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 sys import stdin,stdout
from collections import Counter
def ai(): return list(map(int, stdin.readline().split()))
def ei(): return map(int, stdin.readline().split())
def ip(): return  int(stdin.readline().strip())
def op(ans): return stdout.write(str(ans) + '\n')
from math import ceil

def seive(n,prime):
	prime[0]=prime[1] = False
	for i in range(2,int(math.sqrt(1000000))+1):
		if prime[i]:
			for j in range(2*i,1000001,i):
				prime[j] = False
	return prime

import math 
def isPerfectSquare(x): 
    sr = math.sqrt(x) 
    return ((sr - math.floor(sr)) == 0)

prime = [True for i in range(1000001)]
seive(1000000,prime)
n = ip()
li = ai()
for i in li:
	if isPerfectSquare(i) and prime[int(math.sqrt(i))]:
		print('YES')
	else:
		print('NO')