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 PYTHON3):

Solution:
def isKprime(num, k=3):
	ctr = 2
	sqr = int(num**0.5)
	for i in range(2,sqr+1):
		if(ctr >= 4):
			return False
		if(num%i == 0):
			if(num//i == i):
				ctr+=1
			else:
				ctr+=2
	return (ctr == k)


def seive(n):

	primes= [1]*(n+1)
	i = 2
	primes[0] = 0
	primes[1] = 0
	while(i*i <= n):
		if(primes[i] == 1):
			for j in range(i*i, n+1, i):
				primes[j]= 0
		i+=1
	actualPrime = {}
	for j in range(len(primes)):
		if(primes[j] == 1):
			actualPrime[j] = 1
	return actualPrime

input()
primes = seive(1000000)
lis = list(map(int, input().split()))
for i in lis:
	value1 = i**0.5
	value2 = int(i**0.5)
	if(value1 in primes and value1 == value2):
		print("YES")
	else:
		print("NO")
# print(seive(1000000))