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.
The first line contains a single positive integer, *n* (1 ≤ *n* ≤ 10<sup class="upper-index">5</sup>), showing how many numbers are in the array. The next line contains *n* space-separated integers *x*<sub class="lower-index">*i*</sub> (1 ≤ *x*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">12</sup>).

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.
from math import floor
n=int(input())
num=[int(x)for x in input().split()]
numbers=[[i,num[i]]for i in range(n)]
numbers.sort(key=lambda s:s[1])
primes=[2]
for p in range(3,floor(numbers[-1][1]**(1/2))+1):#创建质数表
    i=0
    while primes[i] <= floor(p**(1/2)):
        if p%primes[i] == 0:
            break
        i+=1
    else:
        primes.append(p)
numbers.sort(key=lambda s:s[0])
for nu in numbers:
    x=nu[1]**(1/2)
    if x%1 == 0 and int(x) in primes:#判断是否为质数平方
        print('YES')
    else:print('NO')