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:
import sys
input = sys.stdin.readline

## BRUTE FORCE CODE
# def if_3_factors(x):
#     dick= {}
#     while(x % 2 == 0):
#         dick[2] = dick.get(2, 0) + 1
#         x = x//2
#     for i in range(3, int(x**0.5)+1,2):
#         while(x % i == 0):
#             dick[i] = dick.get(i, 0) + 1
#             x = x//i
#     if x > 2:
#         dick[x] = dick.get(x, 0) + 1
#     nums = 1
#     for i in dick:
#         nums *= (dick[i]+1)
#     if nums == 3:
#         return True
#     return False

# n = int(input()); array = list(map(int, input().split()));
# for x in array:
#     if if_3_factors(x):
#         print('YES')
#     else:
#         print('NO')

## INTELLIGENT EFFICEINT CODE USING THE PRINCIPLE THAT ONLY SQUARE'S OF PRIME NUMBERS
## CAN BE T-Primes

caps = 1000009
sieve = [True for _ in range(caps)]
sieve[0] = sieve[1] = False
primes = set()
p = 2
while(p*p <= caps):
    if sieve[p] == True:
        for i in range(p*p, caps, p):
            sieve[i] = False
    p += 1
    
for i in range(caps):
    if sieve[i]:
        primes.add(i)    
    
n = int(input())
array = list(map(int, input().split()))
for i in array:
    this = i**0.5
    if this in primes:
        print('YES')
    else:
        print('NO')