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 sqrt

def is_prime(n):
    if n <= 1:
        return False
    if n <= 3:
        return True
    if n % 2 == 0 or n % 3 == 0:
        return False
    i = 5
    while i * i <= n:
        if n % i == 0 or n % (i + 2) == 0:
            return False
        i += 6
    return True

n = int(input())
numbers = list(map(int, input().split()))
numset = {1, 9, 21, 41, 61, 81, 29, 49, 69, 89}

for number in numbers:
    if number in [4, 9, 25]:
        print("YES")
        continue
    if len(str(number)) == 1 and number not in [4, 9]:
        print("NO")
        continue
    last_two_digits = int(str(number)[-2:])
    if last_two_digits in numset:
        sqrt_number = int(sqrt(number))
        if sqrt_number * sqrt_number == number and is_prime(sqrt_number):
            print("YES")
        else:
            print("NO")
    else:
        print("NO")