**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Perform Operations to Maximize Score
time_limit_ms: 3000
memory_limit_mb: 256
Description: I see satyam343. I'm shaking. Please more median problems this time. I love those. Please satyam343 we believe in you.

— satyam343's biggest fan

You are given an array $a$ of length $n$ and an integer $k$. You are also given a binary array $b$ of length $n$.

You can perform the following operation at most $k$ times:

  * Select an index $i$ ($1 \leq i \leq n$) such that $b_i = 1$. Set $a_i = a_i + 1$ (i.e., increase $a_i$ by $1$).

Your score is defined to be $\max\limits_{i = 1}^{n} \left( a_i + \operatorname{median}(c_i) \right)$, where $c_i$ denotes the array of length $n-1$ that you get by deleting $a_i$ from $a$. In other words, your score is the maximum value of $a_i + \operatorname{median}(c_i)$ over all $i$ from $1$ to $n$.

Find the maximum score that you can achieve if you perform the operations optimally.

For an arbitrary array $p$, $\operatorname{median}(p)$ is defined as the $\left\lfloor \frac{|p|+1}{2} \right\rfloor$-th smallest element of $p$. For example, $\operatorname{median} \left( [3,2,1,3] \right) = 2$ and $\operatorname{median} \left( [6,2,4,5,1] \right) = 4$.

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

Each test case begins with two integers $n$ and $k$ ($2 \leq n \leq 2 \cdot 10^5$, $0 \leq k \leq 10^9$) — the length of the $a$ and the number of operations you can perform.

The following line contains $n$ space separated integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — denoting the array $a$.

The following line contains $n$ space separated integers $b_1, b_2, \ldots, b_n$ ($b_i$ is $0$ or $1$) — denoting the array $b$.

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

Output: For each test case, output the maximum value of score you can get on a new line.

Examples:
input:
8
2 10
3 3
1 1
3 10
3 3 3
0 0 0
4 4
2 1 5 1
0 1 0 1
5 4
7 5 2 5 4
0 0 1 0 1
5 1
5 15 15 2 11
1 0 0 1 1
5 2
10 11 4 10 15
1 1 0 1 0
4 4
1 1 2 5
1 1 0 0
2 1000000000
1000000000 1000000000
1 1
output:
16
6
8
13
21
26
8
3000000000

Note: For the first test case, it is optimal to perform $5$ operations on both elements so $a = [8,8]$. So, the maximum score we can achieve is $\max(8 + \operatorname{median}[8], 8 + \operatorname{median}[8]) = 16$, as $c_1 = [a_2] = [8]$. It can be proven that you cannot get a better score.

For the second test case, you are not able to perform operations on any elements, so $a$ remains $[3,3,3]$. So, the maximum score we can achieve is $3 + \operatorname{median}[3, 3] = 6$, as $c_1 = [a_2, a_3] = [3, 3]$. It can be proven that you cannot get a better score.

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

public class Main {
    public static int INF = 0x3f3f3f3f, mod = 1000000007, mod9 = 998244353;
    public static void main(String args[]){
        try {
            PrintWriter o = new PrintWriter(System.out);
            boolean multiTest = true;
            // init
            if(multiTest) {
                int t = nextInt(), loop = 0;
                while (loop < t) {loop++;solve(o);}
            } else solve(o);
            o.close();
        } catch (Exception e) {e.printStackTrace();}
    }
    static void solve(PrintWriter o) {
        try {
            int n = nextInt(), k = nextInt();
            int[] a = new int[n];
            for(int i=0;i<n;i++) a[i] = nextInt();
            int[] b = new int[n];
            int[] mx = new int[2];
            int[] idx = new int[2];
            Arrays.fill(idx, -1);
            for(int i=0;i<n;i++) {
                b[i] = nextInt();
                if(b[i] == 0) {
                    if (a[i] > mx[0]) {
                        mx[0] = a[i];
                        idx[0] = i;
                    }
                }
                else if(a[i] > mx[1]) {
                    mx[1] = a[i];
                    idx[1] = i;
                }
            }
            long res = 0;
            if(idx[1] != -1) {
                long ans = 0l;
                List<Integer> li = new ArrayList<>();
                ans += a[idx[1]]+k;
                for(int i=0;i<n;i++) if(i != idx[1]) li.add(a[i]);
                Collections.sort(li, (x,y)->x-y);
                ans += li.get((li.size()-1)/2);
                res = Math.max(res, ans);
            }
            if(idx[0] != -1) {
                long ans = 0l;
                List<int[]> li = new ArrayList<>();
                ans += a[idx[0]];
                for(int i=0;i<n;i++) if(i != idx[0]) li.add(new int[]{a[i], b[i]});
                Collections.sort(li, (x,y)->y[0]-x[0]);
                int l = 0, r = (int)2e9;
                while(l < r) {
                    int m = (l+r)>>1;
                    if(check(li, m, k)) l = m+1;
                    else r = m;
                }
                ans += l-1;
                res = Math.max(res, ans);
            }
            o.println(res);
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
    static boolean check(List<int[]> li, int m, int k) {
        int cnt = 0;
        for(int[] it: li) {
            int d = it[0]-m;
            if(d >= 0) cnt++;
            else if(it[1] == 1 && k+d >= 0){
                k += d;
                cnt++;
            }
        }
        return cnt >= li.size()/2+1;
    }
    public static int upper_bound(List<Integer> li, int val){
        int l = 0, r = li.size();
        while(l < r){
            int mid = l + (r - l) / 2;
            if(li.get(mid) <= val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int upper_bound(int[] a, int val){
        int l = 0, r = a.length;
        while(l < r){
            int mid = l + (r - l) / 2;
            if(a[mid] <= val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int lower_bound(List<Integer> li, int val){
        int l = 0, r = li.size();
        while(l < r){
            int mid = l + (r - l) / 2;
            if(li.get(mid) < val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static int lower_bound(int[] a, int val){
        int l = 0, r = a.length;
        while(l < r){
            int mid = l + (r - l) / 2;
            if(a[mid] < val) l = mid + 1;
            else r = mid;
        }
        return l;
    }
    public static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a%b);
    }
    public static long[] extgcd(long a, long b) {
        if(b == 0) return new long[]{1, 0};
        long[] it = extgcd(b, a%b);
        long x = it[1], y = it[0];
        y -= a/b*x;
        return new long[]{x, y};
    }
    public static long lcm(long a, long b){
        return a / gcd(a,b)*b;
    }
    public static long qpow(long a, long n, int md){
        a %= md;
        long ret = 1l;
        while(n > 0){
            if((n & 1) == 1) ret = ret * a % md;
            n >>= 1;
            a = a * a % md;
        }
        return ret;
    }
    public static class FenWick {
        int n;
        long[] a;
        long[] tree;
        public FenWick(int n){
            this.n = n;
            a = new long[n+1];
            tree = new long[n+1];
        }
        private void add(int x, long val){
            while(x <= n){
                tree[x] += val;
                x += x&-x;
            }
        }
        private void addMx(int x, long val) {
            a[x] += val;
            tree[x] = a[x];
            while(x <= n) {
                for(int i=1;i<(x&-x);i<<=1) tree[x] = Math.max(tree[x], tree[x-i]);
                x += x&-x;
            }
        }
        private long query(int x){
            long ret = 0l;
            while(x > 0){
                ret += tree[x];
                x -= x&-x;
            }
            return ret;
        }
        private long queryMx(int l, int r) {
            long res = 0l;
            while(l <= r) {
                if(r-(r&-r) >= l) {
                    res = Math.max(res, tree[r]);
                    r -= r&-r;
                }
                else {
                    res = Math.max(res, a[r]);
                    r--;
                }
            }
            return res;
        }
    }
    static class DSU {
        int[] parent;
        public DSU(int n) {
            parent = new int[n];
            for(int i=0;i<n;i++) parent[i] = i;
        }
        public int find(int x) {
            if(parent[x] != x) parent[x] = find(parent[x]);
            return parent[x];
        }
        public void union(int x, int y) {
            int root_x = find(x);
            int root_y = find(y);
            if(root_x == root_y) return;
            parent[root_x] = root_y;
        }
    }
    private static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    private static StringTokenizer tokenizer = new StringTokenizer("");
    private static String next() throws IOException{
        while(!tokenizer.hasMoreTokens()){tokenizer = new StringTokenizer(reader.readLine());}
        return tokenizer.nextToken();
    }
    public static int nextInt() throws IOException {return Integer.parseInt(next());}
    public static Long nextLong() throws IOException {return Long.parseLong(next());}
    public static double nextDouble() throws IOException {return Double.parseDouble(next());}
    public static char nextChar() throws IOException {return next().toCharArray()[0];}
    public static String nextString() throws IOException {return next();}
    public static String nextLine() throws IOException {return reader.readLine();}
}

**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 and scenarios that might cause the code to fail. Here are some potential areas to test:

1. **All elements in `b` are 0**: This should prevent any operations from being performed, and the score should be calculated based on the original array `a`.
2. **All elements in `b` are 1**: This should allow all operations to be performed, and the score should be maximized.
3. **Mixed elements in `b`**: This should test the logic of selecting elements to perform operations on.
4. **Large values for `k`**: This should test the logic of maximizing the score when a large number of operations are allowed.
5. **Small values for `k`**: This should test the logic of selecting the best elements to perform operations on when only a few operations are allowed.

Let's create a small-scale test input that covers these scenarios:

```plaintext
4
3 0
1 2 3
0 0 0
3 1
1 2 3
1 0 0
3 3
1 2 3
1 1 1
3 2
1 2 3
1 0 1
```

Explanation:
1. **First Test Case**: All elements in `b` are 0, so no operations can be performed. The score should be calculated based on the original array `a`.
2. **Second Test Case**: Only the first element in `b` is 1, so only one operation can be performed on the first element of `a`.
3. **Third Test Case**: All elements in `b` are 1, so all operations can be performed. The score should be maximized.
4. **Fourth Test Case**: Mixed elements in `b`, with a small number of operations allowed. This tests the logic of selecting the best elements to perform operations on.

This test input should help expose any bugs 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: Perform Operations to Maximize Score
time_limit_ms: 3000
memory_limit_mb: 256
Description: I see satyam343. I'm shaking. Please more median problems this time. I love those. Please satyam343 we believe in you.

— satyam343's biggest fan

You are given an array $a$ of length $n$ and an integer $k$. You are also given a binary array $b$ of length $n$.

You can perform the following operation at most $k$ times:

  * Select an index $i$ ($1 \leq i \leq n$) such that $b_i = 1$. Set $a_i = a_i + 1$ (i.e., increase $a_i$ by $1$).

Your score is defined to be $\max\limits_{i = 1}^{n} \left( a_i + \operatorname{median}(c_i) \right)$, where $c_i$ denotes the array of length $n-1$ that you get by deleting $a_i$ from $a$. In other words, your score is the maximum value of $a_i + \operatorname{median}(c_i)$ over all $i$ from $1$ to $n$.

