Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter y Koa selects must be strictly greater alphabetically than x (read statement for better understanding). You can make hacks in these problems independently.

Koa the Koala has two strings A and B of the same length n (|A|=|B|=n) consisting of the first 20 lowercase English alphabet letters (ie. from a to t).

In one move Koa:

  1. selects some subset of positions p_1, p_2, …, p_k (k ≥ 1; 1 ≤ p_i ≤ n; p_i ≠ p_j if i ≠ j) of A such that A_{p_1} = A_{p_2} = … = A_{p_k} = x (ie. all letters on this positions are equal to some letter x).

  2. selects a letter y (from the first 20 lowercase letters in English alphabet) such that y>x (ie. letter y is strictly greater alphabetically than x).

  3. sets each letter in positions p_1, p_2, …, p_k to letter y. More formally: for each i (1 ≤ i ≤ k) Koa sets A_{p_i} = y.

Note that you can only modify letters in string A.




Koa wants to know the smallest number of moves she has to do to make strings equal to each other (A = B) or to determine that there is no way to make them equal. Help her!

Input

Each test contains multiple test cases. The first line contains t (1 ≤ t ≤ 10) — the number of test cases. Description of the test cases follows.

The first line of each test case contains one integer n (1 ≤ n ≤ 10^5) — the length of strings A and B.

The second line of each test case contains string A (|A|=n).

The third line of each test case contains string B (|B|=n).

Both strings consists of the first 20 lowercase English alphabet letters (ie. from a to t).

It is guaranteed that the sum of n over all test cases does not exceed 10^5.

Output

For each test case:

Print on a single line the smallest number of moves she has to do to make strings equal to each other (A = B) or -1 if there is no way to make them equal.

Example

Input


5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda


Output


2
-1
3
2
-1

Note

  * In the 1-st test case Koa: 
    1. selects positions 1 and 2 and sets A_1 = A_2 =  b (\color{red}{aa}b → \color{blue}{bb}b). 
    2. selects positions 2 and 3 and sets A_2 = A_3 =  c (b\color{red}{bb} → b\color{blue}{cc}). 

  * In the 2-nd test case Koa has no way to make string A equal B.

  * In the 3-rd test case Koa: 
    1. selects position 1 and sets A_1 =  t (\color{red}{a}bc → \color{blue}{t}bc). 
    2. selects position 2 and sets A_2 =  s (t\color{red}{b}c → t\color{blue}{s}c). 
    3. selects position 3 and sets A_3 =  r (ts\color{red}{c} → ts\color{blue}{r}). 
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.FileNotFoundException;
import java.util.StringTokenizer;
import java.io.Writer;
import java.io.BufferedReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Asgar Javadov
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        OutputWriter out = new OutputWriter(outputStream);
        CStringTransformation1 solver = new CStringTransformation1();
        int testCount = Integer.parseInt(in.next());
        for (int i = 1; i <= testCount; i++)
            solver.solve(i, in, out);
        out.close();
    }

    static class CStringTransformation1 {
        public void solve(int testNumber, InputReader in, OutputWriter out) {
            int n = in.nextInt();
            char[] a = in.next().toCharArray();
            char[] b = in.next().toCharArray();

            boolean[][] graph = new boolean[20][20];
            for (int i = 0; i < a.length; ++i) {
                int from = a[i] - 'a';
                int to = b[i] - 'a';
                if (from > to) {
                    out.minusOne();
                    return;
                }
                graph[from][to] = true;
            }

            int operation = 0;
            for (int src = 0; src < 20; src++) {
                outer:
                for (int dest = src + 1; dest < 20; dest++) {
                    if (graph[src][dest]) {
                        for (int k = src + 1; k < dest; ++k) {
                            if (graph[src][k]) {
                                graph[src][dest] = false;
                                graph[k][dest] = true;
                                continue outer;
                            }
                        }
                        operation++;
                    }
                }
            }

            out.println(operation);
        }

    }

    static class OutputWriter extends PrintWriter {
        public OutputWriter(OutputStream outputStream) {
            super(outputStream);
        }

        public OutputWriter(Writer writer) {
            super(writer);
        }

        public OutputWriter(String filename) throws FileNotFoundException {
            super(filename);
        }

        public void minusOne() {
            println(-1);
        }

        public void close() {
            super.close();
        }

    }

    static class InputReader extends BufferedReader {
        StringTokenizer tokenizer;

        public InputReader(InputStream inputStream) {
            super(new InputStreamReader(inputStream), 32768);
        }

        public InputReader(String filename) {
            super(new InputStreamReader(Thread.currentThread().getContextClassLoader().getResourceAsStream(filename)));
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(readLine());
                } catch (IOException e) {
                    throw new RuntimeException();
                }
            }
            return tokenizer.nextToken();
        }

        public Integer nextInt() {
            return Integer.valueOf(next());
        }

    }
}

