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.
#include <bits/stdc++.h>
using namespace std;
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  os << '{';
  string sep;
  for (const T &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
const int MAX_N = 1e5 + 5;
const long long MOD = 1e9 + 7;
const long long INF = 1e9;
const long double EPS = 1e-9;
void solve() {
  long long n;
  cin >> n;
  unordered_map<long long, vector<long long>> topc_diff;
  unordered_map<long long, vector<long long>> diff_topc;
  for (int i = 0; i < n; i++) {
    long long t1, t2;
    cin >> t1;
    cin >> t2;
    topc_diff[t1].push_back(t2);
    diff_topc[t2].push_back(t1);
  }
  long long total = (n * (n - 1) * (n - 2)) / 6, ans = 0;
  for (auto &topc : topc_diff) {
    long long temp1 = topc.second.size() - 1;
    for (long long diff : topc.second) {
      long long temp2 = diff_topc[diff].size() - 1;
      ans += temp1 * temp2;
    }
  }
  cout << total - ans << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tc = 1;
  cin >> tc;
  for (int t = 1; t <= tc; t++) {
    solve();
  }
}
