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 Main {
  	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  	static StringTokenizer st;
  	static PrintWriter pr = new PrintWriter(new OutputStreamWriter(System.out));
  	static String readLine() throws IOException {
  		return br.readLine();
  	}
  	static String next() throws IOException {
  		while (st == null || !st.hasMoreTokens())
  			st = new StringTokenizer(readLine());
  		return st.nextToken();
  	}
  	static int readInt() throws IOException {
  		return Integer.parseInt(next());
  	}
  	static long readLong() throws IOException {
  		return Long.parseLong(next());
  	}
  	static double readDouble() throws IOException {
  		return Double.parseDouble(next());
  	}
  	static char readChar() throws IOException {
  		return next().charAt(0);
  	}
  	static class Pair implements Comparable<Pair> {
  		int f, s;
  		Pair(int f, int s) {
  			this.f = f; this.s = s;
  		}
  		public int compareTo(Pair other) {
  			if (this.f != other.f) return this.f - other.f;
  			return this.s - other.s;
  		}
  	}
  	static void solve() throws IOException {
  		int n = readInt(), ans = 0;
  		String s = " " + readLine();
  		char c[] = s.toCharArray();
  		Arrays.sort(c, 1, n + 1);
  		for (int i = 1; i <= n; ++i)
  			if (s.charAt(i) != c[i]) ++ans;
  		pr.println(ans);
  	}
  	public static void main(String[] args) throws IOException {
  		//solve();
  		for (int t = readInt(); t > 0; --t) solve();
  		pr.close();
  	}
  }
  