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
from random import choice,randint
inp=sys.stdin.readline
out=sys.stdout.write
flsh=sys.stdout.flush
 
sys.setrecursionlimit(10**9)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
 
def MI(): return map(int, inp().strip().split())
def LI(): return list(map(int, inp().strip().split()))
def LLI(): return [list(map(int, l.split())) for l in sys.stdin.readlines().strip()]
def LI_(): return [int(x)-1 for x in inp().strip().split()]
def LF(): return [float(x) for x in inp().strip().split()]
def LS(): return inp().strip().split()
def I(): return int(inp().strip())
def F(): return float(inp().strip())
def S(): return inp().strip()
def pf(s): return out(s+'\n')
def JA(a, sep): return sep.join(map(str, a))
def JAA(a, s, t): return s.join(t.join(map(str, b)) for b in a)

def main():
    n = I()
    a = LI()
    def seive(n,prime):
        p=2
        while p**2<=n:
            if prime[p]:
                for i in range(p*2,n+1,p):
                    prime[i]=False
            p+=1
 
    prime = [True]*((10**6)+1)
    seive(10**6,prime)
    for i in a:
        x = int(i**0.5)
        if i<4:
            pf("NO")
        elif x*x!=i:
            pf("NO")
        elif prime[x]:
            pf("YES")
        else:
            pf("NO")

if __name__ == "__main__":
    main()