#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long binpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long primeFactors(long long n) {
  long long ans = 0;
  long long m = n;
  for (long long i = 2; i * i <= m; i++) {
    while (n % i == 0) {
      n /= i;
      ans++;
    }
  }
  if (n != 1) return ans + 1;
  return ans;
}
long long power(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  long long ans = power(a, b / 2);
  ans *= ans;
  ans %= 1000000007;
  if (b % 2) {
    ans *= a;
  }
  return ans % 1000000007;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long total = n * (n - 1) * (n - 2) / 6;
    pair<long long, long long> p[n];
    map<long long, long long> t, d;
    for (long long i = 0; i < n; i++) {
      long long x, y;
      cin >> x >> y;
      p[i] = {x, y};
      t[x]++;
      d[y]++;
    }
    long long di = 0;
    for (long long i = 0; i < n; i++) {
      di += (t[p[i].first] - 1) * (d[p[i].second] - 1);
    }
    cout << total - di << '\n';
  }
  return 0;
}

### Prompt

Generate a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long binpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long primeFactors(long long n) {
  long long ans = 0;
  long long m = n;
  for (long long i = 2; i * i <= m; i++) {
    while (n % i == 0) {
      n /= i;
      ans++;
    }
  }
  if (n != 1) return ans + 1;
  return ans;
}
long long power(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  long long ans = power(a, b / 2);
  ans *= ans;
  ans %= 1000000007;
  if (b % 2) {
    ans *= a;
  }
  return ans % 1000000007;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long total = n * (n - 1) * (n - 2) / 6;
    pair<long long, long long> p[n];
    map<long long, long long> t, d;
    for (long long i = 0; i < n; i++) {
      long long x, y;
      cin >> x >> y;
      p[i] = {x, y};
      t[x]++;
      d[y]++;
    }
    long long di = 0;
    for (long long i = 0; i < n; i++) {
      di += (t[p[i].first] - 1) * (d[p[i].second] - 1);
    }
    cout << total - di << '\n';
  }
  return 0;
}

```