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 gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long binpow(long long a, long long b, long long m) {
    a %= m;
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = res * a % m;
      a = a * a % m;
      b >>= 1;
    }
    return res;
  }
  long long primeFactors(long long n) {
    long long ans = 0;
    long long m = n;
    for (long long i = 2; i * i <= m; i++) {
      while (n % i == 0) {
        n /= i;
        ans++;
      }
    }
    if (n != 1) return ans + 1;
    return ans;
  }
  long long power(long long a, long long b) {
    if (b == 0) {
      return 1;
    }
    long long ans = power(a, b / 2);
    ans *= ans;
    ans %= 1000000007;
    if (b % 2) {
      ans *= a;
    }
    return ans % 1000000007;
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ;
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      long long total = n * (n - 1) * (n - 2) / 6;
      pair<long long, long long> p[n];
      map<long long, long long> t, d;
      for (long long i = 0; i < n; i++) {
        long long x, y;
        cin >> x >> y;
        p[i] = {x, y};
        t[x]++;
        d[y]++;
      }
      long long di = 0;
      for (long long i = 0; i < n; i++) {
        di += (t[p[i].first] - 1) * (d[p[i].second] - 1);
      }
      cout << total - di << '\n';
    }
    return 0;
  }
  