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;
template <typename T>
void print(const T& v) {
  cerr << v << ' ';
}
template <typename T1, typename... T2>
void print(const T1& first, const T2&... rest) {
  print(first);
  print(rest...);
}
long long BigMod(long long int a, long long int b, int M) {
  long long remainder, answer = 1;
  remainder = a % M;
  while (b != 0) {
    if (b % 2 == 1) answer = (answer * remainder) % M;
    remainder = (remainder * remainder) % M;
    b /= 2;
  }
  return answer;
}
bool cmp(int a, int b) { return a > b; }
vector<int> prime;
char sieve[10007];
void primeSieve(int n) {
  sieve[0] = sieve[1] = 1;
  prime.push_back(2);
  for (int i = 4; i <= n; i += 2) sieve[i] = 1;
  int sqrtn = sqrt(n);
  for (int i = 3; i <= sqrtn; i += 2) {
    if (sieve[i] == 0) {
      for (int j = i * i; j <= n; j += 2 * i) sieve[j] = 1;
    }
  }
  for (int i = 3; i <= n; i += 2)
    if (sieve[i] == 0) prime.push_back(i);
}
int main() {
  primeSieve(10007);
  int n, m;
  cin >> n >> m;
  int tmp = *(lower_bound(prime.begin(), prime.end(), n + 1));
  if (m == tmp) {
    printf("YES\n");
  } else
    printf("NO\n");
  return 0;
}
