#include <bits/stdc++.h>
using namespace std;
long long pw(int x, int y, int z) {
  if (y == 0) return 1 % z;
  long long a = pw(x, y / 2, z);
  if (y % 2 == 0)
    return (a * a) % z;
  else
    return (((a * a) % z) * x % z) % z;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<vector<long long>> arr(n, vector<long long>(2));
    for (int i = 0; i < n; ++i) cin >> arr[i][0] >> arr[i][1];
    map<long long, long long> l, r;
    for (int i = 0; i < n; ++i) {
      l[arr[i][0]]++;
      r[arr[i][1]]++;
    }
    sort(arr.begin(), arr.end());
    long long ans = (n * (n - 1) * (n - 2)) / 6;
    sort(arr.begin(), arr.end());
    for (int i = 0; i < n; ++i) {
      long long curl = arr[i][0];
      long long nos = l[curl];
      long long othel = 0;
      if (nos >= 2) {
        for (int j = i; j < i + nos; ++j) {
          othel += (r[arr[j][1]] - 1);
        }
      }
      ans -= ((nos - 1) * othel);
      i += nos - 1;
    }
    cout << ans << endl;
  }
}

### Prompt

Your challenge is to write 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;
long long pw(int x, int y, int z) {
  if (y == 0) return 1 % z;
  long long a = pw(x, y / 2, z);
  if (y % 2 == 0)
    return (a * a) % z;
  else
    return (((a * a) % z) * x % z) % z;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<vector<long long>> arr(n, vector<long long>(2));
    for (int i = 0; i < n; ++i) cin >> arr[i][0] >> arr[i][1];
    map<long long, long long> l, r;
    for (int i = 0; i < n; ++i) {
      l[arr[i][0]]++;
      r[arr[i][1]]++;
    }
    sort(arr.begin(), arr.end());
    long long ans = (n * (n - 1) * (n - 2)) / 6;
    sort(arr.begin(), arr.end());
    for (int i = 0; i < n; ++i) {
      long long curl = arr[i][0];
      long long nos = l[curl];
      long long othel = 0;
      if (nos >= 2) {
        for (int j = i; j < i + nos; ++j) {
          othel += (r[arr[j][1]] - 1);
        }
      }
      ans -= ((nos - 1) * othel);
      i += nos - 1;
    }
    cout << ans << endl;
  }
}

```