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 maxn = 2e5 + 5, N1 = maxn + 7;
  const long long inf = 0x3f3f3f3f3f3fll;
  template <typename T>
  inline void read(T &x) {
    x = 0;
    T k = 1;
    char c = getchar();
    while (!isdigit(c)) {
      if (c == '-') k = -1;
      c = getchar();
    }
    while (isdigit(c)) {
      x = x * 10 + c - '0';
      c = getchar();
    }
    x *= k;
  }
  int n;
  int a[N1], b[N1];
  long long sa[N1], sb[N1];
  long long calc3(long long x) { return x * (x - 1) * (x - 2) / 6; }
  void solve() {
    read(n);
    long long ans = calc3(1ll * n), tmp;
    for (int i = 1; i <= n; i++) read(a[i]), read(b[i]);
    tmp = 0;
    for (int i = 1; i <= n; i++) {
      tmp += sa[a[i]] * sb[b[i]];
      sa[a[i]]++;
      sb[b[i]]++;
    }
    for (int i = 1; i <= n; i++) sa[i] = 0;
    for (int i = 1; i <= n; i++) {
      sb[b[i]]--;
      tmp += sa[a[i]] * sb[b[i]];
      sa[a[i]]++;
    }
    for (int i = 1; i <= n; i++) {
      sa[a[i]]--;
      tmp += sa[a[i]] * sb[b[i]];
      sb[b[i]]++;
    }
    for (int i = 1; i <= n; i++) sb[i] = 0;
    for (int i = n; i >= 1; i--) {
      tmp += sa[a[i]] * sb[b[i]];
      sa[a[i]]++;
      sb[b[i]]++;
    }
    for (int i = 1; i <= n; i++) sa[i] = sb[i] = 0;
    ans -= tmp;
    printf("%lld\n", ans);
  }
  int main() {
    int Te;
    read(Te);
    while (Te--) solve();
    return 0;
  }
  