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:
from math import sqrt

n,arr = int(input()),list(map(int,input().split(' ')))
M = int(sqrt(max(arr)))+1

def Euler(upper):                                        
    filters=[False]*(upper+2)
    prime=[]
    for num in range(2,upper+1):
        if not filters[num]:
            prime.append(num)
        for each in prime:
            if num*each > upper:
                break
            filters[num*each] = True
            if num%each == 0:      
                break;
    return prime,filters



def binary_search(aim,l):
    ele = len(l)
    mid = ele//2
    if ele>0:
        if l[mid] == aim:
            return True
        elif l[mid] > aim:
            return binary_search(aim,l[:mid])
        else:
            return binary_search(aim,l[mid+1:])
    else:
        return False



p_list,table_all = Euler(M)

for i in range(n):
    if arr[i] in [1,2,3]:
        print('NO')
    else:
        if (int(sqrt(arr[i])))**2 == arr[i] and not table_all[int(sqrt(arr[i]))]:
            print('YES')
        else:
            print('NO')