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.


-----Example-----
Input
4
6 2
abaaba
6 3
abaaba
36 9
hippopotomonstrosesquippedaliophobia
21 7
wudixiaoxingxingheclp

Output
2
0
23
16



-----Note-----

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

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

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'import sys\ninput=sys.stdin.readline\nt=int(input())\ndef find(no):\n    nonlocal par\n    if par[no]==no:\n        return no\n    par[no]=find(par[no])\n    return par[no]\nfor _ in range(t):\n    n,k=list(map(int,input().split()))\n    s=[ord(i)-97 for i in input()]\n    \n    par=[i for i in range(n)]\n    vis=[0 for i in range(n)]\n    for i in range(k):\n        for j in range(i,n,k):\n        \n            vis[j]=1\n            x=n-j-1\n            aa=find(x)\n            bb=find(j)\n            par[bb]=aa\n            if(j>i):\n                aa=find(j-k)\n                bb=find(j)\n                par[bb]=aa\n    tot=n\n    aa=set(par)\n    co=[[0 for i in range(26)] for j in range(n)]\n   # print(par)\n    for i in range(n):\n        co[par[i]][s[i]]+=1\n    tot-=sum([max(i) for i in co])\n    print(tot)\n        \n        \n    \n            \n                \n            \n    \n    \n    \n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: