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 sys
import math

n = int(sys.stdin.readline())
x = [int(x) for x in (sys.stdin.readline()).split()]

n = 1000001
simple = [True] * n
simple[0] = False
simple[1] = True
simple[2] = False

prime = [True] * n
result = [2]
append = result.append
sqrt_n = (int(n ** .5) + 1) | 1    # ensure it's odd

for p in range(3, sqrt_n, 2):
    if prime[p]:
        simple[p] = False
        prime[p*p::2*p] = [False] * ((n - p*p - 1) // (2*p) + 1)

for p in range(sqrt_n, n, 2):
    if prime[p]:
        simple[p] = False

for c in x:
    t = math.sqrt(c)
    if(t - int(t) == 0):
        if(simple[int(t)] == False):
            print("YES")
        else:
            print("NO")
    else:
        print("NO")