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 seive(n):
	b=[i for i in range(n+1)]

	i=2
	while i<len(b):
		if(b[i]==0):
			i+=1
		else:
			k=2
			j=i*k
			while(j<len(b)):
				b[j]=0
				k+=1
				j=i*k

			i+=1

	return b

# print(seive(int(math.sqrt(999966000289))))

n=int(input()) #=len(a)
a=list(map(int,input().split(" ")))

import math

for i in range(len(a)):
	a[i]=[a[i],i]

def myfun(v):
	return v[0]

a.sort(key=myfun)

for i in range(n):
	a[i][0]=math.sqrt(a[i][0])
# print(a)

b=seive(int(a[-1][0]))[2:]
# print(b)

i=0;j=0;
c=[i for i in range(n)]
while(i<len(b) and j<n):
	if(b[i]==a[j][0]):
		c[a[j][1]]="YES"
		j+=1
	elif(b[i]<a[j][0]):
		i+=1
	else:
		c[a[j][1]]="NO"
		j+=1

for k in range(j,n):
	# print("reached")
	c[a[k][1]]="NO"

for o in c:
	print(o)