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 Solution {
  
  	public static void main(String[] args) throws IOException {
  
  		Reader sc=new Reader();
  		sc.init(System.in);
  
  		BufferedWriter out = new BufferedWriter(
  				new OutputStreamWriter(System.out));
  
  		int t=Integer.parseInt(sc.nextLine());
  		
  		for (int i = 0; i < t; i++) {
  			
  			int n=Integer.parseInt(sc.nextLine());
  			
  			char[] c=sc.nextLine().toCharArray();
  			
  			char temp[]=new char[n];
  			
  			for(int j=0;j<n;j++) {
  				temp[j]=c[j];
  			}
  			
  			Arrays.sort(temp);
  			
  			int ans=0;
  			
  			for(int j=0;j<n;j++) {
  				if(temp[j]!=c[j]) ans++;
  			}
  			
  			out.write(Integer.toString(ans)+"\n");
  			
  		}
  		out.flush();
  	}
  
  	static class Reader {
  		static BufferedReader reader;
  		static StringTokenizer tokenizer;
  
  		/** call this method to initialize reader for InputStream */
  		static void init(InputStream input) {
  			reader = new BufferedReader(
  					new InputStreamReader(input) );
  			tokenizer = new StringTokenizer("");
  		}
  
  		/** get next word */
  		static String next() throws IOException {
  			while ( ! tokenizer.hasMoreTokens() ) {		
  				try {
  					tokenizer = new StringTokenizer(reader.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			}
  			return tokenizer.nextToken();
  		}
  
  		String nextLine() {
  			String str = "";
  			try {
  				str = reader.readLine();
  			} catch (IOException e) {
  				e.printStackTrace();
  			}
  			return str;
  		}
  
  		static int nextInt() throws IOException {
  			return Integer.parseInt( next() );
  		}
  
  		static long nextLong() throws IOException {
  			return Long.parseLong( next() );
  		}
  
  		static double nextDouble() throws IOException {
  			return Double.parseDouble( next() );
  		}
  	}
  }