#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n), b(n);
  vector<long long> ad[n + 1], bd[n + 1];
  for (long long i = 0; i < n; i++) {
    cin >> a[i] >> b[i];
  }
  for (long long i = 0; i < n; i++) {
    ad[a[i]].push_back(b[i]);
    bd[b[i]].push_back(a[i]);
  }
  long long ans = (n * (n - 1) * (n - 2)) / 6;
  for (long long i = 1; i <= n; i++) {
    long long as = ad[i].size() - 1;
    for (auto i : ad[i]) {
      long long bs = bd[i].size() - 1;
      ans -= as * bs;
    }
  }
  cout << ans << '\n';
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long _t;
  cin >> _t;
  while (_t--) solve();
  return 0;
}

### Prompt

Please formulate a CPP solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n), b(n);
  vector<long long> ad[n + 1], bd[n + 1];
  for (long long i = 0; i < n; i++) {
    cin >> a[i] >> b[i];
  }
  for (long long i = 0; i < n; i++) {
    ad[a[i]].push_back(b[i]);
    bd[b[i]].push_back(a[i]);
  }
  long long ans = (n * (n - 1) * (n - 2)) / 6;
  for (long long i = 1; i <= n; i++) {
    long long as = ad[i].size() - 1;
    for (auto i : ad[i]) {
      long long bs = bd[i].size() - 1;
      ans -= as * bs;
    }
  }
  cout << ans << '\n';
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long _t;
  cin >> _t;
  while (_t--) solve();
  return 0;
}

```