**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves Inversions
time_limit_ms: 2000
memory_limit_mb: 256
Description: Shohag has an array $$$a$$$ of integers. Initially $$$a = [0, 1]$$$. He can repeatedly perform the following operation any number of times:

- Let $$$k$$$ be the number of inversions$$$^{\text{∗}}$$$ in the current array $$$a$$$.
- Insert $$$k$$$ at any position in $$$a$$$, including the beginning or the end.

For example, if $$$a = [4, 6, 2, 4]$$$, then the number of inversions is $$$k = 3$$$. So Shohag can obtain the following arrays after the operation: $$$[\textbf{3}, 4, 6, 2, 4]$$$, $$$[4, \textbf{3}, 6, 2, 4]$$$, $$$[4, 6, \textbf{3}, 2, 4]$$$, $$$[4, 6, 2, \textbf{3}, 4]$$$, and $$$[4, 6, 2, 4, \textbf{3}]$$$.

Given an integer $$$n$$$, help Shohag count, modulo $$$998\,244\,353$$$, the number of distinct arrays of length $$$n$$$ that can be obtained after performing the operations.

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

The first and only line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 10^6$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output an integer — the number of possible arrays modulo $$$998\,244\,353$$$.

Examples:
input:
4
4
2
7
69
output:
5
1
682
325188814

Note: In the first test case, the following $$$5$$$ arrays can be obtained (the inserted inversion count is shown in bold):

- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, 1, \textbf{0}]$$$,
- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, \textbf{0}, 1]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, 0, \textbf{1}]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, \textbf{1}, 0]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [\textbf{1}, 0, 1, 0]$$$.

Buggy Code:
//#pragma GCC optimize(1,2,3,"Ofast","inline")
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int mod=998244353;
int cnt1[1000006];
void solve(){
    int n;
    cin>>n;
    int sm=1;
    int ans=cnt1[n];
    for(int i=n-1;i>=4;i--){
        int nw=sm*i+1;nw%=mod;
        ans+=nw*cnt1[i];ans%=mod;
        sm+=nw;
    }
    cout<<(ans+n-1)%mod<<endl;
}
signed main(){
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    for(int i=4;i<=1000000;i++)cnt1[i]=(cnt1[i-1]+i-2)%mod;
    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
4
4
2
7
1000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves Inversions
time_limit_ms: 2000
memory_limit_mb: 256
Description: Shohag has an array $$$a$$$ of integers. Initially $$$a = [0, 1]$$$. He can repeatedly perform the following operation any number of times:

- Let $$$k$$$ be the number of inversions$$$^{\text{∗}}$$$ in the current array $$$a$$$.
- Insert $$$k$$$ at any position in $$$a$$$, including the beginning or the end.

For example, if $$$a = [4, 6, 2, 4]$$$, then the number of inversions is $$$k = 3$$$. So Shohag can obtain the following arrays after the operation: $$$[\textbf{3}, 4, 6, 2, 4]$$$, $$$[4, \textbf{3}, 6, 2, 4]$$$, $$$[4, 6, \textbf{3}, 2, 4]$$$, $$$[4, 6, 2, \textbf{3}, 4]$$$, and $$$[4, 6, 2, 4, \textbf{3}]$$$.

Given an integer $$$n$$$, help Shohag count, modulo $$$998\,244\,353$$$, the number of distinct arrays of length $$$n$$$ that can be obtained after performing the operations.

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

The first and only line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 10^6$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output an integer — the number of possible arrays modulo $$$998\,244\,353$$$.

Examples:
input:
4
4
2
7
69
output:
5
1
682
325188814

Note: In the first test case, the following $$$5$$$ arrays can be obtained (the inserted inversion count is shown in bold):

- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, 1, \textbf{0}]$$$,
- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, \textbf{0}, 1]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, 0, \textbf{1}]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, \textbf{1}, 0]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [\textbf{1}, 0, 1, 0]$$$.

Buggy Code:
#include<bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pii;
const int Mod = 998244353;
inline void uadd(int &a, const int &b){ a += b - Mod; a += (a>>31) & Mod; }
inline int add(int a, const int &b){ a += b - Mod; a += (a>>31) & Mod; return a; }
inline void usub(int &a, const int &b){ a -= b; a += (a>>31) & Mod; }
inline int sub(int a, const int &b){ a -= b, a += (a>>31) & Mod; return a; }
inline void umul(int &a, const int &b){ a = (int)(1ll * a * b % Mod); }
inline int mul(const int &a, const int &b){ return (int)(1ll * a * b % Mod); }
int qpow(int b, ll p){ int ret = 1; while(p){ if(p & 1) umul(ret, b); umul(b, b), p >>= 1; } return ret; }
const int fN = 10010;
int fact[fN], invfact[fN], inv[fN];
void initfact(int n){
	fact[0] = 1; for(int i = 1; i <= n; ++i) fact[i] = mul(fact[i - 1], i);
	invfact[n] = qpow(fact[n], Mod - 2); for(int i = n; i > 0; --i) invfact[i - 1] = mul(invfact[i], i);
	for(int i = 1; i <= n; ++i) inv[i] = mul(invfact[i], fact[i - 1]);
}
inline int binom(int n, int m){ return mul(fact[n], mul(invfact[m], invfact[n - m])); }

const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
template<typename T> inline void chmax(T &_a, const T &_b){ (_b>_a) ? (_a=_b) : _a; }
template<typename T> inline void chmin(T &_a, const T &_b){ (_b<_a) ? (_a=_b) : _a; }

int n;
inline int g(int i){
	return (i >= 4 ? sub((int)((1ll * (i - 1) * (i - 2) / 2) % Mod), 1) : 0);
}
void solve(){
	cin >> n;
	static int f[100100];
	f[4] = 0;
	for(int i = 5; i <= n; ++i) f[i] = add(mul(f[i - 1], i), mul(g(i - 1), i - 1));
	int ans = 0;
	for(int i = 1; i <= n; ++i){
		uadd(ans, (i != 2));
		uadd(ans, g(i));
		uadd(ans, f[i]);
	}
	cout << ans << "\n";
}

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

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

	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves Inversions
time_limit_ms: 2000
memory_limit_mb: 256
Description: Shohag has an array $$$a$$$ of integers. Initially $$$a = [0, 1]$$$. He can repeatedly perform the following operation any number of times:

- Let $$$k$$$ be the number of inversions$$$^{\text{∗}}$$$ in the current array $$$a$$$.
- Insert $$$k$$$ at any position in $$$a$$$, including the beginning or the end.

For example, if $$$a = [4, 6, 2, 4]$$$, then the number of inversions is $$$k = 3$$$. So Shohag can obtain the following arrays after the operation: $$$[\textbf{3}, 4, 6, 2, 4]$$$, $$$[4, \textbf{3}, 6, 2, 4]$$$, $$$[4, 6, \textbf{3}, 2, 4]$$$, $$$[4, 6, 2, \textbf{3}, 4]$$$, and $$$[4, 6, 2, 4, \textbf{3}]$$$.

Given an integer $$$n$$$, help Shohag count, modulo $$$998\,244\,353$$$, the number of distinct arrays of length $$$n$$$ that can be obtained after performing the operations.

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

The first and only line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 10^6$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output an integer — the number of possible arrays modulo $$$998\,244\,353$$$.

Examples:
input:
4
4
2
7
69
output:
5
1
682
325188814

Note: In the first test case, the following $$$5$$$ arrays can be obtained (the inserted inversion count is shown in bold):

- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, 1, \textbf{0}]$$$,
- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, \textbf{0}, 1]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, 0, \textbf{1}]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, \textbf{1}, 0]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [\textbf{1}, 0, 1, 0]$$$.

Buggy Code:
M=998244353
exec("n=int(input());a,b=n-1,0\nfor i in range(2,n):a,b=(a+b)%M,(b*(i+2)+i)%M\nprint(a)\n"*int(input()))

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider edge cases and scenarios that maximize the failure probability. The buggy code has a specific structure, and we need to identify where it might fail.

### Analysis of the Buggy Code
The buggy code performs the following operations:
1. It initializes `a` and `b` with `n-1` and `0` respectively.
2. It iterates from `2` to `n-1`, updating `a` and `b` in each iteration.
3. It prints `a` after the loop.

### Potential Issues
1. **Initialization and Loop Range**: The loop starts from `2` and goes up to `n-1`. This might be incorrect if `n` is small.
2. **Modulo Operations**: The modulo operations might not be applied correctly, leading to incorrect results.
3. **Edge Cases**: Small values of `n` (e.g., `2`, `3`, `4`) and large values of `n` (e.g., `10^6`) should be tested.

### Challenging Test Input
To expose the bug, we should test the following cases:
1. **Small Values of `n`**: These can help identify issues in the loop range and initialization.
2. **Large Values of `n`**: These can help identify issues in the modulo operations and performance.

### Test Input
```plaintext
4
2
3
4
1000000
```

### Explanation
1. **`n = 2`**: This is the smallest valid input. The expected output is `1`.
2. **`n = 3`**: This is a slightly larger input. The expected output is `2`.
3. **`n = 4`**: This is a small input where the number of possible arrays is `5`.
4. **`n = 1000000`**: This is a large input to test the performance and correctness of the modulo operations.

