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;
  int T;
  int n, m, cnt;
  int a[1000010], b[1000010];
  int tmp[2000010];
  int c[8000010], tag[8000010];
  inline void up2(int x) { c[x] = min(c[x << 1], c[x << 1 | 1]) + tag[x]; }
  inline void down(int x) {
    if (tag[x] == 0) return;
    tag[x << 1] += tag[x];
    tag[x << 1 | 1] += tag[x];
    c[x << 1] += tag[x];
    c[x << 1 | 1] += tag[x];
    tag[x] = 0;
  }
  void build(int x, int l, int r) {
    tag[x] = 0;
    c[x] = l;
    if (l == r) return;
    int mid = l + r >> 1;
    build(x << 1, l, mid);
    build(x << 1 | 1, mid + 1, r);
  }
  inline void add(int x) {
    for (; x <= cnt; x += x & -x) c[x]++;
  }
  inline int query(int x) {
    int ans = 0;
    for (; x; x -= x & -x) ans += c[x];
    return ans;
  }
  inline long long calca() {
    for (int i = 1; i <= cnt; i++) c[i] = 0;
    long long ans = 0;
    for (int i = n; i > 0; i--) {
      add(a[i]);
      ans += query(a[i] - 1);
    }
    return ans;
  }
  int pos[1000010];
  void add(int x, int l, int r, int L, int R, int v) {
    if (l > R || L > r) return;
    if (L <= l && r <= R) {
      tag[x] += v, c[x] += v;
      return;
    }
    int mid = l + r >> 1;
    add(x << 1, l, mid, L, R, v);
    add(x << 1 | 1, mid + 1, r, L, R, v);
    up2(x);
  }
  inline long long calc() {
    build(1, 0, n);
    for (int i = 1; i <= n; i++) pos[i] = i;
    sort(pos + 1, pos + 1 + n, [](int x, int y) { return a[x] < a[y]; });
    long long ans = 0;
    for (int i = 1, j = 1, k = 1; i <= m; i++) {
      while (j <= n && a[pos[j]] < b[i]) {
        add(1, 0, n, 0, pos[j] - 1, 1);
        j++;
      }
      while (k <= n && a[pos[k]] <= b[i]) {
        add(1, 0, n, pos[k], n, -1);
        k++;
      }
      ans += c[1];
    }
    return ans;
  }
  inline void rmain() {
    scanf("%d%d", &n, &m);
    cnt = 0;
    for (int i = 1; i <= n; i++) scanf("%d", a + i), tmp[++cnt] = a[i];
    for (int i = 1; i <= m; i++) scanf("%d", b + i);
    sort(b + 1, b + 1 + m);
    long long ans = calc();
    sort(tmp + 1, tmp + 1 + cnt);
    cnt = unique(tmp + 1, tmp + 1 + cnt) - tmp - 1;
    for (int i = 1; i <= n; i++)
      a[i] = lower_bound(tmp + 1, tmp + 1 + cnt, a[i]) - tmp;
    ans += calca();
    printf("%lld\n", ans);
  }
  int main() {
    scanf("%d", &T);
    while (T--) rmain();
  }
  