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.
#等差数列法
n=int(input())
s=list(map(int,input().split()))
maxnum=max(s)
import math
first=int(math.sqrt(maxnum))     #找到需要检索的最大质数范围
l, x, y=[True]*(first+1), 2, int(math.sqrt(first))+1
l[0]=l[1]=False
while x<y:        #遍历到开方后的开方
    if l[x]:
        for j in range(x*x, first+1, x):   #只需检索x方后的倍数，之前的已检索过
            l[j]=False
    x+=1
for i in s:
    if not math.sqrt(i)%1==0:
        print("NO")
    else:
        r=int(math.sqrt(i))
        if l[r]:
            print("YES")
        else:
            print("NO")            