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:



#calculate 1~n prime
def calcPrime(n):
	dicPrime = [0 for i in range(n)]
	for i in xrange(2, n):
		if dicPrime[i] == 0:
			for j in xrange(2*i, n, i):
				dicPrime[j] = 1
		else:
			pass
	return dicPrime
	
dicPrime =  calcPrime(1000000)   

def decStr(x):
	x = int(x)
	x_m = int(x ** 0.5)
	x_m_m = int(x_m ** 0.5)

	if 1 == x or x_m == x or x_m ** 2 != x:
		return 'NO' 
	elif x_m % 2 == 0 and x_m != 2:
		return 'NO'   
	
	elif dicPrime[x_m] == 0:
		
		return 'YES'
	else:
		return "NO"
	
	

input()
N = map(decStr, raw_input().split())
print '\n'.join(N)