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;
  using ll = long long;
  using ld = long double;
  using pii = pair<int, int>;
  using pll = pair<ll, ll>;
  int n, m;
  int A[1000005];
  int B[1000005];
  int T[1000005];
  int S[1000005];
  ll inverse(int s, int e) {
    if (s == e) {
      T[s] = A[s];
      return 0;
    }
    int m = (s + e) / 2;
    ll ans = inverse(s, m) + inverse(m + 1, e);
    for (int i = s; i <= e; i++) {
      S[i] = T[i];
    }
    int i = s, j = m + 1, k = s;
    while (k <= e) {
      if (j > e || (i <= m && S[i] <= S[j]))
        T[k++] = S[i++];
      else
        T[k++] = S[j++], ans += m - i + 1;
    }
    return ans;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int T;
    cin >> T;
    for (int t = 1; t <= T; t++) {
      cin >> n >> m;
      for (int i = 1; i <= n; i++) cin >> A[i];
      for (int i = 1; i <= m; i++) cin >> B[i];
      sort(B + 1, B + m + 1);
      priority_queue<int> pq;
      ll mx = inverse(1, n);
      for (int i = 1; i <= n; i++) {
        int x = lower_bound(B + 1, B + m + 1, A[i]) - B - 1;
        int y = upper_bound(B + 1, B + m + 1, A[i]) - B - 1;
        pq.push(x);
        pq.push(y);
        pq.pop();
        mx += x;
      }
      while (!pq.empty()) {
        mx -= pq.top();
        pq.pop();
      }
      cout << mx << '\n';
    }
    return 0;
  }
  