#include <bits/stdc++.h>
using namespace std;
long long bigmod(long long a, int p, long long int m) {
  if (p == 0) return 1;
  if (p & 1) {
    return ((a % m) * (bigmod(a, p - 1, m))) % m;
  } else {
    long long tmp = bigmod(a, p / 2, m);
    return (tmp * tmp) % m;
  }
}
vector<long long int> v1;
void SieveOfEratosthenes(long long int n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (long long int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (long long int p = 2; p <= n; p++)
    if (prime[p]) v1.push_back(p);
}
string s;
long long int n;
char c;
long long int solve(long long int i, long long int len, char d) {
  if (len == 1) {
    if (s[i] == d)
      return 0;
    else
      return 1;
  }
  long long int j;
  long long int ct1 = 0, ct2 = 0;
  for (j = i; j < i + len / 2; j++) {
    if (s[j] != d) ct1++;
  }
  for (j = i + len / 2; j < i + len; j++) {
    if (s[j] != d) ct2++;
  }
  long long int ans1 = ct1 + solve(i + len / 2, len / 2, d + 1);
  long long int ans2 = ct2 + solve(i, len / 2, d + 1);
  return min(ans1, ans2);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int chu;
  cin >> chu;
  while (chu--) {
    cin >> n >> s;
    c = 'a';
    cout << solve(0, n, 'a');
    cout << "\n";
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
long long bigmod(long long a, int p, long long int m) {
  if (p == 0) return 1;
  if (p & 1) {
    return ((a % m) * (bigmod(a, p - 1, m))) % m;
  } else {
    long long tmp = bigmod(a, p / 2, m);
    return (tmp * tmp) % m;
  }
}
vector<long long int> v1;
void SieveOfEratosthenes(long long int n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (long long int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (long long int p = 2; p <= n; p++)
    if (prime[p]) v1.push_back(p);
}
string s;
long long int n;
char c;
long long int solve(long long int i, long long int len, char d) {
  if (len == 1) {
    if (s[i] == d)
      return 0;
    else
      return 1;
  }
  long long int j;
  long long int ct1 = 0, ct2 = 0;
  for (j = i; j < i + len / 2; j++) {
    if (s[j] != d) ct1++;
  }
  for (j = i + len / 2; j < i + len; j++) {
    if (s[j] != d) ct2++;
  }
  long long int ans1 = ct1 + solve(i + len / 2, len / 2, d + 1);
  long long int ans2 = ct2 + solve(i, len / 2, d + 1);
  return min(ans1, ans2);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int chu;
  cin >> chu;
  while (chu--) {
    cin >> n >> s;
    c = 'a';
    cout << solve(0, n, 'a');
    cout << "\n";
  }
  return 0;
}

```