You are given two strings s and t, each of length n and consisting of lowercase Latin alphabets. You want to make s equal to t. 

You can perform the following operation on s any number of times to achieve it — 

  * Choose any substring of s and rotate it clockwise once, that is, if the selected substring is s[l,l+1...r], then it becomes s[r,l,l + 1 ... r - 1]. All the remaining characters of s stay in their position. 

For example, on rotating the substring [2,4] , string "abcde" becomes "adbce". 




A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.

Find the minimum number of operations required to convert s to t, or determine that it's impossible.

Input

The first line of the input contains a single integer t (1≤ t ≤ 2000) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer n (1≤ n ≤ 2000) — the length of the strings. 

The second and the third lines contain strings s and t respectively.

The sum of n over all the test cases does not exceed 2000.

Output

For each test case, output the minimum number of operations to convert s to t. If it is not possible to convert s to t, output -1 instead.

Example

Input


6
1
a
a
2
ab
ba
3
abc
cab
3
abc
cba
4
abab
baba
4
abcc
aabc


Output


0
1
1
2
1
-1

Note

For the 1-st test case, since s and t are equal, you don't need to apply any operation.

For the 2-nd test case, you only need to apply one operation on the entire string ab to convert it to ba.

For the 3-rd test case, you only need to apply one operation on the entire string abc to convert it to cab.

For the 4-th test case, you need to apply the operation twice: first on the entire string abc to convert it to cab and then on the substring of length 2 beginning at the second character to convert it to cba.

For the 5-th test case, you only need to apply one operation on the entire string abab to convert it to baba.

For the 6-th test case, it is not possible to convert string s to t.
import java.io.*;
import java.util.*;

/*
polyakoff
*/

public class Main {

    static FastReader in;
    static PrintWriter out;
    static Random rand = new Random();
    static final int oo = (int) 2e9;
    static final long OO = (long) 4e18;
    static final int MOD = (int) 1e9 + 7;

    static int n;
    static char[] s, t;
    static int[][] sSuf, tSuf, memo;

    static int dp(int i, int j) {
        if (i < 0)
            return 0;
        if (memo[i][j] != -1)
            return memo[i][j];

        memo[i][j] = oo;
        if (s[i] == t[j])
            memo[i][j] = Math.min(memo[i][j], dp(i - 1, j - 1));
        if (sSuf[t[j] - 'a'][i + 1] > tSuf[t[j] - 'a'][j + 1])
            memo[i][j] = Math.min(memo[i][j], dp(i, j - 1));
        memo[i][j] = Math.min(memo[i][j], dp(i - 1, j) + 1);
        return memo[i][j];
    }

    static void solve() {
        n = in.nextInt();
        s = in.next().toCharArray();
        t = in.next().toCharArray();

        sSuf = new int[26][n + 1];
        tSuf = new int[26][n + 1];
        for (int j = n - 1; j >= 0; j--) {
            for (int i = 0; i < 26; i++) {
                sSuf[i][j] = sSuf[i][j + 1];
                tSuf[i][j] = tSuf[i][j + 1];
            }
            sSuf[s[j] - 'a'][j]++;
            tSuf[t[j] - 'a'][j]++;
        }

        for (int i = 0; i < 26; i++) {
            if (sSuf[i][0] != tSuf[i][0]) {
                out.println(-1);
                return;
            }
        }

        memo = new int[n][n];
        for (int i = 0; i < n; i++)
            Arrays.fill(memo[i], -1);

        out.println(dp(n - 1, n - 1));


    }

    public static void main(String[] args) throws FileNotFoundException {
        in = new FastReader(System.in);
//        in = new FastReader(new FileInputStream("sometext.txt"));
        out = new PrintWriter(System.out);
//        out = new PrintWriter(new FileOutputStream("output.txt"));

        int T = 1;
        T = in.nextInt();
        while (T-- > 0)
            solve();

//        Thread thread = new Thread(null, () -> {
//            solve();
//        }, "Main", 1 << 28);
//        thread.start();
//        try {
//            thread.join();
//        } catch (InterruptedException e) {
//            e.printStackTrace();
//        }

//        out.flush();
        out.close();
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        FastReader(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

        Integer nextInt() {
            return Integer.parseInt(next());
        }

        Long nextLong() {
            return Long.parseLong(next());
        }

        Double nextDouble() {
            return Double.parseDouble(next());
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(nextLine());
            }
            return st.nextToken();
        }

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

    static void dbg(Object... args) {
        for (Object arg : args)
            out.print(arg + " ");
        out.println();
    }

}