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.io.*;
public class threedivisor{
   static int n,k;
   static boolean sf[]=new boolean[1000001];
   static StringBuilder ans; 
   static HashMap<Integer,Integer> map=new HashMap<>(); 
   public static void main(String[] args) throws IOException{ 
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); 
        ans=new StringBuilder(); 
        PrintWriter out=new PrintWriter(System.out); 
        for(int i=2;i*i<=sf.length;i++){
            if(!sf[i]){
                for(int j=i*i;j<sf.length;j+=i){
                    sf[j]=true;
                }
            }
        }
        
        int t=1;
        // Integer.parseInt(br.readLine().trim()); 
        while(t-->0){ 
            n=Integer.parseInt(br.readLine().trim()); 
            StringTokenizer tok=new StringTokenizer(br.readLine()," ");
            for(int i=0;i<n;i++){ 
                if(call(Long.parseLong(tok.nextToken()))){
                    ans.append("YES");

                }
                else{
                    ans.append("NO");
                }ans.append("\n");
            }

        }
        out.println(ans); 
        out.close(); 
    }
    static boolean call(long x){
        int sq=(int)Math.sqrt(x);
        long te=sq;
        if(te*te!=x || x==1)return false;
        if(!sf[sq]){
            return true;

        }
        return false;
    }
}