description: There is a city that can be represented as a square grid with corner points in (0, 0) and (10^6, 10^6).
  
  The city has n vertical and m horizontal streets that goes across the whole city, i. e. the i-th vertical streets goes from (x_i, 0) to (x_i, 10^6) and the j-th horizontal street goes from (0, y_j) to (10^6, y_j). 
  
  All streets are bidirectional. Borders of the city are streets as well.
  
  There are k persons staying on the streets: the p-th person at point (x_p, y_p) (so either x_p equal to some x_i or y_p equal to some y_j, or both).
  
  Let's say that a pair of persons form an inconvenient pair if the shortest path from one person to another going only by streets is strictly greater than the Manhattan distance between them.
  
  Calculate the number of inconvenient pairs of persons (pairs (x, y) and (y, x) are the same pair).
  
  Let's recall that Manhattan distance between points (x_1, y_1) and (x_2, y_2) is |x_1 - x_2| + |y_1 - y_2|.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
  
  The first line of each test case contains three integers n, m and k (2 ≤ n, m ≤ 2 ⋅ 10^5; 2 ≤ k ≤ 3 ⋅ 10^5) — the number of vertical and horizontal streets and the number of persons.
  
  The second line of each test case contains n integers x_1, x_2, ..., x_n (0 = x_1 < x_2 < ... < x_{n - 1} < x_n = 10^6) — the x-coordinates of vertical streets.
  
  The third line contains m integers y_1, y_2, ..., y_m (0 = y_1 < y_2 < ... < y_{m - 1} < y_m = 10^6) — the y-coordinates of horizontal streets.
  
  Next k lines contains description of people. The p-th line contains two integers x_p and y_p (0 ≤ x_p, y_p ≤ 10^6; x_p ∈ \\{x_1, ..., x_n\} or y_p ∈ \\{y_1, ..., y_m\}) — the coordinates of the p-th person. All points are distinct.
  
  It guaranteed that sum of n doesn't exceed 2 ⋅ 10^5, sum of m doesn't exceed 2 ⋅ 10^5 and sum of k doesn't exceed 3 ⋅ 10^5.
  
  Output
  
  For each test case, print the number of inconvenient pairs.
  
  Example
  
  Input
  
  
  2
  2 2 4
  0 1000000
  0 1000000
  1 0
  1000000 1
  999999 1000000
  0 999999
  5 4 9
  0 1 2 6 1000000
  0 4 8 1000000
  4 4
  2 5
  2 2
  6 3
  1000000 1
  3 8
  5 8
  8 8
  6 8
  
  
  Output
  
  
  2
  5
  
  Note
  
  The second test case is pictured below: 
  
  <image>
  
  For example, points 3 and 4 form an inconvenient pair, since the shortest path between them (shown red and equal to 7) is greater than its Manhattan distance (equal to 5).
  
  Points 3 and 5 also form an inconvenient pair: the shortest path equal to 1000001 (shown green) is greater than the Manhattan distance equal to 999999.
  
  But points 5 and 9 don't form an inconvenient pair, since the shortest path (shown purple) is equal to its Manhattan distance.
solution: #include <bits/stdc++.h>
  using namespace std;
  const int N = 3e5 + 10;
  struct Point {
    int x, y, axis;
  };
  int X[N], Y[N];
  vector<Point> npx, npy;
  bool cmp1(Point a, Point b) { return a.x < b.x; }
  bool cmp2(Point a, Point b) { return a.y < b.y; }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
      int n, m, k;
      cin >> n >> m >> k;
      npx.clear();
      npy.clear();
      for (int i = 1; i <= n; i++) {
        cin >> X[i];
      }
      for (int i = 1; i <= m; i++) {
        cin >> Y[i];
      }
      for (int i = 1; i <= k; i++) {
        int x, y;
        cin >> x >> y;
        auto it1 = lower_bound(X + 1, X + n + 1, x);
        auto it2 = lower_bound(Y + 1, Y + m + 1, y);
        if (*it1 == x && *it2 == y) {
          continue;
        }
        if (*it1 == x) {
          npx.push_back({x, y, *it1});
        } else {
          npy.push_back({x, y, *it2});
        }
      }
      sort(npx.begin(), npx.end(), cmp2);
      sort(npy.begin(), npy.end(), cmp1);
      int idx = 0;
      long long ans = 0;
      for (int i = 2; i <= n; i++) {
        long long ct = 0;
        map<int, long long> mp;
        while (idx < npy.size() && X[i - 1] < npy[idx].x && npy[idx].x < X[i]) {
          mp[npy[idx].axis]++;
          idx++;
          ct++;
        }
        for (auto [it, v] : mp) {
          ans -= v * (v - 1) / 2;
        }
        ans += ct * (ct - 1) / 2;
      }
      idx = 0;
      for (int i = 2; i <= m; i++) {
        long long ct = 0;
        map<int, long long> mp;
        while (idx < npx.size() && Y[i - 1] < npx[idx].y && npx[idx].y < Y[i]) {
          mp[npx[idx].axis]++;
          idx++;
          ct++;
        }
        for (auto [it, v] : mp) {
          ans -= v * (v - 1) / 2;
        }
        ans += ct * (ct - 1) / 2;
      }
      cout << ans << endl;
    }
    return 0;
  }
  