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.

import java.util.HashMap;
import java.util.Scanner;

public class TrainingSession {

    public static void main(String args[]) {
        Scanner scan = new Scanner(System.in);
        int t = scan.nextInt();
        while (t-- > 0) {
            long n = scan.nextInt();
            long res = (n * (n - 1) * (n - 2)) / 6;
            int[] a = new int[(int) n];
            int[] b = new int[(int) n];
            long[] fa = new long[(int) (n + 1)];
            long[] fb = new long[(int) (n + 1)];
            for (int i = 0; i < n; i++) {
                int x = scan.nextInt();
                int y = scan.nextInt();
                ++fa[x];
                ++fb[y];
                a[i] = x;
                b[i] = y;
            }
            for (int i = 0; i < n; i++) {
                res -= (fa[a[i]] - 1) * fb[b[i]-1];
            }
            System.out.println(res);
        }

    }

}
