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.
n=int(input())
nums=input().split()
nums=[int(i) for i in nums]
sorted(nums)
flag=False
for i in reversed(range(n)):
    if pow(nums[i],1/2)==int(pow(nums[i],1/2)):
        if flag==False:
            max=nums[i]
        flag=True
    else:
        nums[i]=-1
#连平方数都不是就直接否决

def prime_generate(x):
    for j in prime:
        if x%j==0:
            return
    prime.append(x)
prime=[2]

for i in range(3,int(pow(max,1/4)+1)):
    prime_generate(i)

for i in range(n):
    flag=True
    if nums[i]==-1:
        print('NO')
        continue
    sqrt=pow(nums[i],1/2)
    if sqrt!=int(sqrt):
        print('NO')
        continue
    for j in prime:
        if sqrt%j==0 and sqrt!=j:#发现平方根不是质数
            print('NO')
            flag=False
            break
    if flag==True:
        print('YES')