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 MOD = 998244353;
  int n, m, a[1000005], b[1000005];
  pair<int, int> perm[1000005];
  struct fwt {
    int dat[1000005];
    void clear() {
      for (int i = 1; i <= (int)(n); i++) dat[i] = 0;
    }
    void add(int id, int val) {
      for (id++; id <= n; id += id & -id) dat[id] += val;
    }
    int query(int id) {
      int ret = 0;
      for (id++; id > 0; id -= id & -id) ret += dat[id];
      return ret;
    }
  } tre;
  long long ans;
  priority_queue<int> que;
  void addnew(int x) {
    if (que.empty()) {
      que.push(x);
    } else if (x >= que.top())
      que.push(x);
    else {
      ans += que.top() - x;
      que.push(x);
      que.push(x);
      que.pop();
    }
  }
  void solve() {
    scanf("%d%d", &n, &m);
    for (int i = 0; i < (int)(n); i++) scanf("%d", &a[i]);
    for (int i = 0; i < (int)(m); i++) scanf("%d", &b[i]);
    sort(b, b + m);
    for (int i = 0; i < (int)(n); i++) perm[i] = make_pair(a[i], i);
    sort(perm, perm + n);
    ans = 0;
    tre.clear();
    for (int i = 0; i < (int)(n); i++) {
      ans += 2 * (i - tre.query(perm[i].second));
      tre.add(perm[i].second, 1);
    }
    while (!que.empty()) que.pop();
    for (int i = 0; i < (int)(n); i++) {
      int r = upper_bound(b, b + m, a[i]) - b,
          l = lower_bound(b, b + m, a[i]) - b;
      addnew(r);
      addnew(l);
      ans -= r - l;
    }
    assert(ans % 2 == 0);
    printf("%lld\n", ans / 2);
  }
  int main() {
    int T;
    scanf("%d", &T);
    while (T--) solve();
    return 0;
  }
  