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 n, m, a[2000100], b[2000100], c[2000100], len, p[2000100];
  long long ans;
  class BIT {
    int b[2000100];
    inline int lowbit(int x) { return x & (-x); }
  
   public:
    void Clear(int n) {
      for (int i = 1; i <= len; ++i) {
        b[i] = 0;
      }
    }
    inline void Add(int x, int d) {
      while (x <= len) b[x] += d, x += lowbit(x);
    }
    inline int Ask(int x) {
      int ans = 0;
      while (x) ans += b[x], x -= lowbit(x);
      return ans;
    }
  } B;
  class Segment_Tree {
    struct node {
      int mx, add;
      inline void Add(int d) { add += d, mx += d; }
    } t[2000100 << 2];
    inline void pushdown(int u) {
      if (t[u].add) {
        t[u << 1].Add(t[u].add);
        t[u << 1 | 1].Add(t[u].add);
        t[u].add = 0;
      }
    }
    inline void update(int u) { t[u].mx = max(t[u << 1].mx, t[u << 1 | 1].mx); }
  
   public:
    void build(int u, int L, int R) {
      t[u].mx = t[u].add = 0;
      if (L == R) {
        t[u].mx = n - L;
        return;
      }
      int mid = (L + R) >> 1;
      build(u << 1, L, mid);
      build(u << 1 | 1, mid + 1, R);
      update(u);
    }
    void Plus(int u, int L, int R, int l, int r, int d) {
      if (L >= l && R <= r) {
        t[u].Add(d);
        return;
      }
      pushdown(u);
      int mid = (L + R) >> 1;
      if (l <= mid) Plus(u << 1, L, mid, l, r, d);
      if (r > mid) Plus(u << 1 | 1, mid + 1, R, l, r, d);
      update(u);
    }
    int get_ans() { return t[1].mx; }
  } T;
  void Solve() {
    cin >> n >> m;
    len = 0;
    for (int i = 1; i <= n; ++i) {
      cin >> a[i];
      c[++len] = a[i];
      p[i] = i;
    }
    for (int i = 1; i <= m; ++i) {
      cin >> b[i];
      c[++len] = b[i];
    }
    sort(c + 1, c + len + 1);
    len = unique(c + 1, c + len + 1) - c - 1;
    for (int i = 1; i <= n; ++i) {
      a[i] = lower_bound(c + 1, c + len + 1, a[i]) - c;
    }
    for (int i = 1; i <= m; ++i) {
      b[i] = lower_bound(c + 1, c + len + 1, b[i]) - c;
    }
    ans = 1LL * m * (m - 1) / 2;
    B.Clear(len);
    for (int i = 1; i <= n; ++i) {
      ans += B.Ask(a[i]);
      B.Add(a[i], 1);
    }
    sort(b + 1, b + m + 1);
    sort(p + 1, p + n + 1, [&](int i, int j) { return a[i] < a[j]; });
    T.build(1, 0, n);
    for (int i = 1, j = 1, k = 1; i <= m; ++i) {
      while (j <= n && a[p[j]] <= b[i]) T.Plus(1, 0, n, p[j], n, 1), ++j;
      while (k <= n && a[p[k]] < b[i]) T.Plus(1, 0, n, 0, p[k] - 1, -1), ++k;
      ans += T.get_ans();
    }
    cout << 1LL * (n + m) * (n + m - 1) / 2 - ans << '\n';
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin >> T;
    while (T--) Solve();
    return 0;
  }
  