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.util.*;
  import java.lang.*;
  import java.io.*;
  
  //author: ankit_bit
  
  public class Main {
  
      static FastReader fr = new FastReader();
      static StringBuilder sb = new StringBuilder();
  
      public static void main(String[] args) throws java.lang.Exception {
  
          int testcases = fr.nextInt();
          while (testcases-- > 0) {
  
              solve();
          }
          System.out.println(sb);
  
      }
  
      public static void solve() {
          int n=fr.nextInt();
          String s=fr.nextLine();
          int count=0;
          char c[]=s.toCharArray();
          Arrays.sort(c);
          for(int i=0;i<n;i++)
          {
              if(s.charAt(i)!=c[i])
                  count++;
          }
          sb.append(count);
          sb.append("\n");
      }
  
  
  
      static boolean isPrime(int x) {
          for (int d = 2; d * d <= x; d++) {
              if (x % d == 0)
                  return false;
          }
          return true;
      }
  
      static int lowerBound(int a[], int low, int high, int x) {
          int pos = -1;
          while (low <= high) {
              int mid = (low + high) / 2;
              if (a[mid] == x) {
                  pos = mid;
                  high = mid - 1;
              } else if (a[mid] < x) low = mid + 1;
              else high = mid - 1;
          }
          if (pos != -1) return pos;
          return high + 1;
      }
  
      static int upperBound(int a[], int low, int high, int x) {
          int pos = -1;
          while (low <= high) {
              int mid = (low + high) / 2;
              if (a[mid] == x) {
                  pos = mid;
                  low = mid + 1;
              } else if (a[mid] < x) low = mid + 1;
              else high = mid - 1;
          }
          if (pos != -1) return pos;
          return high + 1;
      }
  
      static long mult(long a, long b, long mod) {
          long res = (a * b) % mod;
          if (res < 0) {
              res += mod;
          }
          return res;
      }
  
      static long add(long a, long b, long mod) {
          long res = (a + b) % mod;
          if (res < 0) {
              res += mod;
          }
          return res;
      }
  
      static long gcd(long a, long b) {
          if (b == 0) return a;
          return gcd(b, a % b);
      }
  
      static long lcm(long a, long b) {
          return (a / gcd(a, b)) * b;
      }
  
      static long fact[] = new long[100001], M = 1000000007;
  
      static long ncr(int n, int r) {
          if (r > n) return 0;
          return fact[n] * power(fact[r], M - 2) % M * power(fact[n - r], M - 2) % M;
      }
  
      static long power(long a, long b) {
          long res = 1;
          while (b != 0) {
              if (b % 2 == 1) res = res * a % M;
              b >>= 1;
              a = a * a % M;
          }
          return res;
      }
  
  
      static void sort(int[] a) {
          ArrayList<Integer> l = new ArrayList<>();
          for (int i = 0; i < a.length; i++) {
              l.add(a[i]);
          }
          Collections.sort(l);
          for (int i = 0; i < l.size(); i++)
              a[i] = l.get(i);
      }
  
  }
  
  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;
      }
  }
  