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 MAXN = 1e6 + 6;
  int p[MAXN], a[MAXN], b[MAXN], t, n, m, vec[MAXN];
  void solve(int li, int ri, int lp, int rp) {
    if (li >= ri) {
      return;
    }
    int mid = (li + ri) >> 1;
    long long res = 0;
    for (int i = lp; i < rp; i++) {
      if (a[i] < b[mid]) res++;
    }
    int pos = lp;
    long long mn = res;
    for (int i = lp + 1; i <= rp; i++) {
      if (a[i - 1] < b[mid]) res--;
      if (a[i - 1] > b[mid]) res++;
      if (res < mn) {
        mn = res;
        pos = i;
      }
    }
    p[mid] = pos;
    solve(li, mid, lp, pos);
    solve(mid + 1, ri, pos, rp);
  }
  pair<vector<int>, long long> invs(vector<int> vec) {
    if (int(vec.size()) == 1) return {vec, 0};
    int mid = int(vec.size()) >> 1;
    vector<int> v1, v2;
    for (int i = 0; i < mid; i++) {
      v1.push_back(vec[i]);
    }
    for (int i = mid; i < int(vec.size()); i++) {
      v2.push_back(vec[i]);
    }
    auto p1 = invs(v1);
    auto p2 = invs(v2);
    v1 = p1.first, v2 = p2.first;
    vec.clear();
    int j = 0;
    long long res = p1.second + p2.second;
    long long cnt = 0;
    for (int i = 0; i < int(v1.size()); i++) {
      while (j < int(v2.size()) && v2[j] < v1[i]) {
        vec.push_back(v2[j]);
        cnt++;
        j++;
      }
      res += cnt;
      vec.push_back(v1[i]);
    }
    while (j < int(v2.size())) {
      vec.push_back(v2[j]);
      j++;
    }
    return {vec, res};
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> t;
    while (t--) {
      cin >> n >> m;
      for (int i = 1; i <= n; i++) cin >> a[i];
      for (int i = 1; i <= m; i++) cin >> b[i];
      sort(b + 1, b + 1 + m);
      solve(1, m + 1, 1, n + 1);
      vector<int> vec;
      int j = 1;
      for (int i = 1; i <= n; i++) {
        while (j <= m && p[j] == i) {
          vec.push_back(b[j]);
          j++;
        }
        vec.push_back(a[i]);
      }
      while (j <= m) {
        vec.push_back(b[j]);
        j++;
      }
      auto p = invs(vec);
      cout << p.second << '\n';
    }
    return 0;
  }
  