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;
  struct STree {
    vector<int> st, lazy;
    int n;
    STree(int n) : st(4 * n + 5, 0), lazy(4 * n + 5, 0), n(n) {}
    void push(int k, int s, int e) {
      if (!lazy[k]) return;
      st[k] += lazy[k];
      if (s + 1 < e) {
        lazy[2 * k] += lazy[k];
        lazy[2 * k + 1] += lazy[k];
      }
      lazy[k] = 0;
    }
    void upd(int k, int s, int e, int a, int b, int v) {
      push(k, s, e);
      if (s >= b || e <= a) return;
      if (s >= a && e <= b) {
        lazy[k] += v;
        push(k, s, e);
        return;
      }
      int m = (s + e) / 2;
      upd(2 * k, s, m, a, b, v);
      upd(2 * k + 1, m, e, a, b, v);
      st[k] = min(st[2 * k], st[2 * k + 1]);
    }
    int query(int k, int s, int e, int a, int b) {
      if (s >= b || e <= a) return 1e9;
      push(k, s, e);
      if (s >= a && e <= b) return st[k];
      int m = (s + e) / 2;
      return min(query(2 * k, s, m, a, b), query(2 * k + 1, m, e, a, b));
    }
    void upd(int a, int b, int v) { upd(1, 0, n, a, b, v); }
    int query(int a, int b) { return query(1, 0, n, a, b); }
  };
  vector<long long> oc[2000006];
  long long cant = 0;
  long long n, m;
  void upd(long long x, STree &st) {
    for (auto i : oc[x - 1]) {
      st.upd(i, n, -1);
      cant++;
    }
    for (auto i : oc[x]) {
      st.upd(i, n, -1);
    }
  }
  struct STrees {
    vector<int> t;
    int n;
    STrees(int n) : t(2 * n + 5, 0), n(n) {}
    void upd(int p, int value) {
      for (t[p += n] += value; p > 1; p >>= 1) t[p >> 1] = t[p] + t[p ^ 1];
    }
    int query(int l, int r) {
      int res = 0;
      for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
        if (l & 1) res = res + t[l++];
        if (r & 1) res = res + t[--r];
      }
      return res;
    }
  };
  STrees str(2000009);
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t;
    cin >> t;
    while (t--) {
      cin >> n >> m;
      vector<long long> a(n), b(m);
      for (int i = 0, ggdem = n; i < ggdem; ++i) cin >> a[i];
      for (int i = 0, ggdem = m; i < ggdem; ++i) cin >> b[i];
      vector<long long> v = {0};
      for (auto i : a) v.push_back(i);
      for (auto i : b) v.push_back(i);
      sort(v.begin(), v.end());
      v.erase(unique(v.begin(), v.end()), v.end());
      for (int i = 0, ggdem = n; i < ggdem; ++i)
        a[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin();
      for (int i = 0, ggdem = m; i < ggdem; ++i)
        b[i] = lower_bound(v.begin(), v.end(), b[i]) - v.begin();
      sort(b.begin(), b.end());
      for (int i = 0, ggdem = ((int)v.size()) + 1; i < ggdem; ++i) oc[i].clear();
      for (int i = 0, ggdem = n; i < ggdem; ++i) oc[a[i]].push_back(i);
      STree st(n + 10);
      for (int i = 0, ggdem = n; i < ggdem; ++i) st.upd(i, i + 1, i + 1);
      long long ul = 0, res = 0;
      vector<pair<long long, long long>> ao;
      for (int i = 0, ggdem = n; i < ggdem; ++i) ao.push_back({a[i], i});
      sort(ao.begin(), ao.end());
      for (auto i : ao) {
        res += str.query(i.second, 2000009);
        str.upd(i.second, 1);
      }
      for (auto i : ao) str.upd(i.second, -1);
      cant = 0;
      for (auto i : b) {
        while (ul < i) {
          ul++;
          upd(ul, st);
        }
        res += cant + min(0, st.query(0, n));
      }
      cout << res << "\n";
    }
    return 0;
  }
  