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;
  vector<int> L[2000010];
  int a[2000010], b[2000010], v[2000010], aib[2000010], w[2000010],
      sp_left[2000010], sp_right[2000010], poz[2000010];
  int t, n, m, i, j, el;
  void solve(int st, int dr, int x, int y) {
    if (st > dr) return;
    int mid = (st + dr) >> 1;
    sp_left[x - 1] = sp_right[y + 1] = 0;
    for (int i = x; i <= y; i++) sp_left[i] = sp_left[i - 1] + (a[i] > b[mid]);
    for (int i = y; i >= x; i--) sp_right[i] = sp_right[i + 1] + (a[i] < b[mid]);
    int mini = 2000000000;
    for (int i = x; i <= y; i++) {
      if (sp_left[i - 1] + sp_right[i] < mini) {
        mini = sp_left[i - 1] + sp_right[i];
        poz[mid] = i;
      }
    }
    solve(st, mid - 1, x, poz[mid]);
    solve(mid + 1, dr, poz[mid], y);
  }
  void update(int p, int val) {
    for (; p <= el; p += (p & -p)) aib[p] += val;
  }
  int query(int p) {
    int sol = 0;
    for (; p; p -= (p & -p)) sol += aib[p];
    return sol;
  }
  int cautare_binara(int v[], int n, int val) {
    int st = 1, dr = n;
    while (st <= dr) {
      int mid = (st + dr) >> 1;
      if (v[mid] == val) return mid;
      if (v[mid] < val)
        st = mid + 1;
      else
        dr = mid - 1;
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin >> t;
    for (; t--;) {
      cin >> n >> m;
      for (i = 1; i <= n; i++) cin >> a[i];
      for (i = 1; i <= m; i++) cin >> b[i];
      sort(b + 1, b + m + 1);
      solve(1, m, 1, n + 1);
      for (i = 1; i <= n + 1; i++) L[i].clear();
      for (i = 1; i <= m; i++) L[poz[i]].push_back(i);
      int k = 0;
      for (i = 1; i <= n + 1; i++) {
        for (auto it : L[i]) v[++k] = b[it];
        if (i <= n) v[++k] = a[i];
      }
      el = 0;
      for (i = 1; i <= k; i++) w[++el] = v[i];
      sort(w + 1, w + el + 1);
      int el2 = 1;
      for (i = 2; i <= el; i++)
        if (w[i] != w[i - 1]) w[++el2] = w[i];
      el = el2;
      for (i = 1; i <= k; i++) v[i] = cautare_binara(w, el, v[i]);
      for (i = 1; i <= el; i++) aib[i] = 0;
      long long sol = 0;
      for (i = 1; i <= k; i++) {
        sol += query(el) - query(v[i]);
        update(v[i], 1);
      }
      cout << sol << "\n";
    }
    return 0;
  }
  