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.
#include <bits/stdc++.h>
using namespace std;
int main(void) {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    string S;
    cin >> S;
    int ans = 0;
    bool used[200010] = {};
    for (int i = 0; i < n; i++) {
      if (used[i]) continue;
      int cnt[26] = {};
      int sum = 0;
      set<int> st;
      for (int j = i; j < n; j += k) {
        used[j] = true;
        used[n - j - 1] = true;
        st.insert(j);
        st.insert(n - j - 1);
      }
      int mx = 0;
      for (auto itr = st.begin(); itr != st.end(); itr++) {
        cnt[S[*itr] - 'a']++;
        mx = max(mx, cnt[S[*itr] - 'a']);
      }
      ans += (int)(st.size()) - mx;
    }
    cout << ans << endl;
  }
  return 0;
}
