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.
#刚好有三个因数：必须为完全平方数
#分析：只有一个质因数
import math
n=int(input())
#埃氏筛
def is_prime(num):
   #初始默认为素数
    global bool_p
    bool_p=[0]*(num+1)
    i=2
    while i*i<=num:
        if bool_p[i]==0:
            for j in range(i+i,num+1,i):
                bool_p[j]=1
        i+=1

bool_t=[0]*n#默认都是NO
nums=[int(i) for i in input().split()]
max_n=max(nums)
##square={i**2:i for i in range(2,math.ceil(math.sqrt(max_n))+1)} #本行可优化

#print(square)

#直接用最大的数字开根
is_prime(int(math.sqrt(max_n))+1)

for i in range(n):
    #判断是否为完全平方数
    sqrt_num=math.sqrt(nums[i])
    #print(sqrt_num)
    if (int(sqrt_num))**2 == nums[i]:
        if not bool_p[int(sqrt_num)]:
            print("YES")
        else:
            print("NO")
    else:
        print("NO")






