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").
import java.io.*;
import java.util.*;

public class Solution {

	public static void main(String[] args) throws IOException {

		Reader sc=new Reader();
		sc.init(System.in);

		BufferedWriter out = new BufferedWriter(
				new OutputStreamWriter(System.out));

		int t=Integer.parseInt(sc.nextLine());
		
		for (int i = 0; i < t; i++) {
			
			int n=Integer.parseInt(sc.nextLine());
			
			char[] c=sc.nextLine().toCharArray();
			
			char temp[]=new char[n];
			
			for(int j=0;j<n;j++) {
				temp[j]=c[j];
			}
			
			Arrays.sort(temp);
			
			int ans=0;
			
			for(int j=0;j<n;j++) {
				if(temp[j]!=c[j]) ans++;
			}
			
			out.write(Integer.toString(ans)+"\n");
			
		}
		out.flush();
	}

	static class Reader {
		static BufferedReader reader;
		static StringTokenizer tokenizer;

		/** call this method to initialize reader for InputStream */
		static void init(InputStream input) {
			reader = new BufferedReader(
					new InputStreamReader(input) );
			tokenizer = new StringTokenizer("");
		}

		/** get next word */
		static String next() throws IOException {
			while ( ! tokenizer.hasMoreTokens() ) {		
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return tokenizer.nextToken();
		}

		String nextLine() {
			String str = "";
			try {
				str = reader.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}

		static int nextInt() throws IOException {
			return Integer.parseInt( next() );
		}

		static long nextLong() throws IOException {
			return Long.parseLong( next() );
		}

		static double nextDouble() throws IOException {
			return Double.parseDouble( next() );
		}
	}
}