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>
  void print(vector<T> a) {
    int n = a.size();
    for (long long i = 0; i < n; ++i) {
      cout << a[i] << (i == n - 1 ? "\n" : " ");
    }
  }
  int sum_vector(vector<int> v) { return accumulate(v.begin(), v.end(), 0); }
  void sort_vector(vector<int> &v) { sort(v.begin(), v.end()); }
  void sort_comp(vector<int> &v, bool func(int, int)) {
    sort(v.begin(), v.end(), func);
  }
  bool comp(array<long long, 2> a, array<long long, 2> b) { return a[1] < b[1]; }
  long long gcd(long long a, long long b) {
    a = abs(a);
    b = abs(b);
    while (a) {
      long long temp = a;
      a = b % a;
      b = temp;
    }
    return abs(b);
  }
  long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
  string binary(long long num) {
    string ans = "";
    do {
      ans = to_string(num % 2) + ans;
      num /= 2;
    } while (num);
    return ans;
  }
  const int mxn = 2e5 + 7;
  const int di = 18;
  const int mill = 1e6 + 3;
  const long long mod = 1e9 + 7;
  long long pwr(long long num, long long p) {
    long long res = 1;
    while (p > 0) {
      if (p & 1) res = (res * num) % mod;
      num = (num * num) % mod;
      p /= 2;
    }
    return res;
  }
  long long inverse(long long num) { return pwr(num, mod - 2); }
  long long get_rand(long long n) { return ((rand() << 15) + rand()) % n; }
  void solve() {
    long long n;
    cin >> n;
    vector<long long> top(n + 2);
    vector<long long> dif(n + 2);
    vector<long long> a(n);
    vector<long long> b(n);
    for (long long i = 0; i < n; i++) {
      cin >> a[i] >> b[i];
      top[a[i]]++;
      dif[b[i]]++;
    }
    long long ans = n * (n - 1) * (n - 2) / 6;
    for (int i = 0; i < n; i++) {
      ans -= (top[a[i]] - 1) * (dif[b[i]] - 1);
    }
    cout << ans << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0), cin.tie(0);
    int t = 1;
    cin >> t;
    for (int ii = 1; ii <= t; ii++) {
      solve();
    }
    return 0;
  }
  