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 binarySearch(arr, l, r, x): 
  
    while l <= r: 
  
        mid = l + (r - l)//2; 
          
        if arr[mid] == x: 
            return 1
  
        elif arr[mid] < x: 
            l = mid + 1
  
        else: 
            r = mid - 1
      
    return 0

def Sieve(n):
    prime = [True for i in range(n+1)]
    p = 2
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    list1=[]
    for p in range(2, n): 
        if prime[p]: 
            list1.append(p**2)
    return list1
    
list1=Sieve(1000000)
n=int(input())
list2=list(map(int,input().split()))
for i in range(n):
    if binarySearch(list1,0,len(list1)-1,list2[i]):
        print("YES")
    else:
        print("NO")