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;
  template <class T>
  T read() {
    T re = 0;
    char c = getchar();
    bool negt = false;
    while (c < '0' || c > '9') negt |= (c == '-'), c = getchar();
    while (c >= '0' && c <= '9')
      re = (re << 1) + (re << 3) + c - '0', c = getchar();
    return negt ? -re : re;
  }
  const int N = 2e5 + 10;
  int a[N], b[N];
  long long cnt1[N], cnt2[N];
  map<pair<int, int>, int> cnt3;
  long long n;
  void solve() {
    long long ans = 0;
    n = read<int>();
    for (int i = 1; i <= n; i++) a[i] = read<int>(), b[i] = read<int>();
    for (int i = 1; i <= n; i++) cnt1[i] = cnt2[i] = 0;
    for (int i = 1; i <= n; i++)
      ++cnt1[a[i]], ++cnt2[b[i]], ++cnt3[make_pair(a[i], b[i])];
    long long add = 0;
    add += n * (n - 1) * (n - 2) / 6;
    for (int i = 1; i <= n; i++)
      if (cnt1[i] != 0) {
        add -= (cnt1[i] - 1) * cnt1[i] * (n - cnt1[i]) / 2;
        add -= (cnt1[i] - 2) * (cnt1[i] - 1) * cnt1[i] / 6;
      }
    add += n * (n - 1) * (n - 2) / 6;
    for (int i = 1; i <= n; i++)
      if (cnt2[i] != 0) {
        add -= (cnt2[i] - 1) * cnt2[i] * (n - cnt2[i]) / 2;
        add -= (cnt2[i] - 2) * (cnt2[i] - 1) * cnt2[i] / 6;
      }
    long long minus = n * (n - 1) * (n - 2) / 6;
    for (int i = 1; i <= n; i++) minus -= (cnt1[a[i]] - 1) * (cnt2[b[i]] - 1);
    minus -= add;
    printf("%lld\n", add + minus);
  }
  int main() {
    int T = read<int>();
    while (T--) {
      solve();
    }
    return 0;
  }
  