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.*;
import java.lang.*;
import java.io.*;

public class Main
{
	PrintWriter out = new PrintWriter(System.out);
	BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer tok = new StringTokenizer("");
    String next() throws IOException {
        if (!tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); }
        return tok.nextToken();
    }
    int ni() throws IOException { return Integer.parseInt(next()); }
    long nl() throws IOException { return Long.parseLong(next()); }
    
    long mod=1000000007;
    
    void solve() throws IOException {
        for (int tests=ni();tests>0;tests--) {
            int n=ni();
            int[][]C=new int[n][2];
            for (int i=0;i<n;i++) { C[i][0]=ni(); C[i][1]=ni(); }
            long[]D=new long[n+1];
            long[]TT=new long[n+1];
            long[]TC=new long[n+1];
            //for (int i=1;i<=n;i++) A[i]=new ArrayList();
            for (int i=0;i<n;i++) {
                TT[C[i][0]]++;
                TC[C[i][0]]++;
                D[C[i][1]]++;
            }
            
            long[]B=new long[n+1];
            
            long tc=0;
            for (int i=1;i<=n;i++) {
                long sz=TT[i];
                B[i]=(sz*(sz-1))/2;
                tc+=B[i];
            }
            
            long ans=0;
            long below=n;
            for (int i=1;i<=n;i++) {
                long sz=TT[i];
                below-=sz;
                long lt=below;
                tc-=B[i];
                lt=(lt*(lt-1))/2-tc;
                
                ans+=(lt*sz);
                //out.println(i+" "+sz+" "+lt+" "+ans);
            }
            
            for (int i=0;i<n;i++) {
                int d=C[i][1];
                int t=C[i][0];
                ans+=((TC[t]-1)*(TC[t]-2))/2;
                ans+=(TC[t]-1)*(n-TT[t]-D[d]+1);
                TC[t]--;
            }
            
            out.println(ans);
        }
        out.flush();
    }
    
    int gcd(int a,int b) { return(b==0?a:gcd(b,a%b)); }
    long gcd(long a,long b) { return(b==0?a:gcd(b,a%b)); }
    long mp(long a,long p) { long r=1; while(p>0) { if ((p&1)==1) r=(r*a)%mod; p>>=1; a=(a*a)%mod; } return r; }
    
    public static void main(String[] args) throws IOException {
        new Main().solve();
    }
}