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 M = 2e6 + 5;
  int n, m, tmp2[M];
  int a[M], b[M], c[M], pos[M];
  void mymid(int ta[], int tb[], int tpos[], int l, int r, int L, int R) {
    if (L > R) return;
    int mid = (L + R) / 2;
    int now = 0, po, mmin;
    for (int i = l; i < r; i++)
      if (ta[i] < tb[mid]) now++;
    mmin = now, po = l;
    for (int i = l + 1; i <= r; i++) {
      if (ta[i - 1] < tb[mid]) now--;
      if (ta[i - 1] > tb[mid]) now++;
      if (now < mmin) mmin = now, po = i;
    }
    tpos[mid] = po;
    mymid(ta, tb, tpos, l, po, L, mid - 1);
    mymid(ta, tb, tpos, po, r, mid + 1, R);
  }
  void print(int tmp[], int len) {
    for (int i = 1; i <= len; i++) printf("%d ", tmp[i]);
    printf("\n");
  }
  long long mysort(int tmp[], int l, int r) {
    if (l >= r) return 0;
    long long mid = (l + r) / 2, res = 0;
    res += mysort(tmp, l, mid);
    res += mysort(tmp, mid + 1, r);
    long long t1 = l, t2 = mid + 1, t3 = 1;
    while (t1 <= mid && t2 <= r) {
      if (tmp[t1] <= tmp[t2])
        tmp2[t3++] = tmp[t1++], res += t2 - mid - 1;
      else
        tmp2[t3++] = tmp[t2++];
    }
    while (t1 <= mid) tmp2[t3++] = tmp[t1++], res += t2 - mid - 1;
    while (t2 <= r) tmp2[t3++] = tmp[t2++];
    for (int i = l; i <= r; i++) tmp[i] = tmp2[i - l + 1];
    return res;
  }
  int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
      scanf("%d%d", &n, &m);
      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);
      mymid(a, b, pos, 1, n + 1, 1, m);
      int t1 = 1, t2 = 1, t3 = 1;
      while (t1 <= n + m) {
        while (pos[t2] <= t3 && t2 <= m) c[t1++] = b[t2++];
        if (t1 > n + m) break;
        c[t1++] = a[t3++];
      }
      printf("%lld\n", mysort(c, 1, n + m));
    }
    return 0;
  }
  