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:
z,zz=input,lambda:list(map(int,z().split()));
szz,dgraphs,mod=lambda:sorted(zz()),{},10**9+7
from string import *
from collections import *
from queue import *
from sys import *
from collections import *
from math import *
from heapq import *
from itertools import *
from bisect import *
from collections import Counter as cc
from math import factorial as f
from bisect import bisect as bs
from bisect import bisect_left as bsl
from itertools import accumulate as ac
def lcd(xnum1,xnum2):return (xnum1*xnum2//gcd(xnum1,xnum2))
def prime(x):
    p=ceil(x**.5)+1
    for i in range(2,p):
        if x%i==0 and x!=2:return 0
    return 1
def dfs(u,visit,graph):
    visit[u]=True
    for i in graph[u]:
        if not visit[i]:
            dfs(i,visit,graph)

###########################---Test-Case---#################################
"""




"""
###########################---START-CODING---##############################

s=[True]*1000001
s[0],s[1]=False,False

for i in range(2,1000001):
    if s[i]:
        for j in range(i*i,1000001,i):
            s[j]=False




t=int(z())
l=zz()

for i in l:
    a=int(i**.5)
    if a*a==i and s[a]:
        print('YES')
    else:
        print('NO')