1332_C. 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 ≤ i ≤ n; 
  * s has a period of k, i.e. s_i=s_{k+i} for all 1 ≤ i ≤ 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 ≤ i ≤ 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 ≤ t≤ 10^5) — the number of test cases.

The first line of each test case contains two integers n and k (1 ≤ k < n ≤ 2 ⋅ 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 ⋅ 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
import sys
input=sys.stdin.readline
t=int(input())
def find(no):
    global par
    if par[no]==no:
        return no
    par[no]=find(par[no])
    return par[no]
for _ in range(t):
    n,k=map(int,input().split())
    s=[ord(i)-97 for i in input()]
    
    par=[i for i in range(n)]
    vis=[0 for i in range(n)]
    for i in range(k):
        for j in range(i,n,k):
        
            vis[j]=1
            x=n-j-1
            aa=find(x)
            bb=find(j)
            par[bb]=aa
            if(j>i):
                aa=find(j-k)
                bb=find(j)
                par[bb]=aa
    tot=n
    aa=set(par)
    co=[[0 for i in range(26)] for j in range(n)]
   # print(par)
    for i in range(n):
        co[par[i]][s[i]]+=1
    tot-=sum([max(i) for i in co])
    print(tot)
        
        
    
            
                
            
    
    
    

```