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 init_code() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  }
  long long int M = 1000000007;
  long long int power(long long int x, long long int y) {
    if (y == 0)
      return 1;
    else if (y == 1)
      return x % M;
    else {
      long long int z = power(x, y / 2);
      long long int w = ((z % M) * (z % M)) % M;
      if (y % 2 == 1)
        return ((x % M) * (w % M)) % M;
      else
        return w;
    }
  }
  int main() {
    init_code();
    long long int t = 1;
    cin >> t;
    long long int o = 1;
    while (t--) {
      long long int m, n, a, i, j, k, l = 0, ans = 0;
      double sum = 0.0;
      cin >> n;
      map<long long int, long long int> mp, np;
      vector<pair<long long int, long long int>> v;
      for (i = 0; i < n; i++) {
        long long int b, c;
        cin >> b >> c;
        v.push_back({b, c});
        mp[b]++;
        np[c]++;
      }
      l = n;
      ans = (n * (n - 1) * (n - 2)) / 6;
      map<pair<long long int, long long int>, long long int> vb;
      for (i = 0; i < n; i++) {
        long long int x = v[i].first, y = v[i].second;
        ans = ans - (mp[x] - 1) * (np[y] - 1);
      }
      cout << ans << endl;
    }
  }
  