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 *
# def pri(number):
#     if (number < 2):
#         return 0
#     if (number % 2 == 0):
#         return (number == 2);
#     root = int(sqrt(number));
#     for i in range(3, root+1, 2):
#         if (number % i == 0):
#             return 0
#     return 1
# nu=int(input())
# l=list(map(int,input().split()))
# for i in l:
#     z=sqrt(i)
#     if z==int(z):
#         if pri(z):
#             print("YES")
#         else:
#             print("NO")
#     else:
#         print("NO")
import math
t = int(input())
arr = input().split()
def isp(n):
    for i in range(2,int(math.sqrt(n))+1):
        if n%i==0:
            return False
    return True
for n in arr:
    n = int(n)
    sqrt = int(math.sqrt(n))
    if n==sqrt*sqrt and n!=1 and isp(sqrt):
        print("YES")
    else:
        print("NO")