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;
  const int DIR4[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
  const int DIR8[8][2] = {{1, 0},   {-1, 0}, {0, 1},  {0, -1},
                          {-1, -1}, {-1, 1}, {1, -1}, {1, 1}};
  const int MOD197 = 1e9 + 7;
  const int MOD19 = 1e9;
  const int MOD353 = 998244353;
  const int INF = 0x3f3f3f3f;
  const long long LINF = 0x3f3f3f3f3f3f3f3f;
  const int NINF = 0xc0c0c0c0;
  const long long LNINF = 0xc0c0c0c0c0c0c0c0;
  inline int LT(int x) { return x << 1; }
  inline int RT(int x) { return (x << 1) | 1; }
  inline int LB(int x) { return x & -x; }
  inline int MSK(int x) { return 1 << x; }
  inline int HF(int x) { return x >> 1; }
  inline int DB(int x) { return x << 1; }
  inline int GT(int x, int i) { return (x >> i) & 1; }
  inline int ST(int x, int i) { return x | (1 << i); }
  inline int ODD(int x) { return x & 1; }
  inline int SGN(int x) { return x >= 0 ? (x == 0 ? 0 : 1) : -1; }
  template <int mod>
  long long fpow(long long a, long long b) {
    long long ans = 1;
    for (; b; b >>= 1, a = (a * a) % mod)
      if (b & 1) ans = (ans * a) % mod;
    return ans;
  }
  template <int mod>
  struct MODINT {
    long long val;
    MODINT() : val(0) {}
    MODINT(long long v) : val(v % mod) {}
    MODINT operator+(const MODINT &o) const { return MODINT(val + o.val); }
    MODINT &operator+=(const MODINT &o) {
      val = (val + o.val) % mod;
      return *this;
    }
    MODINT operator*(const MODINT &o) const { return MODINT(val * o.val); }
    MODINT &operator*=(const MODINT &o) {
      val = (val * o.val) % mod;
      return *this;
    }
    MODINT operator-(const MODINT &o) const {
      return MODINT((val - o.val) % mod + mod);
    }
    MODINT &operator-=(const MODINT &o) {
      val = ((val - o.val) % mod + mod) % mod;
      return *this;
    }
    MODINT operator/(const MODINT &o) const {
      return MODINT(val * fpow<mod>(o.val, mod - 2));
    }
    MODINT &operator/=(const MODINT &o) {
      val = (val * fpow<mod>(o.val, mod - 2)) % mod;
      return *this;
    }
  };
  template <int mod>
  ostream &operator<<(ostream &out, const MODINT<mod> &val) {
    out << val.val;
    return out;
  }
  long long fdiv(long long a, long long b) {
    return a / b - ((a ^ b) < 0 && a % b);
  }
  long long cdiv(long long a, long long b) {
    return a / b + ((a ^ b) > 0 && a % b);
  }
  int a[1000005];
  int b[1000005];
  int p[1000005];
  int big[1000005];
  int sml[1000005];
  long long tree[3000005];
  int N;
  void add(int x, long long val) {
    while (x <= N) {
      tree[x] += val;
      x += LB(x);
    }
  }
  long long query(int x) {
    long long val = 0;
    while (x) {
      val += tree[x];
      x -= LB(x);
    }
    return val;
  }
  void conquer(int bl, int br, int al, int ar) {
    if (bl > br) return;
    int mid = (bl + br) / 2;
    big[al] = 0;
    for (int i = al + 1; i <= ar; i++) {
      big[i] = big[i - 1];
      if (a[i - 1] > b[mid]) big[i]++;
    }
    sml[ar + 1] = 0;
    for (int i = ar; i >= al; i--) {
      sml[i] = sml[i + 1];
      if (a[i] < b[mid]) sml[i]++;
    }
    p[mid] = al;
    for (int i = al; i <= ar; i++) {
      if (big[i] + sml[i] < big[p[mid]] + sml[p[mid]]) p[mid] = i;
    }
    conquer(bl, mid - 1, al, p[mid]);
    conquer(mid + 1, br, p[mid], ar);
  }
  void solve() {
    int n, m;
    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);
    a[n + 1] = INT_MAX;
    conquer(1, m, 1, n + 1);
    vector<int> lis;
    int ptr = 1;
    for (int i = 1; i <= n + 1; i++) {
      while (ptr <= m && p[ptr] == i) lis.push_back(b[ptr++]);
      lis.push_back(a[i]);
    }
    vector<int> decre;
    for (int i : lis) decre.push_back(i);
    sort(decre.begin(), decre.end());
    decre.erase(unique(decre.begin(), decre.end()), decre.end());
    for (int i = 0; i < lis.size(); i++)
      lis[i] =
          lower_bound(decre.begin(), decre.end(), lis[i]) - decre.begin() + 1;
    N = decre.size();
    for (int i = 0; i <= N; i++) tree[i] = 0;
    long long ans = 0;
    for (int i = 0; i < lis.size(); i++) {
      ans += i - query(lis[i]);
      add(lis[i], 1);
    }
    cout << ans << endl;
  }
  int main() {
    ;
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin >> T;
    while (T--) solve();
    return 0;
  }
  