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:
#!/usr/bin/env python
import math
n = int(raw_input())
alist = map(int, raw_input().strip().split())

x = 1000000
numlist = range(3, x+1, 2)
counter = 0 # Keeps count of index in while loop
backup = 0 # Used to reset count after each iteration and keep count outside of while loop
for num in numlist:
    counter = backup
    if num != 0:
        counter += num
        while counter <= len(numlist)-1: # Sifts through multiples of num, setting them all to 0
                numlist[counter] = 0
                counter += num
    else: # If number is 0 already, skip
        pass
    backup += 1 # Increment backup to move on to next index
primes = [2] + [x for x in numlist if x]
d = {}
for p in primes:
    d[p] = 1

for a in alist:
    t = int(math.sqrt(a))
    if t*t==a and d.get(t, False):
        print 'YES'
    else:
        print 'NO'