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 <class T>
  using pq = priority_queue<T>;
  template <class T>
  using pqg = priority_queue<T, vector<T>, greater<T>>;
  int mod(long long int a, long long int b) { return ((a % b) + b) % b; }
  long long int MODnumber(long long int a) {
    return (a % (long long int)1000000007 + (long long int)1000000007) %
           (long long int)1000000007;
  }
  long long int MODsum(long long int a, long long int b) {
    return ((a % (long long int)1000000007 + b % (long long int)1000000007) %
                (long long int)1000000007 +
            (long long int)1000000007) %
           (long long int)1000000007;
  }
  long long int MODproduct(long long int a, long long int b) {
    return (((a % (long long int)1000000007) * (b % (long long int)1000000007)) %
                (long long int)1000000007 +
            (long long int)1000000007) %
           (long long int)1000000007;
  }
  void solve() {
    long long int n;
    cin >> n;
    map<long long int, long long int> difficulties;
    map<long long int, vector<long long int>> topicandDifficulty;
    for (long long int i = 0; i < n; i++) {
      long long int x, y;
      cin >> x >> y;
      difficulties[y]++;
      topicandDifficulty[x].push_back(y);
    }
    long long int ans = 0;
    for (auto itr : topicandDifficulty) {
      long long int sz = itr.second.size();
      sz--;
      for (auto it : itr.second) {
        ans += sz * (difficulties[it] - 1);
      }
    }
    long long int countSame = 0;
    long long int countDiff = 0;
    for (auto it : difficulties) {
      countSame += (n - it.second) * (n - it.second - 1);
      countDiff += (it.second * (it.second - 1));
    }
    long long int temp = (n * (n - 1) * (n - 2)) / 6;
    temp -= ans;
    cout << temp << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long int t;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  