This test input will help identify any issues in the buggy code, especially in handling small and large values of `n`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves Inversions
time_limit_ms: 2000
memory_limit_mb: 256
Description: Shohag has an array $$$a$$$ of integers. Initially $$$a = [0, 1]$$$. He can repeatedly perform the following operation any number of times:

- Let $$$k$$$ be the number of inversions$$$^{\text{∗}}$$$ in the current array $$$a$$$.
- Insert $$$k$$$ at any position in $$$a$$$, including the beginning or the end.

For example, if $$$a = [4, 6, 2, 4]$$$, then the number of inversions is $$$k = 3$$$. So Shohag can obtain the following arrays after the operation: $$$[\textbf{3}, 4, 6, 2, 4]$$$, $$$[4, \textbf{3}, 6, 2, 4]$$$, $$$[4, 6, \textbf{3}, 2, 4]$$$, $$$[4, 6, 2, \textbf{3}, 4]$$$, and $$$[4, 6, 2, 4, \textbf{3}]$$$.

Given an integer $$$n$$$, help Shohag count, modulo $$$998\,244\,353$$$, the number of distinct arrays of length $$$n$$$ that can be obtained after performing the operations.

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

The first and only line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 10^6$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output an integer — the number of possible arrays modulo $$$998\,244\,353$$$.

Examples:
input:
4
4
2
7
69
output:
5
1
682
325188814

Note: In the first test case, the following $$$5$$$ arrays can be obtained (the inserted inversion count is shown in bold):

- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, 1, \textbf{0}]$$$,
- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, \textbf{0}, 1]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, 0, \textbf{1}]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, \textbf{1}, 0]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [\textbf{1}, 0, 1, 0]$$$.

Buggy Code:
M=998244353
exec("n=int(input());a,b=n-1,0\nfor i in range(2,n):a,b=(a+b)%M,(b*(i+2)+i)%M\nprint(a)\n"*int(input()))

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on edge cases or scenarios that can highlight issues in the algorithm. The code appears to be using a dynamic programming approach to count the number of distinct arrays, but there might be issues with how it handles the initial conditions or the transition between states.

Let's consider the following scenarios:

1. **Small values of \( n \)**: These can help us verify the basic logic and initial conditions.
2. **Intermediate values of \( n \)**: These can help us check the transition logic.
3. **Large values of \( n \)**: These can help us verify the performance and correctness for larger inputs.

Given the constraints, a small-scale test input that is complete and valid would be:

```plaintext
4
4
2
7
10
```

This input includes:
- \( n = 4 \) (a small value to verify basic logic)
- \( n = 2 \) (the smallest valid input to verify initial conditions)
- \( n = 7 \) (an intermediate value to verify transition logic)
- \( n = 10 \) (a slightly larger value to verify correctness for larger inputs)

This test input should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Shohag Loves Inversions
time_limit_ms: 2000
memory_limit_mb: 256
Description: Shohag has an array $$$a$$$ of integers. Initially $$$a = [0, 1]$$$. He can repeatedly perform the following operation any number of times:

- Let $$$k$$$ be the number of inversions$$$^{\text{∗}}$$$ in the current array $$$a$$$.
- Insert $$$k$$$ at any position in $$$a$$$, including the beginning or the end.

For example, if $$$a = [4, 6, 2, 4]$$$, then the number of inversions is $$$k = 3$$$. So Shohag can obtain the following arrays after the operation: $$$[\textbf{3}, 4, 6, 2, 4]$$$, $$$[4, \textbf{3}, 6, 2, 4]$$$, $$$[4, 6, \textbf{3}, 2, 4]$$$, $$$[4, 6, 2, \textbf{3}, 4]$$$, and $$$[4, 6, 2, 4, \textbf{3}]$$$.

Given an integer $$$n$$$, help Shohag count, modulo $$$998\,244\,353$$$, the number of distinct arrays of length $$$n$$$ that can be obtained after performing the operations.

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

The first and only line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 10^6$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$.

Output: For each test case, output an integer — the number of possible arrays modulo $$$998\,244\,353$$$.

Examples:
input:
4
4
2
7
69
output:
5
1
682
325188814

Note: In the first test case, the following $$$5$$$ arrays can be obtained (the inserted inversion count is shown in bold):

- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, 1, \textbf{0}]$$$,
- $$$[0, 1] \rightarrow [0, \textbf{0}, 1] \rightarrow [0, 0, \textbf{0}, 1]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, 0, \textbf{1}]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [0, 1, \textbf{1}, 0]$$$,
- $$$[0, 1] \rightarrow [0, 1, \textbf{0}] \rightarrow [\textbf{1}, 0, 1, 0]$$$.

Buggy Code:
import java.io.*;
import java.util.*;
public class P6 {
    public static int MOD = 998244353;
    public static void main(String[] args) throws IOException {
        BufferedReader scan = new BufferedReader(new InputStreamReader(System.in)); PrintWriter out = new PrintWriter(System.out);
        StringTokenizer l1 = new StringTokenizer(scan.readLine());
        int t = Integer.parseInt(l1.nextToken());
        for (int q = 0;q<t;q++) {
            StringTokenizer st = new StringTokenizer(scan.readLine());
            int num = Integer.parseInt(st.nextToken());

            long[] dp = new long[num]; dp[num-1] = 1; //Number of final arrays we can make with a current array of length i+1 and having # inversion>max(a_i)
            long[] ss = new long[num]; ss[num-1] = 1; //Suffix Sum
            for (int i = num-2;i>=0;i--) {
                dp[i] = ((i+1)*ss[i+1]+1)%MOD;
                ss[i] = (ss[i+1]+dp[i])%MOD;
            }

            long ans = 0;
            for (int i = 2;i<num;i++) {
                int l = i+1;
                ans = (ans+((long) (l-2)*(l-1)/2-1)%MOD*dp[i])%MOD;
                //If the 0,1,0 is at the front, then 2 places. If it's at the back, then total length is l-1 (-1 since we've yet to place the 1) so l-2 total places to put it.
            }
            ans += num-1; //If the final number of inversions is <= 1.
            out.println(ans);
        }
        out.close();
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Satyam and Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: Satyam is given $$$n$$$ distinct points on the 2D coordinate plane. It is guaranteed that $$$0 \leq y_i \leq 1$$$ for all given points $$$(x_i, y_i)$$$. How many different nondegenerate right triangles$$$^{\text{∗}}$$$ can be formed from choosing three different points as its vertices?

Two triangles $$$a$$$ and $$$b$$$ are different if there is a point $$$v$$$ such that $$$v$$$ is a vertex of $$$a$$$ but not a vertex of $$$b$$$.

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

The first line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 2 \cdot 10^5$$$) — the number of points.

The following $$$n$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \leq x_i \leq n$$$, $$$0 \leq y_i \leq 1$$$) — the $$$i$$$'th point that Satyam can choose from. It is guaranteed that all $$$(x_i, y_i)$$$ are pairwise distinct.

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

Output: Output an integer for each test case, the number of distinct nondegenerate right triangles that can be formed from choosing three points.

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

Note: The four triangles in question for the first test case:

Buggy Code:
#include <iostream>
#include <algorithm>
using namespace std;
long long int n,ans=0;
struct point
{
    int x=100,y=100;
};
bool cmp(point a,point b)
{
    if(a.x!=b.x)
        return a.x<b.x;
    else return a.y>b.y;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        cin>>n;
        point p[n+5];
        for(int i=1;i<=n;i++)
            cin>>p[i].x>>p[i].y;
        sort(p+1,p+n+1,cmp);
        for(int i=1;i<=n;i++)
        {
            if(p[i+1].x==p[i].x)
            {
                ans+=n-2;
            }
            if(i>=2&&i<=n-1)
            {
                if(p[i].y==1)
                {
                    if(p[i-1].x==p[i].x-1&&p[i-1].y==p[i].y-1)
                    {
                        if(p[i+1].x==p[i].x+1&&p[i+1].y==p[i].y-1||p[i+2].x==p[i].x+1&&p[i+2].y==p[i].y-1||p[i+3].x==p[i].x+1&&p[i+3].y==p[i].y-1)
                            ans++;
                    }
                }
                else
                {
                    if(p[i-1].x==p[i].x-1&&p[i-1].y==p[i].y+1||p[i-2].x==p[i].x-1&&p[i-2].y==p[i].y+1||p[i-3].x==p[i].x-1&&p[i-3].y==p[i].y+1)
                    {
                        if(p[i+1].x==p[i].x+1&&p[i+1].y==p[i].y+1||p[i+2].x==p[i].x+1&&p[i+2].y==p[i].y+1)
                            ans++;
                    }
                }
            }
        }
        cout<<ans<<endl;
        ans=0;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases or scenarios that can maximize the failure probability. The key issues in the code include incorrect counting of right triangles and improper handling of points with the same x-coordinate.

Here are the steps to create the test input:

1. **Points with the same x-coordinate**: The code incorrectly adds `n-2` to the answer when it finds points with the same x-coordinate. This can lead to an overcounting of triangles.

2. **Handling of points with y = 0 and y = 1**: The code has specific conditions to check for right triangles, but these conditions are not comprehensive and can miss valid triangles.

3. **Small-scale test case**: We need a small-scale test case to ensure it fits within the constraints and is easy to validate.

