String Transformation 1
Contest: Codeforces Round 659 (Div. 1)
Difficulty rating: 1700
Tags: dsu, graphs, greedy, sortings, strings, trees, two pointers

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 $$$p_1, p_2, \ldots, p_k$$$ ($$$k \ge 1; 1 \le p_i \le n; p_i \neq p_j$$$ if $$$i \neq j$$$) of $$$A$$$ such that $$$A_{p_1} = A_{p_2} = \ldots = A_{p_k} = x$$$ (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 $$$p_1, p_2, \ldots, p_k$$$ to letter $$$y$$$. More formally: for each $$$i$$$ ($$$1 \le i \le k$$$) 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 format:
Each test contains multiple test cases. The first line contains $$$t$$$ ($$$1 \le t \le 10$$$) — the number of test cases. Description of the test cases follows.

The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — 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 format:
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.

Constraints:
Time limit: 1.0 s
Memory limit: 256.0 MB

Note:
- In the $$$1$$$-st test case Koa:   selects positions $$$1$$$ and $$$2$$$ and sets $$$A_1 = A_2 = $$$ b ($$$\color{red}{aa}b \rightarrow \color{blue}{bb}b$$$).  selects positions $$$2$$$ and $$$3$$$ and sets $$$A_2 = A_3 = $$$ c ($$$b\color{red}{bb} \rightarrow b\color{blue}{cc}$$$).
- In the $$$2$$$-nd test case Koa has no way to make string $$$A$$$ equal $$$B$$$.
- In the $$$3$$$-rd test case Koa:   selects position $$$1$$$ and sets $$$A_1 = $$$ t ($$$\color{red}{a}bc \rightarrow \color{blue}{t}bc$$$).  selects position $$$2$$$ and sets $$$A_2 = $$$ s ($$$t\color{red}{b}c \rightarrow t\color{blue}{s}c$$$).  selects position $$$3$$$ and sets $$$A_3 = $$$ r ($$$ts\color{red}{c} \rightarrow ts\color{blue}{r}$$$).

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

Example output:
2
-1
3
2
-1

Editorial:
First of all, if there exists some $$$i$$$ such that $$$A_i > B_i$$$ there isn't a solution.
Otherwise, create a graph where every character is a node, and put a directed edge between node $$$u$$$ and node $$$v$$$ if character $$$u$$$ must be transformed into character $$$v$$$ (ie. from $$$A_i$$$ to $$$B_i$$$ for all $$$i$$$). We must select a list with minimum number of operations such that if there is an edge from node $$$u$$$ to node $$$v$$$, then it must exist a subsequence of the operations in the list that transforms $$$u$$$ into $$$v$$$.
Each weakly connected component $$$C$$$ can be solved independently and the answer for each component is $$$|C| - 1$$$. So total answer is $$$|ALP| - k$$$ where $$$k$$$ is the number of weakly connected components in the graph.
Proof:
Each weakly connected component $$$C$$$ requires at least $$$|C| - 1$$$ operations (because they are connected). Since there are no cycles in the graph a topological order exists. Find one and select each pair of consecutive nodes in this order as the list of operations.
Time complexity: $$$O(|A| + |B| + |ALP|)$$$ per test case where $$$|ALP| = 20$$$ denotes size of alphabet