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
def rl():
    return list(map(int,input().split()))
def isperfectsquare(num):
    if ceil(sqrt(num))*ceil(sqrt(num))==num:
        return (True,int(sqrt(num)))
    return (False,-1)
    
input()
inp = rl()
global sethai,arr
sethai = set()
arr = [True]*int(1e6+1)
def seive():
    arr[0],arr[1]=False,False
    
    for i in range(2,len(arr)):
        if arr[i]:
            for j in range(2*i, len(arr),i):
                arr[j] = False
    for i in range(len(arr)):
        if arr[i]:
            sethai.add(i*i)

seive()
for i in inp:
    if i==1:
        print("NO")
    elif i in sethai:
        print("YES")
    else:
        print("NO")