Find the maximum score that you can achieve if you perform the operations optimally.

For an arbitrary array $p$, $\operatorname{median}(p)$ is defined as the $\left\lfloor \frac{|p|+1}{2} \right\rfloor$-th smallest element of $p$. For example, $\operatorname{median} \left( [3,2,1,3] \right) = 2$ and $\operatorname{median} \left( [6,2,4,5,1] \right) = 4$.

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

Each test case begins with two integers $n$ and $k$ ($2 \leq n \leq 2 \cdot 10^5$, $0 \leq k \leq 10^9$) — the length of the $a$ and the number of operations you can perform.

The following line contains $n$ space separated integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — denoting the array $a$.

The following line contains $n$ space separated integers $b_1, b_2, \ldots, b_n$ ($b_i$ is $0$ or $1$) — denoting the array $b$.

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

Output: For each test case, output the maximum value of score you can get on a new line.

Examples:
input:
8
2 10
3 3
1 1
3 10
3 3 3
0 0 0
4 4
2 1 5 1
0 1 0 1
5 4
7 5 2 5 4
0 0 1 0 1
5 1
5 15 15 2 11
1 0 0 1 1
5 2
10 11 4 10 15
1 1 0 1 0
4 4
1 1 2 5
1 1 0 0
2 1000000000
1000000000 1000000000
1 1
output:
16
6
8
13
21
26
8
3000000000

Note: For the first test case, it is optimal to perform $5$ operations on both elements so $a = [8,8]$. So, the maximum score we can achieve is $\max(8 + \operatorname{median}[8], 8 + \operatorname{median}[8]) = 16$, as $c_1 = [a_2] = [8]$. It can be proven that you cannot get a better score.

For the second test case, you are not able to perform operations on any elements, so $a$ remains $[3,3,3]$. So, the maximum score we can achieve is $3 + \operatorname{median}[3, 3] = 6$, as $c_1 = [a_2, a_3] = [3, 3]$. It can be proven that you cannot get a better score.

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

public class Usaco {
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static StringTokenizer in;
	public static void main(String[] args) throws IOException {
		int t = shit();
		l:
		while (t-->0){
			int n = shit(), k = shit();
			int[] a = new int[n];
			Pair[] arr = new Pair[n];
			for (int i = 0; i < n; i++) {
				a[i]=shit();
			}

			for (int i = 0; i < n; i++) {
				arr[i]=new Pair(a[i], shit());
			}
			Arrays.sort(arr);

			long ans = 0;
			for (int i = 0; i < n; i++) {
				int idx = n/2-1;
				if (i <= idx) {
					idx++;
				}
				if (arr[i].b==1){
					ans = Math.max(ans, 1L*arr[i].a+k+arr[idx].a);
				}
			}
			if (arr[n-1].b==0) {

				int l = 0, r = (int)2e9;

				while (l<=r) {
					int m = (l+r)/2;
					int left = k;
					int cur = 0;
					for (int i = n-2; i >=0; i--) {
						if (arr[i].a >= m) cur++;
						else if (m-arr[i].a <= left && arr[i].b == 1) {

							left-=m-arr[i].a;
							cur++;
						}
					}

					if (cur >= (n+1)/2) {

						ans = Math.max(ans, 1L*arr[n-1].a+m);

						l = m+1;
					} else {
						r = m-1;
					}

				}
			}
			System.out.println(ans);

		}

	}

	static String next() throws IOException {
		while (in==null||!in.hasMoreTokens()) {
			in = new StringTokenizer(br.readLine());
		}
		return in.nextToken();
	}
	static int shit() throws IOException {
		return Integer.parseInt(next());
	}
	static long ll() throws IOException {
		return Long.parseLong(next());
	}
}

class Pair implements Comparable<Pair> {
  int a;
  int b;

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

  @Override
  public int compareTo(Pair other) {
	  if (a> other.a) return 1;
	  if (a < other.a) return -1;
	  return b-other.b;
  }

  public String toString() {
	  return a + " " + b;
  }

}

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

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

Think step by step.

Response:
```plaintext
1
5 3
1 2 3 4 5
0 1 0 1 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Variable Damage
time_limit_ms: 5000
memory_limit_mb: 512
Description: Monocarp is gathering an army to fight a dragon in a videogame.

The army consists of two parts: the heroes and the defensive artifacts. Each hero has one parameter — his health. Each defensive artifact also has one parameter — its durability.

Before the battle begins, Monocarp distributes artifacts to the heroes so that each hero receives at most one artifact.

The battle consists of rounds that proceed as follows:

  * first, the dragon deals damage equal to $\frac{1}{a + b}$ (a real number without rounding) to each hero, where $a$ is the number of heroes alive and $b$ is the number of active artifacts;    * after that, all heroes with health $0$ or less die;    * finally, some artifacts are deactivated. An artifact with durability $x$ is deactivated when one of the following occurs: the hero holding the artifact either dies or receives $x$ total damage (from the start of the battle). If an artifact is not held by any hero, it is inactive from the beginning of the battle.

The battle ends when there are no heroes left alive.

Initially, the army is empty. There are $q$ queries: add a hero with health $x$ or an artifact with durability $y$ to the army. After each query, determine the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Input: The first line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$) — the number of queries.

In the $i$-th of the following $q$ lines, there are two integers $t_i$ and $v_i$ ($t_i \in \\{1, 2\\}$; $1 \le v_i \le 10^9$) — the type of the query and the value of the query parameter. If the type is $1$, a hero with health $v_i$ is added. If the type is $2$, an artifact with durability $v_i$ is added.

Output: Print $q$ integers. After each query, output the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Examples:
input:
3
2 5
1 4
1 10
output:
0
8
19

input:
10
1 9
1 6
2 4
1 8
1 3
2 10
1 3
1 6
1 10
2 6
output:
9
15
19
27
30
39
42
48
59
65

Note: Let's consider the first example.

  * An artifact with durability $5$ is added. Since there are no heroes yet, the battle ends immediately.    * A hero with health $4$ is added. Monocarp can give him an artifact with durability $5$. First, there are rounds in which the hero takes $\frac{1}{1 + 1} = \frac{1}{2}$ damage. After $8$ such rounds, a total of $4$ damage will have been dealt, and the hero will die, while the artifact will deactivate. There are no more heroes alive, so the battle ends after $8$ rounds.    * A hero with health $10$ is added. Now let the artifact with durability $5$ be with this hero. Then, in the first $12$ rounds, the heroes will take $12 \cdot \frac{1}{2 + 1} = 4$ damage, and the first hero will die. The second hero has $6$ health left, and the artifact has $1$ durability. Now the damage is $\frac{1}{2}$, so after another $2$ rounds, the artifact will deactivate. The second hero has $5$ health left. After another $5$ rounds, the second hero will die. Therefore, the answer is $12 + 2 + 5 = 19$.

Buggy Code:
#include <iostream>
#include <vector>
#include <map>
#include <algorithm>
#include <math.h>

using namespace std;
const double EPSILON = 1e-9;
int main()
{
    int nTests;

    vector<int> artifacts;
    vector<int> heros;
    cin >> nTests;
    for (int n = 0; n < nTests; n++)
    {
        int item, health;
        cin >> item >> health;
        if (item == 2)
            artifacts.push_back(health);
        else
            heros.push_back(health);

        int nHeros = heros.size();
        sort(heros.begin(), heros.end(), [](int a, int b)
             { return b < a; });
        sort(artifacts.begin(), artifacts.end(), [](int a, int b)
             { return b < a; });
        int nArts = artifacts.size();
        if (nArts > nHeros)
        {
            nArts = nHeros;
        }

        map<int, int> equip;
        for (int i = 0; i < nArts; i++)
        {
            equip[heros[i]] = artifacts[i];
        }
        int rounds = 0;
        double damage = 0;
        vector<int> popHeros;
        while (nHeros > 0)
        {
            damage += 1 / (double)(nHeros + nArts);
            // cout << "Round: " << rounds << endl;
            for (int i = 0; i < heros.size(); i++)
            {
                if (heros[i] == 0)
                {
                    continue;
                }
                double temp = heros[i] - damage;
                if (temp <= 0)
                {
                    popHeros.push_back(heros[i]);
                    heros[i] = 0;
                    nHeros--;
                    if (equip.find(heros[i]) != equip.end())
                    {
                        if (equip[heros[i]] != 0)
                        {
                            nArts--;
                            equip[heros[i]] = 0;
                        }
                    }
                }
                if (equip.find(heros[i]) != equip.end())
                {
                    temp = equip[heros[i]] - damage;
                    if (equip[heros[i]] > 0 && temp <= 0)
                    {
                        nArts--;
                        equip[heros[i]] = 0;
                    }
                }
            }
            rounds++;
        }
        heros.assign(popHeros.begin(), popHeros.end());
        cout << rounds << endl;
    }
}

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

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

Think step by step.

Response:
```plaintext
5
1 1000000000
2 1
2 1
2 1
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Variable Damage
time_limit_ms: 5000
memory_limit_mb: 512
Description: Monocarp is gathering an army to fight a dragon in a videogame.

The army consists of two parts: the heroes and the defensive artifacts. Each hero has one parameter — his health. Each defensive artifact also has one parameter — its durability.

Before the battle begins, Monocarp distributes artifacts to the heroes so that each hero receives at most one artifact.

The battle consists of rounds that proceed as follows:

  * first, the dragon deals damage equal to $\frac{1}{a + b}$ (a real number without rounding) to each hero, where $a$ is the number of heroes alive and $b$ is the number of active artifacts;    * after that, all heroes with health $0$ or less die;    * finally, some artifacts are deactivated. An artifact with durability $x$ is deactivated when one of the following occurs: the hero holding the artifact either dies or receives $x$ total damage (from the start of the battle). If an artifact is not held by any hero, it is inactive from the beginning of the battle.

The battle ends when there are no heroes left alive.

Initially, the army is empty. There are $q$ queries: add a hero with health $x$ or an artifact with durability $y$ to the army. After each query, determine the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Input: The first line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$) — the number of queries.

In the $i$-th of the following $q$ lines, there are two integers $t_i$ and $v_i$ ($t_i \in \\{1, 2\\}$; $1 \le v_i \le 10^9$) — the type of the query and the value of the query parameter. If the type is $1$, a hero with health $v_i$ is added. If the type is $2$, an artifact with durability $v_i$ is added.

Output: Print $q$ integers. After each query, output the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Examples:
input:
3
2 5
1 4
1 10
output:
0
8
19

input:
10
1 9
1 6
2 4
1 8
1 3
2 10
1 3
1 6
1 10
2 6
output:
9
15
19
27
30
39
42
48
59
65

Note: Let's consider the first example.

