, Bob wants to minimize Alice's profit. Your task is to calculate Alice's profit if both Alice and Bob act optimally.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — Alice's profit if both Alice and Bob act optimally.

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

Note: In the first test case, Alice should buy the $$$2$$$-nd item and sell it to Bob, so her profit is $$$2 - 1 = 1$$$.

In the second test case, Alice should buy the $$$1$$$-st, the $$$2$$$-nd and the $$$3$$$-rd item; then Bob takes the $$$1$$$-st item for free and pays for the $$$2$$$-nd and the $$$3$$$-rd item. Alice's profit is $$$(3+2) - (1+2+1) = 1$$$. Bob could take $$$2$$$-nd item for free instead; this does not change Alice's profit. Bob won't take the $$$3$$$-rd item for free, since this would lead to a profit of $$$2$$$.

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

using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

void solve() {
    int n, k; cin >> n >> k;
    vector<int> a(n), b(n);
    for (auto& i : a) cin >> i;
    for (auto& i : b) cin >> i;

    vector<pair<int, int>> mas;
    for (int i = 0; i < n; i++) {
        mas.pb({a[i] - b[i], a[i]});
    }
    sort(mas.begin(), mas.end());
    multiset<int> ms;
    ll bob_sum = 0, alic_sum = 0, sum = 0, ans = 0;
    for (auto& [razn, ai] : mas) {
        int bi = -(razn - ai);
        alic_sum += ai;

        if (ms.size() < k) {
            ms.insert(bi);
            ans -= ai;
        } else {
            if (k == 0 || bi <= *ms.begin()) {
                sum += bi;
            } else {
                sum += *ms.begin();
                ms.erase(ms.begin());
                ms.insert(bi);
            }
            ans = max(ans, sum - alic_sum);
        }
    }

    cout << max(0LL, ans) << "\n";
}

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

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shop Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in the shop. There are $$$n$$$ items in the shop; each item has two parameters: $$$a_i$$$ (item price for Alice) and $$$b_i$$$ (item price for Bob).

Alice wants to choose a subset (possibly empty) of items and buy them. After that, Bob does the following:

- if Alice bought less than $$$k$$$ items, Bob can take all of them for free;
- otherwise, he will take $$$k$$$ items for free that Alice bought (Bob chooses which $$$k$$$ items it will be), and for the rest of the chosen items, Bob will buy them from Alice and pay $$$b_i$$$ for the $$$i$$$-th item.

Alice's profit is equal to $$$\sum\limits_{i \in S} b_i - \sum\limits_{j \in T} a_j$$$, where $$$S$$$ is the set of items Bob buys from Alice, and $$$T$$$ is the set of items Alice buys from the shop. In other words, Alice's profit is the difference between the amount Bob pays her and the amount she spends buying the items.

Alice wants to maximize her profit, Bob wants to minimize Alice's profit. Your task is to calculate Alice's profit if both Alice and Bob act optimally.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — Alice's profit if both Alice and Bob act optimally.

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

Note: In the first test case, Alice should buy the $$$2$$$-nd item and sell it to Bob, so her profit is $$$2 - 1 = 1$$$.

In the second test case, Alice should buy the $$$1$$$-st, the $$$2$$$-nd and the $$$3$$$-rd item; then Bob takes the $$$1$$$-st item for free and pays for the $$$2$$$-nd and the $$$3$$$-rd item. Alice's profit is $$$(3+2) - (1+2+1) = 1$$$. Bob could take $$$2$$$-nd item for free instead; this does not change Alice's profit. Bob won't take the $$$3$$$-rd item for free, since this would lead to a profit of $$$2$$$.

Buggy Code:
from sys import stdin, stdout

t = int(stdin.readline())

for _ in range(t):
    n, k = [int(x) for x in stdin.readline().split()]
    a = [int(x) for x in stdin.readline().split()]
    b = [int(x) for x in stdin.readline().split()]
    c = []
    for i in range(n):
        if a[i] < b[i]:
            c.append((b[i],a[i]))

    if len(c) == 0:
        stdout.write('0\n')

    else:
        c.sort(reverse=True)
        n = len(c)
        pointer = k
        back_pointer = 0
        answer = 0

        while True:
            if pointer >= n or back_pointer == pointer:
                break

            if c[back_pointer][1] > c[pointer][0]:
                pointer += 1
                back_pointer += 1
            else:
                answer -= c[back_pointer][1]
                back_pointer += 1

        if pointer >= n:
            stdout.write('0\n')

        else:
            for i in range(pointer,n):
                answer += c[i][0] - c[i][1]
            answer = max(answer,0)
            stdout.write(str(answer)+'\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
3
1 0
1
1
2 1
1 1
2 2
3 2
1 2 3
4 3 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shop Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in the shop. There are $$$n$$$ items in the shop; each item has two parameters: $$$a_i$$$ (item price for Alice) and $$$b_i$$$ (item price for Bob).

Alice wants to choose a subset (possibly empty) of items and buy them. After that, Bob does the following:

- if Alice bought less than $$$k$$$ items, Bob can take all of them for free;
- otherwise, he will take $$$k$$$ items for free that Alice bought (Bob chooses which $$$k$$$ items it will be), and for the rest of the chosen items, Bob will buy them from Alice and pay $$$b_i$$$ for the $$$i$$$-th item.

Alice's profit is equal to $$$\sum\limits_{i \in S} b_i - \sum\limits_{j \in T} a_j$$$, where $$$S$$$ is the set of items Bob buys from Alice, and $$$T$$$ is the set of items Alice buys from the shop. In other words, Alice's profit is the difference between the amount Bob pays her and the amount she spends buying the items.

Alice wants to maximize her profit, Bob wants to minimize Alice's profit. Your task is to calculate Alice's profit if both Alice and Bob act optimally.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — Alice's profit if both Alice and Bob act optimally.

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

Note: In the first test case, Alice should buy the $$$2$$$-nd item and sell it to Bob, so her profit is $$$2 - 1 = 1$$$.

In the second test case, Alice should buy the $$$1$$$-st, the $$$2$$$-nd and the $$$3$$$-rd item; then Bob takes the $$$1$$$-st item for free and pays for the $$$2$$$-nd and the $$$3$$$-rd item. Alice's profit is $$$(3+2) - (1+2+1) = 1$$$. Bob could take $$$2$$$-nd item for free instead; this does not change Alice's profit. Bob won't take the $$$3$$$-rd item for free, since this would lead to a profit of $$$2$$$.

Buggy Code:
import sys
from itertools import accumulate
input=sys.stdin.readline
for test_case in range(1,int(input())+1):
    n,k=map(int,input().split())
    a=list(map(int,input().split()))
    b = list(map(int, input().split()))

    if test_case == 14 and ([n,k] != [1,1]):
        print("n,k=",n,k)
        print("a=",a)
        print("b=",b)
        continue
    arr = []
    for i in range(n):
        if a[i] < b[i]:
            arr.append([a[i],b[i]])
    #
    arr.sort(key=lambda x:[ x[1],-(x[1]-x[0]) ] )
    #
    a = [x[0] for x in arr]
    b = [x[1] for x in arr]
    diff = [x[1] - x[0] for x in arr]

    b_prefix_sum = [0] + list(accumulate(b))
    diff_prefix_sum = [0] + list(accumulate(diff))
    #
    ans = 0
    for i in range(k-1,len(arr)):
        diff_value = diff_prefix_sum[i+1]
        b_value = b_prefix_sum[i+1] - b_prefix_sum[i+1-k]

        curr_ans = diff_value - b_value
        ans = max(ans,curr_ans)

    print(ans)

'''
1
19 2
9 7 4 8 9 4 8 6 6 5 10 2 4 4 2 4 5 2 1
1 6 7 6 1 5 10 2 7 7 6 5 10 6 7 6 5 5 6

'''

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

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

Think step by step.

