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 + 5;
  namespace SGT {
  int tag[N << 2], mn[N << 2];
  inline void up(int rt) { mn[rt] = min(mn[(rt << 1)], mn[(rt << 1 | 1)]); }
  inline void cov(int rt, int v) { mn[rt] += v, tag[rt] += v; }
  inline void dn(int rt) {
    if (!tag[rt]) return;
    cov((rt << 1), tag[rt]), cov((rt << 1 | 1), tag[rt]);
    tag[rt] = 0;
  }
  void build(int rt, int l, int r) {
    tag[rt] = 0;
    if (l == r) return mn[rt] = l, void();
    int mid = (l + r) >> 1;
    build((rt << 1), l, mid), build((rt << 1 | 1), mid + 1, r);
    up(rt);
  }
  void update(int rt, int l, int r, int ql, int qr, int v) {
    if (ql <= l && r <= qr) return cov(rt, v);
    dn(rt);
    int mid = (l + r) >> 1;
    if (ql <= mid) update((rt << 1), l, mid, ql, qr, v);
    if (mid < qr) update((rt << 1 | 1), mid + 1, r, ql, qr, v);
    up(rt);
  }
  }  // namespace SGT
  pair<int, int> a[N];
  int n, m, b[N], tmp[N];
  namespace Fenwick {
  int c[N];
  inline void clear() { fill(c + 1, c + n + 1, 0); }
  inline void upd(int x) {
    for (; x; x -= (x & (-x))) ++c[x];
  }
  inline int ask(int x) {
    int res = 0;
    for (; x <= n; x += (x & (-x))) res += c[x];
    return res;
  }
  }  // namespace Fenwick
  using namespace Fenwick;
  int main() {
    int T;
    scanf("%d", &T);
    while (T-- > 0) {
      scanf("%d%d", &n, &m);
      long long ans = 0;
      for (int i = 1; i <= n; ++i)
        scanf("%d", tmp + i), a[i] = make_pair(tmp[i], i);
      sort(tmp + 1, tmp + n + 1);
      int tot = unique(tmp + 1, tmp + n + 1) - tmp - 1;
      for (int i = 1; i <= n; ++i) {
        int now = lower_bound(tmp + 1, tmp + tot + 1, a[i].first) - tmp;
        ans += ask(now + 1);
        upd(now);
      }
      clear();
      sort(a + 1, a + n + 1);
      for (int i = 1; i <= m; ++i) scanf("%d", b + i);
      sort(b + 1, b + m + 1);
      SGT::build(1, 0, n);
      int p1 = 1, p2 = 1;
      for (int i = 1; i <= m; ++i) {
        while (p1 <= n && a[p1].first < b[i])
          SGT::update(1, 0, n, 0, a[p1++].second - 1, 1);
        while (p2 <= n && a[p2].first <= b[i])
          SGT::update(1, 0, n, a[p2++].second, n, -1);
        ans += SGT::mn[1];
      }
      printf("%lld\n", ans);
    }
    return 0;
  }
  