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.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++;
					
				}
			}
			System.out.println(Arrays.toString(alpha));
			System.out.println(Arrays.toString(chars));
			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++;
							System.out.println(iter);
						}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++;
							System.out.println(iter);
						}
						before--;
					}
				}
			}
			
			System.out.println(sum);
		}
	}
}
