因为所有人都向右走，速度相同，因此如果两个人都还没有到达终点，那么这两个人不会相遇。只有一种情况会相遇，$i, j$ 两个人满足：$a_i > a_j, b_i < b_j$。可以把所有区间按照左端点从小到大排序，对于每一个 $b_i$，寻找之前所有 $> b_i$ 的区间的个数。把端点离散化之后，把点的值作为下标放到数组中，也就是求数组的后缀和，使用 $n + 1 - b_i$ 转换成求前缀和，可以使用树状数组，点更新、区间查询。

// Date: Thu Dec 28 23:25:57 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

template <typename t> istream &operator>>(istream &in, vector<t> &vec) {
  for (t &x : vec)
    in >> x;
  return in;
}

template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) {
  int n = SZ(vec);
  For(i, 0, n) {
    out << vec[i];
    if (i < n - 1)
      out << ' ';
  }
  return out;
}

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

const int N = 200010;
int t, n, fen[N * 4];

inline int lowbit(int x) { return x & -x; }

void update(int pos, int n) {
  while (pos <= n) {
    fen[pos]++;
    pos += lowbit(pos);
  }
}

int sum(int pos) {
  int result = 0;

  while (pos > 0) {
    result += fen[pos];
    pos -= lowbit(pos);
  }

  return result;
}

int main(void) {
#ifdef _DEBUG
  freopen("f.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  cin >> t;

  while (t--) {
    cin >> n;

    memset(fen, 0, sizeof fen);
    unordered_map<int, int> mm;
    int idx = 0;

    vector<PII> a;
    VI bv;
    For1(i, 1, n) {
      int x, y;
      cin >> x >> y;
      a.pb({x, y});
      bv.pb(y);
    }

    sort(all(bv));
    For(i, 0, n) { mm[bv[i]] = ++idx; }

    sort(all(a));

    set<int> s;

    ll res{};

    for (auto &[x, y] : a) {
      int v = mm[y];
      v = idx + 1 - v;

      int cnt = sum(v);

      res += cnt;
      update(v, idx);
    }
    cout << res << '\n';
  }

  return 0;
}