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 N = 1e6;
  int n, m, a[N + 10], b[N + 10], p[N + 10];
  int mi[N * 4 + 10], add[N * 4 + 10];
  long long ans;
  void pushdown(int p) {
    add[p * 2] += add[p], add[p * 2 + 1] += add[p];
    mi[p * 2] += add[p], mi[p * 2 + 1] += add[p];
    add[p] = 0;
  }
  void build(int p, int l, int r) {
    add[p] = 0;
    if (l == r) {
      mi[p] = l;
      return;
    }
    int mid = (l + r) >> 1;
    build(p * 2, l, mid);
    build(p * 2 + 1, mid + 1, r);
    mi[p] = min(mi[p * 2], mi[p * 2 + 1]);
  }
  void update(int p, int l, int r, int x, int y, int v) {
    if (x <= l && r <= y) {
      mi[p] += v, add[p] += v;
      return;
    }
    pushdown(p);
    int mid = (l + r) >> 1;
    if (x <= mid) update(p * 2, l, mid, x, y, v);
    if (y > mid) update(p * 2 + 1, mid + 1, r, x, y, v);
    mi[p] = min(mi[p * 2], mi[p * 2 + 1]);
  }
  bool cmp(int x, int y) { return a[x] < a[y]; }
  int bit[N + 10], lsh[N + 10];
  void Add(int x) {
    for (; x <= n; x += (x & -x)) bit[x]++;
  }
  int Ask(int x) {
    int ans = 0;
    for (; x; x -= (x & -x)) ans += bit[x];
    return ans;
  }
  void solve() {
    cin >> n >> m;
    ans = 0;
    for (int i = 1; i <= n; i++) bit[i] = 0;
    for (int i = 1; i <= n; i++) cin >> a[i], p[i] = i, lsh[i] = a[i];
    sort(lsh + 1, lsh + n + 1);
    int tot = unique(lsh + 1, lsh + n + 1) - lsh - 1;
    for (int i = 1; i <= n; i++) {
      int x = lower_bound(lsh + 1, lsh + tot + 1, a[i]) - lsh;
      ans += i - 1 - Ask(x);
      Add(x);
    }
    for (int j = 1; j <= m; j++) cin >> b[j];
    build(1, 0, n);
    sort(b + 1, b + m + 1);
    sort(p + 1, p + n + 1, cmp);
    for (int i = 1, j = 1, k = 1; i <= m; i++) {
      while (j <= n && a[p[j]] <= b[i]) update(1, 0, n, p[j++], n, -1);
      while (k <= n && a[p[k]] < b[i]) update(1, 0, n, 0, p[k++] - 1, 1);
      ans += mi[1];
    }
    printf("%lld\n", ans);
  }
  int main() {
    std::ios::sync_with_stdio(false);
    int t;
    cin >> t;
    while (t--) solve();
  }
  