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 = 1000005, M = 998244353, L = 1000000000;
  const long long inf = 100000000000000000ll;
  int n, m, k, i, j, t, b[N];
  struct node {
    int a, x;
  } a[N];
  bool cmp(node a, node b) {
    if (a.a == b.a) return a.x < b.x;
    return a.a < b.a;
  }
  struct str {
    int tree[N * 4], lazy[N * 4];
    void clear() {
      for (int i = 1; i <= n * 4; ++i) lazy[i] = tree[i] = 0;
      k = 1;
    }
    void pushup(int i) {
      tree[i] = min(tree[i << 1], tree[i << 1 | 1]) + lazy[i];
    }
    void modify(int i, int l, int r, int ll, int rr, int x) {
      if (l >= ll && r <= rr) {
        tree[i] += x;
        lazy[i] += x;
        return;
      }
      int mid = l + r >> 1;
      if (mid >= ll) modify(i << 1, l, mid, ll, rr, x);
      if (mid < rr) modify(i << 1 | 1, mid + 1, r, ll, rr, x);
      pushup(i);
    }
    int Query(int i, int l, int r, int ll, int rr) {
      if (l >= ll && r <= rr) return tree[i];
      int mid = l + r >> 1, s = 1 << 30;
      if (mid >= ll) s = min(s, Query(i << 1, l, mid, ll, rr));
      if (mid < rr) s = min(s, Query(i << 1 | 1, mid + 1, r, ll, rr));
      return s + lazy[i];
    }
  } x;
  int main() {
    scanf("%d", &t);
    while (t--) {
      scanf("%d %d", &n, &m);
      long long ans = 0;
      for (i = 1; i <= n; ++i) {
        scanf("%d", &a[i].a);
        a[i].x = i;
        x.modify(1, 0, n, i, n, 1);
      }
      sort(a + 1, a + 1 + n, cmp);
      for (i = 1; i <= m; ++i) scanf("%d", &b[i]);
      sort(b + 1, b + 1 + m);
      int l1 = 1, l2 = 1;
      for (i = 1; i <= m; ++i) {
        while (l1 <= n && a[l1].a <= b[i]) {
          x.modify(1, 0, n, a[l1].x, n, -1);
          ++l1;
        }
        while (l2 <= n && a[l2].a < b[i]) {
          x.modify(1, 0, n, 0, a[l2].x - 1, 1);
          ++l2;
        }
        ans += x.tree[1];
      }
      x.clear();
      for (i = 1; i <= n; ++i) {
        ans += x.Query(1, 0, n, a[i].x, a[i].x);
        x.modify(1, 0, n, 0, a[i].x - 1, 1);
      }
      x.clear();
      printf("%lld\n", ans);
    }
  }
  