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 N = 4e6 + 99;
  int n, m, p, t, cnt, a[N], b[N], c[N];
  vector<int> v[N];
  map<int, int> T;
  long long ans;
  int Fen[N];
  void update(int x, int val) {
    while (x <= n + m) {
      Fen[x] += val;
      x += (x & -x);
    }
  }
  int query(int x) {
    int res = 0;
    while (x > 0) {
      res += Fen[x];
      x -= (x & -x);
    }
    return res;
  }
  void Add(int x) {
    ans += query(n + m) - query(x);
    update(x, 1);
  }
  void make() {
    cnt = 1;
    sort(c, c + n + m);
    for (int i = 0; i < n + m; i++) {
      if (i == 0 || c[i] != c[i - 1]) {
        T[c[i]] = cnt++;
      }
    }
    for (int i = 1; i < n + 1; i++) a[i] = T[a[i]];
    for (int i = 1; i < m + 1; i++) b[i] = T[b[i]];
    sort(b + 1, b + 1 + m);
    for (int i = 1; i < n + 1; i++) {
      if (a[i] > n + m) cout << "99" << '=' << (99) << endl;
    }
    for (int i = 1; i < m + 1; i++) {
      if (b[i] > n + m) cout << "90" << '=' << (90) << endl;
    }
  }
  void solve(int L, int R, int l, int r) {
    if (R <= L) return;
    int Mid = (L + R) / 2, x = b[Mid], ans = 0, mn = N, mid = 0;
    for (int i = l; i < r; i++) {
      if (a[i] < x) ans--;
      if (a[i] > x) ans++;
      if (ans < mn) {
        mn = ans, mid = i;
      }
    }
    v[mid].push_back(x);
    solve(L, Mid, l, mid + 1);
    solve(Mid + 1, R, mid, r);
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while (t--) {
      ans = p = 0;
      cin >> n >> m;
      for (int i = 1; i < n + 1; i++) {
        cin >> a[i];
        c[i - 1] = a[i];
      }
      for (int i = 1; i < m + 1; i++) {
        cin >> b[i];
        c[n + i - 1] = b[i];
      }
      make();
      solve(1, m + 1, 0, n + 1);
      for (int i = 0; i < n + 1; i++) {
        if (i) Add(a[i]);
        sort(v[i].begin(), v[i].end());
        for (int j = 0; j < v[i].size(); j++) {
          Add(v[i][j]);
        }
        v[i].clear();
      }
      cout << ans << endl;
      T.clear();
      fill(Fen, Fen + n + m + 9, 0);
    }
  }
  