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 valid(n):
  x = int(sqrt(n))
  return x*x == n and prime(x)

prime = [True] * 1000001
i = 3
while i < len(prime):
  if prime[i]:
    j = i * i
    while j < len(prime):
      prime[j] = False
      j += i
  i += 2
primes = set()
primes.add(2)
i = 3
while i < len(prime):
  if prime[i]:
    primes.add(i)
  i += 2
  
qq = int(raw_input())
line = raw_input()
qqqq = line.split()
for qqq in qqqq:
  n = int(qqq)
  x = int(sqrt(n))
  if x * x != n:
    print "NO"
  else:
    if x in primes:
      print "YES"
    else:
      print "NO"