#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1e9 + 7;
const long double error = 2e-6;
const long double PI = acosl(-1);
inline long long int MOD(long long int x, long long int m = mod) {
  long long int y = x % m;
  return (y >= 0) ? y : y + m;
}
const int inf = 1e9;
const long long int infl = 1061109567;
const int nmax = 1000 + 10;
string ss;
int DP(int start, int finish, char let) {
  if (start == finish) {
    if (ss[start - 1] == let)
      return 0;
    else
      return 1;
  }
  int t1 = start, t2 = (start + finish) / 2, t3 = t2 + 1, t4 = finish;
  int cnt = t4 - t3 + 1 - count(ss.begin() + t3 - 1, ss.begin() + t4, let) +
            DP(t1, t2, let + 1);
  int cntt = t2 - t1 + 1 - count(ss.begin() + t1 - 1, ss.begin() + t2, let) +
             DP(t3, t4, let + 1);
  return min(cnt, cntt);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    cin >> ss;
    int i;
    if (n == 1 && ss == "a") {
      cout << 0 << endl;
      continue;
    }
    if (n == 1) {
      cout << 1 << endl;
      continue;
    }
    cout << DP(1, n, 'a') << endl;
  }
  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;
const long long int mod = 1e9 + 7;
const long double error = 2e-6;
const long double PI = acosl(-1);
inline long long int MOD(long long int x, long long int m = mod) {
  long long int y = x % m;
  return (y >= 0) ? y : y + m;
}
const int inf = 1e9;
const long long int infl = 1061109567;
const int nmax = 1000 + 10;
string ss;
int DP(int start, int finish, char let) {
  if (start == finish) {
    if (ss[start - 1] == let)
      return 0;
    else
      return 1;
  }
  int t1 = start, t2 = (start + finish) / 2, t3 = t2 + 1, t4 = finish;
  int cnt = t4 - t3 + 1 - count(ss.begin() + t3 - 1, ss.begin() + t4, let) +
            DP(t1, t2, let + 1);
  int cntt = t2 - t1 + 1 - count(ss.begin() + t1 - 1, ss.begin() + t2, let) +
             DP(t3, t4, let + 1);
  return min(cnt, cntt);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    cin >> ss;
    int i;
    if (n == 1 && ss == "a") {
      cout << 0 << endl;
      continue;
    }
    if (n == 1) {
      cout << 1 << endl;
      continue;
    }
    cout << DP(1, n, 'a') << endl;
  }
  return 0;
}

```