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[1001000], b[1001000], buf[1001000];
  long long cnt_inv(int f[], int l, int r) {
    if (r - l <= 1) return 0;
    int m = (l + r) / 2;
    long long ans = cnt_inv(f, l, m) + cnt_inv(f, m, r);
    int tl = l, tm = m, s = l;
    while (s < r) {
      if (tl < m && (tm == r || f[tl] <= f[tm])) {
        ans += tm - m;
        buf[s++] = f[tl++];
      } else
        buf[s++] = f[tm++];
    }
    for (int i = l; i < r; i++) f[i] = buf[i];
    return ans;
  }
  int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
      int n, m;
      scanf("%d%d", &n, &m);
      for (int i = 0; i < n; i++) scanf("%d", &a[i]);
      for (int i = 0; i < m; i++) scanf("%d", &b[i]);
      sort(b, b + m);
      priority_queue<int> pq;
      long long ans = 0;
      for (int i = 0; i < n; i++) {
        int l = lower_bound(b, b + m, a[i]) - b;
        int r = upper_bound(b, b + m, a[i]) - b;
        ans += l;
        pq.push(l);
        pq.push(r);
        pq.pop();
      }
      for (int i = 0; i < n; i++) {
        ans -= pq.top();
        pq.pop();
      }
      ans += cnt_inv(a, 0, n);
      printf("%lld\n", ans);
    }
    return 0;
  }
  