  * An artifact with durability $5$ is added. Since there are no heroes yet, the battle ends immediately.    * A hero with health $4$ is added. Monocarp can give him an artifact with durability $5$. First, there are rounds in which the hero takes $\frac{1}{1 + 1} = \frac{1}{2}$ damage. After $8$ such rounds, a total of $4$ damage will have been dealt, and the hero will die, while the artifact will deactivate. There are no more heroes alive, so the battle ends after $8$ rounds.    * A hero with health $10$ is added. Now let the artifact with durability $5$ be with this hero. Then, in the first $12$ rounds, the heroes will take $12 \cdot \frac{1}{2 + 1} = 4$ damage, and the first hero will die. The second hero has $6$ health left, and the artifact has $1$ durability. Now the damage is $\frac{1}{2}$, so after another $2$ rounds, the artifact will deactivate. The second hero has $5$ health left. After another $5$ rounds, the second hero will die. Therefore, the answer is $12 + 2 + 5 = 19$.

Buggy Code:
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int MAXN=3e5+5;
int n,m,a[MAXN],op[MAXN],st[MAXN],w[MAXN];
ll ans[MAXN],f[MAXN<<1],g[MAXN<<1];
void build(int l,int r) {
	for(int j=l;j<=r;++j) {
		f[m-w[j]]+=st[j]-st[j-1];
		g[m-w[j]]+=1ll*(st[j]-st[j-1])*w[j];
	}
}
signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0),cout.tie(0);
	cin>>m;
	for(int i=1;i<=m;++i) cin>>op[i]>>a[i],st[++n]=a[i];
	sort(st+1,st+n+1),n=unique(st+1,st+n+1)-st-1;
	for(int i=1;i<=m;++i) a[i]=lower_bound(st+1,st+n+1,a[i])-st;
	for(int i=1;i<=m;++i) ans[i]=ans[i-1]+st[a[i]];
	for(int l=1,r,B=sqrt(n);l<=n;l=r+1) {
		memset(f,0,sizeof(f));
		memset(g,0,sizeof(g));
		r=min(n,l+B-1),build(l,r);
		ll cnt=f[m],sum=g[m];
		for(int i=1,tg=0;i<=m;++i) {
			int x=a[i],c=(op[i]==1?-1:1);
			if(x>=r) {
				if(c==1) ++tg,cnt+=f[m+tg],sum+=g[m+tg];
				else cnt-=f[m+tg],sum-=g[m+tg],--tg;
			} else if(x>=l) {
				cnt=0,sum=0;
				for(int j=l;j<=r;++j) f[m-w[j]]=g[m-w[j]]=0;
				for(int j=l;j<=r;++j) {
					w[j]+=tg+(j<=x?c:0);
					if(w[j]>=tg) {
						cnt+=st[j]-st[j-1];
						sum+=1ll*(st[j]-st[j-1])*w[j];
					}
				}
				tg=0,build(l,r);
			}
			ans[i]-=sum+tg*cnt;
		}
	}
	for(int i=1;i<=m;++i) cout<<ans[i]<<"\n";
	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Variable Damage
time_limit_ms: 5000
memory_limit_mb: 512
Description: Monocarp is gathering an army to fight a dragon in a videogame.

The army consists of two parts: the heroes and the defensive artifacts. Each hero has one parameter — his health. Each defensive artifact also has one parameter — its durability.

Before the battle begins, Monocarp distributes artifacts to the heroes so that each hero receives at most one artifact.

The battle consists of rounds that proceed as follows:

  * first, the dragon deals damage equal to $\frac{1}{a + b}$ (a real number without rounding) to each hero, where $a$ is the number of heroes alive and $b$ is the number of active artifacts;    * after that, all heroes with health $0$ or less die;    * finally, some artifacts are deactivated. An artifact with durability $x$ is deactivated when one of the following occurs: the hero holding the artifact either dies or receives $x$ total damage (from the start of the battle). If an artifact is not held by any hero, it is inactive from the beginning of the battle.

The battle ends when there are no heroes left alive.

Initially, the army is empty. There are $q$ queries: add a hero with health $x$ or an artifact with durability $y$ to the army. After each query, determine the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Input: The first line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$) — the number of queries.

In the $i$-th of the following $q$ lines, there are two integers $t_i$ and $v_i$ ($t_i \in \\{1, 2\\}$; $1 \le v_i \le 10^9$) — the type of the query and the value of the query parameter. If the type is $1$, a hero with health $v_i$ is added. If the type is $2$, an artifact with durability $v_i$ is added.

Output: Print $q$ integers. After each query, output the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Examples:
input:
3
2 5
1 4
1 10
output:
0
8
19

input:
10
1 9
1 6
2 4
1 8
1 3
2 10
1 3
1 6
1 10
2 6
output:
9
15
19
27
30
39
42
48
59
65

Note: Let's consider the first example.

  * An artifact with durability $5$ is added. Since there are no heroes yet, the battle ends immediately.    * A hero with health $4$ is added. Monocarp can give him an artifact with durability $5$. First, there are rounds in which the hero takes $\frac{1}{1 + 1} = \frac{1}{2}$ damage. After $8$ such rounds, a total of $4$ damage will have been dealt, and the hero will die, while the artifact will deactivate. There are no more heroes alive, so the battle ends after $8$ rounds.    * A hero with health $10$ is added. Now let the artifact with durability $5$ be with this hero. Then, in the first $12$ rounds, the heroes will take $12 \cdot \frac{1}{2 + 1} = 4$ damage, and the first hero will die. The second hero has $6$ health left, and the artifact has $1$ durability. Now the damage is $\frac{1}{2}$, so after another $2$ rounds, the artifact will deactivate. The second hero has $5$ health left. After another $5$ rounds, the second hero will die. Therefore, the answer is $12 + 2 + 5 = 19$.

Buggy Code:
/*
2024-10-15
q409257934
*/
#include<bits/stdc++.h>
using namespace std;

/**
 * 题意：每个英雄有个生命值，且每个英雄最多带1个神器，神器有个耐久度
 * 战斗每回合会
 * 1. 对每个英雄造成1/(a+b)伤害，其中a为存活的英雄，b为激活的神器
 * 2. 没血量的英雄死亡
 * 3. 一些神器失效。拥有神器的英雄死亡或者英雄受到的伤害达到神器耐久度
 *
 * 初始军队为0，每次查询会加一个英雄或者神器
 * 需要输出每个查询后军队最多能存活多少回合
 *
 *
 *
 * 思路：
 * 可以每1单位的生命考虑，大于等于这个生命的英雄个数a,大于等于这个耐久度的神器个数b，
 * 贡献的回合数就是a+min(a,b),即2*a+min(0,b-a),也可以a+b+min(a-b,0)
 * 这里我们用a+b+min(a-b,0)
 * a和b的总和很容易，现在要考虑的是min(a-b,0)
 * 可以考虑按v排序，每sqrt(n)个为一块进行处理，可以知道这一块中a-b的最大最小值相差最大为sqn-1，所以对每一块的更新复杂度为O(sqn)
 * 总复杂度O(n*sqn)
 */
void solve() {
  int n;
  cin >> n;
  vector<array<int, 2>> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i][0] >> a[i][1];
  }
  vector<int> ord(n);
  iota(ord.begin(), ord.end(), 0);
  sort(ord.begin(), ord.end(), [&](int x, int y) {
    return a[x][1] < a[y][1];
  });
  vector<int> pos(n);
  for (int i = 0; i < n; ++i) {
    pos[ord[i]] = i;
  }
  int B = sqrt(n);
  int bnb = (n + B - 1) / B;
  vector<int> len(n), blen(bnb);
  blen[0] = len[0] = a[ord[0]][1];
  for (int i = 1; i < n; ++i) {
    len[i] = a[ord[i]][1] - a[ord[i - 1]][1];
    blen[i / B] += len[i];
  }
  vector<int> mi(bnb), lazy(bnb), w(n);
  vector<vector<int64_t>> f(bnb, vector<int64_t>(B));
  for (int i = 0; i < bnb; ++i) {
    for (int j = 0; j < B; ++j) {
      f[i][j] = 1LL * (j - B) * blen[i];
    }
  }
  int64_t ans = 0;
  auto upans = [&](int bid, int o) {
    if (mi[bid] >= 0) return;
    int c = B + mi[bid];
    if (c < 0) {
      ans += o * c * blen[bid];
      c = 0;
    }
    ans += o * f[bid][c];
  };
  auto updateblock = [&](int bid) {
    int bl = B * bid, br = min((B + 1) * bid, n);
    mi[bid] = n;
    for (int i = bl; i < br; ++i) {
      w[i] += lazy[bid];
      mi[bid] = min(mi[bid], w[i]);
    }
    lazy[bid] = 0;
    auto &dp = f[bid];
    dp.assign(B, 0);
    for (int i = bl; i < br; ++i) {
      int c = B - w[i] + mi[bid] - 1;
      if (c < 0) continue;
      dp[c] -= len[i];
    }
    for (int i = B - 2; i >= 0; --i) {
      dp[i] += dp[i + 1];
    }
    for (int i = B - 2; i >= 0; --i) {
      dp[i] += dp[i + 1];
    }
  };
  for (int i = 0; i < n; ++i) {
    ans += a[i][1];
    int p = pos[i];
    int bid = p / B;
    int ad = a[i][0] == 1 ? 1 : -1;
    for (int j = 0; j < bid; ++j) {
      upans(j, -1);
      lazy[j] += ad;
      mi[j] += ad;
      upans(j, 1);
    }
    upans(bid, -1);
    for (int j = B * bid; j <= p; ++j) {
      w[j] += ad;
    }
    updateblock(bid);
    upans(bid, 1);
    cout << ans << "\n";
  }
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int T = 1;
  // cin >> T;
  while (T--) solve();
  return 0;
}

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

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

Think step by step.

Response:
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 maximize the failure probability. Here are the steps to consider:

1. **Small Scale**: The test input should be small enough to fit the output limit but still be effective in exposing the bug.
2. **Edge Cases**: Consider scenarios where the number of heroes and artifacts are equal, or where the health and durability values are extreme.
3. **Complex Interactions**: Create a scenario where the distribution of artifacts and the sequence of queries can lead to unexpected behavior.

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

```plaintext
6
1 1
2 1
1 2
2 2
1 3
2 3
```

### Explanation:
1. **Initial State**: No heroes or artifacts.
2. **Query 1**: Add a hero with health 1.
   - Output: 0 (No artifacts, so the hero dies immediately).
3. **Query 2**: Add an artifact with durability 1.
   - Output: 0 (Still no heroes to hold the artifact).
4. **Query 3**: Add a hero with health 2.
   - Output: 2 (The hero can survive for 2 rounds with the artifact).
5. **Query 4**: Add an artifact with durability 2.
   - Output: 3 (The hero with health 2 can now survive for 3 rounds with the new artifact).
6. **Query 5**: Add a hero with health 3.
   - Output: 5 (The hero with health 3 can survive for 5 rounds if the artifacts are distributed optimally).
7. **Query 6**: Add an artifact with durability 3.
   - Output: 6 (The hero with health 3 can now survive for 6 rounds with the new artifact).

