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;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
bool sort2(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second > b.second);
}
const long long INF = 1e18;
const int mod = 1e9 + 7;
inline int add(int x, int y) {
  x += y;
  if (x >= mod) return x - mod;
  return x;
}
inline int sub(int x, int y) {
  x -= y;
  if (x < 0) return x + mod;
  return x;
}
inline int mul(int x, int y) { return (x * 1ll * y) % mod; }
inline int power(int x, int y) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = mul(ans, x);
    x = mul(x, x);
    y >>= 1;
  }
  return ans;
}
inline int inv(int x) { return power(x, mod - 2); }
const long long FACTORIAL_SIZE = 1.1e6;
long long fact[FACTORIAL_SIZE], ifact[FACTORIAL_SIZE];
void gen_factorial(long long n) {
  fact[0] = fact[1] = ifact[0] = ifact[1] = 1;
  for (long long i = 2; i <= n; i++) {
    fact[i] = (i * fact[i - 1]) % mod;
  }
  ifact[n] = inv(fact[n]);
  for (long long i = n - 1; i >= 2; i--) {
    ifact[i] = ((i + 1) * ifact[i + 1]) % mod;
  }
}
long long nck(long long n, long long k) {
  long long den = (ifact[k] * ifact[n - k]) % mod;
  return (den * fact[n]) % mod;
}
const int N = 2e5 + 5;
vector<long long> diff[N], topic[N];
long long n;
void solve() {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    diff[i].clear();
    topic[i].clear();
  }
  for (long long i = 0; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    diff[b].push_back(a);
    topic[a].push_back(b);
  }
  long long ans = n * (n - 1) * (n - 2) / 6;
  for (long long i = 1; i <= n; i++) {
    if (diff[i].size() < 2) continue;
    for (auto it : diff[i]) {
      if (topic[it].size() < 2) continue;
      ans -= (diff[i].size() - 1) * (topic[it].size() - 1);
    }
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long tc = 1;
  cin >> tc;
  long long i = 1;
  while (tc--) {
    solve();
    i++;
  }
  return 0;
}
