Name Reduction

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In an attempt to reduce the growing population, Archer was asked to come up with a plan. Archer being as intelligent as he is, came up with the following plan:
If N children, with names C_{1}, C_{2}, ..., C_{N}, are born to parents with names A and B, and you consider C to be the concatenation of all the names of the children, i.e. C = C_{1} + C_{2} + ... + C_{N} (where + is concatenation operator), then C should be a substring of one of the permutations of A + B.
You are given the task to verify whether the names parents propose to give their children are in fact permissible by Archer's plan or not.
Input
The first line contains an integer T, the number of test cases. T test cases follow. Each test case stats with a line containing two space separated strings A and B, denoting the names of the parents. The next line contains a single integer N denoting the number of children A and B are planning to have. Following this are N lines, the i'th line containing C_{i}, the proposed name for the i'th child.
Output
For each test case output a single line containing "YES"
if the names are permissible by Archer's plan, otherwise print "NO"
. (quotes are meant for clarity, please don't print them)
Constraints
 1 ≤ T ≤ 100
 1 ≤ N ≤ 1000
 The lengths of all the strings including A, B, and all C_{i} will be in the range [1, 40000], both inclusive. All these strings will contain only lowercase English letters.
 The combined lengths of all names of children will not exceed the combined length of the names of their parents.
Example
Input: 3 tom marvoloriddle 2 lord voldemort cheap up 1 heapcup bruce wayne 2 bat man Output: YES YES NO
Explanation:
Let Y denote the concatenation of names of all the children, and X denote the concatenation of the names of the parents.
Case 1: Here X = "tommarvoloriddle", and Y = "lordvoldemort". Consider Z = "iamlordvoldemort". It is not difficult to see that Z is a permutation of X and Y is a substring of Z. Hence Y is a substring of a permutation of X, so the answer is "YES".
Case 2: Here X = "cheapup", and Y = "heapcup". Since Y in itself is a permutation of X, and as every string is a substring of itself, Y is a substring of X and also a permutation of X. Hence "YES".
Case 3: Here X = "brucewayne", and Y = "batman". As "t" is not present in X, "t" wont be present in any permutation of X, hence the answer is "NO".
Author:  kaushik_iska 
Tester:  pieguy 
Editorial  http://discuss.codechef.com/problems/NAME1 
Tags  adhoc, cakewalk, kaushik_iska, may13 
Date Added:  27032013 
Time Limit:  1 sec 
Source Limit:  50000 Bytes 
Languages:  C, CPP14, JAVA, PYTH, PYTH 3.6, CS2, PAS fpc, PAS gpc, RUBY, PHP, GO, NODEJS, HASK, SCALA, D, PERL, FORT, WSPC, ADA, CAML, ICK, BF, ASM, CLPS, PRLG, ICON, SCM qobi, PIKE, ST, NICE, LUA, BASH, NEM, LISP sbcl, LISP clisp, SCM guile, JS, ERL, TCL, PERL6, TEXT, PYP3, CLOJ, FS 
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