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.
// Created by ...
#include <bits/stdc++.h>

#define vll vector< ll >
#define M 100000
#define MD 1000000007
#define pb push_back
#define eb emplace_back
#define sz(x) (int)(x).size()
#define pii pair<ll,ll>
#define vec(a) vector<a >
#define all(a) (a).begin(),(a).end()
#define se second
#define fi first
#define inf 0xffffffff
#define int long long
#define endl '\n'
#define rep(i,a,b) for(ll i=a;i<=b;++i)
#define per(i,b,a) for(ll i=b;i>=a;--i)
#define IOS ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define EACH(x, a) auto _Itr=begin(a);for (auto x=*_Itr;_Itr!=end(a);_Itr=next(_Itr),x=*_Itr)

using namespace std;
using ll = long long;
using ld = long double;
ll md = MD;

template<class... Args> auto Vec(size_t n, Args&&... args) {
	if constexpr(sizeof...(args) == 1)return vector(n, args...);
	else return vector(n, Vec(args...));
}
template<class T, class... Args> void DBS(T t, Args... args) {
	cerr << t;
	if constexpr(sizeof...(args) == 0) cerr << "]\n";
	else cerr << ", ", DBS(args...);
}
#ifdef _DEBUG
#define db(...) cerr << "[" << #__VA_ARGS__ << "]: [", DBS(__VA_ARGS__)
#else
#define db(...) 0
#endif

ll exp(ll a, ll b) {ll r = 1ll; while (b > 0) {if (b & 1) {r = r * (a % md); r = (r + md) % md;} b >>= 1; a = (a % md) * (a % md); a = (a + md) % md;} return (r + md) % md;}
ll gcd(ll a, ll b) {if (b == 0)return a; return gcd(b, a % b);}
template<class T> T Min(T a, T b) {if (a < b)return a; return b;}
template<class T> T Max(T a, T b) {if (a > b)return a; return b;}
int nc3(int x) {
	return x * (x - 1) * (x - 2);
}
void solve() {
	int n; cin >> n;
	map<int, int> mx, my;
	vll x(n), y(n);
	rep(i, 0, n - 1) {
		cin >> x[i] >> y[i];
		++mx[x[i]]; ++my[y[i]];
	}
	int ans = nc3(n);
	rep(i, 0, n - 1) {
		int cnt = mx[x[i]] + my[y[i]] - 1;
		ans -= nc3(cnt);
	}
	cout << ans / 6 << endl;
}

int32_t main() {
	IOS;

	int t = 1;
	cin >> t;
	while (t--) {
		solve();
	}
	return 0;
}