#include <bits/stdc++.h>
using namespace std;
long long int F(const vector<long long int> &v) {
  map<long long int, long long int> first;
  for (auto el : v) first[el]++;
  vector<long long int> a(first.size() + 2), dp(first.size() + 2),
      dp2(first.size() + 2), dp3(first.size() + 2);
  {
    int z = 0;
    for (auto temp : first) {
      a[z++] = temp.second;
    }
  }
  for (auto a : a) cout << a << " ";
  cout << endl;
  ;
  for (int i = (int)first.size() - 1; i >= 0; i--) {
    dp[i] = dp[i + 1] + a[i];
  }
  for (auto a : dp) cout << a << " ";
  cout << endl;
  ;
  for (int i = (int)first.size() - 1; i >= 0; i--) {
    dp2[i] = dp2[i + 1] + a[i] * dp[i + 1];
    dp3[i] = dp3[i + 1] + a[i] * dp2[i + 1];
  }
  return dp3[0];
}
long long int nc3(long long int n) { return ((n * (n - 1)) / 2 * (n - 2)) / 3; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int T;
  cin >> T;
  for (int tc = 1; tc <= T; tc++) {
    int n;
    cin >> n;
    vector<long long int> a(n), b(n);
    map<long long int, long long int> fA, fB;
    vector<pair<long long, long long> > Temp;
    for (int i = 0; i < n; i++) {
      cin >> a[i] >> b[i];
      fA[a[i]]++;
      fB[b[i]]++;
      Temp.push_back({a[i], b[i]});
    }
    long long int res = nc3(n);
    for (auto temp : Temp) {
      res -= (fA[temp.first] - 1) * (fB[temp.second] - 1);
    }
    cout << res << "\n";
  }
}

### 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;
long long int F(const vector<long long int> &v) {
  map<long long int, long long int> first;
  for (auto el : v) first[el]++;
  vector<long long int> a(first.size() + 2), dp(first.size() + 2),
      dp2(first.size() + 2), dp3(first.size() + 2);
  {
    int z = 0;
    for (auto temp : first) {
      a[z++] = temp.second;
    }
  }
  for (auto a : a) cout << a << " ";
  cout << endl;
  ;
  for (int i = (int)first.size() - 1; i >= 0; i--) {
    dp[i] = dp[i + 1] + a[i];
  }
  for (auto a : dp) cout << a << " ";
  cout << endl;
  ;
  for (int i = (int)first.size() - 1; i >= 0; i--) {
    dp2[i] = dp2[i + 1] + a[i] * dp[i + 1];
    dp3[i] = dp3[i + 1] + a[i] * dp2[i + 1];
  }
  return dp3[0];
}
long long int nc3(long long int n) { return ((n * (n - 1)) / 2 * (n - 2)) / 3; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int T;
  cin >> T;
  for (int tc = 1; tc <= T; tc++) {
    int n;
    cin >> n;
    vector<long long int> a(n), b(n);
    map<long long int, long long int> fA, fB;
    vector<pair<long long, long long> > Temp;
    for (int i = 0; i < n; i++) {
      cin >> a[i] >> b[i];
      fA[a[i]]++;
      fB[b[i]]++;
      Temp.push_back({a[i], b[i]});
    }
    long long int res = nc3(n);
    for (auto temp : Temp) {
      res -= (fA[temp.first] - 1) * (fB[temp.second] - 1);
    }
    cout << res << "\n";
  }
}

```