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 PYTHON):

Solution:
MX = 1000000

vis = [False]
prime = [2]
DP = []

def sieve():
    for i in range (1,MX+1):
        vis.append(True)

    vis[0] = vis[1] = False

    i = 3

    while (i<=MX):
        if (vis[i]==True):

            prime.append(i)

            j = 3*i
            
            while (j<=MX):
                vis[j] = False

                j += 2*i

        i += 2

def solve():
    for i in range (0,len(prime)):
        DP.append(prime[i]*prime[i])

def BS(n):
    l=0
    h=len(DP)-1

    while (l<=h):
        mid = (l+h)/2

        if (DP[mid] == n):
            return True

        elif (DP[mid] > n):
            h = mid - 1

        else : l = mid + 1

    return False

sieve()
solve()

n = input ()

tmp = raw_input()

num = map(int, tmp.split())

for i in range (0,n):
    if (BS(num[i]) == True):
        print "YES"

    else : print "NO"