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 getdiv(n):
##    factors = []
##    for i in range(1,n+1):
##        if n%i == 0:
##            factors.append(i)
##    return len(factors)
##l=[]
##for i in range(2,1000):
##   if getdiv(i)==3:
##      l.append(i)
##print(*l)
##4 9 25 49 121 169 289 361 529 841 961 Square of primes
a = [True] * (10**6+1000)
def primes_sieve2(limit):
    a[0] = a[1] = False
    l=[]
    for (i, isprime) in enumerate(a):
        if isprime:
            l.append(i)
            for n in range(i*i, limit, i):
                a[n] = False
    return l
primes_sieve2(10**6+1000)
from collections import *
vis=defaultdict(lambda:False)
for i in range(1,10**6+1000):
   if a[i]:
      vis[i**2]=True
n=int(input())
l=[int(i) for i in input().split()]
for i in l:
   if vis[i]:
      print("YES")
   else:
      print("NO")