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:
from math import sqrt

def get_prime(n):
	if n == 2:
		return [2]
	else:
		prime_sqrt_n = get_prime(int(sqrt(n - 1)) + 1)
		is_prime = [1] * (n - 1)
		for p in prime_sqrt_n:
			q = p
			while q < n // p + 1:
				is_prime[p * q - 2] = 0
				q += 1
		return [i for i, j in enumerate(is_prime, start=2) if j == 1]



prime_min = get_prime(10**3)
input()
t_list = [*map(int, input().split())]
for squared_t in t_list:
	t = int(sqrt(squared_t))
	if squared_t == 4:
		print('YES')
		continue
	elif squared_t % 2 == 0 or t**2 != squared_t or t == 1:
		print('NO')
		continue
	sqrt_t = int(sqrt(t)) + 1
	j = 0
	prime = True
	while prime_min[j] < sqrt_t:
		if t % prime_min[j] == 0:
			print('NO')
			prime = False
			break
		if j < len(prime_min) - 1:
			j += 1
		else:
			break
	if prime:
		print('YES')