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 = 200010;
int T, n, m;
long long a[MAXN];
map<int, int> cnt[MAXN];
int cnt1[MAXN], cnt2[MAXN];
long long C3(long long a) { return a * (a - 1) * (a - 2) / 6; }
long long C2(long long a) { return a * (a - 1) / 2; }
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) cnt[i].clear(), cnt1[i] = cnt2[i] = 0;
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      int a, b;
      scanf("%d%d", &a, &b);
      ans += cnt2[b] * (i - 1 - cnt2[b]) + cnt1[a] * (i - 1 - cnt1[a]);
      ans += C2(cnt2[b]) + C2(cnt1[a]);
      ans -= cnt[a][b] * (i - 1 - cnt[a][b]) +
             (cnt1[a] - cnt[a][b]) * (cnt2[b] - cnt[a][b]);
      ans -= C2(cnt[a][b]);
      cnt[a][b]++;
      cnt1[a]++;
      cnt2[b]++;
    }
    printf("%lld\n", C3(n) - ans);
  }
  return 0;
}
