#include <bits/stdc++.h>
using namespace std;
const int M = 1000000007;
const int MM = 998244353;
long long newmod(long long a, long long b) { return ((a % b) + b) % b; }
long long powM(long long a, long long b, long long m) {
  if (m <= 1) return 0;
  a %= m;
  long long ans = 1LL;
  while (b) {
    if (b & 1) ans = ans * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return ans;
}
long long poww(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a;
    a = a * a;
    b >>= 1;
  }
  return ans;
}
template <typename T, typename F>
void chmax(T &a, F b) {
  if (b > a) a = b;
}
template <typename T, typename F>
void chmin(T &a, F b) {
  if (b < a) a = b;
}
const long long N = 1e6 + 5;
long long sp[N];
void fill() {
  for (int i = 1; i < N; i++) sp[i] = i;
  for (int i = 2; i < N; i++) {
    for (int j = i; j < N; j += i) {
      if (sp[j] == j) sp[j] = i;
    }
  }
}
bool prime(long long x) {
  if (x == 2) return true;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) return false;
  }
  return true;
}
long long lala(string s, char x, long long n) {
  if (n == 1) {
    if (s[0] != x)
      return 1;
    else
      return 0;
  }
  long long cnt1 = 0, cnt2 = 0;
  string s1 = "", s2 = "";
  for (int i = 0; i < n / 2; i++) {
    if (s[i] != x) cnt1++;
    s1 += s[i];
  }
  for (int i = n / 2; i < n; i++) {
    if (s[i] != x) cnt2++;
    s2 += s[i];
  }
  return min(cnt1 + lala(s2, char(x + 1), n / 2),
             cnt2 + lala(s1, char(x + 1), n / 2));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    string s;
    cin >> s;
    cout << lala(s, 'a', n) << "\n";
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 M = 1000000007;
const int MM = 998244353;
long long newmod(long long a, long long b) { return ((a % b) + b) % b; }
long long powM(long long a, long long b, long long m) {
  if (m <= 1) return 0;
  a %= m;
  long long ans = 1LL;
  while (b) {
    if (b & 1) ans = ans * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return ans;
}
long long poww(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a;
    a = a * a;
    b >>= 1;
  }
  return ans;
}
template <typename T, typename F>
void chmax(T &a, F b) {
  if (b > a) a = b;
}
template <typename T, typename F>
void chmin(T &a, F b) {
  if (b < a) a = b;
}
const long long N = 1e6 + 5;
long long sp[N];
void fill() {
  for (int i = 1; i < N; i++) sp[i] = i;
  for (int i = 2; i < N; i++) {
    for (int j = i; j < N; j += i) {
      if (sp[j] == j) sp[j] = i;
    }
  }
}
bool prime(long long x) {
  if (x == 2) return true;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) return false;
  }
  return true;
}
long long lala(string s, char x, long long n) {
  if (n == 1) {
    if (s[0] != x)
      return 1;
    else
      return 0;
  }
  long long cnt1 = 0, cnt2 = 0;
  string s1 = "", s2 = "";
  for (int i = 0; i < n / 2; i++) {
    if (s[i] != x) cnt1++;
    s1 += s[i];
  }
  for (int i = n / 2; i < n; i++) {
    if (s[i] != x) cnt2++;
    s2 += s[i];
  }
  return min(cnt1 + lala(s2, char(x + 1), n / 2),
             cnt2 + lala(s1, char(x + 1), n / 2));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    string s;
    cin >> s;
    cout << lala(s, 'a', n) << "\n";
  }
  return 0;
}

```