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;
  using namespace chrono;
  template <class T>
  void show(vector<vector<T>> &a) {
    for (auto &i : a) {
      for (T &j : i) cout << j << ' ';
      cout << endl;
    }
  }
  template <class T>
  void show(vector<T> &a) {
    for (T i : a) cout << i << ' ';
    cout << endl;
  }
  void show(vector<string> &a) {
    for (auto &i : a) cout << i << endl;
  }
  template <class T>
  void take(vector<T> &a) {
    for (auto &i : a) cin >> i;
  }
  long long digits(long long n) {
    long long k = 0;
    while (n) k++, n /= 10;
    return k;
  }
  bool isprime(long long n) {
    long long i;
    if (n == 1) return 0;
    if (n == 2) return 1;
    if (n % 2 == 0) return 0;
    for (i = 3; i * i <= n; i += 2) {
      if (n % i == 0) return 0;
    }
    return 1;
  }
  long long power(long long x, long long y, long long MOD) {
    long long ans = 1;
    x %= MOD;
    while (y > 0) {
      if (y % 2 == 1) ans = (ans * x) % MOD;
      y /= 2;
      x = (x * x) % MOD;
    }
    return ans;
  }
  void sieve(vector<bool> &a, long long n) {
    long long i, j;
    a[1] = 0;
    a[0] = 0;
    long long h = sqrt(n + 1);
    for (i = 2; i < h; i++) {
      if (!a[i]) continue;
      for (j = 2 * i; j < n + 1; j += i) a[j] = 0;
    }
  }
  void load(unordered_map<long long, long long> &g) { g.max_load_factor(0.25); }
  void sub() {}
  const long long N = 1e9 + 7;
  const long long M = 998244353;
  void solve() {
    long long n, i, m, k, j;
    cin >> n;
    vector<long long> a(n), b(n);
    for (i = 0; i < n; i++) cin >> a[i] >> b[i];
    long long ans = n * (n - 1) * (n - 2) / 6;
    vector<long long> a1(n + 1), b1(n + 1);
    for (i = 0; i < n; i++) {
      a1[a[i]]++;
      b1[b[i]]++;
    }
    for (i = 0; i < n; i++) {
      ans -= max(0ll, a1[a[i]] - 1) * max(0ll, b1[b[i]] - 1);
    }
    cout << ans << '\n';
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    sub();
    auto start1 = high_resolution_clock::now();
    long long t = 1, s;
    cin >> t;
    for (s = 1; s < t + 1; s++) {
      solve();
    }
    auto stop1 = high_resolution_clock::now();
    auto duration = duration_cast<microseconds>(stop1 - start1);
    cerr << "Time: " << (double)duration.count() / 1000 << " ms" << endl;
    return 0;
  }
  