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
L=lambda:list(map(int,input().split()))
M=lambda:map(int,input().split())
I=lambda:int(input())
IN=lambda:input()
def s(a):
    print("".join(list(map(str,a))))
def soe(n):
    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
    return prime
n=I()
a=L()
l=soe(1000000)
for i in a:
    if i==4:
        print("YES")
    elif i%2==0 or i<4:
        print("NO")
    elif math.sqrt(i)==int(math.sqrt(i)):
        if l[int(math.sqrt(i))]==True:
            print("YES")
        else:
            print("NO")
    else:
        print("NO")