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:
  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 $x$).
  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 $x$).
  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-----
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-----
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:   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 ($ts\color{red}{c} \rightarrow ts\color{blue}{r}$).
import sys
input = lambda: sys.stdin.readline().rstrip()
T = int(input())
for _ in range(T):
    N = int(input())
    A = [ord(a) - 97 for a in input()]
    B = [ord(a) - 97 for a in input()]
    X = [[0] * 20 for _ in range(20)]
    for a, b in zip(A, B):
        X[a][b] = 1
        if a > b:
            print(-1)
            break
    else:
        ans = 0
        for i in range(20):
            for j in range(i+1, 20):
                if X[i][j]:
                    ans += 1
                    for jj in range(j+1, 20):
                        if X[i][jj]:
                            X[j][jj] = 1
                    break
        print(ans)

import sys
input = lambda: sys.stdin.readline().rstrip()
T = int(input())
for _ in range(T):
    N = int(input())
    A = [ord(a) - 97 for a in input()]
    B = [ord(a) - 97 for a in input()]
    X = [[0] * 20 for _ in range(20)]
    for a, b in zip(A, B):
        X[a][b] = 1
        if a > b:
            print(-1)
            break
    else:
        ans = 0
        for i in range(20):
            for j in range(i+1, 20):
                if X[i][j]:
                    ans += 1
                    for jj in range(j+1, 20):
                        if X[i][jj]:
                            X[j][jj] = 1
                    break
        print(ans)
