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
#分析：只有一个质因数
#埃氏筛
n=int(input())

def is_prime(num):
   #初始默认为素数
    global bool_p
    bool_p=[0]*(num+1)
    for i in range(2,num+1):
        if bool_p[i]==0:
            for j in range(i+i,num+1,i):
                bool_p[j]=1


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

    if nums[i] in square:
        bool_t[i]=1
        sqrt_n=square[nums[i]]
        if sqrt_n>max_s:
            max_s=sqrt_n
#print(max_s)
is_prime(max_s)
for i in range(n):
    if bool_t[i]:
        if not bool_p[square[nums[i]]]:
            print("YES")
        else:
            print("NO")
    else:
        print("NO")






