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 MAXN = 1000010;
  int n, m;
  int a[MAXN];
  int b[MAXN];
  int p[MAXN];
  void solve(int l, int r, int L, int R) {
    if (l > r) return;
    int mid = (l + r) / 2;
    int tmp = 0;
    for (int i = L + 1; i <= R; ++i)
      if (b[mid] > a[i]) tmp++;
    int minn = tmp, pos = L;
    for (int i = L + 1; i <= R; ++i) {
      if (b[mid] > a[i]) tmp--;
      if (b[mid] < a[i]) tmp++;
      if (tmp < minn) {
        minn = tmp, pos = i;
      }
    }
    p[mid] = pos;
    solve(l, mid - 1, L, pos);
    solve(mid + 1, r, pos, R);
  }
  struct BIT {
    int len;
    int c[MAXN << 1];
    inline int lowbit(int x) { return x & (-x); }
    void clr(int l) {
      len = l;
      for (int i = 1; i <= len; ++i) c[i] = 0;
    }
    inline void upd(int x, int v) {
      while (x <= len) {
        c[x] += v;
        x += lowbit(x);
      }
    }
    inline int qry(int x) {
      int ret = 0;
      while (x) {
        ret += c[x];
        x -= lowbit(x);
      }
      return ret;
    }
  } T;
  struct node {
    int pos, val;
    node(int pp = 0, int vv = 0) { pos = pp, val = vv; }
  } t[MAXN << 1];
  int tot;
  bool operator<(node a, node b) {
    return a.val == b.val ? a.pos > b.pos : a.val > b.val;
  }
  int main() {
    int testcase;
    scanf("%d", &testcase);
    while (testcase--) {
      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);
      solve(1, m, 0, n);
      T.clr(n + m);
      int now = 1;
      tot = 0;
      for (int i = 1; i <= n; ++i) {
        while (p[now] < i && now <= m) tot++, t[tot] = node(tot, b[now]), now++;
        tot++, t[tot] = node(tot, a[i]);
      }
      while (now <= m) tot++, t[tot] = node(tot, b[now]), now++;
      sort(t + 1, t + 1 + tot);
      long long ans = 0;
      for (int i = 1; i <= tot; ++i) {
        ans += T.qry(t[i].pos);
        T.upd(t[i].pos, 1);
      }
      printf("%lld\n", ans);
    }
  }
  