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

# 质数生成器
def prime_generator():
    yield 2
    prime_list = [2]
    num = 3
    while True:
        is_prime = True
        sqrt_num = int(math.sqrt(num)) + 1
        for prime_num in prime_list:
            if prime_num > sqrt_num:
                break
            if num % prime_num == 0:
                is_prime = False
                break
        if is_prime:
            prime_list.append(num)
            yield num
        num += 2

n = int(input())
nums = list(map(int, input().split()))

prime_gen = prime_generator()
for t in nums:
    sqrt_t = int(math.sqrt(t))
    if sqrt_t * sqrt_t == t:
        print('YES')
    else:
        # 判断是否为质数
        is_prime = True
        prime_num = next(prime_gen)
        while prime_num <= sqrt_t:
            if t % prime_num == 0:
                is_prime = False
                break
            prime_num = next(prime_gen)
        
        if is_prime:
            print('YES')
        else:
            print('NO')