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 = 2e6 + 10;
  struct UPD {
    int a, pos;
  } sta[maxn];
  int top = 0;
  int n, m;
  int a[maxn], b[maxn], t[maxn << 1];
  int mn[maxn << 2], lz[maxn << 2];
  inline void pushup(int id) { mn[id] = min(mn[id << 1], mn[id << 1 | 1]); }
  inline void change(int id, int v) {
    lz[id] += v;
    mn[id] += v;
  }
  inline void pushdown(int id) {
    if (lz[id]) {
      change(id << 1, lz[id]);
      change(id << 1 | 1, lz[id]);
      lz[id] = 0;
    }
  }
  inline void update(int id, int l, int r, int x, int y, int v) {
    if (x <= l && r <= y) {
      change(id, v);
      return;
    }
    pushdown(id);
    int mid = (l + r) >> 1;
    if (x <= mid) update(id << 1, l, mid, x, y, v);
    if (y > mid) update(id << 1 | 1, mid + 1, r, x, y, v);
    pushup(id);
  }
  bool cmp(UPD A, UPD B) { return A.a < B.a; }
  int mm;
  inline void init() {
    scanf("%d%d", &n, &m);
    int cnt = 0;
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]), t[++cnt] = a[i];
    for (int i = 1; i <= m; i++) scanf("%d", &b[i]), t[++cnt] = b[i];
    sort(t + 1, t + cnt + 1);
    mm = unique(t + 1, t + cnt + 1) - t - 1;
    for (int i = 1; i <= n; i++) a[i] = lower_bound(t + 1, t + mm + 1, a[i]) - t;
    for (int i = 1; i <= m; i++) b[i] = lower_bound(t + 1, t + mm + 1, b[i]) - t;
    top = 0;
    for (int i = 1; i <= n; i++) sta[++top].a = a[i], sta[top].pos = i;
    sort(sta + 1, sta + top + 1, cmp);
  }
  long long c[maxn << 1];
  inline void upd(int x, int v) {
    for (int i = x; i <= mm + 1; i += (i & (-i))) c[i] += v;
  }
  inline long long qur(int x) {
    long long res = 0;
    for (int i = x; i > 0; i -= (i & (-i))) res += c[i];
    return res;
  }
  inline long long init_ans(int mm) {
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      ans += i - 1 - qur(a[i]);
      upd(a[i], 1);
    }
    for (int i = 1; i <= n; i++) {
      upd(a[i], -1);
    }
    return ans;
  }
  inline long long work(long long ans) {
    for (int i = m + 1; i <= m + 10; i++) b[i] = 0;
    sort(b + 1, b + m + 1);
    for (int i = 1; i <= n; i++) update(1, 1, n + 1, i + 1, n + 1, +1);
    for (int i = 1, j = 1; i <= top, j <= m; j++) {
      while (i <= top && sta[i].a < b[j]) {
        update(1, 1, n + 1, sta[i].pos + 1, n + 1, -1);
        update(1, 1, n + 1, 1, sta[i].pos, +1);
        i++;
        if (i > top) break;
      }
      int ii = i, val = b[j];
      while (i <= top && sta[i].a == val) {
        update(1, 1, n + 1, sta[i].pos + 1, n + 1, -1);
        i++;
        if (i > top) break;
      }
      ans += mn[1];
      while (b[j] == b[j + 1]) j++, ans += mn[1];
      while (ii <= top && sta[ii].a == val) {
        update(1, 1, n + 1, 1, sta[ii].pos, +1);
        ii++;
        if (ii > top) break;
      }
    }
    for (int i = 1; i <= ((n + 1) << 2); i++) mn[i] = lz[i] = 0;
    return ans;
  }
  int main() {
    int C03;
    scanf("%d", &C03);
    while (C03--) {
      init();
      long long ans = init_ans(mm);
      printf("%lld\n", work(ans));
    }
    return 0;
  }
  