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 int maxn = 2e5, N1 = maxn + 7;
const long long inf = 0x3f3f3f3f3f3fll;
template <typename T>
inline void read(T &x) {
  x = 0;
  T k = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') k = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= k;
}
int n;
int a[N1], b[N1];
long long sa[N1], sb[N1];
long long calc3(long long x) { return x * (x - 1) * (x - 2) / 6; }
void solve() {
  read(n);
  long long ans = calc3(1ll * n), tmp;
  for (int i = 1; i <= n; i++) read(a[i]), read(b[i]);
  tmp = 0;
  for (int i = 1; i <= n; i++) {
    tmp += sa[a[i]] * sb[b[i]];
    sa[a[i]]++;
    sb[b[i]]++;
  }
  for (int i = 1; i <= n; i++) sa[i] = 0;
  for (int i = 1; i <= n; i++) {
    sb[b[i]]--;
    tmp += sa[a[i]] * sb[b[i]];
    sa[a[i]]++;
  }
  for (int i = 1; i <= n; i++) {
    sa[a[i]]--;
    tmp += sa[a[i]] * sb[b[i]];
    sb[b[i]]++;
  }
  for (int i = 1; i <= n; i++) sb[i] = 0;
  for (int i = n; i >= 1; i--) {
    tmp += sa[a[i]] * sb[b[i]];
    sa[a[i]]++;
    sb[b[i]]++;
  }
  for (int i = 1; i <= n; i++)
    tmp += calc3(sa[i]) + calc3(sb[i]), sa[i] = sb[i] = 0;
  ans -= tmp;
  printf("%lld\n", ans);
}
int main() {
  int Te;
  read(Te);
  while (Te--) solve();
  return 0;
}
