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;
  inline int read() {
    int w = 0, f = 1;
    char ch = getchar();
    while (ch < '0' or ch > '9') {
      if (ch == '-') f = -f;
      ch = getchar();
    }
    while (ch >= '0' and ch <= '9') w = w * 10 + ch - '0', ch = getchar();
    return w * f;
  }
  const int maxn = 1e6 + 5, INF = 1e9 + 7;
  int n, m, a[maxn], b[maxn], p[maxn], sum1[maxn], sum2[maxn], uni[maxn << 1],
      cnt, t[maxn << 1];
  vector<int> c[maxn];
  void solve(int l1, int r1, int l2, int r2) {
    if (l1 > r1) return;
    int mid = (l1 + r1) >> 1;
    sum1[l2 - 1] = sum2[l2 - 1] = 0;
    for (int i = l2; i <= r2; i++) {
      sum1[i] = sum2[i] = 0;
      if (a[i] > b[mid]) sum1[i]++;
      if (a[i] < b[mid]) sum2[i]++;
    }
    for (int i = l2 + 1; i <= r2; i++) sum1[i] += sum1[i - 1];
    for (int i = r2 - 1; i >= l2; i--) sum2[i] += sum2[i + 1];
    p[mid] = 0;
    for (int i = l2; i <= r2; i++)
      if (sum1[i - 1] + sum2[i] < sum1[p[mid] - 1] + sum2[p[mid]] or p[mid] == 0)
        p[mid] = i;
    solve(l1, mid - 1, l2, p[mid]), solve(mid + 1, r1, p[mid], r2);
  }
  int Find(int x) { return lower_bound(uni + 1, uni + cnt + 1, x) - uni; }
  void add(int x, int y) {
    for (; x <= cnt; x += x & -x) t[x] += y;
  }
  int ask(int x) {
    int ans = 0;
    for (; x; x -= x & -x) ans += t[x];
    return ans;
  }
  int main() {
    int _ = read();
    while (_--) {
      n = read(), m = read();
      cnt = 0;
      for (int i = 1; i <= n; i++) a[i] = read(), uni[++cnt] = a[i], c[i].clear();
      for (int i = 1; i <= m; i++) b[i] = read(), uni[++cnt] = b[i];
      sort(b + 1, b + m + 1);
      sort(uni + 1, uni + cnt + 1);
      cnt = unique(uni + 1, uni + cnt + 1) - uni - 1;
      solve(1, m, 1, n + 1);
      int tot = m;
      for (int i = n + 1; i >= 0; i--) {
        c[i].clear();
        if (i <= n) c[i].emplace_back(Find(a[i]));
        while (p[tot] == i and tot) c[i].emplace_back(Find(b[tot--]));
      }
      for (int i = 0; i <= cnt; i++) t[i] = 0;
      long long ans = 0;
      for (int i = n + 1; i >= 1; i--)
        for (auto it : c[i]) {
          ans += ask(it - 1);
          add(it, 1);
        }
      printf("%lld\n", ans);
    }
    return 0;
  }
  