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.
#include <bits/stdc++.h>
using namespace std;
void init_code() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
long long int M = 1000000007;
long long int power(long long int x, long long int y) {
  if (y == 0)
    return 1;
  else if (y == 1)
    return x % M;
  else {
    long long int z = power(x, y / 2);
    long long int w = ((z % M) * (z % M)) % M;
    if (y % 2 == 1)
      return ((x % M) * (w % M)) % M;
    else
      return w;
  }
}
int main() {
  init_code();
  long long int t = 1;
  cin >> t;
  long long int o = 1;
  while (t--) {
    long long int m, n, a, i, j, k, l = 0, ans = 0;
    double sum = 0.0;
    cin >> n;
    map<long long int, long long int> mp, np;
    vector<pair<long long int, long long int>> v;
    for (i = 0; i < n; i++) {
      long long int b, c;
      cin >> b >> c;
      v.push_back({b, c});
      mp[b]++;
      np[c]++;
    }
    l = n;
    ans = (n * (n - 1) * (n - 2)) / 6;
    for (i = 0; i < n; i++) {
      long long int x = v[i].first, y = v[i].second;
      l--;
      mp[x]--;
      np[y]--;
      long long int p = l - mp[x] - np[y];
      if (mp[x] > 0 && np[y] > 0) ans = ans - mp[x] * np[y];
    }
    cout << ans << endl;
  }
}