Response:
```plaintext
1
5 3
10 20 30 40 50
15 25 35 45 55
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shop Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in the shop. There are $$$n$$$ items in the shop; each item has two parameters: $$$a_i$$$ (item price for Alice) and $$$b_i$$$ (item price for Bob).

Alice wants to choose a subset (possibly empty) of items and buy them. After that, Bob does the following:

- if Alice bought less than $$$k$$$ items, Bob can take all of them for free;
- otherwise, he will take $$$k$$$ items for free that Alice bought (Bob chooses which $$$k$$$ items it will be), and for the rest of the chosen items, Bob will buy them from Alice and pay $$$b_i$$$ for the $$$i$$$-th item.

Alice's profit is equal to $$$\sum\limits_{i \in S} b_i - \sum\limits_{j \in T} a_j$$$, where $$$S$$$ is the set of items Bob buys from Alice, and $$$T$$$ is the set of items Alice buys from the shop. In other words, Alice's profit is the difference between the amount Bob pays her and the amount she spends buying the items.

Alice wants to maximize her profit, Bob wants to minimize Alice's profit. Your task is to calculate Alice's profit if both Alice and Bob act optimally.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — Alice's profit if both Alice and Bob act optimally.

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

Note: In the first test case, Alice should buy the $$$2$$$-nd item and sell it to Bob, so her profit is $$$2 - 1 = 1$$$.

In the second test case, Alice should buy the $$$1$$$-st, the $$$2$$$-nd and the $$$3$$$-rd item; then Bob takes the $$$1$$$-st item for free and pays for the $$$2$$$-nd and the $$$3$$$-rd item. Alice's profit is $$$(3+2) - (1+2+1) = 1$$$. Bob could take $$$2$$$-nd item for free instead; this does not change Alice's profit. Bob won't take the $$$3$$$-rd item for free, since this would lead to a profit of $$$2$$$.

Buggy Code:
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class main3 {
    static PrintWriter pw = new PrintWriter(new BufferedOutputStream(System.out));
    static FastReader in = new FastReader();
    public static void main(String[] args) {
        int t = in.nextInt();
        while(t-- > 0){
            int n = in.nextInt();
            int k = in.nextInt();
            ArrayList<Long> a = input(n);
            ArrayList<Long> b = input(n);
            ArrayList<long[]> arr = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                long l = a.get(i);
                long r = b.get(i);
                arr.add(new long[]{l, r});
            }
            Collections.sort(arr, (long[] c, long[] d) -> {
                if(c[0] == d[0]){
                    return Long.compare(d[1], c[1]);
                }
                return Long.compare(c[0], d[0]);
            });
            long ans = 0;
            long curr = 0;
            PriorityQueue<Long> q = new PriorityQueue<>();
            for (int i = 0; i < n; i++) {
                long l = arr.get(i)[0];
                long r = arr.get(i)[1];
                curr -= l;
                q.add(r);
                if(q.size() > k){
                    curr += q.poll();
                }
                ans = Math.max(ans, curr);
            }
            pw.println(ans);
        }
	pw.close();
    }
    static class Pair implements Comparable<Pair>{
        int key;
        int value;
        Pair(int key, int value){
            this.key = key;
            this.value = value;
        }
        @Override
        public int compareTo(Pair that){
            if(this.key == that.key){
                return that.value - this.value;
            }
            return this.key - that.key;
        }
	    @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Pair pair = (Pair) o;
            return key == pair.key && value == pair.value;
        }

        @Override
        public int hashCode() {
            return Objects.hash(key, value);
        }
    }
    static class BIT{
        int bit[], n;
        public BIT(int s){
            bit = new int[s+1];
            n = s;
        }
        void add(int i, int diff){
            while(i<=n){
                bit[i] += diff;
                i += (i&(-i));
            }
        }
        int query(int i){
            int sum = 0;
            while(i>0){
                sum += bit[i];
                i -= (i&(-i));
            }
            return sum;
        }
        int rangeQuery(int l, int r){
            return query(r)-query(l-1);
        }
    }
    static class DisjointUnionSets {
        int[] rank, parent;
        int n;
        int count;
        public DisjointUnionSets(int n){
            rank = new int[n];
            parent = new int[n];
            count = n;
            this.n = n;
            makeSet();
        }
        void makeSet(){
            for (int i = 0; i < n; i++) {
                parent[i] = i;
            }
        }
        int find(int x){
            if (parent[x] != x) {
                parent[x] = find(parent[x]);
            }

            return parent[x];
        }
        void union(int x, int y){
            int xRoot = find(x), yRoot = find(y);

            if (xRoot == yRoot)
                return;

            if (rank[xRoot] < rank[yRoot])
                parent[xRoot] = yRoot;

            else if (rank[yRoot] < rank[xRoot])
                parent[yRoot] = xRoot;

            else{
                parent[yRoot] = xRoot;
                rank[xRoot] = rank[xRoot] + 1;
            }
            count -= 1;
        }
        int num(){
            return count;
        }
    }
    public static void kmp(int[] lps, String s){
        int length = 0;
        int i = 1;
        while(i < s.length()){
            if(s.charAt(i) == s.charAt(length)){
                length++;
                lps[i] = length;
                i++;
            }else{
                if(length != 0){
                    length = lps[length - 1];
                }else{
                    lps[i] = 0;
                    i++;
                }
            }
        }
    }
    public static int calcMex(List<Integer> list) {
        Set<Integer> set = new HashSet<>(list);
        for (int i = 0; i <= list.size(); i++) {
            if (!set.contains(i)) {
                return i;
            }
        }
        return list.size();
    }
    public static long gcd(long a, long b){
        if(a == 0){
            return b;
        }
        return gcd(b % a, a);
    }
    public static long sqrt(long x){
        long i = 0;
        long j = Integer.MAX_VALUE;
        while(i < j){
            long mid = i + (j - i) / 2;
            if(mid * mid <= x){
                i = mid + 1;
            }else{
                j = mid;
            }
        }
        return i - 1;
    }
    static long lcm(long a, long b){
        return (a / gcd(a, b)) * b;
    }
    static class FastReader {

        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    public static ArrayList<Long> input(int n){
        ArrayList<Long> a = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            a.add(in.nextLong());
        }
        return a;
    }
    public static ArrayList<ArrayList<Integer>> adj(int n){
        ArrayList<ArrayList<Integer>> arr = new ArrayList<>();
        for (int i = 0; i <= n; i++) {
            arr.add(new ArrayList<>());
        }
        for (int i = 0; i < n - 1; i++) {
            int l = in.nextInt() - 1;
            int r = in.nextInt() - 1;
            arr.get(l).add(r);
            arr.get(r).add(l);
        }
        return arr;
    }
    public static void printlist(ArrayList<Long> a){
        for (int i = 0; i < a.size(); i++) {
            pw.print(a.get(i) + " ");
        }
        pw.println();
    }
    public static void printlist2d(ArrayList<ArrayList<Long>> a){
        for (int i = 0; i < a.size(); i++) {
            for (int j = 0; j < a.get(0).size(); j++) {
                pw.print(a.get(i).get(j) + " ");
            }
            pw.println();
        }
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases where the algorithm's logic might fail. The code sorts the items based on Alice's price and then processes them, but it does not correctly handle the scenario where the difference between Bob's price and Alice's price is significant. This can lead to suboptimal choices for Alice's profit.

Let's consider a scenario where:
- Alice's prices are all very high.
- Bob's prices are all very low.
- The value of `k` is such that it influences the selection of items in a way that the current code might not handle correctly.

Here's a small-scale test input that could expose the bug:

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

### Explanation:
- **Number of test cases:** 1
- **Number of items (n):** 5
- **Number of items Bob can take for free (k):** 3
- **Alice's prices (a):** 10 10 10 10 10
- **Bob's prices (b):** 1 2 3 4 5

### Expected Behavior:
- Alice will choose all 5 items because the total profit from Bob's prices is maximized.
- Bob will take the 3 items with the highest Bob's prices (4, 5, 3) for free.
- Alice's profit will be calculated as follows:
  - Bob pays for items with prices 1 and 2: \(1 + 2 = 3\)
  - Alice pays for all items: \(10 + 10 + 10 + 10 + 10 = 50\)
  - Alice's profit: \(3 - (50 - 30) = 3 - 20 = -17\)

However, the current code might not handle this scenario correctly because it sorts based on Alice's prices and might not consider the optimal combination of items to maximize profit.

### Correct Output:
```plaintext
-17
```

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shop Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in the shop. There are $$$n$$$ items in the shop; each item has two parameters: $$$a_i$$$ (item price for Alice) and $$$b_i$$$ (item price for Bob).

Alice wants to choose a subset (possibly empty) of items and buy them. After that, Bob does the following:

- if Alice bought less than $$$k$$$ items, Bob can take all of them for free;
- otherwise, he will take $$$k$$$ items for free that Alice bought (Bob chooses which $$$k$$$ items it will be), and for the rest of the chosen items, Bob will buy them from Alice and pay $$$b_i$$$ for the $$$i$$$-th item.

Alice's profit is equal to $$$\sum\limits_{i \in S} b_i - \sum\limits_{j \in T} a_j$$$, where $$$S$$$ is the set of items Bob buys from Alice, and $$$T$$$ is the set of items Alice buys from the shop. In other words, Alice's profit is the difference between the amount Bob pays her and the amount she spends buying the items.

Alice wants to maximize her profit, Bob wants to minimize Alice's profit. Your task is to calculate Alice's profit if both Alice and Bob act optimally.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$0 \le k \le n$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

The third line contains $$$n$$$ integers $$$b_1, b_2, \dots, b_n$$$ ($$$1 \le b_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print a single integer — Alice's profit if both Alice and Bob act optimally.

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

Note: In the first test case, Alice should buy the $$$2$$$-nd item and sell it to Bob, so her profit is $$$2 - 1 = 1$$$.

In the second test case, Alice should buy the $$$1$$$-st, the $$$2$$$-nd and the $$$3$$$-rd item; then Bob takes the $$$1$$$-st item for free and pays for the $$$2$$$-nd and the $$$3$$$-rd item. Alice's profit is $$$(3+2) - (1+2+1) = 1$$$. Bob could take $$$2$$$-nd item for free instead; this does not change Alice's profit. Bob won't take the $$$3$$$-rd item for free, since this would lead to a profit of $$$2$$$.

Buggy Code:
// #define winner formidablechief27
import java.io.*;
import java.util.*;

public class Main {

    long mod = (long) 1e9 + 7;
    long inf = (long) (9e18);
    PrintWriter out = new PrintWriter(System.out);
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st;
    Scanner SC = new Scanner(System.in);

    String ins() {while (st == null || !st.hasMoreElements()) {try {st = new StringTokenizer(br.readLine());} catch (IOException e) {}} return st.nextToken();}
    char[] inc() {return ins().toCharArray();}
    ArrayList<Character> incl() {ArrayList<Character> ch = new ArrayList<>(); String s = ins(); for(int i=0;i<s.length();i++) ch.add(s.charAt(i)); return ch;}
    int ini() {return Integer.parseInt(ins());}
    long in() {return Long.parseLong(ins());}
    double ind() {return Double.parseDouble(ins());}

    void p(String s) {out.print(s);}
    void p(int s) {out.print(s);}
    void p(long s) {out.print(s);}
    void p(double s) {out.print(s);}
    void p(char s) {out.print(s);}
    void pl(String s) {out.println(s);}
    void pl(int s) {out.println(s);}
    void pl(long s) {out.println(s);}
    void pl(double s) {out.println(s);}
    void pl(char c) {out.println(c);}
    void pl() {out.println();}
    void ps(int s) {out.print(s + " ");}
    void ps(long s) {out.print(s + " ");}
    void ps(double s) {out.print(s + " ");}
    void ps(char c) {out.print(c + " ");}
    void ps(String s) {out.print(s + " ");}
    void yes() {pl("Yes");}
    void no() {pl("No");}

    long max(long a, long b) {return a > b ? a : b;}
    long min(long a, long b) {return a > b ? b : a;}
    long ceil(long a, long b) {return ((long) Math.ceil(((double) (a) / b)));}
    long abs(long a) {if (a < 0) return (-1 * a);return a;}
    long log(long a) {return (long) (Math.log(a));}
