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;
  void solve();
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    solve();
    return 0;
  }
  const long long N = 1e6 + 100, K = 31, INF = 2e18, MOD = 1e9 + 7;
  long long t, n, m, a[N], b[N], p[N], s[N], fen[2 * N], ll;
  vector<long long> ad[N];
  long long get(long long v) {
    long long res = 0;
    for (; v < ll; v |= v + 1) {
      res += fen[v];
    }
    return res;
  }
  void add(long long v) {
    for (; v >= 0; v = (v & (v + 1)) - 1) {
      fen[v]++;
    }
  }
  void f(long long ln, long long rn, long long lm, long long rm) {
    long long mm = (rm + lm) / 2;
    p[ln] = 0;
    for (long long i = ln + 1; i <= rn; i++) {
      p[i] = p[i - 1];
      if (a[i - 1] > b[mm]) {
        p[i]++;
      }
    }
    s[rn + 1] = 0;
    for (long long i = rn; i >= ln; i--) {
      s[i] = s[i + 1];
      if (a[i] < b[mm]) {
        s[i]++;
      }
    }
    long long mn = INF, pos = -1;
    for (long long i = ln; i <= rn; i++) {
      if (mn > p[i] + s[i]) {
        mn = p[i] + s[i];
        pos = i;
      }
    }
    ad[pos].push_back(b[mm]);
    if (lm == rm) return;
    if (mm != lm) {
      f(ln, pos, lm, mm - 1);
    }
    if (mm != rm) {
      f(pos, rn, mm + 1, rm);
    }
  }
  void solve() {
    cin >> t;
    while (t--) {
      cin >> n >> m;
      for (long long i = 0; i < n; i++) {
        cin >> a[i];
      }
      for (long long i = 0; i < m; i++) {
        cin >> b[i];
      }
      sort(b, b + m);
      vector<long long> v;
      for (long long i = 0; i <= n; i++) {
        ad[i].clear();
      }
      f(0, n, 0, m - 1);
      for (long long i = 0; i <= n; i++) {
        sort(ad[i].begin(), ad[i].end());
      }
      for (long long i = 0; i < n; i++) {
        for (long long x : ad[i]) {
          v.push_back(x);
        }
        v.push_back(a[i]);
      }
      for (long long x : ad[n]) {
        v.push_back(x);
      }
      vector<long long> e = v;
      sort(e.begin(), e.end());
      map<long long, long long> ma;
      for (long long i = 0; i < e.size(); i++) {
        if (!ma.count(e[i])) {
          ma[e[i]] = ma.size();
        }
      }
      long long ans = 0;
      for (long long i = 0; i < v.size(); i++) {
        v[i] = ma[v[i]];
      }
      ll = ma.size() + 1;
      fill(fen, fen + ll, 0);
      for (long long i = 0; i < v.size(); i++) {
        ans += get(v[i] + 1);
        add(v[i]);
      }
      cout << ans << '\n';
    }
  }
  