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;
  string to_string(string s) { return '"' + s + '"'; }
  string to_string(const char* s) { return to_string((string)s); }
  string to_string(bool b) { return (b ? "true" : "false"); }
  string to_string(char ch) { return string("'") + ch + string("'"); }
  template <typename A, typename B>
  string to_string(pair<A, B> p) {
    return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
  }
  template <class InputIterator>
  string to_string(InputIterator first, InputIterator last) {
    bool start = true;
    string res = "{";
    while (first != last) {
      if (!start) {
        res += ", ";
      }
      start = false;
      res += to_string(*first);
      ++first;
    }
    res += "}";
    return res;
  }
  template <typename A>
  string to_string(A v) {
    bool first = true;
    string res = "{";
    for (const auto& x : v) {
      if (!first) {
        res += ", ";
      }
      first = false;
      res += to_string(x);
    }
    res += "}";
    return res;
  }
  void debug_out() { cerr << '\n'; }
  template <typename Head, typename... Tail>
  void debug_out(Head H, Tail... T) {
    cerr << " " << to_string(H);
    debug_out(T...);
  }
  template <typename A, typename B>
  istream& operator>>(istream& input, pair<A, B>& x) {
    input >> x.first >> x.second;
    return input;
  }
  template <typename A>
  istream& operator>>(istream& input, vector<A>& x) {
    for (auto& i : x) input >> i;
    return input;
  }
  void solve() {
    int n;
    cin >> n;
    vector<pair<int, int> > v(n);
    cin >> v;
    long long ans = 1LL * n * (n - 1) * (n - 2) / 6;
    map<int, int> topic_cnt;
    map<int, int> diff_cnt;
    for (int i = 0; i < n; ++i) {
      topic_cnt[v[i].first]++;
      diff_cnt[v[i].second]++;
    }
    for (int i = 0; i < n; ++i) {
      ans -= 1LL * (topic_cnt[v[i].first] - 1) * (diff_cnt[v[i].second] - 1);
    }
    cout << ans << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  