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.
/*
If you want to aim high, aim high
Don't let that studying and grades consume you
Just live life young
******************************
What do you think? What do you think?
1st on Billboard, what do you think of it
Next is a Grammy, what do you think of it
However you think, I’m sorry, but shit, I have no fcking interest
*******************************
I'm standing on top of my Monopoly board
That means I'm on top of my game and it don't stop
til my hip don't hop anymore
https://www.a2oj.com/Ladder16.html
*******************************
300IQ as writer = Sad!
*/
import java.util.*;
import java.io.*;
import java.math.*;

   public class x1363F
   {
      public static void main(String hi[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));  
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int T = Integer.parseInt(st.nextToken());
         StringBuilder sb = new StringBuilder();
         while(T-->0)
         {
            int N = Integer.parseInt(infile.readLine());
            arr = infile.readLine().toCharArray();
            target = infile.readLine().toCharArray();
            if(!equal(arr, target))
               sb.append("-1\n"); 
            else
            {
               int[][] suff1 = new int[N][26];
               int[][] suff2 = new int[N][26];
               for(int c=0; c < 26; c++)
               {
                  int curr = 0;  int curr2 = 0;
                  for(int i=N-1; i >= 0; i--) 
                  {
                     if(arr[i]-'a' == c)
                        curr++;
                     if(target[i]-'a' == c)
                        curr2++;
                     suff1[i][c] += curr;
                     suff2[i][c] += curr2;
                  }
               }
               //one indexed
               dp = new int[N+1][N+1];
               for(int i=1; i <= N; i++)
                  Arrays.fill(dp[i], -1);
               for(int i=0; i <= N; i++)
                  dp[i][0] = 0;
               sb.append(dfs(N, N, suff1, suff2)+"\n");
            }
         }
         System.out.print(sb);
      }
      static char[] arr;
      static char[] target;
      static int[][] dp;
      public static int dfs(int a, int b, int[][] suff1, int[][] suff2)
      {
         if(dp[a][b] != -1)
            return dp[a][b];
         int N = suff1.length;
         int res = Integer.MAX_VALUE;
         if(a > 0 && arr[a-1] == target[b-1])
            res = dfs(a-1, b-1, suff1, suff2);
         int curr = target[b-1]-'a';
         int ac = 0; int bc = 0;
         if(a < N)   ac = suff1[a][curr];
         if(b < N)   bc = suff2[b][curr];
         if(ac > bc)
            res = Math.min(res, dfs(a, b-1, suff1, suff2));
         if(a > 0)
            res = Math.min(res, 1+dfs(a-1, b, suff1, suff2));
         return dp[a][b] = res;
      }
      public static boolean equal(char[] a, char[] b)
      {
         int[] freq = new int[26];
         for(char c: a)
            freq[c-'a']++;
         for(char c: b)
            freq[c-'a']--;
         for(int x: freq)
            if(x != 0)
               return false;
         return true;
      }
   }