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 as mt 
  
MAXN = 1000001
spf = [0 for i in range(MAXN)] 

def sieve(): 
    spf[1] = 1
    for i in range(2, MAXN): 

        spf[i] = i 

    for i in range(4, MAXN, 2): 
        spf[i] = 2
  
    for i in range(3, mt.ceil(mt.sqrt(MAXN))): 
          
        if (spf[i] == i): 
            for j in range(i * i, MAXN, i):  
                
                if (spf[j] == j): 
                    spf[j] = i 
  
def getFactorization(x): 
    ret = list() 
    while (x != 1): 
        ret.append(spf[x]) 
        x = x // spf[x] 
  
    return ret 
  

n = int(input())
x = [int(a) for a in input().split()]

sieve() 

for e in x:
	c = e ** 0.5
	if c % 1 == 0:
		iri = getFactorization(int(c))
		if len(iri) == 1 and iri[0] % 1 == 0:
			print("YES")
		else:
			print("NO")
	else:
		print("NO")