Let's create a test case that includes points with the same x-coordinate and points that form right triangles:

```plaintext
3
5
1 0
1 1
3 0
5 0
2 1
3
0 0
1 0
3 0
9
1 0
2 0
3 0
4 0
5 0
2 1
7 1
8 1
9 1
```

This test case includes:
- The first test case with points (1,0), (1,1), (3,0), (5,0), (2,1) which should form 4 right triangles.
- The second test case with points (0,0), (1,0), (3,0) which should form 0 right triangles.
- The third test case with points (1,0), (2,0), (3,0), (4,0), (5,0), (2,1), (7,1), (8,1), (9,1) which should form 8 right triangles.

This test input will help expose the bug in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Satyam and Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: Satyam is given $$$n$$$ distinct points on the 2D coordinate plane. It is guaranteed that $$$0 \leq y_i \leq 1$$$ for all given points $$$(x_i, y_i)$$$. How many different nondegenerate right triangles$$$^{\text{∗}}$$$ can be formed from choosing three different points as its vertices?

Two triangles $$$a$$$ and $$$b$$$ are different if there is a point $$$v$$$ such that $$$v$$$ is a vertex of $$$a$$$ but not a vertex of $$$b$$$.

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

The first line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 2 \cdot 10^5$$$) — the number of points.

The following $$$n$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \leq x_i \leq n$$$, $$$0 \leq y_i \leq 1$$$) — the $$$i$$$'th point that Satyam can choose from. It is guaranteed that all $$$(x_i, y_i)$$$ are pairwise distinct.

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

Output: Output an integer for each test case, the number of distinct nondegenerate right triangles that can be formed from choosing three points.

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

Note: The four triangles in question for the first test case:

Buggy Code:
//#include <bits\stdc++.h>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdio>
#define int long long
using namespace std;

bool vis[200050][2];

void solve()
{
	int n;
	cin >> n;

	for (int i = 0; i <= 200050; i++)
	{
		vis[i][0] = 0;
		vis[i][1] = 0;
	}

	for (int i = 1; i <= n; i++)
	{
		int x, y;
		cin >> x >> y;
		vis[x][y] = 1;
	}

	int ans = 0;

	for (int i = 1; i <= n; i++)
	{
		if (vis[i][0] == 1 && vis[i][1] == 1)
			ans += (n - 2);

		if (vis[i][1] == 1 && vis[i - 1][0] == 1 && vis[i + 1][0] == 1)
			ans++;

		if (vis[i][0] == 1 && vis[i - 1][1] == 1 && vis[i + 1][1] == 1)
			ans++;
	}

	cout << ans << "\n";
}

signed main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.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 maximize the failure probability. The bug in the code seems to be in how it counts the number of right triangles. Specifically, the code incorrectly counts triangles by checking only a few specific conditions, which are not comprehensive enough to cover all valid right triangles.

### Step-by-Step Analysis:
1. **Identify the Bug**:
   - The code checks for right triangles by looking at specific patterns of points.
   - It only checks for triangles where the points are adjacent in the x-coordinate and differ in the y-coordinate.
   - This approach is too restrictive and will miss many valid right triangles.

2. **Generate Test Cases**:
   - **Test Case 1**: Points that form right triangles in a way that the code does not check.
   - **Test Case 2**: Points that are not adjacent in the x-coordinate but still form right triangles.
   - **Test Case 3**: Points that form right triangles with a large gap in the x-coordinate.

### Challenging Test Input:
```plaintext
3
5
1 0
2 0
3 0
2 1
4 1
4
1 0
3 0
2 1
4 1
7
1 0
2 0
3 0
4 0
1 1
2 1
4 1
```

### Explanation:
1. **Test Case 1**:
   - Points: (1, 0), (2, 0), (3, 0), (2, 1), (4, 1)
   - Expected Output: 4
   - Explanation: The valid right triangles are:
     - (1, 0), (2, 0), (2, 1)
     - (2, 0), (3, 0), (2, 1)
     - (2, 0), (2, 1), (4, 1)
     - (3, 0), (2, 1), (4, 1)

2. **Test Case 2**:
   - Points: (1, 0), (3, 0), (2, 1), (4, 1)
   - Expected Output: 2
   - Explanation: The valid right triangles are:
     - (1, 0), (3, 0), (2, 1)
     - (1, 0), (3, 0), (4, 1)

3. **Test Case 3**:
   - Points: (1, 0), (2, 0), (3, 0), (4, 0), (1, 1), (2, 1), (4, 1)
   - Expected Output: 6
   - Explanation: The valid right triangles are:
     - (1, 0), (2, 0), (1, 1)
     - (1, 0), (2, 0), (2, 1)
     - (1, 0), (3, 0), (1, 1)
     - (1, 0), (3, 0), (2, 1)
     - (1, 0), (4, 0), (1, 1)
     - (1, 0), (4, 0), (2, 1)

This test input should expose the bug in the code by showing that it fails to count all valid right triangles.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Satyam and Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: Satyam is given $$$n$$$ distinct points on the 2D coordinate plane. It is guaranteed that $$$0 \leq y_i \leq 1$$$ for all given points $$$(x_i, y_i)$$$. How many different nondegenerate right triangles$$$^{\text{∗}}$$$ can be formed from choosing three different points as its vertices?

Two triangles $$$a$$$ and $$$b$$$ are different if there is a point $$$v$$$ such that $$$v$$$ is a vertex of $$$a$$$ but not a vertex of $$$b$$$.

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

The first line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 2 \cdot 10^5$$$) — the number of points.

The following $$$n$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \leq x_i \leq n$$$, $$$0 \leq y_i \leq 1$$$) — the $$$i$$$'th point that Satyam can choose from. It is guaranteed that all $$$(x_i, y_i)$$$ are pairwise distinct.

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

Output: Output an integer for each test case, the number of distinct nondegenerate right triangles that can be formed from choosing three points.

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

Note: The four triangles in question for the first test case:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
#define int long long
#define loop(i, n, x) for(int i = 0; i < n; i += x)
int fact(int n, int m) {
    int ans = 1;
    for(int i = 1; i <= n; i++) {
        ans = (ans % m) * (i % m);
        ans = ans % m;
    }
    return ans;
}
int expo(int x, int n, int m) {
    if (n == 0) return 1;
    if (n % 2 == 0) return expo(((x % m) * (x % m)) % m, n / 2, m);
    else return (x % m) * expo(((x % m) * (x % m)) % m, (n - 1) / 2, m) % m;
}
int gcd(int a, int b) {
    if (b == 0) return a;
    if (a == 0) return b;
    else return gcd(b, a % b);
}
bool c(pair<int, int> p1, pair<int, int> p2) {
    return p1.second < p2.second;
}
bool d(pair<int, int> p1, pair<int, int> p2) {
    return p1.first > p2.first;
}
struct cmp{
    bool operator()(int a, int b) const { return a > b;}
};
int ans(int  n,int k,int x){
    int sum = 0;
    int s = 0;
    loop(i,x,1) sum+=(k+i);
    for(int i = x ; i<n;i++) s += (k+i);
    return  (sum - s);
}
void solve() {
   int n; cin >> n ;
   vector<pair<int,int>>xy;
   int ans = 0;
   loop(i,n,1){
    int  x , y;
    cin >> x >> y;
    if(y == 1)ans++;
   }
   cout << ans*2 << "\n";
}
int32_t main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    // solve();
    int t;
    cin >> t;
    while(t--) {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
3
4
1 0
2 0
1 1
2 1
5
0 0
1 0
2 0
3 0
4 0
6
0 0
1 0
2 0
0 1
1 1
2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Satyam and Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: Satyam is given $$$n$$$ distinct points on the 2D coordinate plane. It is guaranteed that $$$0 \leq y_i \leq 1$$$ for all given points $$$(x_i, y_i)$$$. How many different nondegenerate right triangles$$$^{\text{∗}}$$$ can be formed from choosing three different points as its vertices?

Two triangles $$$a$$$ and $$$b$$$ are different if there is a point $$$v$$$ such that $$$v$$$ is a vertex of $$$a$$$ but not a vertex of $$$b$$$.

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

The first line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 2 \cdot 10^5$$$) — the number of points.

The following $$$n$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \leq x_i \leq n$$$, $$$0 \leq y_i \leq 1$$$) — the $$$i$$$'th point that Satyam can choose from. It is guaranteed that all $$$(x_i, y_i)$$$ are pairwise distinct.

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

Output: Output an integer for each test case, the number of distinct nondegenerate right triangles that can be formed from choosing three points.

