#include <bits/stdc++.h>
using namespace std;
void enter(vector<long long int> &ar) {
  long long int n = ar.size();
  for (long long int i = 0; i < n; i++) {
    cin >> ar[i];
  }
}
void show(vector<long long int> &a) {
  long long int n = a.size();
  for (long long int i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << "\n";
}
void SieveOfEratosthenes(int n, vector<long long int> &pri) {
  vector<bool> prime(n + 1, true);
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      pri.push_back(p);
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
long long int min_change(vector<long long int> &s,
                         vector<vector<long long int>> &cha, long long int i,
                         long long int e, long long int size,
                         long long int currentChar) {
  if (size == 1) {
    if (s[i] == currentChar) {
      return 0;
    } else {
      return 1;
    }
  }
  if (i == 0) {
    long long int first = size / 2 - (cha[(i + e) / 2][currentChar]);
    long long int second =
        size / 2 - (cha[e][currentChar] - cha[(i + e) / 2][currentChar]);
    return min(
        first +
            min_change(s, cha, (i + e) / 2 + 1, e, size / 2, currentChar + 1),
        second + min_change(s, cha, i, (i + e) / 2, size / 2, currentChar + 1));
  } else {
    long long int first =
        size / 2 - (cha[(i + e) / 2][currentChar] - cha[i - 1][currentChar]);
    long long int second =
        size / 2 - (cha[e][currentChar] - cha[(i + e) / 2][currentChar]);
    return min(
        first +
            min_change(s, cha, (i + e) / 2 + 1, e, size / 2, currentChar + 1),
        second + min_change(s, cha, i, (i + e) / 2, size / 2, currentChar + 1));
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    string s;
    cin >> ws;
    cin >> s;
    cin >> ws;
    vector<long long int> st(n);
    for (long long int i = 0; i < n; i++) {
      st[i] = s[i] - 'a';
    }
    vector<vector<long long int>> cha(n, vector<long long int>(26, 0));
    cha[0][s[0] - 'a']++;
    for (long long int i = 1; i < n; i++) {
      for (long long int j = 0; j < 26; j++) {
        if (j == s[i] - 'a') {
          cha[i][j] = cha[i - 1][j] + 1;
        } else {
          cha[i][j] = cha[i - 1][j];
        }
      }
    }
    long long int res = min_change(st, cha, 0, n - 1, n, 0);
    cout << res << "\n";
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
void enter(vector<long long int> &ar) {
  long long int n = ar.size();
  for (long long int i = 0; i < n; i++) {
    cin >> ar[i];
  }
}
void show(vector<long long int> &a) {
  long long int n = a.size();
  for (long long int i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << "\n";
}
void SieveOfEratosthenes(int n, vector<long long int> &pri) {
  vector<bool> prime(n + 1, true);
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      pri.push_back(p);
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
long long int min_change(vector<long long int> &s,
                         vector<vector<long long int>> &cha, long long int i,
                         long long int e, long long int size,
                         long long int currentChar) {
  if (size == 1) {
    if (s[i] == currentChar) {
      return 0;
    } else {
      return 1;
    }
  }
  if (i == 0) {
    long long int first = size / 2 - (cha[(i + e) / 2][currentChar]);
    long long int second =
        size / 2 - (cha[e][currentChar] - cha[(i + e) / 2][currentChar]);
    return min(
        first +
            min_change(s, cha, (i + e) / 2 + 1, e, size / 2, currentChar + 1),
        second + min_change(s, cha, i, (i + e) / 2, size / 2, currentChar + 1));
  } else {
    long long int first =
        size / 2 - (cha[(i + e) / 2][currentChar] - cha[i - 1][currentChar]);
    long long int second =
        size / 2 - (cha[e][currentChar] - cha[(i + e) / 2][currentChar]);
    return min(
        first +
            min_change(s, cha, (i + e) / 2 + 1, e, size / 2, currentChar + 1),
        second + min_change(s, cha, i, (i + e) / 2, size / 2, currentChar + 1));
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    string s;
    cin >> ws;
    cin >> s;
    cin >> ws;
    vector<long long int> st(n);
    for (long long int i = 0; i < n; i++) {
      st[i] = s[i] - 'a';
    }
    vector<vector<long long int>> cha(n, vector<long long int>(26, 0));
    cha[0][s[0] - 'a']++;
    for (long long int i = 1; i < n; i++) {
      for (long long int j = 0; j < 26; j++) {
        if (j == s[i] - 'a') {
          cha[i][j] = cha[i - 1][j] + 1;
        } else {
          cha[i][j] = cha[i - 1][j];
        }
      }
    }
    long long int res = min_change(st, cha, 0, n - 1, n, 0);
    cout << res << "\n";
  }
  return 0;
}

```