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 per() { cerr << endl; }
  template <typename Head, typename... Tail>
  void per(Head H, Tail... T) {
    cerr << H << ' ';
    per(T...);
  }
  template <class T>
  bool uin(T& a, T b) {
    return a > b ? (a = b, true) : false;
  }
  template <class T>
  bool uax(T& a, T b) {
    return a < b ? (a = b, true) : false;
  }
  template <class U, class V>
  ostream& operator<<(ostream& out, const pair<U, V>& a) {
    return out << "(" << a.first << ", " << a.second << ")";
  }
  template <class U, class V>
  istream& operator>>(istream& in, pair<U, V>& a) {
    return in >> a.first >> a.second;
  }
  template <typename W,
            typename T = typename enable_if<!is_same<W, string>::value,
                                            typename W::value_type>::type>
  ostream& operator<<(ostream& out, const W& v) {
    out << "{ ";
    for (const auto& first : v) out << first << ", ";
    return out << '}';
  }
  template <class T>
  void readArr(T from, T to) {
    for (auto i = from; i != to; ++i) cin >> *i;
  }
  mt19937 mrand(1337);
  unsigned int myRand32() { return mrand() & (unsigned int)(-1); }
  unsigned long long myRand64() {
    return ((unsigned long long)myRand32() << 32) ^ myRand32();
  }
  const int mod = 1000000007;
  void add(int& a, int b) {
    a += b;
    if (a >= mod) a -= mod;
  }
  void dec(int& a, int b) {
    a -= b;
    if (a < 0) a += mod;
  }
  int mult(int a, int b) { return a * (long long)b % mod; }
  int bp(int a, int b) {
    int res = 1;
    while (b > 0) {
      if (b & 1) res = mult(res, a);
      a = mult(a, a);
      b >>= 1;
    }
    return res;
  }
  struct Vertex {
    int sum;
    int min_prefix_sum;
  };
  vector<Vertex> rmq;
  void build(int i, int l, int r) {
    if (r - l == 1) {
      rmq[i].sum = 1;
      rmq[i].min_prefix_sum = 0;
      return;
    }
    int mid = (l + r) / 2;
    build(2 * i + 1, l, mid);
    build(2 * i + 2, mid, r);
    rmq[i].sum = rmq[2 * i + 1].sum + rmq[2 * i + 2].sum;
    rmq[i].min_prefix_sum =
        min(rmq[2 * i + 1].min_prefix_sum,
            rmq[2 * i + 1].sum + rmq[2 * i + 2].min_prefix_sum);
  }
  void upd(int i, int l, int r, int index, int w) {
    if (r - l == 1) {
      rmq[i].sum = w;
      rmq[i].min_prefix_sum = w;
      return;
    }
    int mid = (l + r) / 2;
    if (index < mid)
      upd(2 * i + 1, l, mid, index, w);
    else
      upd(2 * i + 2, mid, r, index, w);
    rmq[i].sum = rmq[2 * i + 1].sum + rmq[2 * i + 2].sum;
    rmq[i].min_prefix_sum =
        min(rmq[2 * i + 1].min_prefix_sum,
            rmq[2 * i + 1].sum + rmq[2 * i + 2].min_prefix_sum);
  }
  struct FenwickTree1 {
    vector<int> t;
    int n;
    int get(int v) {
      int res = 0;
      for (; v >= 0; v = (v & (v + 1)) - 1) {
        res += t[v];
      }
      return res;
    }
    int get(int vl, int vr) {
      if (vl > vr) return 0;
      int res = get(vr);
      if (vl > 0) {
        res -= get(vl - 1);
      }
      return res;
    }
    void up(int v, int val) {
      for (; v < n; v = v | (v + 1)) {
        t[v] += val;
      }
    }
    FenwickTree1(int n_) {
      n = n_;
      t.assign(n, 0);
    }
  };
  void solve() {
    int n, m;
    cin >> n >> m;
    FenwickTree1 f(n);
    vector<pair<int, int> > farr;
    vector<int> sarr(m);
    for (int i = 0; i < n; ++i) {
      int first;
      cin >> first;
      farr.push_back(make_pair(first, i));
    }
    for (auto& first : sarr) cin >> first;
    sort(farr.begin(), farr.end());
    sort(sarr.begin(), sarr.end());
    rmq.assign(4 * n, {});
    build(0, 0, n);
    int cur = 0;
    long long ans = 0;
    int u = 0;
    int tt = 0;
    for (auto first : sarr) {
      while (u < farr.size() && farr[u].first < first) {
        cur++;
        upd(0, 0, n, farr[u++].second, -1);
      }
      tt = max(tt, u);
      while (tt < farr.size() && farr[tt].first == first) {
        upd(0, 0, n, farr[tt++].second, 0);
      }
      ans += cur + rmq[0].min_prefix_sum;
    }
    for (int i = 0; i < farr.size(); ++i) {
      ans += f.get(farr[i].second, n - 1);
      f.up(farr[i].second, 1);
    }
    cout << ans << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    for (int i = 0; i < t; ++i) solve();
  }
  