Problem: 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.

Input

The first line contains a single positive integer, n (1 ≤ n ≤ 105), showing how many numbers are in the array. The next line contains n space-separated integers xi (1 ≤ xi ≤ 1012).

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.

Output

Print n lines: the i-th line should contain "YES" (without the quotes), if number xi is Т-prime, and "NO" (without the quotes), if it isn't.

Examples

Input

3
4 5 6


Output

YES
NO
NO

Note

The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO".

Solution (in PYTHON3):

Solution:
def primes2(n):
    """ Input n>=6, Returns a list of primes, 2 <= p < n """
    n, correction = n-n%6+6, 2-(n%6>1)
    sieve = [True] * (n//3)
    for i in range(1,int(n**0.5)//3+1):
      if sieve[i]:
        k=3*i+1|1
        sieve[      k*k//3      ::2*k] = [False] * ((n//6-k*k//6-1)//k+1)
        sieve[k*(k-2*(i&1)+4)//3::2*k] = [False] * ((n//6-k*(k-2*(i&1)+4)//6-1)//k+1)
    return [2,3] + [3*i+1|1 for i in range(1,n//3-correction) if sieve[i]]
n=1000000
y=[0]*1000001
t=primes2(n)
for i in t:
    y[i]=1

q=int(input())
import math
w=list(map(int,input().split()))
for i in w:
    if i==1:
        print('NO')
    else:
        e=int(math.sqrt(i))
        if e*e==i and y[e]==1:
            print('YES')
        else:
            print('NO')