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:
import atexit

@atexit.register
def main():
	from sys import stdout, stdin
	
	def is_prime(numero,crivo):
		if numero == int(numero):
			return crivo[int(numero)]
		return False
	
	################## CRIVO #####################################
	crivo = [True] * int(pow(pow(10,12),0.5)+1)
	crivo[0] = False
	crivo[1] = False
	for j in xrange(2,int(pow(pow(10,12),0.5)+1)):
		if crivo[j]:
			for m in xrange(2*j,int(pow(pow(10,12),0.5)+1),j):
				crivo[m] = False
	################## CRIVO #####################################
	
	numero_entradas = int(stdin.readline())
	numeros = map(int,stdin.readline().split())

	for i in xrange(numero_entradas):
		if is_prime(numeros[i] ** 0.5,crivo):
			stdout.write("YES\n")
		else:
			stdout.write("NO\n")
			
if __name__ == "__main__":
	main