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.
#pragma GCC optimize("Ofast")
//#pragma target("avx2")
///*/////////////////////////////////////////////////////////*///
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
///*/////////////////////////////////////////////////////////*///
#define pb push_back
#define ll long long
#define ld long double
#define pii pair<int,int>
#define int long long
#define F first
#define S second
#define endl "\n"
#define usk ios_base::sync_with_stdio(0);
#define ore cin.tie(0);
#define nie cout.tie(0);
///*/////////////////////////////////////////////////////////*///
using namespace std;
using namespace __gnu_pbds;
///*/////////////////////////////////////////////////////////*///
template <typename T>
using ordered_set = tree <T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
///*/////////////////////////////////////////////////////////*///
const int N=2e5+7;
///*/////////////////////////////////////////////////////////*///
const int MOD=1e9+7;
int32_t main() {
 usk ore nie
    int e;
    cin>>e;
    while(e--)
    {
     int n;
     cin>>n;
     int x,y;
     //cin>>x>>y;
     set <int> st1;
     map <int,int> mp1;
     set <int> st2;
     map <int,int> mp2;
     for(int i=0;i<n;++i)
     {
    cin>>x>>y;
    mp1[x]++;
    st1.insert(x);
    mp2[y]++;
    st2.insert(y);
     }
     int ans=n*(n-1)*(n-2)/6;
     for(auto to:st1)
        ans-=mp1[to]*(mp1[to]-1)/2;
        for(auto to:st2)
            ans-=mp1[to]*(mp1[to]-1)/2;
     cout<<ans<<endl;
    }
    return 0;
 }


