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 n, m;
  int a[1000005], b[1000005];
  long long ans;
  vector<int> lsh;
  int mi[4000005], tag[4000005];
  void build(int x, int l, int r) {
    mi[x] = tag[x] = 0;
    if (l == r) return;
    int mid = (l + r) / 2;
    build(x * 2, l, mid);
    build(x * 2 + 1, mid + 1, r);
  }
  void update(int x, int l, int r, int L, int R, int v) {
    if (L > R) return;
    if (L <= l && r <= R) {
      mi[x] += v;
      tag[x] += v;
      return;
    }
    int mid = (l + r) / 2;
    if (L <= mid) update(x * 2, l, mid, L, R, v);
    if (R > mid) update(x * 2 + 1, mid + 1, r, L, R, v);
    mi[x] = min(mi[x * 2], mi[x * 2 + 1]);
    mi[x] += tag[x];
  }
  int c[2000005];
  void add(int x, int y) {
    while (x) c[x] += y, x -= (x & -x);
  }
  int query(int x) {
    int s = 0;
    while (x <= lsh.size()) s += c[x], x += (x & -x);
    return s;
  }
  struct opt {
    int h, l, r, v;
  } d[2000005];
  bool cmp(opt x, opt y) { return x.h < y.h; }
  void solve() {
    int i;
    scanf("%d%d", &n, &m);
    for ((i) = 1; (i) <= (n); (i)++) scanf("%d", &a[i]);
    for ((i) = 1; (i) <= (m); (i)++) scanf("%d", &b[i]);
    lsh.clear();
    for ((i) = 1; (i) <= (n); (i)++) lsh.push_back(a[i]);
    for ((i) = 1; (i) <= (m); (i)++) lsh.push_back(b[i]);
    sort(lsh.begin(), lsh.end());
    lsh.resize(unique(lsh.begin(), lsh.end()) - lsh.begin());
    for ((i) = 1; (i) <= (n); (i)++)
      a[i] = upper_bound(lsh.begin(), lsh.end(), a[i]) - lsh.begin();
    for ((i) = 1; (i) <= (m); (i)++)
      b[i] = upper_bound(lsh.begin(), lsh.end(), b[i]) - lsh.begin();
    for ((i) = 1; (i) <= (n); (i)++)
      d[i] = (opt){a[i], i, n, -1}, d[i + n] = (opt){a[i] + 1, 0, i - 1, 1};
    ans = 0;
    for ((i) = 1; (i) <= (lsh.size()); (i)++) c[i] = 0;
    for ((i) = 1; (i) <= (n); (i)++) ans += query(a[i] + 1), add(a[i], 1);
    build(1, 0, n);
    for ((i) = 1; (i) <= (n); (i)++) update(1, 0, n, i, n, 1);
    sort(d + 1, d + n + n + 1, cmp);
    sort(b + 1, b + m + 1);
    int p = 1;
    for ((i) = 1; (i) <= (m); (i)++) {
      while (p <= n + n && d[p].h <= b[i]) {
        update(1, 0, n, d[p].l, d[p].r, d[p].v);
        p++;
      }
      ans += mi[1];
    }
    printf("%lld\n", ans);
  }
  int main() {
    int t;
    scanf("%d", &t);
    while (t--) solve();
    return 0;
  }
  