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;
  int n, m;
  vector<int> a, b;
  vector<int> p;
  vector<int> pref, suff;
  void dnc(int ln, int rn, int lm, int rm) {
    if (ln > rn or lm > rm) return;
    int pm = (lm + rm) / 2;
    for (int i = ln; i <= rn; i++) {
      pref[i] = 0;
      suff[i] = 0;
    }
    for (int i = ln; i <= rn; i++) {
      if (a[i] > b[pm]) pref[i]++;
      if (i - 1 >= ln) pref[i] = pref[i] + pref[i - 1];
    }
    for (int i = rn; i >= ln; i--) {
      if (a[i] < b[pm]) suff[i]++;
      if (i + 1 <= rn) suff[i] = suff[i] + suff[i + 1];
    }
    int pos = -1, ans = INT_MAX;
    for (int i = ln; i <= rn; i++) {
      int r = 0;
      if (i - 1 >= ln) r = r + pref[i - 1];
      r = r + suff[i];
      if (r <= ans) {
        ans = r;
        pos = i;
      }
    }
    p[pm] = pos;
    dnc(ln, pos, lm, pm - 1);
    dnc(pos, rn, pm + 1, rm);
  }
  template <class T>
  struct fenwick_tree {
   private:
    int _n;
    vector<T> arr;
    T query(long long x) {
      long long ans = 0;
      for (long long i = x; i >= 0; i = (i & (i + 1)) - 1) {
        ans += arr[i];
      }
      return ans;
    }
  
   public:
    fenwick_tree() : _n(0) {}
    fenwick_tree(int n) : _n(n), arr(n) {}
    void update(long long x, T val) {
      long long ans = 0;
      for (long long i = x; i < arr.size(); i = (i | (i + 1))) {
        arr[i] += (long long)(val);
      }
    }
    T query(int l, int r) { return query(r) - query(l - 1); }
  };
  void solve() {
    a.clear();
    b.clear();
    p.clear();
    cin >> n >> m;
    n++;
    a.resize(n, INT_MAX);
    b.resize(m);
    for (int i = 0; i < n - 1; i++) cin >> a[i];
    for (int i = 0; i < m; i++) cin >> b[i];
    pref.resize(n);
    suff.resize(n);
    p.resize(m);
    sort(b.begin(), b.end());
    dnc(0, n - 1, 0, m - 1);
    vector<int> arr;
    for (int i = 0, j = 0; i < n; i++) {
      while (j < m and p[j] <= i) {
        arr.push_back(b[j]);
        j++;
      }
      arr.push_back(a[i]);
    }
    arr.pop_back();
    n--;
    vector<pair<int, int>> sorted(n + m);
    for (int i = 0; i < n + m; i++) {
      sorted[i] = {arr[i], i};
    }
    sort(sorted.begin(), sorted.end());
    vector<int> res(n + m);
    for (int i = 0; i < n + m; i++) {
      res[sorted[i].second] = i;
    }
    fenwick_tree<long long> f(n + m + 1);
    long long ans = 0;
    for (int i = 0; i < n + m; i++) {
      ans = ans + f.query(res[i], n + m);
      f.update(res[i], 1);
    }
    cout << ans << '\n';
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--) solve();
  }
  