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:
import math

def sieve_of_eratosthenes(n): 
    global prime
    prime = [True for i in range(n + 1)] 
    p = 2
    while (p * p <= n): 
          
        # If prime[p] is not changed, then it is a prime 
        if (prime[p] == True): 
              
            # Update all multiples of p 
            for i in range(p * 2, n + 1, p): 
                prime[i] = False
        p += 1
    prime[0]= False
    prime[1]= False 
  
sieve_of_eratosthenes( 1000001 )

test_case = int( input() )
vals = [ int(x) for x in input().split() ]
for val in vals:
    sqr_val = int( math.sqrt( val ) )
    if sqr_val * sqr_val != val:
        print( "NO" )
    elif prime[ sqr_val ] == True:
        print( "YES" )
    else:
        print( "NO" )