#include <bits/stdc++.h>
using namespace std;
void fs(long long int &number) {
  bool negative = false;
  register long long int c;
  number = 0;
  c = getchar();
  if (c == '-') {
    negative = true;
    c = getchar();
  }
  for (; (c > 47 && c < 58); c = getchar()) number = number * 10 + c - 48;
  if (negative) number *= -1;
}
long long int power(long long int a, long long int x) {
  a %= 1000000007;
  long long int res = 1;
  while (x > 0) {
    if (x & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
    x >>= 1;
  }
  return res;
}
long long int max(long long int a, long long int b) {
  if (a > b) {
    return a;
  } else {
    return b;
  }
}
long long int min(long long int a, long long int b) {
  if (a > b) {
    return b;
  } else {
    return a;
  }
}
long long int madd(long long int a, long long int b) {
  long long int ans;
  ans = (a % 1000000007 + b % 1000000007) % 1000000007;
  return ans;
}
long long int mmult(long long int a, long long int b) {
  long long int ans = 1;
  ans = ((a % 1000000007) * (b % 1000000007)) % 1000000007;
  return ans % 1000000007;
}
long long int modinverse(long long int a) { return power(a, 1000000007 - 2); }
string sub(string s, long long int start, long long int end) {
  string temp = "";
  for (long long int i = start; i <= end; i++) {
    temp += s[i];
  }
  return temp;
}
long long int recurse(string s, long long int id) {
  if (s.length() == 1) {
    if (s[0] == 'a' + id) {
      return 0;
    } else {
      return 1;
    }
  }
  long long int x = s.length() / 2;
  string s1 = sub(s, 0, x - 1);
  string s2 = sub(s, x, s.length() - 1);
  char rc = 'a' + id;
  long long int c1 = 0, c2 = 0;
  for (long long int i = 0; i < s1.length(); i++) {
    if (s1[i] != rc) {
      c1++;
    }
  }
  for (long long int i = 0; i < s2.length(); i++) {
    if (s2[i] != rc) {
      c2++;
    }
  }
  return min(c1 + recurse(s2, id + 1), c2 + recurse(s1, id + 1));
}
void its_her_choice() {
  long long int n;
  cin >> n;
  string s;
  cin >> s;
  cout << recurse(s, 0);
  cout << "\n";
}
signed main() {
  long long int t;
  cin >> t;
  for (long long int casn = 1; casn < t + 1; casn++) {
    its_her_choice();
  }
}

### Prompt

Please formulate 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;
void fs(long long int &number) {
  bool negative = false;
  register long long int c;
  number = 0;
  c = getchar();
  if (c == '-') {
    negative = true;
    c = getchar();
  }
  for (; (c > 47 && c < 58); c = getchar()) number = number * 10 + c - 48;
  if (negative) number *= -1;
}
long long int power(long long int a, long long int x) {
  a %= 1000000007;
  long long int res = 1;
  while (x > 0) {
    if (x & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
    x >>= 1;
  }
  return res;
}
long long int max(long long int a, long long int b) {
  if (a > b) {
    return a;
  } else {
    return b;
  }
}
long long int min(long long int a, long long int b) {
  if (a > b) {
    return b;
  } else {
    return a;
  }
}
long long int madd(long long int a, long long int b) {
  long long int ans;
  ans = (a % 1000000007 + b % 1000000007) % 1000000007;
  return ans;
}
long long int mmult(long long int a, long long int b) {
  long long int ans = 1;
  ans = ((a % 1000000007) * (b % 1000000007)) % 1000000007;
  return ans % 1000000007;
}
long long int modinverse(long long int a) { return power(a, 1000000007 - 2); }
string sub(string s, long long int start, long long int end) {
  string temp = "";
  for (long long int i = start; i <= end; i++) {
    temp += s[i];
  }
  return temp;
}
long long int recurse(string s, long long int id) {
  if (s.length() == 1) {
    if (s[0] == 'a' + id) {
      return 0;
    } else {
      return 1;
    }
  }
  long long int x = s.length() / 2;
  string s1 = sub(s, 0, x - 1);
  string s2 = sub(s, x, s.length() - 1);
  char rc = 'a' + id;
  long long int c1 = 0, c2 = 0;
  for (long long int i = 0; i < s1.length(); i++) {
    if (s1[i] != rc) {
      c1++;
    }
  }
  for (long long int i = 0; i < s2.length(); i++) {
    if (s2[i] != rc) {
      c2++;
    }
  }
  return min(c1 + recurse(s2, id + 1), c2 + recurse(s1, id + 1));
}
void its_her_choice() {
  long long int n;
  cin >> n;
  string s;
  cin >> s;
  cout << recurse(s, 0);
  cout << "\n";
}
signed main() {
  long long int t;
  cin >> t;
  for (long long int casn = 1; casn < t + 1; casn++) {
    its_her_choice();
  }
}

```