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:
from array import *

s = {}
isPrime = None
XMAX = 10 ** 6 + 100

def makePrimes():
	global isPrime
	isPrime = [1] * XMAX
	isPrime[0] = isPrime[1] = 0
	i = 1
	while i + 1 < XMAX:
		i += 1
		if isPrime[i]:
			cur = i ** 2
			while cur < XMAX:
				isPrime[cur] = 0
				cur += i

def run(a):
	#if a in s:
	#	print (s[a])
	#	return None
	if a == 1:
		print ("NO")
		return None	
	x = int(a ** 0.5 + 1e-9)
	if x ** 2 != a:
		print ("NO")
		return None
	if not isPrime[x]:
	#	s[a] = "NO"
		print ("NO")
		return None
	#s[a] = "YES"
	print ("YES")
	return None

makePrimes()
n = int(raw_input())
a = [int(i) for i in raw_input().split()]

for i in a:
	run(i)