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CF2217B.Flip the Bit (Easy Version)
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 (). You can hack only if you solved all versions of this problem.
You are given a binary array of length and special indices (). It is given that the values of all elements at special indices are the same (i. e., ).
In one operation, you can choose a range () such that the range contains at least one special index () and flip all bits for . Flipping a bit changes to and to .
Let 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 .
Input
Each test contains multiple test cases. The first line contains the number of test cases (). The description of the test cases follows.
The first line of each test case contains two integers and (; ) — the length of the array and the number of special indices.
The second line contains integers () — the elements of the array.
The third line contains integers () — the special indices. It is guaranteed that .
It is guaranteed that the sum of over all test cases does not exceed .
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 and flip all the bits to get . Then you can choose the range and flip the second bit to get .
For the second test case, all the bits already match the value at the special index. You do not need any operations.
Samples
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
2
0
4
10
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