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:
def SieveOfEratosthenes(n, prime):#from 2,3,4... change its multiples as false.
    p = 2
    while (p * p <= n):
        # If prime[p] is not changed, then it is a prime
        if (prime[p] == True):
            # Update all multiples of p:
            for i in range(p * 2, n + 1, p):#(start,stop,step)
                prime[i] = False
        p += 1
        

n = int(input())
x = [int(i) for i in input().split()]
s = [True] * (10 ** 6 + 1)#creat a long list
SieveOfEratosthenes(10 ** 6, s)#save primes as true/false.
for i in x:
    if i < 4:
        print('NO')
        continue
    elif int(i ** 0.5) ** 2 != i:
        print('NO')
        continue
    if s[int(i ** 0.5)]:
        print('YES')
    else:
        print('NO')