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 a[1000000 + 5], b[1000000 + 5];
  int p[2000000 + 5];
  void solve(int l1, int r1, int l2, int r2) {
    int i, j, k;
    if (l2 > r2) return;
    if (l1 == r1) {
      for (i = l2; i <= r2; i++) p[i] = l1;
      return;
    }
    int minn = 0, id = l1, num = 0, mid = (l2 + r2) >> 1;
    for (i = l1 + 1; i <= r1; i++) {
      if (a[i] < b[mid])
        num--;
      else if (a[i] > b[mid])
        num++;
      if (num < minn) minn = num, id = i;
    }
    p[mid] = id;
    solve(l1, id, l2, mid - 1);
    solve(id, r1, mid + 1, r2);
  }
  int N, c[2000000 + 5], s[2000000 + 5];
  int lowbit(int x) { return x & (-x); }
  void modify(int x) {
    for (int i = x; i <= N; i += lowbit(i)) s[i]++;
  }
  int query(int x) {
    int sum = 0;
    for (int i = x; i > 0; i -= lowbit(i)) sum += s[i];
    return sum;
  }
  int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
      int n, m, i, j, k = 0;
      scanf("%d%d", &n, &m);
      for (i = 1; i <= n; i++) scanf("%d", &a[i]);
      for (i = 1; i <= m; i++) scanf("%d", &b[i]);
      sort(b + 1, b + m + 1);
      solve(0, n, 1, m);
      j = 1;
      p[m + 1] = 1e9;
      for (i = 1; i <= n; i++) {
        while (p[j] < i) c[++k] = b[j++];
        c[++k] = a[i];
      }
      while (j <= m) c[++k] = b[j++];
      n = k;
      for (i = 1; i <= n; i++) p[i] = c[i];
      sort(p + 1, p + n + 1);
      m = unique(p + 1, p + n + 1) - p - 1;
      for (i = 1; i <= n; i++) c[i] = lower_bound(p + 1, p + m + 1, c[i]) - p;
      N = m;
      long long ans = 0;
      for (i = 1; i <= n; i++) {
        modify(c[i]);
        ans += i - query(c[i]);
      }
      for (i = 1; i <= N; i++) s[i] = 0;
      printf("%lld\n", ans);
    }
  }
  