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

Note: The four triangles in question for the first test case:

Buggy Code:
T = int(input())
for tt in range(T):
    try:
        num_pts = int(input())
        x0 = []
        x1 = []
        for _ in range(num_pts):
            x,y = list(map(lambda a: int(a), input().split(" ")))
            if y == 1:
                x1.append(x)
            else:
                x0.append(x)
        ans = 0
        x0 = sorted(x0)
        x1 = sorted(x1)
        t0 = len(x0)
        t1 = len(x1)

        idx = 0
        p = x0[idx]
        for idex, x in enumerate(x1):
            while p <= x+1 and idx < t0:
                if p == x+1:
                    if idex+1 < t1:
                        if x1[idex+1] == p:
                            if idex+2 < t1:
                                if x1[idex+2] == p+1:
                                    ans += 1
                        elif x1[idex+1] == p+1:
                            ans += 1

                idx += 1
                if idx < t0:
                    p = x0[idx]

        idx = 0
        p = x0[idx]
        for idex, x in enumerate(x1):
            while p <= x and idx < t0:
                if p == x:
                    ans += t0+t1-2

                idx += 1
                if idx < t0:
                    p = x0[idx]

        idx = 0
        p = x0[idx]
        for idex, x in enumerate(x1):
            while p <= x-1 and idx < t0:
                if p == x-1:
                    if idx+1 < t0:
                        if x0[idx+1] == p+1:
                            if idx+2 < t0:
                                if x0[idx+2] == p+2:
                                    ans += 1
                        elif x0[idx+1] == p+2:
                            ans += 1

                idx += 1
                if idx < t0:
                    p = x0[idx]

        print(ans)

    except Exception as e:
        print(e)

**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
7
0 0
1 0
2 0
3 0
0 1
1 1
2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Satyam and Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: Satyam is given $$$n$$$ distinct points on the 2D coordinate plane. It is guaranteed that $$$0 \leq y_i \leq 1$$$ for all given points $$$(x_i, y_i)$$$. How many different nondegenerate right triangles$$$^{\text{∗}}$$$ can be formed from choosing three different points as its vertices?

Two triangles $$$a$$$ and $$$b$$$ are different if there is a point $$$v$$$ such that $$$v$$$ is a vertex of $$$a$$$ but not a vertex of $$$b$$$.

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

The first line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 2 \cdot 10^5$$$) — the number of points.

The following $$$n$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \leq x_i \leq n$$$, $$$0 \leq y_i \leq 1$$$) — the $$$i$$$'th point that Satyam can choose from. It is guaranteed that all $$$(x_i, y_i)$$$ are pairwise distinct.

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

Output: Output an integer for each test case, the number of distinct nondegenerate right triangles that can be formed from choosing three points.

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

Note: The four triangles in question for the first test case:

Buggy Code:
def solve():
    n = int(input())
    xc =  set()
    yc = set()
    for i in range(n):
        x,y = map(int,input().split())
        xc.add(x)
        yc.add(y)
    if(len(yc) ==1):
        print(0)
    else:
        print(len(xc))

t = int(input())

while t:
    solve()
    t -= 1

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

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

Think step by step.

Response:
```plaintext
3
4
1 0
2 0
3 0
4 1
5
1 0
2 0
3 0
4 1
5 1
6
1 0
2 0
3 0
4 1
5 1
6 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Satyam and Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: Satyam is given $$$n$$$ distinct points on the 2D coordinate plane. It is guaranteed that $$$0 \leq y_i \leq 1$$$ for all given points $$$(x_i, y_i)$$$. How many different nondegenerate right triangles$$$^{\text{∗}}$$$ can be formed from choosing three different points as its vertices?

Two triangles $$$a$$$ and $$$b$$$ are different if there is a point $$$v$$$ such that $$$v$$$ is a vertex of $$$a$$$ but not a vertex of $$$b$$$.

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

The first line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 2 \cdot 10^5$$$) — the number of points.

The following $$$n$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \leq x_i \leq n$$$, $$$0 \leq y_i \leq 1$$$) — the $$$i$$$'th point that Satyam can choose from. It is guaranteed that all $$$(x_i, y_i)$$$ are pairwise distinct.

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

Output: Output an integer for each test case, the number of distinct nondegenerate right triangles that can be formed from choosing three points.

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

Note: The four triangles in question for the first test case:

Buggy Code:
from sys import stdin, stdout
from collections import defaultdict,deque
input = stdin.readline
inp = lambda :  list(map(int,input().split()))
inps= lambda :	input().strip()
inpsl= lambda :	list(input().strip().split())
import math,random
MOD = 1000000007
class CustomSet:
    def __init__(self, arr = []):
        self.RANDOM = random.randrange(2**62)
        self.set = set(arr)
    def wrapper(self,num):
        return num ^ self.RANDOM
    def add(self, key):
        self.set.add(self.wrapper(key))
    def remove(self, key):
        self.set.remove(self.wrapper(key))
    def __contains__(self, key):
        return self.wrapper(key) in self.set
    def __iter__(self):
        return iter(self.wrapper(i) for i in self.set)
    def __len__(self):
        return len(self.set)
    def __repr__(self):
        return repr({self.wrapper(i) for i in self.set})
for _ in range(*inp()):
    n,=inp()
    y=[[0]*n for i in range(2)]
    for i in range(n):
        a,b=inp()
        y[b-1][a-1]+=1
  #  print(y)
    y0=sum(y[0])
    y1=sum(y[1])
    ans=0
   # print(y0,y1)
    for i in range(n):
        if y[0][i] and y[1][i]:
            ans+=y[0][i]*y[1][i]*(y0+y1-y[0][i]-y[1][i])
    for i in range(n-2):
        ans+=y[0][i]*y[0][i+2]*y[1][i+1]
        ans+=y[1][i]*y[1][i+2]*y[0][i+1]
    print(ans)

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

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

Think step by step.

Response:
```plaintext
3
4
1 0
2 0
1 1
2 1
5
1 0
2 0
3 0
1 1
3 1
6
1 0
2 0
3 0
1 1
2 1
3 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Satyam and Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: Satyam is given $$$n$$$ distinct points on the 2D coordinate plane. It is guaranteed that $$$0 \leq y_i \leq 1$$$ for all given points $$$(x_i, y_i)$$$. How many different nondegenerate right triangles$$$^{\text{∗}}$$$ can be formed from choosing three different points as its vertices?

Two triangles $$$a$$$ and $$$b$$$ are different if there is a point $$$v$$$ such that $$$v$$$ is a vertex of $$$a$$$ but not a vertex of $$$b$$$.

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

The first line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 2 \cdot 10^5$$$) — the number of points.

The following $$$n$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \leq x_i \leq n$$$, $$$0 \leq y_i \leq 1$$$) — the $$$i$$$'th point that Satyam can choose from. It is guaranteed that all $$$(x_i, y_i)$$$ are pairwise distinct.

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

Output: Output an integer for each test case, the number of distinct nondegenerate right triangles that can be formed from choosing three points.

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

Note: The four triangles in question for the first test case:

Buggy Code:
import java.util.*;
public class cp{
	public static void main(String [] args) {
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0) {
			Set<Integer> s1=new HashSet<>();
			Set<Integer> s2=new HashSet<>();
			int n=sc.nextInt();
			for(int i=0;i<n;i++) {
				int xi=sc.nextInt();
				int yi=sc.nextInt();
				if(yi==0) {
					s1.add(xi);
				}
				else {
					s2.add(xi);
				}
			}
			Set<Integer> s=new HashSet<>();
			for(int a:s1) {
				s.add(a);
			}
			for(int a:s2) {
				s.add(a);
			}
			int ans=(s1.size()+s2.size()-s.size())*s.size();
			System.out.println(ans);
		}
	}
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, let's analyze the code and identify the potential issues:

1. **Buggy Code Analysis:**
   - The code uses two sets, `s1` and `s2`, to store the x-coordinates of points with y=0 and y=1, respectively.
   - It then combines these sets into a single set `s` and calculates the number of distinct nondegenerate right triangles using the formula `(s1.size() + s2.size() - s.size()) * s.size()`.

2. **Potential Issues:**
   - The formula used to calculate the number of right triangles is incorrect. The correct approach should involve counting the number of pairs of points with y=0 and y=1 that can form the legs of the right triangle.

3. **Edge Cases:**
   - **All points on y=0 or y=1:** This can cause the formula to fail because it will not correctly count the number of right triangles.
   - **Points with the same x-coordinate:** This can cause the formula to incorrectly count the number of points.

4. **Challenging Test Input:**
   - We need a test case where the points have the same x-coordinates but different y-coordinates to expose the bug in the formula.

Given these considerations, here is a small-scale test input that should expose the bug:

```plaintext
1
6
1 0
1 1
2 0
2 1
3 0
3 1
```

This test case includes points with the same x-coordinates but different y-coordinates, which should cause the buggy code to produce an incorrect result. The correct number of right triangles that can be formed from these points is 3, but the buggy code will likely produce a different result.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Satyam and Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: Satyam is given $$$n$$$ distinct points on the 2D coordinate plane. It is guaranteed that $$$0 \leq y_i \leq 1$$$ for all given points $$$(x_i, y_i)$$$. How many different nondegenerate right triangles$$$^{\text{∗}}$$$ can be formed from choosing three different points as its vertices?

Two triangles $$$a$$$ and $$$b$$$ are different if there is a point $$$v$$$ such that $$$v$$$ is a vertex of $$$a$$$ but not a vertex of $$$b$$$.

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

The first line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 2 \cdot 10^5$$$) — the number of points.

The following $$$n$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \leq x_i \leq n$$$, $$$0 \leq y_i \leq 1$$$) — the $$$i$$$'th point that Satyam can choose from. It is guaranteed that all $$$(x_i, y_i)$$$ are pairwise distinct.

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

Output: Output an integer for each test case, the number of distinct nondegenerate right triangles that can be formed from choosing three points.

