Imp is watching a documentary about cave painting.

<image>

Some numbers, carved in chaotic order, immediately attracted his attention. Imp rapidly proposed a guess that they are the remainders of division of a number n by all integers i from 1 to k. Unfortunately, there are too many integers to analyze for Imp.

Imp wants you to check whether all these remainders are distinct. Formally, he wants to check, if all <image>, 1 ≤ i ≤ k, are distinct, i. e. there is no such pair (i, j) that: 

  * 1 ≤ i < j ≤ k, 
  * <image>, where <image> is the remainder of division x by y. 

Input

The only line contains two integers n, k (1 ≤ n, k ≤ 1018).

Output

Print "Yes", if all the remainders are distinct, and "No" otherwise.

You can print each letter in arbitrary case (lower or upper).

Examples

Input

4 4


Output

No


Input

5 3


Output

Yes

Note

In the first sample remainders modulo 1 and 4 coincide.
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.*;
 
public class Main {
	public static void main(String[] args) {
		InputReader in = new InputReader();
		PrintWriter out = new PrintWriter(System.out);
		Thread th = new Thread(null, new Runnable(){public void run(){new Task1().solve(in, out);}},"Task1",1<<24);
		try{
			th.start();
			th.join();
		} catch(InterruptedException e){}
		out.close();
	}

	static class Task1{
		static final int MAX = 1000001;

		public void solve(InputReader in, PrintWriter out){
			long n = in.nextLong(), k = in.nextLong();
			HashSet<Long> rem = new HashSet<Long>();
			for(long i=1; i<=Math.min(k, MAX); i++){
				if(rem.contains(n%i)){
					out.println("No");
					return;
				}
				rem.add(n%i);
			}

			out.println("Yes");
		}
		
		long gcd(long a, long b){
			while(a>0){
				long rem = b%a;
				b = a;
				a = rem;
			}
			return b;
		}
		long expo(long a, long b, long MOD){
			long result = 1;
			while (b>0){
				if (b%2==1) result=(result*a)%MOD;
				b>>=1;
				a=(a*a)%MOD;
			}
			return result%MOD;
		}
		long inverseModullo(long numerator, long denominator, long MOD){
			return ((numerator )*(expo(denominator, MOD-2, MOD))) ;
		}
	}
 	
	static class InputReader{
		final InputStream stream;
		final byte[] buf = new byte[8192];
		int curChar, numChars;
		SpaceCharFilter filter;
		public InputReader(){
			this.stream = System.in;
		}
		public int read(){
			if(numChars == -1) throw new InputMismatchException();
			if(curChar >= numChars){
				curChar = 0;
				try{
					numChars = stream.read(buf);
				} catch(IOException e){
					throw new InputMismatchException();
				}
				if(numChars <= 0) return -1;
			}
			return buf[curChar++];
		}
		public int nextInt(){
			int c = read();
			while(isSpaceChar(c)) c = read();
			int sgn = 1;
			if(c == '-'){
				sgn = -1;
				c = read();
			}
			int res = 0;
			do{
				if(c<'0' || c>'9') throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			} while(!isSpaceChar(c));
			return res*sgn;
		}
		public long nextLong(){
			int c = read();
			while(isSpaceChar(c)) c = read();
			int sgn = 1;
			if(c == '-'){
				sgn = -1;
				c = read();
			}
			long res = 0;
			do{
				if(c<'0' || c>'9') throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			} while(!isSpaceChar(c));
			return res*sgn;
		}
		public String next(){
			int c = read();
			while(isSpaceChar(c)) c = read();
			StringBuilder res = new StringBuilder();
			do{
				res.appendCodePoint(c);
				c = read();
			}while(!isSpaceChar(c));
			return res.toString();
		}
		public boolean isSpaceChar(int c){
			if(filter != null) return filter.isSpaceChar(c);
			return c==' ' || c=='\n' || c=='\r' || c=='\t' || c==-1;
		}
		public interface SpaceCharFilter{
			public boolean isSpaceChar(int ch);
		}
	}
}