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]. 
#include <bits/stdc++.h>
using namespace std;
const int max_n = 2000111, inf = 1000111222;
int a[max_n];
int b[max_n];
int n, m;
vector<int> ccc;
void clr() { ccc.clear(); }
int get_small(int large) {
  return lower_bound(ccc.begin(), ccc.end(), large) - ccc.begin();
}
struct segment_tree {
  int sum[max_n];
  int mn[max_n * 4];
  void build(int cur, int l, int r) {
    if (l == r) {
      mn[cur] = 0;
      sum[cur] = 1;
      return;
    }
    int mid = (l + r) / 2;
    build(cur * 2, l, mid);
    build(cur * 2 + 1, mid + 1, r);
    sum[cur] = sum[cur * 2] + sum[cur * 2 + 1];
    mn[cur] = min(mn[cur * 2], sum[cur * 2] + mn[cur * 2 + 1]);
  }
  void upd(int cur, int l, int r, int pos, int val) {
    if (l == r) {
      sum[cur] = val;
      if (val < 0)
        mn[cur] = val;
      else
        mn[cur] = 0;
      return;
    }
    int mid = (l + r) / 2;
    if (pos <= mid) {
      upd(cur * 2, l, mid, pos, val);
    } else {
      upd(cur * 2 + 1, mid + 1, r, pos, val);
    }
    sum[cur] = sum[cur * 2] + sum[cur * 2 + 1];
    mn[cur] = min(mn[cur * 2], sum[cur * 2] + mn[cur * 2 + 1]);
  }
} st;
int ft[max_n];
void add(int pos, int val) {
  for (int i = pos; i < n; i |= (i + 1)) {
    ft[i] += val;
  }
}
int get(int pos) {
  int ans = 0;
  for (int i = pos; i >= 0; i = (i & (i + 1)) - 1) {
    ans += ft[i];
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> n >> m;
    clr();
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
      ccc.push_back(a[i]);
    }
    for (int i = 0; i < m; ++i) {
      cin >> b[i];
      ccc.push_back(b[i]);
    }
    st.build(1, 0, n - 1);
    sort(ccc.begin(), ccc.end());
    ccc.erase(unique(ccc.begin(), ccc.end()), ccc.end());
    vector<pair<int, int>> v;
    for (int i = 0; i < n; ++i) {
      ft[i] = 0;
    }
    long long ans2 = 0;
    for (int i = 0; i < n; ++i) {
      a[i] = get_small(a[i]);
      ans2 += (i - get(a[i]));
      add(a[i], 1);
      v.emplace_back(a[i], i);
    }
    for (int i = 0; i < m; ++i) {
      b[i] = get_small(b[i]);
    }
    long long ans = 0;
    sort(b, b + m);
    int cnt_less = 0;
    sort(v.begin(), v.end());
    reverse(v.begin(), v.end());
    vector<pair<int, int>> v2;
    for (int i = 0; i < m; ++i) {
      int cur = b[i];
      while (!v2.empty() && v2.back().first < cur) {
        st.upd(1, 0, n - 1, v2.back().second, -1);
        v2.pop_back();
        ++cnt_less;
      }
      while (!v.empty() && v.back().first < cur) {
        st.upd(1, 0, n - 1, v.back().second, -1);
        v.pop_back();
        ++cnt_less;
      }
      while (!v.empty() && v.back().first == cur) {
        st.upd(1, 0, n - 1, v.back().second, 0);
        v2.push_back(v.back());
        v.pop_back();
      }
      ans += cnt_less + st.mn[1];
    }
    cout << ans2 + ans << "\n";
  }
  return 0;
}
