A. Panoramix's Prediction

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Sample test(s)

input

3 5

output

YES

input

7 11

output

YES

input

7 9

output

NO

解题说明：此题就是暴力穷举即可，判断n到m之间的数是否为质数，为了简单，编写一个子函数专门进行判断

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;

bool Check(int n)
{
	int i;
	bool flag;
	flag=true;
	for(i=2;i<=n/2;i++)
	{
		if(n%i==0)
		{
			flag=false;
			break;
		}
	}
	if(flag==false)
	{
		return false;
	}
	else
	{
		return true;
	}
}

int main()
{
	int n,m;
	int i,j;
	bool flag;
	scanf("%d %d",&n,&m);
	for(i=n+1;i<=m;i++)
	{
		flag=Check(i);
		if(flag==true)
		{
			break;
		}
	}
	if(i==m&&flag==true)
	{
		printf("YES\n");
	}
	else
	{
		printf("NO\n");
	}
	
	return 0;
}