#include <bits/stdc++.h>
using namespace std;
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
void file() {}
void fast() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
}
void test_case() {
  int n;
  cin >> n;
  vector<vector<int> > adjL(n + 1);
  vector<pair<int, int> > v1(n);
  for (auto& it : v1) {
    cin >> it.first >> it.second;
    adjL[it.first].push_back(it.second);
  }
  long long ans = 1LL * n * (n - 1) * (n - 2) / 6;
  vector<long long> fre(n + 1);
  for (int i = 1; i <= n; i++) {
    for (auto& it : adjL[i]) {
      ans -= fre[it];
    }
    for (auto& it : adjL[i]) fre[it] += (int)(adjL[i].size()) - 1;
  }
  fre = vector<long long>(n + 1);
  for (int i = n; i >= 1; i--) {
    for (auto& it : adjL[i]) ans -= fre[it];
    for (auto& it : adjL[i]) fre[it] += (int)(adjL[i].size()) - 1;
  }
  cout << ans << '\n';
}
int main() {
  file();
  fast();
  int t;
  cin >> t;
  while (t--) {
    test_case();
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
void file() {}
void fast() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
}
void test_case() {
  int n;
  cin >> n;
  vector<vector<int> > adjL(n + 1);
  vector<pair<int, int> > v1(n);
  for (auto& it : v1) {
    cin >> it.first >> it.second;
    adjL[it.first].push_back(it.second);
  }
  long long ans = 1LL * n * (n - 1) * (n - 2) / 6;
  vector<long long> fre(n + 1);
  for (int i = 1; i <= n; i++) {
    for (auto& it : adjL[i]) {
      ans -= fre[it];
    }
    for (auto& it : adjL[i]) fre[it] += (int)(adjL[i].size()) - 1;
  }
  fre = vector<long long>(n + 1);
  for (int i = n; i >= 1; i--) {
    for (auto& it : adjL[i]) ans -= fre[it];
    for (auto& it : adjL[i]) fre[it] += (int)(adjL[i].size()) - 1;
  }
  cout << ans << '\n';
}
int main() {
  file();
  fast();
  int t;
  cin >> t;
  while (t--) {
    test_case();
  }
  return 0;
}

```