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;
  int T;
  int n, m;
  int a[2000010], b[2000010];
  int pos[2000010];
  long long ans;
  void find_ans(int L, int R, int l, int r) {
    if (l > r) return;
    int mid = (l + r) >> 1;
    int x = b[mid], cnt = 0;
    for (int i = L; i <= R; i++)
      if (a[i] < x) cnt++;
    int mn = cnt;
    pos[mid] = L - 1;
    for (int i = L; i <= R; i++) {
      if (a[i] > x) cnt++;
      if (a[i] < x) cnt--;
      if (cnt < mn) {
        mn = cnt;
        pos[mid] = i;
      }
    }
    find_ans(L, pos[mid], l, mid - 1);
    find_ans(pos[mid], R, mid + 1, r);
  }
  struct Node {
    int pos, val, tp;
  } node[2000010 << 1];
  int tot;
  int tmp[2000010 << 1];
  int N;
  void lisan() {
    tmp[0] = 0;
    for (int i = 1; i <= n; i++) tmp[++tmp[0]] = a[i];
    for (int i = 1; i <= m; i++) tmp[++tmp[0]] = b[i];
    sort(tmp + 1, tmp + tmp[0] + 1);
    N = unique(tmp + 1, tmp + tmp[0] + 1) - tmp - 1;
    for (int i = 1; i <= n; i++)
      a[i] = lower_bound(tmp + 1, tmp + N + 1, a[i]) - tmp;
    for (int i = 1; i <= m; i++)
      b[i] = lower_bound(tmp + 1, tmp + N + 1, b[i]) - tmp;
  }
  int sum[2000010 << 1];
  int lowbit(int x) { return x & -x; }
  void Upd(int x) {
    for (int i = x; i <= N; i += lowbit(i)) sum[i]++;
  }
  long long query(int x) {
    long long ret = 0;
    for (int i = x; i; i -= lowbit(i)) ret += sum[i];
    return ret;
  }
  bool cmp1(Node a, Node b) {
    if (a.pos != b.pos) return a.pos < b.pos;
    if (a.val != b.val) return a.val < b.val;
    return a.tp < b.tp;
  }
  void work() {
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= m; i++) scanf("%d", &b[i]);
    lisan();
    sort(b + 1, b + m + 1);
    find_ans(1, n, 1, m);
    tot = 0;
    for (int i = 1; i <= n; i++) {
      node[++tot] = (Node){i, a[i], 0};
    }
    for (int i = 1; i <= m; i++) {
      node[++tot] = (Node){pos[i], b[i], 1};
    }
    sort(node + 1, node + tot + 1, cmp1);
    for (int i = 0; i <= N; i++) sum[i] = 0;
    long long ans = 0;
    for (int i = tot; i >= 1; i--) {
      Upd(node[i].val);
      int now = query(node[i].val - 1);
      ans += now;
    }
    printf("%lld\n", ans);
  }
  int main() {
    scanf("%d", &T);
    while (T--) work();
  }
  