This test input is designed to:
- Test the initial state with no heroes or artifacts.
- Test the addition of a single hero and a single artifact.
- Test the addition of multiple heroes and artifacts in a sequence that requires optimal distribution.
- Ensure the code handles the dynamic addition and optimal distribution of artifacts correctly.

This should help expose any bugs in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Variable Damage
time_limit_ms: 5000
memory_limit_mb: 512
Description: Monocarp is gathering an army to fight a dragon in a videogame.

The army consists of two parts: the heroes and the defensive artifacts. Each hero has one parameter — his health. Each defensive artifact also has one parameter — its durability.

Before the battle begins, Monocarp distributes artifacts to the heroes so that each hero receives at most one artifact.

The battle consists of rounds that proceed as follows:

  * first, the dragon deals damage equal to $\frac{1}{a + b}$ (a real number without rounding) to each hero, where $a$ is the number of heroes alive and $b$ is the number of active artifacts;    * after that, all heroes with health $0$ or less die;    * finally, some artifacts are deactivated. An artifact with durability $x$ is deactivated when one of the following occurs: the hero holding the artifact either dies or receives $x$ total damage (from the start of the battle). If an artifact is not held by any hero, it is inactive from the beginning of the battle.

The battle ends when there are no heroes left alive.

Initially, the army is empty. There are $q$ queries: add a hero with health $x$ or an artifact with durability $y$ to the army. After each query, determine the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Input: The first line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$) — the number of queries.

In the $i$-th of the following $q$ lines, there are two integers $t_i$ and $v_i$ ($t_i \in \\{1, 2\\}$; $1 \le v_i \le 10^9$) — the type of the query and the value of the query parameter. If the type is $1$, a hero with health $v_i$ is added. If the type is $2$, an artifact with durability $v_i$ is added.

Output: Print $q$ integers. After each query, output the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Examples:
input:
3
2 5
1 4
1 10
output:
0
8
19

input:
10
1 9
1 6
2 4
1 8
1 3
2 10
1 3
1 6
1 10
2 6
output:
9
15
19
27
30
39
42
48
59
65

Note: Let's consider the first example.

  * An artifact with durability $5$ is added. Since there are no heroes yet, the battle ends immediately.    * A hero with health $4$ is added. Monocarp can give him an artifact with durability $5$. First, there are rounds in which the hero takes $\frac{1}{1 + 1} = \frac{1}{2}$ damage. After $8$ such rounds, a total of $4$ damage will have been dealt, and the hero will die, while the artifact will deactivate. There are no more heroes alive, so the battle ends after $8$ rounds.    * A hero with health $10$ is added. Now let the artifact with durability $5$ be with this hero. Then, in the first $12$ rounds, the heroes will take $12 \cdot \frac{1}{2 + 1} = 4$ damage, and the first hero will die. The second hero has $6$ health left, and the artifact has $1$ durability. Now the damage is $\frac{1}{2}$, so after another $2$ rounds, the artifact will deactivate. The second hero has $5$ health left. After another $5$ rounds, the second hero will die. Therefore, the answer is $12 + 2 + 5 = 19$.

Buggy Code:
l=[]
art=[]
hero=[]
def damagetaken(hero,art):
    la=len(art)
    lh=len(hero)
    if la!=0 and lh!=0:
        return 1/(la+lh)

    if lh==0:
        if la==0:
            return 0
        else:
            return 1/la
    else:
        return 1/lh

for _ in range(int(input())):
    r=0
    add=list(map(int, input().split()))
    if add[0]==1:
        hero.append(add)
    else:
        art.append(add)
    nhero=[x for x in hero]
    nart=[x for x in art]
    while(len(nhero)>0):
        r+=1
        dt=damagetaken(nhero,nart)
        i=0
        while(i<len(nhero)):
            nhero[i]=[nhero[i][0],nhero[i][1]-dt]
            if nhero[i][1]<=0:
                nhero.remove(nhero[i])
                i-=1
            i+=1
        i=0
        while(i<len(nart)):
            nart[i]=[nart[i][0],nart[i][1]-dt]
            if int(nart[i][1])<=0:
                nart.remove(nart[i])
                i-=1
            i+=1
        if not nhero :
            break
    print(r)

**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 1
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Variable Damage
time_limit_ms: 5000
memory_limit_mb: 512
Description: Monocarp is gathering an army to fight a dragon in a videogame.

The army consists of two parts: the heroes and the defensive artifacts. Each hero has one parameter — his health. Each defensive artifact also has one parameter — its durability.

Before the battle begins, Monocarp distributes artifacts to the heroes so that each hero receives at most one artifact.

The battle consists of rounds that proceed as follows:

  * first, the dragon deals damage equal to $\frac{1}{a + b}$ (a real number without rounding) to each hero, where $a$ is the number of heroes alive and $b$ is the number of active artifacts;    * after that, all heroes with health $0$ or less die;    * finally, some artifacts are deactivated. An artifact with durability $x$ is deactivated when one of the following occurs: the hero holding the artifact either dies or receives $x$ total damage (from the start of the battle). If an artifact is not held by any hero, it is inactive from the beginning of the battle.

The battle ends when there are no heroes left alive.

Initially, the army is empty. There are $q$ queries: add a hero with health $x$ or an artifact with durability $y$ to the army. After each query, determine the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Input: The first line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$) — the number of queries.

In the $i$-th of the following $q$ lines, there are two integers $t_i$ and $v_i$ ($t_i \in \\{1, 2\\}$; $1 \le v_i \le 10^9$) — the type of the query and the value of the query parameter. If the type is $1$, a hero with health $v_i$ is added. If the type is $2$, an artifact with durability $v_i$ is added.

Output: Print $q$ integers. After each query, output the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Examples:
input:
3
2 5
1 4
1 10
output:
0
8
19

input:
10
1 9
1 6
2 4
1 8
1 3
2 10
1 3
1 6
1 10
2 6
output:
9
15
19
27
30
39
42
48
59
65

Note: Let's consider the first example.

  * An artifact with durability $5$ is added. Since there are no heroes yet, the battle ends immediately.    * A hero with health $4$ is added. Monocarp can give him an artifact with durability $5$. First, there are rounds in which the hero takes $\frac{1}{1 + 1} = \frac{1}{2}$ damage. After $8$ such rounds, a total of $4$ damage will have been dealt, and the hero will die, while the artifact will deactivate. There are no more heroes alive, so the battle ends after $8$ rounds.    * A hero with health $10$ is added. Now let the artifact with durability $5$ be with this hero. Then, in the first $12$ rounds, the heroes will take $12 \cdot \frac{1}{2 + 1} = 4$ damage, and the first hero will die. The second hero has $6$ health left, and the artifact has $1$ durability. Now the damage is $\frac{1}{2}$, so after another $2$ rounds, the artifact will deactivate. The second hero has $5$ health left. After another $5$ rounds, the second hero will die. Therefore, the answer is $12 + 2 + 5 = 19$.

Buggy Code:
def max_rounds(heroes, artifacts):
    # Sort heroes and artifacts in descending order
    heroes.sort(reverse=True)
    artifacts.sort(reverse=True)

    n = len(heroes)
    m = len(artifacts)

    # Initialize the total rounds
    total_rounds = 0

    # Pair heroes with artifacts
    paired_heroes = []
    for i in range(min(n, m)):
        paired_heroes.append((heroes[i], artifacts[i]))

    # Remaining unpaired heroes
    remaining_heroes = heroes[min(n, m):]

    # Simulate the battle
    while paired_heroes or remaining_heroes:
        a = len(paired_heroes) + len(remaining_heroes)
        b = min(len(paired_heroes), len(artifacts))

        damage = 1 / (a + b)

        new_paired_heroes = []
        for h, d in paired_heroes:
            h -= damage
            d -= damage
            if h > 0 and d > 0:
                new_paired_heroes.append((h, d))
            elif h > 0:
                remaining_heroes.append(h)

        paired_heroes = new_paired_heroes

        remaining_heroes = [h - damage for h in remaining_heroes if h - damage > 0]

        total_rounds += 1

    return total_rounds

# Read input
import sys
input = sys.stdin.read
data = input().split()

q = int(data[0])
queries = [(int(data[2*i+1]), int(data[2*i+2])) for i in range(q)]

heroes = []
artifacts = []

results = []

for t, v in queries:
    if t == 1:
        heroes.append(v)
    else:
        artifacts.append(v)

    results.append(max_rounds(heroes.copy(), artifacts.copy()))

# Print results
print("\n".join(map(str, results)))

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

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

Think step by step.

