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.*;

public class Main{
	static int n;
	static char[]s,t;
	static int[][]sufS,sufT,memo;
	static int dp(int i,int j){
		if(i<0)return 0;
		if(j<0)return (int)1e9;
		if(memo[i][j]!=-1)return memo[i][j];
		if(s[i]==t[j]) {
			return memo[i][j]=dp(i-1, j-1);
		}
		//remove s[i] to be matched later
		int option1=dp(i-1, j)+1;
		int option2=(int)1e9;
		if(sufS[i+1][t[j]-'a']>sufT[j+1][t[j]-'a']) {
			//match t[j] with previously remove element from s
			option2=dp(i, j-1);
		}
		return memo[i][j]=Math.min(option1, option2);
	}
   	static void main() throws Exception{
   		n=sc.nextInt();
   		s=sc.nextLine().toCharArray();
   		t=sc.nextLine().toCharArray();
   		
   		sufS=new int[n+1][26];
   		sufT=new int[n+1][26];
   		
   		for(int i=n-1;i>=0;i--) {
   			sufS[i]=sufS[i+1].clone();
   			sufS[i][s[i]-'a']++;
   			
   			sufT[i]=sufT[i+1].clone();
   			sufT[i][t[i]-'a']++;
   		}
   		boolean can=true;
   		for(int c=0;c<26;c++) {
   			can&=(sufS[0][c]==sufT[0][c]);
   		}
   		if(!can) {
   			pw.println(-1);
   			return;
   		}
   		memo=new int[n][n];
   		for(int i=0;i<n;i++) {
   			Arrays.fill(memo[i], -1);
   		}
   		pw.println(dp(n-1, n-1));
	}
	public static void main(String[] args) throws Exception{
		pw=new PrintWriter(System.out);
		sc = new MScanner(System.in);
		int tc=sc.nextInt();
		while(tc-->0)main();
		pw.flush();
	}
	static PrintWriter pw;
	static MScanner  sc;
	static class MScanner {
		StringTokenizer st;
		BufferedReader br;
		public MScanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}
 
		public MScanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}
 
		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public int[] intArr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public long[] longArr(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public int[] intSortedArr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        shuffle(in);
	        Arrays.sort(in);
	        return in;
		}
		public long[] longSortedArr(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        shuffle(in);
	        Arrays.sort(in);
	        return in;
		}
		public Integer[] IntegerArr(int n) throws IOException {
	        Integer[]in=new Integer[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public Long[] LongArr(int n) throws IOException {
	        Long[]in=new Long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public String nextLine() throws IOException {
			return br.readLine();
		}
 
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
 
		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
 
		public char nextChar() throws IOException {
			return next().charAt(0);
		}
 
		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}
 
		public boolean ready() throws IOException {
			return br.ready();
		}
 
		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
	static void shuffle(int[]in) {
		for(int i=0;i<in.length;i++) {
			int idx=(int)(Math.random()*in.length);
			int tmp=in[i];
			in[i]=in[idx];
			in[idx]=tmp;
		}
	}
	static void shuffle(long[]in) {
		for(int i=0;i<in.length;i++) {
			int idx=(int)(Math.random()*in.length);
			long tmp=in[i];
			in[i]=in[idx];
			in[idx]=tmp;
		}
	}
}
