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;
  long long MOD = 1e9 + 7;
  int setBit(int num, int idx, int value = 1) {
    return (value) ? (num | (1 << idx)) : (num & ~(1 << idx));
  }
  int getBit(int num, int idx) { return ((num >> idx) & 1) == 1; }
  int cntBits(int num) {
    int ret = 0;
    while (num) {
      if (num % 2) ret++;
      num /= 2;
    }
    return ret;
  }
  int is_prime[1000001];
  void Sieve() {
    for (int i = 0; i <= 1000000; i++) {
      is_prime[i] = 1;
    }
    is_prime[0] = 0;
    is_prime[1] = 0;
    for (int i = 2; i * i <= 1000000; i++) {
      if (is_prime[i]) {
        for (int j = i * i; j <= 1000000; j += i) {
          is_prime[j] = 0;
        }
      }
    }
  }
  vector<int> get_bin(int n) {
    vector<int> ret;
    while (n > 0) {
      if (n % 2 == 0)
        ret.push_back(0);
      else
        ret.push_back(1);
      n /= 2;
    }
    return ret;
  }
  unordered_map<long long, long long> factorise(long long n) {
    unordered_map<long long, long long> map;
    if (n % 2 == 0) {
      while (n % 2 == 0) {
        if (map.find(2) != map.end())
          map[2]++;
        else
          map[2] = 1;
        n = n / 2;
      }
    }
    for (long long i = 3; i * i <= n; i += 2) {
      if (n % i == 0) {
        while (n % i == 0) {
          if (map.find(i) != map.end())
            map[i]++;
          else
            map[i] = 1;
          n = n / i;
        }
      }
    }
    if (n > 1) map[n] = 1;
    return map;
  }
  unsigned long long power(unsigned long long x, int y, int p) {
    unsigned long long res = 1;
    x = x % p;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  int main() {
    int t;
    t = 1;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      vector<pair<long long, long long> > arr;
      map<long long, long long> mpa;
      map<long long, long long> mpb;
      for (int i = 0; i < n; i++) {
        int a, b;
        cin >> a >> b;
        arr.push_back({a, b});
        mpa[a]++;
        mpb[b]++;
      }
      long long ans = (n * (n - 1) * (n - 2)) / 6;
      for (auto &val : arr) {
        long long cand1 = mpa[val.first] - 1;
        long long cand2 = mpb[val.second] - 1;
        ans -= cand1 * cand2;
      }
      cout << ans << endl;
    }
    return 0;
  }
  