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;
  const int mxn = 200000, m = 3;
  long long n, k;
  long long a[mxn][2], b[mxn], d[mxn], h[mxn], dp[2][m + 1];
  vector<int> v[mxn][2];
  map<int, vector<int>> first[mxn];
  void answer() {
    cin >> n;
    for (int i = 0; i < n; i++) {
      first[i].clear();
      for (int j = 0; j < 2; j++) v[i][j].clear();
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < 2; j++) {
        cin >> a[i][j];
        v[--a[i][j]][j].push_back(i);
      }
      first[a[i][0]][a[i][1]].push_back(i);
    }
    memset(dp, 0, sizeof(dp));
    dp[0][0] = dp[1][0] = 1;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < 2; j++)
        for (int l = m; l; l--) {
          dp[j][l] += v[i][j].size() * dp[j][l - 1];
        }
    k = 0;
    for (int i = 0; i < n; i++) {
      b[i] = n + first[a[i][0]][a[i][1]].size(), d[i] = 1;
      for (int j = 0; j < 2; j++) b[i] -= v[a[i][j]][j].size();
      k += b[i];
    }
    k /= 2;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < 2; j++) {
        for (int l : v[i][j]) h[a[l][!j]]++;
        for (int l : v[i][j]) d[l] *= v[i][j].size() - h[a[l][!j]];
        for (int l : v[i][j]) h[a[l][!j]]--;
      }
    for (int i = 0; i < n; i++)
      for (auto j : first[i]) {
        long long x = 0;
        for (int l : j.second) x += b[l];
        for (int l : j.second) d[l] += k + x;
      }
    for (int i = 0; i < n; i++)
      for (int j = 0; j < 2; j++) {
        long long x = 0;
        for (int l : v[i][j]) x += b[l];
        for (int l : v[i][j]) d[l] -= x;
      }
    long long ret = dp[0][m] + dp[1][m], x = 0;
    for (int i = 0; i < n; i++) x += d[i];
    ret -= x / 3;
    cout << ret << '\n';
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    for (int i = 0; i < t; i++) answer();
    return 0;
  }
  