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>
  const long double eps = 1e-9;
  const long long mod = 1e9 + 7, ll_max = 1e18;
  const int MX = 2e6 + 10, int_max = 0x3f3f3f3f;
  using namespace std;
  template <class T>
  void read(T &x) {
    cin >> x;
  }
  template <class H, class T>
  void read(pair<H, T> &p) {
    cin >> p.first >> p.second;
  }
  template <class T, size_t S>
  void read(array<T, S> &a) {
    for (T &i : a) read(i);
  }
  template <class T>
  void read(vector<T> &v) {
    for (T &i : v) read(i);
  }
  template <class H, class... T>
  void read(H &h, T &...t) {
    read(h);
    read(t...);
  }
  template <class H, class T>
  ostream &operator<<(ostream &o, pair<H, T> &p) {
    o << p.first << " " << p.second;
    return o;
  }
  template <class T, size_t S>
  ostream &operator<<(ostream &o, array<T, S> &a) {
    string s;
    for (T i : a) o << s << i, s = " ";
    return o;
  }
  template <class T>
  ostream &operator<<(ostream &o, vector<T> &v) {
    string s;
    for (T i : v) o << s << i, s = " ";
    return o;
  }
  template <class T>
  void write(T x) {
    cout << x;
  }
  template <class H, class... T>
  void write(const H &h, const T &...t) {
    write(h);
    write(t...);
  }
  void print() { write('\n'); }
  template <class H, class... T>
  void print(const H &h, const T &...t) {
    write(h);
    if (sizeof...(t)) write(' ');
    print(t...);
  }
  template <typename T>
  void ckmin(T &a, const T &b) {
    a = min(a, b);
  }
  template <typename T>
  void ckmax(T &a, const T &b) {
    a = max(a, b);
  }
  vector<int> adj[MX];
  int a[MX], b[MX], c[MX], BIT[MX], srt[MX];
  int n, m;
  void U(int k) {
    for (; k <= MX - 1; k += (k & (-k))) BIT[k]++;
  }
  int S(int k) { return (!k) ? 0 : (BIT[k] + S(k - (k & (-k)))); }
  void dnq(int lo, int hi, int l, int r) {
    if (r < l) return;
    if (lo == hi) {
      for (int i = l; i <= r; i++) {
        adj[lo].push_back(i);
      }
      return;
    }
    int mid = (l + r) / 2;
    int suf = 0, pre = 0, ans = int_max, ind = lo;
    for (int i = lo; i <= hi; i++) {
      if (a[i] < b[mid]) suf++;
    }
    for (int i = lo; i <= hi; i++) {
      if (a[i] > b[mid]) pre++;
      if (a[i] < b[mid]) suf--;
      if (pre + suf < ans) {
        ckmin(ans, pre + suf);
        ind = i;
      }
    }
    adj[ind].push_back(mid);
    dnq(lo, ind, l, mid - 1);
    dnq(ind, hi, mid + 1, r);
  }
  void solve() {
    read(n, m);
    for (int i = 0; i <= (n + m + 5); i++) {
      adj[i].clear();
      a[i] = b[i] = c[i] = BIT[i] = srt[i] = 0;
    }
    for (int i = 1; i <= n; i++) read(i[a]);
    for (int i = 0; i < m; i++) read(b[i]);
    sort(b, b + m);
    a[0] = 0;
    n++;
    dnq(0, n - 1, 0, m - 1);
    int nm = 0;
    for (int i = 0; i < n; i++) {
      c[nm++] = a[i];
      sort(adj[i].begin(), adj[i].end());
      for (int v : adj[i]) {
        c[nm++] = b[v];
      }
    }
    assert(nm == n + m);
    for (int i = 0; i < nm; i++) srt[i] = c[i];
    sort(srt, srt + nm);
    for (int i = 0; i < nm; i++)
      c[i] = lower_bound(srt, srt + nm, c[i]) - srt + 1;
    long long inv = 0;
    for (int i = nm - 1; ~i; i--) {
      inv += (long long)S(c[i] - 1);
      U(c[i]);
    }
    print(inv);
  }
  int main() {
    cin.tie(0)->sync_with_stdio(0);
    int T = 1;
    cin >> T;
    while (T--) {
      solve();
    }
    return 0;
  }
  