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 _mak_() {
    int n;
    cin >> n;
    vector<pair<int, int> > a;
    vector<vector<long long> > s(n + 1, vector<long long>()),
        g(n + 1, vector<long long>());
    for (int i = 0; i < n; i++) {
      int r, f;
      cin >> r >> f;
      a.push_back({r, f});
      s[r].push_back(i);
      g[f].push_back(i);
    }
    long long r = n, x = n - 1, z = n - 2;
    if (r % 3 == 0) {
      r /= 3;
    } else if (x % 3 == 0) {
      x /= 3;
    } else if (z % 3 == 0) {
      z /= 3;
    }
    if (r % 2 == 0) {
      r /= 2;
    } else if (x % 2 == 0) {
      x /= 2;
    } else if (z % 2 == 0) {
      z /= 2;
    }
    long long ans = r * x * z;
    for (int i = 0; i < n; i++) {
      long long v = s[a[i].first].size();
      v--;
      long long d = g[a[i].second].size();
      d--;
      ans -= (v * d);
    }
    cout << ans << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
      _mak_();
    }
    return 0;
  }
  