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;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<bool> vbool;
typedef vector<vbool> vvbool;
typedef pair<ll, ll> pll;
#define ff first
#define ss second
#define FAST_IO	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define fileInput freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);
#ifdef largeInteger
	#include <boost/multiprecision/cpp_int.hpp>
	using namespace boost::multiprecision;
	// Use 'cpp_int' instead of 'int'
#endif
#ifdef useOrderedSet
	#include <ext/pb_ds/assoc_container.hpp>
	#include <ext/pb_ds/tree_policy.hpp>
	using namespace __gnu_pbds;
	#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
	// os.find_by_order(int order), os.order_of_key(int val) {Order is 0-based index} 
	// NOTE: Using less_equal<int> comparator makes lower_bound function like upper_bound and vice-versa.
	// Same with greater_equal<int>
#endif

int main(){
	FAST_IO
	
	int TT;
	cin >> TT;
	for(int tt = 1; tt <= TT; tt++){
		#ifdef caseInfo
			cout << "Case #" << tt << ":\n";
		#endif
		ll n;
		cin >> n;
		int a, b;
		int cntT[n + 1] = {}, cntD[n + 1] = {};
		for(int i = 0; i < n; i++)	cin >> a >> b, cntT[a]++, cntD[b]++;
		ll sumT = 0, sumD = 0;
		for(int i = 1; i <= n; i++){
			sumT += max(0, cntT[i] - 1);
			sumD += max(0, cntD[i] - 1);
		}
		ll res = (n * (n - 1ll) * (n - 2ll)) / 6ll - sumT * sumD;
		cout << res << "\n";
	} 
	
	return 0;
}
