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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.StringTokenizer;

/**
 * Accomplished using the EduTools plugin by JetBrains https://plugins.jetbrains.com/plugin/10081-edutools
 * <p>
 * To modify the template, go to Preferences -> Editor -> File and Code Templates -> Other
 */

public class Main {
    static class Solution {
        private int n;
        private int[] C;
        private int[] N;
        private long[][] ch;
        private long[] P;
        private int[] match;
        public Solution(int n, List<List<Integer>> points) {
            this.n = n;
            C = new int[n];
            N = new int[n];
            for (List<Integer> pr: points) {
                int i = pr.get(0), j = pr.get(1);
                C[j]++;
                N[i]++;
            }
            precomputeP();
            precomputeChoose();
            precomputeMatch(points);
        }
        private void precomputeP() {
            P = new long[n];
            int tot = 0;
            for (int i = n-1; i>=0; i--) {
                if (i==n-1) {
                    P[i] = 0;
                } else  {
                    P[i] = P[i+1];
                }
                P[i] += (long)(tot)*(long)N[i];
                tot += N[i];
            }
        }
        private long choose(int n, int k) {
            if (k>n) return 0;
            return ch[n][k];
        }
        private void precomputeChoose() {
            ch = new long[n+1][4];

            ch[0][0] = 1;
            for (int num = 1; num<=n; num++) {
                ch[num][0] = 1;
                for (int k = 1; k<=3; k++) {
                    ch[num][k] = ch[num-1][k-1] + ch[num-1][k];
                }
            }
        }
        private void precomputeMatch(List<List<Integer>> points) {
            match = new int[n];
            for (List<Integer> pr: points) {
                int i = pr.get(0), j = pr.get(1);
//                assert(C[j]>0);
                match[i] += C[j]-1;
            }
        }
        public long solve() {
            long ans = 0;

            for (int i = 0; i<n; i++) {
                long Tiii = choose(N[i],3);
                long Ti = 0;
                if (i+1<n) Ti = N[i] * P[i+1];

//                assert (n-N[i]-match[i] >= 0);
                long Tii = (long)(n-N[i]-match[i])*choose(N[i],2) + (long)(match[i])*choose(N[i]-1,2);
                ans += Tiii+Tii+Ti;
            }
            return ans;
        }
        /* The hard case is
         * 2 of the same topic "i"
         * Note if 2 are of the same topic, we MUST
         * satisfy the 3 unique difficulties constraint
         * We can NOT do Choose(N[i],2) * (n - N[i])
         * Because any random (n-N[i]) choose for the third
         * might have the same difficulty as one of the 2 from row i
         *
         *
         * */


    }

    public static void main(String[] args) {
        // Write your solution here
        FastScanner fs = new FastScanner();
        int T = fs.nextInt();
        while (T > 0) {
            int n = fs.nextInt();
            final List<List<Integer>> points = new ArrayList<>();
            for (int t=0; t<n; t++) {
                int i = fs.nextInt(), j = fs.nextInt();
                i--; j--;
                points.add(Arrays.asList(i,j));
            }
            System.out.println((new Solution(n,points)).solve());
            T--;
        }
    }


    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        String next() {
            while (!st.hasMoreTokens())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        long nextLong() {
            return Long.parseLong(next());
        }
    }

}