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 N = 1e6 + 5;
  int T, n, m, a[N], b[N], cnt, res[N << 1], v[N << 1], t, c[N << 1], pr[N],
      sf[N], p[N];
  long long ans;
  void add(int x, int y) {
    for (int i = x; i; i -= i & (-i)) c[i] += y;
  }
  int query(int x) {
    int ans = 0;
    for (int i = x; i <= t; i += i & (-i)) ans += c[i];
    return ans;
  }
  void solve(int l, int r, int nl, int nr) {
    if (l > r || nl > nr) return;
    int mid = (l + r) / 2, ans = 1e9;
    pr[nl - 1] = sf[nr + 1] = 0;
    for (int i = nl; i <= nr; i++) pr[i] = pr[i - 1] + (a[i] > b[mid]);
    for (int i = nr; i >= nl; i--) sf[i] = sf[i + 1] + (a[i] < b[mid]);
    for (int i = nl; i <= nr; i++)
      if (ans > pr[i - 1] + sf[i]) ans = pr[i - 1] + sf[i], p[mid] = i;
    solve(l, mid - 1, nl, p[mid]), solve(mid + 1, r, p[mid], nr);
  }
  signed main() {
    scanf("%d", &T);
    while (T--) {
      scanf("%d%d", &n, &m), cnt = ans = 0;
      for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
      for (int i = 1; i <= m; i++) scanf("%d", &b[i]);
      sort(b + 1, b + 1 + m), solve(1, m + 1, 1, n + 1);
      for (int i = 1, j = 1; i <= n + 1; i++) {
        while (j <= m && p[j] <= i) res[++cnt] = b[j], j++;
        if (i <= n) res[++cnt] = a[i];
      }
      for (int i = 1; i <= cnt; i++) v[i] = res[i];
      sort(v + 1, v + 1 + cnt), t = unique(v + 1, v + 1 + cnt) - v - 1,
                                fill(c + 1, c + 1 + t, 0);
      for (int i = 1; i <= cnt; i++) {
        res[i] = lower_bound(v + 1, v + 1 + t, res[i]) - v;
        ans += query(res[i] + 1), add(res[i], 1);
      }
      printf("%lld\n", ans);
    }
    return 0;
  }
  