#include <bits/stdc++.h>
using namespace std;
int xx[4] = {0, 0, 1, -1};
int yy[4] = {-1, 1, 0, 0};
int main() {
  long long t;
  scanf("%lld", &t);
  ;
  while (t--) {
    long long n;
    scanf("%lld", &n);
    string s;
    cin >> s;
    if (s.size() == 1) {
      if (s[0] == 'a')
        cout << 0 << endl;
      else
        cout << 1 << endl;
      continue;
    }
    long long size = log2(n + 1) + 1;
    vector<long long> vec[size];
    char c = 'a';
    long long ind = 0;
    for (int i = n / 2; i >= 1;) {
      long long cnt = 0;
      for (int j = 0; j < n; j++) {
        if (s[j] != c) cnt++;
        if ((j + 1) % i == 0) {
          vec[ind].push_back(cnt);
          cnt = 0;
        }
      }
      i /= 2;
      ind++;
      c++;
    }
    for (int i = 0; i < n; i++) {
      if (s[i] != c)
        vec[ind].push_back(1);
      else
        vec[ind].push_back(0);
    }
    for (int i = 1; i < size - 1; i++) {
      long long cnt = 1;
      for (int j = 0; j < vec[i].size(); j += 4) {
        vec[i][j] += vec[i - 1][cnt];
        vec[i][j + 1] += vec[i - 1][cnt];
        vec[i][j + 2] += vec[i - 1][cnt - 1];
        vec[i][j + 3] += vec[i - 1][cnt - 1];
        cnt += 2;
      }
    }
    long long res = LLONG_MAX;
    for (int i = 0; i < vec[size - 1].size(); i += 2) {
      vec[size - 1][i] += vec[size - 2][i + 1];
      vec[size - 1][i + 1] += vec[size - 2][i];
      res = min(res, vec[size - 1][i]);
      res = min(res, vec[size - 1][i + 1]);
    }
    cout << res << endl;
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

You are given a string s[1 ... n] consisting of lowercase Latin letters. It is guaranteed that n = 2^k for some integer k ≥ 0.

The string s[1 ... n] is called c-good if at least one of the following three conditions is satisfied:

  * The length of s is 1, and it consists of the character c (i.e. s_1=c);
  * The length of s is greater than 1, the first half of the string consists of only the character c (i.e. s_1=s_2=...=s_{n/2}=c) and the second half of the string (i.e. the string s_{n/2 + 1}s_{n/2 + 2} ... s_n) is a (c+1)-good string; 
  * The length of s is greater than 1, the second half of the string consists of only the character c (i.e. s_{n/2 + 1}=s_{n/2 + 2}=...=s_n=c) and the first half of the string (i.e. the string s_1s_2 ... s_{n/2}) is a (c+1)-good string. 



For example: "aabc" is 'a'-good, "ffgheeee" is 'e'-good.

In one move, you can choose one index i from 1 to n and replace s_i with any lowercase Latin letter (any character from 'a' to 'z').

Your task is to find the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string for c= 'a'). It is guaranteed that the answer always exists.

You have to answer t independent test cases.

Another example of an 'a'-good string is as follows. Consider the string s = "cdbbaaaa". It is an 'a'-good string, because:

  * the second half of the string ("aaaa") consists of only the character 'a'; 
  * the first half of the string ("cdbb") is 'b'-good string, because: 
    * the second half of the string ("bb") consists of only the character 'b'; 
    * the first half of the string ("cd") is 'c'-good string, because: 
      * the first half of the string ("c") consists of only the character 'c'; 
      * the second half of the string ("d") is 'd'-good string. 

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1 ≤ n ≤ 131~072) — the length of s. It is guaranteed that n = 2^k for some integer k ≥ 0. The second line of the test case contains the string s consisting of n lowercase Latin letters.

It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer — the minimum number of moves required to obtain an 'a'-good string from s (i.e. c-good string with c = 'a'). It is guaranteed that the answer exists.

Example

Input


6
8
bbdcaaaa
8
asdfghjk
8
ceaaaabb
8
bbaaddcc
1
z
2
ac


Output


0
7
4
5
1
1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int xx[4] = {0, 0, 1, -1};
int yy[4] = {-1, 1, 0, 0};
int main() {
  long long t;
  scanf("%lld", &t);
  ;
  while (t--) {
    long long n;
    scanf("%lld", &n);
    string s;
    cin >> s;
    if (s.size() == 1) {
      if (s[0] == 'a')
        cout << 0 << endl;
      else
        cout << 1 << endl;
      continue;
    }
    long long size = log2(n + 1) + 1;
    vector<long long> vec[size];
    char c = 'a';
    long long ind = 0;
    for (int i = n / 2; i >= 1;) {
      long long cnt = 0;
      for (int j = 0; j < n; j++) {
        if (s[j] != c) cnt++;
        if ((j + 1) % i == 0) {
          vec[ind].push_back(cnt);
          cnt = 0;
        }
      }
      i /= 2;
      ind++;
      c++;
    }
    for (int i = 0; i < n; i++) {
      if (s[i] != c)
        vec[ind].push_back(1);
      else
        vec[ind].push_back(0);
    }
    for (int i = 1; i < size - 1; i++) {
      long long cnt = 1;
      for (int j = 0; j < vec[i].size(); j += 4) {
        vec[i][j] += vec[i - 1][cnt];
        vec[i][j + 1] += vec[i - 1][cnt];
        vec[i][j + 2] += vec[i - 1][cnt - 1];
        vec[i][j + 3] += vec[i - 1][cnt - 1];
        cnt += 2;
      }
    }
    long long res = LLONG_MAX;
    for (int i = 0; i < vec[size - 1].size(); i += 2) {
      vec[size - 1][i] += vec[size - 2][i + 1];
      vec[size - 1][i + 1] += vec[size - 2][i];
      res = min(res, vec[size - 1][i]);
      res = min(res, vec[size - 1][i + 1]);
    }
    cout << res << endl;
  }
  return 0;
}

```