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 bisect import bisect_left


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


def get_sieve():
    len_nums = 10**6+1
    data = list(range(len_nums+2))

    for i in range(2, int(len_nums**0.5)+1):
        if data[i]:
            for j in range(i*i, len_nums, i):
                data[j] = 0

    return list(map(lambda i: i**2, filter(lambda i: i, data[2:])))


def binary_search(a, x, lo=0, hi=None):
    hi = hi if hi is not None else len(a)
    pos = bisect_left(a, x, lo, hi)

    return (pos if pos != hi and a[pos] == x else -1)

data = get_sieve()
print('\n'.join(['YES' if binary_search(data, n) != -1 else 'NO' for n in nums]))