Ashley likes playing with strings.

She gives Mojo a fun problem to solve. In her imaginary string world, a string of even length is called as "Doublindrome" if both halves of the string are palindromes (both halves have length equal to half of original string).

She gives Mojo a string and asks him if he can form a "Doublindrome" by rearranging the characters of the given string or keeping the string as it is.

As Mojo is busy playing with cats, solve the problem for him.

Print "YES" (without quotes) if given string can be rearranged to form a "Doublindrome" else print "NO" (without quotes).

-----Input:-----
- First line will contain a single integer $T$, the number of testcases. 
- Each testcase consists of two lines, first line consists of an integer $N$ (length of the string) and second line consists of  the string $S$. 

-----Output:-----
For each testcase, print "YES"(without quotes) or "NO"(without quotes) on a new line.

-----Constraints-----
- $1 \leq T \leq 10^5$
- $1 \leq N \leq 100$
- $N$ is always even.
- String $S$ consists only of lowercase English alphabets.

-----Sample Input:-----
1
8
abbacddc

-----Sample Output:-----
YES

-----EXPLANATION:-----
The given string is a Doublindrome as its 2 halves "abba" and "cddc" are palindromes.

Source: codechef | Difficulty: UNKNOWN_DIFFICULTY


Solutions:
for _ in range(int(input())):
	n = int(input())
	s = input()
	d = {}
	for j in s:
		if j not in d:
			d[j] = 1
		else:
			d[j] += 1
	f = 0
	for j in d:
		if d[j] % 2 == 1:
			f = f + 1
	if n // 2 % 2 == 0 and f == 0:
		print('YES')
		continue
	if n // 2 % 2 == 1 and f <= 2 and (f % 2 == 0):
		print('YES')
		continue
	print('NO')


T = int(input())
for _ in range(T):
	N = int(input())
	S = input()
	if N <= 2:
		print('YES')
	else:
		unique_c = {}
		for c in S:
			if c in list(unique_c.keys()):
				unique_c[c] += 1
			else:
				unique_c[c] = 1
		safe = 0
		no_odd = 0
		for c in list(unique_c.keys()):
			if unique_c[c] % 2 != 0:
				no_odd += 1
		if N // 2 % 2 == 1:
			if no_odd == 2 or no_odd == 0:
				print('YES')
			else:
				print('NO')
		elif no_odd != 0:
			print('NO')
		else:
			print('YES')


try:
	t = int(input())
	for i in range(t):
		n = int(input())
		s = input()
		flag = 1
		d = dict()
		for j in range(n):
			d[s[j]] = 0
		for j in range(n):
			d[s[j]] = d[s[j]] + 1
		flag = 1
		if n // 2 % 2 == 0:
			for j in list(d.keys()):
				if d[j] % 2 != 0:
					flag = 0
					break
			if flag == 1:
				print('YES')
			else:
				print('NO')
		else:
			count = 0
			for j in list(d.keys()):
				if d[j] % 2 != 0:
					count = count + 1
			if count == 2 or count == 0:
				print('YES')
			else:
				print('NO')
except:
	pass


from collections import defaultdict
for t in range(int(input())):
	n = int(input())
	s = input()
	d = defaultdict(int)
	for i in s:
		d[i] += 1
	count = 0
	for (i, j) in d.items():
		if j % 2 == 1:
			count += 1
	if n // 2 % 2 == 0:
		if count == 0:
			print('YES')
		else:
			print('NO')
	elif count == 2 or count == 0:
		print('YES')
	else:
		print('NO')


for _ in range(int(input())):
	n = int(input())
	s = list(input())
	d = dict()
	for i in s:
		d[i] = d.get(i, 0) + 1
	if n // 2 % 2 == 0:
		for i in list(d.values()):
			if i % 2 == 1:
				print('NO')
				break
		else:
			print('YES')
	else:
		c = 0
		for i in list(d.values()):
			if i % 2 == 1:
				c += 1
				if c == 3:
					print('NO')
					break
		else:
			print('YES')


for t in range(int(input())):
	n = int(input()) // 2
	s = input()
	d = {}
	for i in s:
		if i not in d:
			d[i] = 0
		d[i] += 1
	res = 0
	for i in d:
		if d[i] % 2:
			res += 1
	if n % 2:
		if res == 0 or res == 2:
			print('YES')
		else:
			print('NO')
	elif res == 0:
		print('YES')
	else:
		print('NO')