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.io.*;
import java.util.*;

public class Main {
    public static void main(String[] args) throws IOException {
        //br = new BufferedReader(new InputStreamReader(new FileInputStream("palindrome.in")));
        //PrintWriter pw = new PrintWriter(new File("u.in"));
        int t = nextInt();
        while (t-- != 0) {
            int n = nextInt();
            long[] a = new long[n];
            long[] b = new long[n];
            long[]counta=new long[n];
            long[]countb=new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = nextInt();
                counta[(int) (a[i]-1)]++;
                b[i] = nextInt();
                countb[(int) (b[i]-1)]++;
            }
            long ans = (long)n * (long)(n - 1) * (long)(n - 2) / 6;
            for (int i = 0; i < n; i++) {
                ans-= (counta[(int) a[i]-1] - 1) *(countb[(int) b[i]-1]-1);
            }
            out.println(ans);
        }
        out.close();
    }

    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter out = new PrintWriter(System.out);
    static StringTokenizer in = new StringTokenizer("");


    public static boolean hasNext() throws IOException {
        if (in.hasMoreTokens()) return true;
        String s;
        while ((s = br.readLine()) != null) {
            in = new StringTokenizer(s);
            if (in.hasMoreTokens()) return true;
        }
        return false;
    }

    public static String nextToken() throws IOException {
        while (!in.hasMoreTokens()) {
            in = new StringTokenizer(br.readLine());
        }
        return in.nextToken();
    }

    public static int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    public static long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    public static double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
}