    long log2(long a) {return (long) (Math.log(a) / Math.log(2));}
    long gcd(long a, long b) {if (b == 0)return a;return gcd(b, a % b);}
    long lcm(long a, long b) {return ((a * b) / gcd(a, b));}
    long add(long a, long b) {return (((a + mod) % mod + (b + mod) % mod) % mod);}
    long sub(long a, long b) {return (((a + mod) % mod + ((-b) + mod) % mod) % mod);}
    long mul(long a, long b) {return ((a % mod * b % mod) % mod);}
    long inv(long x) {return pow(x, mod - 2);}
    long div(long x, long y) {return mul(x, inv(y));}
    long pow(long a, long b) {a %= mod;long res = 1;while (b > 0) {if ((b & 1) != 0)res = mul(res, a);a = mul(a, a);b /= 2;}return res;}
    long sqrt(long x) {long start = 0, end = (long) 3e9, ans = 1; while (start <= end) {long mid = (start + end) / 2; if (mid * mid <= x) {ans = mid;start = mid + 1;} else end = mid - 1;} return ans;}
    boolean isSet(long x, long bit) {return ((x >> bit) & 1L) == 1L;}

    ArrayList<Long> lin(int n) {ArrayList<Long> a = new ArrayList<Long>(n);for (int i = 0; i < n; i++) a.add(in()); return a;}
    long sum(ArrayList<Long> a) {long sum = 0; for(long ele : a) sum += ele; return sum;}
    void sort(ArrayList<Long> a) {Collections.sort(a);}
    void rsort(ArrayList<Long> a) {Collections.sort(a, Collections.reverseOrder());}
    void ps(ArrayList<Long> a) {for(long ele : a) ps(ele); pl();}

    long[] in(int n) {long a[] = new long[n]; for (int i = 0; i < n; i++) a[i] = in();return a;}
    int[] ini(int n) {int a[] = new int[n]; for (int i = 0; i < n; i++) a[i] = ini();return a;}
    long sum(long a[]) {long sum = 0; for(long ele : a) sum += ele; return sum;}
    void ps(long a[]) {for(long ele : a) ps(ele); pl();}
    int max(long a[]) {int in = 0;long m = a[0];for (int i = 1; i < a.length; i++) {if (a[i] > m) {m = a[i];in = i;}}return in;}
    int min(long a[]) {int in = 0;long m = a[0];for (int i = 1; i < a.length; i++) {if (a[i] < m) {m = a[i];in = i;}}return in;}
    void sort(long a[]) {divide(a, 0, a.length - 1, true);}
    void rsort(long a[]) {divide(a, 0, a.length - 1, false);}
    void divide(long a[], int l, int r, boolean order) {if (l < r) {int m = l + (r - l) / 2;divide(a, l, m, order);divide(a, m + 1, r, order);merge(a, l, m, r, order);}}
    void merge(long a[], int l, int m, int r, boolean order) {int n1 = m - l + 1;int n2 = r - m;long L[] = new long[n1];long R[] = new long[n2];for (int i = 0; i < n1; ++i) L[i] = a[l + i];for (int j = 0; j < n2; ++j) R[j] = a[m + 1 + j];int i = 0, j = 0; int k = l;while (i < n1 && j < n2) {if ((L[i] <= R[j] && order) || (L[i] >= R[j] && !order)) {a[k] = L[i];i++;} else {a[k] = R[j];j++;}k++;}while (i < n1) {a[k] = L[i];i++;k++;}while (j < n2) {a[k] = R[j];j++;k++;}}
    void sort(long a[][]) {divide(a, 0, a[0].length - 1, true);}
    void rsort(long a[][]) {divide(a, 0, a[0].length - 1, false);}
    void divide(long a[][], int l, int r, boolean order) {if (l < r) {int m = l + (r - l) / 2;divide(a, l, m, order);divide(a, m + 1, r,order);merge(a, l, m, r, order);}}
    void merge(long a[][], int l, int m, int r, boolean order) {int n1 = m - l + 1;int n2 = r - m;long L[] = new long[n1]; long R[] = new long[n2];long b1[][] = new long[a.length][n1]; long b2[][] = new long[a.length][n2];for (int i = 0; i < n1; ++i) {L[i] = a[0][l + i];for (int p = 1; p < a.length; p++) b1[p][i] = a[p][l + i];}for (int j = 0; j < n2; ++j) {R[j] = a[0][m + 1 + j];for (int p = 1; p < a.length; p++) b2[p][j] = a[p][m + 1 + j];}int i = 0, j = 0; int k = l;while (i < n1 && j < n2) {if ((L[i] <= R[j] && order) || (L[i] >= R[j] && !order)) {a[0][k] = L[i]; for (int p = 1; p < a.length; p++) a[p][k] = b1[p][i];i++;} else {a[0][k] = R[j];for (int p = 1; p < a.length; p++) a[p][k] = b2[p][j];j++;}k++;}while (i < n1) {a[0][k] = L[i];for (int p = 1; p < a.length; p++) a[p][k] = b1[p][i];i++;k++;}while (j < n2) {a[0][k] = R[j];for (int p = 1; p < a.length; p++) a[p][k] = b2[p][j];j++;k++;}}

    void fillmap(Map<Long, Long> map, long a[]) {for (int i = 0; i < a.length; i++) {fill(map, a[i]);}}
    void fill(Map<Long, Long> map, long val) {if (map.containsKey(val))map.put(val, map.get(val) + 1); else map.put(val, (long) 1);}
    void fill(Map<Long, Long> map, long val, long count) {if(map.containsKey(val)) map.put(val, map.get(val) + count); else map.put(val, count);}
    void remove(Map<Long, Long> map, long val) {if (map.get(val) == 1) map.remove(val); else map.replace(val, map.get(val) - 1);}
    void replace(Map<Long, Long> map, long val, long newval) {remove(map, val);fill(map, newval);}
    void fill(Map<Long, ArrayList<Long>> map, long a[]) {for (int i = 0; i < a.length; i++) {fill(map, a[i], i);}}
    void fill(Map<Long, ArrayList<Long>> map, long val, int i) {if (map.containsKey(val)) map.get(val).add((long) i); else {ArrayList<Long> index = new ArrayList<>();index.add((long) i);map.put(val, index);}}
    void lfill(Map<Long, Long> map, ArrayList<Long> a) {for (Map.Entry<Long, Long> entry : map.entrySet()) {long val = entry.getValue(); long key = entry.getKey();for (long j = 0; j < val; j++) a.add(key);}}
    int MAXN = 10000001;int spf[] = new int[MAXN];
    public void sieve() {spf[1] = 1; for (int i = 2; i < MAXN; i++) spf[i] = i;for (int i = 4; i < MAXN; i += 2) spf[i] = 2;for (int i = 3; i * i < MAXN; i++) if (spf[i] == i) for (int j = i * i; j < MAXN; j += i) if (spf[j] == j) spf[j] = i;}
    public void primefactors(HashMap<Long, Long> map, int x) {while (x != 1) {fill(map, spf[x]);x /= spf[x];}}

    char[] substring(char a[], int start, int end) {char ch[] = new char[end - start];int index = 0;for (int i = start; i < end; i++) ch[index++] = a[i];return ch;}
    boolean equals(char a[], char b[]) {for (int i = 0; i < a.length; i++) if (a[i] != b[i]) return false;return a.length == b.length;}
    char[] append(char a[], char[] b) {char ch[] = new char[a.length + b.length];int i = 0; for (i = 0; i < a.length; i++) ch[i] = a[i]; for (int j = 0; j < b.length; j++) ch[i++] = b[j];return ch;}
    void p(char[] c) {for (char ch : c) p(ch);}
    void ps(char[] c) {for (char ch : c) ps(ch); pl();}
    void pl(char[] c) {for (char ch : c) p(ch); pl();}

    int upper_bound(long a[], long val) {int start = 0;int end = a.length - 1;while (start <= end) {int mid = (start + end) / 2;if (a[mid] >= val) {if (mid == 0 || a[mid-1] < val)return mid;end = mid - 1;} else start = mid + 1;}return -1;}
    int upper_bound(ArrayList<Long> a, long val) {if (a.size() == 0)return -1;int start = 0;int end = a.size() - 1;while (start <= end) {int mid = (start + end) / 2;if (a.get(mid) >= val) {if (mid == 0 || a.get(mid-1) < val) return mid;end = mid - 1;} else start = mid + 1;}return -1;}
    int lower_bound(ArrayList<Long> a, long val) {if (a.size() == 0)return -1;int start = 0;int end = a.size() - 1;while (start <= end) {int mid = (start + end) / 2;if (a.get(mid) > val) {end = mid - 1;continue;} else {if (mid == a.size() - 1 || a.get(mid+1) > val) return mid; start = mid + 1;}}return -1;}
    int lower_bound(long a[], long val) {int start = 0;int end = a.length - 1;while (start <= end) {int mid = (start + end) / 2;if (a[mid] > val) {end = mid - 1;continue;} else {if (mid == a.length - 1 || a[mid + 1] > val) return mid; start = mid + 1;}}return -1;}

	PriorityQueue<long[]> queue(boolean order) {if(order) {PriorityQueue<long[]> queue = new PriorityQueue<long[]>((a, b) -> {if (a[0] > b[0])return 1;else if (a[0] == b[0] && a[1] < b[1])return 1;else return -1;});return queue;}else {PriorityQueue<long[]> queue = new PriorityQueue<long[]>((a, b) -> {if (a[0] < b[0])return 1;else if (a[0] == b[0] && a[1] > b[1])return 1;else return -1;});return queue;}}
    class Pair {long f; long s; Pair(long f, long s) {this.f = f;this.s = s;}@Override public boolean equals(Object o) {if (this == o) return true; if (o == null || getClass() != o.getClass()) return false;Pair pair = (Pair) o;return f == pair.f && s == pair.s;}@Override public int hashCode() {return Objects.hash(f, s);}}
    class SegmentTree {long[] tree; long[] nums;int n; SegmentTree(long[] nums) {this.nums = nums;this.n = nums.length;this.tree = new long[4 * n];buildTree(1, 0, n - 1);} void buildTree(int node, int start, int end) {if (start == end) tree[node] = nums[start]; else {int mid = (start + end) / 2; buildTree(2 * node, start, mid); buildTree(2 * node + 1, mid + 1, end);tree[node] = tree[2 * node] + tree[2 * node + 1];}}long query(int left, int right) {return query(1, 0, n - 1, left-1, right-1);} long query(int node, int start, int end, int left, int right) {if (right < start || left > end)  return 0; else if (left <= start && right >= end) return tree[node]; else return query(2 * node, start, (start + end) / 2, left, right) + query(2 * node + 1, ((start + end) / 2) + 1, end, left, right);} void update(int index, long value) {update(1, 0, n - 1, index-1, value);} void update(int node, int start, int end, int index, long value) {if (start == end) {nums[index] = value;tree[node] = value;} else {int mid = (start + end) / 2; if (index >= start && index <= mid) update(2 * node, start, mid, index, value); else update(2 * node + 1, mid + 1, end, index, value);tree[node] = tree[2 * node] + tree[2 * node + 1];}}}
    class SegmentTreeMax {long[] tree; long[] nums;int n; SegmentTreeMax(long[] nums) {this.nums = nums;this.n = nums.length;this.tree = new long[4 * n];for(int i=0;i<4*n;i++) tree[i] = -inf;buildTree(1, 0, n - 1);} void buildTree(int node, int start, int end) {if (start == end) tree[node] = nums[start]; else {int mid = (start + end) / 2; buildTree(2 * node, start, mid); buildTree(2 * node + 1, mid + 1, end);tree[node] = max(tree[2 * node] , tree[2 * node + 1]);}}long query(int left, int right) {return query(1, 0, n - 1, left-1, right-1);} long query(int node, int start, int end, int left, int right) {if (right < start || left > end)  return -inf; else if (left <= start && right >= end) return tree[node]; else return max(query(2 * node, start, (start + end) / 2, left, right) , query(2 * node + 1, ((start + end) / 2) + 1, end, left, right));} void update(int index, long value) {update(1, 0, n - 1, index-1, value);} void update(int node, int start, int end, int index, long value) {if (start == end) {nums[index] = value;tree[node] = value;} else {int mid = (start + end) / 2; if (index >= start && index <= mid) update(2 * node, start, mid, index, value); else update(2 * node + 1, mid + 1, end, index, value);tree[node] = max(tree[2 * node] , tree[2 * node + 1]);}}}
    class SegmentTreeMin {long[] tree; long[] nums;int n; SegmentTreeMin(long[] nums) {this.nums = nums;this.n = nums.length;this.tree = new long[4 * n];for(int i=0;i<4*n;i++) tree[i] = inf; buildTree(1, 0, n - 1);} void buildTree(int node, int start, int end) {if (start == end) tree[node] = nums[start]; else {int mid = (start + end) / 2; buildTree(2 * node, start, mid); buildTree(2 * node + 1, mid + 1, end);tree[node] = min(tree[2 * node] , tree[2 * node + 1]);}}long query(int left, int right) {return query(1, 0, n - 1, left-1, right-1);} long query(int node, int start, int end, int left, int right) {if (right < start || left > end)  return inf; else if (left <= start && right >= end) return tree[node]; else return min(query(2 * node, start, (start + end) / 2, left, right) , query(2 * node + 1, ((start + end) / 2) + 1, end, left, right));} void update(int index, long value) {update(1, 0, n - 1, index-1, value);} void update(int node, int start, int end, int index, long value) {if (start == end) {nums[index] = value;tree[node] = value;} else {int mid = (start + end) / 2; if (index >= start && index <= mid) update(2 * node, start, mid, index, value); else update(2 * node + 1, mid + 1, end, index, value);tree[node] = min(tree[2 * node] , tree[2 * node + 1]);}}}
    class FenwickTree {long[] tree;int n;FenwickTree(int size) {this.tree = new long[size + 1];this.n = size;}void update(int index, long delta) {for (;index <= n; index += index & -index) tree[index] += delta;  }long query(int index) {long sum = 0;for (; index > 0; index -= index & -index) sum += tree[index];return sum;}long query(int left, int right) {return query(right) - query(left - 1);}}
    class FenwickTreeMod {long[] tree;int n; FenwickTreeMod(int size) {this.tree = new long[size + 1];this.n = size;}void update(int index, long delta) {for (;index <= n; index += index & -index) tree[index] = add(tree[index], delta);  }long query(int index) {long sum = 0;for (; index > 0; index -= index & -index) sum = add(sum, tree[index]);return sum;}long query(int left, int right) {return sub(query(right), query(left - 1));}}
    class FenwickTree2D {long[][] bit;int n;int m;FenwickTree2D(int n, int m) {this.bit = new long[n + 1][m + 1];this.n = n;this.m = m;}void update(int x, int y, long val) {for (; x <= n; x += (x & (-x))) {for (int i = y; i <= m; i += (i & (-i))) {bit[x][i] += val;}}}long query(int x1, int y1, int x2, int y2) {return query(x2, y2) - query(x2, y1 - 1) - query(x1 - 1, y2) + query(x1 - 1, y1 - 1);}long query(int x, int y) {long ans = 0;for (int i = x; i > 0; i -= (i & (-i))) {for (int j = y; j > 0; j -= (j & (-j))) {ans += bit[i][j];}}return ans;}}
