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 N = 4e6 + 99;
int n, m, p, t, cnt, a[N], b[N], c[N];
vector<int> v[N];
map<int, int> T;
long long ans;
int Fen[N];
void update(int x, int val) {
  while (x <= n + m) {
    Fen[x] += val;
    x += (x & -x);
  }
}
int query(int x) {
  int res = 0;
  while (x > 0) {
    res += Fen[x];
    x -= (x & -x);
  }
  return res;
}
void Add(int x) {
  ans += query(n + m) - query(x);
  update(x, 1);
}
void make() {
  cnt = 1;
  sort(c, c + n + m);
  for (int i = 0; i < n + m; i++) {
    if (i == 0 || c[i] != c[i - 1]) {
      T[c[i]] = cnt++;
    }
  }
  for (int i = 1; i < n + 1; i++) a[i] = T[a[i]];
  for (int i = 1; i < m + 1; i++) b[i] = T[b[i]];
  sort(b + 1, b + 1 + m);
  for (int i = 1; i < n + 1; i++) {
    if (a[i] > n + m) cout << "99" << '=' << (99) << endl;
  }
  for (int i = 1; i < m + 1; i++) {
    if (b[i] > n + m) cout << "90" << '=' << (90) << endl;
  }
}
void solve(int L, int R, int l, int r) {
  if (R <= L) return;
  int Mid = (L + R) / 2, x = b[Mid], ans = 0, mn = N, mid = 0;
  for (int i = l; i < r; i++) {
    if (a[i] < x) ans--;
    if (a[i] > x) ans++;
    if (ans < mn) {
      mn = ans, mid = i;
    }
  }
  v[mid].push_back(x);
  solve(L, Mid, l, mid + 1);
  solve(Mid + 1, R, mid, r);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    ans = p = 0;
    cin >> n >> m;
    for (int i = 1; i < n + 1; i++) {
      cin >> a[i];
      c[i - 1] = a[i];
    }
    for (int i = 1; i < m + 1; i++) {
      cin >> b[i];
      c[n + i - 1] = b[i];
    }
    make();
    solve(1, m + 1, 0, n + 1);
    for (int i = 0; i < n + 1; i++) {
      if (i) Add(a[i]);
      sort(v[i].begin(), v[i].end());
      for (int j = 0; j < v[i].size(); j++) {
        Add(v[i][j]);
      }
      v[i].clear();
    }
    cout << ans << endl;
    T.clear();
    fill(Fen, Fen + n + m + 9, 0);
  }
}
