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 javax.swing.plaf.synth.SynthEditorPaneUI;
import java.io.*;
import java.util.*;

public class Contest1 {
    static int[][]cntS,cntT,memo;
    static char[]s;
    static char[]t;
    static int dp(int i,int j){
        if (i<0)
            return 0;
        if (memo[i][j]!=-1)
            return memo[i][j];
        int ans = dp(i-1,j)+1;
        if (s[i]==t[j]){
            ans=Math.min(ans,dp(i-1,j-1));
        }
        if (cntS[t[j]-'a'][i+1]>cntT[t[j]-'a'][j+1]){
            ans=Math.min(ans,dp(i,j-1));
        }
        return memo[i][j]=ans;
    }
    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int tc = sc.nextInt();
        while (tc-->0) {
            int n = sc.nextInt();
            s = sc.nextLine().toCharArray();
            t = sc.nextLine().toCharArray();
            memo= new int[n][n];
            cntS= new int[26][n+1];
            cntT= new int[26][n+1];
            for (int i =n-1;i>=0;i--){
                Arrays.fill(memo[i],-1);
                for (int j =0;j<26;j++)
                {
                    cntS[j][i]=cntS[j][i+1];
                    cntT[j][i]=cntT[j][i+1];
                }
                cntS[s[i]-'a'][i]++;
                cntT[t[i]-'a'][i]++;
            }
            boolean is =true;
            for (int i =0;i<26;i++){
                is&=(cntS[i][0]==cntT[i][0]);
            }
            if (is)
            pw.println(dp(n-1,n-1));
            else pw.println(-1);
        }
        pw.flush();
    }
    static long gcd(long a,long b){
        if (a==0)
            return b;
        return gcd(b%a,a);
    }
    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(FileReader r) {
            br = new BufferedReader(r);
        }

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

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

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

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++)
                if (x.charAt(i) == '.') {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec)
                        f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

        public boolean ready() throws IOException {
            return br.ready();
        }
    }
}