Problem: We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer t Т-prime, if t has exactly three distinct positive divisors.

You are given an array of n positive integers. For each of them determine whether it is Т-prime or not.

Input

The first line contains a single positive integer, n (1 ≤ n ≤ 105), showing how many numbers are in the array. The next line contains n space-separated integers xi (1 ≤ xi ≤ 1012).

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier.

Output

Print n lines: the i-th line should contain "YES" (without the quotes), if number xi is Т-prime, and "NO" (without the quotes), if it isn't.

Examples

Input

3
4 5 6


Output

YES
NO
NO

Note

The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO".

Solution (in PYTHON):

Solution:
'''                      
   	
   			 ||Sri:||                 

                      __|
 ______________|_________________________
      |    |   ___|    |    |    ___|      |      |    |___
 /\  /\    |   |___    |    |   |___|      |      |    |___|
/  \/  \   |   ___|    |    |         |_/    |___|    |___

I am a beginner. Feel free to send tutorials/help to srinivasan1995@gmail.com.

Codeforces, SPOJ handle: desikachariar.
Hackerrank, Codechef handle: Prof_Calculus.

Advises/Help are welcome. You will definitely get a thanks in return. Comments to be avoided.

College: International Institute of Information Technology, Bangalore.

'''
import math
def primes2(n):
    """ Input n>=6, Returns a list of primes, 2 <= p < n """
    n, correction = n-n%6+6, 2-(n%6>1)
    sieve = [True] * (n/3)
    for i in xrange(1,int(n**0.5)/3+1):
      if sieve[i]:
        k=3*i+1|1
        sieve[      k*k/3      ::2*k] = [False] * ((n/6-k*k/6-1)/k+1)
        sieve[k*(k-2*(i&1)+4)/3::2*k] = [False] * ((n/6-k*(k-2*(i&1)+4)/6-1)/k+1)
    return [2,3] + [3*i+1|1 for i in xrange(1,n/3-correction) if sieve[i]]

if __name__ == '__main__':
	t = input()
	ls = [float(math.sqrt(int(x))) for x in raw_input().split(' ')]
	primes = primes2(1000001)
	mp = {}
	for i in primes:
		mp[i] = True
	for i in ls:
		if i in mp:
			print "YES"
		else:
			print "NO"