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;
const long long Mod = 1000000007;
const long long INF = 999999999999999999;
const long long NN = (long long)(1e6 + 5);
template <typename T>
T Min(T x, T y) {
  if (x < y) return x;
  return y;
}
template <typename T>
T Max(T x, T y) {
  if (x > y) return x;
  return y;
}
long long power(long long x, unsigned long long y, long long mod = Mod) {
  long long res = 1;
  x = x % mod;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1LL;
    x = (x * x) % mod;
  }
  return res % mod;
}
template <typename T>
T hcf(T a, T b) {
  if (b == 0) return a;
  return hcf(b, a % b);
}
long long c3(long long n) { return (n * (n - 1) * (n - 2)) / 6; }
long long c2(long long n) { return (n * (n - 1)) / 2; }
void solve() {
  long long n;
  cin >> n;
  vector<pair<long long, long long> > v(n);
  map<long long, long long> x, y;
  for (auto &u : v) cin >> u.first >> u.second, x[u.first]++, y[u.second]++;
  long long i, ans = 0, tot = c3(n);
  for (i = 0; i < n; i++) {
    tot -= c2(x[v[i].first] + y[v[i].second] - 2);
  }
  cout << tot << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long TT = 1;
  cin >> TT;
  for (long long tt = 1; tt <= TT; tt++) {
    solve();
  }
}