Response:
```plaintext
5
1 1
1 1
2 1
1 1
2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Variable Damage
time_limit_ms: 5000
memory_limit_mb: 512
Description: Monocarp is gathering an army to fight a dragon in a videogame.

The army consists of two parts: the heroes and the defensive artifacts. Each hero has one parameter — his health. Each defensive artifact also has one parameter — its durability.

Before the battle begins, Monocarp distributes artifacts to the heroes so that each hero receives at most one artifact.

The battle consists of rounds that proceed as follows:

  * first, the dragon deals damage equal to $\frac{1}{a + b}$ (a real number without rounding) to each hero, where $a$ is the number of heroes alive and $b$ is the number of active artifacts;    * after that, all heroes with health $0$ or less die;    * finally, some artifacts are deactivated. An artifact with durability $x$ is deactivated when one of the following occurs: the hero holding the artifact either dies or receives $x$ total damage (from the start of the battle). If an artifact is not held by any hero, it is inactive from the beginning of the battle.

The battle ends when there are no heroes left alive.

Initially, the army is empty. There are $q$ queries: add a hero with health $x$ or an artifact with durability $y$ to the army. After each query, determine the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Input: The first line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$) — the number of queries.

In the $i$-th of the following $q$ lines, there are two integers $t_i$ and $v_i$ ($t_i \in \\{1, 2\\}$; $1 \le v_i \le 10^9$) — the type of the query and the value of the query parameter. If the type is $1$, a hero with health $v_i$ is added. If the type is $2$, an artifact with durability $v_i$ is added.

Output: Print $q$ integers. After each query, output the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Examples:
input:
3
2 5
1 4
1 10
output:
0
8
19

input:
10
1 9
1 6
2 4
1 8
1 3
2 10
1 3
1 6
1 10
2 6
output:
9
15
19
27
30
39
42
48
59
65

Note: Let's consider the first example.

  * An artifact with durability $5$ is added. Since there are no heroes yet, the battle ends immediately.    * A hero with health $4$ is added. Monocarp can give him an artifact with durability $5$. First, there are rounds in which the hero takes $\frac{1}{1 + 1} = \frac{1}{2}$ damage. After $8$ such rounds, a total of $4$ damage will have been dealt, and the hero will die, while the artifact will deactivate. There are no more heroes alive, so the battle ends after $8$ rounds.    * A hero with health $10$ is added. Now let the artifact with durability $5$ be with this hero. Then, in the first $12$ rounds, the heroes will take $12 \cdot \frac{1}{2 + 1} = 4$ damage, and the first hero will die. The second hero has $6$ health left, and the artifact has $1$ durability. Now the damage is $\frac{1}{2}$, so after another $2$ rounds, the artifact will deactivate. The second hero has $5$ health left. After another $5$ rounds, the second hero will die. Therefore, the answer is $12 + 2 + 5 = 19$.

Buggy Code:
q=int(input())
aux1=[]
aux2=[]
lr=[]
for aaaa in range(q):
    rounds=0
    typ,value=map(int,input().split())
    if(typ==1):
        aux1.append(value)
    else:
        aux2.append(value)
    heroes=[x for x in aux1]
    artifacts=[x for x in aux2]
    while(len(heroes)>0):
        if(len(heroes)<=len(artifacts)):
            dmg=1/(len(heroes)*2)
        else:
            dmg=1/(len(heroes)+len(artifacts))
        i = 0
        while i < len(heroes):
            heroes[i] -= dmg
            if heroes[i] <= 0:
                heroes.remove(heroes[i])
            else:
                i += 1

        i = 0
        while i < min(len(heroes), len(artifacts)):
            artifacts[i] -= dmg
            if artifacts[i] <= 0:
                artifacts.remove(artifacts[i])
            else:
                i += 1
        rounds=rounds+1
    if(rounds<15):
        lr.append(rounds)
    else:
        lr.append(rounds-1)
for i in lr:
    print(i)

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Variable Damage
time_limit_ms: 5000
memory_limit_mb: 512
Description: Monocarp is gathering an army to fight a dragon in a videogame.

The army consists of two parts: the heroes and the defensive artifacts. Each hero has one parameter — his health. Each defensive artifact also has one parameter — its durability.

Before the battle begins, Monocarp distributes artifacts to the heroes so that each hero receives at most one artifact.

The battle consists of rounds that proceed as follows:

  * first, the dragon deals damage equal to $\frac{1}{a + b}$ (a real number without rounding) to each hero, where $a$ is the number of heroes alive and $b$ is the number of active artifacts;    * after that, all heroes with health $0$ or less die;    * finally, some artifacts are deactivated. An artifact with durability $x$ is deactivated when one of the following occurs: the hero holding the artifact either dies or receives $x$ total damage (from the start of the battle). If an artifact is not held by any hero, it is inactive from the beginning of the battle.

The battle ends when there are no heroes left alive.

Initially, the army is empty. There are $q$ queries: add a hero with health $x$ or an artifact with durability $y$ to the army. After each query, determine the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Input: The first line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$) — the number of queries.

In the $i$-th of the following $q$ lines, there are two integers $t_i$ and $v_i$ ($t_i \in \\{1, 2\\}$; $1 \le v_i \le 10^9$) — the type of the query and the value of the query parameter. If the type is $1$, a hero with health $v_i$ is added. If the type is $2$, an artifact with durability $v_i$ is added.

Output: Print $q$ integers. After each query, output the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Examples:
input:
3
2 5
1 4
1 10
output:
0
8
19

input:
10
1 9
1 6
2 4
1 8
1 3
2 10
1 3
1 6
1 10
2 6
output:
9
15
19
27
30
39
42
48
59
65

Note: Let's consider the first example.

  * An artifact with durability $5$ is added. Since there are no heroes yet, the battle ends immediately.    * A hero with health $4$ is added. Monocarp can give him an artifact with durability $5$. First, there are rounds in which the hero takes $\frac{1}{1 + 1} = \frac{1}{2}$ damage. After $8$ such rounds, a total of $4$ damage will have been dealt, and the hero will die, while the artifact will deactivate. There are no more heroes alive, so the battle ends after $8$ rounds.    * A hero with health $10$ is added. Now let the artifact with durability $5$ be with this hero. Then, in the first $12$ rounds, the heroes will take $12 \cdot \frac{1}{2 + 1} = 4$ damage, and the first hero will die. The second hero has $6$ health left, and the artifact has $1$ durability. Now the damage is $\frac{1}{2}$, so after another $2$ rounds, the artifact will deactivate. The second hero has $5$ health left. After another $5$ rounds, the second hero will die. Therefore, the answer is $12 + 2 + 5 = 19$.

Buggy Code:
import java.util.Scanner;

public class VariableDamage {
  public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int n = in.nextInt();
    int sum = 0;
    boolean firstType = false;
    for (int i = 0; i < n; i++) {
      int type = in.nextInt();
      int val = in.nextInt();
      if (i == 0 && type == 2) {
        System.out.println(0);
        sum += val - 1;
        firstType = true;
      } else {
        sum += val;
        if (firstType && i == n - 1) {
          sum++;
        }
        System.out.println(sum);
      }
    }
  }
}

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

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

Think step by step.

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

This test input is designed to expose the bug in the given buggy code. Here's the reasoning:

1. **First Query (2 5):** Adds an artifact with durability 5.
   - The buggy code incorrectly prints `0` and updates `sum` to `4` (5 - 1).
   - The correct output should be `0` because there are no heroes yet, and the battle ends immediately.

2. **Second Query (1 4):** Adds a hero with health 4.
   - The buggy code incorrectly adds `4` to `sum`, making `sum` equal to `8`.
   - The correct output should be `8` because the hero with health 4, when given the artifact with durability 5, can survive for 8 rounds (4 * 2 = 8).

3. **Third Query (1 10):** Adds another hero with health 10.
   - The buggy code incorrectly adds `10` to `sum`, making `sum` equal to `18`.
   - The correct output should be `19` because the hero with health 10, when given the artifact with durability 5, can survive for 12 rounds (10 - 4 = 6, 6 / (1/3) = 18, 18 - 6 = 12), and the remaining 7 rounds (5 - 1 = 4, 4 / (1/2) = 8, 8 - 3 = 5) for the second hero.

This test input highlights the issues with the buggy code's handling of the initial state and the accumulation of values in `sum`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Variable Damage
time_limit_ms: 5000
memory_limit_mb: 512
Description: Monocarp is gathering an army to fight a dragon in a videogame.

The army consists of two parts: the heroes and the defensive artifacts. Each hero has one parameter — his health. Each defensive artifact also has one parameter — its durability.

Before the battle begins, Monocarp distributes artifacts to the heroes so that each hero receives at most one artifact.

The battle consists of rounds that proceed as follows:

  * first, the dragon deals damage equal to $\frac{1}{a + b}$ (a real number without rounding) to each hero, where $a$ is the number of heroes alive and $b$ is the number of active artifacts;    * after that, all heroes with health $0$ or less die;    * finally, some artifacts are deactivated. An artifact with durability $x$ is deactivated when one of the following occurs: the hero holding the artifact either dies or receives $x$ total damage (from the start of the battle). If an artifact is not held by any hero, it is inactive from the beginning of the battle.

The battle ends when there are no heroes left alive.

Initially, the army is empty. There are $q$ queries: add a hero with health $x$ or an artifact with durability $y$ to the army. After each query, determine the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Input: The first line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$) — the number of queries.

In the $i$-th of the following $q$ lines, there are two integers $t_i$ and $v_i$ ($t_i \in \\{1, 2\\}$; $1 \le v_i \le 10^9$) — the type of the query and the value of the query parameter. If the type is $1$, a hero with health $v_i$ is added. If the type is $2$, an artifact with durability $v_i$ is added.

Output: Print $q$ integers. After each query, output the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Examples:
input:
3
2 5
1 4
1 10
output:
0
8
19

input:
10
1 9
1 6
2 4
1 8
1 3
2 10
1 3
1 6
1 10
2 6
output:
9
15
19
27
30
39
42
48
59
65

Note: Let's consider the first example.

  * An artifact with durability $5$ is added. Since there are no heroes yet, the battle ends immediately.    * A hero with health $4$ is added. Monocarp can give him an artifact with durability $5$. First, there are rounds in which the hero takes $\frac{1}{1 + 1} = \frac{1}{2}$ damage. After $8$ such rounds, a total of $4$ damage will have been dealt, and the hero will die, while the artifact will deactivate. There are no more heroes alive, so the battle ends after $8$ rounds.    * A hero with health $10$ is added. Now let the artifact with durability $5$ be with this hero. Then, in the first $12$ rounds, the heroes will take $12 \cdot \frac{1}{2 + 1} = 4$ damage, and the first hero will die. The second hero has $6$ health left, and the artifact has $1$ durability. Now the damage is $\frac{1}{2}$, so after another $2$ rounds, the artifact will deactivate. The second hero has $5$ health left. After another $5$ rounds, the second hero will die. Therefore, the answer is $12 + 2 + 5 = 19$.

Buggy Code:
import java.util.Arrays;

public class JavaApplication4 {

       static double [] heros ;
       static int herosNumber;
       static double [] artifacts ;
       static int artifactsNumber;
       static int [] result;
       static int resultNumber=0;
       static int rounds;

    public static void startRound (double [] heros,int herosNumber,double [] artifacts,int artifactsNumber){
        rounds=0;
        if(herosNumber==0){
            result[resultNumber]=rounds;

        }else{
            while(true){
                if(herosNumber==0){
                    result[resultNumber]=rounds;

                    break;}
                //if(artifactsNumber>0){System.out.print(artifacts[artifactsNumber-1]);}
                //System.out.print(" ; herosNumber "+ herosNumber+" ; artifactsNumber "+artifactsNumber +" ;;;  "+ rounds);

                double damage= calculateDamege(herosNumber,artifactsNumber);
                //System.out.println("  ;; damage " +damage);
                rounds++;
                if(herosNumber>1){

                for(int c =0;c<herosNumber;c++){
                    heros[c]     -= damage;
                }
                for(int i =herosNumber-1;i>=0;i--){
                    if(heros[i]<=0.0000000000001){
                        if(i==herosNumber-1){herosNumber--;}else{
                        for(int j=i; j<herosNumber-1;j++){
                            heros[j]=heros[j+1];
                        }
                        herosNumber--;
                        }
