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 std::cin;
using std::cout;
using std::endl;
using std::max;
using std::min;
using std::pair;
using std::swap;
using std::vector;
using ll = long long;
ll solve(const vector<pair<int, int> > &v) {
  int n = v.size();
  std::map<int, int> mp;
  for (auto [ix, iy] : v) mp[ix]++;
  ll pc = 0;
  for (auto [ix, iy] : v) pc += n - mp[ix];
  pc >>= 1;
  ll res = 0;
  for (auto [ix, iy] : v) res += pc - mp[ix] * ll(n - mp[ix]);
  return res / 3;
}
ll solve2(const vector<pair<int, int> > &v) {
  int n = v.size();
  std::map<int, ll> sx, sy, sx2, sy2;
  for (auto [ix, iy] : v) {
    sx[ix]++;
    sy[iy]++;
  }
  ll pc = 0;
  for (auto [ix, iy] : v) {
    ll d = n - sx[ix] - sy[iy] + 1;
    sx2[ix] += d;
    sy2[iy] += d;
    pc += d;
  }
  pc >>= 1;
  ll res = 0;
  for (auto [ix, iy] : v) {
    ll r = sx2[ix] + sy2[iy] - (n - sx[ix] - sy[iy] + 1) -
           ll(sx[ix] - 1) * (sy[iy] - 1);
    res += pc - r;
  }
  return res / 3;
}
void testc() {
  int n;
  std::cin >> n;
  vector<pair<int, int> > v(n);
  ll ans = 0;
  for (auto &i : v) std::cin >> i.first >> i.second;
  ans += solve(v);
  for (auto &i : v) swap(i.first, i.second);
  ans += solve(v);
  for (auto &i : v) swap(i.first, i.second);
  ans -= solve2(v);
  cout << ans;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    testc();
    if (t) putchar('\n');
  }
  return 0;
}
