#include <bits/stdc++.h>
using namespace std;
long long a[26][131073];
map<pair<long long, pair<long long, long long>>, long long> memo;
long long findans(long long st, long long en, long long req, long long n,
                  string &s) {
  if (st > en) {
    return 0;
  }
  if (st == en) {
    if (s[st] == (req + 'a')) {
      return 0;
    } else {
      return 1;
    }
  }
  if (memo.find(make_pair(st, make_pair(en, req))) != memo.end()) {
    return memo[make_pair(st, make_pair(en, req))];
  }
  long long tn = (en - st + 1);
  long long reqc = (en - st + 1) / 2;
  long long op1 = 0;
  if (st - 1 >= 0) {
    op1 = a[req][st + reqc - 1] - a[req][st - 1];
    op1 = reqc - op1;
  } else {
    op1 = a[req][st + reqc - 1];
    op1 = reqc - op1;
  }
  long long choice1 = op1 + findans(st + (reqc), en, req + 1, n, s);
  long long ts = st + (tn) / 2;
  if (ts - 1 >= 0) {
    op1 = a[req][en] - a[req][ts - 1];
    op1 = reqc - op1;
  } else {
    op1 = a[req][en];
    op1 = reqc - op1;
  }
  long long choice2 = op1 + findans(st, en - (reqc), req + 1, n, s);
  return memo[make_pair(st, make_pair(en, req))] = min(choice1, choice2);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  long long t1;
  cin >> t1;
  while (t1--) {
    long long n;
    cin >> n;
    string s;
    cin >> s;
    for (long long i = 0; i < 26; i++) {
      for (long long j = 0; j < n; j++) {
        a[i][j] = 0;
      }
    }
    for (long long i = 0; i < n; i++) {
      a[s[i] - 'a'][i] = 1;
    }
    for (long long i = 0; i < 26; i++) {
      for (long long j = 1; j < n; j++) {
        a[i][j] = a[i][j - 1] + a[i][j];
      }
    }
    memo.clear();
    long long ans = findans(0, n - 1, 0, n, s);
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 a[26][131073];
map<pair<long long, pair<long long, long long>>, long long> memo;
long long findans(long long st, long long en, long long req, long long n,
                  string &s) {
  if (st > en) {
    return 0;
  }
  if (st == en) {
    if (s[st] == (req + 'a')) {
      return 0;
    } else {
      return 1;
    }
  }
  if (memo.find(make_pair(st, make_pair(en, req))) != memo.end()) {
    return memo[make_pair(st, make_pair(en, req))];
  }
  long long tn = (en - st + 1);
  long long reqc = (en - st + 1) / 2;
  long long op1 = 0;
  if (st - 1 >= 0) {
    op1 = a[req][st + reqc - 1] - a[req][st - 1];
    op1 = reqc - op1;
  } else {
    op1 = a[req][st + reqc - 1];
    op1 = reqc - op1;
  }
  long long choice1 = op1 + findans(st + (reqc), en, req + 1, n, s);
  long long ts = st + (tn) / 2;
  if (ts - 1 >= 0) {
    op1 = a[req][en] - a[req][ts - 1];
    op1 = reqc - op1;
  } else {
    op1 = a[req][en];
    op1 = reqc - op1;
  }
  long long choice2 = op1 + findans(st, en - (reqc), req + 1, n, s);
  return memo[make_pair(st, make_pair(en, req))] = min(choice1, choice2);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  long long t1;
  cin >> t1;
  while (t1--) {
    long long n;
    cin >> n;
    string s;
    cin >> s;
    for (long long i = 0; i < 26; i++) {
      for (long long j = 0; j < n; j++) {
        a[i][j] = 0;
      }
    }
    for (long long i = 0; i < n; i++) {
      a[s[i] - 'a'][i] = 1;
    }
    for (long long i = 0; i < 26; i++) {
      for (long long j = 1; j < n; j++) {
        a[i][j] = a[i][j - 1] + a[i][j];
      }
    }
    memo.clear();
    long long ans = findans(0, n - 1, 0, n, s);
    cout << ans << '\n';
  }
  return 0;
}

```