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 main():
    num_ints = int(input())
    nums = list(map(int, input().split()))
    num2output = {}

    for i in range(num_ints):
        if nums[i] in num2output:
            print(num2output[nums[i]])
        else:
            output = 'YES' if is_t_prime(nums[i]) else 'NO'
            print(output)
            num2output[nums[i]] = output


def is_t_prime(n):
    """ Returns True if `n` has 3 unique factors. Else returns False """
    the_sqrt = int(sqrt(n))
    if sqrt(n) - the_sqrt != 0:
        return False

    return is_prime(the_sqrt)


def is_prime(n):
    the_sqrt = int(sqrt(n))
    num_factors = 0
    for divisor in range(1, the_sqrt + 1):
        if n % divisor == 0:
            num_factors += 2 if n // divisor != divisor else 1
    return num_factors == 2


if __name__ == '__main__':
    main()