#include <bits/stdc++.h>
using namespace std;
int a[205000], b[200500];
long long int p[200500] = {}, q[200500] = {}, r[200500] = {}, sum[205000] = {},
              sumsq[200500] = {};
int main() {
  int tc;
  scanf("%d", &tc);
  for (int t = 1; t <= tc; t++) {
    long long int n;
    scanf("%lld", &n);
    for (int i = 1; i <= n; i++) {
      p[i] = 0;
      q[i] = 0;
      r[i] = 0;
      sum[i] = 0;
      sumsq[i] = 0;
    }
    for (int i = 1; i <= n; i++) {
      scanf("%d %d", &a[i], &b[i]);
      p[a[i]]++;
      q[b[i]]++;
    }
    long long int ans = n * (n - 1) * (n - 2) / 6;
    for (int i = 1; i <= n; i++) {
      ans -= (p[a[i]] - 1) * (q[b[i]] - 1);
    }
    printf("%lld\n", ans);
  }
}

### 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;
int a[205000], b[200500];
long long int p[200500] = {}, q[200500] = {}, r[200500] = {}, sum[205000] = {},
              sumsq[200500] = {};
int main() {
  int tc;
  scanf("%d", &tc);
  for (int t = 1; t <= tc; t++) {
    long long int n;
    scanf("%lld", &n);
    for (int i = 1; i <= n; i++) {
      p[i] = 0;
      q[i] = 0;
      r[i] = 0;
      sum[i] = 0;
      sumsq[i] = 0;
    }
    for (int i = 1; i <= n; i++) {
      scanf("%d %d", &a[i], &b[i]);
      p[a[i]]++;
      q[b[i]]++;
    }
    long long int ans = n * (n - 1) * (n - 2) / 6;
    for (int i = 1; i <= n; i++) {
      ans -= (p[a[i]] - 1) * (q[b[i]] - 1);
    }
    printf("%lld\n", ans);
  }
}

```