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 math

def prime():
	num=1000001
	sqnum=int(math.ceil(math.sqrt(num))+1.0000)
	mark[1]=1
	mark[2]=0
	for i in range(4,num,2):
		mark[i]=1
	for i in range(1,num,2):
		if mark[i]!=1:
			if i<=num:
				for j in range(i*i,num,i*2):
					mark[j]=1


mark=1000001*[0]
prime()
#print(mark)
#for i in range(988000,988080):
	#if mark[i]==0:
		#print(i," ")
n=int(input())
d=list(map(int,input().split()))
#print(d)
for i in range(n):
	if math.ceil(math.sqrt(d[i]))!=math.floor(math.sqrt(d[i])):
		print("NO")
	else:
	    s=int(math.sqrt(d[i]))
	    if mark[s]==0:
	    	print("YES")
	    else:
	    	print('NO')