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;
  long long calc3(long long n) { return n * (n - 1) * (n - 2) / 6; }
  long long calc(map<int, long long> &mp) {
    vector<long long> vec;
    long long sum = 0;
    for (auto pr : mp) {
      vec.push_back(pr.second);
      sum += pr.second;
    }
    long long res = 0;
    for (int i = 0; i < vec.size(); i++) {
      sum -= vec[i];
      res += (sum * vec[i] * (vec[i] - 1)) / 2;
      sum += vec[i];
    }
    return res;
  }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      int arr[n];
      int brr[n];
      map<int, long long> mp1;
      map<int, long long> mp2;
      for (int i = 0; i < n; i++) {
        cin >> arr[i];
        cin >> brr[i];
        mp1[arr[i]]++;
        mp2[brr[i]]++;
      }
      long long ans = calc3(n);
      for (int i = 0; i < n; i++) {
        ans -= (mp1[arr[i]] - 1) * (mp2[brr[i]] - 1);
      }
      cout << ans << endl;
    }
    return 0;
  }
  