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

# 埃氏筛算法生成范围内的所有素数
def sieveOfEratosthenes(n):
    primes = [True] * (n+1)
    primes[0] = primes[1] = False
    
    p = 2
    while p * p <= n:
        if primes[p]:
            for i in range(p * p, n+1, p):
                primes[i] = False
        p += 1
    
    return primes

# 判断数是否为"T-prime"
def isTPrime(primes, number):
    sqrt_number = int(math.sqrt(number))
    if sqrt_number * sqrt_number != number:  # 不是完全平方数
        return False
    if sqrt_number < len(primes) and primes[sqrt_number]:  # 平方根是素数
        return True
    return False

# 读取输入
n = int(input())
numbers = list(map(int, input().split()))

# 生成范围内的素数列表
max_number = max(numbers)
primes = sieveOfEratosthenes(int(math.sqrt(max_number)) + 1)

# 对每个数判断是否为T-prime并输出结果
for number in numbers:
    if isTPrime(primes, number):
        print("YES")
    else:
        print("NO")