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

Note: The four triangles in question for the first test case:

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

public class Main {

    static ArrayList<Integer>[] adjList;

    public static void doJob() throws Exception {

            int n = sc.nextInt();
            pair[] arr = new pair[n];
            for(int i=0; i<n; i++){
                int a = sc.nextInt();
                int b = sc.nextInt();
                pair p = new pair(a, b);
                arr[i] = p;
            }
            Arrays.sort(arr);

            boolean[] zeros = new boolean[200001];
            boolean[] ones = new boolean[200001];

            for(int i=0; i<n; i++) {
                if(arr[i].y==0){
                    zeros[arr[i].x] = true;
                }
                else{
                    ones[arr[i].x] = true;
                }
            }

             int count =0;
             for(int i=0;i<n-1;i++){
                 if(arr[i].x==arr[i+1].x){
                     count += n-2;
                 }

//                 if(i>1 && arr[i].x==arr[i-1].x){
//                     //|| (arr[i-2].y!=arr[i].y && arr[i-2].y!=arr[i+1].y) || (arr[i+1].y!=arr[i].y && arr[i+1].y!=arr[i-2].y)
//                     if(arr[i].x==arr[i-2].x+1 && arr[i].x==arr[i+1].x-1 && ((arr[i].y==0 && ones[arr[i-2].x] && ones[arr[i+1].x]) || (arr[i].y==1 && zeros[arr[i-2].x] && zeros[arr[i+1].x]))){
//                         count++;
//                     }
//                 }
//                 else if(i>1 && i<n-2 && arr[i].x==arr[i+1].x){
////                     || (arr[i-1].y!=arr[i].y && arr[i-1].y!=arr[i+2].y) || (arr[i+2].y!=arr[i].y && arr[i+2].y!=arr[i-1].y)
//                     if(i!=0 && arr[i].x==arr[i-1].x+1 && arr[i].x==arr[i+2].x-1 && ((arr[i].y==0 && ones[arr[i-1].x] && ones[arr[i+2].x]) || (arr[i].y==1 && zeros[arr[i-1].x] && zeros[arr[i+2].x]))){
//                         count++;
//                     }
//                 }
//                 //|| (arr[i-1].y!=arr[i].y && arr[i-1].y!=arr[i+1].y) || (arr[i+1].y!=arr[i].y && arr[i+1].y!=arr[i-1].y)
//                 else if(i!=0 && arr[i].x==arr[i-1].x+1 && arr[i].x==arr[i+1].x-1 && ((zeros[arr[i].x] && ones[arr[i-1].x] && ones[arr[i+1].x]) || (ones[arr[i].x] && zeros[arr[i-1].x] && zeros[arr[i+1].x]))){
//                     count ++;
//                 }

             }

             for(int i=1;i<n;i++){
                 if(arr[i].y==0 && ones[arr[i].x-1] && ones[arr[i].x+1]){
                     count++;
                 }
                 else if(arr[i].y==1 && zeros[arr[i].x-1] && zeros[arr[i].x+1]){
                     count++;
                 }
             }

             pw.println(count);

    }

    public static class pair implements Comparable<pair> {
        int x;
        int y;

        public pair(int x, int y) {
            this.x = x;
            this.y = y;
        }

        public int compareTo(pair o) {
            if (this.x == o.x)
                return this.y - o.y;
            return this.x - o.x;

        }

        public String toString() {
            return x + " " + y;
        }
    }

    public static void doJobT() throws Exception {
        int t = sc.nextInt();

        while (t-- > 0) {
            doJob();
        }
    }

    public static void main(String[] args) throws Exception {
        sc = new Scanner(System.in);
        pw = new PrintWriter(System.out);
        // sc = new Scanner("in.txt");
        // pw = new PrintWriter("out.txt");

        //doJob();
        doJobT();
        //System.out.print(Character.getNumericValue('1'));

        pw.flush();
        pw.close();
    }

    /*-----------------------------------------Helpers--------------------------------------------*/

    public static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a % b);
    }

    public static int getLength(Stack s){
        int count = 0;
        Stack temp = new Stack<>();
        while(!s.isEmpty()){
            temp.push(s.pop());
            count++;
        }
        while(!temp.isEmpty()){
            s.push(temp.pop());
        }
        return count;
    }

    public static long lcm(long a, long b) {
        return (a / gcd(a, b)) * b;
    }

    public static long fact(long n) {
        if (n == 0)
            return 1;
        int res = 1;
        for (int i = 2; i <= n; i++)
            res = res * i;
        return res;
    }

    public static int count(int[] arr, int c) {

        int count = 0;
        for (int j : arr) {
            if (j == c)
                count++;
        }
        return count;

    }

    /*-----------------------------------------Scanner--------------------------------------------*/

    static Scanner sc;

    static PrintWriter pw;

    static class Scanner {

        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(String r) throws Exception {
            br = new BufferedReader(new FileReader(new File(r)));
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

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

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

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++)
                if (x.charAt(i) == '.') {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec)
                        f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

        public Long[] nextLongArray(int n) throws IOException {
            Long[] a = new Long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        public int[] nextIntArray(int n) throws IOException {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public Integer[] nextIntegerArray(int n) throws IOException {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public boolean ready() throws IOException {
            return br.ready();
        }
    }

}

**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
5
1 0
1 1
2 0
2 1
3 0
3
0 0
1 0
1 1
9
1 0
2 0
3 0
4 0
5 0
2 1
3 1
4 1
5 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Satyam and Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: Satyam is given $$$n$$$ distinct points on the 2D coordinate plane. It is guaranteed that $$$0 \leq y_i \leq 1$$$ for all given points $$$(x_i, y_i)$$$. How many different nondegenerate right triangles$$$^{\text{∗}}$$$ can be formed from choosing three different points as its vertices?

Two triangles $$$a$$$ and $$$b$$$ are different if there is a point $$$v$$$ such that $$$v$$$ is a vertex of $$$a$$$ but not a vertex of $$$b$$$.

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

The first line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 2 \cdot 10^5$$$) — the number of points.

The following $$$n$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \leq x_i \leq n$$$, $$$0 \leq y_i \leq 1$$$) — the $$$i$$$'th point that Satyam can choose from. It is guaranteed that all $$$(x_i, y_i)$$$ are pairwise distinct.

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

Output: Output an integer for each test case, the number of distinct nondegenerate right triangles that can be formed from choosing three points.

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

Note: The four triangles in question for the first test case:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define lld long double
#define ull unsigned long long
#define all(x) x.begin(),x.end()
#define vl vector<ll>
#define fr(v) for(ll i=0;i<v.size();i++)
#define fi(i,v) for(auto &i:v)
#define pb push_back
#define ppb pop_back
#define um unordered_map<ll,ll>
#define us unordered_set<ll>
#define vpl vector<pair<ll,ll>>
#define endl "\n"

#ifndef ONLINE_JUDGE
    #define debug(x) cerr << #x <<" "; _print(x); cerr << endl;
    #else
    #define debug(x)
    #endif

    void _print(ll t) {cerr << t;}
    void _print(int t) {cerr << t;}
    void _print(string t) {cerr << t;}
    void _print(char t) {cerr << t;}
    void _print(lld t) {cerr << t;}
    void _print(double t) {cerr << t;}
    void _print(ull t) {cerr << t;}

    template <class T, class V> void _print(pair <T, V> p);
    template <class T> void _print(vector <T> v);
    template <class T> void _print(set <T> v);
    template <class T, class V> void _print(map <T, V> v);
    template <class T> void _print(multiset <T> v);
    template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.ff); cerr << ","; _print(p.ss); cerr << "}";}
    template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
    template <class T> void _print(set <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
    template <class T> void _print(multiset <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
    template <class T, class V> void _print(map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}

const ll mod=1000000007;
ll power(ll x,ll y)
{
    ll res = 1;
    // x%=k;0
    while (y > 0) {
        if (y & 1)
            res = (res%mod * x)%mod;
        y = y >> 1;
        x = (x * x)%mod;
    }
    return res%mod;
}
ll signum(ll x)
{
    if(x>0) return 1;
    else if(x==0) return 0;
    else return -1;
}
void fast_io()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
}
ll tc=1;
#define gcc(x) cout<<"Case #"<<tc++<<": "<<x<<endl
//////////////////////////////////////////////////////////
///////////////////Solving area is Below//////////////////
/////////////////////////////////////////////////////////

void solve()
{
    int n;
    cin>>n;
    int x,y;
    int yCnt=0;
    us ans;
    for(int i=0;i<n;i++) {
        cin>>x>>y;
        ans.insert(x);
        if(y!=0) yCnt++;
    }
    if(yCnt>0) cout<<ans.size()<<endl;
    else cout<<0<<endl;
}

int main()
{
    #ifndef ONLINE_JUDGE
        freopen("error.txt", "w", stderr);
    #endif
    fast_io();
    ll 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
3
5
1 0
1 1
2 0
2 1
3 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Satyam and Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: Satyam is given $$$n$$$ distinct points on the 2D coordinate plane. It is guaranteed that $$$0 \leq y_i \leq 1$$$ for all given points $$$(x_i, y_i)$$$. How many different nondegenerate right triangles$$$^{\text{∗}}$$$ can be formed from choosing three different points as its vertices?

Two triangles $$$a$$$ and $$$b$$$ are different if there is a point $$$v$$$ such that $$$v$$$ is a vertex of $$$a$$$ but not a vertex of $$$b$$$.

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

The first line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 2 \cdot 10^5$$$) — the number of points.

