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(string t) { cerr << t; }
  void _print(char t) { cerr << t; }
  void _print(long double t) { cerr << t; }
  void _print(double t) { cerr << t; }
  void _print(unsigned long long 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 << "]";
  }
  void init_code() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  void solve() {
    long long n;
    cin >> n;
    map<long long, long long> m1, m2;
    map<pair<long long, long long>, long long> m3;
    vector<long long> a(n), b(n);
    vector<long long> mt[n + 1];
    for (long long i = 0; i < n; i++) {
      cin >> a[i] >> b[i];
      m1[a[i]]++;
      m2[b[i]]++;
      m3[{a[i], b[i]}]++;
      mt[a[i]].push_back(b[i]);
    }
    long long ans = 0;
    for (long long i = 1; i <= n; i++) {
      vector<long long> temp = mt[i];
      map<long long, long long> mt1;
      for (long long j : temp) {
        mt1[j]++;
        m2[j]--;
      }
      for (auto j : mt1) {
        ans += (((j.second) * (j.second - 1) / 2) *
                (n - ((long long)(temp).size()) - m2[j.first]));
        ans +=
            (j.second * (((long long)(temp).size()) - j.second) * (m2[j.first]));
        ans += (((j.second) * (j.second - 1) / 2) * (m2[j.first]));
        ans += ((j.second) * (j.second - 1) / 2 *
                (((long long)(temp).size()) - j.second));
        ans += ((j.second) * (j.second - 1) * (j.second - 2) / 6);
      }
      for (long long j : temp) {
        mt1[j]++;
        m2[j]++;
      }
    }
    cout << (n * (n - 1) * (n - 2) / 6 - ans) << "\n";
  }
  signed main() {
    init_code();
    long long t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  