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>
  #pragma GCC optimize("Ofast")
  #pragma GCC target("avx,avx2,fma")
  #pragma GCC optimization("unroll-loops")
  using namespace std;
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int testcase = 1;
    cin >> testcase;
    int __testcase = testcase;
    while (testcase--) {
      long long int n, m, i, j, k;
      cin >> n;
      int a[n], b[n];
      for (i = 0; i < n; i++) cin >> a[i] >> b[i];
      vector<long long int> _a(n + 1, 0), _b(n + 1, 0);
      vector<int> ta[n + 1], tb[n + 1];
      long long int ans = 0;
      for (i = 0; i < n; i++) {
        _a[a[i]]++;
        _b[b[i]]++;
      }
      for (i = 0; i < n; i++) {
        ta[a[i]].push_back(b[i]);
        tb[b[i]].push_back(a[i]);
      }
      vector<int> ca(n + 1, 0), cb(n + 1, 0);
      for (i = 1; i <= n; i++) {
        for (auto j : ta[i]) cb[j] += ta[i].size() - 1;
        for (auto j : tb[i]) ca[j] += tb[i].size() - 1;
      }
      long long int cm = 0;
      for (i = 1; i <= n; i++)
        cm += _a[i] * (_a[i] - 1) / 2 + _b[i] * (_b[i] - 1) / 2;
      for (i = 0; i < n; i++) {
        long long int t = n - _a[a[i]] - _b[b[i]] + 1;
        j = _a[a[i]];
        k = _b[b[i]];
        j--;
        k--;
        ans += t * (t - 1) / 2 - cm + j * (j - 1) / 2 + k * (k - 1) / 2 +
               ca[a[i]] + cb[b[i]];
      }
      ans = -ans / 3;
      long long int suma = 0, sumb = 0;
      for (i = 1; i <= n; i++) {
        ans += suma * _a[i] * (n - suma - _a[i]);
        ans += sumb * _b[i] * (n - sumb - _b[i]);
        suma += _a[i];
        sumb += _b[i];
      }
      cout << ans << endl;
    }
  }
  