The following $$$n$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \leq x_i \leq n$$$, $$$0 \leq y_i \leq 1$$$) — the $$$i$$$'th point that Satyam can choose from. It is guaranteed that all $$$(x_i, y_i)$$$ are pairwise distinct.

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

Output: Output an integer for each test case, the number of distinct nondegenerate right triangles that can be formed from choosing three points.

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

Note: The four triangles in question for the first test case:

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

#define int long long
#define fast ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);

void solve(){
    int n;
    cin >> n;
    int z = 0, o = 0;
    map <int, int> ol, zl;
    for(int i = 0; i < n; i++){
        int x, y;
        cin >> x >> y;
        if(y == 1){
            o ++;
            ol[x] ++;
        }
        else{
            z ++;
            zl[x] ++;
        }
    }
    int ans = 0;
    for(auto i : ol){
        if(zl[i.first -1] > 0 && zl[i.first + 1] > 0){
            ans ++;
        }
    }
    for(auto i : ol){
        if(zl[i.first] > 0){
            ans += z + o -2;
        }
    }
    cout << ans << endl;
}

signed main() {
    fast
    int T = 1;
    cin >> T;
    while (T--) {
        solve();
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider the following:

1. **Edge Cases**: Points that are aligned in such a way that the code might incorrectly count the number of right triangles.
2. **Degenerate Cases**: Points that do not form a right triangle.
3. **Maximum and Minimum Values**: Points at the boundaries of the allowed range.
4. **Distribution of Points**: Points that are spread out or concentrated in specific areas.

The buggy code has a few issues:
- It incorrectly counts the number of right triangles by not considering all possible combinations of points.
- It does not handle the cases where the points are aligned in a way that forms right triangles correctly.

Let's create a small-scale test input that includes points that are likely to expose these issues.

### Test Input
```plaintext
3
5
1 0
1 1
2 0
2 1
3 0
3
0 0
1 0
2 0
7
0 0
1 0
2 0
0 1
1 1
2 1
3 1
```

### Explanation
1. **First Test Case (5 points)**:
   - Points: (1, 0), (1, 1), (2, 0), (2, 1), (3, 0)
   - This set of points includes two vertical lines and one horizontal line. The correct number of right triangles should be 4 (two from each vertical line with the horizontal line).

2. **Second Test Case (3 points)**:
   - Points: (0, 0), (1, 0), (2, 0)
   - All points are on the same horizontal line, so no right triangles can be formed. The correct output should be 0.

3. **Third Test Case (7 points)**:
   - Points: (0, 0), (1, 0), (2, 0), (0, 1), (1, 1), (2, 1), (3, 1)
   - This set of points includes three horizontal lines and one vertical line. The correct number of right triangles should be 8 (each combination of two points on a horizontal line with one point on the vertical line).

This test input should help expose the bugs in the provided code by covering various scenarios where the code might fail.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Satyam and Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: Satyam is given $$$n$$$ distinct points on the 2D coordinate plane. It is guaranteed that $$$0 \leq y_i \leq 1$$$ for all given points $$$(x_i, y_i)$$$. How many different nondegenerate right triangles$$$^{\text{∗}}$$$ can be formed from choosing three different points as its vertices?

Two triangles $$$a$$$ and $$$b$$$ are different if there is a point $$$v$$$ such that $$$v$$$ is a vertex of $$$a$$$ but not a vertex of $$$b$$$.

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

The first line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 2 \cdot 10^5$$$) — the number of points.

The following $$$n$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \leq x_i \leq n$$$, $$$0 \leq y_i \leq 1$$$) — the $$$i$$$'th point that Satyam can choose from. It is guaranteed that all $$$(x_i, y_i)$$$ are pairwise distinct.

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

Output: Output an integer for each test case, the number of distinct nondegenerate right triangles that can be formed from choosing three points.

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

Note: The four triangles in question for the first test case:

Buggy Code:
t = int(input())
for i in range(t):
    points = []
    no_pts = int(input())
    for j in range(no_pts):
        a = tuple(map(int,input().split()))
        points.append(a)
    count = 0
    for k in points:
        if k[1] == 0 and (k[0],1) in points:
            count += (len(points)-2)
        elif k[1] == 0 and (((k[0]-1,1) in points and (k[0]+1,1)) in points):
            count +=1
        elif k[1] == 1 and (((k[0]-1,0) in points and (k[0]+1,0)) in points):
            count += 1
    print(count)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to consider the following:

1. **Identify the Bug**: The buggy code incorrectly counts the number of right triangles. It seems to count all possible combinations without properly checking for the right angle condition.

2. **Edge Cases**:
   - **Collinear Points**: Points that lie on the same horizontal or vertical line should not form a right triangle.
   - **Minimal Set of Points**: Use a small set of points to make it easier to verify the output.
   - **Specific Configurations**: Points that can form right triangles in specific configurations to test the logic.

3. **Construct the Test Case**:
   - **Test Case 1**: A set of points where no right triangles can be formed.
   - **Test Case 2**: A set of points where exactly one right triangle can be formed.
   - **Test Case 3**: A set of points where multiple right triangles can be formed.

Let's construct a test case that includes a minimal set of points to expose the bug:

```plaintext
3
3
1 0
2 0
3 0
4
1 0
2 0
1 1
2 1
6
1 0
2 0
3 0
1 1
2 1
3 1
```

**Explanation**:
- **Test Case 1**: All points are collinear (on the same horizontal line), so no right triangles can be formed.
- **Test Case 2**: Points form one right triangle at (1,0), (2,0), (1,1).
- **Test Case 3**: Points form multiple right triangles, specifically:
  - (1,0), (2,0), (1,1)
  - (1,0), (2,0), (2,1)
  - (1,0), (3,0), (1,1)
  - (1,0), (3,0), (2,1)
  - (2,0), (3,0), (2,1)
  - (2,0), (3,0), (1,1)

This test case should help expose the bug in the code by showing incorrect counts for the number of right triangles.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Satyam and Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: Satyam is given $$$n$$$ distinct points on the 2D coordinate plane. It is guaranteed that $$$0 \leq y_i \leq 1$$$ for all given points $$$(x_i, y_i)$$$. How many different nondegenerate right triangles$$$^{\text{∗}}$$$ can be formed from choosing three different points as its vertices?

Two triangles $$$a$$$ and $$$b$$$ are different if there is a point $$$v$$$ such that $$$v$$$ is a vertex of $$$a$$$ but not a vertex of $$$b$$$.

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

The first line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 2 \cdot 10^5$$$) — the number of points.

The following $$$n$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \leq x_i \leq n$$$, $$$0 \leq y_i \leq 1$$$) — the $$$i$$$'th point that Satyam can choose from. It is guaranteed that all $$$(x_i, y_i)$$$ are pairwise distinct.

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

Output: Output an integer for each test case, the number of distinct nondegenerate right triangles that can be formed from choosing three points.

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

Note: The four triangles in question for the first test case:

Buggy Code:
import sys

import sys

def main():
    import sys
    import sys
    input = sys.stdin.read
    data = input().split()
    idx = 0
    t = int(data[idx])
    idx +=1
    results = []
    for _ in range(t):
        n = int(data[idx])
        idx +=1
        set_Y0 = set()
        set_Y1 = set()
        for _ in range(n):
            x = int(data[idx])
            y = int(data[idx+1])
            idx +=2
            if y ==0:
                set_Y0.add(x)
            else:
                set_Y1.add(x)
        intersection = set_Y0 & set_Y1
        m = len(set_Y0)
        k = len(set_Y1)
        count_A = len(intersection) * (m -1)
        count_C_Y1 = len(intersection) * (k -1)
        count_C_Y0 = 0
        for x in set_Y1:
            if (x-1) in set_Y0 and (x+1) in set_Y0:
                count_C_Y0 +=1
        total = count_A + count_C_Y1 + count_C_Y0
        results.append(str(total))
    print('\n'.join(results))

if __name__ == "__main__":
    main()

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases that can highlight issues in the logic. The key points to consider are:

1. **Points on the same horizontal line**: This can test the ability of the code to handle cases where no right triangles can be formed.
2. **Points on the same vertical line**: This is not possible given the constraint \(0 \leq y_i \leq 1\), but we can still test points with the same \(x\) value on different \(y\) values.
3. **Points that form right triangles**: Ensure the code correctly counts all possible right triangles.
4. **Points that form degenerate triangles**: Ensure the code correctly handles cases where the points form degenerate triangles (e.g., collinear points).

