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 double pi = 3.14159265358979323846, eps = 1e-9;
  const long long M = 1e9 + 7, I = 1e18;
  const int mxn = 1e6, mxn1 = 2e6, lg = 20;
  int t, n, m, a[mxn], b[mxn], sml[mxn + 1], smr[mxn + 1], tree[mxn1 * 4],
      pos[mxn];
  vector<int> v;
  unordered_map<int, int> mp;
  void build(int v, int l, int r) {
    if (r - l == 1) {
      tree[v] = 0;
      return;
    }
    build(v * 2 + 1, l, (r + l) / 2);
    build(v * 2 + 2, (r + l) / 2, r);
    tree[v] = 0;
  }
  void upd(int v, int l, int r, int pos) {
    if (l > pos or r <= pos) return;
    if (r - l == 1) {
      tree[v]++;
      return;
    }
    upd(v * 2 + 1, l, (r + l) / 2, pos);
    upd(v * 2 + 2, (r + l) / 2, r, pos);
    tree[v] = tree[v * 2 + 1] + tree[v * 2 + 2];
  }
  int get(int v, int l, int r, int ql, int qr) {
    if (l >= qr or r <= ql) return 0;
    if (l >= ql and r <= qr) return tree[v];
    return get(v * 2 + 1, l, (r + l) / 2, ql, qr) +
           get(v * 2 + 2, (r + l) / 2, r, ql, qr);
  }
  void f(int al, int ar, int bl, int br) {
    if (bl == br) return;
    int cur = b[(bl + br) / 2], ind = -1, mn = M, curl = -1, curr = -1;
    sml[al] = 0;
    for (int i = al + 1; i <= ar; i++) {
      sml[i] = sml[i - 1];
      if (a[i - 1] > cur) sml[i]++;
    }
    smr[ar] = 0;
    for (int i = ar - 1; i >= al; i--) {
      smr[i] = smr[i + 1];
      if (cur > a[i]) smr[i]++;
    }
    for (int i = al; i <= ar; i++) {
      if (smr[i] + sml[i] < mn) {
        mn = smr[i] + sml[i];
        curl = sml[i];
        curr = smr[i];
        ind = i;
      }
    }
    pos[(bl + br) / 2] = ind;
    f(al, ind, bl, (bl + br) / 2);
    f(ind, ar, (bl + br) / 2 + 1, br);
  }
  void solve() {
    v.clear();
    mp.clear();
    cin >> n >> m;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      v.push_back(a[i]);
    }
    for (int i = 0; i < m; i++) {
      cin >> b[i];
      v.push_back(b[i]);
    }
    sort(v.begin(), v.end());
    int ln = (int)(unique(v.begin(), v.end()) - v.begin());
    for (int i = 0; i < ln; i++) {
      mp[v[i]] = i;
    }
    for (int i = 0; i < n; i++) a[i] = mp[a[i]];
    for (int i = 0; i < m; i++) b[i] = mp[b[i]];
    sort(b, b + m);
    f(0, n, 0, m);
    long long ans = 0;
    build(0, 0, ln);
    int ind = 0;
    for (int i = 0; i < n; i++) {
      while (ind < m and pos[ind] == i) {
        ans += (long long)get(0, 0, ln, b[ind] + 1, ln);
        ind++;
      }
      ans += (long long)get(0, 0, ln, a[i] + 1, ln);
      upd(0, 0, ln, a[i]);
    }
    while (ind < m) {
      ans += (long long)get(0, 0, ln, b[ind] + 1, ln);
      ind++;
    }
    build(0, 0, ln);
    ind = m - 1;
    while (pos[ind] == n) ind--;
    for (int i = n - 1; i >= 0; i--) {
      upd(0, 0, ln, a[i]);
      while (ind >= 0 and pos[ind] == i) {
        ans += (long long)get(0, 0, ln, 0, b[ind]);
        ind--;
      }
    }
    cout << ans << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ;
    cin >> t;
    while (t--) solve();
  }
  