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.InputStreamReader;
import java.util.LinkedList;
import java.util.StringTokenizer;

public class B230Alt {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        String next(){
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e){
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt(){
            return Integer.parseInt(next());
        }
        long nextLong(){
            return Long.parseLong(next());
        }
        double nextDouble(){
            return Double.parseDouble(next());
        }
        String nextLine(){
            String str = "";
            try{
                str = br.readLine();
            }
            catch (IOException e){
                e.printStackTrace();
            }
            return str;
        }
    }
    static FastReader in=new FastReader();

    static int[] sieveOfEratosthenes(int n){
        boolean prime[] = new boolean[n+1];
        LinkedList<Integer> primes=new LinkedList<>();
        int[] tPrimes;
        for(int i=0;i<n;i++)
            prime[i] = true;

        for(int p = 2; p*p <=n; p++){
            if(prime[p]){
                for(int i = p*p; i <= n; i += p)
                    prime[i] = false;
            }
        }

        for(int i = 2; i <= n; i++){
            if(prime[i])
                primes.add(i);
        }
        tPrimes=new int[primes.size()];
        int l=0;
        for(Integer i:primes){
            tPrimes[l]=i;
            l++;
        }
        return tPrimes;
    }

    static int binarySearch(int arr[], int l, int r,int x){
        if (r >= l) {
            int mid = l + (r - l) / 2;
            if (arr[mid] == x)
                return mid;
            if (arr[mid] > x)
                return binarySearch(arr, l, mid - 1, x);
            return binarySearch(arr, mid + 1, r, x);
        }
        return -1;
    }

    public static void main(String[] args) {
        int limit=(int)(Math.sqrt(999999999999L));
        Long limites=new Long(limit*limit);
        int[] tPrimes=sieveOfEratosthenes(limit);
        int n=in.nextInt();
        for(int i=0;i<n;++i) {
            Long t = in.nextLong();
            double test=Math.sqrt(t);
            if (test%1==0) {
                if (binarySearch(tPrimes,0,tPrimes.length-1,(int)test)>=0) {
                    System.out.println("YES");
                } else System.out.println("NO");
            } else System.out.println("NO");
        }
    }
}