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 <iostream>
    #include <vector>
    #include <cmath>
    #include <string>
    #include <algorithm>
    #include <iomanip>
    #include <map>
    #include <set>
    #include <bitset>
    #include <fstream>
    #include <unordered_set>
    #include <unordered_map>
    #include <ext/pb_ds/assoc_container.hpp>
    #include <stack>
    #include <queue>
    #include <complex>
    #include <random>


    using namespace std;
    using namespace __gnu_pbds;



    /*#pragma GCC optimize("Ofast")
    #pragma GCC optimize("no-stack-protector")
    #pragma GCC optimize("unroll-loops")
    #pragma GCC target("sse,sse2,sse3,ssse3,popcnt,abm,mmx,tune=native")
    #pragma GCC optimize("fast-math")*/

    /*#pragma GCC optimize("Ofast")
    #pragma GCC target("sse,sse2,sse3,ssse3,sse4,avx,tune=native")
    #pragma GCC target("avx2")
    #pragma GCC optimization("O3")
    #pragma GCC optimization("unroll-loops")*/

    #define int long long
    #define ll long long
    //#define int short int
    #define eb emplace_back
    #define pb push_back
    #define ld long double
    //#define ld double
    #define mp make_pair
    #define f first
    #define s second
    #define gleg(x) return cout << x, 0
    #define pii pair <int, int>
    #define deb(a) cerr << #a << " = " << a << '\n';
    //#define sz(x) (int)s.size();
    #define fast() {     ios_base::sync_with_stdio(0);     cin.tie(0);     cout << fixed << setprecision(9);     cerr << fixed << setprecision(12); }

    template < typename firstType, typename secondType = null_type, class compare = less < firstType > >
    struct sett {
        typedef tree <
                firstType,
                secondType,
                compare,
                rb_tree_tag,
                tree_order_statistics_node_update
        > _ ;
    };


    const int INF = 1e18 + 7;
    const ld EPS = 1e-6;
    const int MAXI = 350;
    const int MOD = 998244353;
    const int MOD1 = 1e9 + 7;
    const int MAXST = 2000000;
    const int P = 31;
    const int P1 = 37;
    const int N = 5100000;
    const ld PI = 3.14159265358979323;
    const int bp = 23;
    //const int sz = 1 << bp;


    ostream &operator<<(ostream &stream, const pair <int, int> &p) {
        stream << p.first << ' ' << p.second << ' ';
        return stream;
    }

    int C(int n, int k){
        int ans = 1;
        for (int i = n - k + 1; i <= n; i++){
            ans *= i;
        }

        for (int i = 1; i <= k; i++)
            ans /= i;
        return ans;
    }

    signed main(){
        fast();
        srand(time(0));
    //    #ifdef LOCAL
        //freopen("stacks.in", "r", stdin);
        //freopen("stacks.out", "w", stdout);
    //    #endif
        int t;
        cin >> t;
        while (t--){
            int n;
            cin >> n;
            vector <vector <int>> a(n + 100);
            vector <int> cnt(n + 100);
            for (int i = 0; i < n; i++){
                int l, r;
                cin >> l >> r;
                a[l].eb(r);
                cnt[r]++;
            }

            int ans = C(n, 3);
            for (auto v: a){
                if (v.size() < 2)
                    continue;
                int k = v.size();

                for (int &j: v){
                    ans -= k * (k - 1) / 2 * (cnt[j] - 1);
                }
            }

            cout << ans << '\n';
        }
        return 0;
    }