                    }
                }
                }else{
                    heros[0]-= damage;
                    if(heros[0]<=0.0000000000001){
                    herosNumber--;
                    }
                }
                if(artifactsNumber>1&&artifactsNumber<=herosNumber){
                    for(int i =0;i<artifactsNumber;i++){
                        artifacts[i] -=damage;
                    }
                    for(int i =artifactsNumber-1;i>=0;i--){
                    if(artifacts[i]<=0.0000000000001){
                        if(i==artifactsNumber-1){artifactsNumber--;}else{
                        for(int j=i; j<artifactsNumber-1;j++){
                            artifacts[j]=artifacts[j+1];
                        }
                        artifactsNumber--;
                        }
                    }
                }
                }else if(artifactsNumber==1) {
                    if(artifacts[0]!=0){
                    artifacts[0] -= damage;
                    if(artifacts[0]<=0.0000000000001){
                        artifactsNumber--;
                    }}
                }else if(artifactsNumber>herosNumber){
                    for(int k = 0;k<herosNumber;k++){
                        artifacts[k] -=damage;
                    }
                    for(int i =artifactsNumber-1;i>=0;i--){
                    if(artifacts[i]<=0.0000000000001){
                        if(i==artifactsNumber-1){artifactsNumber--;}else{
                        for(int j=i; j<artifactsNumber-1;j++){
                            artifacts[j]=artifacts[j+1];
                        }
                        artifactsNumber--;
                        }
                    }
                }
                }
            }
        }
    }

    public static double calculateDamege(double herosNumber,double artifactsNumber){
        return (1.0/(herosNumber+artifactsNumber));
    }

    public static void main(String[] args){

        java.util.Scanner s = new java.util.Scanner(System.in);
        int q =s.nextInt();
        heros  = new double[q];
        artifacts = new double[q];
        result =new int[q];
        for(int i=0;i<q;i++){
            int r = s.nextInt();
            int value = s.nextInt();
            if(r==1){
                heros[herosNumber++]=value;
            }else{
                artifacts[artifactsNumber++]=value;
            }

            double[] herosCopy=Arrays.copyOf(heros, herosNumber);
            double[] artifactsCopy=Arrays.copyOf(artifacts, artifactsNumber);
            startRound(herosCopy,herosNumber,artifactsCopy,artifactsNumber);

            //System.out.println(result[resultNumber]);
            resultNumber++;
        }
        for(int r:result){
            System.out.println(r);
        }
    }
}

**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 1
1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Variable Damage
time_limit_ms: 5000
memory_limit_mb: 512
Description: Monocarp is gathering an army to fight a dragon in a videogame.

The army consists of two parts: the heroes and the defensive artifacts. Each hero has one parameter — his health. Each defensive artifact also has one parameter — its durability.

Before the battle begins, Monocarp distributes artifacts to the heroes so that each hero receives at most one artifact.

The battle consists of rounds that proceed as follows:

  * first, the dragon deals damage equal to $\frac{1}{a + b}$ (a real number without rounding) to each hero, where $a$ is the number of heroes alive and $b$ is the number of active artifacts;    * after that, all heroes with health $0$ or less die;    * finally, some artifacts are deactivated. An artifact with durability $x$ is deactivated when one of the following occurs: the hero holding the artifact either dies or receives $x$ total damage (from the start of the battle). If an artifact is not held by any hero, it is inactive from the beginning of the battle.

The battle ends when there are no heroes left alive.

Initially, the army is empty. There are $q$ queries: add a hero with health $x$ or an artifact with durability $y$ to the army. After each query, determine the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Input: The first line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$) — the number of queries.

In the $i$-th of the following $q$ lines, there are two integers $t_i$ and $v_i$ ($t_i \in \\{1, 2\\}$; $1 \le v_i \le 10^9$) — the type of the query and the value of the query parameter. If the type is $1$, a hero with health $v_i$ is added. If the type is $2$, an artifact with durability $v_i$ is added.

Output: Print $q$ integers. After each query, output the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Examples:
input:
3
2 5
1 4
1 10
output:
0
8
19

input:
10
1 9
1 6
2 4
1 8
1 3
2 10
1 3
1 6
1 10
2 6
output:
9
15
19
27
30
39
42
48
59
65

Note: Let's consider the first example.

  * An artifact with durability $5$ is added. Since there are no heroes yet, the battle ends immediately.    * A hero with health $4$ is added. Monocarp can give him an artifact with durability $5$. First, there are rounds in which the hero takes $\frac{1}{1 + 1} = \frac{1}{2}$ damage. After $8$ such rounds, a total of $4$ damage will have been dealt, and the hero will die, while the artifact will deactivate. There are no more heroes alive, so the battle ends after $8$ rounds.    * A hero with health $10$ is added. Now let the artifact with durability $5$ be with this hero. Then, in the first $12$ rounds, the heroes will take $12 \cdot \frac{1}{2 + 1} = 4$ damage, and the first hero will die. The second hero has $6$ health left, and the artifact has $1$ durability. Now the damage is $\frac{1}{2}$, so after another $2$ rounds, the artifact will deactivate. The second hero has $5$ health left. After another $5$ rounds, the second hero will die. Therefore, the answer is $12 + 2 + 5 = 19$.

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

#define ll long long

const int N = 300005;
const int B = 300;

pair<int, int> tmp[N];
int rk[N];
int _opt[N], _v[N], q;

ll ans[N];

// buc1: 人(+1), buc1[suf[*]] += health
// buc2: 器(-1), buc2[suf[*]] += damage
ll pool_buc[N << 2], *buc1 = pool_buc + N, *buc2 = pool_buc + 3 * N;
int suf[N], lazy;
void solve(int L, int R) {
    memset(pool_buc, 0, sizeof(pool_buc));
    ll res = 0;
    for (int i = 1; i <= q; i++) {
        if (L <= rk[i] && rk[i] <= R) { // rebuild block
            // [L, rk[i]] suf
            for (int t = L; t <= R; t++) {
                // if (_opt[tmp[t].second] == 1) buc1[suf[t]] = 0;
                // else buc2[suf[t]] = 0;
                buc1[suf[t]] = buc2[suf[t]] = 0;
                suf[t] += lazy;
                if (t <= rk[i]) {
                    if (_opt[i] == 1) suf[t]++;
                    else suf[t]--;
                }
            }
            lazy = res = 0;
            for (int t = L; t <= R; t++) {
                if (_opt[tmp[t].second] == 1 && tmp[t].second <= i) {
                    buc1[suf[t]] += tmp[t].first;
                    if (suf[t] <= 0) res += tmp[t].first;
                }
                if (_opt[tmp[t].second] == 2 && tmp[t].second <= i) {
                    buc2[suf[t]] += tmp[t].first;
                    if (suf[t] >= 0) res += tmp[t].first;
                }
            }
        } else if (rk[i] > R) {
            if (_opt[i] == 1) { // all suf += 1
                res -= buc1[-lazy];
                lazy += 1;
                res += buc2[-lazy];
            } else { // all suf -= 1
                res -= buc2[-lazy];
                lazy -= 1;
                res += buc1[-lazy];
            }
        }
        ans[i] += res;
    }
}

int main() {
    scanf("%d", &q);
    for (int i = 1; i <= q; i++) {
        scanf("%d%d", &_opt[i], &_v[i]);
        tmp[i] = {_v[i], i};
    }
    sort(tmp + 1, tmp + q + 1);
    for (int i = 1; i <= q; i++) {
        rk[tmp[i].second] = i;
    }
    for (int L = 1, R = B; L <= q; L += B, R += B) {
        R = min(R, q);
        solve(L, R);
    }
    ll cur = 0;
    for (int i = 1; i <= q; i++) {
        if (_opt[i] == 1) cur += _v[i];
        printf("%lld\n", cur + ans[i]);
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Variable Damage
time_limit_ms: 5000
memory_limit_mb: 512
Description: Monocarp is gathering an army to fight a dragon in a videogame.

The army consists of two parts: the heroes and the defensive artifacts. Each hero has one parameter — his health. Each defensive artifact also has one parameter — its durability.

Before the battle begins, Monocarp distributes artifacts to the heroes so that each hero receives at most one artifact.

The battle consists of rounds that proceed as follows:

  * first, the dragon deals damage equal to $\frac{1}{a + b}$ (a real number without rounding) to each hero, where $a$ is the number of heroes alive and $b$ is the number of active artifacts;    * after that, all heroes with health $0$ or less die;    * finally, some artifacts are deactivated. An artifact with durability $x$ is deactivated when one of the following occurs: the hero holding the artifact either dies or receives $x$ total damage (from the start of the battle). If an artifact is not held by any hero, it is inactive from the beginning of the battle.

The battle ends when there are no heroes left alive.

Initially, the army is empty. There are $q$ queries: add a hero with health $x$ or an artifact with durability $y$ to the army. After each query, determine the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Input: The first line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$) — the number of queries.

In the $i$-th of the following $q$ lines, there are two integers $t_i$ and $v_i$ ($t_i \in \\{1, 2\\}$; $1 \le v_i \le 10^9$) — the type of the query and the value of the query parameter. If the type is $1$, a hero with health $v_i$ is added. If the type is $2$, an artifact with durability $v_i$ is added.

Output: Print $q$ integers. After each query, output the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Examples:
input:
3
2 5
1 4
1 10
output:
0
8
19

input:
10
1 9
1 6
2 4
1 8
1 3
2 10
1 3
1 6
1 10
2 6
output:
9
15
19
27
30
39
42
48
59
65

Note: Let's consider the first example.

  * An artifact with durability $5$ is added. Since there are no heroes yet, the battle ends immediately.    * A hero with health $4$ is added. Monocarp can give him an artifact with durability $5$. First, there are rounds in which the hero takes $\frac{1}{1 + 1} = \frac{1}{2}$ damage. After $8$ such rounds, a total of $4$ damage will have been dealt, and the hero will die, while the artifact will deactivate. There are no more heroes alive, so the battle ends after $8$ rounds.    * A hero with health $10$ is added. Now let the artifact with durability $5$ be with this hero. Then, in the first $12$ rounds, the heroes will take $12 \cdot \frac{1}{2 + 1} = 4$ damage, and the first hero will die. The second hero has $6$ health left, and the artifact has $1$ durability. Now the damage is $\frac{1}{2}$, so after another $2$ rounds, the artifact will deactivate. The second hero has $5$ health left. After another $5$ rounds, the second hero will die. Therefore, the answer is $12 + 2 + 5 = 19$.

Buggy Code:
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define N 1000010
#define int long long
#define i128 __int128
#define db long double
#define pii pair<int,int>
#define st first
#define ed second
#define mkp make_pair
#define pb push_back
#define eps 1e-9
#define mod 998244353
#define mod2 1000000007
#define bs 13131
#define bs2 131
#define INF 0x3f3f3f3f3f3f3f3f
#define il inline
#define vi vector<int>
#define ins insert
#define umap unordered_map
#define uset unordered_set
#define R(x) x.begin(),x.end()
#define B(x) x.begin()
#define E(x) x.end()
#define lb lower_bound
#define ub upper_bound
#define prq priority_queue
il int read(){
	int x=0,f=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){
		if(ch=='-')f=-1;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9'){
		x=(x<<1)+(x<<3)+(ch^48);
		ch=getchar();
	}
	return x*f;
}
int T=1,n,m,q,k;
struct node{
	int op,x;
}s[N];
int p[N],ps[N];
il bool cmp(int i,int j){
	return s[i].x>s[j].x;
}
il bool cmp2(node a,node b){
	return a.x>b.x;
}
int B=500,L[N],R[N],id[N];
il void build(){
	m=(n-1)/B+1;
	for(int i=1;i<=m;i++){
		L[i]=(i-1)*B+1,R[i]=min(n,i*B);
		for(int j=L[i];j<=R[i];j++)id[j]=i;
	}
}
int w[610][1010],add[610],a[1010],b[1010];
int vis[N];
il void mdf(int id){
	for(int i=0;i<=(B<<1);i++)a[i]=b[i]=0;
	int now=0;
	for(int i=L[id];i<=R[id];i++){
		if(!vis[i])continue;
		if(s[i].op==1)a[-now-1+B]+=s[i].x;
		else b[-now+1+B]+=s[i].x,b[0]+=s[i].x;
		now+=s[i].op;
	}
	for(int i=1;i<=(B<<1);i++)b[i]+=b[i-1];
	for(int i=(B<<1)-1;i>=0;i--)a[i]+=a[i+1];
	for(int i=0;i<=(B<<1);i++)w[id][i]=a[i]+b[i];
}
il int calc(){
	int now=0,sum=0;
	for(int i=1;i<=m;i++){
		int k=now+B;
		k=max(0ll,k),k=min(k,(B<<1));
		sum+=w[i][k];
		now+=add[i];
	}
	return sum;
}
signed main(){
	n=read();
	for(int i=1;i<=n;i++)
		s[i].op=(read()<<1)-3,s[i].x=read(),p[i]=i;
	sort(p+1,p+1+n,cmp);
	for(int i=1;i<=n;i++)ps[p[i]]=i;
	sort(s+1,s+1+n,cmp2);
	build();
	for(int i=1;i<=n;i++){
		vis[ps[i]]=1;
		mdf(id[ps[i]]);
		printf("%lld\n",calc());
	}
	return 0;
}

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

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

Think step by step.

Response:
```plaintext
5
1 1
2 1
1 1
2 2
1 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Variable Damage
time_limit_ms: 5000
memory_limit_mb: 512
Description: Monocarp is gathering an army to fight a dragon in a videogame.

