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 isPerfectSquare(x): 
    sr = math.sqrt(x)  
    if((sr - math.floor(sr)) == 0):
        return sr
n1=int(input())
l1=list(map(int,input().split()))
n=1000000 
l=[False,False]
prime = [True for i in range(n+1)] 
p = 2
while (p * p <= n):  
    if (prime[p] == True):
        for i in range(p * p, n+1, p):
            prime[i] = False
    p += 1 
for i in range(n1):
    if(l1[i]==1 or l1[i]==2 or l1[i]==3):
        print("NO")
    elif(isPerfectSquare(l1[i])):
        k=int(isPerfectSquare(l1[i]))
        if(prime[k]==True):
            print("YES")
        else:
            print("NO")
    else:
        print("NO")