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 MAXN = 1e6 + 6;
int p[MAXN], a[MAXN], b[MAXN], t, n, m;
long long solve(int li, int ri, int lp, int rp) {
  if (li >= ri) {
    return 0;
  }
  int mid = (li + ri) >> 1;
  long long res = 0;
  for (int i = lp; i < rp; i++) {
    if (a[i] < b[mid]) res++;
  }
  int pos = lp;
  long long mn = res;
  for (int i = lp + 1; i <= rp; i++) {
    if (a[i - 1] < b[mid]) res--;
    if (a[i - 1] > b[mid]) res++;
    if (res < mn) {
      mn = res;
      pos = i;
    }
  }
  return mn + solve(li, mid, lp, pos) + solve(mid + 1, ri, pos, rp);
}
long long invs(int l, int r) {
  if (l == r) return 0;
  int mid = (l + r) >> 1;
  long long res = invs(l, mid) + invs(mid + 1, r);
  long long cnt = 0;
  for (int i = l, j = mid + 1; i <= mid; i++) {
    while (j <= r && a[i] > a[j]) {
      cnt++;
      j++;
    }
    res += cnt;
  }
  sort(a + l, a + r + 1);
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> t;
  while (t--) {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= m; i++) cin >> b[i];
    sort(b + 1, b + 1 + m);
    cout << solve(1, m + 1, 1, n + 1) + invs(1, n) << '\n';
  }
  return 0;
}
