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 _print(long long t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(double t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
long long po(long long a, long long b, long long m) {
  long long t = 1;
  while (b > 0) {
    if (b % 2 != 0) {
      t = (t * a) % m;
    }
    a = (a * a) % m;
    b = b / 2;
  }
  return t;
}
long long inv(long long a, long long m) { return po(a, m - 2, m); }
bool sortbysec(const pair<char, long long> &a, const pair<char, long long> &b) {
  return (a.second < b.second);
}
long long get(long long n) {
  long long ans = 1;
  for (long long i = 2; i <= n; i++)
    ans = (ans % 1000000007 * i % 1000000007) % 1000000007;
  return ans % 1000000007;
}
long long nPr(long long n, long long r) {
  return get(n) * inv(get(n - r), 1000000007);
}
long long modularExponentiation(long long x, long long n) {
  long long result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % 1000000007;
    x = (x * x) % 1000000007;
    n = n / 2;
  }
  return result;
}
void solve() {
  long long i, j, flag = 0;
  long long n;
  cin >> n;
  map<long long, long long> M, N;
  map<pair<long long, long long>, long long> x;
  vector<pair<long long, long long>> v;
  for (i = 0; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    v.push_back({a, b});
    M[a]++;
    N[b]++;
    x[{a, b}]++;
  }
  long long sum = (n * (n - 1) * (n - 2)) / 6;
  ;
  for (i = 0; i < n; i++) {
    if (M[v[i].first] >= 2 && N[v[i].second] >= 2) {
      sum -= (M[v[i].first] - 1) * (N[v[i].second] - 1);
    }
  }
  cout << sum << '\n';
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}
