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.*;
import java.math.*;

public class F {
    static final boolean RUN_TIMING = false;
    static PushbackReader in = new PushbackReader(new BufferedReader(new InputStreamReader(System.in)), 1024);
    static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

    public void go() throws IOException {
        // in = new PushbackReader(new BufferedReader(new FileReader(new File("test.txt"))), 1024);
        // out = new PrintWriter(new FileWriter(new File("output.txt")));
        int zzz = ipar();
        for (int zz = 0; zz < zzz; zz++) {
            int n = ipar();
            char[] s = spar().toCharArray();
            char[] t = spar().toCharArray();
            int[][] sFreq = new int[n+1][26];
            sFreq[0] = new int[26];
            for (int i = 0; i < n; i++) {
                sFreq[i+1] = Arrays.copyOf(sFreq[i], 26);
                sFreq[i+1][s[i]-'a']++;
            }
            int[][] tFreq = new int[n+1][26];
            tFreq[0] = new int[26];
            for (int i = 0; i < n; i++) {
                tFreq[i+1] = Arrays.copyOf(tFreq[i], 26);
                tFreq[i+1][t[i]-'a']++;
            }
            if (!Arrays.equals(sFreq[n], tFreq[n])) {
                out.println(-1);
                continue;
            }
            int[][] dp = new int[n+1][n+1];
            // for (int i = 0; i <= n; i++) {
            //     dp[0][i] = i;
            // }
            for (int i = 1; i <= n; i++) {
                for (int j = i; j <= n; j++) {
                    dp[i][j] = dp[i-1][j] + 1;
                    if (s[i-1] == t[j-1]) {
                        dp[i][j] = Math.min(dp[i][j], dp[i-1][j-1]);
                    }
                    // out.printf("%d %d = %d %d%n", i, j, sFreq[n][t[j-1]-'a'] - sFreq[i][t[j-1]-'a'], tFreq[n][t[j-1]-'a'] - tFreq[j][t[j-1]-'a']);
                    if (sFreq[n][t[j-1]-'a'] - sFreq[i][t[j-1]-'a'] > tFreq[n][t[j-1]-'a'] - tFreq[j][t[j-1]-'a']) {
                        dp[i][j] = Math.min(dp[i][j], dp[i][j-1]);
                    }
                }
            }
            // out.println(Arrays.toString(sFreq[n]));
            // out.println(Arrays.toString(tFreq[n]));
            // for (int[] arr : dp) {
            //     out.println(Arrays.toString(arr));
            // }
            out.println(dp[n][n]);
        }
    }

    public int ipar() throws IOException {
        return Integer.parseInt(spar());
    }

    public int[] iapar(int n) throws IOException {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = ipar();
        }
        return arr;
    }

    public long lpar() throws IOException {
        return Long.parseLong(spar());
    }

    public long[] lapar(int n) throws IOException {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++) {
            arr[i] = lpar();
        }
        return arr;
    }

    public double dpar() throws IOException {
        return Double.parseDouble(spar());
    }

    public String spar() throws IOException {
        StringBuilder sb = new StringBuilder(1024);
        int c;
        do {
            c = in.read();
        } while (Character.isWhitespace(c) && c != -1);
        if (c == -1) {
            throw new NoSuchElementException("Reached EOF");
        }
        do {
            sb.append((char)c);
            c = in.read();
        } while (!Character.isWhitespace(c) && c != -1);
        while (c != '\n' && Character.isWhitespace(c) && c != -1) {
            c = in.read();
        }
        if (c != -1 && c != '\n') {
            in.unread(c);
        }
        return sb.toString();
    }

    public String linepar() throws IOException {
        StringBuilder sb = new StringBuilder(1024);
        int c;
        while ((c = in.read()) != '\n' && c != -1) {
            if (c == '\r') {
                continue;
            }
            sb.append((char)c);
        }
        return sb.toString();
    }

    public boolean haspar() throws IOException {
        String line = linepar();
        if (line.isEmpty()) {
            return false;
        }
        in.unread('\n');
        in.unread(line.toCharArray());
        return true;
    }

    public static void main(String[] args) throws IOException {
        long time = 0;
        time -= System.nanoTime();
        new F().go();
        time += System.nanoTime();
        if (RUN_TIMING) {
            System.out.printf("%.3f ms%n", time / 1000000.0);
        }
        out.flush();
        in.close();
    }
}
