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 PYTHON):

Solution:
import math
m=1000001
#list can be conctanated as strings... also can be multiplied to 
#have dat within list as times d given in a single list(no sublists but all elements)
#
s=[0]*2+[1]*m
#>>[0,0,1,1,1,1,1,1,1,1,1,1,...(1000001 times 1)]
#
for i in range(2,1001):
     if s[i]:
          for j in range(i*2,m,i):
               s[j]=0
#at first step all even positions from 4 onwards are set to 0,so not in our checking list anymore(i.e if statement False)
#iske baad 3 ke saare multiples 0 ko set ho jaayenge(6 pehle hi ho chuka hai)
#
#kisi no ke square se uske factors delete karna 
#is a step to find pime no, since n*m(m<n) means dat no is simutaneously divisible by m too
#so definitely not a prime no
#is tarah se sirf prime numbered positions hongi jo non zero hongi
#it follows from d fact dat that very number hadn't any divisor before itself
#
input()
for a in map(int,raw_input().split(' ')):
     b=int(math.sqrt(a))
#we might get a not perfect square 'a' as input
#then ssqrt will return float which cant be inputted as index
#so in and condition to avoid error check for perfect square condition first
#
     print ['NO','YES'][b*b==a and s[b]]
#it is avery beautifu way of writing list[index] where index is 0 for False condition and 1 for True