A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.

The next prime number after x is the smallest prime number greater than x. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.

One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.

Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly x Roman soldiers, where x is a prime number, and next day they beat exactly y Roman soldiers, where y is the next prime number after x, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.

Yesterday the Gauls beat n Roman soldiers and it turned out that the number n was prime! Today their victims were a troop of m Romans (m > n). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix?

Input

The first and only input line contains two positive integers — n and m (2 ≤ n < m ≤ 50). It is guaranteed that n is prime.

Pretests contain all the cases with restrictions 2 ≤ n < m ≤ 4.

Output

Print YES, if m is the next prime number after n, or NO otherwise.

Examples

Input

3 5


Output

YES

Input

7 11


Output

YES

Input

7 9


Output

NO

import java.util.Scanner;
import java.util.TreeSet;

public class Main {

    static TreeSet<Integer> primesBeforeMaxSqrt = new TreeSet();

    public static void seive(int max) {
        //1: Declare/Define variables
        int sqrt = max, j, x, index;

        //Make the size of the array half the sqrt of max, to save only odd numbers
        int[] primesBeforesqrt = new int[(sqrt % 2 == 0) ? sqrt / 2 - 1 : sqrt / 2];

        //2: Insert odd numbers into the array
        if (3 <= sqrt) {
            for (int p = 3, i = 0; i < primesBeforesqrt.length; p += 2, i++) {
                primesBeforesqrt[i] = p;
            }
        }

        //3: Filter the multiples of odd numbers
        for (int p = 3; p <= sqrt; p += 2) {

            //Check if the number hasn't been filtered
            if (primesBeforesqrt[(p - 1) / 2 - 1] != 0) {
                //A: Add it to the list of prime numbers
                primesBeforeMaxSqrt.add(p);

                //B: Skip to the next odd multiple of p
                x = p;
                j = 2 * x + p;
                index = (j - 1) / 2 - 1;

                //C: Eliminate all the odd multiples of p
                while (index < primesBeforesqrt.length) {
                    //1- Assign location of an odd multiple of p to 0
                    primesBeforesqrt[index] = 0;

                    //2- Iterate to the next odd multiple of p
                    x += p;
                    j = 2 * x + p;
                    index = (j - 1) / 2 - 1;
                }
            }

        }

    }

    public static void main(String[] args) {
        //1: Declare and initialize variables
        Scanner in = new Scanner(System.in);
        int n = in.nextInt(), m = in.nextInt();
        Integer higherPrime;
        String answer = "NO";

        //2: Take in the ranges and give them a sequence number
        seive(m);

        primesBeforeMaxSqrt.add(2);

        //3: Check if m is the next prime number after n
        higherPrime = primesBeforeMaxSqrt.higher(n);
        if (higherPrime != null && (primesBeforeMaxSqrt.higher(n)) == m) {
            answer = "YES";
        }

        //4: Print the result
        System.out.println(answer);

    }
}
