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>
  const int N = 1e6 + 10;
  const int M = 9699690;
  const int mod = 1e9 + 7, Gi = 332748118;
  const int inf = 0x3f3f3f3f;
  const long long INF = 0x3f3f3f3f3f3f3f3fll;
  const int SIZE = 1 << 21;
  char ibuf[SIZE], *iS, *iT;
  using std::pair;
  template <class T>
  void read(T &x) {
    x = 0;
    int f = 0;
    char c = getchar();
    while (!isdigit(c)) f |= c == '-', c = getchar();
    while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
    if (f) x = -x;
  }
  int random(int i) { return rand() % i + 1; }
  template <class T>
  void ckmax(T &x, T y) {
    x = x > y ? x : y;
  }
  template <class T>
  void ckmin(T &x, T y) {
    x = x < y ? x : y;
  }
  template <class T>
  T Abs(T x) {
    return x > 0 ? x : -x;
  }
  inline int plus(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
  int qp(int d, int k) {
    int f = 1;
    while (k) {
      if (k & 1) f = (1ll * (f) * (d) % mod);
      d = (1ll * (d) * (d) % mod), k >>= 1;
    }
    return f;
  }
  int n, m, a[N], b[N], tmp[N];
  pair<int, int> p[N];
  long long ans;
  void msort(int l, int r) {
    if (l == r) return;
    int mid = l + r >> 1;
    msort(l, mid), msort(mid + 1, r);
    int lp = l, rp = mid + 1, ct = 0;
    while (lp <= mid && rp <= r) {
      if (a[lp] <= a[rp])
        tmp[++ct] = a[lp++], ans += rp - 1 - mid;
      else
        tmp[++ct] = a[rp++];
    }
    while (lp <= mid) tmp[++ct] = a[lp++], ans += rp - 1 - mid;
    while (rp <= r) tmp[++ct] = a[rp++];
    for (int i = (l); i <= (r); i++) a[i] = tmp[i + 1 - l];
  }
  int tag[N << 2], mi[N << 2];
  void build(int id, int l, int r) {
    tag[id] = 0;
    if (l == r) {
      mi[id] = l - 1;
      return;
    }
    int mid = l + r >> 1;
    build(id << 1, l, mid), build(id << 1 | 1, mid + 1, r);
    mi[id] = std::min(mi[id << 1], mi[id << 1 | 1]);
  }
  void pushdown(int id) {
    if (tag[id]) {
      tag[id << 1] += tag[id];
      tag[id << 1 | 1] += tag[id];
      mi[id << 1] += tag[id];
      mi[id << 1 | 1] += tag[id];
      tag[id] = 0;
    }
  }
  void upt(int id, int L, int R, int l, int r, int d) {
    if (l <= L && R <= r) {
      tag[id] += d;
      mi[id] += d;
      return;
    }
    if (l > R || r < L) return;
    int Mid = L + R >> 1;
    pushdown(id);
    upt(id << 1, L, Mid, l, r, d);
    upt(id << 1 | 1, Mid + 1, R, l, r, d);
    mi[id] = std::min(mi[id << 1], mi[id << 1 | 1]);
  }
  void work() {
    read(n), read(m);
    for (int i = (1); i <= (n); i++) read(a[i]), p[i] = std::make_pair(a[i], i);
    ans = 0;
    msort(1, n);
    for (int i = (1); i <= (m); i++) read(b[i]);
    std::sort(p + 1, p + 1 + n);
    std::sort(b + 1, b + 1 + m);
    build(1, 1, n + 1);
    int ld = 1, lu = 1;
    for (int i = (1); i <= (m); i++) {
      while (ld <= n && p[ld].first <= b[i])
        upt(1, 1, n + 1, p[ld++].second + 1, n + 1, -1);
      while (lu <= n && p[lu].first < b[i])
        upt(1, 1, n + 1, 1, p[lu++].second, 1);
      ans += mi[1];
    }
    printf("%lld\n", ans);
  }
  int main() {
    int T = 1;
    read(T);
    while (T--) work();
    return 0;
  }
  