The army consists of two parts: the heroes and the defensive artifacts. Each hero has one parameter — his health. Each defensive artifact also has one parameter — its durability.

Before the battle begins, Monocarp distributes artifacts to the heroes so that each hero receives at most one artifact.

The battle consists of rounds that proceed as follows:

  * first, the dragon deals damage equal to $\frac{1}{a + b}$ (a real number without rounding) to each hero, where $a$ is the number of heroes alive and $b$ is the number of active artifacts;    * after that, all heroes with health $0$ or less die;    * finally, some artifacts are deactivated. An artifact with durability $x$ is deactivated when one of the following occurs: the hero holding the artifact either dies or receives $x$ total damage (from the start of the battle). If an artifact is not held by any hero, it is inactive from the beginning of the battle.

The battle ends when there are no heroes left alive.

Initially, the army is empty. There are $q$ queries: add a hero with health $x$ or an artifact with durability $y$ to the army. After each query, determine the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Input: The first line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$) — the number of queries.

In the $i$-th of the following $q$ lines, there are two integers $t_i$ and $v_i$ ($t_i \in \\{1, 2\\}$; $1 \le v_i \le 10^9$) — the type of the query and the value of the query parameter. If the type is $1$, a hero with health $v_i$ is added. If the type is $2$, an artifact with durability $v_i$ is added.

Output: Print $q$ integers. After each query, output the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Examples:
input:
3
2 5
1 4
1 10
output:
0
8
19

input:
10
1 9
1 6
2 4
1 8
1 3
2 10
1 3
1 6
1 10
2 6
output:
9
15
19
27
30
39
42
48
59
65

Note: Let's consider the first example.

  * An artifact with durability $5$ is added. Since there are no heroes yet, the battle ends immediately.    * A hero with health $4$ is added. Monocarp can give him an artifact with durability $5$. First, there are rounds in which the hero takes $\frac{1}{1 + 1} = \frac{1}{2}$ damage. After $8$ such rounds, a total of $4$ damage will have been dealt, and the hero will die, while the artifact will deactivate. There are no more heroes alive, so the battle ends after $8$ rounds.    * A hero with health $10$ is added. Now let the artifact with durability $5$ be with this hero. Then, in the first $12$ rounds, the heroes will take $12 \cdot \frac{1}{2 + 1} = 4$ damage, and the first hero will die. The second hero has $6$ health left, and the artifact has $1$ durability. Now the damage is $\frac{1}{2}$, so after another $2$ rounds, the artifact will deactivate. The second hero has $5$ health left. After another $5$ rounds, the second hero will die. Therefore, the answer is $12 + 2 + 5 = 19$.

Buggy Code:
def damagetaken(hero,art):
    la=len(art)
    lh=len(hero)
    if la>=lh:
        return 1/(2*lh)
    return 1/(la+lh)
art=[]
hero=[]
for _ in range(int(input())):
    add=list(map(int, input().split()))
    if add[0]==1:
        hero.append(add[1])
    else:
        art.append(add[1])
    nhero=[x for x in hero]
    nart=[x for x in art]
    r=0
    while(nhero):
        dt=damagetaken(nhero,nart)
        mina=-1
        nart.sort()
        if nart:
            mina=nart[0]
        nhero.sort()
        minh=nhero[0]
        if nart:
            if mina!=-1 and minh<mina:
                nr=minh/dt
                r+=nr
            else:
                nr=mina/dt
                r+=nr
        else:
            nr=minh/dt
            r+=nr
        i=0
        while(i<len(nhero)):
            nhero[i]=nhero[i]-(nr*dt)
            if nhero[i]<=0:
                nhero.remove(nhero[i])
                i-=1
            i+=1
        if not nhero:
            continue
        i=0
        if nart:
            while(i<len(nart) and nart):
                nart[i]=nart[i]-(nr*dt)
                if nart[i]<=0:
                    nart.remove(nart[i])
                    i-=1
                i+=1
    print(int(r))

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

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

Think step by step.

