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 sun.awt.windows.WBufferStrategy;

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


public class MainSolve {
    public static void main(String[] args) throws IOException {
        BufferedReader reader=new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter out=new BufferedWriter(new OutputStreamWriter(System.out));
        int testNum=Integer.parseInt(reader.readLine());
        for (int i = 0; i < testNum; i++) {
            int n=Integer.parseInt(reader.readLine());
            int[] countA=new int[n+1];
            int[] countB=new int[n+1];
            int[][] arr=new int[n][2];
            for (int j = 0; j < n; j++) {
                String aAndB=reader.readLine();
                String []AB=aAndB.split(" ");
                arr[j][0]=Integer.parseInt(AB[0]);
                arr[j][1]=Integer.parseInt(AB[1]);
                countA[Integer.parseInt(AB[0])]++;
                countB[Integer.parseInt(AB[1])]++;
            }
            out.write(solve(countA,countB,n,arr)+"\n");
            ans.clear();
            indexArray.clear();
        }
        out.flush();
        out.close();
        reader.close();
    }

    public static String solve(int[]a,int[] b,int n,int[][] arr){
        long sum=n*1L*(n-1)*(n-2)/6;
        for (int i = 0; i < n; i++) {
            sum-=(long) (a[arr[i][0]]-1)*(b[arr[i][1]]-1);
        }
        return String.valueOf(sum);
    }

    private static List<List<int []>> ans=new ArrayList<>();
    private static List<int []> indexArray=new ArrayList<>();

    private static void dfs(int[][] a,int index){
        if (indexArray.size()==3){
            boolean zero=indexArray.get(0)[0]!=indexArray.get(1)[0]&&indexArray.get(1)[0]!=indexArray.get(2)[0]&&indexArray.get(0)[0]!=indexArray.get(2)[0];
            boolean one=indexArray.get(0)[1]!=indexArray.get(1)[1]&&indexArray.get(1)[1]!=indexArray.get(2)[1]&&indexArray.get(0)[1]!=indexArray.get(2)[1];
            if ((zero||one)&&!ans.contains(new ArrayList<>(indexArray) )){
                ans.add(new ArrayList<>(indexArray));
                return;
            }
        }
        for (int i = index; i < a.length; i++) {
            indexArray.add(a[i]);
            dfs(a,i+1);
            indexArray.remove(indexArray.size()-1);
        }
    }

}

