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:
"""
██╗ ██████╗ ██╗    ██████╗  ██████╗  ██╗ █████╗
██║██╔═══██╗██║    ╚════██╗██╔═████╗███║██╔══██╗
██║██║   ██║██║     █████╔╝██║██╔██║╚██║╚██████║
██║██║   ██║██║    ██╔═══╝ ████╔╝██║ ██║ ╚═══██║
██║╚██████╔╝██║    ███████╗╚██████╔╝ ██║ █████╔╝
╚═╝ ╚═════╝ ╚═╝    ╚══════╝ ╚═════╝  ╚═╝ ╚════╝
"""                                        
__author__ = "Dilshod"
p = [True] * 1000100
p[1] = False
for i in range(2, 1100):
	if p[i]:
		for j in range(i + i, 1000100, i):
			p[j] = False
from math import sqrt as root
n = int(input())
m = list(map(int, input().split()))
ans = ""
answer = []
for i in range(n):
	if m[i] == 1:
		answer += ["NO"]
		continue
	m[i] = root(m[i])
	a = str(m[i])
	if a[-2] != "." or m[i] % 2 == 0 and m[i] != 2:
		answer += ["NO"]
		continue
	m[i] = int(m[i])
	if p[m[i]]:
		answer += ["YES"]
	else:
		answer += ["NO"]
print("\n".join(answer))