Response:
```plaintext
5
1 1
1 1
2 1
1 2
2 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Variable Damage
time_limit_ms: 5000
memory_limit_mb: 512
Description: Monocarp is gathering an army to fight a dragon in a videogame.

The army consists of two parts: the heroes and the defensive artifacts. Each hero has one parameter — his health. Each defensive artifact also has one parameter — its durability.

Before the battle begins, Monocarp distributes artifacts to the heroes so that each hero receives at most one artifact.

The battle consists of rounds that proceed as follows:

  * first, the dragon deals damage equal to $\frac{1}{a + b}$ (a real number without rounding) to each hero, where $a$ is the number of heroes alive and $b$ is the number of active artifacts;    * after that, all heroes with health $0$ or less die;    * finally, some artifacts are deactivated. An artifact with durability $x$ is deactivated when one of the following occurs: the hero holding the artifact either dies or receives $x$ total damage (from the start of the battle). If an artifact is not held by any hero, it is inactive from the beginning of the battle.

The battle ends when there are no heroes left alive.

Initially, the army is empty. There are $q$ queries: add a hero with health $x$ or an artifact with durability $y$ to the army. After each query, determine the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Input: The first line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$) — the number of queries.

In the $i$-th of the following $q$ lines, there are two integers $t_i$ and $v_i$ ($t_i \in \\{1, 2\\}$; $1 \le v_i \le 10^9$) — the type of the query and the value of the query parameter. If the type is $1$, a hero with health $v_i$ is added. If the type is $2$, an artifact with durability $v_i$ is added.

Output: Print $q$ integers. After each query, output the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Examples:
input:
3
2 5
1 4
1 10
output:
0
8
19

input:
10
1 9
1 6
2 4
1 8
1 3
2 10
1 3
1 6
1 10
2 6
output:
9
15
19
27
30
39
42
48
59
65

Note: Let's consider the first example.

  * An artifact with durability $5$ is added. Since there are no heroes yet, the battle ends immediately.    * A hero with health $4$ is added. Monocarp can give him an artifact with durability $5$. First, there are rounds in which the hero takes $\frac{1}{1 + 1} = \frac{1}{2}$ damage. After $8$ such rounds, a total of $4$ damage will have been dealt, and the hero will die, while the artifact will deactivate. There are no more heroes alive, so the battle ends after $8$ rounds.    * A hero with health $10$ is added. Now let the artifact with durability $5$ be with this hero. Then, in the first $12$ rounds, the heroes will take $12 \cdot \frac{1}{2 + 1} = 4$ damage, and the first hero will die. The second hero has $6$ health left, and the artifact has $1$ durability. Now the damage is $\frac{1}{2}$, so after another $2$ rounds, the artifact will deactivate. The second hero has $5$ health left. After another $5$ rounds, the second hero will die. Therefore, the answer is $12 + 2 + 5 = 19$.

Buggy Code:
def damagetaken(hero,art):
    la=len(art)
    lh=len(hero)
    if la>=lh:
        return 1/(2*lh)
    return 1/(la+lh)
art=[]
hero=[]
for _ in range(int(input())):
    add=list(map(int, input().split()))
    if add[0]==1:
        hero.append(add[1])
    else:
        art.append(add[1])
    nhero=hero.copy()
    nart=art.copy()
    r=0
    while(nhero):
        dt=damagetaken(nhero,nart)
        mina=-1
        nart.sort()
        if nart:
            mina=nart[0]
        nhero.sort()
        minh=nhero[0]
        if nart:
            if mina!=-1 and minh<mina:
                nr=minh/dt
                r+=nr
            else:
                nr=mina/dt
                r+=nr
        else:
            nr=minh/dt
            r+=nr
        i=0
        while(i<len(nhero)):
            nhero[i]=nhero[i]-(nr*dt)
            if nhero[i]<=0:
                nhero.remove(nhero[i])
                i-=1
            i+=1
        if not nhero:
            continue
        i=0
        if nart:
            while(i<len(nart) and nart):
                nart[i]=nart[i]-(nr*dt)
                if nart[i]<=0:
                    nart.remove(nart[i])
                    i-=1
                i+=1
    print(int(r))

**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 1
1 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Variable Damage
time_limit_ms: 5000
memory_limit_mb: 512
Description: Monocarp is gathering an army to fight a dragon in a videogame.

The army consists of two parts: the heroes and the defensive artifacts. Each hero has one parameter — his health. Each defensive artifact also has one parameter — its durability.

Before the battle begins, Monocarp distributes artifacts to the heroes so that each hero receives at most one artifact.

The battle consists of rounds that proceed as follows:

  * first, the dragon deals damage equal to $\frac{1}{a + b}$ (a real number without rounding) to each hero, where $a$ is the number of heroes alive and $b$ is the number of active artifacts;    * after that, all heroes with health $0$ or less die;    * finally, some artifacts are deactivated. An artifact with durability $x$ is deactivated when one of the following occurs: the hero holding the artifact either dies or receives $x$ total damage (from the start of the battle). If an artifact is not held by any hero, it is inactive from the beginning of the battle.

The battle ends when there are no heroes left alive.

Initially, the army is empty. There are $q$ queries: add a hero with health $x$ or an artifact with durability $y$ to the army. After each query, determine the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Input: The first line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$) — the number of queries.

In the $i$-th of the following $q$ lines, there are two integers $t_i$ and $v_i$ ($t_i \in \\{1, 2\\}$; $1 \le v_i \le 10^9$) — the type of the query and the value of the query parameter. If the type is $1$, a hero with health $v_i$ is added. If the type is $2$, an artifact with durability $v_i$ is added.

Output: Print $q$ integers. After each query, output the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Examples:
input:
3
2 5
1 4
1 10
output:
0
8
19

input:
10
1 9
1 6
2 4
1 8
1 3
2 10
1 3
1 6
1 10
2 6
output:
9
15
19
27
30
39
42
48
59
65

Note: Let's consider the first example.

  * An artifact with durability $5$ is added. Since there are no heroes yet, the battle ends immediately.    * A hero with health $4$ is added. Monocarp can give him an artifact with durability $5$. First, there are rounds in which the hero takes $\frac{1}{1 + 1} = \frac{1}{2}$ damage. After $8$ such rounds, a total of $4$ damage will have been dealt, and the hero will die, while the artifact will deactivate. There are no more heroes alive, so the battle ends after $8$ rounds.    * A hero with health $10$ is added. Now let the artifact with durability $5$ be with this hero. Then, in the first $12$ rounds, the heroes will take $12 \cdot \frac{1}{2 + 1} = 4$ damage, and the first hero will die. The second hero has $6$ health left, and the artifact has $1$ durability. Now the damage is $\frac{1}{2}$, so after another $2$ rounds, the artifact will deactivate. The second hero has $5$ health left. After another $5$ rounds, the second hero will die. Therefore, the answer is $12 + 2 + 5 = 19$.

Buggy Code:
#pragma GCC optimize("Ofast,unroll-loops")

#include <iostream>
#include <cmath>
#include <iomanip>
#include <vector>
#include <stack>
#include <queue>
#include <list>
#include <map>
#include <bitset>
#include <string>
#include <algorithm>
#include <iterator>
#include <unordered_map>
#include <unordered_set>
#include <sstream>
#include <fstream>
#include <numeric>
#include <tuple>
#include <ctime>
#include <random>
#include <array>
#include <cassert>
#include <complex>
#include <valarray>
#include <set>
#include <chrono>

#pragma GCC target("avx2,bmi,bmi2,popcnt,lzcnt")

using namespace std;

#define sorti(a) sort((a).begin(), (a).end());
#define reversi(a) reverse((a).begin(), (a).end());
#define yes cout << "Yes\n";
#define no cout << "No\n";
#define ans(a) cout << a << '\n';
#define read(a, n) {(a).resize(n); for (auto &i : a) cin >> i;}
//#define int long long

using ll = long long;
using ld = long double;
using ull = unsigned long long;

const int INF = 2147483647;
const ll LONGINF = 9223372036854775807;
const ll MOD = 1e9 + 7;

vector<bool> active;

void solve()
{
    int q;
    cin >> q;
    constexpr int d = 400;
    active.resize(q);
    struct Block
    {
        int balance = 0;
        ll sum = 0;
        array<int, 2 * d + 1> artSum;
        vector<tuple<ll, int, int>> vals;

        Block()
        {
            artSum.fill(0);
        }

        void update()
        {
            sum = 0;
            balance = 0;
            for (auto [v, t, id] : vals)
            {
                if (!active[id])
                    continue;
                sum += v * (t < 0);
                if (balance * t < 0)
                    sum += v;
                balance += t;
            }
        }

        ll get(ll entry = 0)
        {
            return sum + artSum[min<ll>(max<ll>(entry, -d), d) + d];
        }
    };
    vector<Block> blocks;
    vector<tuple<ll, int, int>> queries;
    for (int i = 0; i < q; ++i)
    {
        int t, v;
        cin >> t >> v;
        queries.emplace_back(v, 2 * t - 3, i);
    }
    sorti(queries)
    reversi(queries)
    vector<int> block(q);
    for (auto [val, t, id] : queries)
    {
        if (blocks.empty() || blocks.back().vals.size() == d)
            blocks.push_back(Block());
        blocks.back().vals.emplace_back(val, t, id);
        block[id] = (int)blocks.size() - 1;
    }
    for (int i = 0; i < q; ++i)
    {
        active[i] = true;
        blocks[block[i]].update();
        int balance = 0;
        ll res = 0;
        for (Block &j : blocks)
        {
            res += j.get(balance);
            balance += j.balance;
        }
        ans(res)
    }
}

signed main()
{
    cin.tie(0)->sync_with_stdio(0);
    int t = 1;
    //cin >> t;
    while (t--)
        solve();
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Variable Damage
time_limit_ms: 5000
memory_limit_mb: 512
Description: Monocarp is gathering an army to fight a dragon in a videogame.

The army consists of two parts: the heroes and the defensive artifacts. Each hero has one parameter — his health. Each defensive artifact also has one parameter — its durability.

Before the battle begins, Monocarp distributes artifacts to the heroes so that each hero receives at most one artifact.

The battle consists of rounds that proceed as follows:

  * first, the dragon deals damage equal to $\frac{1}{a + b}$ (a real number without rounding) to each hero, where $a$ is the number of heroes alive and $b$ is the number of active artifacts;    * after that, all heroes with health $0$ or less die;    * finally, some artifacts are deactivated. An artifact with durability $x$ is deactivated when one of the following occurs: the hero holding the artifact either dies or receives $x$ total damage (from the start of the battle). If an artifact is not held by any hero, it is inactive from the beginning of the battle.

The battle ends when there are no heroes left alive.

Initially, the army is empty. There are $q$ queries: add a hero with health $x$ or an artifact with durability $y$ to the army. After each query, determine the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Input: The first line contains one integer $q$ ($1 \le q \le 3 \cdot 10^5$) — the number of queries.

In the $i$-th of the following $q$ lines, there are two integers $t_i$ and $v_i$ ($t_i \in \\{1, 2\\}$; $1 \le v_i \le 10^9$) — the type of the query and the value of the query parameter. If the type is $1$, a hero with health $v_i$ is added. If the type is $2$, an artifact with durability $v_i$ is added.

Output: Print $q$ integers. After each query, output the maximum number of rounds that Monocarp can survive if he distributes the artifacts optimally.

Examples:
input:
3
2 5
1 4
1 10
output:
0
8
19

input:
10
1 9
1 6
2 4
1 8
1 3
2 10
1 3
1 6
1 10
2 6
output:
9
15
19
27
30
39
42
48
59
65

Note: Let's consider the first example.

  * An artifact with durability $5$ is added. Since there are no heroes yet, the battle ends immediately.    * A hero with health $4$ is added. Monocarp can give him an artifact with durability $5$. First, there are rounds in which the hero takes $\frac{1}{1 + 1} = \frac{1}{2}$ damage. After $8$ such rounds, a total of $4$ damage will have been dealt, and the hero will die, while the artifact will deactivate. There are no more heroes alive, so the battle ends after $8$ rounds.    * A hero with health $10$ is added. Now let the artifact with durability $5$ be with this hero. Then, in the first $12$ rounds, the heroes will take $12 \cdot \frac{1}{2 + 1} = 4$ damage, and the first hero will die. The second hero has $6$ health left, and the artifact has $1$ durability. Now the damage is $\frac{1}{2}$, so after another $2$ rounds, the artifact will deactivate. The second hero has $5$ health left. After another $5$ rounds, the second hero will die. Therefore, the answer is $12 + 2 + 5 = 19$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
struct apos{
	int a;
	int pos;
	friend bool operator<(apos a,apos b){
		return a.a<b.a;
	}
}ap[300000];
long long t[300000],v[300000];
long long ans[300000],lval[300001];
const long long sqn=550;
long long blksum[sqn],blkval[sqn][sqn*2+1],curval[300000];
int main(){
	ios::sync_with_stdio(false),cin.tie(0);
	long long n,i,j,x;
	long long cans;
	cin>>n;
	cans=0;
	for(i=0;i<n;i++)
	{
		cin>>t[i]>>v[i];
		ap[i].a=v[i];
		ap[i].pos=i;
		cans+=v[i];
		ans[i]=cans;
	}
	sort(ap,ap+n);
	for(i=0;i<n;i++)
	{
		lval[i+1]=ap[i].a;
		v[ap[i].pos]=i;
	}
	for(i=0;i<n;i++)lval[i]=lval[i+1]-lval[i];
	for(i=0;i<n;i++)blkval[i/sqn][sqn]+=lval[i];
	for(i=0;i<sqn;i++)
	{
		for(j=1;j<=sqn*2;j++)blkval[i][j]+=blkval[i][j-1];
	}
	cans=0;
	for(i=0;i<n;i++)
	{
		if(t[i]==1)
		{
			for(j=0;j<v[i]/sqn;j++)
			{
				blksum[j]++;
				if(sqn-blksum[j]>=0)cans-=blkval[j][min(sqn*2,sqn-blksum[j])];
			}
			x=v[i]/sqn;
			for(j=sqn*2;j>0;j--)blkval[x][j]-=blkval[x][j-1];
			for(j=v[i]/sqn*sqn;j<=v[i];j++)
			{
				if(curval[j]+blksum[j/sqn]<=-1)cans-=lval[j];
				blkval[j/sqn][curval[j]]-=lval[j];
				curval[j]++;
				blkval[j/sqn][curval[j]]+=lval[j];
			}
			for(j=1;j<=sqn*2;j++)blkval[x][j]+=blkval[x][j-1];
		}
		else
		{
			for(j=0;j<v[i]/sqn;j++)
			{
				if(sqn-blksum[j]>=0)cans-=blkval[j][min(sqn*2,sqn-blksum[j])];
				blksum[j]--;
			}
			x=v[i]/sqn;
			for(j=sqn*2;j>0;j--)blkval[x][j]-=blkval[x][j-1];
			for(j=v[i]/sqn*sqn;j<=v[i];j++)
			{
				if(curval[j]+blksum[j/sqn]<=0)cans+=lval[j];
				blkval[j/sqn][curval[j]]-=lval[j];
				curval[j]--;
				blkval[j/sqn][curval[j]]+=lval[j];
			}
			for(j=1;j<=sqn*2;j++)blkval[x][j]+=blkval[x][j-1];
		}
		cout<<ans[i]-cans<<'\n';
	}
	return 0;
}

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

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

Think step by step.

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