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;
  template <typename A, typename B>
  ostream &operator<<(ostream &os, const pair<A, B> &p) {
    return os << '(' << p.first << ", " << p.second << ')';
  }
  template <typename T_container, typename T = typename enable_if<
                                      !is_same<T_container, string>::value,
                                      typename T_container::value_type>::type>
  ostream &operator<<(ostream &os, const T_container &v) {
    os << '{';
    string sep;
    for (const T &x : v) os << sep << x, sep = ", ";
    return os << '}';
  }
  void dbg_out() { cerr << endl; }
  template <typename Head, typename... Tail>
  void dbg_out(Head H, Tail... T) {
    cerr << ' ' << H;
    dbg_out(T...);
  }
  const int MAX_N = 1e5 + 5;
  const long long MOD = 1e9 + 7;
  const long long INF = 1e9;
  const long double EPS = 1e-9;
  void solve() {
    long long n;
    cin >> n;
    unordered_map<long long, vector<long long>> topc_diff;
    unordered_map<long long, vector<long long>> diff_topc;
    for (int i = 0; i < n; i++) {
      long long t1, t2;
      cin >> t1;
      cin >> t2;
      topc_diff[t1].push_back(t2);
      diff_topc[t2].push_back(t1);
    }
    long long total = (n * (n - 1) * (n - 2)) / 6, ans = 0;
    for (auto &topc : topc_diff) {
      long long temp1 = topc.second.size() - 1;
      for (long long diff : topc.second) {
        long long temp2 = diff_topc[diff].size() - 1;
        ans += temp1 * temp2;
      }
    }
    cout << total - ans << endl;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int tc = 1;
    cin >> tc;
    for (int t = 1; t <= tc; t++) {
      solve();
    }
  }
  