Statement : B. Flip the Bit (Easy Version)

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 $x$.

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 $2 \cdot 10^5$.

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

Input

4
3 1
0 1 0
2
5 1
1 1 1 1 1
1
6 1
0 1 0 1 0 1
3
17 1
0 1 1 0 1 1 0 1 0 0 1 0 1 0 1 0 1
5

Output

2
0
4
10

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.