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;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
int n, m;
int A[1000005];
int B[1000005];
int T[1000005];
int S[1000005];
ll inverse(int s, int e) {
  if (s == e) {
    T[s] = A[s];
    return 0;
  }
  int m = (s + e) / 2;
  ll ans = inverse(s, m) + inverse(m + 1, e);
  for (int i = s; i <= e; i++) {
    S[i] = T[i];
  }
  int i = s, j = m + 1, k = s;
  while (k <= e) {
    if (j > e || (i <= m && S[i] <= S[j]))
      T[k++] = S[i++];
    else
      T[k++] = S[j++], ans += m - i + 1;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T;
  cin >> T;
  for (int t = 1; t <= T; 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 + m + 1);
    priority_queue<int> pq;
    ll mx = inverse(1, n);
    for (int i = 1; i <= n; i++) {
      int x = lower_bound(B + 1, B + m + 1, A[i]) - B - 1;
      int y = upper_bound(B + 1, B + m + 1, A[i]) - B - 1;
      pq.push(x);
      pq.push(y);
      pq.pop();
      mx += x;
    }
    while (!pq.empty()) {
      mx -= pq.top();
      pq.pop();
    }
    cout << mx << '\n';
  }
  return 0;
}
