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());
  inline int rand(int x, int y) {
    ++y;
    return (rng() % (y - x)) + x;
  }
  int TC, N, M;
  int A[1000010], B[1000010];
  pair<int, int> C[1000010];
  int D[1000010];
  struct node {
    int second, e, m, lazy = 0;
    pair<int, int> val;
    node *l, *r;
    node(int ss, int ee) {
      second = ss;
      e = ee;
      m = (second + e) / 2;
      if (second != e) {
        l = new node(second, m);
        r = new node(m + 1, e);
        val = min(l->val, r->val);
      } else
        val = pair<int, int>(0, second);
    }
    void prop() {
      if (lazy == 0 || second == e) return;
      l->lazy += lazy;
      r->lazy += lazy;
      l->val.first += lazy;
      r->val.first += lazy;
      lazy = 0;
    }
    void upd(int a, int b, int c) {
      prop();
      if (a <= second && e <= b) {
        val.first += c;
        lazy += c;
      } else if (a > e || second > b)
        return;
      else {
        l->upd(a, b, c);
        r->upd(a, b, c);
        val = min(l->val, r->val);
      }
    }
  } * root;
  void upd(int x, int p) {
    root->upd(x, N, -D[x]);
    D[x] = p;
    root->upd(x, N, D[x]);
  }
  vector<int> ans[1000010];
  int fw[1000010 * 2];
  void fwupd(int x, int p) {
    for (int i = x; i <= N + M; i += i & (-i)) fw[i] += p;
  }
  int fwqry(int x) {
    int ans = 0;
    for (int i = x; i > 0; i -= i & (-i)) ans += fw[i];
    return ans;
  }
  void solve() {
    cin >> N >> M;
    for (int i = 1; i <= int(N); ++i) {
      cin >> A[i];
      C[i] = pair<int, int>(A[i], i);
    }
    for (int i = 1; i <= int(M); ++i) cin >> B[i];
    sort(B + 1, B + M + 1);
    sort(C + 1, C + N + 1);
    root = new node(0, N);
    for (int i = 1; i <= int(N); ++i) D[i] = 0;
    for (int i = 1; i <= int(N); ++i) upd(i, 1);
    for (int i = 0; i <= int(N); ++i) ans[i].clear();
    int k = 1;
    for (int i = 1; i <= int(M); ++i) {
      while (k <= N && C[k].first < B[i]) {
        upd(C[k].second, -1);
        k++;
      }
      if (i == 1 || B[i] != B[i - 1]) {
        for (int j = k; j <= N && C[j].first == B[i]; j++) upd(C[j].second, 0);
      }
      ans[root->val.second].push_back(B[i]);
    }
    vector<int> v;
    for (auto i : ans[0]) v.push_back(i);
    for (int i = 1; i <= int(N); ++i) {
      v.push_back(A[i]);
      for (auto j : ans[i]) v.push_back(j);
    }
    vector<int> v2;
    for (auto i : v) v2.push_back(i);
    sort((v2).begin(), (v2).end());
    v2.erase(unique((v2).begin(), (v2).end()), v2.end());
    for (int i = 1; i <= int(N + M); ++i) fw[i] = 0;
    for (int i = 0; i <= int(v.size() - 1); ++i)
      v[i] = lower_bound((v2).begin(), (v2).end(), v[i]) - v2.begin() + 1;
    long long ans = 0;
    for (auto i : v) {
      ans += fwqry(N + M) - fwqry(i);
      fwupd(i, 1);
    }
    cout << ans << '\n';
  }
  int32_t main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    cin >> TC;
    while (TC--) solve();
  }
  