CF1332C.K-Complete Word

Word $s$ of length $n$ is called $k$ -complete if

• $s$ is a palindrome, i.e. $s_i=s_{n+1-i}$ for all $1 \le i \le n$ ;
• $s$ has a period of $k$ , i.e. $s_i=s_{k+i}$ for all $1 \le i \le n-k$ .

For example, "abaaba" is a $3$ -complete word, while "abccba" is not.

Bob is given a word $s$ of length $n$ consisting of only lowercase Latin letters and an integer $k$ , such that $n$ is divisible by $k$ . He wants to convert $s$ to any $k$ -complete word.

To do this Bob can choose some $i$ ( $1 \le i \le n$ ) and replace the letter at position $i$ with some other lowercase Latin letter.

So now Bob wants to know the minimum number of letters he has to replace to convert $s$ to any $k$ -complete word.

Note that Bob can do zero changes if the word $s$ is already $k$ -complete.

You are required to answer $t$ test cases independently.

The first line contains a single integer $t$ ( $1 \le t\le 10^5$ ) — the number of test cases.

The first line of each test case contains two integers $n$ and $k$ ( $1 \le k < n \le 2 \cdot 10^5$ , $n$ is divisible by $k$ ).

The second line of each test case contains a word $s$ of length $n$ .

It is guaranteed that word $s$ only contains lowercase Latin letters. And it is guaranteed that the sum of $n$ over all test cases will not exceed $2 \cdot 10^5$ .

For each test case, output one integer, representing the minimum number of characters he has to replace to convert $s$ to any $k$ -complete word.

In the first test case, one optimal solution is aaaaaa.

In the second test case, the given word itself is $k$ -complete.