#include <bits/stdc++.h>
using namespace std;
string s;
vector<vector<long long int>> A;
long long int fn(long long int l, long long int r, char x, long long int ld,
                 long long int rd) {
  if (l == r) {
    long long int a = 0, b = 0;
    if (s[l] != x) a++;
    if (s[ld] != x + 1) a++;
    if (s[l] != x + 1) b++;
    if (s[ld] != x) b++;
    return (min(a, b));
  }
  long long int coun;
  if (l != 0)
    coun = A[r][x - 97] - A[l - 1][x - 97];
  else
    coun = A[r][x - 97];
  return (coun + min(fn(ld, (ld + rd) / 2, x + 1, ((ld + rd) / 2) + 1, rd),
                     fn(((ld + rd) / 2) + 1, rd, x + 1, ld, (ld + rd) / 2)));
}
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(NULL);
  cin.tie(NULL);
  long long int t, i, j, n;
  cin >> t;
  while (t--) {
    cin >> n >> s;
    vector<long long int> temp(26, 1);
    A.clear();
    for (i = 0; i < n; i++) {
      A.push_back(temp);
    }
    if (n == 1) {
      long long int ans = 0;
      if (s[0] != 'a') ans++;
      cout << ans << "\n";
      continue;
    }
    for (i = 0; i < n; i++) {
      A[i][s[i] - 97] = 0;
    }
    for (j = 0; j < 26; j++) {
      for (i = 1; i < n; i++) {
        A[i][j] += A[i - 1][j];
      }
    }
    cout << min(fn(0, (n / 2) - 1, 'a', (n / 2), n - 1),
                fn((n / 2), n - 1, 'a', 0, (n / 2) - 1))
         << "\n";
  }
  return (0);
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
string s;
vector<vector<long long int>> A;
long long int fn(long long int l, long long int r, char x, long long int ld,
                 long long int rd) {
  if (l == r) {
    long long int a = 0, b = 0;
    if (s[l] != x) a++;
    if (s[ld] != x + 1) a++;
    if (s[l] != x + 1) b++;
    if (s[ld] != x) b++;
    return (min(a, b));
  }
  long long int coun;
  if (l != 0)
    coun = A[r][x - 97] - A[l - 1][x - 97];
  else
    coun = A[r][x - 97];
  return (coun + min(fn(ld, (ld + rd) / 2, x + 1, ((ld + rd) / 2) + 1, rd),
                     fn(((ld + rd) / 2) + 1, rd, x + 1, ld, (ld + rd) / 2)));
}
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(NULL);
  cin.tie(NULL);
  long long int t, i, j, n;
  cin >> t;
  while (t--) {
    cin >> n >> s;
    vector<long long int> temp(26, 1);
    A.clear();
    for (i = 0; i < n; i++) {
      A.push_back(temp);
    }
    if (n == 1) {
      long long int ans = 0;
      if (s[0] != 'a') ans++;
      cout << ans << "\n";
      continue;
    }
    for (i = 0; i < n; i++) {
      A[i][s[i] - 97] = 0;
    }
    for (j = 0; j < 26; j++) {
      for (i = 1; i < n; i++) {
        A[i][j] += A[i - 1][j];
      }
    }
    cout << min(fn(0, (n / 2) - 1, 'a', (n / 2), n - 1),
                fn((n / 2), n - 1, 'a', 0, (n / 2) - 1))
         << "\n";
  }
  return (0);
}

```