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 <typename T1, typename T2>
  inline ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
    return os << "(" << p.first << ", " << p.second << ")";
  }
  template <typename T, size_t N>
  inline ostream &operator<<(ostream &os, const array<T, N> &a) {
    os << "[";
    int cnt = 0;
    for (auto &val : a) {
      if (cnt++) os << ", ";
      os << val;
    }
    os << "]";
    return os;
  }
  template <typename T>
  inline ostream &operator<<(ostream &os, const vector<T> &v) {
    os << "[";
    int cnt = 0;
    for (auto &val : v) {
      if (cnt++) os << ", ";
      os << val;
    }
    return os << "]";
  }
  template <typename T>
  inline ostream &operator<<(ostream &os, const set<T> &v) {
    os << "{";
    int cnt = 0;
    for (auto &val : v) {
      if (cnt++) os << ", ";
      os << val;
    }
    return os << "}";
  }
  template <typename T>
  inline ostream &operator<<(ostream &os, const unordered_set<T> &v) {
    os << "{";
    int cnt = 0;
    for (auto &val : v) {
      if (cnt++) os << ", ";
      os << val;
    }
    return os << "}";
  }
  template <typename T>
  inline ostream &operator<<(ostream &os, const multiset<T> &v) {
    os << "{";
    int cnt = 0;
    for (auto &val : v) {
      if (cnt++) os << ", ";
      os << val;
    }
    return os << "}";
  }
  template <typename T1, typename T2>
  inline ostream &operator<<(ostream &os, const map<T1, T2> &v) {
    os << "[";
    int cnt = 0;
    for (auto &val : v) {
      if (cnt++) os << ", ";
      os << val;
    }
    return os << "]";
  }
  template <typename Arg1>
  void __f(const char *name, Arg1 &&arg1) {
    cerr << name << ": " << arg1 << endl;
  }
  template <typename Arg1, typename... Args>
  void __f(const char *names, Arg1 &&arg1, Args &&...args) {
    const char *comma = strchr(names + 1, ',');
    cerr.write(names, comma - names) << ": " << arg1 << " |";
    __f(comma + 1, args...);
  }
  using ll = long long;
  using ull = unsigned long long;
  using vi = vector<int>;
  using vll = vector<long long>;
  using vvi = vector<vector<int>>;
  using vs = vector<string>;
  using vb = vector<bool>;
  using vd = vector<double>;
  using vvb = vector<vector<bool>>;
  using vvc = vector<vector<char>>;
  using vc = vector<char>;
  using si = set<int>;
  using mpii = map<int, int>;
  using pii = pair<int, int>;
  using pll = pair<ll, ll>;
  const char nl = '\n';
  const int dx[] = {-1, 0, 1, 0};
  const int dy[] = {0, 1, 0, -1};
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  template <class T>
  inline bool umin(T &a, const T b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  inline bool umax(T &a, const T b) {
    return a < b ? a = b, 1 : 0;
  }
  ll solve() {
    int n;
    cin >> n;
    vector<pii> t(n);
    vi a(n + 1, 0), b(n + 1, 0);
    for (int i = 0; i < (n); i++) {
      cin >> t[i].first >> t[i].second;
      a[t[i].first]++;
      b[t[i].second]++;
    }
    ll ans = 1LL * n * (n - 1) * (n - 2) / 6;
    for (int i = 0; i < (n); i++) {
      ans -= 1LL * (a[t[i].first] - 1) * (b[t[i].second] - 1);
    }
    return ans;
  }
  void solve(int _cas) { cout << solve() << endl; }
  int main() {
    srand(time(NULL));
    ios_base::sync_with_stdio(false);
    cin.tie(NULL), cout.tie(NULL), cout.precision(12);
    int cas;
    cin >> cas;
    for (int i = 0; i < (cas); i++) solve(i + 1);
  }
  