Alice has just learnt about primeStrings. A string is a primeString if the number of distinct alphabets used in the string is a prime and also the number of occurrences of each alphabet in the string is also a prime.
Given a String you need to tell if it is a primeString or not.

Input:
First line contains T which is the number of test cases.
T lines follow each containing a string of characters 'a' to 'z'.

Output:
For each input, output "YES" if the number is a primeString or "NO" if not.

Constraints:

1 ≤ T ≤ 10
 1 ≤ Length\; of\; string ≤ 10^5

Scoring:

1 ≤ T ≤ 10, 1 ≤ Length\; of\; string  ≤ 10\; (20 pts)
1 ≤ T ≤ 10, 1 ≤ Length\; of\; string  ≤ 1000 \;(30 pts)
1 ≤ T ≤ 10, 1 ≤ Length\; of\; string  ≤ 10^5 (50 pts)

SAMPLE INPUT
3
ababb
abcab
aabbccdd

SAMPLE OUTPUT
YES
NO
NO

Explanation

Case 1: 2 different alphabets each occurring 2 and 3 times respectively so string "ababb" is a PrimeString.   

Case 2: In second string char 'a' occurs 2 times, char 'b' occurs 2 times but char 'c' occur only 1 time which is not a prime number that's why string "abcab" is not a PrimeString.  

Case 3: String contains 4 distinct alphabets and 4 is not a prime Number so the string "aabbccdd" is not a PrimeString.
def prime(a):
	if a==1:
		return False
	for i in range(2,a/2+1):
		if a%i==0:
			return False
	return True

t = input()
while t:
	s = raw_input()
	l = len(s)
	freq = {}
	for i in range(l):
		if s[i] not in freq.keys():
			if 'dis' not in freq.keys():
				freq['dis']=1
			else:
				freq['dis']+=1
			freq[s[i]]=1
		else:
			freq[s[i]]+=1
	flag=1
	for i in freq.keys():
		if not prime(freq[i]):
			flag=0
			break
	if not prime(freq['dis']):
		flag=0
	if flag==1:
		print "YES"
	else:
		print "NO"
	t-=1