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;
  int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
  int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
  void file() {}
  void fast() {
    std::ios_base::sync_with_stdio(0);
    cin.tie(NULL);
  }
  void test_case() {
    int n;
    cin >> n;
    vector<vector<int> > adjL(n + 1);
    vector<pair<int, int> > v1(n);
    for (auto& it : v1) {
      cin >> it.first >> it.second;
      adjL[it.first].push_back(it.second);
    }
    long long ans = 1LL * n * (n - 1) * (n - 2) / 6;
    vector<long long> fre(n + 1);
    for (int i = 1; i <= n; i++) {
      for (auto& it : adjL[i]) {
        ans -= fre[it];
      }
      for (auto& it : adjL[i]) fre[it] += (int)(adjL[i].size()) - 1;
    }
    fre = vector<long long>(n + 1);
    for (int i = n; i >= 1; i--) {
      for (auto& it : adjL[i]) ans -= fre[it];
      for (auto& it : adjL[i]) fre[it] += (int)(adjL[i].size()) - 1;
    }
    cout << ans << '\n';
  }
  int main() {
    file();
    fast();
    int t;
    cin >> t;
    while (t--) {
      test_case();
    }
    return 0;
  }
  