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:
n=int(input())
a=[]
s=input().split()
for i in range(n):
    a.append(int(s[i]))

def se(n,prime):#求素数算法
    p=2
    while(p*p<=n):
        if prime[p]==True:
            for i in range(p*2,n+1,p):
                prime[i]=False
        p+=1

prime=[True]*(10**7+1)
se(int(max(a)**0.5)+1,prime)    


for i in range(n):
    if int(a[i]**(0.5))!=a[i]**(0.5) or a[i]==1:
        print('NO')
    else:
        x=int(a[i]**0.5)
        if prime[x]==True:
            print('YES')
        else:
            print('NO')