17A. Noldbach problem

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本文探讨了Noldbach问题，即验证在指定范围内是否存在至少k个素数可以表示为两个相邻素数及1之和。通过预先计算素数及其组合，提供了一种高效的解决方案。

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Nick is interested in prime numbers. Once he read about Goldbach problem. It states that every even integer greater than 2 can be expressed as the sum of two primes. That got Nick's attention and he decided to invent a problem of his own and call it Noldbach problem. Since Nick is interested only in prime numbers, Noldbach problem states that at least k prime numbers from 2 to n inclusively can be expressed as the sum of three integer numbers: two neighboring prime numbers and 1. For example, 19 = 7 + 11 + 1, or 13 = 5+ 7 + 1.

Two prime numbers are called neighboring if there are no other prime numbers between them.

You are to help Nick, and find out if he is right or wrong.

Input

The first line of the input contains two integers n (2 ≤ n ≤ 1000) and k (0 ≤ k ≤ 1000).

Output

Output YES if at least k prime numbers from 2 to n inclusively can be expressed as it was described above. Otherwise output NO.

Examples

input
27 2

output
YES

input
45 7

output
NO

Note

In the first sample the answer is YES since at least two numbers can be expressed as it was described (for example, 13 and 19). In the second sample the answer is NO since it is impossible to express 7 prime numbers from 2 to 45 in the desired form.

给出n，问在[0-n] 区间内是否能找到K个数，要求这k个数都是素数且每个数都能写成三个数的的和，且这三个数中，一个是1，另外两个数是相邻的素数

素数打表+记录

#include<cstdio>
const int maxn=10000;
int prime[maxn],num[maxn*2],result[maxn*2];
void work()
{
	int a=0,b=2;
	for(int i=2;i<maxn;++i)
	{
		if(prime[i]==0)
		{
			if(i>2)
			{
				a=b;b=i;
				num[a+b+1]=1;
			}
			for(int j=2*i;j<maxn;j+=i)
			{
				prime[j]=1;
			}
		}
	}
	for(int i=2;i<maxn;++i)
	{
		if(prime[i]==0&&num[i]==1)
		{
			result[i]=1;
		}
		result[i]+=result[i-1];
	}
}
int main()
{
	work();
	int n,k;
	while(~scanf("%d%d",&n,&k))
	{
		printf("%s\n",result[n]>=k?"YES":"NO");
	}
	return 0;
}