description: Monocarp is the coach of the Berland State University programming teams. He decided to compose a problemset for a training session for his teams.
  
  Monocarp has n problems that none of his students have seen yet. The i-th problem has a topic a_i (an integer from 1 to n) and a difficulty b_i (an integer from 1 to n). All problems are different, that is, there are no two tasks that have the same topic and difficulty at the same time.
  
  Monocarp decided to select exactly 3 problems from n problems for the problemset. The problems should satisfy at least one of two conditions (possibly, both):
  
    * the topics of all three selected problems are different; 
    * the difficulties of all three selected problems are different. 
  
  
  
  Your task is to determine the number of ways to select three problems for the problemset.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 50000) — the number of testcases.
  
  The first line of each testcase contains an integer n (3 ≤ n ≤ 2 ⋅ 10^5) — the number of problems that Monocarp have.
  
  In the i-th of the following n lines, there are two integers a_i and b_i (1 ≤ a_i, b_i ≤ n) — the topic and the difficulty of the i-th problem.
  
  It is guaranteed that there are no two problems that have the same topic and difficulty at the same time.
  
  The sum of n over all testcases doesn't exceed 2 ⋅ 10^5.
  
  Output
  
  Print the number of ways to select three training problems that meet either of the requirements described in the statement.
  
  Example
  
  Input
  
  
  2
  4
  2 4
  3 4
  2 1
  1 3
  5
  1 5
  2 4
  3 3
  4 2
  5 1
  
  
  Output
  
  
  3
  10
  
  Note
  
  In the first example, you can take the following sets of three problems:
  
    * problems 1, 2, 4; 
    * problems 1, 3, 4; 
    * problems 2, 3, 4. 
  
  
  
  Thus, the number of ways is equal to three.
solution: #include <bits/stdc++.h>
  using namespace std;
  void _print(long long t) { cerr << t; }
  void _print(int t) { cerr << t; }
  void _print(string t) { cerr << t; }
  void _print(char t) { cerr << t; }
  void _print(double 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.first);
    cerr << ",";
    _print(p.second);
    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 << "]";
  }
  long long po(long long a, long long b, long long m) {
    long long t = 1;
    while (b > 0) {
      if (b % 2 != 0) {
        t = (t * a) % m;
      }
      a = (a * a) % m;
      b = b / 2;
    }
    return t;
  }
  long long inv(long long a, long long m) { return po(a, m - 2, m); }
  bool sortbysec(const pair<char, long long> &a, const pair<char, long long> &b) {
    return (a.second < b.second);
  }
  long long get(long long n) {
    long long ans = 1;
    for (long long i = 2; i <= n; i++)
      ans = (ans % 1000000007 * i % 1000000007) % 1000000007;
    return ans % 1000000007;
  }
  long long nPr(long long n, long long r) {
    return get(n) * inv(get(n - r), 1000000007);
  }
  long long modularExponentiation(long long x, long long n) {
    long long result = 1;
    while (n > 0) {
      if (n % 2 == 1) result = (result * x) % 1000000007;
      x = (x * x) % 1000000007;
      n = n / 2;
    }
    return result;
  }
  void solve() {
    long long i, j, flag = 0;
    long long n;
    cin >> n;
    map<long long, long long> M, N;
    map<pair<long long, long long>, long long> x;
    vector<pair<long long, long long>> v;
    for (i = 0; i < n; i++) {
      long long a, b;
      cin >> a >> b;
      v.push_back({a, b});
      M[a]++;
      N[b]++;
      x[{a, b}]++;
    }
    long long sum = (n * (n - 1) * (n - 2)) / 6;
    ;
    for (i = 0; i < n; i++) {
      if (M[v[i].first] >= 2 && N[v[i].second] >= 2) {
        sum -= (M[v[i].first] - 1) * (N[v[i].second] - 1);
      }
    }
    cout << sum << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    long long t = 1;
    cin >> t;
    for (int i = 1; i <= t; i++) {
      solve();
    }
    return 0;
  }
  