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:
# T-Primes

def is_prime(n):
    if (n % 2 == 0 and n != 2) or n == 1:
        return False
    
    sqrt_n = int(n ** 0.5) + 1
    for x in range(3, sqrt_n, 2):
        if n % x == 0:
            return False
    return True

def sieve(n):
    prime = [True for i in range(n+1)]
    prime[0] = False
    prime[1] = False
    p = 2
    while p * p <= n:
        if prime[p] == True:
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    return prime

_ = input()
l = map(int, input().split())
s = sieve(10 ** 6)

for n in l:
    if n ** 0.5 == int(n ** 0.5):
        print('YES' if s[int(n ** 0.5)] else 'NO')
    else:
        print('NO')