F. Permutation Edit Distance
time limit per test
2 seconds
memory limit per test
1024 megabytes
input
standard input
output
standard output

You are given two permutations$$$^{\text{∗}}$$$ and must compute the edit distance between them using only the following edits:

  • Given a length $$$n$$$ permutation, insert $$$n+1$$$ anywhere into the array, resulting in a permutation of length $$$n+1$$$.
  • Given a length $$$n$$$ permutation, delete $$$n$$$ from the array, resulting in a permutation of length $$$n-1$$$.

The edit distance between two permutations $$$p$$$ and $$$q$$$ is the minimum possible number of edits it takes to turn $$$p$$$ into $$$q$$$ or vice versa. It can be shown that, given the edits above, it does not matter which permutation you start with.

$$$^{\text{∗}}$$$A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array), and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).

Input

Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 2 \cdot 10^5$$$). The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$1\leq n,\,m\leq 2\cdot10^5$$$) — the lengths of the permutations.

The second line of each test case contains $$$n$$$ integers $$$p_1,\,p_2,\,\ldots,\,p_n$$$ ($$$1\leq p_i\leq n$$$) — the permutation $$$p$$$.

The last line of each test case contains $$$m$$$ integers $$$q_1,\,q_2,\,\ldots,\,q_m$$$ ($$$1\leq q_i\leq m$$$) — the permutation $$$q$$$.

It is guaranteed that the sum of $$$\max(n,\,m)$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output

For each test case, output a single integer — the minimum number of valid edits to transform $$$p$$$ into $$$q$$$.

Example
Input
3
4 5
1 2 3 4
1 5 2 3 4
3 3
1 2 3
3 2 1
1 1
1
1
Output
1
4
0
Note

In the first test case, inserting 5 into $$$p$$$ in the second position will result in $$$q$$$.

In the second test case, first delete $$$3$$$ and $$$2$$$ from the end of $$$p$$$. Then insert $$$2$$$ into the beginning, and after that insert $$$3$$$ into the beginning as well. This will result in $$$q$$$, and it can be shown that this is a minimal solution.

In the third test case, the permutations are already equal.