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 JAVA):

Solution:
import java.util.*;
import java.lang.*;
public class Seive
{
	public static void main(String []args)
	{
		int l=1000001;
		int a[]=new int[10000001];
		int i=0;
		TreeSet al=new TreeSet();
		for( i=0;i<=l;i++)
		{
			a[i]=1;
		}
		a[0]=0;
		a[1]=0;
		for(i=2;i<=l;i++)
		{
			if(a[i]==1)
			{
				for(int j=2;i*j<=l;j++)
				{
					a[i*j]=0;
				}
			}
		}
		for(i=0;i<=l;i++)
			if(a[i]==1){al.add(i);}
		//System.out.println(al.size());
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		for(i=0;i<t;i++)
		{
			long x=sc.nextLong();
			int y=(int)Math.sqrt(x);
		//	System.out.println(al.contains(y));
			
			long r=y;
			r=r*r;
			//System.out.println(r);
			if(r==x && al.contains(y))
			{
				System.out.println("YES");
			}else
			{
				System.out.println("NO");
			}
		}
	}
}