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 TPrimes {
    static boolean[] sieveOfEratosthenes(int n) 
    { 
        boolean prime[] = new boolean[n+1]; 
        for(int i=0;i<n;i++) 
            prime[i] = true; 
          
        for(int p = 2; p*p <=n; p++) 
        { 
            if(prime[p] == true) 
            { 
                for(int i = p*p; i <= n; i += p) 
                    prime[i] = false; 
            } 
        } 
        return prime;
    } 
    public static void main(String args[])throws IOException{
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer(br.readLine());
        int n=Integer.parseInt(st.nextToken());
        double a[]=new double[n];
        StringTokenizer st1=new StringTokenizer(br.readLine());
        for(int i=0;i<n;i++)
            a[i]=Double.parseDouble(st1.nextToken());
        boolean p[]=new boolean[1000001];
        p=sieveOfEratosthenes(1000001);
        p[1]=false;
        StringBuilder ans=new StringBuilder();
        for(int i=0;i<n;i++){
            boolean flag=false;
            if(Math.sqrt(a[i])==Math.floor(Math.sqrt(a[i])) && p[(int)Math.sqrt(a[i])])
                flag=true;
            if(flag)
                ans.append("YES \n");
            else
                ans.append("NO \n"); 
        }
        System.out.println(ans);
    }
}