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.

Input

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 $2 \cdot 10^5$.

Output

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.

见题面 Input 部分

见题面 Output 部分

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

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