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
#include <bits/stdc++.h>
using namespace std;
void fast() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
}
void file() { freopen("largest.in", "r", stdin); }
bool isPrime(long long x) {
  if (x == 1 || x == 0) return 0;
  for (long long i = 2; i <= sqrt(x); i++) {
    if (x % i == 0) return 0;
  }
  return 1;
}
long long gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
bool palindrom(string s) {
  string r = s;
  reverse(r.begin(), r.end());
  if (r == s)
    return 1;
  else
    return 0;
}
long long power(long long x, long long y, long long m) {
  long long z = 1;
  for (int i = 0; i < y; i++) {
    z = ((z % m) * (x % m)) % m;
  }
  return z % m;
}
long long fpower(long long x, long long y, long long mod) {
  long long p = 1;
  while (y) {
    if (y & 1) p = (p * x) % mod;
    y >>= 1;
    x = (x * x) % mod;
  }
  return p;
}
int prfac(int n) {
  long long c = 0;
  for (long long i = 2; i * i <= n; i++) {
    while (n % i == 0) {
      n /= i;
      c++;
    }
  }
  if (n > 1) c++;
  return c;
}
long long div(long long n) {
  long long s = 0;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      if (i != n / i) {
        if (n / i != n) s += n / i;
        s += i;
      }
    } else
      s += i;
  }
  return s;
}
int cnt(string s, int j) {
  int c = 0;
  for (int i = j + 1; i < (int)(s.size()); i++) {
    if (s[i] == '+') break;
    c++;
  }
  return c;
}
int main() {
  fast();
  int n, m, p;
  cin >> n >> m;
  for (int i = n + 1;; i++) {
    if (isPrime(i)) {
      p = i;
      break;
    }
  }
  if (p == m)
    cout << "YES";
  else
    cout << "NO";
}
