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(3)
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC target("avx","sse2")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
*/
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/rope>
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define pb push_back
#define eb emplace_back
#define vec vector

using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;

typedef long long ll;
typedef long double ld;

#define int ll
#define double ld

typedef pair<int, int> pii;
template <class T> using indexed_set = tree <T, null_type, less <T>, rb_tree_tag, tree_order_statistics_node_update>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int SIZE = 2e5, MOD = 1e9 + 7, INF = 2e9;
const ll INFLL = 1e18;

int binpow(int b, int e) {
    int res = 1;
    while(e) {
        if(e & 1) res = (1LL * res * b) % MOD;
        b = (1LL * b * b) % MOD;
        e >>= 1;
    }
    return res;
}

template<typename T> void umax(T& a, T b){ a = max(a, b); }
template<typename T> void umin(T& a, T b){ a = min(a, b); }
template<typename T> T ceildiv(T x, T y) { return (x + y - 1) / y; }


int n, a[SIZE], b[SIZE];

void solve() {
    cin >> n;
    for(int i = 0; i < n; i++) {
        cin >> a[i] >> b[i];
    }

    map<int, int> cnta, cntb;
    int ans = (n * (n - 1) * (n - 2)) / 6;

    for(int i = 0; i < n; i++) {
        if(i > 1) {
            int same_a = cnta[a[i]], same_b = cntb[b[i]];
            ans -= same_a * (i - same_a);
            ans -= same_b * (i - same_b - same_a);
        }

        cnta[a[i]]++;
        cntb[b[i]]++;
    }

    cout << ans;
}



signed main () {
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

    int t = 1;
    cin >> t;

    while(t--) {
        solve();
        cout << endl;
    }

    return 0;
}
