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;
  int t, n, m;
  pair<int, int> a[1000005], b[1000005];
  int bit[2 * 1000005 + 1];
  int sum(int i) {
    int s = 0;
    while (i > 0) {
      s += bit[i];
      i -= i & -i;
    }
    return s;
  }
  void add(int i, int x) {
    while (i <= n) {
      bit[i] += x;
      i += i & -i;
    }
  }
  struct segtree {
    int ans[4 * 1000005], sum[4 * 1000005], rsum[4 * 1000005];
    void pushup(int k) {
      ans[k] = min(ans[k * 2] + rsum[k * 2 + 1], ans[k * 2 + 1] + sum[k * 2]);
      sum[k] = sum[k * 2] + sum[k * 2 + 1];
      rsum[k] = rsum[k * 2] + rsum[k * 2 + 1];
    }
    void build(int k, int l, int r) {
      if (l == r) {
        sum[k] = 1;
        rsum[k] = 0;
        ans[k] = 0;
        return;
      }
      int mid = (l + r) / 2;
      build(k * 2, l, mid);
      build(k * 2 + 1, mid + 1, r);
      pushup(k);
    }
    void update(int k, int l, int r, int p, int type) {
      if (l == r) {
        if (type == 0) {
          sum[k] = 0;
          rsum[k] = 0;
        } else {
          sum[k] = 0;
          rsum[k] = 1;
        }
        return;
      }
      int mid = (l + r) / 2;
      if (p <= mid)
        update(k * 2, l, mid, p, type);
      else
        update(k * 2 + 1, mid + 1, r, p, type);
      pushup(k);
    }
  } seg;
  int main() {
    scanf("%d", &t);
    while (t--) {
      scanf("%d%d", &n, &m);
      for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i].first);
        a[i].second = i;
      }
      for (int i = 1; i <= m; i++) {
        scanf("%d", &b[i].first);
        b[i].second = i;
      }
      long long ans = 0;
      for (int i = 1; i <= n; i++) bit[i] = 0;
      sort(a + 1, a + n + 1);
      sort(b + 1, b + m + 1);
      int l = 1;
      seg.build(1, 1, n);
      vector<int> equal;
      for (int i = 1; i <= m; i++) {
        if (i == 1 || b[i].first != b[i - 1].first) {
          for (auto x : equal) seg.update(1, 1, n, x, 1);
          equal.clear();
        } else {
          ans += seg.ans[1];
          continue;
        }
        while (l <= n && a[l].first <= b[i].first) {
          if (a[l].first == b[i].first) {
            equal.push_back(a[l].second);
            seg.update(1, 1, n, a[l].second, 0);
          } else
            seg.update(1, 1, n, a[l].second, 1);
          l++;
        }
        ans += seg.ans[1];
      }
      int now = 1;
      while (now <= n) {
        int cur = now;
        while (cur + 1 <= n && a[cur].first == a[cur + 1].first) cur++;
        for (int i = now; i <= cur; i++) ans += sum(n + 1 - a[i].second);
        for (int i = now; i <= cur; i++) add(n + 1 - a[i].second, 1);
        now = cur + 1;
      }
      printf("%lld\n", ans);
    }
    return 0;
  }
  