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 = 2e6 + 5;
  int T, n, m, tot, cnt;
  int a[N], b[N], A[N], c[N], li[N], p[N];
  void lisan() {
    sort(li + 1, li + 1 + tot);
    cnt = unique(li + 1, li + 1 + tot) - li - 1;
    for (int i = 1; i <= n; i++)
      a[i] = lower_bound(li + 1, li + 1 + cnt, a[i]) - li;
    for (int i = 1; i <= m; i++)
      b[i] = lower_bound(li + 1, li + 1 + cnt, b[i]) - li;
  }
  int lowbit(int x) { return x & (-x); }
  void upd(int x) {
    for (; x <= cnt; x += lowbit(x)) c[x]++;
  }
  long long qry(int x) {
    long long summ = 0;
    for (; x; x -= lowbit(x)) summ += c[x];
    return summ;
  }
  long long calc() {
    long long summ = 0;
    for (int i = 1; i <= tot; i++) {
      summ += qry(cnt) - qry(A[i]);
      upd(A[i]);
    }
    return summ;
  }
  void solve(int Lb, int Rb, int La, int Ra) {
    if (Lb > Rb) return;
    int mid = (Lb + Rb) >> 1;
    int val = b[mid];
    int minn = 2e9, summ1 = 0, summ2 = 0;
    for (int i = Ra - 1; i >= La; i--)
      if (a[i] < val) summ2++;
    minn = summ2;
    p[mid] = La;
    for (int i = La + 1; i <= Ra; i++) {
      if (a[i - 1] < val) summ2--;
      if (a[i - 1] > val) summ1++;
      if (summ1 + summ2 < minn) minn = summ1 + summ2, p[mid] = i;
    }
    solve(Lb, mid - 1, La, p[mid]);
    solve(mid + 1, Rb, p[mid], Ra);
  }
  int main() {
    scanf("%d", &T);
    while (T--) {
      tot = 0;
      scanf("%d%d", &n, &m);
      for (int i = 1; i <= n; i++) scanf("%d", &a[i]), li[++tot] = a[i];
      for (int i = 1; i <= m; i++) scanf("%d", &b[i]), li[++tot] = b[i];
      sort(b + 1, b + 1 + m);
      lisan();
      for (int i = 1; i <= cnt; i++) c[i] = 0;
      solve(1, m, 1, n + 1);
      tot = 0;
      int nowb = 1;
      for (int i = 1; i <= n; i++) {
        while (p[nowb] == i && nowb <= m) A[++tot] = b[nowb++];
        A[++tot] = a[i];
      }
      while (nowb <= m) A[++tot] = b[nowb++];
      printf("%lld\n", calc());
    }
    return 0;
  }
  