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:
n=input()
def ss(n):
    lst = []
    prime = [True for i in range(n + 1)]
    p = 2
    while (p * p <= n):

        if (prime[p] == True):

            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 1

    for p in range(2, n):
        if prime[p]:
            lst.append(p)
    return lst
lst=ss(1000002)
#sk=set(lst)
hp=[0]*1000004
for i in lst:
    hp[i]=1
from math import sqrt
f=map(int,raw_input().split())
#f=[10**12]*n
fv=[]
for i in f:
    if sqrt(i)==int(sqrt(i)) and hp[int(sqrt(i))]==1:
        fv.append("YES")
    else:
        fv.append("NO")
print "\n".join(fv)