#include <bits/stdc++.h>
using namespace std;
vector<int> findPrimes(int n) {
  vector<bool> prime(n + 1, true);
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  vector<int> pr;
  for (int p = 2; p <= n; p++)
    if (prime[p]) pr.push_back(p);
  return pr;
}
int gcd(int a, int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a(n), b(n), ca(n + 1), cb(n + 1);
    for (int i = 0; i < n; i++) {
      cin >> a[i] >> b[i];
      ca[a[i]]++;
      cb[b[i]]++;
    }
    long long res = 0;
    res = (n) * (n - 1LL) * (n - 2LL);
    res /= 6LL;
    for (int i = 0; i < n; i++) {
      res -= ((ca[a[i]] - 1LL) * (cb[b[i]] - 1LL));
    }
    cout << res << endl;
  }
}

### Prompt

Your task is to create 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;
vector<int> findPrimes(int n) {
  vector<bool> prime(n + 1, true);
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  vector<int> pr;
  for (int p = 2; p <= n; p++)
    if (prime[p]) pr.push_back(p);
  return pr;
}
int gcd(int a, int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a(n), b(n), ca(n + 1), cb(n + 1);
    for (int i = 0; i < n; i++) {
      cin >> a[i] >> b[i];
      ca[a[i]]++;
      cb[b[i]]++;
    }
    long long res = 0;
    res = (n) * (n - 1LL) * (n - 2LL);
    res /= 6LL;
    for (int i = 0; i < n; i++) {
      res -= ((ca[a[i]] - 1LL) * (cb[b[i]] - 1LL));
    }
    cout << res << endl;
  }
}

```