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
from bisect import bisect_left,bisect,bisect_right
def index(a, x):
    'Locate the leftmost value exactly equal to x'
    i = bisect_left(a, x)
    if i != len(a) and a[i] == x:
        return 1
    return 0
def SieveOfEratosthenes(n):
    prime = [True for i in range(n + 1)]
    p = 2
    while (p * p <= n):

        if (prime[p] == True):

            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 1
    prime[0] = False
    prime[1] = False
    l1 = []
    for i in range(n + 1):
        if prime[i] == True:
            l1.append(i)
    return (l1)
primes=SieveOfEratosthenes(1000000)
n=int(input())
l=list(map(int,input().split()))
for i in l:
    p=(math.sqrt(i))
    if p%1!=0:
        print('NO')
        continue
    if index(primes,p):
        print('YES')
    else:
        print("NO")