    class OfflineTree {long a[];int n;long[] tree;long[] nums;TreeMap<Long, Integer> map;OfflineTree(long[] a) {this.a = a;sort(a);map = new TreeMap<>();int ind = 0;for(int i=0;i<a.length;i++) if(!map.containsKey(a[i])) map.put(a[i], ind++);this.n = ind;tree = new long[4 * n];nums = new long[n];}void buildTree(int node, int start, int end) {if (start == end) tree[node] = nums[start];else {int mid = (start + end) / 2;buildTree(2 * node, start, mid);buildTree(2 * node + 1, mid + 1, end);tree[node] = tree[2 * node] + tree[2 * node + 1];}}long query(long a, long b) {Long key1 = map.ceilingKey(a);Long key2 = map.floorKey(b);if(key1 == null || key2 == null) return 0;int index1 = map.get(key1);int index2 = map.get(key2);return query(1, 0, n - 1, index1, index2);}long query(int node, int start, int end, int left, int right) {if (right < start || left > end) return 0;else if (left <= start && right >= end) return tree[node];else return query(2 * node, start, (start + end) / 2, left, right) + query(2 * node + 1, ((start + end) / 2) + 1, end, left, right);}void update(long key, long value) {if(!map.containsKey(key)) return;int index = map.get(key);update(1, 0, n - 1, index, value);}void update(int node, int start, int end, int index, long value) {if (start == end) {nums[index] = value;tree[node] = value;} else {int mid = (start + end) / 2;if (index >= start && index <= mid) update(2 * node, start, mid, index, value);else update(2 * node + 1, mid + 1, end, index, value);tree[node] = tree[2 * node] + tree[2 * node + 1];}}}
    class OfflineTreeMax {long a[];int n;long[] tree;long[] nums;TreeMap<Long, Integer> map;OfflineTreeMax(long[] a) {this.a = a;sort(a);map = new TreeMap<>();int ind = 0;for(int i=0;i<a.length;i++) if(!map.containsKey(a[i])) map.put(a[i], ind++);this.n = ind;tree = new long[4 * n];nums = new long[n];}void buildTree(int node, int start, int end) {if (start == end) tree[node] = nums[start];else {int mid = (start + end) / 2;buildTree(2 * node, start, mid);buildTree(2 * node + 1, mid + 1, end);tree[node] = max(tree[2 * node] , tree[2 * node + 1]);}}long query(long a, long b) {Long key1 = map.ceilingKey(a);Long key2 = map.floorKey(b);if(key1 == null || key2 == null) return 0;int index1 = map.get(key1);int index2 = map.get(key2);return query(1, 0, n - 1, index1, index2);}long query(int node, int start, int end, int left, int right) {if (right < start || left > end) return 0;else if (left <= start && right >= end) return tree[node];else return max(query(2 * node, start, (start + end) / 2, left, right) , query(2 * node + 1, ((start + end) / 2) + 1, end, left, right));}void update(long key, long value) {if(!map.containsKey(key)) return;int index = map.get(key);update(1, 0, n - 1, index, value);}void update(int node, int start, int end, int index, long value) {if (start == end) {nums[index] = value;tree[node] = value;} else {int mid = (start + end) / 2;if (index >= start && index <= mid) update(2 * node, start, mid, index, value);else update(2 * node + 1, mid + 1, end, index, value);tree[node] = max(tree[2 * node] , tree[2 * node + 1]);}}}
    class OfflineTreeMin {long a[];int n;long[] tree;long[] nums;TreeMap<Long, Integer> map;OfflineTreeMin(long[] a) {this.a = a;sort(a);map = new TreeMap<>();int ind = 0;for(int i=0;i<a.length;i++) if(!map.containsKey(a[i])) map.put(a[i], ind++);this.n = ind;tree = new long[4 * n];nums = new long[n];for(int i=0;i<4*n;i++) tree[i]=inf;}void buildTree(int node, int start, int end) {if (start == end) tree[node] = nums[start];else {int mid = (start + end) / 2;buildTree(2 * node, start, mid);buildTree(2 * node + 1, mid + 1, end);tree[node] = min(tree[2 * node] , tree[2 * node + 1]);}}long query(long a, long b) {Long key1 = map.ceilingKey(a);Long key2 = map.floorKey(b);if(key1 == null || key2 == null) return -5;int index1 = map.get(key1);int index2 = map.get(key2);return query(1, 0, n - 1, index1, index2);}long query(int node, int start, int end, int left, int right) {if (right < start || left > end) return inf;else if (left <= start && right >= end) return tree[node];else return min(query(2 * node, start, (start + end) / 2, left, right) , query(2 * node + 1, ((start + end) / 2) + 1, end, left, right));}void update(long key, long value) {if(!map.containsKey(key)) return;int index = map.get(key);update(1, 0, n - 1, index, value);}void update(int node, int start, int end, int index, long value) {if (start == end) {nums[index] = value;tree[node] = value;} else {int mid = (start + end) / 2;if (index >= start && index <= mid) update(2 * node, start, mid, index, value);else update(2 * node + 1, mid + 1, end, index, value);tree[node] = min(tree[2 * node] , tree[2 * node + 1]);}}}
    class SparseTableMax {long[][] table;long[] logTable;long[] arr;public SparseTableMax(long[] arr) {int n = arr.length;int logN = (int) (Math.log(n) / Math.log(2)) + 1;table = new long[n][logN];logTable = new long[n + 1];this.arr = arr;buildSparseTable();}public void buildSparseTable() {int n = arr.length;for (int i = 2; i <= n; i++) logTable[i] = logTable[i / 2] + 1;for (int i = 0; i < n; i++) table[i][0] = arr[i];for (int j = 1; (1 << j) <= n; j++) {for (int i = 0; i + ((1L << j)) <= n; i++) {long leftVal = table[i][j - 1];long rightVal = table[i + (int)((1L << (j - 1)))][j - 1];table[i][j] = max(leftVal, rightVal);}}}public long query(int l, int r) {long k = logTable[r - l + 1];long leftVal = table[l][(int)k];long rightVal = table[r - (int)((1L << k)) + 1][(int)k];return max(leftVal, rightVal);}}
    class SparseTableMin {long[][] table;long[] logTable;long[] arr;public SparseTableMin(long[] arr) {int n = arr.length;int logN = (int) (Math.log(n) / Math.log(2)) + 1;table = new long[n][logN];logTable = new long[n + 1];this.arr = arr;for(int i=0;i<n;i++) for(int j=0;j<logN;j++) table[i][j] = inf;buildSparseTable();}public void buildSparseTable() {int n = arr.length;for (int i = 2; i <= n; i++) logTable[i] = logTable[i / 2] + 1;for (int i = 0; i < n; i++) table[i][0] = arr[i];for (int j = 1; (1 << j) <= n; j++) {for (int i = 0; i + ((1L << j)) <= n; i++) {long leftVal = table[i][j - 1];long rightVal = table[i + (int)((1L << (j - 1)))][j - 1];table[i][j] = min(leftVal, rightVal);}}}public long query(int l, int r) {long k = logTable[r - l + 1];long leftVal = table[l][(int)k];long rightVal = table[r - (int)((1L << k)) + 1][(int)k];return min(leftVal, rightVal);}}
	class MissingSet {TreeSet<Long> set;TreeMap<Long, Long> superset;HashMap<Long, Long> count;MissingSet() {set = new TreeSet<>();superset = new TreeMap<>();count = new HashMap<>();}void add(long ele) {if (!set.contains(ele)) {if (superset.containsKey(ele + 1)) {long end = superset.get(ele + 1L);superset.remove(ele + 1L);Long key = superset.lowerKey(ele);if (key == null) superset.put(ele, end);else if (superset.get(key) == ele - 1)superset.put(key, end);else superset.put(ele, end);} else {Long key = superset.lowerKey(ele);if (key == null)superset.put(ele, ele);else if (superset.get(key) == ele - 1)superset.put(key, ele);else superset.put(ele, ele);}set.add(ele);}fill(count, ele);}boolean contains(long ele) {return set.contains(ele);}long count(long ele) {if(count.containsKey(ele)) return count.get(ele);else return 0;}long findlow(long ele) {Long key = superset.floorKey(ele);if (key == null) return ele;if (superset.get(key) >= ele) return key - 1;else return ele;}long findhigh(long ele) {Long key1 = superset.ceilingKey(ele);Long key2 = superset.floorKey(ele);if(key2 != null && superset.get(key2) >= ele) return superset.get(key2) + 1;if(key1 != null && key1 == ele) return superset.get(key1) + 1;return ele;}}

    /*
     * Remember the pain
     * Remember the promises they broke
     * Remember how worthless they made you feel
     * Turn that pain into anger
     * And show them who you're really are
     */

    public static void main(String args[]) {
        Main FC27 = new Main();
        FC27.start();
    }

    // This is where the fun begins
    public void start() {
    	int t = 1;
    	t = ini();
    	while (t-- > 0) solve();
    	out.close();
    }

    // This one's for you ;)
    public void solve() {
    	int n = ini();
    	int k = ini();
    	long a[] = in(n);
    	long b[] = in(n);
    	PriorityQueue<long[]> temp = queue(false);
    	for(int i=0;i<n;i++) if(a[i] < b[i]) temp.add(new long[] {b[i], a[i], b[i] - a[i]});
    	if(temp.size() < k) {
    		pl(0);
    		return;
    	}
    	long ans = 0;
    	PriorityQueue<Long> bob = new PriorityQueue<>(Collections.reverseOrder());
    	for(int i=0;i<k;i++) bob.add(temp.poll()[1]);
    	for(long ele : bob) ans -= ele;
    	PriorityQueue<long[]> alice = queue(false);
    	while(temp.size() > 0) alice.add(temp.poll());
    	for(long ele[] : alice) ans += ele[2];
    	long fans = ans;
    	while(true) {
    		if(bob.size() == 0 || alice.size() == 0) break;
    		fans += bob.poll();
    		long ele[] = alice.poll();
    		fans -= ele[2];
    		fans -= ele[1];
    		ans = max(ans, fans);
    	}
    	pl(max(0, ans));
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider edge cases and scenarios that might cause the algorithm to fail. Here are the steps to create such a test case:

1. **Edge Cases**:
   - **k = 0**: This means Bob can take all items for free if Alice buys any.
   - **k = n**: This means Bob will take all items for free, and Alice will not get any profit from Bob.
   - **n = 1**: This means there is only one item, and the behavior of the algorithm should be straightforward.

2. **Scenarios**:
   - **All items have \(a_i < b_i\)**: This means Alice can potentially make a profit from each item.
   - **Mixed items with \(a_i < b_i\) and \(a_i > b_i\)**: This can test the algorithm's ability to handle both profitable and non-profitable items.
   - **Items with \(a_i = b_i\)**: This can test the algorithm's behavior when the profit is zero for some items.

3. **Combination of the Above**:
   - **k = 1, n = 3**: This can test the algorithm's ability to handle a small number of items with one free item for Bob.

Given these considerations, let's create a test case that combines some of these edge cases and scenarios:

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

**Explanation**:
- **n = 3**: There are three items.
- **k = 1**: Bob can take one item for free.
- **a = [1, 2, 3]**: Prices for Alice.
- **b = [3, 2, 1]**: Prices for Bob.

**Expected Output**:
- Alice should buy all three items and Bob will take the item with the highest profit for Alice (which is the third item with profit \(1 - 3 = -2\)). The remaining two items will be sold to Bob with profits \(3 - 1 = 2\) and \(2 - 2 = 0\).

**Calculation**:
- Alice's profit = \(2 + 0 - 3 = -1\)

However, the correct optimal strategy should be:
- Alice buys the first and second items, and Bob takes the second item for free.
- Alice's profit = \(3 - 1 = 2\)

This test case should expose the bug in the algorithm if it does not handle the optimal selection of items correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $a+b$ is a multiple of $b \cdot \gcd(a,b)$.

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

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
1
3
4
14
153
1643498

Note: In the first test case, only $(1,1)$ satisfies the conditions.

In the fourth test case, $(1,1),(2,1),(2,2),(3,1),(4,1),(5,1),(6,1),(6,2),(6,3),(7,1),(8,1),(9,1),(10,1),(10,2)$ satisfy the conditions.

Buggy Code:
#include <algorithm>
#include <cmath>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native" )
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("profile-values,profile-reorder-functions,tracer")
#pragma GCC optimize("vpt")
#pragma GCC optimize("rename-registers")
#pragma GCC optimize("move-loop-invariants")
#pragma GCC optimize("unswitch-loops")
#pragma GCC optimize("function-sections")
#pragma GCC optimize("data-sections")
#pragma GCC optimize("branch-target-load-optimize")
#pragma GCC optimize("branch-target-load-optimize2")
#pragma GCC optimize("btr-bb-exclusive")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-fhoist-adjacent-loads")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")

void solv() {
	int n, m;
	cin >> n >> m;
	int ans = 0;
	for (int i = 1; i <= n; i++) {
		ans++;
		for (int j = 2; (j * j <= i + j) && (j <= i); j++) {
			if ((i + j) % (j * j) == 0) {
				ans++;
			}
		}
	}
	cout << ans << '\n';
}

signed main() {
	int t;  cin >> t;
	while (t--) solv();
	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 expose the bug in the provided code, we need to focus on the logic used to count the valid pairs \((a, b)\). The code seems to have an issue with the loop condition and the way it checks the validity of pairs. Specifically, the inner loop's condition and the way it calculates the pairs are suspect.

Let's break down the problem and the code:

1. **Problem Statement**: We need to find pairs \((a, b)\) such that \(1 \le a \le n\), \(1 \le b \le m\), and \(a + b\) is a multiple of \(b \cdot \gcd(a, b)\).

2. **Buggy Code Analysis**:
   - The outer loop iterates over \(a\) from 1 to \(n\).
   - The inner loop iterates over \(b\) from 2 to \(i\) with the condition \(j * j \le i + j\).
   - It checks if \((i + j) % (j * j) == 0\) to increment the answer.

3. **Potential Issues**:
   - The inner loop condition \(j * j \le i + j\) is not the correct condition for checking the pairs.
   - The condition should be more aligned with the problem's requirement, which is \(a + b\) being a multiple of \(b \cdot \gcd(a, b)\).

4. **Test Case to Expose the Bug**:
   - We need a test case where the pairs \((a, b)\) are such that \(a + b\) is a multiple of \(b \cdot \gcd(a, b)\), but the code's logic fails to count them correctly.
   - A small test case with \(n = 3\) and \(m = 5\) can be used to check if the code correctly counts the pairs.

Let's create the test input:

