description: A string s of length n, consisting of lowercase letters of the English alphabet, is given.
  
  You must choose some number k between 0 and n. Then, you select k characters of s and permute them however you want. In this process, the positions of the other n-k characters remain unchanged. You have to perform this operation exactly once.
  
  For example, if s="andrea", you can choose the k=4 characters "a_d_ea" and permute them into "d_e_aa" so that after the operation the string becomes "dneraa".
  
  Determine the minimum k so that it is possible to sort s alphabetically (that is, after the operation its characters appear in alphabetical order).
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
  
  The first line of each test case contains one integer n (1 ≤ n ≤ 40) — the length of the string.
  
  The second line of each test case contains the string s. It is guaranteed that s contains only lowercase letters of the English alphabet.
  
  Output
  
  For each test case, output the minimum k that allows you to obtain a string sorted alphabetically, through the operation described above.
  
  Example
  
  Input
  
  
  4
  3
  lol
  10
  codeforces
  5
  aaaaa
  4
  dcba
  
  
  Output
  
  
  2
  6
  0
  4
  
  Note
  
  In the first test case, we can choose the k=2 characters "_ol" and rearrange them as "_lo" (so the resulting string is "llo"). It is not possible to sort the string choosing strictly less than 2 characters.
  
  In the second test case, one possible way to sort s is to consider the k=6 characters "_o__force_" and rearrange them as "_c__efoor_" (so the resulting string is "ccdeefoors"). One can show that it is not possible to sort the string choosing strictly less than 6 characters.
  
  In the third test case, string s is already sorted (so we can choose k=0 characters).
  
  In the fourth test case, we can choose all k=4 characters "dcba" and reverse the whole string (so the resulting string is "abcd").
solution: import java.io.*;
  import java.util.*;
   
  
  public class FastIO {
  	BufferedReader br;
      StringTokenizer st;
  
      public FastIO()
      {
          br = new BufferedReader(
              new InputStreamReader(System.in));
      }
  
      String next()
      {
          while (st == null || !st.hasMoreElements()) {
              try {
                  st = new StringTokenizer(br.readLine());
              }
              catch (IOException e) {
                  e.printStackTrace();
              }
          }
          return st.nextToken();
      }
  
      int nextInt() { return Integer.parseInt(next()); }
  
      long nextLong() { return Long.parseLong(next()); }
      
      float nextFloat() {
      	return Float.parseFloat(next());
      }
  
      double nextDouble()
      {
          return Double.parseDouble(next());
      }
  
      String nextLine()
      {
          String str = "";
          try {
              str = br.readLine();
          }
          catch (IOException e) {
              e.printStackTrace();
          }
          return str;
      }
   
    
      public static void main(String[] args) throws IOException
      {
         try {
      	  FastIO sc=new FastIO();
      	  int t=sc.nextInt();
      	  while(t-- >0) {
      		 long n=sc.nextLong();
      		 String s=sc.next();
      		 char[] ans=s.toCharArray();
      		 Arrays.sort(ans);
      			 int count=0;
      			 for(int i=0;i<n;i++) {
      				 if(s.charAt(i)!=ans[i]) {
      					 count++;
      				 }
      		      }
      			 System.out.println(count);
    
      		 
      		 
      	  }
         }catch(Exception e) {
      	   return;
         }
      }	
  }