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 = 1.2e6 + 10;
  const int INF = 2e9;
  const double eps = 1e-6;
  const double PI = acos(-1);
  int T, n, m;
  int mn[MAXN << 2], tag[MAXN << 2], BIT[MAXN];
  long long Ans;
  pair<int, int> b[MAXN], a[MAXN];
  vector<int> v;
  template <typename T>
  inline bool read(T &a) {
    a = 0;
    char c = getchar();
    int f = 1;
    while (c < '0' || c > '9') {
      if (c == '-') f = -1;
      c = getchar();
    }
    while (c >= '0' && c <= '9') {
      a = a * 10 + (c ^ 48);
      c = getchar();
    }
    return a *= f, true;
  }
  template <typename A, typename... B>
  inline bool read(A &x, B &...y) {
    return read(x) && read(y...);
  }
  void pushup(int x) { mn[x] = min(mn[x << 1], mn[x << 1 | 1]); }
  void push(int x, int v) {
    tag[x] += v;
    mn[x] += v;
  }
  void pushdown(int x) {
    if (tag[x]) push(x << 1, tag[x]), push(x << 1 | 1, tag[x]), tag[x] = 0;
  }
  void update(int x, int l, int r, int L, int R, int v) {
    if (L <= l && r <= R) return push(x, v), void();
    int mid = (l + r) >> 1;
    pushdown(x);
    if (L <= mid) update(x << 1, l, mid, L, R, v);
    if (R > mid) update(x << 1 | 1, mid + 1, r, L, R, v);
    pushup(x);
  }
  void add(int x, int v) {
    for (; x <= n; x += ((x) & (-(x)))) BIT[x] += v;
  }
  int ask(int x) {
    int res = 0;
    for (; x; x -= ((x) & (-(x)))) res += BIT[x];
    return res;
  }
  void build(int x, int l, int r) {
    mn[x] = tag[x] = 0;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(x << 1, l, mid), build(x << 1 | 1, mid + 1, r);
  }
  void solve() {
    read(n), read(m);
    Ans = 0;
    build(1, 1, n + 1);
    v.clear();
    memset(BIT, 0, sizeof(int) * (n + 5));
    for (int i = 1; i <= n; ++i)
      read(a[i].first), a[i].second = i, v.push_back(a[i].first);
    for (int i = 1; i <= m; ++i) read(b[i].first), b[i].second = i;
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
    for (int i = n, x; i >= 1; --i) {
      x = lower_bound(v.begin(), v.end(), a[i].first) - v.begin() + 1;
      Ans += ask(x - 1);
      add(x, 1);
    }
    sort(a + 1, a + n + 1);
    sort(b + 1, b + m + 1);
    for (int i = 1; i <= n; ++i) update(1, 1, n + 1, i + 1, n + 1, 1);
    for (int L = 1, R, l = 1, r; L <= m; L = R + 1) {
      R = L, r = l;
      while (r <= n && a[r].first < b[L].first) {
        update(1, 1, n + 1, a[r].second + 1, n + 1, -1);
        update(1, 1, n + 1, 1, a[r].second, 1);
        ++r;
      }
      l = r;
      while (r <= n && a[r].first == b[L].first) {
        update(1, 1, n + 1, a[r].second + 1, n + 1, -1);
        ++r;
      }
      while (R < m && b[R + 1].first == b[L].first) R++;
      Ans += 1ll * (R - L + 1) * mn[1];
      for (int i = l; i < r; ++i) update(1, 1, n + 1, 1, a[i].second, 1);
      l = r;
    }
    printf("%lld\n", Ans);
  }
  signed main() {
    read(T);
    while (T--) solve();
    return 0;
  }
  