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.
def fg(n):
    primes = [] #存放质数
    is_prime = [1]*(n+1) #假设所有数都是质数，做标记。1表示是质数，0表示不是质数。
    is_prime[0] = is_prime[1] = 0 #0和1不是质数
    for i in range(2, n+1):
        if is_prime[i]:
            primes.append(i)
        for p in primes:
            if i * p > n:
                break
            is_prime[i * p] = 0
            if i % p == 0:  #保证每个合数只会被它的最小的质因数筛选掉
                break
    return [i for i in range(2, n+1) if is_prime[i]]
list1=fg(100000)
def f(x,list3):
    if x%1==0:
        if list3[int(x)]==1:
            return True
    else:
        return False
        
n=int(input())
list2=list(map(int,input().split()))
for i in list2:
    if f(i**0.5,list1)==True and i!=1:
        print('YES')
    else:
        print('NO')    