Solve the programming task below in a Python markdown code block.
Mario was going to gift Princess Peach a string S of even length N. 

Mario was clumsy and thus, broke the string in the middle. He now has two strings A and B such that A = S[1, \frac{N}{2}] and B = S[\frac{N}{2} + 1, N]. 

Mario is not sure about the order in which he should join the strings A and B to get the string S. Thus, he joins the strings in any random order. Find whether it is guaranteed that Mario will get the same string S if he joins the strings A and B in any order.

Note that S[L, R] denotes a substring of string S starting at index L and having a length (R - L + 1).

------ Input Format ------ 

- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of two lines of input:
- The first line of each test case contains N - the length of the initial string S.
- The second line contains the string S.

------ Output Format ------ 

For each test case, print YES if it is guaranteed that Mario will get the same string S irrespective of the order in which he joins the strings A and B and NO otherwise. 

You may print each character of the string in uppercase or lowercase (for example, the strings YES, yEs, yes, and yeS will all be treated as identical).

------ Constraints ------ 

$1 ≤ T ≤ 3000$
$1 ≤ N ≤ 1000$
$N$ is even.
$S$ consists of lowercase english alphabets.

----- Sample Input 1 ------ 
4
6
abcabc
6
abcdef
4
aaaa
4
baab
----- Sample Output 1 ------ 
YES
NO
YES
NO

----- explanation 1 ------ 
Test case $1$: On breaking, the string $S$ gives $A = abc$ and $B = abc$. Thus, joining it in either way $(AB$ or $BA)$, would give the same string $S$.

Test case $2$: On breaking, the string $S$ gives $A = abc$ and $B = def$. Joining it as $BA$ would give the string $defabc$ which is not equal to string $S$.

Test case $3$: On breaking, the string $S$ gives $A = aa$ and $B = aa$. Thus, joining it in either way $(AB$ or $BA)$, would give the same string $S$.

Test case $4$: On breaking, the string $S$ gives $A = ba$ and $B = ab$. Joining it as $BA$ would give the string $abba$ which is not equal to string $S$.