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 static java.lang.Math.max;
  import static java.lang.Math.min;
  import static java.lang.Math.abs;
  import static java.lang.System.out;
  import java.util.* ;
  import java.math.*;
  import java.io.*;
  
  
   public class javaTemplate {
  	
  	public static final int M = 1000000007 ;
  	static FastReader sc = new FastReader();
  	static int N = 86028121 ;
  	static boolean globB[] = new boolean[N+1] ;
  	static int globA[] = new int[10000000] ;
  	
  	
  	public static void main(String[] args) {
  		
  	int q = sc.nextInt() ;
  		
  		while(q -- != 0) {
  			int n = sc.nextInt() ;
  			String s = sc.next() ;
  			char tempArray[] = s.toCharArray();
  	        Arrays.sort(tempArray);
  	         int c = 0 ;
  	        for(int i = 0 ; i< n ; i++) {
  	        	if(tempArray[i] != s.charAt(i)) c ++ ;
  	        }
  	        
  	        System.out.println(c);
  	        
  	        
  		}
  
  	}
  
  
  	/*
  	 @ Observation is imp. for problem A and B
  	 1->> Read the problem properly take time to read the problem 
  	 if it is too lengthy.
  	 2->> Draw the test cases of A and B 
  	 3->> Don't over think the problem 
  	 4->> Try to take sum variable as long
  	 5->> Avoid Nested if else statement
  	 6->> If it looks like a Greedy problem try to figure out the ans. 
  	 for the smallest test case.
  	 7->> If it is a binary String check no. of 0's and 1's. 
  	 If it is a even try to pair it (may be not applicable for every questions).
  	 8->> Always check the constrains and try BruteForce
  	 @ Implement it quickly 
  	 */
  	
  
  	
  	
  			
  	
  	
  	
  	
  
  
  	
  	
  //_________________________//Template//_____________________________________________________________________
  	
  	
  		//		FAST I/O
  		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;
  	        }
  		}
  	
  	
  		
  		
  		
  		// Sorting
  		public static void sort(int[] arr) {
  	        //because Arrays.sort() uses quicksort which is dumb
  	        //Collections.sort() uses merge sort
  	        ArrayList<Integer> ls = new ArrayList<Integer>();
  	        for(int x: arr)
  	            ls.add(x);
  	        Collections.sort(ls);
  	        for(int i=0; i < arr.length; i++)
  	            arr[i] = ls.get(i);
  	    }
  		
  		
  		
  		
  		
  		// Modulo
  		static int mod(int a) {
  		   return (a%M + M) % M;
  		}
  		
  		
  		
  		
  		
  		
  		
  		//		Palindrom or Not
  		static boolean isPalindrome(String str, int low, int high) {
  					while (low < high) {
  						if (str.charAt(low) != str.charAt(high)) return false;
  					 	low++;
  					 	high--;
  					}
  					return true;
  				}
  	
  	
  
  	
  	
  	
  	
  	
  	
  	
  	
  		// check if sorted or not	
  		public static boolean isSorted(int[] a)
  	    {
  	        for (int i = 0; i < a.length - 1; i++)
  	        {
  	            if (a[i] > a[i + 1]) {
  	                return false;
  	            }
  	        }
  	        return true;
  	    }
  	
  	
  	
  		
  		
  	
  		// GCD
  		public static int GCD(int a , int b) {
  				if(b == 0) return a ;
  				
  				return GCD(b, a%b) ;
  			}
  	
  		
  		
  		
  		
  		// sieve
  		public static void sieve() {
  			Arrays.fill(globB, true);
  			globB[0] = false ;
  			globB[1] = false ;
  			for(int i = 2 ; i * i <= N ; i++) {
  				if(globB[i]) {
  				for(int j = i * i ; j<= N ; j+= i) {
  					globB[j] = false ;
  				 }
  				}
  			}
  			
  		}
  
  		
  		
  		
  		// fastPower
  		public static long fastPowerModulo(long a, long b, long n) {
  			long res = 1 ;
  			while(b > 0) {
  			if((b&1) != 0) {
  				res = (res * a % n) % n ;
  			}
  			a = (a % n * a % n) % n ;
  			b = b >> 1 ;
  			}
  			return res ;
  		}
  	
  		
  		
          // LCM
  		static int LCM(int a, int b)
  	    {
  	        return (a / GCD(a, b)) * b;
  	    }
  		
  
  }
  
  
   
   
   
   
   
   
  