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;
template <class T>
void print(vector<T> a) {
  int n = a.size();
  for (long long i = 0; i < n; ++i) {
    cout << a[i] << (i == n - 1 ? "\n" : " ");
  }
}
int sum_vector(vector<int> v) { return accumulate(v.begin(), v.end(), 0); }
void sort_vector(vector<int> &v) { sort(v.begin(), v.end()); }
void sort_comp(vector<int> &v, bool func(int, int)) {
  sort(v.begin(), v.end(), func);
}
bool comp(array<long long, 2> a, array<long long, 2> b) { return a[1] < b[1]; }
long long gcd(long long a, long long b) {
  a = abs(a);
  b = abs(b);
  while (a) {
    long long temp = a;
    a = b % a;
    b = temp;
  }
  return abs(b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
string binary(long long num) {
  string ans = "";
  do {
    ans = to_string(num % 2) + ans;
    num /= 2;
  } while (num);
  return ans;
}
const int mxn = 2e5 + 7;
const int di = 18;
const int mill = 1e6 + 3;
const long long mod = 1e9 + 7;
long long pwr(long long num, long long p) {
  long long res = 1;
  while (p > 0) {
    if (p & 1) res = (res * num) % mod;
    num = (num * num) % mod;
    p /= 2;
  }
  return res;
}
long long inverse(long long num) { return pwr(num, mod - 2); }
long long get_rand(long long n) { return ((rand() << 15) + rand()) % n; }
void solve() {
  long long n;
  cin >> n;
  vector<long long> top(n + 2);
  vector<long long> dif(n + 2);
  vector<pair<long long, long long> > v(n);
  long long a, b;
  for (long long i = 0; i < n; i++) {
    cin >> a >> b;
    top[a]++;
    dif[b]++;
    v[i] = {a, b};
  }
  long long ans = n * (n - 1) * (n - 2) / 6;
  for (pair<long long, long long> p : v) {
    ans -= (top[p.first] - 1) * (dif[p.second] - 1);
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int t = 1;
  cin >> t;
  for (int ii = 1; ii <= t; ii++) {
    solve();
  }
  return 0;
}
