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.
//package cn.itcast.day05.demp;

import java.util.Scanner;

public class CodeforcesF {
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        int t = cin.nextInt();
        while (t-- > 0) {
//            System.out.println("s");
                int n=cin.nextInt();
                String s=cin.next();
                String st=cin.next();
                char []p=s.toCharArray();
                char []q=st.toCharArray();
                int [][]dp=new int [n+7][n+7];
                int [][]num1=new int[30][n+7];
                int [][]num2=new int[30][n+7];
//                System.out.println(s);
            for(int i=1;i<=n+1;i++)
            {
                dp[0][i]=0;
                dp[i][0]=0;
                for(int j=0;j<=26;j++)num1[j][i]=num2[j][i]=0;
            }
            for(int i=n;i>=1;i--)
            {
                for(int j=0;j<26;j++)
                {
                    num1[j][i]=num1[j][i+1];
                    num2[j][i]=num2[j][i+1];
                }
                num1[p[i-1]-'a'][i]++;
                num2[q[i-1]-'a'][i]++;
            }
            int tmp=0;
            for(int i=0;i<26;i++)
            {
                if(num1[i][1]!=num2[i][1])
                {
                    System.out.println("-1");
                    tmp++;
                    break;
                    //return ;
                }
            }
            if(tmp>0)continue;
            for(int i=1;i<=n;i++)
            {
                for(int j=i;j<=n;j++)
                {
                    dp[i][j]=dp[i-1][j]+1;
                    if(p[i-1]==q[j-1])
                    {
                        dp[i][j]=min(dp[i][j],dp[i-1][j-1]);
                    }
                    int ch=q[j-1]-'a';
                    if(num1[ch][i+1]>num2[ch][j+1]) {
                        dp[i][j] = min(dp[i][j - 1], dp[i][j]);
                    }
                    ch=p[i-1]-'a';
                    if(num1[ch][i+1]<num2[ch][j+1])
                    {
                        dp[i][j]=min(dp[i-1][j-1],dp[i][j]);
                    }
                }
            }

            System.out.println(dp[n][n]);

        }
    }

    private static int min(int i, int i1) {
        if (i > i1) return i1;
        else return i;
    }

}
