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:
# http://codeforces.com/problemset/problem/230/B

n = int(input())
l = map(int, input().split())

def sieve_of_eratosthenes(n): 
	prime_squares = set()  
	# Create a boolean array "prime[0..n]" and initialize 
	# all entries it as true. A value in prime[i] will 
	# finally be false if i is Not a prime, else true. 
	prime = [True for i in range(n + 1)] 
	p = 2
	while (p * p <= n): 
		  
		# If prime[p] is not changed, then it is a prime 
		if (prime[p] == True): 
			  
			# Update all multiples of p 
			for i in range(p * 2, n + 1, p): 
				prime[i] = False
		p += 1
	prime[0]= False
	prime[1]= False
	# Print all prime numbers 
	for p in range(n + 1): 
		if prime[p]:
			prime_squares.add(p ** 2)
	return prime_squares

prime_squares = sieve_of_eratosthenes(1000001)

for i in l:
	if i in prime_squares:
		print('YES')
	else:
		print('NO')