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 math

input()
nums = list(map(int, input().split()))


def SieveOfEratosthenes(n):
    # Create a boolean array "prime[0..n]" and initialize
    # all entries it as true. A value in prime[i] will
    # finally be false if i is Not a prime, else true.
    prime = [True for i in range(n + 1)]
    p = 2
    while p * p <= n:

        # If prime[p] is not changed, then it is a prime
        if (prime[p] == True):

            # Update all multiples of p
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 1
    prime[0] = False
    prime[1] = False
    # Print all prime numbers
    p_set = set()
    for p in range(n + 1):
        if prime[p]:
            p_set.add(p)
    return p_set


big_p = SieveOfEratosthenes(int(math.sqrt(sorted(nums)[-1])))


def is_prime(n):
    return n in big_p


for num in nums:
    prime = True
    sqrt = math.sqrt(num)

    if sqrt % 1 != 0:
        prime = False

    if prime and (sqrt < 2):
        prime = False

    if prime and not is_prime(int(sqrt)):
        prime = False

    if prime:
        print("YES")
    else:
        print("NO")