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;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int d4x[4] = {1, 0, -1, 0};
  int d4y[4] = {0, 1, 0, -1};
  int d8x[8] = {0, 1, 1, 1, 0, -1, -1, -1};
  int d8y[8] = {1, 1, 0, -1, -1, -1, 0, 1};
  const long long oo = 1e18;
  const long long maxN = 1e6;
  void maximize(int &a, int b) { a = max(a, b); }
  void minimize(int &a, int b) { a = min(a, b); }
  struct FenwickTree {
    int n;
    vector<int> Node;
    FenwickTree(int _n = 0) {
      n = _n;
      Node.resize(n + 5);
    }
    void update(int x, int val) {
      for (x; x <= n; x += x & (-x)) Node[x] += val;
    }
    int get(int x) {
      int ans = 0;
      for (x; x > 0; x -= x & (-x)) ans += Node[x];
      return ans;
    }
  };
  struct SegemntTree {
    int n;
    vector<int> Node;
    vector<int> Lazy;
    SegemntTree(int _n = 0) {
      n = _n;
      Node.resize(4 * n + 7);
      Lazy.resize(4 * n + 7);
      for (int i = 1; i <= 4 * n; i++) Node[i] = Lazy[i] = 0;
    }
  
   private:
    void Down(int id) {
      if (!Lazy[id]) return;
      for (int i = 2 * id; i <= 2 * id + 1; i++) {
        Node[i] += Lazy[id];
        Lazy[i] += Lazy[id];
      }
      Lazy[id] = 0;
    }
    void Update(int L, int R, int lo, int hi, int id, int val) {
      if (R < lo || L > hi) return;
      if (lo <= L && R <= hi) {
        Node[id] += val;
        Lazy[id] += val;
        return;
      }
      int mid = (L + R) >> 1;
      Down(id);
      Update(L, mid, lo, hi, id << 1, val);
      Update(mid + 1, R, lo, hi, id << 1 | 1, val);
      Node[id] = max(Node[id << 1], Node[id << 1 | 1]);
    }
  
   public:
    void update(int L, int R, int val) { Update(1, n, L, R, 1, val); }
  };
  const int N = 2e6 + 10;
  int a[N], b[N];
  pair<int, int> c[N];
  vector<int> pos[N];
  void solve() {
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n + m; i++) pos[i].clear();
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      c[i] = make_pair(a[i], i);
    }
    for (int i = 1; i <= m; i++) {
      cin >> b[i];
      c[i + n] = make_pair(b[i], i + n);
    }
    sort(c + 1, c + 1 + n + m);
    int cnt = 0;
    for (int i = 1; i <= n + m; i++) {
      if (c[i].first != c[i - 1].first) ++cnt;
      if (c[i].second > n)
        b[c[i].second - n] = cnt;
      else
        a[c[i].second] = cnt;
    }
    sort(b + 1, b + 1 + m);
    long long ans = 0;
    FenwickTree bit = FenwickTree(cnt);
    SegemntTree it = SegemntTree(n);
    for (int i = 1; i <= n; i++) {
      int x = a[i];
      ans += i - 1 - bit.get(x);
      bit.update(x, 1);
      pos[x].push_back(i);
    }
    for (int i = 1; i <= n; i++) it.update(i, n, -1);
    int lower = 0;
    for (int i = 1; i <= m; i++) {
      while (lower + 1 < b[i]) {
        ++lower;
        for (auto v : pos[lower]) {
          it.update(v, n, (lower == b[i - 1] ? 1 : 2));
        }
      }
      if (b[i] != b[i - 1])
        for (auto v : pos[b[i]]) {
          it.update(v, n, 1);
        }
      ans += bit.get(b[i] - 1) - max(0, it.Node[1]);
    }
    cout << ans << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int tc = 1, ddd = 0;
    cin >> tc;
    while (tc--) {
      solve();
    }
  }
  