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 inf = 1 << 30;
  const long long Inf = 1ll << 60ll;
  namespace io {
  char buf[(1 << 23)], *p1 = buf, *p2 = buf, c;
  int f;
  template <typename T>
  T read() {
    T x = 0;
    f = 0;
    c = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, (1 << 23), stdin), p1 == p2)
             ? EOF
             : *p1++);
    while (!isdigit(c))
      (c == '-') && (f = 1),
          c = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, (1 << 23), stdin),
                            p1 == p2)
                   ? EOF
                   : *p1++);
    while (isdigit(c))
      x = x * 10 + (c & 15),
      c = (p1 == p2 &&
                   (p2 = (p1 = buf) + fread(buf, 1, (1 << 23), stdin), p1 == p2)
               ? EOF
               : *p1++);
    return f ? -x : x;
  }
  }  // namespace io
  const int N = 1000005;
  int pre[N], suf[N], a[N], b[N];
  int ans[N];
  void solve(int la, int ra, int lb, int rb) {
    if (lb > rb) return;
    int mid = lb + rb >> 1;
    pre[la] = 0;
    for (int i = la + 1; i <= ra; ++i) pre[i] = pre[i - 1] + (a[i - 1] > b[mid]);
    suf[ra] = 0;
    for (int i = ra - 1; i >= la; --i) suf[i] = suf[i + 1] + (a[i] < b[mid]);
    int mn = inf, pos = -1;
    for (int i = la; i <= ra; ++i) {
      if (pre[i] + suf[i] < mn) {
        mn = pre[i] + suf[i];
        pos = i;
      }
    }
    ans[mid] = pos;
    solve(la, pos, lb, mid - 1);
    solve(pos, ra, mid + 1, rb);
  }
  int c[N * 2];
  bool cmp(int A, int B) { return c[A] != c[B] ? c[A] > c[B] : A > B; }
  int t[N * 2], tot;
  void update(int x, int v) {
    while (x <= tot) {
      t[x] += v;
      x += x & (-x);
    }
  }
  int query(int x) {
    int res = 0;
    while (x) {
      res += t[x];
      x -= x & (-x);
    }
    return res;
  }
  int n, m, id[N * 2];
  int main() {
    int T = io::read<int>();
    while (T--) {
      n = io::read<int>();
      m = io::read<int>();
      for (int i = 1; i <= n; ++i) a[i] = io::read<int>();
      for (int i = 1; i <= m; ++i) b[i] = io::read<int>();
      sort(b + 1, b + m + 1);
      solve(1, n + 1, 1, m);
      int it = 0;
      tot = 0;
      for (int i = 1; i <= n; ++i) {
        while (it < m && ans[it + 1] <= i) {
          ++it;
          c[++tot] = b[it];
        }
        c[++tot] = a[i];
      }
      while (it < m) {
        ++it;
        c[++tot] = b[it];
      }
      for (int i = 1; i <= tot; ++i) id[i] = i, t[i] = 0;
      sort(id + 1, id + tot + 1, cmp);
      long long res = 0;
      for (int i = 1; i <= tot; ++i) {
        update(id[i], 1);
        res += query(id[i] - 1);
      }
      printf("%lld\n", res);
    }
    return 0;
  }
  