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;
static constexpr int Maxn = 1e6 + 5;
int n, m;
int a[Maxn], b[Maxn], buf[Maxn];
int64_t calc(int l, int r) {
  if (l == r) return 0;
  int mid = (l + r) >> 1;
  int64_t ans = calc(l, mid) + calc(mid + 1, r);
  for (int s = 0, i = l, j = mid + 1; s < r - l + 1;) {
    if (i <= mid && (j == r + 1 || a[i] <= a[j]))
      ans += j - (mid + 1), buf[s++] = a[i++];
    else
      buf[s++] = a[j++];
  }
  for (int i = l; i <= r; ++i) a[i] = buf[i - l];
  return ans;
}
int main(void) {
  int tests;
  scanf("%d", &tests);
  while (tests--) {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    for (int i = 1; i <= m; ++i) scanf("%d", &b[i]);
    sort(b + 1, b + m + 1);
    priority_queue<int> pq;
    pq.push(0);
    int64_t ans = 0;
    for (int i = 1; i <= n; ++i) {
      int l = lower_bound(b + 1, b + m + 1, a[i]) - b;
      int r = upper_bound(b + 1, b + m + 1, a[i]) - b;
      pq.push(l);
      if (pq.top() > r) {
        ans += pq.top() - r;
        pq.pop();
        pq.push(r);
      }
    }
    ans += calc(1, n);
    printf("%lld\n", ans);
  }
  exit(EXIT_SUCCESS);
}
