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;
  int T, n, m, a[N], b[N], v[N], t, cnt, tag[N << 2], mn[N << 2], c[N];
  long long ans;
  struct node {
    int x, l, r, op;
  } p[N << 1];
  void add(int x, int y) {
    for (int i = x; i; i -= i & (-i)) c[i] += y;
  }
  int query(int x) {
    int ans = 0;
    for (int i = x; i <= t; i += i & (-i)) ans += c[i];
    return ans;
  }
  void build(int p, int l, int r) {
    tag[p] = 0;
    if (l == r) {
      mn[p] = l - 1;
      return;
    }
    int mid = (l + r) / 2;
    build(p << 1, l, mid);
    build(p << 1 | 1, mid + 1, r);
    mn[p] = min(mn[p << 1], mn[p << 1 | 1]);
  }
  void pushdown(int p) {
    if (!tag[p]) return;
    tag[p << 1] += tag[p], tag[p << 1 | 1] += tag[p], mn[p << 1] += tag[p],
        mn[p << 1 | 1] += tag[p];
    tag[p] = 0;
  }
  void modify(int p, int l, int r, int lx, int rx, int v) {
    if (l >= lx && r <= rx) {
      mn[p] += v, tag[p] += v;
      return;
    }
    int mid = (l + r) / 2;
    pushdown(p);
    if (lx <= mid) modify(p << 1, l, mid, lx, rx, v);
    if (rx > mid) modify(p << 1 | 1, mid + 1, r, lx, rx, v);
    mn[p] = min(mn[p << 1], mn[p << 1 | 1]);
  }
  signed main() {
    scanf("%d", &T);
    while (T--) {
      scanf("%d%d", &n, &m), cnt = ans = 0;
      for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]), v[i] = a[i];
        p[++cnt] = {a[i], i + 1, n + 1, -1}, p[++cnt] = {a[i] + 1, 1, i, 1};
      }
      for (int i = 1; i <= m; i++) scanf("%d", &b[i]);
      sort(v + 1, v + 1 + n), t = unique(v + 1, v + 1 + n) - v - 1,
                              fill(c + 1, c + 1 + t, 0);
      for (int i = 1; i <= n; i++) {
        a[i] = lower_bound(v + 1, v + 1 + t, a[i]) - v;
        ans += query(a[i] + 1), add(a[i], 1);
      }
      sort(b + 1, b + 1 + m),
          sort(p + 1, p + 1 + cnt, [](node x, node y) { return x.x < y.x; }),
          build(1, 1, n + 1);
      for (int i = 1, j = 1; i <= m; i++) {
        while (j <= cnt && p[j].x <= b[i])
          modify(1, 1, n + 1, p[j].l, p[j].r, p[j].op), j++;
        ans += mn[1];
      }
      printf("%lld\n", ans);
    }
    return 0;
  }
  