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.math.BigInteger;
  import java.util.*;
  import java.io.*;
  
  public class A_Subsequence_Permutation {
      public static void main (String[] args) {
          FastReader scan = new FastReader();
          int t = scan.nextInt();
          while (t-- > 0) {
              int n = scan.nextInt();
              String str = scan.nextLine();
              char[] arr = str.toCharArray();
              Arrays.sort(arr);
              String str1 = new String(arr);
              int count = 0;
              for (int i=0; i<n; i++) {
                  if (str.charAt(i) != str1.charAt(i)) {
                      count++;
                  }
              }
              System.out.println(count);
          }
      }
      public static void inputArray (int[] arr, int n, FastReader scan) {
          for (int i=0; i<n; i++) {
              arr[i] = scan.nextInt();
          }
      }
      public static void printArray (int[] arr) {
          for (int i=0; i<arr.length; i++) {
              System.out.print(arr[i] + " ");
          }
          System.out.println();
      }
      public static boolean isPrime (int n) {
          if (n <= 1)
              return false;
          if (n <= 3)
              return true;
          if (n % 2 == 0 || n % 3 == 0)
              return false;
          for (int i = 5; i * i <= n; i = i + 6)
              if (n % i == 0 || n % (i + 2) == 0)
                  return false;
          return true;
      }
      public static int gcd (int a, int b) {
          BigInteger a1 = BigInteger.valueOf(a);
          BigInteger b1 = BigInteger.valueOf(b);
          return a1.gcd(b1).intValue();
      }
      public static int numberOfDigitsInNumber (int n) {
          return (int)Math.floor(Math.log10(n)) + 1;
      }
      public static boolean isPowerOfTwo (int x) {
          return x!=0 && ((x&(x-1)) == 0);   
      }
      public static String toBinaryString (int n) {
          return Integer.toString(n, 2);
          // for other bases change 2 to other bases
      }
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
   
          public FastReader()
          {
              br = new BufferedReader(
                  new InputStreamReader(System.in));
          }
   
          String next()
          {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  }
                  catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
   
          int nextInt() { return Integer.parseInt(next()); }
   
          long nextLong() { return Long.parseLong(next()); }
   
          double nextDouble()
          {
              return Double.parseDouble(next());
          }
   
          String nextLine()
          {
              String str = "";
              try {
                  str = br.readLine();
              }
              catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
  }
  
  // For IO using Files
  // public static void main throws IOException
  // BufferedReader br = new BufferedReader(new FileReader("input.txt"));
  // PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("output.txt")));
  // String st[]=br.readLine().split(" ");
  // int n=Integer.parseInt(st[0]);
  // int m=Integer.parseInt(st[1]);
  // at the end
  // br.close(); pw.close();
  // for printing pw.println(), pw.print(), etc...
  // for arrays : int arr[] = Arrays.stream(br.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();