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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.StringTokenizer;
  import java.util.Arrays;
  
  public class subPerm {
  	public static void main(String[] args) throws IOException {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		
  		int cases = Integer.parseInt(br.readLine());
  		
  		for(int i = 0; i< cases; i++) {
  			int letters = Integer.parseInt(br.readLine());
  			int sum = 0;
  			int[] alpha = new int[26];
  			String str = br.readLine();
  			char[] chars = str.toCharArray();
  			Arrays.sort(chars);
  			int count = 0;
  			for(int j = 0; j<letters-1; j++) {
  				if(j == letters-2) {
  					if(chars[j] == chars[j+1]) {
  						alpha[count]+=2;
  					}else {
  						alpha[count]++;
  						alpha[count+1]++;
  					}
  				}
  				else if(chars[j] == chars[j+1]) {
  					alpha[count]++;
  				}else {
  					alpha[count]++;
  					count++;
  					
  				}
  			}
  
  			int loci = 0;
  			for(int k = 0;k<26; k++) {
  				int before = alpha[k];
  				int iter = -1;
  				int sum1 = 0;
  				for(int a =0;a<k;a++) {
  					sum1+=alpha[a];
  				}
  				if(before!=0) {
  					while(before>0) {
  						if(iter == -1) {
  							if(str.indexOf(chars[loci])>sum1-1 && str.indexOf(chars[loci])<sum1+alpha[k]) {
  								iter = str.indexOf(chars[loci])+1;
  							}else {
  								iter = str.indexOf(chars[loci])+1;
  								sum++;
  								
  							}
  							loci++;
  
  						}else {
  							if(str.indexOf(chars[loci], iter)>sum1-1 && str.indexOf(chars[loci], iter)<sum1+alpha[k]) {
  								iter = str.indexOf(chars[loci], iter)+1;
  							}else {
  								iter = str.indexOf(chars[loci], iter)+1;
  								sum++;
  								
  							}
  							loci++;
  	
  						}
  						before--;
  					}
  				}
  			}
  			
  			System.out.println(sum);
  		}
  	}
  }