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 lint = long long;
  using pi = pair<lint, lint>;
  const int MAXN = 1000005;
  int n, m, a[MAXN], b[MAXN];
  struct bit {
    int tree[MAXN];
    void add(int x, int v) {
      for (int i = x; i < MAXN; i += i & -i) tree[i] += v;
    }
    int query(int x) {
      int ret = 0;
      for (int i = x; i; i -= i & -i) ret += tree[i];
      return ret;
    }
  } bit;
  lint inv() {
    vector<int> v(a, a + n);
    sort((v).begin(), (v).end());
    v.resize(unique((v).begin(), (v).end()) - v.begin());
    lint ret = 0;
    for (int i = 0; i < n; i++) {
      a[i] = lower_bound((v).begin(), (v).end(), a[i]) - v.begin() + 1;
      ret += bit.query(n) - bit.query(a[i]);
      bit.add(a[i], 1);
    }
    for (int i = 0; i < n; i++) {
      bit.add(a[i], -1);
    }
    return ret;
  }
  void solve() {
    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;
    lint ret = 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;
      if (((int)(pq).size()) && pq.top() > r) {
        ret += pq.top() - r;
        pq.pop();
        pq.push(r);
      }
      pq.push(l);
    }
    ret += inv();
    printf("%lld\n", ret);
  }
  int main() {
    int tc;
    scanf("%d", &tc);
    while (tc--) solve();
  }
  