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 std::cin;
  using std::cout;
  using std::endl;
  using std::max;
  using std::min;
  using std::pair;
  using std::swap;
  using std::vector;
  using ll = long long;
  ll solve(const vector<pair<int, int> > &v) {
    int n = v.size();
    std::map<int, int> mp;
    for (auto [ix, iy] : v) mp[ix]++;
    ll pc = 0;
    for (auto [ix, iy] : v) pc += n - mp[ix];
    pc >>= 1;
    ll res = 0;
    for (auto [ix, iy] : v) res += pc - mp[ix] * ll(n - mp[ix]);
    return res / 3;
  }
  ll solve2(const vector<pair<int, int> > &v) {
    int n = v.size();
    std::map<int, ll> sx, sy, sx2, sy2;
    for (auto [ix, iy] : v) {
      sx[ix]++;
      sy[iy]++;
    }
    ll pc = 0;
    for (auto [ix, iy] : v) {
      ll d = n - sx[ix] - sy[iy] + 1;
      sx2[ix] += d;
      sy2[iy] += d;
      pc += d;
    }
    pc >>= 1;
    ll res = 0;
    for (auto [ix, iy] : v) {
      ll r = sx2[ix] + sy2[iy] - (n - sx[ix] - sy[iy] + 1) -
             ll(sx[ix] - 1) * (sy[iy] - 1);
      res += pc - r;
    }
    return res / 3;
  }
  void testc() {
    int n;
    std::cin >> n;
    vector<pair<int, int> > v(n);
    ll ans = 0;
    for (auto &i : v) std::cin >> i.first >> i.second;
    ans += solve(v);
    for (auto &i : v) swap(i.first, i.second);
    ans += solve(v);
    for (auto &i : v) swap(i.first, i.second);
    ans -= solve2(v);
    cout << ans;
  }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      testc();
      if (t) putchar('\n');
    }
    return 0;
  }
  