```plaintext
1
3 5
```

This test input will help us verify if the code correctly counts the pairs \((a, b)\) for \(1 \le a \le 3\) and \(1 \le b \le 5\). The correct answer for this test case should be 4, which includes the pairs \((1,1)\), \((2,2)\), \((3,3)\), and \((3,5)\).

If the code does not output 4, it will indicate a bug in the logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $a+b$ is a multiple of $b \cdot \gcd(a,b)$.

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

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
1
3
4
14
153
1643498

Note: In the first test case, only $(1,1)$ satisfies the conditions.

In the fourth test case, $(1,1),(2,1),(2,2),(3,1),(4,1),(5,1),(6,1),(6,2),(6,3),(7,1),(8,1),(9,1),(10,1),(10,2)$ satisfy the conditions.

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

#define ll long long
#define mod 1000000007
#define mod2 998244353
#define all(a) a.begin(), a.end()
#define pb push_back
#define ins insert
#define fr(i, n) for(int i = 0; i < n; ++ i)
#define si(x) (int)(x).size()
#define mset(m,v) memset(m,v,sizeof(m))
#define mp make_pair
#define F first
#define S second
#define lb lower_bound
#define ub upper_bound

#define max2(a, b) ((a) > (b) ? (a) : (b))
#define max3(a, b, c) max2(max2(a, b), c)
#define max4(a, b, c, d) max2(max2(a, b), max2(c, d))
#define max5(a, b, c, d, e) max2(max4(a, b, c, d), e)
#define max6(a, b, c, d, e, f) max2(max4(a, b, c, d), max2(e, f))
#define max7(a, b, c, d, e, f, g) max2(max4(a, b, c, d), max3(e, f, g))

#define min2(a, b) ((a) < (b) ? (a) : (b))
#define min3(a, b, c) min2(min2(a, b), c)
#define min4(a, b, c, d) min2(min2(a, b), min2(c, d))
#define min5(a, b, c, d, e) min2(min4(a, b, c, d), e)
#define min6(a, b, c, d, e, f) min2(min4(a, b, c, d), min2(e, f))
#define min7(a, b, c, d, e, f, g) min2(min4(a, b, c, d), min3(e, f, g))

typedef vector<int> vi;
typedef vector<ll> vl;

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<int, ll> pil;
typedef pair<ll, int> pli;

typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<pli> vpli;
typedef vector<pil> vpil;

template <typename T>
void get_unique(vector<T>& a) {
    sort(a.begin(), a.end());
    a.erase(unique(a.begin(), a.end()), a.end());
}

template <class S, class T> ostream& operator<<(ostream& os, const vector<pair<S, T>>& p) { os << "[ "; for (auto it = p.begin(); it != p.end(); ++it) { os << *it; if (it != prev(p.end())) { os << ", "; } } os << " ]"; return os; }
template <class T> ostream& operator<<(ostream& os, const vector<vector<T>>& p) { os << "[ "; for (auto it = p.begin(); it != p.end(); ++it) { os << "[ "; for (auto inner_it = it->begin(); inner_it != it->end(); ++inner_it) { os << *inner_it; if (inner_it != prev(it->end())) { os << ", "; } } os << " ]"; if (it != prev(p.end())) { os << ", "; } } os << " ]"; return os; }
template <class S, class T> ostream& operator<<(ostream& os, const pair<S, T>& p) { return os << "(" << p.first << ", " << p.second << ")"; }
template <class T> ostream& operator<<(ostream& os, const vector<T>& p) { os << "[ "; for (auto it = p.begin(); it != p.end(); ++it) { os << *it; if (it != prev(p.end())) { os << ", "; } } os << " ]"; return os; }
template <class T> ostream& operator<<(ostream& os, const unordered_set<T>& p) { os << "[ "; for (auto it = p.begin(); it != p.end(); ++it) { os << *it; if (next(it) != p.end()) { os << ", "; } } return os << " ]"; }
template <class S, class T> ostream& operator<<(ostream& os, const unordered_map<S, T>& p) { os << "[ "; for (auto it = p.begin(); it != p.end(); ++it) { os << "( " << it->first << ": " << it->second << " )"; if (next(it) != p.end()) { os << ", "; } } os << " ]"; return os; }
template <class T> ostream& operator<<(ostream& os, const set<T>& p) { os << "[ "; for (auto it = p.begin(); it != p.end(); ++it) { os << *it; if (it != prev(p.end())) { os << ", "; } } os << " ]"; return os; }
template <class T> ostream& operator<<(ostream& os, const multiset<T>& p) { os << "[ "; for (auto it = p.begin(); it != p.end(); ++it) { os << *it; if (it != prev(p.end())) { os << ", "; } } os << " ]"; return os; }
template <class S, class T> ostream& operator<<(ostream& os, const map<S, T>& p) { os << "[ "; for (auto it = p.begin(); it != p.end(); ++it) { os << "( " << it->first << ": " << it->second << " )" << (it != prev(p.end()) ? ", " : ""); } os << " ]"; return os; }
template <class T> ostream& operator<<(ostream& os, const unordered_multiset<T>& p) { os << "[ "; for (auto it = p.begin(); it != p.end(); ++it) { os << *it; if (next(it) != p.end()) { os << ", "; } } os << " ]"; return os; }

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

    int t;
    cin >> t;
    fr(_, t){
        ll n, m;
        cin >> n >> m;
        ll res = 0;
        for (ll b = 1; b <= m; ++ b){
            ll val = 0;
            ll s = 1, e = 1e7;
            while (s <= e){
                ll m = (s + e) / 2;
                if (b * (b * m - 1) <= n){
                    val = m;
                    s = m + 1;
                } else e = m - 1;
            }
            res += val;
        }
        cout << res - 1 << "\n";
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can cause the binary search logic to fail. Specifically, we should look at values of \( n \) and \( m \) that can lead to incorrect calculations in the binary search loop.

### Step-by-Step Analysis:

1. **Binary Search Logic**:
   - The binary search is used to find the maximum value \( m \) such that \( b \cdot (b \cdot m - 1) \leq n \).
   - If \( b \cdot (b \cdot m - 1) \) is very close to \( n \), the binary search might not correctly handle the upper limit \( e \).

2. **Edge Cases**:
   - **Small Values of \( n \) and \( m \)**: These can cause the binary search to terminate early or incorrectly.
   - **Large Values of \( n \) and \( m \)**: These can cause the binary search to run for a longer time and might expose issues in the logic.
   - **Specific Values of \( n \) and \( m \)**: Values that make \( b \cdot (b \cdot m - 1) \) very close to \( n \) can cause the binary search to fail.

3. **Test Case Selection**:
   - **Small Values**: \( n = 1 \), \( m = 1 \)
   - **Medium Values**: \( n = 10 \), \( m = 10 \)
   - **Values Making \( b \cdot (b \cdot m - 1) \) Close to \( n \)**: \( n = 100 \), \( m = 10 \)
   - **Large Values**: \( n = 1000000 \), \( m = 1000000 \)

### Chosen Test Input:
```plaintext
4
1 1
10 10
100 10
1000000 1000000
```

### Explanation:
- **Test Case 1**: \( n = 1 \), \( m = 1 \) - This is a simple edge case where only the pair \((1, 1)\) is valid.
- **Test Case 2**: \( n = 10 \), \( m = 10 \) - This is a medium-sized test case to check the binary search logic.
- **Test Case 3**: \( n = 100 \), \( m = 10 \) - This test case is designed to make \( b \cdot (b \cdot m - 1) \) close to \( n \) for some values of \( b \).
- **Test Case 4**: \( n = 1000000 \), \( m = 1000000 \) - This is a large test case to check the performance and correctness of the binary search logic.

This test input should help expose any issues in the binary search logic and the overall algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $a+b$ is a multiple of $b \cdot \gcd(a,b)$.

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

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
1
3
4
14
153
1643498

Note: In the first test case, only $(1,1)$ satisfies the conditions.

In the fourth test case, $(1,1),(2,1),(2,2),(3,1),(4,1),(5,1),(6,1),(6,2),(6,3),(7,1),(8,1),(9,1),(10,1),(10,2)$ satisfy the conditions.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define gc getchar_unlocked
#define fo(i, n) for (i = 0; i < n; i++)
#define Fo(i, k, n) for (i = k; k < n ? i < n : i > n; k < n ? i += 1 : i -= 1)
#define ll long long
#define deb(x) cout << #x << "=" << x << endl
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define clr(x) memset(x, false, sizeof(x))
#define sortall(x) sort(all(x))
#define tr(it, a) for (auto it = a.begin(); it != a.end(); it++)
#define PI 3.1415926535897932384626
#define mod 1000000007
typedef pair<int, int> pii;
typedef pair<ll, ll> pl;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpii;
typedef vector<pl> vpl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;

long long count_pairs(int n, int m) {
    long long pairs = 0;
    for (int a = n; a > 0; a--) {
        for (int b = 1; b * b <= a + b; b++) {
            if ((a % b == 0) && ((a / b + 1) % __gcd(a, b) == 0)) {
                pairs++;
            }
        }
    }
    return pairs;
}

int main()
{
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    // who takes the last U wins
    while (t--) {
        int n, m;
        cin >> n >> m;
        cout << count_pairs(n, m) << 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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases or scenarios where the code might fail. Let's analyze the conditions and constraints:

1. **Edge Cases:**
   - Small values of \( n \) and \( m \) (e.g., \( n = 1 \), \( m = 1 \)).
   - Large values of \( n \) and \( m \) (e.g., \( n = 2 \cdot 10^6 \), \( m = 2 \cdot 10^6 \)).
   - Values where \( n \) and \( m \) are equal.
   - Values where \( n \) and \( m \) are powers of 2.
   - Values where \( n \) and \( m \) are prime numbers.
   - Values where \( n \) and \( m \) are multiples of each other.

2. **Buggy Code Analysis:**
   - The code iterates over \( a \) from \( n \) to 1.
   - It iterates over \( b \) from 1 to \( \sqrt{a + b} \).
   - It checks if \( a \) is divisible by \( b \) and if \( (a / b + 1) \) is divisible by \( \gcd(a, b) \).

3. **Potential Issues:**
   - The condition \( b * b \leq a + b \) might be too restrictive, leading to missed valid pairs.
   - The logic for checking the divisibility and gcd might be incorrect.

4. **Challenging Test Input:**
   - A test case where \( n \) and \( m \) are both large and have a complex relationship (e.g., \( n = 100 \), \( m = 100 \)).
   - A test case where \( n \) and \( m \) are equal and large (e.g., \( n = 1000 \), \( m = 1000 \)).
   - A test case where \( n \) and \( m \) are prime numbers (e.g., \( n = 97 \), \( m = 97 \)).
   - A test case where \( n \) and \( m \) are powers of 2 (e.g., \( n = 64 \), \( m = 64 \)).

Given these considerations, a small-scale test input that is complete and valid is:

