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;
  inline void chmax(long long &x, long long y) { x = max(x, y); }
  inline void chmin(long long &x, long long y) { x = min(x, y); }
  inline void pls(long long &x, long long y, long long Mod) {
    x = (x + y >= Mod) ? (x + y - Mod) : (x + y);
  }
  inline void sub(long long &x, long long y, long long Mod) {
    x = (x - y < 0) ? (x - y + Mod) : (x - y);
  }
  long long max(long long a, long long b) { return a > b ? a : b; }
  long long min(long long a, long long b) { return a < b ? a : b; }
  long long lowbit(long long x) { return x & (-x); }
  long long prime(long long x) {
    if (x <= 1) return 0;
    for (int i = 2; i <= int(sqrt(x)); i++) {
      if (x % i == 0) return 0;
    }
    return 1;
  }
  bool cmp(long long a, long long b) { return a > b; }
  long long gcd(long long a, long long b) {
    long long r;
    while (b > 0) {
      r = a % b;
      a = b;
      b = r;
    }
    return a;
  }
  long long powmod(long long a, long long b, long long mod) {
    long long res = 1;
    a %= mod;
    assert(b >= 0);
    for (; b; b >>= 1) {
      if (b & 1) res = res * a % mod;
      a = a * a % mod;
    }
    return res;
  }
  inline int readint() {
    int f = 1, x = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    return f * x;
  }
  inline long long readll() {
    long long f = 1, x = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    return f * x;
  }
  void fl(string name) {
    freopen((name + ".in").c_str(), "r", stdin);
    freopen((name + ".out").c_str(), "w", stdout);
  }
  const int dx[8] = {-1, 1, 0, 0, -1, 1, -1, 1},
            dy[8] = {0, 0, -1, 1, -1, -1, 1, 1};
  int main() {
    long long q;
    cin >> q;
    while (q--) {
      long long n;
      cin >> n;
      long long a[n + 1], b[n + 1];
      long long res = 0, t1[n + 1], t2[n + 1];
      for (int i = 1; i <= n; i++) t1[i] = t2[i] = 0;
      res = n * (n - 1) * (n - 2) / 6;
      for (int i = 1; i <= n; i++) {
        cin >> a[i] >> b[i];
        t1[a[i]]++;
        t2[b[i]]++;
      }
      for (int i = 1; i <= n; i++) {
        res = res - (t1[a[i]] - 1) * (t2[b[i]] - 1);
      }
      cout << res << endl;
    }
    return 0;
  }
  