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.OutputStream;
  import java.io.IOException;
  import java.io.InputStream;
  import java.io.PrintWriter;
  import java.util.*;
  import java.io.IOException;
  import java.io.BufferedReader;
  import java.io.InputStreamReader;
  import java.io.InputStream;
  
  public class java1 {
      public static void main(String[] args) {
          InputStream inputStream = System.in;
          OutputStream outputStream = System.out;
          InputReader in = new InputReader(inputStream);
          PrintWriter out = new PrintWriter(outputStream);
          TaskB solver = new TaskB();
          solver.solve(1, in, out);
          out.close();
      }
      static class TaskB {
  
          public void solve(int testNumber, InputReader in, PrintWriter out) {
              int t=in.nextInt();
            tt:  while(t-- >0)
              {
              	int n=in.nextInt();
              	String s=in.next();
              	char ch[]=s.toCharArray();
              	
              	Arrays.parallelSort(ch);
              	
              	int count=0;
              	for(int x=0;x<n;x++)
              	{
              		//out.println(ch1[x]);
              		if(ch[x]!=s.charAt(x)) {
              			count++;
              		}
              	}
              	out.println(count);
              	
              	
              }
          }
          static void sort(int[] a) {
      		ArrayList<Integer> l=new ArrayList<>();
      		for (int i:a) l.add(i);
      		Collections.sort(l);
      		for (int i=0; i<a.length; i++) a[i]=l.get(i);
      	}
          
      }
      static class InputReader {
          public BufferedReader reader;
          public StringTokenizer tokenizer;
  
          public InputReader(InputStream stream) {
              reader = new BufferedReader(new InputStreamReader(stream), 32768);
              tokenizer = null;
          }
          public String next() {
              while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                  try {
                      tokenizer = new StringTokenizer(reader.readLine());
                  } catch (IOException e) {
                      throw new RuntimeException(e);
                  }
              }
              return tokenizer.nextToken();
          }
          public int nextInt() {
              return Integer.parseInt(next());
          }
          public long nextLong() {
              return Long.parseLong(next());
          }
          public int[] inputarInt(int n)
          {
          	int ar[]=new int[n];
          	for(int x=0;x<n;x++)
          	{
          		ar[x]=nextInt();
          	}
          	return ar;
          }
          public long[] inputarLong(int n)
          {
          	long ar[]=new long[n];
          	for(int x=0;x<n;x++)
          	{
          		ar[x]=nextLong();
          	}
          	return ar;
          }
          
      }
  }