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:
# -*- coding: utf-8 -*-
"""
Created on Wed Oct 24 14:14:32 2018

@author: Quaint Sun
"""

'''
import math

n=int(input())
array=[int(i) for i in input().split()]

for i in array:
    if i==1:
        print('NO')
    else:
        if math.sqrt(i)%1==0:
            t=0
            for j in range(1,int(math.sqrt(i))+1):
                if int(math.sqrt(i))%j==0:
                    t=t+1
            if t==2:
                print('YES')
            else:
                print('NO')
        else:
            print('NO')
'''
'''
import math

n=int(input())
array=[int(i) for i in input().split()]

def prime(x):
    if x==4:
        print('NO')
    else:
        for i in range(2,x):
            if (x % i) == 0:
                print('NO')
                break
        else:
            print('YES')
        
        
medium=[4]*n
def sq(x):       
    if math.sqrt(array[i])%1==0 and array[i]!=1:
        medium[i]=(math.sqrt(array[i]))

for i in range(n):
    sq(i)
    
for k in medium:
    prime(int(k))'''
    
    
from math import sqrt

n = 1000000
a = list(range(n+1))

a[1] = 0
lst = []
i = 2
while i <= n:
    if a[i] != 0:
        for j in range(i*i, n+1, i):
            a[j] = 0
    i += 1
input()
nums = [int(x) for x in input().split()]

for n in nums:
    if n > 10 and n%2 == 0:
        print("NO")
        continue
    r = sqrt(n)
    if r - int(r) == 0:
        if a[int(r)] != 0:
            print("YES")
        else:
            print("NO")
    else:
        print("NO")