#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
const double pi = acos(-1.0);
const double eps = 1e-6;
int dfs(string s, int n, char ch) {
  if (n == 1) return ch != s[0];
  int tmp1 = 0, tmp2 = 0;
  for (int i = 0; i < n / 2; i++)
    if (s[i] != ch) tmp1++;
  for (int i = n / 2; i < n; i++)
    if (s[i] != ch) tmp2++;
  int ans = tmp2 + dfs(s.substr(0, n / 2), n / 2, ch + 1);
  ans = min(ans, tmp1 + dfs(s.substr(n / 2, n), n / 2, ch + 1));
  return ans;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    string s;
    cin >> s;
    cout << dfs(s, n, 'a') << endl;
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const int MAXN = 1e5 + 10;
const double pi = acos(-1.0);
const double eps = 1e-6;
int dfs(string s, int n, char ch) {
  if (n == 1) return ch != s[0];
  int tmp1 = 0, tmp2 = 0;
  for (int i = 0; i < n / 2; i++)
    if (s[i] != ch) tmp1++;
  for (int i = n / 2; i < n; i++)
    if (s[i] != ch) tmp2++;
  int ans = tmp2 + dfs(s.substr(0, n / 2), n / 2, ch + 1);
  ans = min(ans, tmp1 + dfs(s.substr(n / 2, n), n / 2, ch + 1));
  return ans;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    string s;
    cin >> s;
    cout << dfs(s, n, 'a') << endl;
  }
  return 0;
}

```