This is the easy version of the problem. The difference between the versions is that in this version, there is exactly one special index ($$$k=1$$$). You can hack only if you solved all versions of this problem.

You are given a binary array $$$a$$$ of length $$$n$$$ and $$$k$$$ special indices $$$p_1, p_2, \ldots, p_k$$$ ($$$1 \le p_i \le n$$$). It is given that the values $$$a_i$$$ of all elements at special indices are the same (i. e., $$$a_{p_1} = a_{p_2} = \ldots = a_{p_k}$$$).

In one operation, you can choose a range $$$[l, r]$$$ ($$$1 \le l \le r \le n$$$) such that the range contains at least one special index ($$$l \le p_i \le r$$$) and flip all bits $$$a_j$$$ for $$$l \le j \le r$$$. Flipping a bit changes $$$0$$$ to $$$1$$$ and $$$1$$$ to $$$0$$$.

Let $$$x$$$ denote the value at special indices before any operations are applied. Find the minimum number of operations required to make all elements of the array equal to $$$x$$$.

Input

Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$k=1$$$) — the length of the array and the number of special indices.

The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i \le 1$$$) — the elements of the array.

The third line contains $$$k$$$ integers $$$p_1, p_2, \ldots, p_k$$$ ($$$1 \le p_1 \lt p_2 \lt \ldots \lt p_k \le n$$$) — the special indices. It is guaranteed that $$$a_{p_1} = a_{p_2} = \ldots = a_{p_k}$$$.

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

Output

For each test case, output a single integer — the minimum number of operations required.

Note

For the first test case, you can choose the range $$$[1, 3]$$$ and flip all the bits to get $$$[1, 0, 1]$$$. Then you can choose the range $$$[2, 2]$$$ and flip the second bit to get $$$[1, 1, 1]$$$.

For the second test case, all the bits already match the value at the special index. You do not need any operations.

Flip the Bit (Easy Version)

Examples