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 Solution1 {

    public static void main(String[] args) throws Exception {
        int t = io.nextInt();
        for (int i = 0; i < t; i++) {
            solve();
        }
        io.close();
    }

    private static void solve() throws Exception {
        io.nextInt();
        String s = io.nextLine();
        char[] c1 = s.toCharArray();
        char[] c2 = s.toCharArray();
        Arrays.sort(c2);
        int d = 0;
        for (int i = 0; i < c1.length; i++) {
            if (c1[i] != c2[i]) {
                d++;
            }
        }
        io.println(d);
    }

    static void sort(int[] a) {
        ArrayList<Integer> l = new ArrayList<>(a.length);
        for (int i : a) l.add(i);
        Collections.sort(l);
        for (int i = 0; i < a.length; i++) a[i] = l.get(i);
    }

    //-----------PrintWriter for faster output---------------------------------
    public static FastIO io = new FastIO();

    //-----------MyScanner class for faster input----------
    static class FastIO extends PrintWriter {
        private InputStream stream;
        private byte[] buf = new byte[1 << 16];
        private int curChar, numChars;

        // standard input
        public FastIO() {
            this(System.in, System.out);
        }

        public FastIO(InputStream i, OutputStream o) {
            super(o);
            stream = i;
        }

        // file input
        public FastIO(String i, String o) throws IOException {
            super(new FileWriter(o));
            stream = new FileInputStream(i);
        }

        // throws InputMismatchException() if previously detected end of file
        private int nextByte() {
            if (numChars == -1) throw new InputMismatchException();
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (numChars == -1) return -1; // end of file
            }
            return buf[curChar++];
        }

        // to read in entire lines, replace c <= ' '
        // with a function that checks whether c is a line break
        public String next() {
            int c;
            do {
                c = nextByte();
            } while (c <= ' ');
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = nextByte();
            } while (c > ' ');
            return res.toString();
        }

        public String nextLine() {
            int c;
            do {
                c = nextByte();
            } while (c < '\n');
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = nextByte();
            } while (c > '\n');
            return res.toString();
        }

        public int nextInt() { // nextLong() would be implemented similarly
            int c;
            do {
                c = nextByte();
            } while (c <= ' ');
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = nextByte();
            }
            int res = 0;
            do {
                if (c < '0' || c > '9')
                    throw new InputMismatchException();
                res = 10 * res + c - '0';
                c = nextByte();
            } while (c > ' ');
            return res * sgn;
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }
    }
//--------------------------------------------------------
}