```plaintext
4
100 100
1000 1000
97 97
64 64
```

This input includes a mix of large values, equal values, prime numbers, and powers of 2, which should help in exposing any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $a+b$ is a multiple of $b \cdot \gcd(a,b)$.

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

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
1
3
4
14
153
1643498

Note: In the first test case, only $(1,1)$ satisfies the conditions.

In the fourth test case, $(1,1),(2,1),(2,2),(3,1),(4,1),(5,1),(6,1),(6,2),(6,3),(7,1),(8,1),(9,1),(10,1),(10,2)$ satisfy the conditions.

Buggy Code:
import math

num = int(input())
for _ in range(num):
    n, m = map(int, input().split())

    if n == 1 and m == 1:
        print(1)
        continue

    counter = 0
    for x in range(1, m):
        t = math.floor((n + x) / x / x)
        for k in range(0, t):
            A = k * x * x - x
            B = x
            if k * x * x - x != 0:
                counter += 1
    print(counter)

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases and scenarios that might cause the code to fail. The buggy code has a nested loop structure that iterates over possible values of `x` and `k`, and it uses a formula to determine valid pairs `(a, b)`. The key points to consider are:

1. **Edge Cases for Small Inputs**: The code has a special case for `n == 1` and `m == 1`, which is correctly handled. However, other small values of `n` and `m` might reveal issues.
2. **Large Values of `n` and `m`**: The code might have performance issues or logical errors when dealing with larger values.
3. **Special Values of `x` and `k`**: The formula `A = k * x * x - x` and the condition `if k * x * x - x != 0` might not handle all valid pairs correctly.

Let's generate a small-scale test input that is complete and valid, focusing on these points:

1. **Small Values of `n` and `m`**: Test with `n = 2` and `m = 2` to see if the code correctly handles small inputs.
2. **Values that Might Cause Division Issues**: Test with `n = 3` and `m = 3` to see if the code handles division and modulo operations correctly.
3. **Values that Might Cause Looping Issues**: Test with `n = 4` and `m = 4` to see if the nested loops work as expected.

Here is the test input:

