<cf>C. Cave Painting

注意：多读一下 题目：
这是精巧的归纳一个观点 ：
 a%b =c ===> c <b
 所以当 i== 1 c == 0
     2    1
    ....    ..
 我只想说 这个题目太皮了
 

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Imp is watching a documentary about cave painting.

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 , 1 ≤ i ≤ k, are distinct, i. e. there is no such pair (i, j) that:

• 1 ≤ i < j ≤ k,
• , where 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.

#include<iostream>
using namespace std;
typedef long long ll ;
#define f(i,l,r) for(int i=l;i<=r;++i)
#define g(i,l,r) for(int i=l;i>=r;--i)
int main(){
	ll a,b;
	cin>>a>>b;
	bool flag = true;
	for(ll i = 1;flag && i<=b;++i)
	 
		if(a%i!=i-1)flag =false;
	 
	if(flag)
			cout<<"Yes"<<endl;  
	else 
		cout<<"No"<<endl; 
}