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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class T_Primes {
	static ArrayList<Integer> primes;
	static int[] isComposite;
		
	static void sieve(int N)	 
	{
		isComposite = new int[N+1];					
		isComposite[0] = isComposite[1] = 1;			
		primes = new ArrayList<Integer>();

		for (int i = 2; i <= N; ++i) 					 
			if (isComposite[i] == 0) 					
			{
				
				primes.add(i);
				if(1l * i * i <= N)
					for (int j = i * i; j <= N; j += i)	// j = i * 2 will not affect performance too much, may alter in modified sieve
						isComposite[j] = 1;
			}   
	}
	
	public static void main(String[] args) throws IOException {
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		StringBuilder sb=new StringBuilder();
		sieve(1000000);
		//TreeSet <Long> Tprime=new TreeSet();
		long [] Tprime=new long[primes.size()];
		for (int i=0;i<primes.size();i++)
			Tprime[i]=(long)primes.get(i)*(long)primes.get(i);
		
		while(n-->0)
		{
			long x=sc.nextLong();
			sb.append((Arrays.binarySearch(Tprime, x)>=0)?"YES":"NO");
			sb.append("\n");
		}
			
				
		
		System.out.print(sb);
		
	}

}


class Scanner 
{
	StringTokenizer st;
	BufferedReader br;

	public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}

	public String next() throws IOException 
	{
		while (st == null || !st.hasMoreTokens()) 
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}

	public int nextInt() throws IOException {return Integer.parseInt(next());}
	
	public long nextLong() throws IOException {return Long.parseLong(next());}

	public String nextLine() throws IOException {return br.readLine();}
	
	public double nextDouble() throws IOException
	{
		String x = next();
		StringBuilder sb = new StringBuilder("0");
		double res = 0, f = 1;
		boolean dec = false, neg = false;
		int start = 0;
		if(x.charAt(0) == '-')
		{
			neg = true;
			start++;
		}
		for(int i = start; i < x.length(); i++)
			if(x.charAt(i) == '.')
			{
				res = Long.parseLong(sb.toString());
				sb = new StringBuilder("0");
				dec = true;
			}
			else
			{
				sb.append(x.charAt(i));
				if(dec)
					f *= 10;
			}
		res += Long.parseLong(sb.toString()) / f;
		return res * (neg?-1:1);
	}
	
	public boolean ready() throws IOException {return br.ready();}


}