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 heapq as hp
import math


def primeSeive(n):
    """return prime numbers <=n"""
    l = [True] * (n + 1)
    l[0] = l[1] = False
    l[2] = l[3] = True
    for i in range(4, n + 1, 2):
        l[i] = False
    for i in range(9, n + 1, 3):
        l[i] = False
    for i in range(10, n + 1, 5):
        l[i] = False
    for i in range(7, int(math.sqrt(n)) + 1, 2):
        if l[i]:
            for j in range(i * i, n + 1, i):
                l[j] = False
    return [i for i in range(n + 1) if l[i]]


def unpack(func=int):
    return map(func, input().split())


def range_n():
    return range(int(input()))


def main():
    n = int(input())
    prime = set(primeSeive(10 ** 6 + 1))
    for n in unpack():
        sqrt = math.sqrt(n)
        print("YES" if n != 1 and int(sqrt) == sqrt and int(sqrt) in prime else 'NO')

main()