ENG  RUSTimus Online Judge
Online Judge
Problems
Authors
Online contests
About Online Judge
Frequently asked questions
Site news
Webboard
Links
Problem set
Submit solution
Judge status
Guide
Register
Update your info
Authors ranklist
Current contest
Scheduled contests
Past contests
Rules

1158. Censored!

Time limit: 2.0 second
Memory limit: 64 MB
The alphabet of Freeland consists of exactly N letters. Each sentence of Freeland language (also known as Freish) consists of exactly M letters without word breaks. So, there exist exactly NM different Freish sentences.
But after recent election of Mr. Grass Jr. as Freeland president some words offending him were declared unprintable and all sentences containing at least one of them were forbidden. The sentence S contains a word W if W is a substring of S i.e. exists such k >= 1 that S[k] = W[1], S[k+1] = W[2], ... , S[k+len(W)-1] = W[len(W)], where k+len(W)-1 <= M and len(W) denotes length of W. Everyone who uses a forbidden sentence is to be put to jail for 10 years.
Find out how many different sentences can be used now by freelanders without risk to be put to jail for using it.

Input

The first line contains three integer numbers: N - the number of letters in Freish alphabet, M - the length of all Freish sentences and P - the number of forbidden words (1 ≤ N ≤ 50, 1 ≤ M ≤ 50, 0 ≤ P ≤ 10).
The second line contains exactly N different characters - the letters of the Freish alphabet (all with ASCII code greater than 32).
The following P lines contain forbidden words, each not longer than min(M, 10) characters, all containing only letters of Freish alphabet.

Output

Output the only integer number - the number of different sentences freelanders can safely use.

Sample

inputoutput
3 3 3
QWE
QQ
WEE
Q
7

Notes

Note that tests may contain characters with ASCII codes more than 127.
Problem Author: Nick Durov
Problem Source: ACM ICPC 2001. Northeastern European Region, Northern Subregion