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;
  template <typename T>
  T max3(T a, T b, T c) {
    return max(a, max(b, c));
  }
  template <typename T>
  T max4(T a, T b, T c, T d) {
    return max(max(a, d), max(b, c));
  }
  template <typename T>
  T min3(T a, T b, T c) {
    return min(a, min(b, c));
  }
  template <typename T>
  T min4(T a, T b, T c, T d) {
    return min(min(a, d), min(b, c));
  }
  template <typename T>
  T binex(T a, T b, T mod) {
    T ans = 1;
    a %= mod;
    while (b > 0) {
      if (b & 1) {
        ans *= a;
        ans %= mod;
      }
      a *= a;
      a %= mod;
      b >>= 1;
    }
    return ans;
  }
  template <class T>
  ostream &operator<<(ostream &os, vector<T> a) {
    for (auto x : a) {
      os << x << " ";
    }
    return os;
  }
  template <class T>
  ostream &operator<<(ostream &os, set<T> a) {
    for (auto x : a) {
      os << x << " ";
    }
    return os;
  }
  template <class T>
  ostream &operator<<(ostream &os, multiset<T> a) {
    for (auto x : a) {
      os << x << " ";
    }
    return os;
  }
  template <class T, class Q>
  ostream &operator<<(ostream &os, pair<T, Q> a) {
    os << "| ";
    os << a.first << ", " << a.second << " ";
    return os << "|";
  }
  template <class P, class Q, class T>
  ostream &operator<<(ostream &os, tuple<P, Q, T> a) {
    os << "| " << (get<0>(a)) << ", " << (get<1>(a)) << ", " << (get<2>(a))
       << "|";
    return os;
  }
  void precomp() {}
  void finder(int64_t l, int64_t r, int64_t st, int64_t end, vector<int64_t> &p,
              vector<int64_t> &a, vector<int64_t> &b) {
    if (r < l) {
      return;
    }
    int64_t mid = (l + r) >> 1;
    vector<int64_t> tmp1(end - st + 1);
    vector<int64_t> tmp2(end - st + 1);
    for (int64_t i = 1; i <= end - st; i++) {
      tmp1[i] = tmp1[i - 1];
      if (b[mid] < a[i + st - 1]) {
        tmp1[i]++;
      }
    }
    for (int64_t i = end - st - 1; i >= 0; i--) {
      tmp2[i] = tmp2[i + 1];
      if (b[mid] > a[i + st]) {
        tmp2[i]++;
      }
    }
    int64_t min = tmp1[0] + tmp2[0];
    int64_t minind = 0;
    for (int64_t i = 1; i < end - st + 1; i++) {
      if (min > tmp1[i] + tmp2[i]) {
        min = tmp1[i] + tmp2[i];
        minind = i;
      }
    }
    p[mid] = minind + st;
    finder(l, mid - 1, st, p[mid], p, a, b);
    finder(mid + 1, r, p[mid], end, p, a, b);
  }
  int64_t merge_sort(vector<int64_t> &a, int64_t l, int64_t r) {
    if (l == r) {
      return 0;
    }
    int64_t mid = (l + r) >> 1;
    int64_t ans = 0;
    ans += merge_sort(a, l, mid);
    ans += merge_sort(a, mid + 1, r);
    vector<int64_t> b(r - l + 1);
    int64_t p1 = l;
    int64_t p2 = mid + 1;
    int64_t curr = 0;
    while (p1 <= mid && p2 <= r) {
      if (a[p1] <= a[p2]) {
        b[curr] = a[p1];
        curr++;
        p1++;
      } else {
        b[curr] = a[p2];
        curr++;
        p2++;
        ans += mid - p1 + 1;
      }
    }
    if (p1 > mid) {
      while (p2 <= r) {
        b[curr] = a[p2];
        p2++;
        curr++;
      }
    } else {
      while (p1 <= mid) {
        b[curr] = a[p1];
        p1++;
        curr++;
      }
    }
    for (int64_t i = 0; i < r - l + 1; i++) {
      a[i + l] = b[i];
    }
    return ans;
  }
  void solve() {
    int64_t n, m;
    cin >> n >> m;
    vector<int64_t> a(n);
    vector<int64_t> b(m);
    for (int64_t i = 0; i < n; i++) {
      cin >> a[i];
    }
    for (int64_t i = 0; i < m; i++) {
      cin >> b[i];
    }
    sort(b.begin(), b.end());
    vector<int64_t> p(m);
    finder(0, m - 1, 0, n, p, a, b);
    vector<vector<int64_t>> sto(n + 1);
    for (int64_t i = 0; i < m; i++) {
      sto[p[i]].push_back(b[i]);
    }
    vector<int64_t> fin(n + m);
    int64_t curr = 0;
    for (int64_t i = 0; i <= n; i++) {
      for (auto x : sto[i]) {
        fin[curr] = x;
        curr++;
      }
      if (i == n) {
        break;
      }
      fin[curr] = a[i];
      curr++;
    }
    cout << merge_sort(fin, 0, n + m - 1) << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int64_t T;
    cin >> T;
    precomp();
    for (int64_t I = 1; I <= T; I++) {
      solve();
    }
  }
  