Contents

• # For Solution

You are given a string SS which denotes a padlock consisting of lower case English letters. You are also given a string FF consisting of set of favorite lower case English letters. You are allowed to perform several operations on the padlock. In each operation, you can change one letter of the string to the one following it or preceding it in the alphabetical order. For example: for the letter c, you are allowed to change it to either b or d in an operation. The letters can be considered in a cyclic order, i.e., the preceding letter for letter a would be letter z. Similarly, the following letter for letter z would be letter a. Transform the string solution kickstart

Your aim is to find the minimum number of operations that are required such that each letter in string SS after applying the operations, is present in string FF.

### Input

The first line of the input gives the number of test cases, TTTT test cases follow.

Each test case consists of two lines.
The first line of each test case contains the string SS.
The second line of each test case contains the string FF.

### Output

For each test case, output one line containing Case #xx: yy, where xx is the test case number (starting from 1) and yy is the minimum number of operations that are required such that each letter in string SS after applying the operations, is one of the characters in string FF.

### Limits Transform the string solution kickstart

Memory limit: 1 GB.
1T1001≤T≤100.
11≤ the length of S105S≤105.
SS only consists of lower case English letters.
FF only consists of distinct lower case English letters.
The letters in string FF are lexicographically sorted.

#### Test Set 1

Time limit: 20 seconds.
The length of F=1F=1.

#### Test Set 2

Time limit: 40 seconds.
11≤ the length of F26F≤26.

### Sample

Note: there are additional samples that are not run on submissions down below.

Sample Input
2
abcd
a
pppp
p

Sample Output Transform the string solution kickstart
Case #1: 6
Case #2: 0


In Sample Case #1, all the letters in string SS should be converted to letter a. We can keep on changing the letters to its preceding letter till we reach the letter a. We do not need to change the first letter as it is already a. The second letter needs 11 operation to change it to a. The third letter needs 22 operations to change it to a. The fourth letter needs 33 operation to change it to a. Hence, we need a total of 66 operations to change string SS such that all letters are changed to a.

In Sample Case #2, string SS already contains only the favorite letter from string FF. Hence, we do not require any more operations.

### Additional Sample – Test Set 2

The following additional sample fits the limits of Test Set 2. It will not be run against your submitted solutions.

Sample Input
3
pqrst
ou
abd
abd
aaaaaaaaaaaaaaab
aceg

Sample Output
Case #1: 9
Case #2: 0
Case #3: 1


In Sample Case #1, all the letters in string SS should be converted to either the letter o or the letter u. For the first and second letters it is optimal to change them to preceding letters till they are changed to letter o. The first letter would take 11 operation to change to letter o. The second letter would take 22 operations to change to letter o. For fourth and fifth letters it is optimal to change them to following letters till they are changed to letter u. The fourth letter would take 22 operations to change to letter u. The fifth letter would take 11 operation to change to letter u. We can change the third letter to either o or u as both of them would require 33 operations. Hence, we need a total of 99 operations to change string SS such that all letters are changed to either o or u.

In Sample Case #2, string SS already contains only the favorite letters from string FF. Hence, we do not require any more operations.

In Sample Case #3, we only need to change the last letter b to either a or c. Thus, we