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.Arrays;
  import java.util.StringTokenizer;
  import java.io.IOException;
  import java.io.BufferedReader;
  import java.io.FileReader;
  import java.io.InputStreamReader;
  import java.io.InputStream;
  
  /**
   * Built using CHelper plug-in
   * Actual solution is at the top
   *
   * @author Khater
   */
  public class Main {
      public static void main(String[] args) {
          InputStream inputStream = System.in;
          OutputStream outputStream = System.out;
          Scanner in = new Scanner(inputStream);
          PrintWriter out = new PrintWriter(outputStream);
          ASubsequencePermutation solver = new ASubsequencePermutation();
          solver.solve(1, in, out);
          out.close();
      }
  
      static class ASubsequencePermutation {
          public void solve(int testNumber, Scanner sc, PrintWriter pw) {
              int t = 1;
              t = sc.nextInt();
              while (t-- > 0) {
                  int n = sc.nextInt();
                  char[] arr = sc.nextLine().toCharArray();
                  char[] sort = arr.clone();
                  Arrays.sort(sort);
                  int c = 0;
                  for (int i = 0; i < n; i++) {
                      if (arr[i] != sort[i]) c++;
                  }
                  pw.println(c);
              }
  
          }
  
      }
  
      static class Scanner {
          StringTokenizer st;
          BufferedReader br;
  
          public Scanner(FileReader r) {
              br = new BufferedReader(r);
          }
  
          public Scanner(InputStream s) {
              br = new BufferedReader(new InputStreamReader(s));
          }
  
          public String next() {
              while (st == null || !st.hasMoreTokens()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      throw new RuntimeException(e);
                  }
              }
              return st.nextToken();
          }
  
          public int nextInt() {
              return Integer.parseInt(next());
          }
  
          public String nextLine() {
              try {
                  return br.readLine();
              } catch (Exception e) {
                  throw new RuntimeException(e);
              }
          }
  
      }
  }
  
  