```plaintext
3
2 2
3 3
4 4
```

This input will help identify any issues in the buggy code, especially in handling small values and ensuring that the nested loops and conditions are working correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $a+b$ is a multiple of $b \cdot \gcd(a,b)$.

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

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
1
3
4
14
153
1643498

Note: In the first test case, only $(1,1)$ satisfies the conditions.

In the fourth test case, $(1,1),(2,1),(2,2),(3,1),(4,1),(5,1),(6,1),(6,2),(6,3),(7,1),(8,1),(9,1),(10,1),(10,2)$ satisfy the conditions.

Buggy Code:
from sys import stdin;input = lambda: stdin.readline().strip()
from math import gcd
for _ in range(int(input())):
    n,m = map(int,input().split())
    res = 0
    for b in range(1, m + 1):
        for a in range(b, n + 1, b):
            if (a // b + 1) % (gcd(a, b)) == 0:
                res += 1
    print(res)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $a+b$ is a multiple of $b \cdot \gcd(a,b)$.

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

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
1
3
4
14
153
1643498

Note: In the first test case, only $(1,1)$ satisfies the conditions.

In the fourth test case, $(1,1),(2,1),(2,2),(3,1),(4,1),(5,1),(6,1),(6,2),(6,3),(7,1),(8,1),(9,1),(10,1),(10,2)$ satisfy the conditions.

Buggy Code:
def cop(a, b):
  count = 0
  for j in range(1, b + 1):
    for i in range(j, a + 1, j):
      add = i + j
      gcd = j
      mult = j * gcd
      if add % mult == 0:
        count += 1
  return count

t = int(input())
for _ in range(t):
  a, b = map(int, input().split())
  result = cop(a, b)
  print(result)

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

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

Think step by step.

Response:
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 the code to fail. Let's analyze the buggy code to identify potential issues:

1. **GCD Calculation**: The code assumes that `gcd` is always `j`, which is incorrect. The GCD of `i` and `j` should be calculated correctly.
2. **Range of Loops**: The loops might not cover all valid pairs, especially when `a` and `b` are large.
3. **Edge Cases**: Small values of `a` and `b` can sometimes reveal bugs more clearly.

Given these points, we can create a test input that includes:
- Small values of `a` and `b` to ensure the GCD calculation is tested.
- Values where `a` and `b` are not multiples of each other to test the GCD calculation.
- A case where the GCD is not 1 to test the GCD calculation.

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

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

This input includes:
- The simplest case (1, 1) where only one pair (1, 1) is valid.
- A case where `a` and `b` are equal (2, 2) to test the GCD calculation.
- A case where `a` and `b` are not multiples of each other (3, 4) to test the GCD calculation.
- A slightly larger case (5, 6) to test the loop ranges and GCD calculation.

This test input should help expose the bug in the GCD calculation and loop ranges 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: Reverse Card (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $a+b$ is a multiple of $b \cdot \gcd(a,b)$.

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

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
1
3
4
14
153
1643498

Note: In the first test case, only $(1,1)$ satisfies the conditions.

In the fourth test case, $(1,1),(2,1),(2,2),(3,1),(4,1),(5,1),(6,1),(6,2),(6,3),(7,1),(8,1),(9,1),(10,1),(10,2)$ satisfy the conditions.

Buggy Code:
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int numTestCases = scanner.nextInt();
        for (int i = 0; i < numTestCases; i++) {
            long[] testCase = readTestCase(scanner);
            processTestCase(testCase);
        }
    }

    public static long[] readTestCase(Scanner scanner) {
        long numN = scanner.nextLong();
        long numM = scanner.nextLong();
        return new long[]{numN, numM};
    }

    public static void processTestCase(long[] testCase) {
        long numN = testCase[0];
        long numM = testCase[1];
        System.out.println(countValidPairs(numN, numM));
    }

    public static long countValidPairs(long numN, long numM) {
        long count = 0;
        for (long b = 1; b <= numM; ++b) {
            for (long aVal1 = 1; aVal1 <= numN / (b * findGreatestCommonDivisor(1, b)); ++aVal1) {
                long aVal = aVal1 * b * findGreatestCommonDivisor(1, b);
                long gcdAB = findGreatestCommonDivisor(aVal, b);
                if ((aVal % (b * gcdAB) + b) % (b * gcdAB) == 0) count++;
            }
        }
        return count;
    }

    public static long findGreatestCommonDivisor(long num1, long num2) {
        long minNum = Math.min(num1, num2);
        while (minNum > 0) {
            if (num1 % minNum == 0 && num2 % minNum == 0) break;
            minNum--;
        }
        return minNum;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases or scenarios that are likely to cause issues. The buggy code has a logical error in how it calculates the valid pairs. Specifically, the inner loop and the condition for checking valid pairs are incorrect.

Let's break down the problem and identify potential issues:

1. **Edge Cases:**
   - Small values of \( n \) and \( m \) (e.g., \( n = 1 \), \( m = 1 \)).
   - Large values of \( n \) and \( m \) (e.g., \( n = 2 \times 10^6 \), \( m = 2 \times 10^6 \)).
   - Values where \( n \) and \( m \) are prime numbers.
   - Values where \( n \) and \( m \) are powers of small integers (e.g., \( n = 100 \), \( m = 100 \)).

2. **Logical Error in the Code:**
   - The inner loop incorrectly uses `aVal1` and the condition for checking valid pairs is flawed.

Given these considerations, a small-scale test input that is complete and valid, and likely to expose the bug, could be:

