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

# list = input().split()


n = int(input())
numList = list(int(num) for num in input().strip().split())[:n]

l = []




# True --> prime || False --> not prime
def sieve():
    for i in range(1000001):
        l.append(True)
    l[0] = False
    l[1] = False
    for x in range(2, 1000001):
        if l[x] == True:
            a = x + x
            b = x
            for z in range(a, 1000001, b):
                l[z] = False


sieve()

for i in range(n):
    sq = int(math.sqrt(numList[i]))
    if sq * sq != numList[i] or numList[i] == 1 or ((numList[i] != 4) and (numList[i] % 2 == 0)):
        print("NO")
    elif sq * sq == numList[i] and l[sq]==True:
        print("YES")
    else:
        print("NO")