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;
  using ll = long long;
  const ll maxn = 2e6 + 10;
  const ll inf = 0x3f3f3f3f;
  ll a[maxn], b[maxn], p[maxn];
  ll n, m;
  map<ll, ll> dic;
  ll bits[maxn];
  ll lowbit(ll x) { return x & (-x); }
  void update(ll x, ll pos) {
    while (pos < maxn) {
      bits[pos] += x;
      pos += lowbit(pos);
    }
  }
  ll query(ll pos) {
    ll res = 0;
    while (pos) {
      res += bits[pos];
      pos -= lowbit(pos);
    }
    return res;
  }
  void insert(ll lb, ll rb, ll la, ll ra) {
    if (lb > rb || la > ra) return;
    ll nowb = b[(lb + rb) / 2];
    ll less = 0;
    for (ll pos = la; pos < ra; ++pos) {
      if (a[pos] < nowb) ++less;
    }
    ll pick = la, ans = less;
    ll now = la;
    while (now < ra) {
      if (a[now] != nowb) {
        if (a[now] < nowb)
          --less;
        else
          ++less;
      }
      ++now;
      if (ans > less) {
        ans = less;
        pick = now;
      }
    }
    p[(lb + rb) / 2] = pick;
    insert(lb, (lb + rb) / 2 - 1, la, pick);
    insert((lb + rb) / 2 + 1, rb, pick, ra);
  }
  void solve() {
    dic.clear();
    cin >> n >> m;
    for (ll i = 1; i <= n; ++i) cin >> a[i];
    for (ll i = 1; i <= m; ++i) cin >> b[i];
    for (ll i = 1; i <= m + n + 2; ++i) bits[i] = 0;
    sort(b + 1, b + m + 1);
    insert(1, m, 1, n + 1);
    vector<ll> discrete;
    for (ll i = 1; i <= n; ++i) {
      discrete.emplace_back(a[i]);
    }
    for (ll i = 1; i <= m; ++i) {
      discrete.emplace_back(b[i]);
    }
    sort(discrete.begin(), discrete.end());
    discrete.erase(unique(discrete.begin(), discrete.end()), discrete.end());
    for (ll i = 0; i < discrete.size(); ++i) {
      dic[discrete[i]] = i + 1;
    }
    vector<ll> c;
    ll j = 1;
    for (ll i = 1; i <= n; ++i) {
      while (j <= m && p[j] == i) c.push_back(b[j]), ++j;
      c.push_back(a[i]);
    }
    while (j <= m) c.push_back(b[j]), ++j;
    ll ans = 0;
    for (ll i = 0; i < c.size(); ++i) {
      ll rk = dic[c[i]];
      ans += i - query(rk);
      update(1, rk);
    }
    cout << ans << '\n';
  }
  signed main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    ll t;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  