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:
import sys
from math import sqrt, floor, factorial, gcd
from collections import deque, Counter, defaultdict
inp = sys.stdin.readline
read = lambda: list(map(int, inp().strip().split()))

def is_prime(n):
	if n == 1:return(False)
	for i in range(2, int(sqrt(n))+1):
		if n%i == 0:
			return(False)
	return(True)

def solve():
	is_prime = [1 for i in range(10**6 + 1)]
	is_prime[1] = 0
	i = 2
	while i*i < 10**6 + 1:
		j = i
		if is_prime[j]:
			while j*i < 10**6 +1:
				is_prime[j*i] = 0
				j += 1
		i += 1
	# print(is_prime[:20])
	ans = ""
	n = int(inp()); arr = read()
	for i in arr:
		root = int(sqrt(i))
		if root**2 == i and is_prime[root]:
			ans += ("YES\n")
		else:ans += "NO\n"
	print(ans)





if __name__ == "__main__":
	solve()