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
//package a2oj;

import java.util.Scanner;
import java.util.ArrayList;
import java.util.Arrays;

public class Q1 {

	public static Scanner scn = new Scanner(System.in);

	public static void main(String[] args) {
		int k = 51;
		boolean n[] = new boolean[k];
		n[0] = true;
		n[1] = true;

		for (int i = 2; i <= Math.pow(k, 0.5); i++) {
			if (!n[i]) {
				for (int j = i * i; j < k; j += i) {
					n[j] = true;
				}
			}
		}

		ArrayList<Integer> prime = new ArrayList<Integer>();

		for (int i = 0; i < k; i++) {
			if (!n[i]) {
				prime.add(i);
			}
		}

		int[] p = new int[prime.size()];

		for (int i = 0; i < prime.size(); i++) {
			p[i] = prime.get(i);
		}

		int n1 = scn.nextInt();
		int n2 = scn.nextInt();

		int x = Arrays.binarySearch(p, n1);
		if (x+1 < prime.size() && n2 == p[x + 1]) {
			System.out.println("YES");
		} else {
			System.out.println("NO");
		}

	}

}
