This is the hard version of the problem. The difference between the versions is that in this version, there can be multiple special indices ((1 \le k \le n)). 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 k \le n \le 2 \cdot 10^5)) — 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. For the first test case, you can choose the range (1, 2) and flip both the bits to get (0, 1). Then you can choose the range (1, 1) and flip the firs