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;
using pii = pair<long long, long long>;
const double EPS = 1e-9;
const long long dx[] = {1, -1, 0, 0};
const long long dy[] = {0, 0, 1, -1};
long long cal(long long k) { return (k * (k - 1) * (k - 2)) / 6; }
void solve() {
  long long n;
  cin >> n;
  vector<pii> v(n, {0, 0});
  map<long long, vector<long long>> mpt, mpd;
  for (long long i = 0; i < n; i++) {
    cin >> v[i].first >> v[i].second;
    if (mpt.find(v[i].first) == mpt.end()) {
      mpt[v[i].first] = vector<long long>(1, v[i].second);
    } else {
      mpt[v[i].first].push_back(v[i].second);
    }
    if (mpd.find(v[i].second) == mpd.end()) {
      mpd[v[i].second] = vector<long long>(1, v[i].first);
    } else {
      mpd[v[i].second].push_back(v[i].first);
    }
  }
  if (mpt.size() == n || mpd.size() == n) {
    cout << cal(n) << '\n';
    return;
  }
  long long ans = 0;
  for (auto &K : mpt) {
    long long ksize = K.second.size();
    ans = ksize * (n - ksize) * (n - ksize - 1) / 2;
  }
  cout << ans << '\n';
}
int32_t main() {
  cin.sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
  int32_t tc;
  cin >> tc;
  while (tc--) {
    solve();
  }
  return 0;
}
