description: You are given two arrays of integers a_1, a_2, …, a_n and b_1, b_2, …, b_m.
  
  You need to insert all elements of b into a in an arbitrary way. As a result you will get an array c_1, c_2, …, c_{n+m} of size n + m.
  
  Note that you are not allowed to change the order of elements in a, while you can insert elements of b at arbitrary positions. They can be inserted at the beginning, between any elements of a, or at the end. Moreover, elements of b can appear in the resulting array in any order.
  
  What is the minimum possible number of inversions in the resulting array c? Recall that an inversion is a pair of indices (i, j) such that i < j and c_i > c_j.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
  
  The first line of each test case contains two integers n and m (1 ≤ n, m ≤ 10^6).
  
  The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).
  
  The third line of each test case contains m integers b_1, b_2, …, b_m (1 ≤ b_i ≤ 10^9).
  
  It is guaranteed that the sum of n for all tests cases in one input doesn't exceed 10^6. The sum of m for all tests cases doesn't exceed 10^6 as well.
  
  Output
  
  For each test case, print one integer — the minimum possible number of inversions in the resulting array c.
  
  Example
  
  Input
  
  
  3
  3 4
  1 2 3
  4 3 2 1
  3 3
  3 2 1
  1 2 3
  5 4
  1 3 5 3 1
  4 3 6 1
  
  
  Output
  
  
  0
  4
  6
  
  Note
  
  Below is given the solution to get the optimal answer for each of the example test cases (elements of a are underscored).
  
    * In the first test case, c = [\underline{1}, 1, \underline{2}, 2, \underline{3}, 3, 4]. 
    * In the second test case, c = [1, 2, \underline{3}, \underline{2}, \underline{1}, 3]. 
    * In the third test case, c = [\underline{1}, 1, 3, \underline{3}, \underline{5}, \underline{3}, \underline{1}, 4, 6]. 
solution: #include <bits/stdc++.h>
  using namespace std;
  const int INF = 1 << 30;
  const long long int MAX = 1e9 + 7;
  void array_show(int* array, int array_n, char middle = ' ') {
    for (int i = 0; i < array_n; i++)
      printf("%d%c", array[i], (i != array_n - 1 ? middle : '\n'));
  }
  void array_show(long long int* array, int array_n, char middle = ' ') {
    for (int i = 0; i < array_n; i++)
      printf("%lld%c", array[i], (i != array_n - 1 ? middle : '\n'));
  }
  void array_show(vector<int>& vec_s, int vec_n = -1, char middle = ' ') {
    if (vec_n == -1) vec_n = vec_s.size();
    for (int i = 0; i < vec_n; i++)
      printf("%d%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
  }
  void array_show(vector<long long int>& vec_s, int vec_n = -1,
                  char middle = ' ') {
    if (vec_n == -1) vec_n = vec_s.size();
    for (int i = 0; i < vec_n; i++)
      printf("%lld%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
  }
  template <typename T>
  ostream& operator<<(ostream& os, const vector<T>& v1) {
    int n = v1.size();
    for (int i = 0; i < n; i++) {
      if (i) os << " ";
      os << v1[i];
    }
    return os;
  }
  template <typename T1, typename T2>
  ostream& operator<<(ostream& os, const pair<T1, T2>& p) {
    os << p.first << " " << p.second;
    return os;
  }
  template <typename T>
  istream& operator>>(istream& is, vector<T>& v1) {
    int n = v1.size();
    for (int i = 0; i < n; i++) is >> v1[i];
    return is;
  }
  template <typename T1, typename T2>
  istream& operator>>(istream& is, pair<T1, T2>& p) {
    is >> p.first >> p.second;
    return is;
  }
  template <typename T>
  class bit {
   private:
    vector<T> bit_vec;
    int bit_N = 20;
    T bit_comp(T a, T b) { return a + b; }
  
   public:
    bit(int n) {
      for (bit_N = 0; bit_N < 30; bit_N++) {
        if (n < (1 << bit_N)) break;
      }
      bit_vec.assign((1 << bit_N), 0);
    }
    void set(int pos, T x) {
      bit_vec[pos] = bit_comp(bit_vec[pos], x);
      for (int i = 0; i < bit_N; i++) {
        if (pos & (1 << i)) continue;
        pos |= (1 << i);
        bit_vec[pos] = bit_comp(bit_vec[pos], x);
      }
    }
    T search(int pos) {
      T s = bit_vec[pos];
      for (int i = 0; pos >= (1 << i); i++) {
        if (pos & (1 << i)) continue;
        pos -= (1 << i);
        s = bit_comp(s, bit_vec[pos]);
      }
      return s;
    }
  };
  template <typename T>
  class Compress {
   private:
    vector<T> vec;
    vector<int> id_perm;
  
   public:
    void init(const vector<T>& v1) {
      int n = v1.size();
      int i, j;
      id_perm.assign(n, -1);
      vector<pair<T, int>> va;
      for (i = 0; i < n; i++) {
        va.push_back({v1[i], i});
      }
      sort(va.begin(), va.end());
      vec.clear();
      for (i = 0, j = -1; i < n; i++) {
        if (vec.empty() || vec.back() != va[i].first) {
          vec.push_back(va[i].first);
          j++;
        }
        id_perm[va[i].second] = j;
      }
    }
    Compress(const vector<T> v1) { init(v1); }
    vector<int> get_id_perm() const { return id_perm; }
    int id(const T a) {
      auto itr = lower_bound(vec.begin(), vec.end(), a);
      assert(itr != vec.end());
      assert(*itr == a);
      return itr - vec.begin();
    }
  };
  namespace sol {
  void init() {}
  void solve() {
    int n, m;
    int i, j, k;
    int a, b, c;
    cin >> n >> m;
    vector<int> v1(n), v2(m);
    vector<tuple<int, int, int>> vt;
    cin >> v1 >> v2;
    Compress<int> ca(v1);
    for (i = 0; i < n; i++) {
      vt.push_back({v1[i], 0, i});
      vt.push_back({v1[i], 2, i});
    }
    for (i = 0; i < m; i++) {
      vt.push_back({v2[i], 1, i});
    }
    set<int> s1;
    int t = 0;
    long long int s = 0;
    bit<int> bt(n);
    for (i = 0; i < n; i++) {
      a = ca.id(v1[i]);
      s += i - bt.search(a);
      bt.set(a, 1);
    }
    for (i = 0; i <= n; i++) {
      s1.insert(i);
    }
    sort(vt.begin(), vt.end());
    for (i = 0; i < vt.size(); i++) {
      tie(ignore, a, b) = vt[i];
      if (a == 1) {
        s += t;
        continue;
      }
      if (s1.empty()) {
        if (a == 0) t--;
        continue;
      }
      auto itr = s1.upper_bound(b);
      if (itr == s1.begin()) {
        if (a == 0) t--;
        continue;
      }
      itr--;
      if (a == 2) t++;
      s1.erase(itr);
      continue;
    }
    cout << s << endl;
  }
  }  // namespace sol
  int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, i;
    sol::init();
    cin >> n;
    for (i = 0; i < n; i++) {
      sol::solve();
    }
  }
  