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

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


def rwh_primes2(n):
    correction = (n % 6 > 1)
    n = {0: n, 1: n - 1, 2: n + 4, 3: n + 3, 4: n + 2, 5: n + 1}[n % 6]
    sieve = [True] * (n // 3)
    sieve[0] = False
    for i in range(int(n ** 0.5) // 3 + 1):
        if sieve[i]:
            k = 3 * i + 1 | 1
            sieve[((k * k) // 3)::2 * k] = [False] * ((n // 6 - (k * k) // 6 - 1) // k + 1)
            sieve[(k * k + 4 * k - 2 * k * (i & 1)) // 3::2 * k] = [False] * (
                    (n // 6 - (k * k + 4 * k - 2 * k * (i & 1)) // 6 - 1) // k + 1)
    # return [2,3] + [3*i+1|1 for i in range(1,n//3-correction) if sieve[i]]
    e = [0] * 2 + [1] * 2 + [0] * n
    for i in range(1, n // 3 - correction):
        if sieve[i]: e[3 * i + 1 | 1] = 1
    return e


e = rwh_primes2(10 ** 6 + 5)
for x in a:
    u = int(math.sqrt(x))
    ok = e[u] and (u * u == x)
    print('YES' if ok else 'NO')