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:
n=int(input())
nums=list(map(int,input().split()))
def sqprime(n):
    p=[1]*(2+((n-3)//2))
    i=1
    while i<len(p) and ((1+(2*i))*(1+(2*i)))<=n:
        if p[i]!=0:
            x=1+(2*i)
            j=((x*x)//2)
            while j<len(p):
                p[j]=0
                j=j+x
        i=i+1
    ans=[4]
    i=1
    while i<len(p):
        if p[i]!=0:
            x=1+(2*i)
            ans.append(x*x)
        i=i+1
    return ans

def bsearch(arr,x):
    s=0
    e=len(arr)
    while True:
        m=(s+e)//2
        if x==arr[m]:
            return 'YES'
        elif x<arr[m]:
            e=m
        elif x>arr[m]:
            s=m
        mn=(s+e)//2
        if mn==m:
            break
    return 'NO'

a=sqprime(100+(10**6))
for x in nums:
    print(bsearch(a,x))