#include <bits/stdc++.h>
using namespace std;
using lint = long long;
using pi = pair<lint, lint>;
const int MAXN = 1000005;
int n, m, a[MAXN], b[MAXN];
struct bit {
  int tree[MAXN];
  void add(int x, int v) {
    for (int i = x; i < MAXN; i += i & -i) tree[i] += v;
  }
  int query(int x) {
    int ret = 0;
    for (int i = x; i; i -= i & -i) ret += tree[i];
    return ret;
  }
} bit;
lint inv() {
  vector<int> v(a, a + n);
  sort((v).begin(), (v).end());
  v.resize(unique((v).begin(), (v).end()) - v.begin());
  lint ret = 0;
  for (int i = 0; i < n; i++) {
    a[i] = lower_bound((v).begin(), (v).end(), a[i]) - v.begin() + 1;
    ret += bit.query(n) - bit.query(a[i]);
    bit.add(a[i], 1);
  }
  for (int i = 0; i < n; i++) {
    bit.add(a[i], -1);
  }
  return ret;
}
void solve() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < m; i++) scanf("%d", &b[i]);
  sort(b, b + m);
  priority_queue<int> pq;
  lint ret = 0;
  for (int i = 0; i < n; i++) {
    int l = lower_bound(b, b + m, a[i]) - b;
    int r = upper_bound(b, b + m, a[i]) - b;
    if (((int)(pq).size()) && pq.top() > r) {
      ret += pq.top() - r;
      pq.pop();
      pq.push(r);
    }
    pq.push(l);
  }
  ret += inv();
  printf("%lld\n", ret);
}
int main() {
  int tc;
  scanf("%d", &tc);
  while (tc--) solve();
}

### 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;
using lint = long long;
using pi = pair<lint, lint>;
const int MAXN = 1000005;
int n, m, a[MAXN], b[MAXN];
struct bit {
  int tree[MAXN];
  void add(int x, int v) {
    for (int i = x; i < MAXN; i += i & -i) tree[i] += v;
  }
  int query(int x) {
    int ret = 0;
    for (int i = x; i; i -= i & -i) ret += tree[i];
    return ret;
  }
} bit;
lint inv() {
  vector<int> v(a, a + n);
  sort((v).begin(), (v).end());
  v.resize(unique((v).begin(), (v).end()) - v.begin());
  lint ret = 0;
  for (int i = 0; i < n; i++) {
    a[i] = lower_bound((v).begin(), (v).end(), a[i]) - v.begin() + 1;
    ret += bit.query(n) - bit.query(a[i]);
    bit.add(a[i], 1);
  }
  for (int i = 0; i < n; i++) {
    bit.add(a[i], -1);
  }
  return ret;
}
void solve() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < m; i++) scanf("%d", &b[i]);
  sort(b, b + m);
  priority_queue<int> pq;
  lint ret = 0;
  for (int i = 0; i < n; i++) {
    int l = lower_bound(b, b + m, a[i]) - b;
    int r = upper_bound(b, b + m, a[i]) - b;
    if (((int)(pq).size()) && pq.top() > r) {
      ret += pq.top() - r;
      pq.pop();
      pq.push(r);
    }
    pq.push(l);
  }
  ret += inv();
  printf("%lld\n", ret);
}
int main() {
  int tc;
  scanf("%d", &tc);
  while (tc--) solve();
}

```