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>
  long long n, m;
  const long long MAXN = 1000006;
  long long bit[MAXN];
  void add(long long x, long long v) {
    x += 1;
    while (x < MAXN) {
      bit[x] += v;
      x += x & (-x);
    }
  }
  long long sum(long long x) {
    x += 1;
    long long ret = 0;
    while (x > 0) {
      ret += bit[x];
      x -= x & (-x);
    }
    return ret;
  }
  struct seg {
    std::vector<long long> arr, lazy;
    void push(long long id, long long l, long long r) {
      if (lazy[id] == 0) return;
      arr[(id + 1)] += lazy[id];
      arr[(id + 2 * (((l + r) / 2) - l + 1))] += lazy[id];
      lazy[(id + 1)] += lazy[id];
      lazy[(id + 2 * (((l + r) / 2) - l + 1))] += lazy[id];
      lazy[id] = 0;
    }
    void update(long long x, long long y, long long val, long long id = 0,
                long long l = 0, long long r = n - 1) {
      if (y < l || r < x) return;
      if (x <= l && r <= y) {
        lazy[id] += val;
        arr[id] += val;
        return;
      }
      push(id, l, r);
      update(x, y, val, (id + 1), l, ((l + r) / 2));
      update(x, y, val, (id + 2 * (((l + r) / 2) - l + 1)), ((l + r) / 2) + 1, r);
      arr[id] = std::min(arr[(id + 1)], arr[(id + 2 * (((l + r) / 2) - l + 1))]);
    }
  };
  signed main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    long long t;
    std::cin >> t;
    while (t--) {
      std::cin >> n >> m;
      seg tree;
      tree.arr = tree.lazy = std::vector<long long>(2 * n - 1, 0);
      long long cur[n];
      for (long long i = 0; i < n; i++) {
        tree.update(i, n - 1, 1);
        cur[i] = 1;
      }
      long long nones = 0;
      std::pair<long long, long long> help[n];
      for (long long i = 0; i < n; i++) {
        std::cin >> help[i].first;
        help[i].second = i;
        bit[i + 1] = 0;
      }
      long long b[m];
      for (long long i = 0; i < m; i++) std::cin >> b[i];
      std::sort(help, help + n);
      std::sort(b, b + m);
      long long init = 0;
      for (long long i = 0; i < n; i++) {
        init += sum(n - 1 - help[i].second);
        add(n - 1 - help[i].second, 1);
      }
      long long op = 0, zp = 0;
      long long ans = 0;
      for (long long i = 0; i < m; i++) {
        while (op < n && help[op].first <= b[i]) {
          tree.update(help[op].second, n - 1, -1);
          op++;
        }
        while (zp < n && help[zp].first < b[i]) {
          tree.update(help[zp].second, n - 1, -1);
          nones++;
          zp++;
        }
        ans += nones;
        ans += std::min(tree.arr[0], 0ll);
      }
      std::cout << ans + init << std::endl;
    }
    return 0;
  }
  