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;
  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
  const int nmax = 2e5 + 5;
  int t, n, a[nmax], b[nmax], f[nmax], L[nmax], R[nmax], d[nmax];
  long long ans;
  vector<int> v1[nmax], v2[nmax];
  int main() {
    srand(time(NULL));
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while (t--) {
      ans = 0;
      cin >> n;
      for (int i = 1; i <= n; i++) {
        v1[i].clear();
        v2[i].clear();
        f[i] = 0;
        d[i] = 0;
      }
      R[n + 1] = 0;
      for (int i = 1; i <= n; i++) {
        cin >> a[i] >> b[i];
        f[b[i]]++;
        v1[a[i]].emplace_back(b[i]);
        v2[b[i]].emplace_back(a[i]);
      }
      for (int i = 1; i <= n; i++) L[i] = L[i - 1] + v2[i].size();
      for (int i = n; i >= 1; i--) R[i] = R[i + 1] + v2[i].size();
      for (int i = 2; i < n; i++)
        ans = ans + 1ll * v2[i].size() * L[i - 1] * R[i + 1];
      for (int i = 1; i <= n; i++) L[i] = L[i - 1] + v1[i].size();
      for (int i = n; i >= 1; i--) R[i] = R[i + 1] + v1[i].size();
      for (int i = 2; i < n; i++)
        ans = ans + 1ll * v1[i].size() * L[i - 1] * R[i + 1];
      for (int i = 1; i <= n; i++) sort(v1[i].begin(), v1[i].end());
      for (int i = 1; i <= n; i++)
        if (v1[i].size() > 0) {
          int dem = 1;
          for (int j = 1; j < v1[i].size(); j++)
            if (v1[i][j] != v1[i][j - 1]) {
              int x = v1[i][j - 1];
              ans = ans - 1ll * dem * (L[i - 1] - d[x]) *
                              (R[i + 1] - (f[x] - d[x] - dem));
              dem = 1;
            } else
              dem++;
          int x = v1[i][v1[i].size() - 1];
          ans = ans -
                1ll * dem * (L[i - 1] - d[x]) * (R[i + 1] - (f[x] - d[x] - dem));
          for (auto p : v1[i]) d[p]++;
        }
      for (int i = 1; i <= n; i++) d[i] = 0;
      long long res = 0;
      for (int i = 1; i <= n; i++)
        if (v1[i].size() > 0) {
          int dem = 1;
          for (int j = 1; j < v1[i].size(); j++)
            if (v1[i][j] != v1[i][j - 1]) {
              int x = v1[i][j - 1];
              res = res - 1ll * dem * d[x];
              dem = 1;
            } else
              dem++;
          int x = v1[i][v1[i].size() - 1];
          res = res - 1ll * dem * d[x];
          ans = ans + 1ll * v1[i].size() * res;
          dem = 1;
          for (int j = 1; j < v1[i].size(); j++)
            if (v1[i][j] != v1[i][j - 1]) {
              int x = v1[i][j - 1];
              res = res + 1ll * dem * (f[x] - d[x] - dem);
              dem = 1;
            } else
              dem++;
          x = v1[i][v1[i].size() - 1];
          res = res + 1ll * dem * (f[x] - d[x] - dem);
          for (auto p : v1[i]) d[p]++;
        }
      for (int i = 1; i <= n; i++) d[i] = 0;
      for (int i = 1; i <= n; i++)
        if (v1[i].size() > 0) {
          int dem = 1;
          for (int j = 1; j < v1[i].size(); j++)
            if (v1[i][j] != v1[i][j - 1]) {
              int x = v1[i][j - 1];
              ans = ans - 1ll * d[x] * (f[x] - d[x] - dem);
              dem = 1;
            } else
              dem++;
          int x = v1[i][v1[i].size() - 1];
          ans = ans - 1ll * d[x] * (f[x] - d[x] - dem);
          for (auto p : v1[i]) d[p]++;
        }
      cout << ans << '\n';
    }
  }
  