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

```plaintext
3
3
0 0
1 0
2 0
5
1 0
1 1
3 0
3 1
5 0
7
0 0
0 1
1 0
1 1
2 0
2 1
3 0
```

### Explanation:
1. **First Test Case**:
   - Points: \((0, 0)\), \((1, 0)\), \((2, 0)\)
   - All points are on the same horizontal line, so no right triangles can be formed.
   - Expected Output: \(0\)

2. **Second Test Case**:
   - Points: \((1, 0)\), \((1, 1)\), \((3, 0)\), \((3, 1)\), \((5, 0)\)
   - Points \((1, 0)\) and \((1, 1)\) form a vertical line.
   - Points \((3, 0)\) and \((3, 1)\) form a vertical line.
   - Right triangles can be formed using these points.
   - Expected Output: \(4\) (right triangles: \((1, 0)\), \((1, 1)\), \((3, 0)\); \((1, 0)\), \((1, 1)\), \((3, 1)\); \((1, 0)\), \((3, 0)\), \((3, 1)\); \((1, 1)\), \((3, 0)\), \((3, 1)\))

3. **Third Test Case**:
   - Points: \((0, 0)\), \((0, 1)\), \((1, 0)\), \((1, 1)\), \((2, 0)\), \((2, 1)\), \((3, 0)\)
   - Multiple points form vertical lines and can form right triangles.
   - Expected Output: \(8\) (right triangles: \((0, 0)\), \((0, 1)\), \((1, 0)\); \((0, 0)\), \((0, 1)\), \((1, 1)\); \((0, 0)\), \((1, 0)\), \((1, 1)\); \((0, 1)\), \((1, 0)\), \((1, 1)\); \((1, 0)\), \((1, 1)\), \((2, 0)\); \((1, 0)\), \((1, 1)\), \((2, 1)\); \((1, 0)\), \((2, 0)\), \((2, 1)\); \((1, 1)\), \((2, 0)\), \((2, 1)\))

This test input should help expose any issues in the buggy code, particularly in handling points that form right triangles and points that are on the same horizontal or vertical lines.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Satyam and Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: Satyam is given $$$n$$$ distinct points on the 2D coordinate plane. It is guaranteed that $$$0 \leq y_i \leq 1$$$ for all given points $$$(x_i, y_i)$$$. How many different nondegenerate right triangles$$$^{\text{∗}}$$$ can be formed from choosing three different points as its vertices?

Two triangles $$$a$$$ and $$$b$$$ are different if there is a point $$$v$$$ such that $$$v$$$ is a vertex of $$$a$$$ but not a vertex of $$$b$$$.

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

The first line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 2 \cdot 10^5$$$) — the number of points.

The following $$$n$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \leq x_i \leq n$$$, $$$0 \leq y_i \leq 1$$$) — the $$$i$$$'th point that Satyam can choose from. It is guaranteed that all $$$(x_i, y_i)$$$ are pairwise distinct.

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

Output: Output an integer for each test case, the number of distinct nondegenerate right triangles that can be formed from choosing three points.

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

Note: The four triangles in question for the first test case:

Buggy Code:
import java.util.Scanner;

public class D {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

//---------Code starts here---------
        int t = sc.nextInt();

        while (t-- > 0) {
            int n = sc.nextInt();
            int yn = 0;
            for (int j = 0; j < n; j++) {
                int x = sc.nextInt();
                int y = sc.nextInt();
                if(y == 1) yn++;
            }

            System.out.println(yn*2);
            // AHelper obj = new AHelper();
            // System.out.println(obj.aHelperMethod(strip, n));
        }
        sc.close();
    }
}

// class DHelper{
//     public int dHelperMethod(String strip, int n) {
//     }
// }

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on the logic that calculates the number of distinct nondegenerate right triangles. The buggy code simply multiplies the number of points with \( y = 1 \) by 2, which is incorrect. We need to create a test case where this logic fails to produce the correct number of right triangles.

### Step-by-Step Reasoning:

1. **Understanding the Bug:**
   - The buggy code assumes that each point with \( y = 1 \) can form exactly 2 right triangles, which is not true. The correct number of right triangles depends on the number of points with \( y = 0 \) and the number of points with \( y = 1 \).

2. **Correct Logic:**
   - A right triangle can be formed if one point is on the x-axis (\( y = 0 \)) and the other two points are on the line \( y = 1 \). The number of such triangles is given by the combination of choosing 2 points from the points on \( y = 1 \) and 1 point from the points on \( y = 0 \).

3. **Creating a Test Case:**
   - We need a test case where the number of points with \( y = 1 \) and \( y = 0 \) are such that the correct number of right triangles is not equal to the number of points with \( y = 1 \) multiplied by 2.

### Example Test Case:

Let's consider a test case with 6 points:
- 4 points on the line \( y = 0 \)
- 2 points on the line \( y = 1 \)

The correct number of right triangles can be calculated as follows:
- Choose 1 point from the 4 points on \( y = 0 \) (4 ways)
- Choose 2 points from the 2 points on \( y = 1 \) (1 way)

So, the number of right triangles is \( 4 \times 1 = 4 \).

However, the buggy code will output \( 2 \times 2 = 4 \), which is coincidentally correct for this specific case. To ensure the bug is exposed, we need a case where the number of points with \( y = 1 \) is greater than 2.

### Final Test Case:

Let's consider a test case with 7 points:
- 3 points on the line \( y = 0 \)
- 4 points on the line \( y = 1 \)

The correct number of right triangles can be calculated as follows:
- Choose 1 point from the 3 points on \( y = 0 \) (3 ways)
- Choose 2 points from the 4 points on \( y = 1 \) (6 ways)

So, the number of right triangles is \( 3 \times 6 = 18 \).

The buggy code will output \( 4 \times 2 = 8 \), which is incorrect.

### Output:

```plaintext
1
7
1 0
2 0
3 0
1 1
2 1
3 1
4 1
```

This test case should expose the bug in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Satyam and Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: Satyam is given $$$n$$$ distinct points on the 2D coordinate plane. It is guaranteed that $$$0 \leq y_i \leq 1$$$ for all given points $$$(x_i, y_i)$$$. How many different nondegenerate right triangles$$$^{\text{∗}}$$$ can be formed from choosing three different points as its vertices?

Two triangles $$$a$$$ and $$$b$$$ are different if there is a point $$$v$$$ such that $$$v$$$ is a vertex of $$$a$$$ but not a vertex of $$$b$$$.

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

The first line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 2 \cdot 10^5$$$) — the number of points.

The following $$$n$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \leq x_i \leq n$$$, $$$0 \leq y_i \leq 1$$$) — the $$$i$$$'th point that Satyam can choose from. It is guaranteed that all $$$(x_i, y_i)$$$ are pairwise distinct.

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

Output: Output an integer for each test case, the number of distinct nondegenerate right triangles that can be formed from choosing three points.

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

Note: The four triangles in question for the first test case:

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

public class Main
{
    static FastScanner sc = new FastScanner(System.in);// 快读
    static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));// 快速输出
    static StringTokenizer st;
    public static void main(String[] args)
    {
        while(sc.hasNext())
        {

            //由此输入：

//          int n=sc.nextInt();
//          out.println(sum<<1);
        	int t=sc.nextInt();
			while(t-->0)
			{
				int n=sc.nextInt();
				boolean[][] arr=new boolean[200300][2];
				int max=-1;
				for(int i=1;i<=n;i++)
				{
					int x=sc.nextInt();
					int y=sc.nextInt();
					arr[x][y]=true;
					max=Math.max(max, x);
				}
				int cnt=0;
				for(int i=0;i<=max;i++)
				{
					if(arr[i][0] && arr[i][1])
					{
						cnt=cnt+n-2;
					}
					if(arr[i][0] && arr[i+2][0] && arr[i+1][1])
					{
						cnt++;
					}
					if(arr[i][1] && arr[i+2][1] && arr[i+1][0])
					{
						cnt++;
					}
				}
				out.println(cnt);
			}

            out.flush();

        }

    }

    static class FastScanner// 用于快速读入大量数据
    {
        static BufferedReader br;
        static StringTokenizer st;

        public FastScanner(InputStream in)
        {
            br = new BufferedReader(new InputStreamReader(in), 16384);
            eat("");
        }

        public static void eat(String s)
        {
            st = new StringTokenizer(s);
        }

        public static String nextLine()
        {
            try
            {
                return br.readLine();
            }
            catch (IOException e)
            {
                return null;
            }
        }

        public static boolean hasNext()
        {
            while (!st.hasMoreTokens())
            {
                String s = nextLine();
                if (s == null) return false;
                eat(s);
            }
            return true;
        }

        public static String next()
        {
            hasNext();
            return st.nextToken();
        }

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

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

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

        public BigInteger nextBigInteger()
        {
            return new BigInteger(next());
        }

        public BigDecimal nextBigDecimal()
        {
            return new BigDecimal(next());
        }
    }

}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider the following aspects:

1. **Edge Cases**: Points with the same x-coordinate or y-coordinate.
2. **Nondegenerate Right Triangles**: Ensure that the points can form right triangles.
3. **Constraints**: The input should be within the given constraints and small enough to be manageable.

The buggy code has a few issues:
- It incorrectly counts the number of right triangles.
- It does not handle all possible configurations of points correctly.

Let's construct a test case that can expose these issues:

### Test Case Analysis
- **Points with the same x-coordinate**: This can cause the code to miss some right triangles.
- **Points with the same y-coordinate**: This can also cause the code to miss some right triangles.
- **Mix of points with different x and y coordinates**: This can ensure that the code has to handle a variety of configurations.

### Chosen Test Case
We will use a set of points that can form multiple right triangles, including points with the same x or y coordinates.

```plaintext
1
9
1 0
2 0
3 0
4 0
5 0
2 1
3 1
4 1
5 1
```

### Explanation
- **9 Points**: This is within the constraint of \(3 \leq n \leq 2 \cdot 10^5\).
- **Points with the same x-coordinate**: Points (2, 0), (3, 0), (4, 0), and (5, 0) all have the same y-coordinate.
- **Points with the same y-coordinate**: Points (2, 1), (3, 1), (4, 1), and (5, 1) all have the same y-coordinate.
- **Mix of points**: This ensures that the code has to handle a variety of configurations.

