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>
  const int N = 1e6, M = N * 4;
  inline char gc() {
    static char buf[1 << 20], *ss, *tt;
    if (tt == ss) {
      tt = (ss = buf) + fread(buf, 1, 1 << 20, stdin);
      if (tt == ss) return EOF;
    }
    return *ss++;
  }
  inline int read() {
    int res, sign = 1;
    char c;
    while ((c = gc()) < '0' || c > '9')
      if (c == '-') sign = -1;
    res = c ^ 48;
    while ((c = gc()) >= '0' && c <= '9')
      res = (res << 3) + (res << 1) + (c ^ 48);
    return res * sign;
  }
  int n, m, va[N + 5];
  long long ans;
  std::set<std::pair<int, int> > se;
  int nv[N + 5], bt[N + 5];
  inline void pls1(int w) {
    for (; w <= n; w += w & -w) ++bt[w];
  }
  inline int sum(int w) {
    int x = 0;
    for (; w; w -= w & -w) x += bt[w];
    return x;
  }
  int su[M + 5], mn[M + 5];
  inline void pu(int w) {
    su[w] = su[w * 2] + su[w * 2 + 1];
    mn[w] = std::min(mn[w * 2], su[w * 2] + mn[w * 2 + 1]);
  }
  void bld(int w = 1, int l = 0, int r = n) {
    mn[w] = su[w] = 0;
    if (l == r) return;
    int mi = l + r >> 1;
    bld(w * 2, l, mi), bld(w * 2 + 1, mi + 1, r);
  }
  void pls(int x, int y, int w = 1, int l = 0, int r = n) {
    if (l == r) {
      su[w] += y, mn[w] += y;
      return;
    }
    int mi = l + r >> 1;
    if (x <= mi)
      pls(x, y, w * 2, l, mi);
    else
      pls(x, y, w * 2 + 1, mi + 1, r);
    pu(w);
  }
  int main() {
    int tc;
    for (tc = read(); tc; --tc) {
      se.clear();
      n = read(), m = read();
      for (int i = 1; i <= n; ++i) {
        int x;
        x = read();
        se.insert(std::make_pair(x, i));
      }
      ans = 0;
      for (int i = 1; i <= n; ++i) bt[i] = 0;
      int nw = 0;
      for (auto it : se) nv[it.second] = ++nw;
      for (int i = 1; i <= n; ++i) {
        pls1(nv[i]);
        ans += nv[i] - sum(nv[i]);
      }
      for (int i = 1; i <= m; ++i) va[i] = read();
      bld();
      std::sort(va + 1, va + m + 1);
      for (int i = 1; i <= n; ++i) pls(i, 1);
      int ap = 0;
      auto it = se.begin(), jt = se.begin();
      for (int i = 1; i <= m; ++i) {
        for (; jt != se.end() && jt->first <= va[i]; ++jt) pls(jt->second, -1);
        for (; it != se.end() && it->first < va[i]; ++it) {
          ++ap;
          pls(it->second, -1);
        }
        ans += ap + mn[1];
      }
      printf("%lld\n", ans);
    }
    return 0;
  }
  