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 = 1e6 + 10;
  int n, m, tot, a[MAXN], b[MAXN], d[MAXN];
  map<int, int> to;
  struct Node {
    int val, pos;
    bool operator<(const Node &AI) const { return val < AI.val; }
  } c[MAXN];
  struct Tree {
    int sum[MAXN * 4], lmin[MAXN * 4];
    void pushup(int rt) {
      sum[rt] = sum[(rt << 1)] + sum[(rt << 1 | 1)];
      lmin[rt] = min(lmin[(rt << 1)], sum[(rt << 1)] + lmin[(rt << 1 | 1)]);
    }
    void build(int rt, int l, int r) {
      if (l == r) {
        sum[rt] = lmin[rt] = 1;
        return;
      }
      int mid = (l + r) >> 1;
      build((rt << 1), l, mid);
      build((rt << 1 | 1), mid + 1, r);
      pushup(rt);
    }
    void modify(int rt, int l, int r, int p, int k) {
      if (l == r) {
        sum[rt] = lmin[rt] = k;
        return;
      }
      int mid = (l + r) >> 1;
      if (p <= mid)
        modify((rt << 1), l, mid, p, k);
      else
        modify((rt << 1 | 1), mid + 1, r, p, k);
      pushup(rt);
    }
  } T;
  int lowbit(int x) { return x & -x; }
  int query(int p) {
    int ret = 0;
    while (p > 0) ret += d[p], p -= lowbit(p);
    return ret;
  }
  void modify(int p) {
    while (p <= tot) d[p]++, p += lowbit(p);
  }
  int main() {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
      scanf("%d %d", &n, &m);
      for (int i = 1; i <= n; ++i) {
        scanf("%d", &a[i]), d[i] = a[i];
        c[i] = (Node){a[i], i};
      }
      long long ans = 0;
      sort(d + 1, d + n + 1);
      tot = unique(d + 1, d + n + 1) - d - 1;
      for (int i = 1; i <= tot; ++i) to[d[i]] = i, d[i] = 0;
      for (int i = 1; i <= n; ++i) {
        ans += i - 1 - query(to[a[i]]);
        modify(to[a[i]]);
      }
      for (int i = 1; i <= m; ++i) scanf("%d", &b[i]);
      sort(a + 1, a + n + 1);
      sort(b + 1, b + m + 1);
      sort(c + 1, c + n + 1);
      int cur = 1;
      T.build(1, 1, n);
      for (int i = 1; i <= m; ++i) {
        long long base_ans = lower_bound(a + 1, a + n + 1, b[i]) - a - 1;
        while (cur <= n && c[cur].val < b[i]) {
          T.modify(1, 1, n, c[cur].pos, -1), cur++;
        }
        for (int j = cur; j <= n; ++j) {
          if (c[j].val != b[i]) break;
          T.modify(1, 1, n, c[j].pos, 0);
        }
        if (T.lmin[1] < 0) base_ans += T.lmin[1];
        int tag = b[i];
        while (i <= m && b[i] == tag) {
          ans += base_ans;
          i++;
        }
        i--;
        while (cur <= n && c[cur].val == b[i]) {
          T.modify(1, 1, n, c[cur].pos, -1), cur++;
        }
      }
      printf("%lld\n", ans);
    }
    return 0;
  }
  