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 NM_MAX = 1000000;
  const int AB_MAX = 1000000000;
  int t;
  int n, m;
  int a[NM_MAX + 2];
  int b[NM_MAX + 2];
  struct SGTNode {
    int minVal;
    int lazy;
  };
  SGTNode operator+(const SGTNode &u, const SGTNode &v) {
    return SGTNode{min(u.minVal, v.minVal), 0};
  }
  void operator+=(SGTNode &u, const int &v) {
    u.minVal += v;
    u.lazy += v;
  }
  SGTNode SGT[NM_MAX * 4 + 2];
  void split(int node) {
    int lSon = node * 2, rSon = node * 2 + 1;
    SGT[lSon] += SGT[node].lazy;
    SGT[rSon] += SGT[node].lazy;
    SGT[node].lazy = 0;
  }
  void build(int node, int l, int r) {
    if (l == r) {
      SGT[node].minVal = 0;
      SGT[node].lazy = 0;
      return;
    }
    split(node);
    int mid = (l + r) / 2;
    int lSon = node * 2, rSon = node * 2 + 1;
    build(lSon, l, mid);
    build(rSon, mid + 1, r);
    SGT[node] = SGT[lSon] + SGT[rSon];
  }
  void build() { build(1, 1, n + 1); }
  void update(int node, int l, int r, int ul, int ur, int uval) {
    if (ul <= l && r <= ur) {
      SGT[node] += uval;
      return;
    }
    split(node);
    int mid = (l + r) / 2;
    int lSon = node * 2, rSon = node * 2 + 1;
    if (ul <= mid) update(lSon, l, mid, ul, ur, uval);
    if (mid + 1 <= ur) update(rSon, mid + 1, r, ul, ur, uval);
    SGT[node] = SGT[lSon] + SGT[rSon];
  }
  void update(int ul, int ur, int uval) { update(1, 1, n + 1, ul, ur, uval); }
  int p[NM_MAX + 2];
  int asmall[NM_MAX + 2];
  int BIT[NM_MAX + 2];
  void BITupdate(int val) {
    for (int i = val; i >= 1; i -= i & -i) BIT[i]++;
  }
  int BITquery(int val) {
    int cnt = 0;
    for (int i = val; i <= n; i += i & -i) cnt += BIT[i];
    return cnt;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while (t--) {
      cin >> n >> m;
      for (int i = 1; i <= n; i++) cin >> a[i];
      for (int i = 1; i <= m; i++) cin >> b[i];
      sort(b + 1, b + m + 1);
      build();
      for (int i = 1; i <= n; i++) update(i + 1, n + 1, +1);
      for (int i = 1; i <= n; i++) p[i] = i;
      sort(p + 1, p + n + 1,
           [&](const int &i, const int &j) { return a[i] < a[j]; });
      {
        int curr = 0;
        for (int i = 1; i <= n; i++) {
          if (i == 1 || a[p[i]] > a[p[i - 1]]) curr++;
          asmall[p[i]] = curr;
        }
      }
      long long answer = 0;
      int l = 0, r = 0;
      for (int i = 1; i <= m; i++) {
        while (r < n && a[p[r + 1]] <= b[i]) {
          r++;
          update(p[r] + 1, n + 1, -1);
        }
        while (l < r && a[p[l + 1]] < b[i]) {
          l++;
          update(1, p[l], +1);
        }
        answer += SGT[1].minVal;
      }
      for (int i = 1; i <= n; i++) {
        answer += BITquery(asmall[i] + 1);
        BITupdate(asmall[i]);
      }
      for (int i = 1; i <= n; i++) BIT[i] = 0;
      cout << answer << "\n";
    }
    return 0;
  }
  