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;
void prar(long long *a, long long n) {
  for (long long i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << "\n";
}
long long mgcd(long long a, long long b) {
  return (a == 0) ? (b) : (mgcd(b % a, a));
}
void solve(long long z) {
  long long n;
  cin >> n;
  map<long long, long long> fa, fb;
  vector<pair<long long, long long> > p(n);
  for (long long i = 0; i < n; i++) {
    cin >> p[i].first;
    cin >> p[i].second;
    fa[p[i].first] = 0;
    fb[p[i].second] = 0;
  }
  for (long long i = 0; i < n; i++) {
    fa[p[i].first] += 1;
    fb[p[i].second] += 1;
  }
  long long cnt = 0;
  for (long long i = 0; i < n; i++) {
    cnt += (fa[p[i].first] - 1) * (fb[p[i].second] - 1);
  }
  long long ans = (n * (n - 1) * (n - 2) / 6) - cnt;
  cout << ans << "\n";
  return;
}
signed main() {
  long long t;
  scanf("%lld", &t);
  for (long long z = 1; z <= t; z++) {
    solve(z);
  }
  return 0;
}
