prime = [1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227]
num,num2 = list(map(int,input().split()))
for i in range(50):
if prime[i] == num:
break
if prime[i+1] == num2:
print("YES")
else:
print("NO")
// Problem: A. Panoramix's Prediction
// Contest: Codeforces - Codeforces Beta Round #69 (Div. 2 Only)
// URL: https://codeforces.com/problemset/problem/80/A
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define mod 1e9 + 7
// #define for(a, c) for (int(a) = 0; (a) < (c); (a)++)
// #define forl(a, b, c) for (int(a) = (b); (a) <= (c); (a)++)
// #define forr(a, b, c) for (int(a) = (b); (a) >= (c); (a)--)
#define INF 1000000000000000003
typedef vector<int> vi;
typedef pair<int, int> pi;
#define F first
#define S second
#define pb push_back
#define pob pop_back
#define MP make_pair
void prime(vi &vec)
{
for(int i=2;i<=50;i++)
{
int flag=1;
for(int j=2;j<i;j++)
{
if(i%j==0)
{
flag=0;
break;
}
}
if(flag)
vec.pb(i);
}
}
void solve()
{
int n,m;
cin>>n>>m;
vi vec;
prime(vec);
int cnt=0;
for(int i=0;i<=n;i++)
{
if(vec[i]==n && vec[i+1]==m)
{
cnt=1;
break;
}
}
if(cnt)
{
cout<<"YES"<<endl;
}
else{
cout<<"NO"<<endl;
}
}
int32_t main()
{
solve();
}
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