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
import sys
input = sys.stdin.readline

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())
l = list(map(int, input().split()))
l1 = []

# Create an array to store prime status
is_prime_cache = [False] * (int(math.sqrt(max(l)))+1)
for i in range(len(is_prime_cache)):
    is_prime_cache[i] = is_prime(i)

for i in l:
    if i == 1:
        l1.append("NO")
    elif i ** 0.5 == math.floor(i ** 0.5):
        if is_prime_cache[int(i**0.5)]:
            l1.append("YES")
        else:
            l1.append("NO")
    else:
        l1.append("NO")

for i in l1:
    print(i)
