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:
# Initially I thought prime factorization was needed but
# actually i only need to check if the number is a square
# of a prime or not
import math

n=int(input())

x=list(map(int, input().split()))

limit=int(math.sqrt(max(x)))+2

l2=[0 for i in range(0,limit)]

p=2

while (p<limit):
    y=p*p
    while (y<limit):
        l2[y]=1
        y=y+p
    
    for i in range(p+1,limit):
        if l2[i]==0:
            p=i
            # print (p)
            break
    else:
        p=limit

l3=[0 for i in range(0,n)]

for i in range(0,n):
    b=math.sqrt(x[i])
    if x[i]==1:
        l3[i]='NO'
    elif b-int(b)==0 and l2[int(b)]==0:
        l3[i]='YES'
    else:
        l3[i]='NO'

print ("\n".join(l3))