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 MOD = 998244353;
int n, m, a[1000005], b[1000005];
pair<int, int> perm[1000005];
struct fwt {
  int dat[1000005];
  void clear() {
    for (int i = 1; i <= (int)(n); i++) dat[i] = 0;
  }
  void add(int id, int val) {
    for (id++; id <= n; id += id & -id) dat[id] += val;
  }
  int query(int id) {
    int ret = 0;
    for (id++; id > 0; id -= id & -id) ret += dat[id];
    return ret;
  }
} tre;
long long ans;
priority_queue<int> que;
void addnew(int x) {
  if (que.empty()) {
    que.push(x);
  } else if (x >= que.top())
    que.push(x);
  else {
    ans += que.top() - x;
    que.push(x);
    que.push(x);
    que.pop();
  }
}
void solve() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < (int)(n); i++) scanf("%d", &a[i]);
  for (int i = 0; i < (int)(m); i++) scanf("%d", &b[i]);
  sort(b, b + m);
  for (int i = 0; i < (int)(n); i++) perm[i] = make_pair(a[i], i);
  sort(perm, perm + n);
  ans = 0;
  tre.clear();
  for (int i = 0; i < (int)(n); i++) {
    ans += 2 * (i - tre.query(perm[i].second));
    tre.add(perm[i].second, 1);
  }
  while (!que.empty()) que.pop();
  for (int i = 0; i < (int)(n); i++) {
    int r = upper_bound(b, b + m, a[i]) - b,
        l = lower_bound(b, b + m, a[i]) - b;
    addnew(r);
    addnew(l);
    ans -= r - l;
  }
  assert(ans % 2 == 0);
  printf("%lld\n", ans / 2);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}
