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 PYTHON):

Solution:
from math import sqrt


def rwh_primes(n):
    # http://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
    """ Returns  a list of primes < n """
    sieve = [True] * n
    for i in xrange(3,int(n**0.5)+1,2):
        if sieve[i]:
            sieve[i*i::2*i]=[False]*((n-i*i-1)/(2*i)+1)
    return [2] + [i for i in xrange(3,n,2) if sieve[i]]


from bisect import bisect_left

def bs(a, x, lo=0, hi=None):   # can't use a to specify default for hi
    hi = hi if hi is not None else len(a) # hi defaults to len(a)   
    pos = bisect_left(a,x,lo,hi)          # find insertion position
    return (True if pos != hi and a[pos] == x else False) # don't walk off the end

primes = rwh_primes(10**6)
#print len(primes),bs(primes,2)
t = input()

arr = map(int, raw_input().split())



for i in arr:
       if i==1:
              print 'NO'
              continue
       z = sqrt(i)
       z = int(z)
       #print z,i
       if z*z==i and bs(primes,z):
              print 'YES'
       else:
              print 'NO'