#include <bits/stdc++.h>
using namespace std;
template <class T>
using pq = priority_queue<T>;
template <class T>
using pqg = priority_queue<T, vector<T>, greater<T>>;
int mod(long long int a, long long int b) { return ((a % b) + b) % b; }
long long int MODnumber(long long int a) {
  return (a % (long long int)1000000007 + (long long int)1000000007) %
         (long long int)1000000007;
}
long long int MODsum(long long int a, long long int b) {
  return ((a % (long long int)1000000007 + b % (long long int)1000000007) %
              (long long int)1000000007 +
          (long long int)1000000007) %
         (long long int)1000000007;
}
long long int MODproduct(long long int a, long long int b) {
  return (((a % (long long int)1000000007) * (b % (long long int)1000000007)) %
              (long long int)1000000007 +
          (long long int)1000000007) %
         (long long int)1000000007;
}
void solve() {
  long long int n;
  cin >> n;
  map<long long int, long long int> difficulties;
  map<long long int, vector<long long int>> topicandDifficulty;
  for (long long int i = 0; i < n; i++) {
    long long int x, y;
    cin >> x >> y;
    difficulties[y]++;
    topicandDifficulty[x].push_back(y);
  }
  long long int ans = 0;
  for (auto itr : topicandDifficulty) {
    long long int sz = itr.second.size();
    sz--;
    for (auto it : itr.second) {
      ans += sz * (difficulties[it] - 1);
    }
  }
  long long int countSame = 0;
  long long int countDiff = 0;
  for (auto it : difficulties) {
    countSame += (n - it.second) * (n - it.second - 1);
    countDiff += (it.second * (it.second - 1));
  }
  long long int temp = (n * (n - 1) * (n - 2)) / 6;
  temp -= ans;
  cout << temp << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

In CPP, your task is to solve 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;
template <class T>
using pq = priority_queue<T>;
template <class T>
using pqg = priority_queue<T, vector<T>, greater<T>>;
int mod(long long int a, long long int b) { return ((a % b) + b) % b; }
long long int MODnumber(long long int a) {
  return (a % (long long int)1000000007 + (long long int)1000000007) %
         (long long int)1000000007;
}
long long int MODsum(long long int a, long long int b) {
  return ((a % (long long int)1000000007 + b % (long long int)1000000007) %
              (long long int)1000000007 +
          (long long int)1000000007) %
         (long long int)1000000007;
}
long long int MODproduct(long long int a, long long int b) {
  return (((a % (long long int)1000000007) * (b % (long long int)1000000007)) %
              (long long int)1000000007 +
          (long long int)1000000007) %
         (long long int)1000000007;
}
void solve() {
  long long int n;
  cin >> n;
  map<long long int, long long int> difficulties;
  map<long long int, vector<long long int>> topicandDifficulty;
  for (long long int i = 0; i < n; i++) {
    long long int x, y;
    cin >> x >> y;
    difficulties[y]++;
    topicandDifficulty[x].push_back(y);
  }
  long long int ans = 0;
  for (auto itr : topicandDifficulty) {
    long long int sz = itr.second.size();
    sz--;
    for (auto it : itr.second) {
      ans += sz * (difficulties[it] - 1);
    }
  }
  long long int countSame = 0;
  long long int countDiff = 0;
  for (auto it : difficulties) {
    countSame += (n - it.second) * (n - it.second - 1);
    countDiff += (it.second * (it.second - 1));
  }
  long long int temp = (n * (n - 1) * (n - 2)) / 6;
  temp -= ans;
  cout << temp << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```