#include <bits/stdc++.h>
using namespace std;
vector<int> L[2000010];
int a[2000010], b[2000010], v[2000010], aib[2000010], w[2000010],
    sp_left[2000010], sp_right[2000010], poz[2000010];
int t, n, m, i, j, el;
void solve(int st, int dr, int x, int y) {
  if (st > dr) return;
  int mid = (st + dr) >> 1;
  sp_left[x - 1] = sp_right[y + 1] = 0;
  for (int i = x; i <= y; i++) sp_left[i] = sp_left[i - 1] + (a[i] > b[mid]);
  for (int i = y; i >= x; i--) sp_right[i] = sp_right[i + 1] + (a[i] < b[mid]);
  int mini = 2000000000;
  for (int i = x; i <= y; i++) {
    if (sp_left[i - 1] + sp_right[i] < mini) {
      mini = sp_left[i - 1] + sp_right[i];
      poz[mid] = i;
    }
  }
  solve(st, mid - 1, x, poz[mid]);
  solve(mid + 1, dr, poz[mid], y);
}
void update(int p, int val) {
  for (; p <= el; p += (p & -p)) aib[p] += val;
}
int query(int p) {
  int sol = 0;
  for (; p; p -= (p & -p)) sol += aib[p];
  return sol;
}
int cautare_binara(int v[], int n, int val) {
  int st = 1, dr = n;
  while (st <= dr) {
    int mid = (st + dr) >> 1;
    if (v[mid] == val) return mid;
    if (v[mid] < val)
      st = mid + 1;
    else
      dr = mid - 1;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> t;
  for (; t--;) {
    cin >> n >> m;
    for (i = 1; i <= n; i++) cin >> a[i];
    for (i = 1; i <= m; i++) cin >> b[i];
    sort(b + 1, b + m + 1);
    solve(1, m, 1, n + 1);
    for (i = 1; i <= n + 1; i++) L[i].clear();
    for (i = 1; i <= m; i++) L[poz[i]].push_back(i);
    int k = 0;
    for (i = 1; i <= n + 1; i++) {
      for (auto it : L[i]) v[++k] = b[it];
      if (i <= n) v[++k] = a[i];
    }
    el = 0;
    for (i = 1; i <= k; i++) w[++el] = v[i];
    sort(w + 1, w + el + 1);
    int el2 = 1;
    for (i = 2; i <= el; i++)
      if (w[i] != w[i - 1]) w[++el2] = w[i];
    el = el2;
    for (i = 1; i <= k; i++) v[i] = cautare_binara(w, el, v[i]);
    for (i = 1; i <= el; i++) aib[i] = 0;
    long long sol = 0;
    for (i = 1; i <= k; i++) {
      sol += query(el) - query(v[i]);
      update(v[i], 1);
    }
    cout << sol << "\n";
  }
  return 0;
}

### Prompt

Generate a cpp solution to the following problem:

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]. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
vector<int> L[2000010];
int a[2000010], b[2000010], v[2000010], aib[2000010], w[2000010],
    sp_left[2000010], sp_right[2000010], poz[2000010];
int t, n, m, i, j, el;
void solve(int st, int dr, int x, int y) {
  if (st > dr) return;
  int mid = (st + dr) >> 1;
  sp_left[x - 1] = sp_right[y + 1] = 0;
  for (int i = x; i <= y; i++) sp_left[i] = sp_left[i - 1] + (a[i] > b[mid]);
  for (int i = y; i >= x; i--) sp_right[i] = sp_right[i + 1] + (a[i] < b[mid]);
  int mini = 2000000000;
  for (int i = x; i <= y; i++) {
    if (sp_left[i - 1] + sp_right[i] < mini) {
      mini = sp_left[i - 1] + sp_right[i];
      poz[mid] = i;
    }
  }
  solve(st, mid - 1, x, poz[mid]);
  solve(mid + 1, dr, poz[mid], y);
}
void update(int p, int val) {
  for (; p <= el; p += (p & -p)) aib[p] += val;
}
int query(int p) {
  int sol = 0;
  for (; p; p -= (p & -p)) sol += aib[p];
  return sol;
}
int cautare_binara(int v[], int n, int val) {
  int st = 1, dr = n;
  while (st <= dr) {
    int mid = (st + dr) >> 1;
    if (v[mid] == val) return mid;
    if (v[mid] < val)
      st = mid + 1;
    else
      dr = mid - 1;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> t;
  for (; t--;) {
    cin >> n >> m;
    for (i = 1; i <= n; i++) cin >> a[i];
    for (i = 1; i <= m; i++) cin >> b[i];
    sort(b + 1, b + m + 1);
    solve(1, m, 1, n + 1);
    for (i = 1; i <= n + 1; i++) L[i].clear();
    for (i = 1; i <= m; i++) L[poz[i]].push_back(i);
    int k = 0;
    for (i = 1; i <= n + 1; i++) {
      for (auto it : L[i]) v[++k] = b[it];
      if (i <= n) v[++k] = a[i];
    }
    el = 0;
    for (i = 1; i <= k; i++) w[++el] = v[i];
    sort(w + 1, w + el + 1);
    int el2 = 1;
    for (i = 2; i <= el; i++)
      if (w[i] != w[i - 1]) w[++el2] = w[i];
    el = el2;
    for (i = 1; i <= k; i++) v[i] = cautare_binara(w, el, v[i]);
    for (i = 1; i <= el; i++) aib[i] = 0;
    long long sol = 0;
    for (i = 1; i <= k; i++) {
      sol += query(el) - query(v[i]);
      update(v[i], 1);
    }
    cout << sol << "\n";
  }
  return 0;
}

```