#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
int n, m, a[N], b[N], t, ans[N], c[2 * N], eq[2 * N];
long long bt[2 * N];
pair<int, int> d[2 * N];
void ded(int l1, int r1, int l2, int r2) {
  if (l2 > r2) return;
  int mid = (l2 + r2) / 2;
  int pb = l1;
  long long cl = 0, cg = 0, claux = 0, cgaux = 0;
  for (int i = l1; i < r1; i++) cgaux += (a[i] < b[mid]);
  cg = cgaux;
  for (int i = l1 + 1; i <= r1; i++) {
    if (a[i - 1] < b[mid]) cgaux--;
    if (a[i - 1] > b[mid]) claux++;
    if (cgaux + claux < cg + cl) {
      pb = i;
      cg = cgaux;
      cl = claux;
    }
  }
  ans[mid] = pb;
  ded(l1, pb, l2, mid - 1);
  ded(pb, r1, mid + 1, r2);
}
long long get(int p) {
  long long r = 0;
  while (p) {
    r += bt[p];
    p -= p & (-p);
  }
  return r;
}
void ins(int p) {
  while (p <= n + m) {
    bt[p]++;
    p += p & (-p);
  }
}
long long cnt() {
  long long r = 0;
  for (int i = 0; i < n + m; i++) d[i] = {c[i], i};
  sort(d, d + n + m);
  for (int i = 0; i < n + m; i++) eq[d[i].second] = i + 1;
  for (int i = 0; i < n + m; i++) {
    r += i - get(eq[i]);
    ins(eq[i]);
  }
  return r;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d %d", &n, &m);
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    for (int i = 0; i <= n + m; i++) bt[i] = 0;
    for (int i = 0; i < m; i++) scanf("%d", &b[i]);
    sort(b, b + m);
    ded(0, n, 0, m - 1);
    int l = 0;
    for (int i = 0; i < n; i++) {
      while (l < m && ans[l] == i) {
        c[i + l] = b[l];
        l++;
      }
      c[i + l] = a[i];
    }
    while (l < m) {
      c[n + l] = b[l];
      l++;
    }
    printf("%lld\n", cnt());
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
const int N = 1e6 + 5;
int n, m, a[N], b[N], t, ans[N], c[2 * N], eq[2 * N];
long long bt[2 * N];
pair<int, int> d[2 * N];
void ded(int l1, int r1, int l2, int r2) {
  if (l2 > r2) return;
  int mid = (l2 + r2) / 2;
  int pb = l1;
  long long cl = 0, cg = 0, claux = 0, cgaux = 0;
  for (int i = l1; i < r1; i++) cgaux += (a[i] < b[mid]);
  cg = cgaux;
  for (int i = l1 + 1; i <= r1; i++) {
    if (a[i - 1] < b[mid]) cgaux--;
    if (a[i - 1] > b[mid]) claux++;
    if (cgaux + claux < cg + cl) {
      pb = i;
      cg = cgaux;
      cl = claux;
    }
  }
  ans[mid] = pb;
  ded(l1, pb, l2, mid - 1);
  ded(pb, r1, mid + 1, r2);
}
long long get(int p) {
  long long r = 0;
  while (p) {
    r += bt[p];
    p -= p & (-p);
  }
  return r;
}
void ins(int p) {
  while (p <= n + m) {
    bt[p]++;
    p += p & (-p);
  }
}
long long cnt() {
  long long r = 0;
  for (int i = 0; i < n + m; i++) d[i] = {c[i], i};
  sort(d, d + n + m);
  for (int i = 0; i < n + m; i++) eq[d[i].second] = i + 1;
  for (int i = 0; i < n + m; i++) {
    r += i - get(eq[i]);
    ins(eq[i]);
  }
  return r;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d %d", &n, &m);
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    for (int i = 0; i <= n + m; i++) bt[i] = 0;
    for (int i = 0; i < m; i++) scanf("%d", &b[i]);
    sort(b, b + m);
    ded(0, n, 0, m - 1);
    int l = 0;
    for (int i = 0; i < n; i++) {
      while (l < m && ans[l] == i) {
        c[i + l] = b[l];
        l++;
      }
      c[i + l] = a[i];
    }
    while (l < m) {
      c[n + l] = b[l];
      l++;
    }
    printf("%lld\n", cnt());
  }
  return 0;
}

```