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 CFGlobal15A {
  	public static void solve(IO io) {
  
  		int k = io.getInt();
  		String s = io.getWord();
  		
  		char [] lol = s.toCharArray();
  		char [] copy = lol.clone();
  		
  		Arrays.sort(copy);
  		
  		
  		int count = 0;
  		for (int i = 0; i < copy.length; i++) {
  			//System.out.print(copy[i]);
  			if (copy[i] != lol[i]) {
  				count++;
  			}
  			
  		}
  		System.out.println(count);
  		
  		
  	}
  
  	public static void main(String[] args) {
  		IO io = new IO();
  		int t = io.getInt();
  		for (int testIndex = 0; testIndex < t; testIndex++) {
  			solve(io);
  		}
  		io.close();
  	}
  
  	static class IO extends PrintWriter {
  
  		public IO() {
  			super(new BufferedOutputStream(System.out));
  			r = new BufferedReader(new InputStreamReader(System.in));
  		}
  
  		public IO(InputStream i) {
  			super(new BufferedOutputStream(System.out));
  			r = new BufferedReader(new InputStreamReader(i));
  		}
  
  		public IO(InputStream i, OutputStream o) {
  			super(new BufferedOutputStream(o));
  			r = new BufferedReader(new InputStreamReader(i));
  		}
  
  		public boolean hasMoreTokens() {
  			return peekToken() != null;
  		}
  
  		public int getInt() {
  			return Integer.parseInt(nextToken());
  		}
  
  		public double getDouble() {
  			return Double.parseDouble(nextToken());
  		}
  
  		public long getLong() {
  			return Long.parseLong(nextToken());
  		}
  
  		public String getWord() {
  			return nextToken();
  		}
  
  		private BufferedReader r;
  		private String line;
  		private StringTokenizer st;
  		private String token;
  
  		private String peekToken() {
  			if (token == null)
  				try {
  					while (st == null || !st.hasMoreTokens()) {
  						line = r.readLine();
  						if (line == null)
  							return null;
  						st = new StringTokenizer(line);
  					}
  					token = st.nextToken();
  				} catch (IOException e) {
  				}
  			return token;
  		}
  
  		private String nextToken() {
  			String ans = peekToken();
  			token = null;
  			return ans;
  		}
  	}
  }
  