```plaintext
6
1 1
2 3
3 5
10 8
100 100
200 200
```

This test input includes:
- A simple case with \( n = 1 \) and \( m = 1 \).
- A case with \( n = 2 \) and \( m = 3 \).
- A case with \( n = 3 \) and \( m = 5 \).
- A case with \( n = 10 \) and \( m = 8 \).
- A case with \( n = 100 \) and \( m = 100 \).
- A case with \( n = 200 \) and \( m = 200 \).

These cases cover a range of values and are small enough to be easily tested. The larger cases (100 and 200) are particularly useful for exposing any performance or logical issues in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $a+b$ is a multiple of $b \cdot \gcd(a,b)$.

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

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
1
3
4
14
153
1643498

Note: In the first test case, only $(1,1)$ satisfies the conditions.

In the fourth test case, $(1,1),(2,1),(2,2),(3,1),(4,1),(5,1),(6,1),(6,2),(6,3),(7,1),(8,1),(9,1),(10,1),(10,2)$ satisfy the conditions.

Buggy Code:
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int test = sc.nextInt();
        while(test-- > 0){
            long n = sc.nextLong();
            long m = sc.nextLong();
//            String str = sc.next();
//            long[] arr = new long[n];
//            for(int i=0; i<n; i++){
//                arr[i] = sc.nextLong();
//            }
            long ans = n;
            for(int i=2; i<=Math.sqrt(m)+1; i++){
                long mul = i*i;
                for(int j=1; mul*j-i<=n; j++){
                    long val = mul * j - i;
                    if(val <= n) ans++;
                    else break;
                }
            }
            System.out.println(ans);
        }
    }
}

**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
2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Reverse Card (Easy Version)
time_limit_ms: 2000
memory_limit_mb: 256
Description: The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

  * $1\le a\le n$, $1\le b\le m$;    * $a+b$ is a multiple of $b \cdot \gcd(a,b)$.

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

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output: For each test case, print a single integer: the number of valid pairs.

Examples:
input:
6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141
output:
1
3
4
14
153
1643498

Note: In the first test case, only $(1,1)$ satisfies the conditions.

In the fourth test case, $(1,1),(2,1),(2,2),(3,1),(4,1),(5,1),(6,1),(6,2),(6,3),(7,1),(8,1),(9,1),(10,1),(10,2)$ satisfy the conditions.

Buggy Code:
import java.util.*;

public class Main {

    static Scanner sc = new Scanner(System.in);

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

    private static void solve() {
        int a = sc.nextInt();
        int b = sc.nextInt();
        long sum = a;
        if (a < 4){
            if (b >= 2){
                System.out.println(sum+1);
            }else {
                System.out.println(sum);
            }
            return;
        }
        for (int i = 2; i*i <= a && i <= b; i++) {
            sum += (a+i)/(i*i);
        }
        System.out.println(sum);
    }
}

class Node{

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider the conditions under which the code might fail. The code has a few key areas where it might go wrong:

1. **Handling small values of `a` and `b`**: The code has a special case for `a < 4`, but it might not handle all edge cases correctly.
2. **Loop conditions**: The loop condition `i*i <= a && i <= b` might not cover all valid pairs.
3. **Sum calculation**: The way the sum is calculated in the loop might be incorrect.

Let's focus on a few specific cases to test these areas:

1. **Small values of `a` and `b`**: Test with `a = 3` and `b = 2` to see if the special case is handled correctly.
2. **Large values of `a` and `b`**: Test with `a = 10` and `b = 8` to see if the loop and sum calculation are correct.
3. **Boundary values**: Test with `a = 4` and `b = 4` to see if the transition from the special case to the general case is handled correctly.
4. **Equal values of `a` and `b`**: Test with `a = 5` and `b = 5` to see if the code handles equal values correctly.

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

```plaintext
4
3 2
10 8
4 4
5 5
```

This input includes:
- A small case with `a = 3` and `b = 2`.
- A larger case with `a = 10` and `b = 8`.
- A boundary case with `a = 4` and `b = 4`.
- A case with equal values `a = 5` and `b = 5`.

These cases should help expose any issues in the buggy code.