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: //---#ON_MY_WAY---
  import static java.lang.Math.*;
  import java.io.*;
  import java.math.*;
  import java.util.*;
  
  public class A {
  	static FastReader  x= new FastReader();
  	static OutputStream outputStream = System.out;
  	static PrintWriter out = new PrintWriter(outputStream);
  
  	/*---------------------------------------CODE STARTS HERE-------------------------*/
  	public static void main(String[] args) {
  
  		int t = x.nextInt();
  		StringBuilder str = new StringBuilder();
  		while (t > 0) {
  			int n=x.nextInt();
  			String s=x.next();
  			char a[]=s.toCharArray();
  			int c=0;
  			Arrays.sort(a);
  			for (int i = 0; i < n; i++) {
  				if(s.charAt(i)!=a[i]) {
  					c++;
  				}
  			}
  			str.append(c);
  			str.append("\n");
  			t--;
  		}
  		out.println(str);
  		out.flush();
  	}
  
  	/*--------------------------------------------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;
  		}
  
  		char nextchar() {
  			char ch = ' ';
  			try {
  				ch = (char) br.read();
  			} catch (IOException e) {
  				e.printStackTrace();
  			}
  			return ch;
  		}
  	}
  
  	/*--------------------------------------------BOILER PLATE---------------------------*/
  	static int[] readarr(int n) {
  		int arr[] = new int[n];
  		for (int i = 0; i < n; i++) {
  			arr[i] = x.nextInt();
  		}
  		return arr;
  	}
  
  	static int[] sort(int a[]) {
  		ArrayList<Integer> al = new ArrayList<>();
  		for (int i : a) {
  			al.add(i);
  		}
  		Collections.sort(al);
  		for (int i = 0; i < a.length; i++) {
  			a[i] = al.get(i);
  		}
  		return a;
  	}
  
  	static int[] revsort(int a[]) {
  		ArrayList<Integer> al = new ArrayList<>();
  		for (int i : a) {
  			al.add(i);
  		}
  		Collections.sort(al, Comparator.reverseOrder());
  		for (int i = 0; i < a.length; i++) {
  			a[i] = al.get(i);
  		}
  		return a;
  	}
  
  	static int[] gcd(int a, int b, int ar[]) {
  		if (b == 0) {
  			ar[0] = a;
  			ar[1] = 1;
  			ar[2] = 0;
  			return ar;
  		}
  		ar = gcd(b, a % b, ar);
  		int t = ar[1];
  		ar[1] = ar[2];
  		ar[2] = t - (a / b) * ar[2];
  		return ar;
  	}
  
  	static boolean[] esieve(int n) {
  		boolean p[] = new boolean[n + 1];
  		Arrays.fill(p, true);
  		for (int i = 2; i * i <= n; i++) {
  			if (p[i] == true) {
  				for (int j = i * i; j <= n; j += i) {
  					p[j] = false;
  				}
  			}
  		}
  		return p;
  	}
  
  	static ArrayList<Integer> primes(int n) {
  		boolean p[] = new boolean[n + 1];
  		ArrayList<Integer> al = new ArrayList<>();
  		Arrays.fill(p, true);
  		int i = 0;
  		for (i = 2; i * i <= n; i++) {
  			if (p[i] == true) {
  				al.add(i);
  				for (int j = i * i; j <= n; j += i) {
  					p[j] = false;
  				}
  			}
  		}
  		for (i = i; i <= n; i++) {
  			if (p[i] == true) {
  				al.add(i);
  			}
  		}
  		return al;
  	}
  
  	static int etf(int n) {
  		int res = n;
  		for (int i = 2; i * i <= n; i++) {
  			if (n % i == 0) {
  				res /= i;
  				res *= (i - 1);
  				while (n % i == 0) {
  					n /= i;
  				}
  			}
  		}
  		if (n > 1) {
  			res /= n;
  			res *= (n - 1);
  		}
  		return res;
  	}
  }