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;
  using ll = long long;
  constexpr int N = 1000000 + 10;
  int n, m, a[N], b[N], p[N], c[N * 2], d[N * 2];
  ll ans;
  void divide(int l, int r, int L, int R) {
    if (l > r) return;
    int mid = (l + r) >> 1, cnt = 0, pos, mi = n + 2;
    for (int i = L; i <= R; i++) cnt += a[i] < b[mid];
    for (int i = L; i <= R; i++) {
      if (cnt < mi) {
        mi = cnt;
        pos = i;
      }
      if (a[i] > b[mid]) {
        cnt++;
      } else if (a[i] < b[mid]) {
        cnt--;
      }
    }
    p[mid] = pos;
    divide(l, mid - 1, L, pos);
    divide(mid + 1, r, pos, R);
  }
  void merge_sort(int l, int r) {
    if (l == r) return;
    int mid = (l + r) >> 1;
    merge_sort(l, mid);
    merge_sort(mid + 1, r);
    int i = l, j = mid + 1, k = l;
    while (i <= mid && j <= r) {
      if (c[i] <= c[j]) {
        d[k++] = c[i++];
      } else {
        ans += mid - i + 1;
        d[k++] = c[j++];
      }
    }
    while (i <= mid) d[k++] = c[i++];
    while (j <= r) d[k++] = c[j++];
    for (int i = l; i <= r; i++) c[i] = d[i];
  }
  int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    cout << fixed << setprecision(20);
    int T;
    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);
      divide(1, m, 1, n + 1);
      for (int i = 1, j = 1, k = 1; i <= n + 1; i++) {
        while (j <= m && p[j] == i) c[k++] = b[j++];
        if (i <= n) c[k++] = a[i];
      }
      ans = 0;
      merge_sort(1, n + m);
      cout << ans << '\n';
    }
    return 0;
  }
  