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.

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 a aaa aa.

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

Solution:

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
typedef double db;
mt19937 mrand(random_device{}());
const ll mod=1000000007;
int rnd(int x) { return mrand() % x;}
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
// head

const int N=201000;
int _,n,k,f[N];
char s[N];
map<char,int> hs[N];

int find(int x) {
return f[x]==x?x:f[x]=find(f[x]);
}

int main() {
for (scanf("%d",&_);_;_--) {
scanf("%d%d",&n,&k);
scanf("%s",s);
for (int i=0;i<n;i++) f[i]=i;
		for (int i=0;i<n;i++) f[find(i)]=find(n-1-i);
		for (int i=0;i<n-k;i++) f[find(i)]=find(i+k);
		for (int i=0;i<n;i++) hs[i].clear();
		for (int i=0;i<n;i++) hs[find(i)][s[i]]++;
		int ans=0;
		for (int i=0;i<n;i++) {
			int s=0,mx=0;
			for (auto p:hs[i]) s+=p.se,mx=max(mx,p.se);
			ans+=s-mx;
		}
		printf("%d\n",ans);
	}
}