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
class prnum:
    def calprnum(n):
        '''
        lp = []
        for i in range(2,n+1):
            s = math.sqrt(i)
            k = 0
            for j in range(2,int(s)+1):
                if i % j == 0:
                    k = 1
                    break
            if k == 0:
                list.append(lp, i)
        return lp
        '''
        prime = [2, 3, 5]
        for i in range(6, n+1):
            for j in prime:
                if i % j == 0 and i != j:
                    break
                elif j > math.sqrt(i):
                    list.append(prime, i)
                    break
        return prime
if __name__ == "__main__":
    prime = prnum.calprnum(1010) 
    n = int(input())
    arr = str(input()).split()
    for i in range(n):
        s = int(arr[i])
        ss = math.sqrt(s)
        if s == 1:
            print("NO")
        else:
            if int(ss) != ss:
                print("NO")
            else:
                ss = int(ss)
                for j in prime:
                    if ss % j == 0 and ss != j:
                        print("NO")
                        break
                    elif j > math.sqrt(ss):
                        print("YES")
                        break