# [Solution] Most Similar Words solution codeforces

Most Similar Words solution codeforces – You are given 𝑛n words of equal length 𝑚m, consisting of lowercase Latin alphabet letters. The 𝑖i-th word is denoted 𝑠𝑖si.

## [Solution] Most Similar Words solution codeforces

In one move you can choose any position in any single word and change the letter at that position to the previous or next letter in alphabetical order. For example:

• you can change ‘e‘ to ‘d‘ or to ‘f‘;
• a‘ can only be changed to ‘b‘;
• z‘ can only be changed to ‘y‘.

The difference between two words is the minimum number of moves required to make them equal. For example, the difference between “best” and “cost” is 1+10+0+0=111+10+0+0=11.

Find the minimum difference of 𝑠𝑖si and 𝑠𝑗sj such that (𝑖<𝑗)(i<j). In other words, find the minimum difference over all possible pairs of the 𝑛n words.

Input

The first line of the input contains a single integer 𝑡t (1𝑡1001≤t≤100) — the number of test cases. The description of test cases follows.

The first line of each test case contains 22 integers 𝑛n and 𝑚m (2𝑛502≤n≤501𝑚81≤m≤8) — the number of strings and their length respectively.

Then follows 𝑛n lines, the 𝑖i-th of which containing a single string 𝑠𝑖si of length 𝑚m, consisting of lowercase Latin letters.

## [Solution] Most Similar Words solution codeforces

For each test case, print a single integer — the minimum difference over all possible pairs of the given strings.

Example
input

Copy
6
2 4
best
cost
6 3
abb
zba
bef
cdu
ooo
zzz
2 7
aaabbbc
bbaezfe
3 2
ab
ab
ab
2 8
aaaaaaaa
zzzzzzzz
3 1
a
u
y

output

Copy
11
8
35
0
200
4


## [Solution] Most Similar Words solution codeforces

For the second test case, one can show that the best pair is (“abb“,”bef“), which has difference equal to 88, which can be obtained in the following way: change the first character of the first string to ‘b‘ in one move, change the second character of the second string to ‘b‘ in 33 moves and change the third character of the second string to ‘b‘ in 44 moves, thus making in total 1+3+4=81+3+4=8 moves.

For the third test case, there is only one possible pair and it can be shown that the minimum amount of moves necessary to make the strings equal is 3535.

For the fourth test case, there is a pair of strings which is already equal, so the answer is 00.