This test case should expose the bug in the buggy code, as it will likely miss some of the right triangles that can be formed.

### Output
```plaintext
1
9
1 0
2 0
3 0
4 0
5 0
2 1
3 1
4 1
5 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Satyam and Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: Satyam is given $$$n$$$ distinct points on the 2D coordinate plane. It is guaranteed that $$$0 \leq y_i \leq 1$$$ for all given points $$$(x_i, y_i)$$$. How many different nondegenerate right triangles$$$^{\text{∗}}$$$ can be formed from choosing three different points as its vertices?

Two triangles $$$a$$$ and $$$b$$$ are different if there is a point $$$v$$$ such that $$$v$$$ is a vertex of $$$a$$$ but not a vertex of $$$b$$$.

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

The first line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 2 \cdot 10^5$$$) — the number of points.

The following $$$n$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \leq x_i \leq n$$$, $$$0 \leq y_i \leq 1$$$) — the $$$i$$$'th point that Satyam can choose from. It is guaranteed that all $$$(x_i, y_i)$$$ are pairwise distinct.

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

Output: Output an integer for each test case, the number of distinct nondegenerate right triangles that can be formed from choosing three points.

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

Note: The four triangles in question for the first test case:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main(){
    int t,n,ans,X,Y;
    cin>>t;
    for(int T=0;T<t;T++){
        int p[200000]={0};
        cin>>n;     ans=0;
        for(int i=0;i<n;i++){
            cin>>X>>Y;
            p[X]+=Y+1;
            if(p[X]==3)
                ans+=n-2;
        }
        for(int i=1;i<n-1;i++){
            if(p[i]==3&&p[i-1]==3&&p[i+1]==3)
                ans+=2;
            else if(p[i]%2==1&&p[i-1]>1&&p[i+1]>1)
                ans++;
            else if(p[i]>1&&p[i-1]%2==1&&p[i+1]%2==1)
                ans++;
        }
        cout<<ans<<endl;
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. Here are the steps to think through this:

1. **Understand the Bug**:
   - The buggy code uses an array `p` to count the number of points at each `x` coordinate.
   - It increments `p[X]` by `Y + 1` for each point `(X, Y)`.
   - It then checks for specific conditions to count the number of right triangles.

2. **Identify Potential Issues**:
   - The code might not correctly handle cases where multiple points share the same `x` coordinate.
   - It might not correctly count the number of right triangles when points are aligned in a way that does not form a right triangle.

3. **Construct a Challenging Test Case**:
   - Create a test case with a small number of points to ensure it is valid and complete.
   - Include points that share the same `x` coordinate to test the array `p` logic.
   - Ensure the points are such that they can form right triangles, but the buggy code might miss some.

4. **Test Case**:
   - Consider a test case with 5 points where some points share the same `x` coordinate and form right triangles.

Here is the test input:

```plaintext
1
5
1 0
1 1
2 0
2 1
3 0
```

**Explanation**:
- This test case has 5 points.
- Points `(1, 0)` and `(1, 1)` share the same `x` coordinate.
- Points `(2, 0)` and `(2, 1)` share the same `x` coordinate.
- Point `(3, 0)` is a single point.
- The expected number of right triangles is 4:
  - `(1, 0)`, `(1, 1)`, `(2, 0)`
  - `(1, 0)`, `(1, 1)`, `(2, 1)`
  - `(2, 0)`, `(2, 1)`, `(3, 0)`
  - `(1, 0)`, `(2, 0)`, `(3, 0)`

This test case should expose the bug in the given buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Satyam and Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: Satyam is given $$$n$$$ distinct points on the 2D coordinate plane. It is guaranteed that $$$0 \leq y_i \leq 1$$$ for all given points $$$(x_i, y_i)$$$. How many different nondegenerate right triangles$$$^{\text{∗}}$$$ can be formed from choosing three different points as its vertices?

Two triangles $$$a$$$ and $$$b$$$ are different if there is a point $$$v$$$ such that $$$v$$$ is a vertex of $$$a$$$ but not a vertex of $$$b$$$.

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

The first line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 2 \cdot 10^5$$$) — the number of points.

The following $$$n$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \leq x_i \leq n$$$, $$$0 \leq y_i \leq 1$$$) — the $$$i$$$'th point that Satyam can choose from. It is guaranteed that all $$$(x_i, y_i)$$$ are pairwise distinct.

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

Output: Output an integer for each test case, the number of distinct nondegenerate right triangles that can be formed from choosing three points.

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

Note: The four triangles in question for the first test case:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main()
{
    int t,j;
    cin>>t;
    for(int i=0;i<t;i++)
    {
        int n;
        cin>>n;
        j=0;
        int a[n][2];
        for(int w=0;w<n;w++)
            cin>>a[w][0]>>a[w][1];
        for(int q=0;q<n-2;q++)
        for(int w=q+1;w<n-1;w++)
            for(int e=w+1;e<n;e++)
            if((a[w][0]-a[q][0])*(a[e][0]-a[q][0])+(a[w][1]-a[q][1])*(a[e][1]-a[q][1])==0||(a[w][0]-a[e][0])*(a[q][0]-a[e][0])+(a[w][1]-a[e][1])*(a[q][1]-a[e][1])==0||(a[q][0]-a[w][0])*(a[e][0]-a[w][0])+(a[q][1]-a[w][1])*(a[e][1]-a[w][1])==0)
                j++;
        cout<<j<<endl;
    }
}

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

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

Think step by step.

Response:
```plaintext
3
5
1 0
1 1
2 0
2 1
3 0
4
0 0
0 1
1 0
1 1
6
0 0
1 0
2 0
0 1
1 1
2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Satyam and Counting
time_limit_ms: 2000
memory_limit_mb: 256
Description: Satyam is given $$$n$$$ distinct points on the 2D coordinate plane. It is guaranteed that $$$0 \leq y_i \leq 1$$$ for all given points $$$(x_i, y_i)$$$. How many different nondegenerate right triangles$$$^{\text{∗}}$$$ can be formed from choosing three different points as its vertices?

Two triangles $$$a$$$ and $$$b$$$ are different if there is a point $$$v$$$ such that $$$v$$$ is a vertex of $$$a$$$ but not a vertex of $$$b$$$.

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

The first line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 2 \cdot 10^5$$$) — the number of points.

The following $$$n$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \leq x_i \leq n$$$, $$$0 \leq y_i \leq 1$$$) — the $$$i$$$'th point that Satyam can choose from. It is guaranteed that all $$$(x_i, y_i)$$$ are pairwise distinct.

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

Output: Output an integer for each test case, the number of distinct nondegenerate right triangles that can be formed from choosing three points.

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

Note: The four triangles in question for the first test case:

Buggy Code:
for v in range(int(input())):
    n = int(input())
    L1 = list()
    L2 = list()
    count1 = 0
    S = set(range(0,n+1))
    L = list()
    for i in range(n):
        l =[int(x) for x in input().split()]
        L.append(l)
        L1.append(l[0])
        L2.append(l[1])
    S1 = set(L1)
    count1 += (len(S-S1)-1)*(len(L)-2)
    L.sort()
    # print(L)
    for i in L:
        if i[1]==0:
            if [i[0]-1,1] in L and [i[0]+1,1] in L:
                count1+=1
        elif i[1]==1:
            if [i[0]-1,0] in L and [i[0]+1,0] in L:
                count1+=1

    print(count1)

**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 create a challenging test input that exposes the bug in the provided code, let's consider the following aspects:

1. **Edge Cases:**
   - Points with the same x-coordinate but different y-coordinates.
   - Points with the same y-coordinate but different x-coordinates.
   - Points that are close to each other to form right triangles.
   - Points that do not form any right triangles.

2. **Maximizing Failure Probability:**
   - Ensure that the code's logic for counting right triangles is thoroughly tested.
   - Focus on scenarios where the code might miss valid right triangles or count invalid ones.

3. **Small-Scale Input:**
   - Keep the number of points and test cases manageable to ensure the input is complete and valid.

Given these considerations, let's create a test input that includes points forming right triangles and some that do not, to expose the bug in the code.

```plaintext
3
5
1 0
2 0
3 0
2 1
4 1
4
0 0
1 0
1 1
2 1
6
0 0
1 0
2 0
0 1
1 1
2 1
```

### Explanation:
1. **First Test Case (5 points):**
   - Points: (1, 0), (2, 0), (3, 0), (2, 1), (4, 1)
   - This set of points forms two right triangles: (1, 0), (2, 0), (2, 1) and (2, 0), (3, 0), (2, 1).
   - The code should count these two triangles.

2. **Second Test Case (4 points):**
   - Points: (0, 0), (1, 0), (1, 1), (2, 1)
   - This set of points forms one right triangle: (0, 0), (1, 0), (1, 1).
   - The code should count this one triangle.

3. **Third Test Case (6 points):**
   - Points: (0, 0), (1, 0), (2, 0), (0, 1), (1, 1), (2, 1)
   - This set of points forms three right triangles: (0, 0), (1, 0), (1, 1), (1, 0), (2, 0), (1, 1), and (0, 0), (1, 0), (0, 1).
   - The code should count these three triangles.

This input will help expose any issues in the code's logic for counting right triangles, especially in cases where points are close to each other and form valid right triangles.