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;
  template <class T>
  inline void read(T &x) {
    x = 0;
    char c = getchar();
    while (!isdigit(c)) c = getchar();
    while (isdigit(c)) x = x * 10 + (c & 15), c = getchar();
  }
  int n, m, a[1000005], b[1000005], c[1000005 << 1], tot, t[1000005 << 1],
      s[1000005 << 2], la[1000005 << 2];
  vector<int> k[1000005 << 1];
  long long ans;
  inline void add(int x, int y) {
    for (int i = x; i <= tot; i += i & -i) t[i] += y;
  }
  inline int ask(int x) {
    int s = 0;
    for (int i = x; i; i -= i & -i) s += t[i];
    return s;
  }
  inline void build(int p, int l, int r) {
    s[p] = la[p] = 0;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(p << 1, l, mid), build(p << 1 | 1, mid + 1, r);
  }
  inline void add(int p, int l, int r, int x, int y, int z) {
    if (x <= l && r <= y) return s[p] += z, la[p] += z, void();
    int mid = (l + r) >> 1;
    if (la[p])
      s[p << 1] += la[p], s[p << 1 | 1] += la[p], la[p << 1] += la[p],
          la[p << 1 | 1] += la[p], la[p] = 0;
    if (mid >= x) add(p << 1, l, mid, x, y, z);
    if (mid < y) add(p << 1 | 1, mid + 1, r, x, y, z);
    s[p] = min(s[p << 1], s[p << 1 | 1]);
  }
  int main() {
    int T;
    read(T);
    while (T--) {
      read(n), read(m), ans = tot = 0;
      for (int i = 1; i <= n; i++) read(a[i]), c[++tot] = a[i];
      for (int i = 1; i <= m; i++) read(b[i]), c[++tot] = b[i];
      sort(c + 1, c + tot + 1), tot = unique(c + 1, c + tot + 1) - c - 1;
      for (int i = 1; i <= n; i++)
        a[i] = lower_bound(c + 1, c + tot + 1, a[i]) - c;
      for (int i = 1; i <= m; i++)
        b[i] = lower_bound(c + 1, c + tot + 1, b[i]) - c;
      sort(b + 1, b + m + 1);
      for (int i = 1; i <= tot; i++) t[i] = 0, k[i].clear();
      for (int i = 1; i <= n; i++) k[a[i]].push_back(i);
      for (int i = 1; i <= n; i++) ans += ask(tot) - ask(a[i]), add(a[i], 1);
      build(1, 0, n);
      for (int i = 1; i <= n; i++) add(1, 0, n, i, n, 1);
      for (int i = 1, las1 = 0, las2 = 0; i <= m; i++) {
        while (las1 < b[i]) {
          ++las1;
          for (auto &j : k[las1]) add(1, 0, n, j, n, -1);
        }
        while (las2 < b[i] - 1) {
          ++las2;
          for (auto &j : k[las2]) add(1, 0, n, 0, j - 1, 1);
        }
        ans += s[1];
      }
      printf("%lld\n", ans);
    }
    return 0;
  }
  