#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5, M = 1e6 + 6, OO = 0x3f3f3f3f;
int tc;
int n;
string s;
int freq[26][131075];
int mem[131075][26];
int solve(int lo = 0, int hi = n - 1, int strc = 0) {
  if (lo > hi || strc > (n / 2) || strc == 26) return 0;
  if (mem[lo][strc] != -1) return mem[lo][strc];
  int mid = (hi + lo) / 2;
  int freqRight = freq[strc][hi] - freq[strc][mid];
  int freqLeft = freq[strc][mid] - (lo == 0 ? 0 : freq[strc][lo - 1]);
  long long ans = 0;
  if (freqLeft == max((hi - lo + 1) / 2, 1))
    ans += solve(mid + 1, hi, strc + 1);
  else if (freqRight == max((hi - lo + 1) / 2, 1)) {
    ans += solve(lo, mid, strc + 1);
  } else {
    long long tmp1 = 0, tmp2 = 0;
    tmp1 =
        max(((hi - lo + 1) / 2) - freqLeft, 1) + solve(mid + 1, hi, strc + 1);
    ;
    tmp2 = max(((hi - lo + 1) / 2) - freqRight, 1) + solve(lo, mid, strc + 1);
    ans += min(tmp1, tmp2);
  }
  return mem[lo][strc] = ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> tc;
  while (tc--) {
    cin >> n;
    cin >> s;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j <= 'z' - 'a'; j++) {
        freq[j][i] = 0;
        mem[i][j] = -1;
      }
    }
    for (int i = 0; i < n; i++) {
      freq[s[i] - 'a'][i]++;
    }
    for (int i = 0; i <= 'z' - 'a'; i++) {
      for (int j = 1; j < n; j++) freq[i][j] += freq[i][j - 1];
    }
    int hi = n - 1, lo = 0, mid;
    int strc = 0;
    long long ans = 0;
    cout << solve() << endl;
  }
  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;
const int N = 2e5 + 5, M = 1e6 + 6, OO = 0x3f3f3f3f;
int tc;
int n;
string s;
int freq[26][131075];
int mem[131075][26];
int solve(int lo = 0, int hi = n - 1, int strc = 0) {
  if (lo > hi || strc > (n / 2) || strc == 26) return 0;
  if (mem[lo][strc] != -1) return mem[lo][strc];
  int mid = (hi + lo) / 2;
  int freqRight = freq[strc][hi] - freq[strc][mid];
  int freqLeft = freq[strc][mid] - (lo == 0 ? 0 : freq[strc][lo - 1]);
  long long ans = 0;
  if (freqLeft == max((hi - lo + 1) / 2, 1))
    ans += solve(mid + 1, hi, strc + 1);
  else if (freqRight == max((hi - lo + 1) / 2, 1)) {
    ans += solve(lo, mid, strc + 1);
  } else {
    long long tmp1 = 0, tmp2 = 0;
    tmp1 =
        max(((hi - lo + 1) / 2) - freqLeft, 1) + solve(mid + 1, hi, strc + 1);
    ;
    tmp2 = max(((hi - lo + 1) / 2) - freqRight, 1) + solve(lo, mid, strc + 1);
    ans += min(tmp1, tmp2);
  }
  return mem[lo][strc] = ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> tc;
  while (tc--) {
    cin >> n;
    cin >> s;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j <= 'z' - 'a'; j++) {
        freq[j][i] = 0;
        mem[i][j] = -1;
      }
    }
    for (int i = 0; i < n; i++) {
      freq[s[i] - 'a'][i]++;
    }
    for (int i = 0; i <= 'z' - 'a'; i++) {
      for (int j = 1; j < n; j++) freq[i][j] += freq[i][j - 1];
    }
    int hi = n - 1, lo = 0, mid;
    int strc = 0;
    long long ans = 0;
    cout << solve() << endl;
  }
  return 0;
}

```