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 n, m, a[N], b[N], t, ans[N], c[2 * N], eq[2 * N];
  long long bt[2 * N];
  pair<int, int> d[2 * N];
  void ded(int l1, int r1, int l2, int r2) {
    if (l2 > r2) return;
    int mid = (l2 + r2) / 2;
    int pb = l1;
    long long cl = 0, cg = 0, claux = 0, cgaux = 0;
    for (int i = l1; i < r1; i++) cgaux += (a[i] < b[mid]);
    cg = cgaux;
    for (int i = l1 + 1; i <= r1; i++) {
      if (a[i - 1] < b[mid]) cgaux--;
      if (a[i - 1] > b[mid]) claux++;
      if (cgaux + claux < cg + cl) {
        pb = i;
        cg = cgaux;
        cl = claux;
      }
    }
    ans[mid] = pb;
    ded(l1, pb, l2, mid - 1);
    ded(pb, r1, mid + 1, r2);
  }
  long long get(int p) {
    long long r = 0;
    while (p) {
      r += bt[p];
      p -= p & (-p);
    }
    return r;
  }
  void ins(int p) {
    while (p <= n + m) {
      bt[p]++;
      p += p & (-p);
    }
  }
  long long cnt() {
    long long r = 0;
    for (int i = 0; i < n + m; i++) d[i] = {c[i], i};
    sort(d, d + n + m);
    for (int i = 0; i < n + m; i++) eq[d[i].second] = i + 1;
    for (int i = 0; i < n + m; i++) {
      r += i - get(eq[i]);
      ins(eq[i]);
    }
    return r;
  }
  int main() {
    scanf("%d", &t);
    while (t--) {
      scanf("%d %d", &n, &m);
      for (int i = 0; i < n; i++) scanf("%d", &a[i]);
      for (int i = 0; i <= n + m; i++) bt[i] = 0;
      for (int i = 0; i < m; i++) scanf("%d", &b[i]);
      sort(b, b + m);
      ded(0, n, 0, m - 1);
      int l = 0;
      for (int i = 0; i < n; i++) {
        while (l < m && ans[l] == i) {
          c[i + l] = b[l];
          l++;
        }
        c[i + l] = a[i];
      }
      while (l < m) {
        c[n + l] = b[l];
        l++;
      }
      printf("%lld\n", cnt());
    }
    return 0;
  }
  