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;
  int const maxn = 1e6 + 5;
  int a[maxn], b[maxn];
  pair<int, int> c[maxn];
  int all_elem[2 * maxn];
  int cur, inf = 1e9 + 7;
  int f[2 * maxn];
  int psh[8 * maxn];
  pair<int, int> imin[4 * maxn];
  inline pair<int, int> unions(pair<int, int> c, pair<int, int> d) {
    if (c.first < d.first || (c.first == d.first && c.second > d.second))
      return c;
    return d;
  }
  void build(int i, int l, int r) {
    psh[i] = 0;
    if (r - l == 1)
      imin[i] = {0, l};
    else {
      int m = (r + l) / 2;
      build(2 * i + 1, l, m);
      build(2 * i + 2, m, r);
      imin[i] = unions(imin[2 * i + 1], imin[2 * i + 2]);
    }
  }
  inline void push(int i) {
    imin[i].first += psh[i];
    psh[2 * i + 1] += psh[i];
    psh[2 * i + 2] += psh[i];
    psh[i] = 0;
  }
  void adds(int i, int l, int r, int lq, int rq, int x) {
    push(i);
    if (lq >= r || l >= rq) return;
    if (lq <= l && r <= rq) {
      psh[i] += x;
      push(i);
      return;
    }
    int m = (r + l) / 2;
    adds(2 * i + 1, l, m, lq, rq, x);
    adds(2 * i + 2, m, r, lq, rq, x);
    imin[i] = unions(imin[2 * i + 1], imin[2 * i + 2]);
  }
  pair<int, int> get(int i, int l, int r, int lq, int rq) {
    push(i);
    if (lq >= r || l >= rq) return {inf, inf};
    if (lq <= l && r <= rq) return imin[i];
    int m = (r + l) / 2;
    return unions(get(2 * i + 1, l, m, lq, rq), get(2 * i + 2, m, r, lq, rq));
  }
  inline void add(int i) {
    while (i <= cur) {
      f[i]++;
      i = (i | (i + 1));
    }
  }
  inline int get(int i) {
    int ans = 0;
    while (i > 0) {
      ans += f[i];
      i = (i & (i + 1)) - 1;
    }
    return ans;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      int n, m;
      cur = 0;
      cin >> n >> m;
      build(0, 0, n + 1);
      for (int i = 1; i <= n; ++i) cin >> a[i];
      for (int i = 1; i <= m; ++i) cin >> b[i];
      sort(b + 1, b + m + 1);
      for (int i = 1; i <= n; ++i) {
        all_elem[++cur] = a[i];
      }
      for (int i = 1; i <= m; ++i) {
        all_elem[++cur] = b[i];
      }
      sort(all_elem + 1, all_elem + cur + 1);
      for (int i = 1; i <= n; ++i) {
        a[i] = lower_bound(all_elem + 1, all_elem + cur + 1, a[i]) - all_elem;
      }
      for (int i = 1; i <= m; ++i) {
        b[i] = lower_bound(all_elem + 1, all_elem + cur + 1, b[i]) - all_elem;
      }
      long long ans = 0;
      for (int i = 0; i <= cur; ++i) f[i] = 0;
      for (int i = n; i >= 1; --i) {
        ans += get(a[i]);
        add(a[i] + 1);
      }
      for (int i = 1; i <= n; ++i) c[i] = {a[i], i};
      sort(c + 1, c + n + 1);
      int pos = n;
      for (int i = 1; i <= n; ++i) {
        adds(0, 0, n + 1, 0, i, 1);
      }
      int lptr = n, rptr = n;
      for (int i = m; i >= 1; --i) {
        if (b[i] != b[i + 1]) {
          while (rptr > lptr) {
            adds(0, 0, n + 1, c[rptr].second, n + 1, 1);
            rptr--;
          }
        }
        while (rptr > 0 && b[i] < c[rptr].first) {
          adds(0, 0, n + 1, 0, c[rptr].second, -1);
          adds(0, 0, n + 1, c[rptr].second, n + 1, 1);
          rptr--;
        }
        lptr = min(lptr, rptr);
        while (lptr > 0 && b[i] == c[lptr].first) {
          adds(0, 0, n + 1, 0, c[lptr].second, -1);
          lptr--;
        }
        pair<int, int> best = get(0, 0, n + 1, 0, pos + 1);
        ans += best.first;
        pos = best.second;
      }
      cout << ans << '\n';
    }
    return 0;
  }
  