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 sys
import heapq, functools, collections
import math, random
from collections import Counter, defaultdict

# available on Google, not available on Codeforces
# import numpy as np
# import scipy


def solve(arr,brr):  # fix inputs here
    console("----- solving ------")
    console(arr)
    console(brr)


    abc = dict((ab, idx) for idx,ab in enumerate("abcdefghijklmnopqrst"))

    crr = []
    drr = []    
    for a,b in zip(arr,brr):
        if a > b:
            return -1
        if a != b:
            crr.append(abc[a])
            drr.append(abc[b])

    console(crr)
    console(drr)

    if len(crr) == 0:
        return 0

    d = defaultdict(set)

    for a,b in zip(crr,drr):
        d[a].add(b)
        d[b].add(a)

    console(d)
    visited = [False for _ in range(len(abc))]

    idx = 1

    for k in [x for x in d.keys()]:
        if visited[k]:
            continue
        idx += 1
        visited[k] = idx
        stack = [x for x in d[k]]
        while stack:
            cur = stack.pop()
            visited[cur] = idx
            # if not cur in d:
            #     continue
            for nex in d[cur]:
                if visited[nex]:
                    continue
                visited[nex] = idx
                stack.append(nex)
        console(d)

    console(visited)

    return sum([a-1 for a in Counter([x for x in visited if x]).values()])


def console(*args):  # the judge will not read these print statement
    # print('\033[36m', *args, '\033[0m', file=sys.stderr)
    return

# fast read all
# sys.stdin.readlines()

for case_num in range(int(input())):
    # read line as a string
    # strr = input()

    # read line as an integer
    _ = int(input())
    a = input()
    b = input()
    
    # read one line and parse each word as a string
    # lst = input().split()

    # read one line and parse each word as an integer
    # lst = list(map(int,input().split()))

    # read matrix and parse as integers (after reading read nrows)
    # lst = list(map(int,input().split()))
    # nrows = lst[0]  # index containing information, please change
    # grid = []
    # for _ in range(nrows):
    #     grid.append(list(map(int,input().split())))

    res = solve(a,b)  # please change
    
    # Google - case number required
    # print("Case #{}: {}".format(case_num+1, res))

    # Codeforces - no case number required
    print(res)
