Anagram
时间限制：C/C++/Rust/Pascal 1秒，其他语言2秒
空间限制：C/C++/Rust/Pascal 64 M，其他语言128 M
64bit IO Format: %lld

题目描述

    Orz has two strings of the same length: A and B. Now she wants to transform A into an anagram of B (which means, a rearrangement of B) by changing some of its letters. The only operation the girl can make is to “increase” some (possibly none or all) characters in A. E.g., she can change an ‘A’ to a ‘B’, or a ‘K’ to an ‘L’. She can increase any character any times. E.g., she can increment an ‘A’ three times to get a ‘D’. The increment is cyclic: if she increases a ‘Z’, she gets an ‘A’ again.

    For example, she can transform “ELLY” to “KRIS” character by character by shifting ‘E’ to ‘K’ (6 operations), ‘L’ to ‘R’ (again 6 operations), the second ‘L’ to ‘I’ (23 operations, going from ‘Z’ to ‘A’ on the 15-th operation), and finally ‘Y’ to ‘S’ (20 operations, again cyclically going from ‘Z’ to ‘A’ on the 2-nd operation). The total number of operations would be 6 + 6 + 23 + 20 = 55. However, to make “ELLY” an anagram of “KRIS” it would be better to change it to “IRSK” with only 29 operations. You are given the strings A and B. Find the minimal number of operations needed to transform A into some other string X, such that X is an anagram of B.

输入描述:

There will be multiple test cases. For each testcase:

There is two strings A and B in one line.∣A∣=∣B∣≤50. A and B will contain only uppercase letters
from the English alphabet (‘A’-‘Z’).

输出描述:

For each test case, output the minimal number of
operations.
示例1

输入

复制 
ABCA BACA
ELLY KRIS
AAAA ZZZZ

输出

复制 
0
29
100