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 + 10;
  int T, n, m, a[N], b[N], st[N << 1], top;
  vector<int> head[N << 1];
  map<int, int> M;
  struct Bit {
    int tr[N << 1], lim;
    void modify(int x, int y) {
      for (; x <= lim; x += x & -x) tr[x] += y;
    }
    int query(int x) {
      int ret = 0;
      for (; x; x -= x & -x) ret += tr[x];
      return ret;
    }
  } B;
  struct Seg {
    int tag, Min;
  } tr[N << 2];
  void upd(int p) { tr[p].Min = min(tr[p << 1].Min, tr[p << 1 | 1].Min); }
  void build(int p, int l, int r) {
    tr[p].tag = 0;
    if (l == r) {
      tr[p].Min = l;
      return;
    }
    int mid = (l + r) >> 1;
    build(p << 1, l, mid);
    build(p << 1 | 1, mid + 1, r);
    upd(p);
  }
  void f(int p, int k) {
    tr[p].Min += k;
    tr[p].tag += k;
  }
  void pushdown(int p, int l, int r) {
    int mid = (l + r) >> 1;
    if (tr[p].tag) {
      f(p << 1, tr[p].tag);
      f(p << 1 | 1, tr[p].tag);
      tr[p].tag = 0;
    }
  }
  void modify(int p, int l, int r, int nl, int nr, int k) {
    if (l == nl && r == nr) {
      f(p, k);
      return;
    }
    int mid = (l + r) >> 1;
    pushdown(p, l, r);
    if (nr <= mid)
      modify(p << 1, l, mid, nl, nr, k);
    else if (nl > mid)
      modify(p << 1 | 1, mid + 1, r, nl, nr, k);
    else
      modify(p << 1, l, mid, nl, mid, k),
          modify(p << 1 | 1, mid + 1, r, mid + 1, nr, k);
    upd(p);
  }
  int main() {
    ios::sync_with_stdio(0);
    cin >> T;
    while (T--) {
      cin >> n >> m;
      top = 0;
      for (int i = 1; i <= n; ++i) cin >> a[i], st[++top] = a[i];
      for (int i = 1; i <= m; ++i) cin >> b[i], st[++top] = b[i];
      sort(b + 1, b + 1 + m);
      sort(st + 1, st + 1 + top);
      int num = unique(st + 1, st + 1 + top) - st - 1;
      for (int i = 1; i <= num; ++i) M[st[i]] = i;
      for (int i = 1; i <= n; ++i) a[i] = M[a[i]];
      for (int i = 1; i <= m; ++i) b[i] = M[b[i]];
      B.lim = num;
      long long ans = 0;
      for (int i = 1; i <= n; ++i) {
        ans += B.query(num) - B.query(a[i]);
        B.modify(a[i], 1);
      }
      build(1, 1, n);
      for (int i = 1; i <= n; ++i) head[a[i]].push_back(i);
      int now = 0;
      for (int i = 1; i <= m; ++i) {
        while (now < b[i]) {
          for (int i = 0; i < head[now].size(); ++i)
            modify(1, 1, n, head[now][i], n, -1);
          ++now;
          for (int i = 0; i < head[now].size(); ++i)
            modify(1, 1, n, head[now][i], n, -1);
        }
        int tmp = B.query(b[i] - 1);
        ans += B.query(b[i] - 1);
        int Min = min(tr[1].Min, 0);
        ans += Min;
      }
      cout << ans << endl;
      for (int i = 0; i <= num; ++i) head[i].clear(), B.tr[i] = 0;
    }
    return 0;
  }
  