#include <bits/stdc++.h>
using namespace std;
string a;
int f[200005][30];
int t, n;
void nhap() {
  ios_base::sync_with_stdio(0);
  cin >> t;
}
int calc1(int l, int r, char x) {
  return (r - l + 1) - f[r + 1][x - 'a'] + f[l][x - 'a'];
}
int calc2(int l, int r, char x) {
  if (r > l)
    return min(calc1(l, l + (r - l + 1) / 2 - 1, x) +
                   calc2(l + (r - l + 1) / 2, r, x + 1),
               calc1(l + (r - l + 1) / 2, r, x) +
                   calc2(l, l + (r - l + 1) / 2 - 1, x + 1));
  else if (a[l] != x)
    return 1;
  else
    return 0;
}
void xl() {
  for (int i = 0; i <= n; ++i)
    for (int j = 0; j <= 26; ++j) f[i][j] = 0;
  for (int i = 0; i < a.length(); ++i)
    for (int j = 0; j <= 26; ++j) {
      f[i + 1][j] = f[i][j];
      if (a[i] - 'a' == j) ++f[i + 1][j];
    }
  int res = min(calc1(0, n / 2 - 1, 'a') + calc2(n / 2, n - 1, 'b'),
                calc1(n / 2, n - 1, 'a') + calc2(0, n / 2 - 1, 'b'));
  if (n != 1)
    cout << res << '\n';
  else if (a[0] != 'a')
    cout << "1" << '\n';
  else
    cout << "0" << '\n';
}
int main() {
  nhap();
  for (int ii = 1; ii <= t; ++ii) {
    cin >> n;
    cin >> a;
    xl();
  }
  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;
string a;
int f[200005][30];
int t, n;
void nhap() {
  ios_base::sync_with_stdio(0);
  cin >> t;
}
int calc1(int l, int r, char x) {
  return (r - l + 1) - f[r + 1][x - 'a'] + f[l][x - 'a'];
}
int calc2(int l, int r, char x) {
  if (r > l)
    return min(calc1(l, l + (r - l + 1) / 2 - 1, x) +
                   calc2(l + (r - l + 1) / 2, r, x + 1),
               calc1(l + (r - l + 1) / 2, r, x) +
                   calc2(l, l + (r - l + 1) / 2 - 1, x + 1));
  else if (a[l] != x)
    return 1;
  else
    return 0;
}
void xl() {
  for (int i = 0; i <= n; ++i)
    for (int j = 0; j <= 26; ++j) f[i][j] = 0;
  for (int i = 0; i < a.length(); ++i)
    for (int j = 0; j <= 26; ++j) {
      f[i + 1][j] = f[i][j];
      if (a[i] - 'a' == j) ++f[i + 1][j];
    }
  int res = min(calc1(0, n / 2 - 1, 'a') + calc2(n / 2, n - 1, 'b'),
                calc1(n / 2, n - 1, 'a') + calc2(0, n / 2 - 1, 'b'));
  if (n != 1)
    cout << res << '\n';
  else if (a[0] != 'a')
    cout << "1" << '\n';
  else
    cout << "0" << '\n';
}
int main() {
  nhap();
  for (int ii = 1; ii <= t; ++ii) {
    cin >> n;
    cin >> a;
    xl();
  }
  return 0;
}

```