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:
import math
global l
l=[]
for i in range(0,1000000):
        l.append(False)
def Tprime(leng):
    p=2
    p1=p
    l[0]=True
    while(True):
        p=p+p1
        while(p<=leng):
            l[p-1]=True
            p+=p1
        p1+=1
        if(p1>leng):
            break
        while(l[p1-1]==True and p1<leng):
            p1+=1
        p=p1

k=input()
h=map(int,raw_input().strip().split())
Tprime(1000000)
for i in range(0,k):
    n=int(math.sqrt(h[i]))
    if(l[n-1]==False and n*n==h[i]):
        print "YES"
    else:
        print "NO"
    


'''
k=input()
l=map(int,raw_input().strip().split())
for i in l:
    p=i
    cout=2
    fact=[]
    for j in range(2,(p/2)+1):
        if(p%j==0):
            cout+=1
            fact.append(j)
        elif(cout>3):
            break
    if(cout==3):
        print "YES"
    else:
        print "NO"
'''