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 long long Mod = 1000000007;
  const long long INF = 999999999999999999;
  const long long NN = (long long)(1e6 + 5);
  template <typename T>
  T Min(T x, T y) {
    if (x < y) return x;
    return y;
  }
  template <typename T>
  T Max(T x, T y) {
    if (x > y) return x;
    return y;
  }
  long long power(long long x, unsigned long long y, long long mod = Mod) {
    long long res = 1;
    x = x % mod;
    if (x == 0) return 0;
    while (y > 0) {
      if (y & 1) res = (res * x) % mod;
      y = y >> 1LL;
      x = (x * x) % mod;
    }
    return res % mod;
  }
  template <typename T>
  T hcf(T a, T b) {
    if (b == 0) return a;
    return hcf(b, a % b);
  }
  long long c3(long long n) { return (n * (n - 1) * (n - 2)) / 6; }
  long long c2(long long n) { return (n * (n - 1)) / 2; }
  void solve() {
    long long n;
    cin >> n;
    vector<pair<long long, long long> > v(n);
    map<long long, long long> x, y;
    for (auto &u : v) cin >> u.first >> u.second, x[u.first]++, y[u.second]++;
    long long i, ans = 0, tot = c3(n);
    for (i = 0; i < n; i++) {
      tot -= (x[v[i].first] - 1) * (y[v[i].second] - 1);
    }
    cout << tot << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);
    ;
    long long TT = 1;
    cin >> TT;
    for (long long tt = 1; tt <= TT; tt++) {
      solve();
    }
  }
  