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;
  void main0();
  int main() {
    clock_t start, end;
    ios::sync_with_stdio(false);
    cin.tie(0);
    main0();
    return 0;
  }
  const int dx[8] = {0, 1, -1, 0, 1, 1, -1, -1};
  const int dy[8] = {1, 0, 0, -1, 1, -1, -1, 1};
  const int N = 1e6 + 5;
  const int M = 1e5;
  const int INF = 0x3f3f3f3f;
  const long long INFF = 0x3f3f3f3f3f3f3f3f;
  const int mod = 998244353;
  const double eps = 1e-6;
  const double Pi = acos(-1.0);
  mt19937 rnd(
      (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
  int n, m;
  int a[N];
  int b[N];
  vector<pair<int, int> > v;
  struct BIT {
    int n;
    int c[N];
    int lowbit(int x) { return x & (-x); }
    void init(int _n) {
      n = _n;
      for (int i = 1; i <= n; ++i) c[i] = 0;
    }
    void update(int i, int x) {
      while (i <= n) {
        c[i] += x;
        i += lowbit(i);
      }
    }
    int query(int i) {
      int ret = 0;
      while (i) {
        ret += c[i];
        i -= lowbit(i);
      }
      return ret;
    }
  } bit;
  struct SGT {
    struct node {
      int l, r, lc, rc;
      int v, tag;
    } t[N * 2];
    int cnt;
    void init(int n) {
      cnt = 0;
      for (int i = 1; i <= 2 * n; ++i) {
        t[i].l = t[i].r = t[i].lc = t[i].rc = t[i].v = t[i].tag = 0;
      }
    }
    void pushup(int u) { t[u].v = min(t[t[u].lc].v, t[t[u].rc].v); }
    void build(int l, int r) {
      int u = ++cnt;
      t[u].l = l, t[u].r = r;
      t[u].v = t[u].tag = 0;
      if (l == r) {
        t[u].v = 1;
      } else {
        t[u].lc = cnt + 1, build(l, ((t[u].l + t[u].r) / 2));
        t[u].rc = cnt + 1, build(((t[u].l + t[u].r) / 2) + 1, r);
        pushup(u);
      }
    }
    void apply(int u, int x) {
      t[u].v += x;
      t[u].tag += x;
    }
    void pushdown(int u) {
      apply(t[u].lc, t[u].tag);
      apply(t[u].rc, t[u].tag);
      t[u].tag = 0;
    }
    void update(int u, int l, int r, int x) {
      if ((l <= t[u].l) && (t[u].r <= r)) {
        apply(u, x);
      } else {
        pushdown(u);
        if (l <= ((t[u].l + t[u].r) / 2)) update(t[u].lc, l, r, x);
        if (((t[u].l + t[u].r) / 2) + 1 <= r) update(t[u].rc, l, r, x);
        pushup(u);
      }
    }
    int query(int u, int l, int r) {
      if ((l <= t[u].l) && (t[u].r <= r)) {
        return t[u].v;
      } else {
        pushdown(u);
        int ret = INF;
        if (l <= ((t[u].l + t[u].r) / 2)) ret = min(ret, query(t[u].lc, l, r));
        if (((t[u].l + t[u].r) / 2) + 1 <= r)
          ret = min(ret, query(t[u].rc, l, r));
        return ret;
      }
    }
  } sgt;
  void main0() {
    int T;
    cin >> T;
    while (T--) {
      v.clear();
      v.push_back({0, 0});
      cin >> n >> m;
      for (int i = 1; i <= n; ++i) {
        cin >> a[i];
        v.push_back({a[i], i});
      }
      for (int i = 1; i <= m; ++i) {
        cin >> b[i];
      }
      sort((v).begin(), (v).end());
      sort(b + 1, b + m + 1);
      long long ans = 0;
      bit.init(n);
      for (int i = n; i >= 1; --i) {
        int rk = lower_bound((v).begin(), (v).end(), (pair<int, int>){a[i], 0}) -
                 v.begin();
        ans += bit.query(rk - 1);
        bit.update(rk, 1);
      }
      sgt.init(n);
      sgt.build(1, n);
      for (int i = 1; i <= n; ++i) {
        sgt.update(1, i, n, 1);
      }
      for (int i = 1; i <= m; ++i) {
        int l, r;
        l = lower_bound((v).begin(), (v).end(), (pair<int, int>){b[i - 1], INF}) -
            v.begin();
        r = lower_bound((v).begin(), (v).end(), (pair<int, int>){b[i], INF}) -
            v.begin();
        r--;
        for (int j = l; j <= r; ++j) {
          sgt.update(1, v[j].second, n, -1);
        }
        l = lower_bound((v).begin(), (v).end(), (pair<int, int>){b[i - 1], 0}) -
            v.begin();
        r = lower_bound((v).begin(), (v).end(), (pair<int, int>){b[i], 0}) -
            v.begin();
        r--;
        for (int j = l; j <= r; ++j) {
          sgt.update(1, v[j].second, n, -1);
        }
        ans += r + min(0, sgt.query(1, 1, n));
      }
      cout << ans << endl;
    }
  }
  