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.
import math

def generatePrimes(n):
    is_prime = [True] * (n+1)
    primes = []

    is_prime[0] = is_prime[1] = False
    for i in range(2, int(math.sqrt(n))+1):
        if is_prime[i]:
            for j in range(i*i, n+1, i):
                is_prime[j] = False

    for i in range(2, n+1):
        if is_prime[i]:
            primes.append(i)

    return primes

def isPrime(x, primes):
    if x < 2:
        return False
    for prime in primes:
        if prime > int(math.sqrt(x)):
            break
        if x % prime == 0:
            return False
    return True

n = int(input())
array = list(map(int, input().split()))
max_num = max(array)

primes = generatePrimes(int(math.sqrt(max_num)))

for i in range(n):
    sqrt_x = int(math.sqrt(array[i]))
    if sqrt_x * sqrt_x == array[i] and isPrime(sqrt_x, primes):
        print("YES")
    else:
        print("NO")
